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RSS Feed for Wolfram Community showing any discussions from all groups sorted by activeHow to style a Grid with ItemStyle?
https://community.wolfram.com/groups/-/m/t/2534552
I have one question: "**How to combine style definitions within ItemStyle?**" and one explanation: "**How to define cell regions within ItemStyle**".
See the following notebook:
Unfortunately this embedded notebook does not show colors. No idea why. That is why I attach the same notebook to this post.
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/498c3a93-8175-4968-a1cf-667f32a984d9Werner Geiger2022-05-19T11:17:53ZBoussinesq equations solving error
https://community.wolfram.com/groups/-/m/t/2535261
Hi all,
I am working on Boussinesq equation. The notebook can run perfectly for only 0.6 steps and then the calculation starts running slowly after 0.7.
All boundary conditions seemed fine. I am unsure if I used the correct method to solve this problem.
This is the method I am currently using.
>![enter image description here][1]
I have tried several methods. Unfortunately, it does not work. I can't figure out the mistake.
Please let me know if you have any suggestions.
Thank you very much. &[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=code.jpg&userId=2127825
[2]: https://www.wolframcloud.com/obj/e342ea3e-5109-4aa9-be37-bbc83c983582Lion Sahara2022-05-20T09:18:31ZVertex colors and tubes does not work
https://community.wolfram.com/groups/-/m/t/2534605
There seems to be something wrong with the VertexColors when using tubes.
dat = Table[{Cos[x], Sin[x], x/10}, {x, 0., 10}];
col = RGBColor /@ dat
{
Graphics3D[{PointSize[Large], Point[dat, VertexColors -> col],
Line[dat]}],
Graphics3D[{Thick, Line[dat, VertexColors -> col]}],
Graphics3D[{CapForm["Square"], JoinForm["Miter"],
Tube[dat, VertexColors -> col]}, Lighting -> "Neutral"]
}
![enter image description here][1]
This also seems to be the case in the documentation, this is an example from the VertexColors doc.
![enter image description here][2]
other examples in also fail like this one in the Tube doc.
![enter image description here][3]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=tube1.png&userId=1332602
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=vertex.png&userId=1332602
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=9549tube.png&userId=1332602Martijn Froeling2022-05-19T10:55:58ZJohansen Test in Mathematica
https://community.wolfram.com/groups/-/m/t/1397534
A post from five years ago, [How to run Johansen Test in Mathematica][1], requested the code for the Johansen test in Mathematica. However, the verbeia.com code that was offered had problems (incorrectly normalized eigenvectors, computational errors). As a better alternative, I'd like to post my Johansen test code here which I believe is correct. I've compared the output of this code with the output of the Matlab Johansen code in the [Spatial Econometrics][2] library and they agree. I've added my Mathematica code as an attachment to this post, "JohansenTest.nb".
The code includes a few subroutines that allows the output from the Johansen test to be displayed in a nice tabular form, such as:
![Johansen Test Output][3]
This table shows the results for a cointegrated portfolio of three Exchange Traded Funds (ETFs), having two cointegrating relationships (r <= 0 and r <= 1) for both the trace and eigen statistics (at > 99% confidence, except for the eigen statistic for r <= 0, which is > 95% confidence).
I use this code to generate the weights for a cointegrated porfolio of ETFs which I've trading profitably for several months now. I usually set order = 2, and detrend = 1. That seems to give the best results for the portfolios I've looked at. As in Ernie Chan's [Algorithmic Trading: Winning Strategies and Their Rationale][4], I apply a Kalman filter to the ETF data and Johansen weights to improve the trading algorithm performance. If there is interest, I can discuss that in future posts, as well. (Chan's Kalman filter discussion is very incomplete, in my opinion.)
I've left a few optional "debug" statements in the code to allow you to check that the matrices are properly normalized. These lines can be deleted. Note that the Johansen weights are the *rows* of the eigenvector matrix, not the *columns* (as in the Spatial Economentrics code). I feel this is more consistent with the way that Mathematica handles vectors and matrices.
For detail on the equations on which this code is based, see this 2005 article by B.E. Sorenson: [Cointegration][5].
I welcome any feedback.
[1]: http://community.wolfram.com/groups/-/m/t/91915 "How to run Johansen test in Mathematicaquot;"
[2]: https://www.spatial-econometrics.com/ "Spatial Econometrics"
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=7158JohansenOutput.jpg&userId=1397503
[4]: https://www.amazon.com/Algorithmic-Trading-Winning-Strategies-Rationale-ebook/dp/B00CY5HC0U/ref=sr_1_1?ie=UTF8&qid=1533423737&sr=8-1&keywords=chan%20algorithmic%20trading "Chan, Algorithmic Trading"
[5]: http://www.uh.edu/~bsorense/coint.pdf "Cointegration, B.E. Sorensen"Amanda Gerrish2018-08-04T23:12:20ZFind semi-prime factors; program runs but doesn't compute
https://community.wolfram.com/groups/-/m/t/2342528
Clear x
Clear pnp
pnp = 85
f[x] = ((x ^ 3 * pnp^2) / (pnp^2 + x) - x^3)
For [x = 3, x < pnp/2, (x + 2), Print[x] ;
If[ {f[x] > 1}, {x = pnp} ]
________________________________________
different attempt
Clear x
Clear pnp
pnp = 85
f[x] = ((x ^ 3 * pnp^2) / (pnp^2 + x) - x^3)
x = 3; While [f[x] < 1, Print[x]; (x + 2)]Bobby Joe Snyder2021-08-14T20:02:32ZHow to group the labels in BarChart?
https://community.wolfram.com/groups/-/m/t/2534096
I want to generate the following bar chart, and I need to group the labels.
How should it be done?
data = {{RandomInteger[{0, 100}, 100],RandomInteger[{0, 100}, 100],
RandomInteger[{0, 100}, 100]}, {RandomInteger[{0, 100}, 100],
RandomInteger[{0, 100}, 100]}};
BoxWhiskerChart[data,
ChartLabels -> {{"Group1", "Group2"}, {"a", "b", "c", "d", "e"}}]
![enter image description here][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10020fig1.png&userId=1170351Tsai Ming-Chou2022-05-18T19:48:27ZAdding GeoBackground to GeoGraphics increases file size excessively
https://community.wolfram.com/groups/-/m/t/2534189
Adding a GeoBackground to a GeoGraphics increases the file size enormously. Does anybody have any ideas for minimising the resulting size?
Adding a ReliefMap background to a simple plot increase the file size from 4kB to 4.3MB. Adding a selection of plots (5) means the resulting file is too large to send by email.Nigel King2022-05-18T20:49:51ZSaving a notebook as a WL script
https://community.wolfram.com/groups/-/m/t/348631
I would like to develop scripts for execution by working first in a notebook. I tried this by save-as a WL package, but clearly, some special syntax is needed -- all my code was commented out (* *). Is there a way to do this? Is there any documentation on the use of save-as-a package in .wl in Mathematica 10?
Best,
DavidDavid Keith2014-09-19T22:51:32ZUnexpect answer for A x B = C x B where A, B, C are R3
https://community.wolfram.com/groups/-/m/t/2534662
My apologies for this dumb question but I have never gotten such a strange answer from Mathematica before. I was hoping that someone could tell me what Mathematica is doing and why. This actually started with a physics problem: The torque on a dipole (M) in a magnetic field (B) is given by torque (T) T = M x B and the torque applied to dipole particle at a distance (R) from the center of rotation is
T = R x F where (F) acts on the particle then we get a vector algebra problem:
R x F = M x B where each capital letter is a 3 vector in R3.
(1) The first question is how do you solve for the three components of (F)?
(2) I tried right and left multiplications with the other three vectors, but Mathematica always returns a null set.
(3) Next, I tried reducing the complexity by creating what should be a very simple problem:
Solve R x F = R x B. This should have the answer F = B. Mathematica doesn't think so. It returns this silliness (below). Would someone please tell me why my Mathematica code is wrong and what happened here???
So, what happened to (fx) and why doesn't F = B appear as an answer?
In[1523]:= Clear[m, b, r, f, t, mx, my, mz, bx, by, bz, rx, ry, rz,
fx, fy, fz, tx, ty, tz]
mx =.;
my =.;
mz =.;
bx =.;
by =.;
bz =.;
fx =.;
fy =.;
fz =.;
rx =.;
ry =.;
rz =.;
tx =.;
ty =.;
tz =.;
m = {mx, my, mz};
b = {bx, by, bz};
r = {rx, ry, rz};
f = {fx, fy, fz};
t = {tx, ty, tz};
ss = Solve[{Simplify[Cross[r, f] == Cross[r, b]]}, {fx, fy, fz}]
Simplify[ss]
During evaluation of In[1523]:= Solve::svars: Equations may not give solutions for all "solve" variables.
Out[1544]= {{fy -> (fx ry)/rx - (-by rx + bx ry)/rx,
fz -> (fx rz)/rx - (-bz rx + bx rz)/rx}}
Out[1545]= {{fy -> (by rx - bx ry + fx ry)/rx,
fz -> (bz rx - bx rz + fx rz)/rx}}
Many thanks to all the great users that took the the time to read and respond to my questions!!! Muchas Gracias!!Gerald Kamin2022-05-19T19:45:36ZX-ray photoelectron spectroscopy (XPS) data quantification
https://community.wolfram.com/groups/-/m/t/2532589
![the GUI interface for XPS quantification][1]
XPS techniques have been extensively used in scientific researches. the analysis of XPS data does not only include the peak fitting, but also include the quantification of the measured core level peaks, from which one could calculate the atomic density in the measured sample.
Here in my script, I made a GUI that enables one to calculate the atomic ratio between two elements. Depending on the different sample surfaces (bulk or thin films), the quantification process will be different. I have considered the photon-energy dependent photoionization cross sections and the inelastic mean free path (both TPP-2M and a simplified version).
First, the IMFP is defined:
ft1[KE_, Z_, eg_, a_] :=
Module[{KE0 = KE, Z0 = Z, eg0 = eg, a0 = a},
Return[(a0^1.7)*(4 + 0.44 Z0^0.5 +
0.104 KE0^0.872)/((Z0^0.38)*(1 - 0.02 eg0))]]
then a figure-showing function is defined to show different XPS models
figmodel[model_] :=
Module[{mo0 = model, im},
SetDirectory[FileNameJoin[{currentpath, "images"}]];
im = Image[Import[mo0], ImageSize -> 100]]
finally, the main function "calculate" is defined with many parameters, when clicking the calculate, one could get the final result, as shown in the figure in the beginning. In this function, one needs to input what elements you want to compare (e.g., c1s, o1s, or n1s), then input the peak areas, which has to be background subtracted. If the sample contains a thin film on top, one has to input the thickness of the thin film. the cross section is aquired from the website(https://vuo.elettra.eu/services/elements/WebElements.html)
calcalator[element1_, element2_, area1_, area2_, penergy_, model_,
zvag_, eg_, a_, BE1_, BE2_, fthickness_] :=
Module[{element10 = element1, element20 = element2, area10 = area1,
area20 = area2, penergy0 = penergy, model0 = model, zvag0 = zvag,
eg0 = eg, a0 = a, BE10 = BE1, BE20 = BE2, fthickness0 = fthickness,
element1cs, element2cs, element1func, element1csv, element2func,
element2csv},
SetDirectory[FileNameJoin[{currentpath, "cs"}]];
element1cs = Import[element10];
element2cs = Import[element20];
element1func = Interpolation[element1cs[[All, 1 ;; 2]]];
element1csv = element1func[penergy0];
element2func = Interpolation[element2cs[[All, 1 ;; 2]]];
element2csv =
element2func[penergy0];(*crosssection interpolation in the table*)
If[model0 == "bulk",
If[ element10 == element20,
Print["---------------"];
Print["Area=N*fph*crosssection*IMFP*Transmission(E)*Cos(thetA)*A"];
Print["CS and IMFP ignored, only area considered"];
Print["element density ratio is: N1/N2=Area1/Area2= ",
area10/area20],
Print["---------------"];
Print["Area=N*fph*crosssection*IMFP*Transmission(E)*Cos(thetA)*A"];
Print["CS and IMFP considered"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print[
"element density ratio is: \
N1/N2=Area1*CS2*IMFP2/(Area2*CS1*IMFP1)= ",
area10*element2csv*
ft1[penergy0 - BE20, zvag0, eg0,
a0]/(area20*element1csv*
ft1[penergy0 - BE10, zvag0, eg0, a0])]
],
If[model0 == "layer1",
If[element10 == element20,
Print["---------------"];
Print["Area=N*crosssection*IMFP*(1-exp(-d/IMFP))"];
Print["CS and IMFP ignored, only area considered"];
Print["element density ratio is: N1/N2=Area1/Area2= ",
area10/area20],
Print["---------------"];
Print["Area=N*crosssection*IMFP*(1-exp(-d/IMFP))"];
Print["CS and IMFP considered"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print[
"element density ratio is: \
N1/N2=Area1*CS2*IMFP2*(1-exp(-d/IMFP2))/(Area2*CS1*IMFP1*(1-exp(-d/\
IMFP1)))= ",
area10*element2csv*
ft1[penergy0 - BE20, zvag0, eg0,
a0]*(1 -
Exp[-fthickness0/
ft1[penergy0 - BE20, zvag0, eg0, a0]])/(area20*
element1csv*
ft1[penergy0 - BE10, zvag0, eg0,
a0]*(1 -
Exp[-fthickness0/ft1[penergy0 - BE10, zvag0, eg0, a0]]))]
],
If[model0 == "layer2",
If[element10 == element20,
Print["---------------"];
Print[
"Area1(toplayer)=N1*crosssection*IMFP*(1-exp(-d/IMFP1))"];
Print["Area2(substrate)=N2*crosssection*IMFP*(exp(-d/IMFP2))"];
Print["CS ignored, IMFP1=IMFP2"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print[
"element density ratio is: \
N1/N2=Area1*exp(-d/IMFP2)/(Area2*(1-exp(-d/IMFP1)))= ",
area10*Exp[-fthickness0/
ft1[penergy0 - BE20, zvag0, eg0,
a0]]/(area20*(1 -
Exp[-fthickness0/ft1[penergy0 - BE10, zvag0, eg0, a0]]))],
Print["---------------"];
Print[
"Area1(toplayer)=N1*crosssection1*IMFP1*(1-exp(-d/IMFP1))"];
Print["Area2(substrate)=N2*crosssection2*IMFP2*(exp(-d/IMFP2))"];
Print["CS and IMFP considered"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print[
"element density ratio is: \
N1/N2=Area1*CS2*IMFP2*exp(-d/IMFP2)/(Area2*CS1*IMFP1*(1-exp(-d/IMFP1))\
)= ", area10*element2csv*ft1[penergy0 - BE20, zvag0, eg0, a0]*
Exp[-fthickness0/
ft1[penergy0 - BE20, zvag0, eg0, a0]]/(area20*element1csv*
ft1[penergy0 - BE10, zvag0, eg0,
a0]*(1 -
Exp[-fthickness0/
ft1[penergy0 - BE10, zvag0, eg0, a0]]))]
],
If[model0 == "layer3",
If[element10 == element20,
Print["---------------"];
Print["Area=N*crosssection*IMFP*exp(-d/IMFP)"];
Print["CS and IMFP ignored"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print["element density ratio is: N1/N2=Area1/Area2= ",
area10/area20],
Print["---------------"];
Print["Area1=N1*crosssection1*IMFP1*exp(-d/IMFP1)"];
Print["Area2=N2*crosssection2*IMFP2*exp(-d/IMFP2)"];
Print["CS and IMFP considered"];
Print["CS1= ", element1csv, ", IMFP1= ",
ft1[penergy0 - BE10, zvag0, eg0, a0] ];
Print["CS2= ", element2csv , ", IMFP2= ",
ft1[penergy0 - BE20, zvag0, eg0, a0]];
Print[
"element density ratio is: \
N1/N2=Area1*CS2*IMFP2*exp(-d/IMFP2)/(Area2*CS1*IMFP1*exp(-d/IMFP1))= \
", area10*element2csv*ft1[penergy0 - BE20, zvag0, eg0, a0]*
Exp[-fthickness0/
ft1[penergy0 - BE20, zvag0, eg0, a0]]/(area20*element1csv*
ft1[penergy0 - BE10, zvag0, eg0, a0]*
Exp[-fthickness0/ft1[penergy0 - BE10, zvag0, eg0, a0]])]
]
]
]
]
]]
Open problems:
the precision of photoinization cross section needs to be further optimized, current results are based on the data from https://vuo.elettra.eu/services/elements/WebElements.html。
&[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=xpscaculate.png&userId=20103
[2]: https://www.wolframcloud.com/obj/7767cb7d-ed6f-4419-91f3-0c490b91c3fdQiankun Wang2022-05-16T15:11:31ZStiefel Manifolds and Polygons
https://community.wolfram.com/groups/-/m/t/2534932
*WOLFRAM MATERIALS for the ARTICLE:*
> Clayton Shonkwiler (2019).
> *Stiefel Manifolds and Polygons*.
> Proceedings of Bridges 2019: Mathematics, Art, Music, Architecture, Education, Culture, 187–194.
> ISBN: 978-1-938664-30-4
> [Full article in PDF][1]
&[Wolfram Notebook][2]
[1]: https://archive.bridgesmathart.org/2019/bridges2019-187.pdf
[2]: https://www.wolframcloud.com/obj/0cb3faef-c6e8-4021-8ee7-af074e917be9Clayton Shonkwiler2022-05-19T19:10:37Z[WSG22] Daily Study Group: Signals, Systems and Signal Processing
https://community.wolfram.com/groups/-/m/t/2527035
A Wolfram U daily study group on "Signals, Systems and Signal Processing" begins on May 16, 2022.
Join instructors [@Leila Fuladi][at0] and [@Mariusz Jankowski][at1] and a cohort of fellow learners to study the concepts, mathematics, principles and techniques of signal processing. We'll cover methods of analysis for both continuous-time and discrete-time signals and systems, sampling and introductory filter design. The concepts and methods of signals and systems play an important role in many areas of science and engineering and many everyday signal processing examples are included. A basic working knowledge of the Wolfram Language is recommended.
**[REGISTER HERE][1]**
![enter image description here][2]
[1]: https://www.bigmarker.com/series/daily-study-group-signals-systems-and-signal-processing/series_details?utm_bmcr_source=community
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=WolframUBanner.jpeg&userId=130003
[at0]: https://community.wolfram.com/web/leilaf
[at1]: https://community.wolfram.com/web/mariuszjAbrita Chakravarty2022-05-06T22:26:34ZWolfram widget asks for Adobe flash player?
https://community.wolfram.com/groups/-/m/t/2228318
So I wanted to make a Wolfram widget, but it told me I needed flash and flash is no longer supported in 2021. How do I make a Wolfram widget with no flash?
![enter image description here][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Capture.png&userId=11733Andrew C2021-03-24T21:47:11ZPlotting the solution of an integral equation
https://community.wolfram.com/groups/-/m/t/2529631
Hello All,
I tried plotting the solution of the integral equations in this notebook, but the plot function gives empty plot. The solution contains Hypergeometric functions. Please guide what needs correction.
Thanks.
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/76d1ae55-d766-4685-85ce-ea35a67c7555Mahnoor Khan Sumbal2022-05-12T05:43:59ZFinding the similarity transformation between two matrices
https://community.wolfram.com/groups/-/m/t/2532981
I've two matrices `A` and `B`, and I try to find the similarity transformation between them by the following code snippet adopted from the method described [here](https://mathematica.stackexchange.com/questions/226664/finding-the-similarity-transformation-between-two-matrices):
In[74]:= (*Data*)
A = {{-I, 0, 0, 0}, {0, I, 0, 0}, {0, 0, 0, -I}, {0, 0, -I, 0}};
B = {{0, 1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, I}, {0, 0, I, 0}};
(*Search for x and y based on characteristic polynomial*)
n = Length@A;
Id = IdentityMatrix@n;
solxy = SolveAlways[Det[A - l*Id] == Det[B - l*Id], l]
(*Update data*)
A = A /. solxy[[1]];
B = B /. solxy[[1]];
(*Solve for general P*)
P = Array[p, {n, n}];
solP = Solve[P . B == A . P, Flatten@P];
P = P /. solP[[1]]
(*Check*)
B == Inverse@P . A . P // Simplify
Out[78]= {{}}
During evaluation of In[74]:= Solve::svars: Equations may not give solutions for all "solve" variables.
Out[83]= {{p[1, 1], I p[1, 1],
p[1, 3], -p[1, 3]}, {p[2, 1], -I p[2, 1], p[2, 3],
p[2, 3]}, {p[3, 1], p[3, 2], p[3, 3], p[3, 4]}, {-I p[3, 2],
I p[3, 1], -p[3, 4], -p[3, 3]}}
Out[84]= True
In[88]:= P
Out[88]= {{p[1, 1], I p[1, 1],
p[1, 3], -p[1, 3]}, {p[2, 1], -I p[2, 1], p[2, 3],
p[2, 3]}, {p[3, 1], p[3, 2], p[3, 3], p[3, 4]}, {-I p[3, 2],
I p[3, 1], -p[3, 4], -p[3, 3]}}
As you can see, it gives a general form symbol matrix `P` as the result, and with all the symbols setting to 1, the following form will be obtained:
In[95]:= P /. {p[1, 1] -> 1, p[1, 3] -> 1, p[2, 1] -> 1, p[2, 3] -> 1,
p[3, 1] -> 1, p[3, 2] -> 1, p[3, 3] -> 1, p[3, 4] -> 1}
Out[95]= {{1, I, 1, -1}, {1, -I, 1, 1}, {1, 1, 1, 1}, {-I, I, -1, -1}}
But this method is a little complicated and not intuitive. Any enhancements/suggestions/comments about this method will be highly appreciated. I also attached the code snippet shown above.
Regards,
HZHongyi Zhao2022-05-17T07:24:12Z[WSG22] Daily Study Group: A Guide to Programming and Mathematics with WL
https://community.wolfram.com/groups/-/m/t/2504513
A new study group on the topic "A Guide to Programming and Mathematics with the Wolfram Language" will begin soon.
Join a cohort of fellow learners and expand your understanding of core programming topics as well as symbolic and applied mathematics functionality in the Wolfram Language with lessons by veteran instructor and developer [David Withoff][1]. A basic working knowledge of the Wolfram Language or introductory-level skill in any programming language is recommended.
April 18–May 6
11am–12pm US CT (4–5pm GMT)
[**REGISTER HERE**][2]
![enter image description here][3]
[1]: https://www.wolfram.com/wolfram-u/instructors/withoff.html
[2]: https://www.bigmarker.com/series/daily-study-group-a-guide-to-programming-and-mathematics/series_details?utm_bmcr_source=community
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=WolframUBanner.jpeg&userId=130003Abrita Chakravarty2022-04-06T14:49:11ZNeumann Boundary Condition for thermal radiation between two bodies
https://community.wolfram.com/groups/-/m/t/2534173
Hi All,
I'm trying to formulate and solve a pde which models thermal radiation between two bodies using a Neumann boundary condition.
I've created a mesh and want to capture the thermal radiation heat transfer between element 106 and element 19 in the boundary mesh wireframe plot below.
![enter image description here][1]
I've modelled convective BCs as follows:
GconvPistComb = NeumannValue[hPistComb*(TPistHeat - temp[z, r]), ElementMarker == 106];
GconvBlkCool = NeumannValue[hCoolWater1*(TWater1 - temp[z, r]), ElementMarker == 75]
The pde with just these included looks like:
pde = {1/r D[-kMesh r D[temp[z, r], r], r] + D[-kMesh D[temp[z, r], z],
z] == GconvPistComb + GconvBlkCool
If I were modelling the thermal radiation from ambient gas to element 19 I would use the following Neumann BC in which TAmb is a pre-defined fixed value:
GradAmbTest =
NeumannValue[Epsilon Sigma (TAmb - temp[z, r])^4,
ElementMarker == 19];
How can I replace TAmb with an expression that represents the solved temperature field for Element 106?
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=MathematicaThermalRadiationQuestion.png&userId=1933607Archie Watts-Farmer2022-05-18T18:11:52ZHow may we write Nth polynomials and perform their derivatives?
https://community.wolfram.com/groups/-/m/t/2533491
Suppose a multivariable polynomial of Nth degree is given by
$P(x,y)=\displaystyle\sum_{j=0}^{N}\sum_{i=0}^{j} a_{i,\,j-i}x^{i}y^{j-i}$.
I want to express this polynomial using Mathematica. In addition, It is of my interest compute its first and second derivatives.
My attempts:
Previously, I tried to study the derivatives of the $P(x,y)$ through a generator of polynomial using the following code:
poly[vars_List, a_, order_] := Module[{n = Length@vars, idx, z},
idx = Cases[Tuples[Range[0, order], n], x_ /; Plus @@ x <= order]; z = Times @@@ (vars^# & /@ idx); z.((Subscript[a, Row[#]]) & /@ idx)]
poly[{x, y}, a, n]
However, it looks like that this code only works when n is equal to some integer number, for example, n =2. Otherwise, I receive an error message.
Attached below, you may find my notebook with my attempts.
Based on the above,
1. Is there another way to express the general form of $P(x,y)$ as presented above using *Mathematica*
2. How may I express the $P(x,y)$ derivatives?
Thanks in advanceVinícius Hopkins2022-05-17T22:42:26ZWolfram Language Paclet Repository: first look
https://community.wolfram.com/groups/-/m/t/2533977
[![enter image description here][1]](https://resources.wolframcloud.com/PacletRepository)
&[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2022-05-18at9.06.31AM.jpg&userId=11733
[2]: https://www.wolframcloud.com/obj/8f2728a5-2fcb-4f12-bcec-680dbad9a05eBob Sandheinrich2022-05-18T13:14:29ZComputational Art Contest 2022
https://community.wolfram.com/groups/-/m/t/2498984
> *SHARE this contest*: https://wolfr.am/CompArt-22
# WINNERS
Thank you to everyone who submitted entries into this contest! It was a blast to see all the amazing art you created. After deliberation by our judges, these are the winners:
- **Honorable Mention**: ARTIST: [Daniel Hoffmann][1], EXHIBIT: "[The Memory of Persistence][2]"
- **Staff Winner**: ARTIST: [Anton Antonov][3], EXHIBIT: "[Rorschach mask animations projected over 3D surfaces][4]"
- **3rd Place**: ARTIST: [Jacqueline Doan][5], EXHIBIT: "[Kuramoto oscillators with phase lag][6]"
- **2nd Place**: ARTIST: [Tom Verhoeff][7], EXHIBIT: "[Sculpture from 18 congruent pieces][8]"
- **1st Place**: ARTIST: [Frederick Wu][9], EXHIBIT: "[Love heart jewelry IV: the giving tree][10]"
![enter image description here][11]
----------
# CONTEST
Flex your creative and computational skill with Wolfram's Computational Art Contest that kicks off today, Monday, March 28th! Share your work with the community, and potentially win free Wolfram merchandise. Programmers and artists of all skill levels are encouraged to participate!
This contest is inspired by Genuary, an annual project releasing generative art prompts during the month of January. We're elated to see the creative works of our users and engage with the community while exploring the scope of computational art within the Wolfram Language.
## Rules & Guidelines ##
- Submission deadline is April 25th at 9am Central Time. Posts posted
after will not be included in judging
- Participants must fill out a detailed Community profile ( example:
https://community.wolfram.com/web/claytonshonkwiler ) and create a
Community post about their submission. The post must include the code
used to create graphics and the final piece of art placed at the top
of the post. An explanation of how their code works is required,
moreover participants are encouraged to write more about their
creative process.
- Participants submit their entry by commenting on this post with an
image of their art, along with a link to their Community post
- Multiple submissions per participant are allowed, but please keep the
number of submissions under three
- Each participant can only win once. Participants' best-performing
piece, as determined by the judges, will be used when determining
winners
- Both static images and animations can be submitted. Animations are
preferred in a GIF format; if the animation is too large for a GIF,
the post can point to a public YouTube video.
- Submissions will be judged by a handful of Wolfram experts, with the
following parameters:
- Visual aesthetics
- Wolfram Language code
- Creativity
- Explanation of process
- Submissions from all areas of computational art are welcome
- Submissions from former or current Wolfram employees are allowed, but
will be judged as their own category with only one winner
- Submitting previous work/posts is allowed, but must meet the
requirements stated above
## Encouragements ##
- Not sure where to start? We encourage you to look at other user's
submission for inspiration, or look at some of the work in the visual
arts group of Community:
- Artists' group: https://wolfr.am/ART-examples
- Artist (see Staff Picks section): https://community.wolfram.com/web/claytonshonkwiler
- We encourage you to vote and comment on other people's submissions.
- Please spread the word about this competition, among your friends and
other social media!
## Prizes ##
First, second, and third place winners will be featured on all of Wolfram's social media accounts, as well as receiving their choice of free Wolfram merchandise. We are able to ship merchandise to countries listed on the Wolfram Store: https://store.wolfram.com. If your country is not listed on the Wolfram Store, we strongly encourage you to still submit an entry, as we will feature winners submissions regardless of location.
### Important ###
All contest rules have been explained above under Rules & Guidelines. It's encouraged for all participants to read the rules carefully to prevent disqualification. If you have any additional questions, ask directly in the thread comments or contact us by email at t-artcontest@wolfram.com . We recommend reading other people comments as they clarify the nature of the contest as well. Comments deemed by moderators as superfluous to the thread may be removed or transferred by moderators to keep competition professional.
[1]: https://community.wolfram.com/web/danielsanderhoffmann
[2]: https://community.wolfram.com/groups/-/m/t/2518220
[3]: https://community.wolfram.com/web/antononcube
[4]: https://community.wolfram.com/groups/-/m/t/2518279
[5]: https://community.wolfram.com/web/jacquelinengocdoan
[6]: https://community.wolfram.com/groups/-/m/t/2509110
[7]: https://community.wolfram.com/web/tverhoeff
[8]: https://community.wolfram.com/groups/-/m/t/2513265
[9]: https://community.wolfram.com/web/wufei1978
[10]: https://community.wolfram.com/groups/-/m/t/2430827
[11]: https://community.wolfram.com//c/portal/getImageAttachment?filename=news-congrads-kkluyshnik-02-04-19.jpg&userId=11733Eryn Gillam2022-03-28T17:35:43ZImport a downloaded Cactus Graph program from a web?
https://community.wolfram.com/groups/-/m/t/1618730
Hellow everyone
I downloaded a notebook about Cactus Graph downloaded from *http://mathworld.wolfram.com/CactusGraph.html.* **I want to get all cacti Graph of order 8.**
But I don't know how to **use** the program from it. what should I do ?
thanksLicheng Zhang2019-02-23T05:36:44ZHow to avoid recalculation of Graphics with moving object?
https://community.wolfram.com/groups/-/m/t/2532925
Often on has to draw some complex mostly static drawing with some animated object in it. Normally this is done by Manipulate which rebuilds the whole graphics for each step. I would like to have a mechanism which, for a refresh of the moving object, does not recalculate and redraw the hole graphics but only this object.
Neither Show nor Overlay do that. The only solution I found and used is not very elegant:
- Build a Graphics from the static part alone. Rasterize it to get a Raster graphics primitive.
- For each step: Draw the moving object only, use the Raster as background.
Can someone point me to a better solution?Werner Geiger2022-05-16T17:31:22ZPlaying with Gilpin's Proposal for Advection-based Cryptographic Hashing
https://community.wolfram.com/groups/-/m/t/1330785
*WOLFRAM MATERIALS for the ARTICLE:*
> William Gilpin. *Cryptographic hashing using chaotic hydrodynamics*.
> Proceedings of the National Academy of Sciences, 115 (19) (2018), pp. 4869-4874.
> https://doi.org/10.1073/pnas.1721852115
> [Full article in PDF][1]
![enter image description here][2]
##Introduction##
Last week, William Gilpin published a fascinating paper suggesting a physics-based hashing mechanism in the Proceedings of the National Society: Cryptographic hashing using chaotic hydrodynamics: https://doi.org/10.1073/pnas.1721852115
The paper discusses a hashing algorithm based on fluid mechanics: Particles are distributed in a 2D disk. The inside of the disk is filled with an idealized fluid and the fluid is now stirred with a single stirrer. The stirring process is modelled with two possible stirrer positions. A message (say made from 0s and 1s) can now be encoded through which of the two stirrer positions is active. If the bit from the message is 1, use stirrer position 1, if the bit of the message is 0, use stirrer position 2. After some bites are processed, and the fluid is stirred, this process mixes the particles. Dropping some of the information content of the actual particle positions, e.g. by taking only the x-positions of the particle positions allows to build a hash by using the particle indices after sorting the particles along the x-axis.
The Methods section of the paper mentions that all calculations were carried out using Mathematica 11.0. Unfortunately no notebook supplement was given. So, playing around with ideas of the paper, I re-implemented some of the computations of the paper.
The paper is behind a paywall, but fortunately the author's website has a downloadable copy of the paper [here][3].
The potential importance of physics-based models for hashing, and so for cryptocurrencies was pointed out at various sites (e.g. [Stanford News][4], [btcmanager][5]).
It is fun to play around with the model.
##The chaotic advection model model##
This classic model goes back to Hassan Aref (whom I was fortunate to know personally) from his 1984 paper [Stirring by chaotic advection][6].
The Hamiltonian for the movement of a particle with coordinates {ξ,η} in a circle of radius a under the influence of agitating vortex at {x,y} (possibly time-dependent) of strength Γ and its mirror vortex.
H[{ξ_, η_}, {x_, y_}] =
Γ/(2 π) Log[ComplexExpand[ Abs[((ξ + I η) - (x + I y))/((ξ + I η) - a^2/( x - I y))]] ]
![enter image description here][7]
The resulting equations of motion:
odes = {ξ'[t] == -D[H[{ξ[t], η[t]}, {x[t], y[t]}], η[t]],
η'[ t] == +D[H[{ξ[t], η[t]}, {x[t], y[t]}], ξ[t]]} // Simplify
![enter image description here][8]
Solve the equations of motion for randomly selected parameters and plot the particle trajectory. For a position-independent and time-independent vortex, the particle moves in a circle.
nds = NDSolveValue[{
Block[{a = 1, Γ = 1, x = 0.5 &, y = 0 &}, odes],
ξ[0] == 0.3, η[0] == -0.4}, {ξ[t], η[t]}, {t, 0, 10}]
![enter image description here][9]
ParametricPlot[Evaluate[nds], {t, 0, 10}]
![enter image description here][10]
Here the vortex moves around in a periodic manner. The resulting particle trajectory has a high degree of symmetry.
nds2 = NDSolveValue[{
Block[{a = 1, Γ = 1, x = 0.8 Cos[2 Pi #/200] &, y = 0.8 Sin[2 Pi #/200] &}, odes],
ξ[0] == 0.3, η[0] == -0.4}, {ξ[t], η[t]}, {t, 0, 400}]
![enter image description here][11]
ParametricPlot[Evaluate[nds2], {t, 0, 400}, PlotPoints -> 400]
![enter image description here][12]
Letting the vortex move along a random curve results in a chaotic movement of the particle. We color the particle's trajectory with time.
(* vortex movement curve *)
randomCurve = BSplineFunction[RandomPoint[Disk[], {100}]];
X[t_Real] := randomCurve[t/100][[1]]
Y[t_Real] := randomCurve[t/100][[2]]
nds2 = NDSolveValue[{
Block[{a = 1, Γ = 10, x = X, y = Y}, odes],
ξ[0] == 0.3, η[0] == -0.4}, {ξ[t], η[t]}, {t, 0, 100}]
![enter image description here][13]
ParametricPlot[Evaluate[nds2], {t, 0, 100}, PlotPoints -> 1000,
ColorFunction -> Function[{x, y, u}, ColorData["DarkRainbow"][u]]]
![enter image description here][14]
Having two possible positions of the vortex and switching periodically between them results in a particle trajectory that is made of piecewise circle arcs.
nds2 = NDSolveValue[{
Block[{a = 1, Γ = 10, x = Sign[Sin[Pi #]]/2 &, y = 0 &}, odes],
ξ[0] == 2/3, η[0] == 0}, {ξ[t], η[t]}, {t, 0, 100}];
ParametricPlot[Evaluate[nds2], {t, 0, 100}, PlotPoints -> 1000,
ColorFunction -> Function[{x, y, u}, ColorData["DarkRainbow"][u]]]
![enter image description here][15]
Using initially 4320 points on a circle shows how the initial circle gets stretched and ripped apart and the points distribute chaotically over the disk.
nds2 = NDSolveValue[{
Block[{a = 1, Γ = 10, x = Sign[Sin[Pi #]]/2 &, y = 0 &}, odes],
ξ[0] == Table[2/3 Cos[φ], {φ, 0, 2 Pi, 2 Pi/(12 360)}],
η[0] == Table[2/3 Sin[φ], {φ, 0, 2 Pi, 2 Pi/(12 360)}]},
{ξ[t], η[t]}, {t, 0, 2}];
GraphicsGrid[
Partition[Graphics[{LightGray, Disk[], Black, PointSize[0.005],
Point[Transpose[nds2 /. t -> #]]}, ImageSize -> 120] & /@Range[0, 2, 2/19] , 5]]
![enter image description here][16]
With time running upwards, here is a 3D image of this stirring process. The transition from using the right stirrer to using the left stirrer at time 1 is clearly visible.
trajectories3D =
Transpose[Table[ Append[#, N[τ]] & /@ Transpose[nds2 /. t -> τ], {τ, 0, 2, 2/100}]];
Graphics3D[{Thickness[0.001], Opacity[0.2],
BSplineCurve /@ RandomSample[trajectories3D, 1600]},
PlotRange -> All, Axes -> True, BoxRatios -> {1, 1, 2}]
![enter image description here][17]
Here is the corresponding interactive demonstration.
Manipulate[
Graphics[{LightGray, Disk[], Black, PointSize[0.005], Point[Transpose[nds2 /. t -> τ]]}],
{τ, 0, 2, Appearance -> "Labeled"}]
![enter image description here][18]
The x-position of the particles will be used later. Here is a plot of the x-positions of 540 points initially on a circle over time.One observes many crossing of these x-trajectories.
nds2B = NDSolveValue[{
Block[{a = 1, Γ = 4, x = Sign[Sin[Pi #]]/2 &, y = 0 &}, odes],
ξ[0] == Table[2/3 Cos[φ], {φ, 0, 2 Pi, 2 Pi/540}],
η[0] == Table[2/3 Sin[φ], {φ, 0, 2 Pi, 2 Pi/540}]},
{ξ[t], η[t]}, {t, 0, 12}];
ListLinePlot[Transpose[Table[{t, #} & /@ nds2B[[1]], {t, 0., 4, 4/200}] ],
PlotStyle -> Table[Directive[Opacity[0.4], Thickness[0.001], ColorData["DarkRainbow"][j/541]], {j, 541}],
Frame -> True, Axes -> False, FrameLabel -> {t, x}]
![enter image description here][19]
Using a Dynamic[particleGraphics] we can easily model many more particles in real time without having to store large interpolating functions. We solve the the equations of motion over small time increments and the graphic updates dynamically.
rPoints = Select[Flatten[Table[N[{x, y}], {x, -1, 1, 2/101}, {y, -1, 1, 2/101}], 1], Norm[#] < 1 &];
Dynamic[Graphics[{LightGray, Disk[], Black, PointSize[0.001], Point[rPoints]}]]
![enter image description here][20]
rhsξ[X_, ξ_List, η_List] := With[{ Γ = 5}, -(((1 - X^2) Γ η (1 + X^2 - 2 X ξ))/(2 π (X^2 + η^2 - 2 X ξ + ξ^2) (1 - 2 X ξ + X^2 (η^2 + ξ^2))))]
rhsη[X_, ξ_List, η_List] := With[{ Γ = 5}, -(((1 - X^2) Γ (-ξ - X^2 ξ + X (1 - η^2 + ξ^2)))/(2 π (X^2 + η^2 - 2 X ξ + ξ^2) (1 - 2 X ξ + X^2 (η^2 + ξ^2))))]
With[{Δt = 10^-2, T = 0.2},
Monitor[
Do[
nds = NDSolveValue[
Block[{ X = Evaluate[1/2 Sign[Sin[Pi (k + 1/2) Δt/T]]] &},
{ξ'[t] == rhsξ[t, ξ[t], η[t]], η'[t] == rhsη[t, ξ[t], η[t]],
ξ[k Δt] == Transpose[rPoints][[1]],
η[k Δt] == Transpose[rPoints][[2]]}],
{ξ[t], η[t]}, {t, (k + 1) Δt, (k + 1) Δt}] /. t -> (k + 1) Δt;
rPoints = Transpose[nds],
{k, 0, 200}], k]]
Switching irregularly between the two stirrer positions gives qualitatively similar-looking trajectories. We use a sum of three trig functions see here for a detailed account on this type of function).
(* left or right stirrer is on *)
Plot[Sign[Sin[Pi t] + Sin[Pi Sqrt[2] t] + Sin[Pi Sqrt[3] t]], {t, 0, 100}, Exclusions -> None]
![enter image description here][21]
nds3 = NDSolveValue[{
Block[{a = 1, Γ = 10,
x = Sign[Sin[Pi #] + Sin[Pi Sqrt[2] #] + Sin[Pi Sqrt[3] #]]/2 &,
y = 0 &}, odes], ξ[0] == 2/3, η[0] == 0}, {ξ[t], η[t]}, {t, 0, 100}];
ParametricPlot[Evaluate[nds3], {t, 0, 100}, PlotPoints -> 1000,
ColorFunction -> Function[{x, y, u}, ColorData["DarkRainbow"][u]]]
![enter image description here][22]
We can also change the stirring direction.
nds3B = NDSolveValue[{
Block[{a = 1, Γ = 10 Sign[Sin[Pi t] + Sin[Pi Sqrt[2] t] + Sin[Pi Sqrt[3] t]],
x = Sign[Sin[Pi #] + Sin[Pi Sqrt[2] #] + Sin[Pi Sqrt[3] #]]/2 &,
y = 0 &}, odes], ξ[0] == 2/3, η[0] == 0}, {ξ[t], η[
t]}, {t, 0, 100}];
ParametricPlot[Evaluate[nds3B], {t, 0, 100}, PlotPoints -> 1000,
ColorFunction -> Function[{x, y, u}, ColorData["DarkRainbow"][u]]]
![enter image description here][23]
Here are three stirrer positions at the vertices of an equilateral triangle.
st3[t_] = Piecewise[
With[{r = RandomInteger[{0, 2}]}, {1/2. {Cos[2 Pi r/3], Sin[2 Pi r/3]}, #[[1]] <= t <= #[[2]] }] & /@
Partition[FoldList[Plus, 0, RandomReal[{0, 1}, 300]], 2, 1]];
x3[t_Real] := st3[t][[1]]
y3[t_Real] := st3[t][[2]]
nds3 = NDSolveValue[{ Block[{a = 1, Γ = 5, x = x3, y = y3}, odes], ξ[0] == 1/3, η[0] == 0}, {ξ[t], η[t]}, {t, 0, 100}]
![enter image description here][24]
ParametricPlot[Evaluate[nds3], {t, 0, 100}, PlotPoints -> 1000,
ColorFunction -> Function[{x, y, u}, ColorData["DarkRainbow"][u]]]
![enter image description here][25]
##Side note: the linked twist map##
Mathematically, alternating left and right use of the stirrer, is isomorphic to a so-called linked twist map. The following implements a simple realization of a linked twist map from Cairns /Kolganova. https://doi.org/10.1088/0951-7715/9/4/011
Clear[g, h, f];
g[{x_, y_}] := {x, Mod[Piecewise[{{y + 4 x, Abs[x] <= 1/4}}, y], 1, -1/2]}
h[{x_, y_}] := {Mod[Piecewise[{{x + 4 y, Abs[y] <= 1/4}}, x], 1, -1/2], y}
f[{x_, y_}] := g[h[{x, y}]]
f[{x, y}]
![enter image description here][26]
In explicit piecewise form, the map is more complicated. (To fit it, we use a reduce size.)
(f2 = PiecewiseExpand[# , -1/2 < x < 1/2 && -1/2 < y < 1/2] & /@ f[{x, y}])
![enter image description here][27]
Here are the first and second component visualized.
{ContourPlot[Evaluate[f2[[1]]], {x, -1/2, 1/2} , {y, -1/2, 1/2},
Exclusions -> {}, PlotPoints -> 60],
ContourPlot[Evaluate[f2[[2]]], {x, -1/2, 1/2} , {y, -1/2, 1/2},
Exclusions -> {}, PlotPoints -> 60]}
![enter image description here][28]
Applying the linked twist map to a set of points inside a square hints at the ergodic nature of the map.
ptsRG = With[{pp = 80}, Table[{RGBColor[x + 1/2, y + 1/2, 0.5], Point[{x, y}]}, {y, -1/2, 1/2, 1/pp}, {x, -1/2, 1/2, 1/pp}]];
Graphics /@ NestList[Function[p, p /. Point[xy_] :> Point[f@xy]], ptsRG, 3]
![enter image description here][29]
##Closed form mapping##
As mentioned, for a fixed stirrer position, a particle moves along a circle. This means that to move the particle to its final position, we don't have to solve nonlinear differential equations, but rather can calculate the final position through algebraic computations which are unfortunately not fully explicit. Doing the calculations (see Aref's paper for details) shows that we will have to solve of transcendental equation.
In the following we will use a unit disk of radius 1 and a unit strength vortex. The particle is initially at {r,θ} (in polar coordinates) and the stirrer is at {±b,0}.
a = 1;
Γ = 1;
The circle the particle is moving on.
circle[{r_, θ_}, b_] :=
Module[{λ, ξc, ρ},
λ = Sqrt[(b^2 + r^2 - 2 b r Cos[θ])/(a^4/b^2 + r^2 - 2 a^2 r Cos[θ]/b)];
ξc = (b - λ^2 a^2/b)/(1 - λ^2);
ρ = Abs[λ/(1 - λ^2) (a^2/b - b)];
Circle[{ξc, 0}, ρ]]
The period for one revolution on the circle.
period[{r_, θ_}, b_, T_] :=
Module[{λ, ρ, Tλ},
λ = Sqrt[(b^2 + r^2 - 2 b r Cos[θ])/(a^4/b^2 + r^2 - 2 a^2 r Cos[θ]/b)];
ρ = Abs[λ/(1 - λ^2) (a^2/b - b)];
Tλ = (2 Pi)^2 ρ^2/Γ (1 + λ^2)/(1 - λ^2) ]
The final position after rotating for time T.
rotate[{r_, θ_}, b_, T_, opts___] :=
Module[{(*λ,ξc,ρ,θp,Tλ,eq,fr*)},
λ = Sqrt[(b^2 + r^2 - 2 b r Cos[θ])/(a^4/b^2 + r^2 - 2 a^2 r Cos[θ]/b)];
ξc = (b - λ^2 a^2/b)/(1 - λ^2);
ρ = Abs[λ/(1 - λ^2) (a^2/b - b)];
θp = ArcTan[r Cos[θ] - ξc, r Sin[θ]];
Tλ = (2 Pi)^2 ρ^2/Γ (1 + λ^2)/(1 - λ^2);
(* the implicit equation to be solved numerically for θt *)
eq = θt - 2 λ/(1 + λ^2) Sin[θt] == θp - 2 λ/(1 + λ^2) Sin[θp] + 2 Pi T/Tλ;
fr = FindRoot[eq, {θt, θp}, opts] // Quiet; (* could use Solve[eq, θt, Reals] *)
{Sqrt[ρ^2 + ξc^2 + 2 ρ ξc Cos[θt]],
ArcTan[ρ Cos[θt] + ξc, ρ Sin[θt]]} /. fr]
A high-precision version for later use.
rotateHP[{r_, θ_}, b_, T_ ] :=
With[{prec = Precision[{r, θ}]},
rotate[{r, θ}, b, T, WorkingPrecision -> Min[prec, 200] - 1, PrecisionGoal -> prec - 10]]
rotateHP[{2/5, 1}, 1/2, 10^-10]
{0.3999999999735895736749310453217879335729586969724148186668989989685300449569453838180010833652353802304511654926113687832716502416432363896855780225351918314742068365263060199921670958293439964299441,
1.000000000034865509829422032384759003150106866109371061507406908121303212377635808764825759934117372434662086592066738925787825280696681907390281734196810440241159138906978471624124765879791828431390}
Also for later use, rotate many points at once.
rotate[l : {_List ..}, b_, T_ ] := rotate[#, b, T] & /@ l
rotateHP[l : {_List ..}, b_, T_ ] := rotateHP[#, b, T] & /@ l
The locator is the initial particle position; the stirrer position is the purple point. We show the circle and the final particle position (gray point).
Manipulate[
Graphics[{LightGray, Disk[],
Purple, PointSize[0.02], Point[{b, 0}] ,
Gray, circle[ToPolarCoordinates[p], b] ,
PointSize[0.01],
Point[p], Blue ,
Point[ FromPolarCoordinates[ rotate @@ SetPrecision[ {ToPolarCoordinates[p], b, T}, 200] ]]}],
{{T, 0.2}, 0.001, 10},
{{b, 0.5}, -0.999, 0.999},
{{p, {0.3, 0.4}}, Locator}]
![enter image description here][30]
Here is a plot of the period for one revolution as a function of the initial position of the particle. The period can get quite large when points are near the boundary of the disk
ParametricPlot3D[{r Cos[θ], r Sin[θ], period[{r, θ}, 1/2, 1]},
{r, 0, 1}, {θ, -Pi, Pi}, BoxRatios -> {1, 1, 0.3},
PlotPoints -> 40, MeshFunctions -> {#3 &}]
![enter image description here][31]
For the stirrer located at {0.5,0}, here are the possible circles that the particle will move on.
Graphics[Table[circle[{r, θ}, 1/2.] , {r, 1/10, 9/10, 1/10}, {θ, -Pi, Pi, 2 Pi/20}]]
![enter image description here][32]
We use the closed form of the stirring map to visualize the flow. Here are about 20k points in stripes in the unit disk.
stripePoints = Select[RandomPoint[Disk[], 40000], Function[xy, Mod[xy[[1]], 0.2] < 0.1]];
Graphics[{PointSize[0.002], Point[stripePoints]}]
![enter image description here][33]
We repeatedly stir with the left and then the right stirrer on for time 1.
swirled[1] = rotate[ToPolarCoordinates[stripePoints], 0.5, 1];
Graphics[{PointSize[0.002], Point[FromPolarCoordinates[swirled[1]]]}]
![enter image description here][34]
swirled[2] = rotate[swirled[1], -0.5, 1];
Graphics[{PointSize[0.002], Point[FromPolarCoordinates[swirled[2]]]}]
![enter image description here][35]
swirled[3] = rotate[swirled[2], 0.5, 1];
Graphics[{PointSize[0.002], Point[FromPolarCoordinates[swirled[3]]]}]
![enter image description here][36]
swirled[4] = rotate[swirled[3], -0.5, 1];
Graphics[{PointSize[0.002], Point[FromPolarCoordinates[swirled[4]]]}]
![enter image description here][37]
##Now stir and make hashes##
We now use two different initial positions of the points: a sunflower-like one and a random one.
randominitialPositions[n_, prec_] :=
SortBy[Table[ReIm[RandomReal[{0, 1}, WorkingPrecision -> prec] *
Exp[ I RandomReal[{-Pi, Pi}, WorkingPrecision -> prec]]], {n}], First]
SeedRandom[1234];
initialPositionsR = randominitialPositions[100, 250];
Graphics[{LightGray, Disk[],
MapIndexed[Function[{p, pos},
{Red, PointSize[0.002], Point[p], Black, Text[pos[[1]], p]}],
N[initialPositionsR]]}]
![enter image description here][38]
sunflower[n_, R_, prec_] :=
Table[FromPolarCoordinates[ N[{Sqrt[j/n] R, Mod[2 Pi (1 - 1/GoldenRatio) j, 2 Pi, -Pi]}, prec]], {j, n}]
Graphics[{LightGray, Disk[], Black, Point[sunflower[1000, 0.9, 20]]}]
![enter image description here][39]
The message to encode we will tell in the form of a sequence of 0s and 1s. A zero means use the left stirrer position and a 1 means use the right stirrer position. So, we define a stirring process with a given message σ and fixed b and T as follows:
stirringProcess[initialPositions_, bAbs_, T_, σ_List] :=
FromPolarCoordinates /@
FoldList[rotateHP[#1, #2 bAbs, T] &, ToPolarCoordinates /@ initialPositions, 2 σ - 1]
Here is a simple message:
message1 = RandomInteger[{0, 1}, 50]
{1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1,1, 0, 1, 0}
Now we carry out the corresponding stirring process. (We arbitrarily use b=1/2 and T=1/10.)
sP1 = stirringProcess[initialPositionsR, 1/2, 1/10, message1];
To get a feeling for the stirring process, we connect successive positions of each particle with a randomly colored spline interpolation. (Note that the spline interpolation does not exactly represent the particle's trajectory.)
Graphics[{LightGray, Disk[], Thickness[0.001],
Map[Function[c, {RandomColor[], BSplineCurve[c, SplineDegree -> 5]}],
Transpose[N[Table[sP1[[j]], {j, 50}]]]]}]
![enter image description here][40]
Now, can we trust these results? We can check by running the stirring process (numerically) backwards.
sP1Rev = stirringProcess[sP1[[-1]], 1/2, -1/10, Reverse[message1]];
The so-recovered initial positions agree to at least 80 digits with the original point positions.
sP1Rev[[-1]] - initialPositionsR // Abs // Max // Accuracy
86.9777
As mentioned earlier, we define the hashes as the positions of the particles when ordered according to their x-values. So, for a given particle configurations, this is the hash-making function.
makeHash[l_] := Sort[ MapIndexed[{#1, #2[[1]]} &, l]][[All, 2]]
Because the hashes are permutations of the particle numberings, the number of hashes grows factorially with the particle number and 58 particles could model a 256 bit hash and 99 particles a 512 bit hash.
{2^256, {57!, 58!}} // N
{1.15792*10^77, {4.05269*10^76, 2.35056*10^78}}
{2^512, {98!, 99!}} // N
{1.34078*10^154, {9.42689*10^153, 9.33262*10^155}}
hashes1 = makeHash /@ sP1;
Here is a visualization how the hashes evolve with each stir. The coloring is from blue (1) to 100 (red) of the initial particle numbering.
ArrayPlot[hashes1, ColorFunction -> ColorData["DarkRainbow"], PlotRange -> {1, 100}]
![enter image description here][41]
This is how the edit distance of the hashes increases with successive stirrings.
ListPlot[Table[{j, EditDistance[hashes1[[1]], hashes1[[j]]]}, {j, 1, 50}], Filling -> Axis]
![enter image description here][42]
Gilpin defines the rearrangement index of a hash as the sum of the absolute difference between neighboring particle indices.
rearrangementIndex[indexList_] := Total[Abs[Differences[indexList]]]
Here is the growth rate of the rearrangement index for our just-calculated hashes.
ListPlot[Table[{j, rearrangementIndex[hashes1[[j]]]}, {j, 1, 50}], Filling -> Axis]
![enter image description here][43]
Let's make a second message that has one bit flipped, say at position 10.
message2 = MapAt[1 - # &, message1, 10]
{1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0,
0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0}
message1 - message2
{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
The resulting particle movements look quite different.
sP2 = stirringProcess[initialPositionsR, 1/2, 1/10, message2];
Graphics[{LightGray, Disk[], Thickness[0.001],
Map[Function[c, {RandomColor[], BSplineCurve[c, SplineDegree -> 5]}],
Transpose[N[Table[sP2[[j]], {j, 50}]]]]}]
![enter image description here][44]
The edit distances between the two hash sequences increase quickly after the different bit is encountered.
hashes2 = makeHash /@ sP2;
ListPlot[Table[{j, EditDistance[hashes1[[j]], hashes2[[j]]]}, {j, 0, 50}], Filling -> Axis]
![enter image description here][45]
Gilpin carried out many numerical experiments investigating how the hashes behave as a function of the hash length (number of particles), stirring time T, and message length.
The next two examples change only the stirring time, and keep all other parameters: message, stirrer distance, initial particle distance.
sP1B = stirringProcess[initialPositionsR, 1/2, 1/9, message1];
hashes1B = makeHash /@ sP1B;
ListPlot[Table[{j, EditDistance[hashes1[[j]], hashes1B[[j]]]}, {j, 0, 50}], Filling -> Axis]
![enter image description here][46]
With increasing stirring time, the edit distance between the hashes quickly increases.
sP1C = stirringProcess[initialPositionsR, 1/2, 1/2, message1];
hashes1C = makeHash /@ sP1C;
ListPlot[Table[{j, EditDistance[hashes1[[j]], hashes1C[[j]]]}, {j, 0, 50}], Filling -> Axis]
![enter image description here][47]
Next, we use the phyllotaxis-like initial position from above with 200 particles. We also use a longer stirring time (T=1).
message3 = RandomInteger[{0, 1}, 50]
{0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0,1, 0, 1, 0}
sP3 = stirringProcess[sunflower[200, 9/10, 250], 1/2, 1, message3];
A plot of the 200 particle trajectories suggests a good mixing.
Graphics[{LightGray, Disk[], Thickness[0.001],
Map[Function[c, {RandomColor[], BSplineCurve[c, SplineDegree -> 5]}],
Transpose[N[Table[sP3[[j]], {j, 50}]]]]}]
![enter image description here][48]
Already after the first stir, the edit distance between the original hash and the one resulting after stirring is quite large.
hashes3 = makeHash /@ sP3;
ListPlot[Table[{j, EditDistance[hashes3[[1]], hashes3[[j]]]}, {j, 0, 50}], Filling -> Axis]
![enter image description here][49]
We will end here. The interested reader can continue to model, stir a million times, count trajectory exchanges, and statistically analyze other aspects of the proposed hashing scheme from Gilpin's paper.
PS: And one can compare with theoretical hashing probabilities for ideal hashes. Such as: draw \[ScriptCapitalN]s hashes from M! possible hashes. How many different hashes Subscript[\[ScriptCapitalN], U] does one draw in average?
uniqueHashes[M_, \[ScriptCapitalN]s_] :=
M! (1 - (1 - 1/M!)^\[ScriptCapitalN]s)
Plot3D[uniqueHashes[M, \[ScriptCapitalN]s], {M, 1, 60}, {\[ScriptCapitalN]s, 1, 1000}, MeshFunctions -> {#3 &},
PlotPoints -> 100, ScalingFunctions -> "Log", PlotRange -> All,
WorkingPrecision -> 50,
AxesLabel -> {Style["M", Italic], Style["\[ScriptCapitalN]s", Italic], Style["\!\(\*SubscriptBox[\(\[ScriptCapitalN]\), \(U\)]\)", Italic]}]
![enter image description here][50]
For Ns≪M!, this becomes:
Series[Mfac (1 - (1 - 1/Mfac)^\[ScriptCapitalN]s), {Mfac, ∞, 2}] /. Mfac -> M! // Simplify
![enter image description here][51]
uniqueHashesApprox[M_, \[ScriptCapitalN]s_] := \[ScriptCapitalN]s - (\[ScriptCapitalN]s*(\[ScriptCapitalN]s - 1))/(2*M!) + (\[ScriptCapitalN]s*(\[ScriptCapitalN]s - 1)*(\[ScriptCapitalN]s - 2))/(6*M!^2)
Here is a quick numerical modeling with a sample size of 100 from 6!=720 possible hashes.
With[{M = 6, \[ScriptCapitalN]s = 100},
{Table[Length[
Tally[RandomChoice[Range[M!], \[ScriptCapitalN]s]]], {10000}] // Mean,
uniqueHashes[M, \[ScriptCapitalN]s],
uniqueHashesApprox[M, \[ScriptCapitalN]s]}] // N
{93.407, 93.4267, 93.4369}
I definitely suggest reading the whole paper, including the Supplementary Information.
The notebook with all the code and visualizations is attached.
[1]: https://www.pnas.org/doi/epdf/10.1073/pnas.1721852115
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2018-05-02at11.55.34AM.png&userId=20103
[3]: http://www.wgilpin.com/papers/gilpin_pnas_2018.pdf
[4]: https://news.stanford.edu/2018/04/23/swirling-liquids-shed-light-bitcoin-works/
[5]: https://btcmanager.com/stanford-university-physicists-uncover-correlation-between-bitcoin-transactions-and-laws-of-nature/
[6]: https://www.cambridge.org/core/journals/journal-of-fluid-mechanics/article/stirring-by-chaotic-advection/7B32CACE61D5AD79077846D7ACF4A31E
[7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=61661.png&userId=20103
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[51]: http://community.wolfram.com//c/portal/getImageAttachment?filename=1026844.png&userId=20103Michael Trott2018-05-02T11:44:45ZOptimize the code (Shirley algorithm for XPS data)?
https://community.wolfram.com/groups/-/m/t/1274188
Hello dear Community,
Recently I've been working on creating a complex program for XPS analysis and I realized I have trouble making the code for Shirley background removal optimized (the code is pasted below). First of all, I thought of using the nested list to generate the iterations of the background, which in turn forced me to use a slow While loop. Second of all, I thought of compiling the code, since it will be used often, however, I have large block of calculations and I want the compiled function/module to return the final result (background) only. So to sum up, I would like to be able to invoke the code below by providing the data (a list of intensities), number of iterations and two values of energy, which constitute for the boundaries of the calculation.
I tried to be descriptive in the comments in code, however, if something is not clear I will try to clear it up as soon as I can.
energyR = 187; energyL = 196; (* L < E < R - boundaries for which background is generated*)
iterations = 50; (* number of iterations *)
NearestPosition[haystack_, needle_] := Nearest[haystack, needle] /. MapIndexed[Rule, haystack];
(*function which searches the data for closest value in the list and returns the index of the number*)
eLi =
NearestPosition[data59B[[All, 1]], energyL][[1, 1]];
eRi = NearestPosition[data59B[[All, 1]], energyR][[1, 1]];
data = data59B[[eLi ;; eRi,
All]]; (*this line takes the data in x-y format and cuts of \
non-analyzed ends of the spectra*)
max = data // Length;
B0 = Table[
x (-data[[1, 2]] + data[[max, 2]])/max + data[[1, 2]], {x, 1,
max}];(* three lines here set up the initial background (which is \
assumed to be linear) and geenrates the first iteration of \
calculations*)
k1 = (data[[1, 2]] - data[[max, 2]])/
Total[data[[All, 2]] - data[[max, 2]] - B0];
B1 = Table[
k1 Total[(data[[x ;; max, 2]] - data[[max, 2]] -
B0[[x ;; max]])], {x, 1, max}];
Btable = Table[
B1, {i, 1,
iterations}]; (*since i have trouble generating nested list i'm \
making local table to contain the iterative results*)
ktable =
Table[k1, {i, 1, iterations}];
j = 2;
While[j <= iterations,
ktable[[j]] = (data[[1, 2]] - data[[max, 2]])/
Total[data[[All, 2]] - data[[max, 2]] - Btable[[j - 1]]];
Btable[[j]] =
Table[ktable[[j]] Total[(data[[x ;; max, 2]] - data[[max, 2]] -
Btable[[j - 1, x ;; max]])], {x, 1, max}]; j++];
N[ktable[[iterations]] -
ktable[[iterations -
1]]] (*this returns the parameter with which we can estimate the \
convergence of the calculations*)
background =
data[[max, 2]] +
Btable[[iterations]]; (*and here we have a final result - a \
background for further analysis*)Igor Wlasny2018-01-30T09:16:30ZWolfram Problem Generator - downloadable worksheets
https://community.wolfram.com/groups/-/m/t/2531574
Hello,
I was wondering what the format is for the downloadable generated problems? Is it a pdf or jpg? I might like to insert generated problems into a Google Slides presentations as examples? I'm looking for something that can replace MathType (which kind of stinks).
Thank you!John Rogers2022-05-14T20:40:52ZEntering a limit of sum product into Wolfram|Alpha?
https://community.wolfram.com/groups/-/m/t/2532484
Hello,
I'm new to wolframalpha.com, not sure I am in the right place to ask this question, apologies if. I just want to know if Wolfram|Alpha is able to calculate this limit, to know if the sum product converges or not:
$\lim_{m\rightarrow\infty} \sum_{n=1}^{m} n^{-0.5} \cos(\ln(n)) \sum_{n=1}^{m} (-1)^{n} n^{-0.5} \cos(\ln(n))$
The LateX may not work...
This is the "code" I enter, is it correct, please?
lim( sum( n^{-0.5} * cos( ln(n) ) ,n,1,m ) * sum( (-1)^{n} *n^{-0.5} * cos(ln(n)) ,n,1,m),m,inf)
What should I modify, please? Because it gives something else like "Interpreting as: cos", or it may be too much for the machine?B B2022-05-16T11:59:24ZTry to beat these MRB constant records!
https://community.wolfram.com/groups/-/m/t/366628
The MRB constant: ALL ABOARD!
-----------------------------
POSTED BY: Marvin Ray Burns.
![If you see this instead of an image, reload the page.][1]
----------
**C**<sub>*MRB*</sub> is defined at
- [https://en-academic.com/][2], Wikipedia, Mathematical constant;
- http://constant.one/ ;
- Crandall, R. E. "The MRB Constant." §7.5 in [Algorithmic Reflections: Selected Works][3]. PSI Press, pp. 28-29, 2012,ISBN-10 : 193563819X ISBN-13: 978-1935638193;
- Crandall, R. E. "[Unified Algorithms for Polylogarithm, L-Series, and Zeta Variants][4]." 2012;
- Encyclopedia of Mathematics (Series #94);
- [Engineering Tools][5] of the Iran Civil Center, an international community dedicated to the construction industry, ISSN: 1735–2614;
- Etymologie CA Kanada Zahlen" (in German). etymologie.info;
- Finch, S. R. [Mathematical Constants][6], Cambridge, England:
Cambridge University Press, p. 450, 2003, ISBN-13: 978-0521818056, ISBN-10: 0521818052;
- Finch's original essay on [Iterated Exponential Constants][7];
- Finch, Steven & Wimp, Jet. (2004). Mathematical constants. The Mathematical Intelligencer. 26. 70-74. 10.1007/BF02985660;
- Journal of Mathematics Research; [Vol. 11, No. 6; December 2019][8] ISSN 1916-9795 E-ISSN 1916-9809 Published by Canadian Center of Science and Education;
- Mauro Fiorentina’s [math notes][9] (in Italian);
- MATHAR, RICHARD J. "NUMERICAL EVALUATION OF THE OSCILLATORY INTEGRAL OVER exp(iπx) x^(1/x) BETWEEN 1 AND INFINITY" [(PDF)][10]. arxiv. Cornell University;
- [Mathematical Constants and Sequences][11] a selection compiled by
Stanislav Sýkora, Extra Byte, Castano Primo, Italy. Stan’s Library,
ISSN 2421-1230, Vol.II;
- ["Matematıksel Sabıtler"][12] (in Turkish). Türk Biyofizik Derneği.
- [MathWorld Encyclopedia][13];
- [OEIS Encyclopedia (The MRB constant);][14]
- [Plouffe's Inverter;][15]
- the LACM [Inverse Symbolic Calculator;][16]
- The On-Line Encyclopedia of Integer Sequences® (OEIS®) as
[A037077][17], Notices Am. Math. Soc. 50 (2003), no. 8, 912–915, MR 1992789 (2004f:11151);
- [Wikipedia Encyclopedia][18].
Content of this first post, as of May 24, 2022
==========================
1. How it all began
2. The why and what of the **C**<sub>*MRB*</sub> Records,
3. **C**<sub>*MRB*</sub> and its applications,
4. MeijerG Representation for **C**<sub>*MRB*</sub>,
5. **C**<sub>*MRB*</sub> formulas and identities,
6. Primary Proof 1,
7. Primary Proof 2,
8. Primary Proof 3,
9. The relationship between **C**<sub>*MRB*</sub> and its integrated analog,
10. The MRB constant supercomputer 0
Second post:
------------
The following might help anyone serious about breaking my record.
Third post
----------
The following email Crandall sent me before he died might be helpful for anyone checking their results.
Fourth post
-----------
Perhaps some of these speed records will be easier to beat.
Many more interesting posts
---------------------------
...including the MRB constant supercomputers 1 and 2.
...including records of computing the MRB constant from Crandall's eta derivative formulas.
...including all the methods used to compute **C**<sub>*MRB*</sub> and their efficiency.
...including the dispersion of the 0-9th decimals in **C**<sub>*MRB*</sub> decimal expansions.
...including the convergence rate of 3 major different forms of **C**<sub>*MRB*</sub>.
...including complete documentation of all multimillion-digit records with many highlights.
...including arbitrarily close approximation formulas for **C**<sub>*MRB*</sub>.
...including efficient programs to compute the integrated analog (MKB) of **C**<sub>*MRB*</sub>.
...including an incredible 7 million digits on 1/4 of MRB constant supercomputer 3 (10 cores at 5.2 GHz and Ram, 64 GB 3200 MHz RAM dedicated to those 10 cores).
...including a recent discovery that could help in verifying digital expansions of the integrated analog (MKB) of **C**<sub>*MRB*</sub>.
...including an incredible 10 million digits on 3/4 of MRB constant supercomputer 3 (30 cores at 5.2 GHz and 224 GB 4800 MHz RAM dedicated to those 30 cores), using an algorithm with a 4th degree
convergence rate.
.
----------
----------
----------
----------
How it all began
==================
**From these meager beginnings:**
I was a D and F student through 6th grade the second time, but in Jr high, in 1976, we were given a self-paced program. Then I noticed there was more to math than rote multiplication and division of 3 and 4-digit numbers! Instead of repetition, I was able to explore what was out there. The more I researched, the better my grades got! It was amazing!! So, having become proficient in mathematics during my high school years, on my birthday in 1994, I decided to put down the TV remote control and pick up a pencil. I began by writing out the powers of 2, like 2*2, 2*2*2, etc. I started making up algebra problems to work at solving, and even started buying books on introductory calculus.
Then came my first opportunity to attend university. I took care of my mother, who suffered from Alzheimer's, so instead of working my usual 60+ hours a week. I started taking a class or two a semester. After my mom passed away, I went back to working my long number of hours but always kept up on my math hobby!
Occasionally, I make a point of going to school and taking a class or two to enrich myself and my math hobby. This has become such a successful routine that some strangers listed me on Wikipedia as an amateur mathematician alphabetically following Jost Bürgi who constructed a table of progressions that is now understood as antilogarithms independently of John Napier at the behest of Johannes Kepler.
I have studied a few graduate-level topics in Mathematics but have yet to earn my Bachelor's, which I intend to work on in the coming year while still working full-time.
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**From these meager beginnings:**
On January 11 and 23,1999 I wrote,
I have started a search for a new mathematical constant! Does anyone want to help me? Consider, 1^(1/1)-2^(1/2)+3^(1/3)...I will take it apart and examine it "bit by bit." I hope to find connections to all kinds of arithmetical manipulations. I realize I am in "no man's land," but I work best there! If anyone else is foolhardy enough to come along and offer advice, I welcome you.
I came to find out that this constant (CMRB)
> ![enter image description here][19]
from https://mathworld.wolfram.com/MRBConstant.html
was more closely related to other constants than I could have imagined:
For example, consider its relationship to Viswanath's constant (VC)
> ![enter image description here][20]
form https://mathworld.wolfram.com/RandomFibonacciSequence.html
**With both being functions of x<sup>1/x</sup> alone,** we have these approximations:
In 2016 I found
![enter image description here][21]==3.57*10^-11.
In[232]:= (-801233750 + 459067873 \[Pi] -
827447500 CMRB \[Pi])/134832500 - VC
Out[232]= 3.57*10^-11
The following two much simpler ones, with the memory of Zeta(2) = Pi^2/6 and Pi/2≈11/7, bear some resemblance to a would-be limit-sum analog to the Abel-Plana integral-sum equations :
In[2]:= Reduce[w Exp[w] == 1, w]
Out[2]= C[1] \[Element] Integers && w == ProductLog[C[1], 1]
![enter image description here][22]
The Mathematica code and result:
VC = 1.1319882487943061;
CMRB = 0.1878596424620671202485179 ;
VC/(6(ProductLog[E]))-CMRB
(* 0.00080506567032*).
VC/(66/7-6ProductLog[1])-CMRB
(* 3.416388*^-8*).
I believe they are approximations rather than equations. Nonetheless, they are ratios of integers and named constants. It is difficult to be certain of the value of VC beyond less than a dozen decimals.
If we want to "reach for the stars" though, there is this, which is accurate to the limit of well-theorized estimates of the value of VC:
![enter image description here][23]
In[436]:= VC = 1.1319882487943061`14;
In[437]:=
VC/(66/7 - 6 ProductLog[1 - 96 h/73]) - CMRB /.
h -> (2 ArcCos[Sqrt[ModularLambda[I/10]]])/\[Pi]
Out[437]= -0.*10^-14
.
**But wait, there is more than that division relationship to Viswanath's-MRB constant; there is a surd relationship:**
![enter image description here][24]
![enter image description here][25]
and
![enter image description here][26]
Here are those codes:
CMRB = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity},
WorkingPrecision -> 20, Method -> "AlternatingSigns"];
For[n = 1, n < 10,
Print[WolframAlpha[ToString[N[((56/5)*CMRB)^(1/n), 10]],
"PodCells"][[6]]], n++]
and
Clear[a]; VC = 1.1319882487943061`14; CMRB = 0.1878596424620671202485179;
Table[a[n] = CMRB - n VC/56; {a[n], "-CMRB", n, "*VC^6/56=",
p = Rationalize[a[n]/CMRB, 1/100], "CMRB=", N[p CMRB, 10]}, {n, 1, 9}] // TableForm
So, are the division and surd approximations related?
I don't know, but their ratio is approximately [Gelfond's constant][27] $=e^\pi.$
![enter image description here][28]
See [cloud notebook][29].
----------
Here is a relationship CMRB has with a couple of other constants,
Artin constant:
> ![enter image description here][30]
from https://mathworld.wolfram.com/ArtinsConstant.html
and
Foias constant:
> ![enter image description here][31]
from https://mathworld.wolfram.com/FoiasConstant.html
Artin constant/MRB constant≈1.99 .
Foias constant - MRB constant≈ 0.999 .
1 - (MRB constant - 1/2 Artin constant) ≈ 0.999 .
Foias constan - MRB constant ≈0.999
In[445]:=
CArtin = 0.3739558136192;(*https://mathworld.wolfram.com/ArtinsConstant.html*)
CFolis = 1.18745235112;(*https://mathworld.wolfram.com/FoiasConstant.html*)
CMRB = 0.187859642462;(*https://mathworld.wolfram.com/MRBConstant.html*)
CArtin/CMRB
CFolis - CMRB
1 - (CMRB - 1/2 CArtin)
CFolis - CMRB
Out[445]= 1.99061
Out[446]= 0.999593
Out[447]= 0.999118
Out[448]= 0.999593
----------
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**From these meager beginnings:**
On Feb 22, 2009, I wrote,
> It appears that the absolute value, minus 1/2, of the limit(integral of (-1)^x*x^(1/x) from 1 to 2N as N->infinity) would equal the partial sum of (-1)^x*x^(1/x) from 1 to where the upper summation is even and growing without bound. Is anyone interested in improving or disproving this conjecture?
>
> ![enter image description here][32]
I came to find out it is a very slow to converge [oscillatory integral][33] that could later be looked up in [Google Scholar.][34]
After I mentioned it to him, Richard Mathar published his meaningful work on it [here in arxiv][35], where M is the MRB constant and M1 is MKB:
> ![enter image description here][36]
**I developed a lot more theory behind it and ways of computing many more digits in [this linked][37] Wolfram post.**
Here is how my analysis (along with improvements to Mathematica) has improved the speed of calculating that constant's digits:
(digits and seconds)
[ 2015 ] [ 2021 Method ]
V 10.1.2 V10.3 v11.3 V12.0 V12.1 V12.3 V13.0
1000 3.3 3.1
2000 437 256 67 67 58 21 20
3000 889 794 217 211 186 84 ?
4000 1633 514 492 447 253* 259 (248*)
5000 2858 1005 925 854 386 378
10000 17678 8327 7748 7470 2800 2748
20000 121,431 71000 66177
40000 362,945 148,817 134,440
* means from a fresh kernel.
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**From these meager beginnings:**
On **Sat, 6 Feb 2010**, it came to me to ask what value of $a$ gives $$\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-a)=0\text{ ?}$$(For what value of a is the Levin's u-transform's and Cesàro's sum result 0 considering weak convergence?)
$C$MRB is approximately 0.1878596424620671202485179340542732. See [this](http://www.wolframalpha.com/input/?i=0.1878596424620671202485179340542732300559030949001387&lk=1&a=ClashPrefs_*Math-)
and [this.](http://www.wolframalpha.com/input/?i=mrb+constant&t=elga01)
$\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-a)$ is formally convergent only when $a =1$. However, if you extend the meaning of $\sum$ through "summation methods", whereby series that diverge in one sense converge in another sense (e.g. Cesaro, etc.) you get results for other $a$.
The solution I got surprised me: it was $a=1-2\times C\mathrm{MRB}=0.6242807150758657595029641318914535398881938101997224\ldots$.
Where $C\mathrm{MRB}$ is $\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-1)$.
To see this for yourself in Mathematica enter
`FindRoot[NSum[(-1)^n*(n^(1/n) - x), {n, 1, Infinity}], {x, 1}]` where regularization is used so that the sum that formally diverges returns a result that can be interpreted as evaluation of the analytic extension of the series.
I wrote Mathematica to find out what method is used and they replied:
> From Marvin Ray Burns <bmmmburns@sbcglobal.net> Date: Sat, 6 Feb 2010
> 21:03:54 -0600 Subject: Premier Service Help Form To:
> support@wolfram.com
>
Name: Marvin Ray Burns Email: ******.net Organization: IUPUI License: L***** Version: 6 OS: "Windows"
Suggestion or Bug: I am writing about the "AlternatingSigns" Method in NSum. Consider the Divergent series NSum[(-1)^n*(n^(1/n) - a), {n, Infinity}, Method -> "AlternatingSigns"], where Abs[a]<1. Strangely, Mathematica does compute a value for many of those series. Obviously, the algorithms used in the "AlternatingSigns" Method produce a value independently of the fact that the series is divergent. Perhaps one should not be surprised that a value is returned but IT SHOULD BE MADE CLEAR WHAT THAT VALUE REPRESENTS! What is the meaning of that value given that its series is divergent? From my experiments in V7 that value is not reproducible with Sum[, Regularization-> ].`
> Hello,
>
> Thank you for the email.
>
`Note that in V7, you may also try these:b `In[11]:= Sum[(-1)^n*(n^(1/n) - 0.624277766757), {n, Infinity},Method -> "AlternatingSigns"]`b> During evaluation of In[11]:= Sum::div:Sum does not converge. >>b `Out[11]= Sum[(-1)^n*(-0.624277766757 + n^(1/n)), {n, Infinity}, Method -> "AlternatingSigns"]``In[33]:= Sum[(-1)^n*(n^(1/n) - 0.624277766757), {n, Infinity}]` During evaluation of In[33]:= Sum::div:Sum does not converge. >> `Out[33]= Sum[(-1)^n*(-0.624277766757 + n^(1/n)), {n, Infinity}]` Thank you for taking the time to send us this report. I have forwarded your examples to our development group. We apologize for any inconvenience caused by this problem.`
>
> I have included your contact information so that you can be notified
> when this has been resolved.
>
> Sincerely,
>
> *******, Ph.D. Technical Support Wolfram Research, Inc. http://support.wolfram.com
>
`Thank you for your email. If you do not specify a Method for NSum it will try to choose between the EulerMaclaurin or WynnEpsilon methods. In any case, some implicit assumptions about the functions you are summing have to be made. If these assumptions are not correct, you may get inaccurate answers`.
Numerical Sums and Products f the ratio test does not give 1, the Wynn epsilon algorithm is applied to a sequence of partial sums or products. Otherwise Euler\[Dash]Maclaurin summation is used with Integrate or NIntegrate.`
1-2$C$MRB is what I call Ma. Here is the formula for Ma by way of Levin-Type Sequence Transformations: in Maple
`Digits := 20; fsolve(sum((-1)^j*(j^(1/j)-a), j = 1 .. infinity) = 0, a)`,
giving 0.62428071507586575950,
and in Mathematica,
Block[{$MaxExtraPrecision = 10000}, FindRoot[NSum[(-1)^j*(j^(1/j) - a), {j, Infinity}], {a, 0.6}, WorkingPrecision -> 20]]
giving {n -> 0.62428071507608096085}.
There is a discrepancy between the two results, but they do agree that Ma is approximately = 0.624280715076.
Finally let a = Ma = $1-2C$MRB = 0.6242807150758... and the two limit-points of the series $\sum_{n=1}^\infty (-1)^n(n^{1/n}-Ma)$ are +/- $C$MRB with its Levin's u-transform's result being 0.
See the integer sequence [here.](http://oeis.org/A173273)
At about the same time, I noticed the following when I extend the meaning of $\sum$ through "summation methods", whereby series that diverge in one sense converge in another sense (e.g. Cesaro, etc.)
Let $c=$MRB constant; $x$, and $y =$ any number,
then it can be shown that
$$\sum_{n=1}^\infty (-1)^n(x n^{1/n}+y n)=(c-1/2) x-\frac14 y$$
----------
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**From these meager beginnings:**
In October 2016, I wrote the following [here in researchgate][38]:
First, we will follow the path the author took to find out that for
![enter image description here][39]
the limit of the ratio of a to a - 1, as a goes to infinity is Gelfond's Constant, (e ^ pi). We will consider the hypothesis and provide hints for proof using L’ Hospital’s Rule (since we have indeterminate forms as a goes to infinity). We find there is no limit a goes to infinity of the ratio of the previous forms of integrals when the "I" is left out, and give a small proof for their divergence.
That was responsible for the integral-equation-discovery mentioned in one of the following posts, where it is written, "Using those ratios, it looks like" (There m is the MRB constant.)
> ![enter image description here][40]
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----------
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MRB Constant Records,
====================
My inspiration to compute a lot of digits of CMRB came from [the following website by Simon Plouffe][41].
There, computer mathematicians calculate millions, then billions of digits of constants like pi, when with only 65 decimal places of pi, we could determine the size of the observable universe to within a Planck length (where the uncertainty of our measure of the universe would be greater than the universe itself)!
In contrast 65 digits of the MRB constant "measures" the value of -1+ssqrt(2)-3^(1/3) up to n^(1/n) where n is 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, which can be called 1 unvigintillion or just 10^66.
And why compute 65 digits of the MRB constant? Because having that much precision is the only way to solve such a problem as
> 1465528573348167959709563453947173222018952610559967812891154^ m-m,
> where m is the MRB constant, which gives the near integer "to beat
> all,"
> 200799291330.9999999999999999999999999999999999999999999999999999999999999900450...
And why compute millions of digits of it? uhhhhhhhhhh.... "Because it's there!" (...Yeah, thanks George Mallory!)
And why?? (c'est ma raison d'être!!!)
> ![enter image description here][42]
> ![enter image description here][43]
> ![enter image description here][44]
>![enter image description here][45]
So, below you find reproducible results with methods. The utmost care has been taken to assure the accuracy of the record number of digits calculations. These records represent the advancement of consumer-level computers and clever programming over the past 23 years.
Here are some record computations of **C**<sub>*MRB*</sub>. If you know of any others let me know, and I will probably add them!
1 digit of the (additive inverse of ) CMRB with my TI-92s, by adding 1-sqrt(2)+3^(1/3)-4^(1/4)+... as far as I could, was computed. That first digit, by the way, was just 0. Then by using the sum key, to compute $\sum _{n=1}^{1000 } (-1)^n \left(n^{1/n}\right),$ the first correct decimal of $\text{CMRB}=\sum _{n=1}^{\infty } (-1)^n \left(n^{1/n}-1\right)$ i.e. (.1). It gave (.1_91323989714) which is close to what Mathematica gives for summing to only an upper limit of 1000.
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4 decimals(.1878) of CMRB were computed on Jan 11, 1999 with the Inverse Symbolic Calculator, applying the command evalf( 0.1879019633921476926565342538468+sum((-1)^n* (n^(1/n)-1),n=140001..150000)); where 0.1879019633921476926565342538468 was the running total of t=sum((-1)^n* (n^(1/n)-1),n=1..10000), then t= t+the sum from (10001.. 20000), then t=t+the sum from (20001..30000) ... up to t=t+the sum from (130001..140000).
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5 correct decimals (rounded to .18786), in Jan of 1999 were drawn from CMRB using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95.
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9 digits of CMRB shortly afterward using Mathcad 7 professional on the Pentium II mentioned below, by summing (-1)^x x^(1/x) for x=1 to 10,000,000, 20,000,000, and a many more, then linearly approximating the sum to a what a few billion terms would have given.
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500 digits of CMRB with an online tool called Sigma on Jan 23, 1999. See [http://marvinrayburns.com/Original_MRB_Post.html][46] if you can read the printed and scanned copy there.
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5,000 digits of CMRB in September of 1999:on a 350 MHz Pentium II with 64 Mb of RAM using the simple PARI commands \p 5000;sumalt(n=1,((-1)^n*(n^(1/n)-1))), after allocating enough memory.
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6,995 accurate digits of CMRB were computed on June 10-11, 2003 over a period, of 10 hours, on a 450 MHz P3 with an available 512 MB RAM,.
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8000 digits of CMRB completed, using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, at 2:04 PM 3/25/2004,
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11,000 digits of CMRB on March 01, 2006, with a 3 GHz PD with 2 GB RAM available calculated.
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40, 000 digits of CMRB in 33 hours and 26 min via my program written in Mathematica 5.2 on Nov 24, 2006. The computation was run on a 32-bit Windows 3 GHz PD desktop computer using 3.25 GB of Ram.
The program was
Block[{a, b = -1, c = -1 - d, d = (3 + Sqrt[8])^n,
n = 131 Ceiling[40000/100], s = 0}, a[0] = 1;
d = (d + 1/d)/2; For[m = 1, m < n, a[m] = (1 + m)^(1/(1 + m)); m++];
For[k = 0, k < n, c = b - c;
b = b (k + n) (k - n)/((k + 1/2) (k + 1)); s = s + c*a[k]; k++];
N[1/2 - s/d, 40000]]
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60,000 digits of CMRB on July 29, 2007, at 11:57 PM EST in 50.51 hours on a 2.6 GHz AMD Athlon with 64-bit Windows XP. Max memory used was 4.0 GB of RAM.
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65,000 digits of CMRB in only 50.50 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP on Aug 3, 2007, at 12:40 AM EST, Max memory used was 5.0 GB of RAM.
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100,000 digits of CMRB on Aug 12, 2007, at 8:00 PM EST, were computed in 170 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. Max memory used was 11.3 GB of RAM. The typical daily record of memory used was 8.5 GB of RAM.
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150,000 digits of CMRB on Sep 23, 2007, at 11:00 AM EST. Computed in 330 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. Max memory used was 22 GB of RAM. The typical daily record of memory used was 17 GB of RAM.
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200,000 digits of CMRB using Mathematica 5.2 on March 16, 2008, at 3:00 PM EST,. Found in 845 hours, on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. Max memory used was 47 GB of RAM. The typical daily record of memory used was 28 GB of RAM.
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300,000 digits of CMRB were destroyed (washed away by Hurricane Ike ) on September 13, 2008 sometime between 2:00 PM - 8:00 PM EST. Computed for a long 4015. Hours (23.899 weeks or 1.4454*10^7 seconds) on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. Max memory used was 91 GB of RAM. The Mathematica 6.0 code is used follows:
Block[{$MaxExtraPrecision = 300000 + 8, a, b = -1, c = -1 - d,
d = (3 + Sqrt[8])^n, n = 131 Ceiling[300000/100], s = 0}, a[0] = 1;
d = (d + 1/d)/2; For[m = 1, m < n, a[m] = (1 + m)^(1/(1 + m)); m++];
For[k = 0, k < n, c = b - c;
b = b (k + n) (k - n)/((k + 1/2) (k + 1)); s = s + c*a[k]; k++];
N[1/2 - s/d, 300000]]
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225,000 digits of CMRB were started with a 2.66 GHz Core 2 Duo using 64-bit Windows XP on September 18, 2008. It was completed in 1072 hours.
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250,000 digits were attempted but failed to be completed to a serious internal error that restarted the machine. The error occurred sometime on December 24, 2008, between 9:00 AM and 9:00 PM. The computation began on November 16, 2008, at 10:03 PM EST. The Max memory used was 60.5 GB.
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250,000 digits of CMRB on Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, with a multiple-step Mathematica command running on a dedicated 64-bit XP using 4 GB DDR2 RAM onboard and 36 GB virtual. The computation took only 333.102 hours. The digits are at http://marvinrayburns.com/250KMRB.txt. The computation is completely documented.
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300000 digit search of CMRB was initiated using an i7 with 8.0 GB of DDR3 RAM onboard on Sun 28 Mar 2010 21:44:50 (UTC-0500) EST, but it failed due to hardware problems.
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299,998 Digits of CMRB: The computation began Fri 13 Aug 2010 10:16:20 pm EDT and ended 2.23199*10^6 seconds later | Wednesday, September 8, 2010. I used Mathematica 6.0 for Microsoft Windows (64-bit) (June 19, 2007) Which is an average of 7.44 seconds per digit. I used my Dell Studio XPS 8100 i7 860 @ 2.80 GHz with 8GB physical DDR3 RAM. Windows 7 reserved an additional 48.929 GB virtual Ram.
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300,000 digits to the right of the decimal point of CMRB from Sat 8 Oct 2011 23:50:40 to Sat 5 Nov 2011 19:53:42 (2.405*10^6 seconds later). This run was 0.5766 seconds per digit slower than the 299,998 digit computation even though it used 16 GB physical DDR3 RAM on the same machine. The working precision and accuracy goal combination were maximized for exactly 300,000 digits, and the result was automatically saved as a file instead of just being displayed on the front end. Windows reserved a total of 63 GB of working memory of which 52 GB were recorded being used. The 300,000 digits came from the Mathematica 7.0 command`
Quit; DateString[]
digits = 300000; str = OpenWrite[]; SetOptions[str,
PageWidth -> 1000]; time = SessionTime[]; Write[str,
NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]},
WorkingPrecision -> digits + 3, AccuracyGoal -> digits,
Method -> "AlternatingSigns"]]; timeused =
SessionTime[] - time; here = Close[str]
DateString[]
----------
314159 digits of the constant took 3 tries due to hardware failure. Finishing on September 18, 2012, I computed 314159 digits, taking 59 GB of RAM. The digits came from the Mathematica 8.0.4 code`
DateString[]
NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]},
WorkingPrecision -> 314169, Method -> "AlternatingSigns"] // Timing
DateString[]
----------
1,000,000 digits of CMRB for the first time in history in 18 days 9 hours 11 minutes 34.253417 seconds by Sam Noble of the Apple Advanced Computation Group.
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1,048,576 digits in a lightning-fast 76.4 hours. Finishing on Dec 11, 2012 were scored by Dr Richard Crandall, an Apple scientist and head of its advanced computational group. That's on a 2.93 GHz 8-core Nehalem.
To beat that, on Aug of 2018, I computed 1,004,993 digits of CMRB in 53.5 hours 34 hours computation time (from the timing command) with 10 DDR4 RAM (of up to 3000 MHz) supported processor cores overclocked up to 4.7 GHz! Search this post for "53.5" for documentation.
To beat that, on Sept 21, 2018: I computed 1,004,993 digits of CMRB in 50.37 hours of absolute time and 35.4 hours of computation time (from the timing command) with 18 (DDR3 and DDR4) processor cores! Search this post for "50.37 hours" for documentation.**
To beat that, on May 11, 2019, I computed over 1,004,993 digits, in 45.5 hours of absolute time and only 32.5 hours of computation time, using 28 kernels on 18 DDR4 RAM (of up to 3200 MHz) supported cores overclocked up to 5.1 GHz Search 'Documented in the attached ":3 fastest computers together 3.nb." ' for the post that has the attached documenting notebook.
To beat that, I accumulated over 1,004,993 correct digits of CMRB in 44 hours of absolute time and 35.4206 hours of computation time on 10/19/20, using 3/4 of the MRB constant supercomputer 2 -- see [https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb][47] for documentation.
To beat that I did a 1,004,993 correct digits computation in 36 hours of absolute time and only 26 hours of computation time, on Sun 15 May 2022 at 06:10:50, using 3/4 of the MRB constant supercomputer 3. Ram Speed was 4800MHz and all of the 30 cores were clocked at up to 5.2 GHz.
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A little over 1,200,000 digits, previously, of CMRB in 11 days, 21 hours, 17 minutes, and 41 seconds (I finished on March 31, 2013, using a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz.
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2,000,000 or more digit computation of CMRB on May 17, 2013, using only around 10GB of RAM. It took 37 days 5 hours 6 minutes 47.1870579 seconds. I used my six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.
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3,014,991 digits of CMRB, world record computation of CMRB was finished on Sun 21 Sep 2014 at 18:35:06. It took 1 month 27 days 2 hours 45 minutes 15 seconds. The processor time from the 3,000,000+ digit computation was 22 days. I computed the 3,014,991 digits of CMRB with Mathematica 10.0. I Used my new version of Richard Crandall's code in the attached 3M.nb, optimized for my platform and large computations. I also used a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz with 64 GB of RAM of which only 16 GB was used. Can you beat it (in more number of digits, less memory used, or less time taken)? This confirms that my previous "2,000,000 or more digit computation" was accurate to 2,009,993 digits. they were used to check the first several digits of this computation. See attached 3M.nb for the full code and digits.
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Over 4 million digits of CMRB was finished on Wed 16 Jan 2019 19:55:20.
It took 4 years of continuous tries. This successful run took 65.13 days absolute time, with a processor time of 25.17 days, on a 3.7 GHz overclocked up to 4.7 GHz on all cores Intel 6 core computer with 3000 MHz RAM. According to this computation, the previous record, 3,000,000+ digit computation, was accurate to 3,014,871 decimals, as this computation used my algorithm for computing n^(1/n) as found in chapter 3 in the paper at
https://www.sciencedirect.com/science/article/pii/0898122189900242 and the 3 million+ computation used Crandall's algorithm. Both algorithms outperform Newton's method per calculation and iteration.
Example use of M R Burns' algorithm to compute 123456789^(1/123456789) 10,000,000 digits:
ClearSystemCache[]; n = 123456789;
(*n is the n in n^(1/n)*)
x = N[n^(1/n),100];
(*x starts out as a relatively small precision approximation to n^(1/n)*)
pc = Precision[x]; pr = 10000000;
(*pr is the desired precision of your n^(1/n)*)
Print[t0 = Timing[While[pc < pr, pc = Min[4 pc, pr];
x = SetPrecision[x, pc];
y = x^n; z = (n - y)/y;
t = 2 n - 1; t2 = t^2;
x = x*(1 + SetPrecision[4.5, pc] (n - 1)/t2 + (n + 1) z/(2 n t)
- SetPrecision[13.5, pc] n (n - 1)/(3 n t2 + t^3 z))];
(*You get a much faster version of N[n^(1/n),pr]*)
N[n - x^n, 10]](*The error*)];
ClearSystemCache[]; n = 123456789; Print[t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]]
Gives
{25.5469,0.*10^-9999984}
{101.359,0.*10^-9999984}
More information is available upon request.
----------
More than 5 million digits of CMRB were found on Fri 19 Jul 2019 18:49:02, Methods described in the reply below which begins with "Attempts at a 5,000,000 digit calculation ." For this 5 million digit calculation of MRB using the 3 node MRB supercomputer: processor time was 40 days. and the actual time was 64 days. That is in less absolute time than the 4-million-digit computation which used just one node.
----------
6,000,000 digits of the MRB constant after 8 tries in 19 months. (Search "8/24/2019 It's time for more digits!" below.) finishing on Tue 30 Mar 2021 at 22:02:49 in 160 days.
The MRB constant supercomputer 2 said the following:
Finished on Tue 30 Mar 2021 22:02:49. computation and absolute time were
5.28815859375*10^6 and 1.38935720536301*10^7 s. respectively
Enter MRB1 to print 6029991 digits. The error from a 5,000,000 or more-digit calculation that used a different method is
0.*10^-5024993.
That means that the 5,000,000-digit computation Was actually accurate to 5024993 decimals!!!
----------
5,609,880, verified by 2 distinct algorithms for x^(1/x), digits of the MRB constant on Thu 4 Mar 2021 at 08:03:45. The 5,500,000+ digit computation using a totally different method showed that many decimals are in common with the 6,000,000+ digit computation.
----------
6,500,000 digits of the MRB constant on my second try,
The MRB constant supercomputer said,
Finished on Wed 16 Mar 2022 02: 02: 10. computation and absolute time
were 6.26628*10^6 and 1.60264035419592*10^7s respectively Enter MRB1
to print 6532491 digits. The error from a 6, 000, 000 or more digit
calculation that used a different method is
0.*10^-6029992.
"Computation time" 72.526 days
"Absolute time" 185.491 days
----------
----------
----------
----------
----------
----------
----------
The MRB constant and its applications
=====================================
**Definition 1**
**C**<sub>*MRB*</sub> is defined at [https://en.wikipedia.org/wiki/MRB_constant][48] .
From [Wikipedia:][49]
> ![If you see this instead of an image, reload the page][50]
> ![If you see this instead of an image, reload the page][51]
References
Plouffe, Simon. "mrburns". Retrieved 12 January 2015.
Burns, Marvin R. (23 January 1999). "RC". math2.org. Retrieved 5 May 2009.
Plouffe, Simon (20 November 1999). "Tables of Constants" (PDF). Laboratoire de combinatoire et d'informatique mathématique. Retrieved 5 May 2009.
Weisstein, Eric W. "MRB Constant". MathWorld.
Mathar, Richard J. (2009). "Numerical Evaluation of the Oscillatory Integral Over exp(iπx) x^*1/x) Between 1 and Infinity". arXiv:0912.3844 [math.CA].
Crandall, Richard. "Unified algorithms for polylogarithm, L-series, and zeta variants" (PDF). PSI Press. Archived from the original (PDF) on April 30, 2013. Retrieved 16 January 2015.
(sequence A037077 in the OEIS)
(sequence A160755 in the OEIS)
(sequence A173273 in the OEIS)
Fiorentini, Mauro. "MRB (costante)". bitman.name (in Italian). Retrieved 14 January 2015.
Finch, Steven R. (2003). Mathematical Constants. Cambridge, England: Cambridge University Press. p. 450. ISBN 0-521-81805-2.
`
The following equation that was shown in the Wikipedia definition shows how closely the MRB constant is related to root two.
![enter image description here][52]
In[1]:= N[Sum[Sqrt[2]^(1/n)* Sqrt[n]^(1/n) - ((Sqrt[2]^y*Sqrt[2]^x)^(1/Sqrt[2]^x))^Sqrt[2]^(-y)/.
x -> 2*Log2[a^2 + b^2] /.
y -> 2*Log2[-ai^2 - bi^2] /.
a -> 1 - (2*n)^(1/4) /.
b -> 2^(5/8)*Sqrt[n^(1/4)] /.
ai -> 1 - I*(2*n)^(1/4) /.
bi -> 2^(5/8)*Sqrt[I*n^(1/4)], {n, 1, Infinity}], 7]
Out[1]= 0.1878596 + 0.*10^-8 I
The complex roots and powers above are found to be well-defined because
we get all either "integer" and "rational" the first of the following lists only, also by working from the bottom to the top of the above list of equations.
![enter image description here][53]
Code:
In[349]:= Table[
Head[FullSimplify[
Expand[(Sqrt[2])^-y/(Sqrt[2])^x] //.
x -> 2 (Log[1 + Sqrt[2] Sqrt[n]]/Log[2]) /.
y -> 2 (Log[-1 + Sqrt[2] Sqrt[n]]/Log[2])]], {n, 1, 10}]
Out[349]= {Integer, Rational, Rational, Rational, Rational, Rational, \
Rational, Rational, Rational, Rational}
In[369]:= Table[
Head[FullSimplify[
Expand[(Sqrt[2])^-y/(Sqrt[2])^x] //.
x -> 2 (Log[1 + Sqrt[2] Sqrt[n]]/Log[3]) /.
y -> 2 (Log[-1 + Sqrt[2] Sqrt[n]]/Log[2])]], {n, 1, 10}]
Out[369]= {Times, Rational, Times, Times, Times, Times, Times, Times, \
Times, Times}
----------
**Definition 2**
**C**<sub>*MRB*</sub> is defined at [http://mathworld.wolfram.com/MRBConstant.html][54].
From MathWorld:
> ![MathWorld MRB][55] ![MathWorld MRB 2][56]
>
> SEE ALSO:
> Glaisher-Kinkelin Constant, Power Tower, Steiner's Problem
> REFERENCES:
> Burns, M. R. "An Alternating Series Involving n^(th) Roots." Unpublished note, 1999.
>
> Burns, M. R. "Try to Beat These MRB Constant Records!" http://community.wolfram.com/groups/-/m/t/366628.
>
> Crandall, R. E. "Unified Algorithms for Polylogarithm, L-Series, and Zeta Variants." 2012a.
> http://www.marvinrayburns.com/UniversalTOC25.pdf.
>
> Crandall, R. E. "The MRB Constant." §7.5 in Algorithmic Reflections: Selected Works. PSI Press, pp. 28-29, 2012b.
>
> Finch, S. R. Mathematical Constants. Cambridge, England: Cambridge University Press, p. 450, 2003.
>
> Plouffe, S. "MRB Constant." http://pi.lacim.uqam.ca/piDATA/mrburns.txt.
>
> Sloane, N. J. A. Sequences A037077 in "The On-Line Encyclopedia of Integer Sequences."
>
> Referenced on Wolfram|Alpha: MRB Constant
> CITE THIS AS:
> Weisstein, Eric W. "MRB Constant." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/MRBConstant.html
**How would we show that the any of the series in the above MathWorld definition are convergent, or even absolutely convergent?**
----------
For "a"<sub>k</sub>=k<sup>1/k</sup>, gven that the sequence is monotonically decreasing according to [Steiner's Problem][57], next, we would like to show (5) is the alternating sum of a sequence that converges to 0 monotonically and use the Alternating series test to see that it is conditionally convergent
Here is a proof that 1 is the limit of "a" as k goes to infinity:
> ![enter image description here][58]
[Here][59] are many other proofs that 1 is the limit of "a" as k goes to infinity.
Thus, (k<sup>1/k</sup>-1) is a monotonically decreasing and bounded below by 0 **sequence.**
If we want an absolutely convergent **series**, we can use (4).
S<sub>k</sub>![enter image description here][60] which, since the sum of the absolute values of the summands is finite, the sum converges absolutely!
There is no closed-form for **C**<sub>*MRB*</sub> in the MathWorld definition; this could be due to the following: in [Mathematical Constants][61],(
Finch, S. R. Mathematical Constants, Cambridge, England: Cambridge University Press, p. 450), Steven Finch wrote that it is difficult to find an "exact formula" ([closed-form solution][62]) for it.
> ![enter image description here][63]
![enter image description here][64]
----------
----------
----------
----------
Real-World, and beyond, Applications
====================================
CMRB as a Growth Model
-----------------------
Its factor ![a][65]
models the interest rate to multiply an investment k times in k periods, as well as "other growth and decay functions involving the more general expression ![(1+k)^n][66], as in Plot 1A," because ![enter image description here][67]
r=(k^(1/k)-1);Animate[ListPlot[l=Accumulate[Table[(r+1)^n,{k,100}]], PlotStyle->Red,PlotRange->{0,150},PlotLegends->{"\!\(\*UnderscriptBox[\(\[Sum]\), \(\)]\)(r+1\!\(\*SuperscriptBox[\()\), \(n\)]\)/.r->(\!\(\*SuperscriptBox[\(k\), \(1/k\)]\)-1)/.n->"n},AxesOrigin->{0,0}],{n,0,5}]
Plot 1A
![enter image description here][68]
The discrete rates looks like the following.
r = (k^(1/k) - 1); me =
Animate[ListPlot[l = Table[(r + 1)^n, {k, 100}], PlotStyle -> Red,
PlotLegends -> {"(r+1)^n/.r->\!\(\*SuperscriptBox[\(k\), \
\(1/k\)]\)=1/.n->", n}, AxesOrigin -> {0, 0},
PlotRange -> {0, 7}], {n, 1, 5}]
![enter image description here][69]
That factor ![enter image description here][70] models not only discretely compounded rates but continuous too, ie ![Pt=p0e^rt.][71]
By entering
Solve[P*E^(r*t) == P*(t^(1/t) - 1), r]
we see, for ![Pt=p0e^rt,][72] ![t>e][73]
gives an effect of continuous decay of ![enter image description here][74] Here Q1 means the first Quarter form 0 to -1.
The alternating sum of the principal of those continuous rates, i.e. P=(-1)<sup>t</sup> e<sup>r t</sup> is the MRB constant (CMRB):
![enter image description here][75]
In[647]:= NSum[(-1)^t ( E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t, 1,
Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 30]
Out[647]= 0.18785964246206712024857897184
Its integral (MKB) is an analog to **C**<sub>*MRB*</sub> :
![enter image description here][76]
In[1]:= NIntegrate[(-1)^t (E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t,
1, Infinity I}, Method -> "Trapezoidal", WorkingPrecision -> 30] -
2 I/Pi
Out[1]= 0.0707760393115288035395280218303 -
0.6840003894379321291827444599927 I
So, integrating P yields about 1/2 greater of a total than summing:
In[663]:=
CMRB = NSum[(-1)^n ( Power[n, ( n)^-1] - 1), {n, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 30];
In[664]:=
MKB = Abs[
NIntegrate[(-1)^t ( E^(r*t)) /. r -> Log[-1 + t^(1/t)]/t, {t, 1,
Infinity I}, Method -> "Trapezoidal", WorkingPrecision -> 30] -
2 I/Pi];
In[667]:= MKB - CMRB
Out[667]= 0.49979272646562724956073343752
Next:
CMRB from Geometric Series and Power Series
---------------------------------
The MRB constant: ![enter image description here][77] is closely related to geometric series: ![enter image description here][78]
The inverse function of the "term" of the MRB constant, i.e. x^(1/x) within a certain domain is solved for in [this link,][79]
> ![enter image description here][80]
>
> ...
> ---
>
>![enter image description here][81]
Consider the following about a slight generalization of that term.
![enter image description here][82]
**C**<sub>*MRB*</sub> can be written in geometric series form:
**C**<sub>*MRB*</sub>=
![enter image description here][83]
In[240]:= N[Quiet[(Sum[q^k, {x, 1, Infinity}] /.
k -> Log[-E^(I*Pi*x) + E^(x*(I*Pi + Log[x]/x^2))]/Log[q]) -
Sum[E^(I*Pi*x)*(-1 + x^(1/x)), {x, 1, Infinity}]]]
Out[240]= -4.163336342344337*^-16
Here too, with some shifting:
**C**<sub>*MRB*</sub>=
![enter image description here][84]
In[269]:= Quiet[
N[Sum[q^p - 1, {x, 1, Infinity}] -
Sum[E^(I*Pi*x)*(x^(1/x) - 1), {x, 1, Infinity}]] /.
p -> Log[(\!\(TraditionalForm\`
\*SuperscriptBox[\((\(-1\))\), \(x\)] \((\
\*SuperscriptBox[\(q\), \(k\)] - 1)\) + 1\))]/Log[q] /.
k -> Log[x]/(x*Log[q])]
Out[269]= 0.
It is also related to a power series of the form
![sums][85], ![a's][86] shifted by 1,
where we have
**C**<sub>*MRB*</sub>=
![enter image description here][87]
Sum[(-1)^x (q^k - 1) /. k -> Log[x]/(x Log[q]), {x, 1, Infinity}]
![enter image description here][88]
In[70]:= N[%]
Out[70]= 0.18786
Next
The Geometry of the MRB constant
----------------
----------
In 1837 Pierre Wantzel proved that an nth root of a given length cannot be constructed if n is not a power of 2 (as mentioned [here][89] in Wikipedia). However, the following is a little different.
For ![,][90] on November 21, 2010, I coined a multiversal [analog][91] to, [Minkowski space][92] that plots their values from constructions arising from a peculiar non-euclidean geometry, below, and fully in [this vixra draft][93].
As in Diagram 2, we give each n-cube a hyperbolic volume (content) equal to its dimension,![enter image description here][94]
Geometrically, as in Diagram3, on the y,z-plane line up an edge of
each n-cube. The numeric values displayed in the diagram are the partial sums of S[x_] = `Sum[(-1)^n*n^(1/n), {n, 1, 2*u}]` where u is an positive integer. Then M is the MRB constant.
![enter image description here][95]
Join[ Table[N[S[x]], {u, 1, 4}], {"..."}, {NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}]}]
Out[421]= {0.414214, 0.386178, 0.354454, 0.330824, "...", 0.18786}
Next
The publishing of CMRB.
------------------
Google Scholar results on MRB constant are [here,][96] which include the following.
**[Dr. Richard Crandall][97] called the MRB constant a [key fundamental constant][98]**
> ![enter image description here][99]
**in [this linked][100] well-sourced and equally greatly cited Google Scholar promoted paper.**
**[Dr. Richard J. Mathar][101] wrote on the MRB constant [here.][102]**
**Xun Zhou, School of Water Resources and Environment, China University of Geosciences (Beijing), [wrote][103] the following in "on Some Series and Mathematic Constants Arising in Radioactive Decay" for the Journal of Mathematics Research, 2019.**
> A divergent infinite series may also lead to mathematical constants if
> its partial sum is bounded. The Marvin Ray Burns’ (MRB) constant is
> the upper bounded value of the partial sum of the divergent and
> alternating infinite series:
> -1<sup>1/1</sup>+2<sup>1/2</sup>-3<sup>1/3</sup>+4<sup>1/4</sup>-5<sup>1/5</sup>+6<sup>1/6</sup>-···=0.187859···(M. Chen, & S. Chen, 2016). Thus, construction of new infinite series has the possibility
> of leading to new mathematical constants.
----------
CMRB is at the 38th most popular Wikipedia Mathematics page as of April 6, 2022. As of 20 April 2022, there are 6,486,890 articles in the English Wikipedia.
----------
The geometry of the MRB constant's connection to string theory describing black holes results found in Ramanujan's modular equations, a connection to the Quantum Cosmological Constant, MRB constant integrals and relationships to other constants in string theory, in a relationship with the Ramanujan-Nardelli mock general formula, from Ramanujan's Mock Theta Functions to Black Hole Entropies and Symmetry, Supersymmetry, Golden Ratio, the MRB Constant and the linked integrals to it, applied to several equations of Geometric Measure Theory, to the Ramanujan’s equations and connections with some sectors of String Theory, new possible mathematical connections between the MRB constant and various sectors of Theoretical Physics and Cosmology, the study of some integrals concerning the MRB Constant and several equations of Geometric Measure Theory, analyzing new possible mathematical connections with some Cosmological parameters and sectors of String Theory, analyzing some Cosmological parameters and sectors of String Theory, and much more
[in this link.][104]
Concerning his prolific writing on the MRB constant, Michele Nardelli added the following.
In string theory, perturbation methods involve such a high degree of approximation that the theory is unable to identify which of the Calabi - Yau spaces are candidates for describing the universe. The consequence is that it does not describe a single universe, but something like 10^500 universes. In reality, admitting 10^500 different quantum voids would allow the only mechanism known at the moment to explain the present value of the cosmological constant following an idea by Steven Weinberg. Furthermore, a very large value of different voids is typical of any type of matter coupled to gravity and is also obtained when coupling the standard model. I believe that the multiverse is a "space of infinite dimensions" with infinite degrees of freedom and infinite possible potential wave functions that when they collapse, formalize a particle or a universe in a quantum state. The strings vibrating, like the strings of a musical instrument, emit frequencies that are not always precise numbers, indeed, very often they are decimal, irrational, and/or transcendent numbers. The MRB constant serves as a "regularizer" to obtain solutions as precise as possible and this in various sectors of string theory, black holes, and cosmology
In [this physics.stackexchange question][105] his concept of the dimensions in string theory and possible link with number theory is inquired about.
Many MRB constant papers by Michele Nardelli are found [here in Google Scholar][106],which include previous versions of these.
> Hello. Here are the links of my more comprehensive articles describing
> the various applications of the CMRB in various fields of theoretical
> physics and cosmology. Thanks always for your availability, see you
> soon.
>
>
> https://www.academia.edu/75884771/
>
> https://www.academia.edu/76084911/
>
> https://www.academia.edu/76405749/
>
> https://www.academia.edu/76784160/
>
> https://www.academia.edu/77164290/
>
> https://www.academia.edu/77531870/
>
> https://www.academia.edu/77752950/
>
> https://www.academia.edu/77978967/
>
> https://www.academia.edu/78104771/
>
> https://www.academia.edu/72576179/
>
> https://www.academia.edu/72674127/
>
> https://www.academia.edu/73043410/
>
> https://www.academia.edu/73201689/
----------
----------
----------
----------
----------
----------
MeijerG Representation
======================
From its integrated analog, I found a [MeijerG][107] representation for **C**<sub>*MRB*</sub>.
The search for it began with the following:
> On 10/10/2021, I found the following proper definite integral that leads to almost identical
> proper integrals from 0 to 1 for **C**<sub>*MRB*</sub> and its integrated analog.
>
> ![m vs m2 0 to 1][108]
>
> See [notebook in this link][109].
>
>
> Here is a [MeijerG][110] function for the integrated analog. See [(proof)][111] of discovery.
>
>
>
>
> ![enter image description here][112]
>
> f(n)=![enter image description here][113].
>
> `
>
> In[135]:=f[n_]:=MeijerG[{{},Table[1,{n+1}]},{Prepend[Table[0,n+1],-n+1],{}},-\[ImaginaryI]\[Pi]];`
> In[337]:=M2=NIntegrate[E^(I Pi x)(SuperscriptBox["x", FractionBox["1", "x"]]-1),
> {x,1,Infinity I},WorkingPrecision->100]
>
> Out[337]=0.07077603931152880353952802183028200136575469620336302758317278816361845726438203658083188126617723821-0.04738061707035078610720940650260367857315289969317363933196100090256586758807049779050462314770913485 \[ImaginaryI]
>
>
> ![enter image description here][114]
>
> I wonder if there is one for the MRB constant sum (CMRB)?
According to "Primary Proof 1" and "Primary Proof 3" shown below along with the section prefixed by the phrase "So far I came up with,"
it can be proven that for
G being the Wolfram MeijerG function
and f(n)=![enter image description here][115], and
![enter image description here][116]
g[x_] = (-1)^x (1 - (x + 1)^(1/(x + 1)));
In[52]:= (1/2)*
NIntegrate[(g[-t] - g[t])/(Sin[Pi*t]*Cos[Pi*t]*I + Sin[Pi*t]^2), {t,
0, I*Infinity}, WorkingPrecision -> 100,
Method -> "GlobalAdaptive"]
Out[52]= 0.\
1170836031505383167089899122239912286901483986967757585888318959258587\
7430027817712246477316693025869 +
0.0473806170703507861072094065026036785731528996931736393319610009025\
6586758807049779050462314770913485 I
In[57]:= Re[
NIntegrate[
g[-t]/(Sin[Pi*t]*Cos[Pi*t]*I + Sin[Pi*t]^2), {t, 0, I*Infinity},
WorkingPrecision -> 100,
Method -> "GlobalAdaptive"]]
Out[57]= 0.\
1878596424620671202485179340542732300559030949001387861720046840894772\
315646602137032966544331074969
----------
----------
----------
MRB constant formulas and identities
====================================
I developed this informal catalog of formulas for the MRB constant with over 20 years of research and ideas from users like you.
------------------------------------------------------------------------
3/25/2022
---------
Formula (11) =
![enter image description here][117]
As Matheamatica says:
Assuming[Element[c, \[DoubleStruckCapitalZ]], FullSimplify[
E^(t*(r + I*Pi*(2*c + 1))) /. r -> Log[t^(1/t) - 1]/t]]
= E^(I (1 + 2 c) \[Pi] t) (-1 + t^(1/t))
![enter image description here][118]
Where for all integers c, (1+2c) is odd leading to ![enter image description here][119]
Expanding the E^log term gives
![enter image description here][120]
which is ![enter image description here][121],
That is exactly (2) in the above-quoted MathWorld definition:
![enter image description here][122]
----------
2/21/2022
----------
Directly from the formula of 12/29/2021 below,
![enter image description here][123]
In[276]:=
u = (-1)^t; N[
NSum[(t^(1/t) - 1) u, {t, 1, Infinity }, WorkingPrecision -> 24,
Method -> "AlternatingSigns"], 15]
Out[276]= 0.187859642462067
In[278]:=
v = (-1)^-t - (-1)^t; 2 I N[
NIntegrate[Im[(t^(1/t) - 1) v^-1], {t, 1, Infinity I},
WorkingPrecision -> 24], 15]
Out[278]= 0.187859642462067
Likewise,
![enter image description here][124]
Expanding the exponents,
![enter image description here][125]
This can be generalized to
![(x+log/][126]
Building upon that, we get a closed form for the inner integral in the following.
![enter image description here][127]
In[1]:=
CMRB = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity},
WorkingPrecision -> 1000, Method -> "AlternatingSigns"];
In[2]:= CMRB - {
Quiet[Im[NIntegrate[
Integrate[
E^(Log[t]/t + x)/(-E^((-I)*Pi*t + x) + E^(I*Pi*t + x)), {x,
I, -I}], {t, 1, Infinity I}, WorkingPrecision -> 200,
Method -> "Trapezoidal"]]];
Quiet[Im[NIntegrate[
Integrate[
Im[E^(Log[t]/t + x)/(-E^((-I)*Pi*t + x) + E^(I*Pi*t + x))], {x,
-t, t }], {t, 1
, Infinity I}, WorkingPrecision -> 2000,
Method -> "Trapezoidal"]]]}
Out[2]= {3.*10^-998, 3.*10^-998}
**Which after a little analysis, can be shown convergent in the continuum limit at t → ∞ i.**
----------
----------
12/29/2021
----------
From "Primary Proof 1" worked below, it can be shown that
![enter image description here][128]
Mathematica knows that because
m = N[NSum[-E^(I*Pi*t) + E^(I*Pi*t)*t^t^(-1), {t, 1, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 27], 18];
Print[{m -
N[NIntegrate[
Im[(E^(Log[t]/t) + E^(Log[t]/t))/(E^(I \[Pi] t) -
E^(-I \[Pi] t))] I, {t, 1, -Infinity I},
WorkingPrecision -> 20], 18],
m - N[NIntegrate[
Im[(E^(Log[t]/t) + E^(Log[t]/t))/(E^(-I \[Pi] t) -
E^(I \[Pi] t))] I, {t, 1, Infinity I},
WorkingPrecision -> 20], 18],
m + 2 I*NIntegrate[
Im[(E^(I*Pi*t + Log[t]/t))/(-1 + E^((2*I)*Pi*t))], {t, 1,
Infinity I}, WorkingPrecision -> 20]}]
yields
{0.*^-19,0.*^-19,0.*^-19}
Partial sums to an upper limit of (10^n i) give approximations for the MRB constant + the same approximation *10^-(n+1) i.
Example:
-2 I*NIntegrate[
Im[(E^(I*Pi*t + Log[t]/t))/(-1 + E^((2*I)*Pi*t))], {t, 1, 10^7 I},
WorkingPrecision -> 20]
gives
0.18785602000738908694 + 1.878560200074*10^-8 I
where CMRB ≈ 0.187856.
Notice **it is special because** if we integrate only the numerator, we have MKB=![enter image description here][129], which defines the "integrated analog of **C**<sub>*MRB*</sub>" (MKB) described by Richard Mathar in [https://arxiv.org/abs/0912.3844][130]. (He called it M1.)
Like how this:
NIntegrate[(E^(I*Pi*t + Log[t]/t)), {t, 1, Infinity I},
WorkingPrecision -> 20] - I/Pi
converges to
0.070776039311528802981 - 0.68400038943793212890 I.
(The upper limits " i infinity" and " infinity" produce the same result in this integral.)
----------
----------
11/14/2021
----------
Here is a standard notation for the above mentioned
**C**<sub>*MRB*,</sub>![enter image description here][131]
![enter image description here][132].
In[16]:= CMRB = 0.18785964246206712024851793405427323005590332204; \
CMRB - NSum[(Sum[
E^(I \[Pi] x) Log[x]^n/(n! x^n), {x, 1, Infinity}]), {n, 1, 20},
WorkingPrecision -> 50]
Out[16]= -5.8542798212228838*10^-30
In[8]:= c1 =
Activate[Limit[(-1)^m/m! Derivative[m][DirichletEta][x] /. m -> 1,
x -> 1]]
Out[8]= 1/2 Log[2] (-2 EulerGamma + Log[2])
In[14]:= CMRB -
N[-(c1 + Sum[(-1)^m/m! Derivative[m][DirichletEta][m], {m, 2, 20}]),
30]
Out[14]= -6.*10^-30
----------
----------
11/01/2021
----------
: The catalog now appears complete, and can all be proven through Primary Proof 1, and the one with the eta function, Primary Proof 2, both found below.
> a ≠b
> ![enter image description here][133] ![enter image description here][134]
>
> g[x_] = x^(1/x); CMRB =
> NSum[(-1)^k (g[k] - 1), {k, 1, Infinity}, WorkingPrecision -> 100,
> Method -> "AlternatingSigns"]; a = -Infinity I; b = Infinity I;
> g[x_] = x^(1/x); (v = t/(1 + t + t I);
> Print[CMRB - (-I /2 NIntegrate[ Re[v^-v Csc[Pi/v]]/ (t^2), {t, a, b},
> WorkingPrecision -> 100])]); Clear[a, b]
> -9.3472*10^-94
>
> Thus, we find
>
> ![enter image description here][135]
>
> [here,][136] and
> ![enter image description here][137]
> next:
> In[93]:= CMRB =
> NSum[Cos[Pi n] (n^(1/n) - 1), {n, 1, Infinity},
> Method -> "AlternatingSigns", WorkingPrecision -> 100]; Table[
> CMRB - (1/2 +
> NIntegrate[
> Im[(t^(1/t) - t^(2 n))] (-Csc[\[Pi] t]), {t, 1, Infinity I},
> WorkingPrecision -> 100, Method -> "Trapezoidal"]), {n, 1, 5}]
>
> Out[93]= {-9.3472*10^-94, -9.3473*10^-94, -9.3474*10^-94, \
> -9.3476*10^-94, -9.3477*10^-94}
> CNT+F "The following is a way to compute the" for more evidence
For such n, ![enter image description here][138] converges to 1/2+0i.
(How I came across all of those and more example code follow in various replies.)
----------
----------
On 10/18/2021
-------------
, I found the following triad of pairs of integrals summed from -complex infinity to +complex infinity.
![CMRB= -complex infinity to +complex infinity][139]
You can see it worked [ in this link here][140].
In[1]:= n = {1, 25.6566540351058628559907};
In[2]:= g[x_] = x^(n/x);
-1/2 Im[N[
NIntegrate[(g[(1 - t)])/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]
Out[3]= {0.18785964246206712025, 0.18785964246206712025}
In[4]:= g[x_] = x^(n/x);
1/2 Im[N[NIntegrate[(g[(1 + t)])/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]
Out[5]= {0.18785964246206712025, 0.18785964246206712025}
In[6]:= g[x_] = x^(n/x);
1/4 Im[N[NIntegrate[(g[(1 + t)] - (g[(1 - t)]))/(Sin[\[Pi] t]), {t, -Infinity I,
Infinity I}, WorkingPrecision -> 60], 20]]
Out[7]= {0.18785964246206712025, 0.18785964246206712025}
Therefore, bringing
![enter image description here][141]
back to mind, we joyfully find,
![CMRB n and 1][142]
In[1]:= n =
25.65665403510586285599072933607445153794770546058072048626118194900\
97321718621288009944007124739159792146480733342667`100.;
g[x_] = {x^(1/x), x^(n/x)};
CMRB = NSum[(-1)^k (k^(1/k) - 1), {k, 1, Infinity},
WorkingPrecision -> 100, Method -> "AlternatingSigns"];
Print[CMRB -
NIntegrate[Im[g[(1 + I t)]/Sinh[\[Pi] t]], {t, 0, Infinity},
WorkingPrecision -> 100], u = (-1 + t); v = t/u;
CMRB - NIntegrate[Im[g[(1 + I v)]/(Sinh[\[Pi] v] u^2)], {t, 0, 1},
WorkingPrecision -> 100],
CMRB - NIntegrate[Im[g[(1 - I v)]/(Sinh[-\[Pi] v] u^2)], {t, 0, 1},
WorkingPrecision -> 100]]
During evaluation of In[1]:= {-9.3472*10^-94,-9.3472*10^-94}{-9.3472*10^-94,-9.3472*10^-94}{-9.3472*10^-94,-9.3472*10^-94}
In[23]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] -
g[(1 + I v)]/(Sinh[\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]
Out[23]= -3.\
9317890831820506378791034479406121284684487483182042179057328100219696\
20202464096600592983999731376*10^-55
In[21]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] -
g[(1 - I v)]/(Sinh[-\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]
Out[21]= -3.\
9317890831820506378791034479406121284684487483182042179057381396998279\
83065832972052160228141179706*10^-55
In[25]:= Quiet[
NIntegrate[
Im[g[(1 + I t)]/Sinh[\[Pi] t] +
g[(1 + I v)]/(Sinh[-\[Pi] v] u^2)], {t, 1, Infinity},
WorkingPrecision -> 100]]
Out[25]= -3.\
9317890831820506378791034479406121284684487483182042179057328100219696\
20202464096600592983999731376*10^-55
----------
----------
On 9/29/2021
------------
I found the following equation for **C**<sub>*MRB*</sub> (great for integer arithmetic because
(1-1/n)^k=(n-1)^k/n^k. )
![CMRB integers 1][143]
So, using only integers, and sufficiently large ones in place of infinity, we can use
![CMRB integers 2][144]
See
In[1]:= Timing[m=NSum[(-1)^n (n^(1/n)-1),{n,1,Infinity},WorkingPrecision->200,Method->"AlternatingSigns"]][[1]]
Out[1]= 0.086374
In[2]:= Timing[m-NSum[(-1)^n/x! (Sum[((-1 + n)^k) /(k n^(1 + k)), {k, 1, Infinity}])^ x, {n, 2, Infinity}, {x, 1,100}, Method -> "AlternatingSigns", WorkingPrecision -> 200, NSumTerms -> 100]]
Out[2]= {17.8915,-2.2*^-197}
It is very much slower, but it can give a rational approximation (p/q), like in the following.
In[3]:= mt=Sum[(-1)^n/x! (Sum[((-1 + n)^k) /(k n^(1 + k)), {k, 1,500}])^ x, {n, 2,500}, {x, 6}];
In[4]:= N[m-mt]
Out[4]= -0.00602661
In[5]:= Head[mt]
Out[5]= Rational
Compared to the NSum formula for m, we see
In[6]:= Head[m]
Out[6]= Real
----------
----------
On 9/19/2021
------------
I found the following quality of **C**<sub>*MRB*</sub>.
![replace constants for CMRB][145]
----------
----------
On 9/5/2021
-----------
I added the following MRB constant integral over an unusual range.
![strange][146]
See proof [in this link here][147].
----------
----------
On Pi Day, 2021, 2:40 pm EST,
-----------------------------
I added a new MRB constant integral.
![CMRB][148] ![=][149] ![integral to sum][150]
We see many more integrals for **C**<sub>*MRB*</sub>.
We can expand
![1/x][151]
into the following.
![xx = 25.656654035][152]
xx = 25.65665403510586285599072933607445153794770546058072048626118194\
90097321718621288009944007124739159792146480733342667`100.;
g[x_] = x^(xx/
x); I NIntegrate[(g[(-t I + 1)] - g[(t I + 1)])/(Exp[Pi t] -
Exp[-Pi t]), {t, 0, Infinity}, WorkingPrecision -> 100]
(*
0.18785964246206712024851793405427323005590309490013878617200468408947\
72315646602137032966544331074969.*)
**Expanding upon the previously mentioned**
![enMRB sinh][153]
we get the following set of formulas that all equal **C**<sub>*MRB*</sub>:
Let
x= 25.656654035105862855990729 ...
along with the following constants (approximate values given)
{u = -3.20528124009334715662802858},
{u = -1.975955817063408761652299},
{u = -1.028853359952178482391753},
{u = 0.0233205964164237996087020},
{u = 1.0288510656792879404912390},
{u = 1.9759300365560440110320579},
{u = 3.3776887945654916860102506},
{u = 4.2186640662797203304551583} or
$
u = \infty .$
Another set follows.
let
x = 1 and
along with the following {approximations}
{u = 2.451894470180356539050514},
{u = 1.333754341654332447320456} or
$
u = \infty $
then
![enter image description here][154]
See
[this notebook from the wolfram cloud][155]
for justification.
----------
----------
2020 and before:
--------
Also, in terms of the Euler-Riemann zeta function,
**C**<sub>*MRB*</sub> =![enter image description here][156]
Furthermore, as ![enter image description here][157],
according to [user90369][158] at Stack Exchange, **C**<sub>*MRB*</sub> can be written as the sum of zeta derivatives similar to the eta derivatives discovered by Crandall.
![zeta hint ][159] Information about η<sup>(j)</sup>(k) please see e.g. [this link here][160], formulas (11)+(16)+(19).![credit][161]
----------
----------
In the light of the parts above, where
**C**<sub>*MRB*</sub>
= ![k^(1/k)-1][162]
= ![eta'(k)][163]
= ![sum from 0][164] ![enter image description here][165]
as well as ![double equals RHS][166]
an internet scholar going by the moniker "Dark Malthorp" wrote:
> ![eta *z^k][167]
----------
----------
----------
----------
----------
Primary Proof 1
------------------
**C**<sub>*MRB*</sub>=![enter image description here][168], based on
**C**<sub>*MRB*</sub>
![eta equals][169]
![enter image description here][170]
is proven below by an internet scholar going by the moniker "Dark Malthorp."
![Dark Marthorp's proof][171]
----------
----------
Primary Proof 2
------------------
![eta sums][172] denoting the kth derivative of the Dirichlet eta function of k and 0 respectively,
was first discovered in 2012 by Richard Crandall of Apple Computer.
The left half is proven below by Gottfried Helms and it is proven more rigorously![(][173]considering the conditionally convergent sum,![enter image description here][174]![)][175] below that. Then the right half is a Taylor expansion of eta(s) around s = 0.
> ![n^(1/n)-1][176]
At
[https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal][177],
it has been noted that "even though one has cause to be a little bit wary around formal rearrangements of conditionally convergent sums (see the [Riemann series theorem][178]), it's not very difficult to validate the formal manipulation of Helms. The idea is to cordon off a big chunk of the infinite double summation (all the terms from the second column on) that we know is absolutely convergent, which we are then free to rearrange with impunity. (Most relevantly for our purposes here, see pages 80-85 of this [document][179], culminating with the Fubini theorem which is essentially the manipulation Helms is using.)"
> ![argument 1][180] ![argument 2][181]
----------
----------
Primary Proof 3
------------------
Here is proof of a faster converging integral for its integrated analog (The MKB constant) by Ariel Gershon.
g(x)=x^(1/x), M1=![hypothesis][182]
Which is the same as
![enter image description here][183]
because changing the upper limit to 2N + 1 increases MI by 2i/?.
MKB constant calculations have been moved to their discussion at [http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH][184] .
![Iimofg->1][185]
![Cauchy's Integral Theorem][186]
![Lim surface h gamma r=0][187]
![Lim surface h beta r=0][188]
![limit to 2n-1][189]
![limit to 2n-][190]
Plugging in equations [5] and [6] into equation [2] gives us:
![left][191]![right][192]
Now take the limit as N?? and apply equations [3] and [4] :
![QED][193]
He went on to note that
![enter image description here][194]
----------
----------
----------
----------
I wondered about the relationship between CMRB and its integrated analog and asked the following.
![enter image description here][195]
So far I came up with
&[Wolfram Notebook][196]
Another relationship between the sum and integral that remains more unproven than I would like is
![CMRB(1-i)][197]
f[x_] = E^(I \[Pi] x) (1 - (1 + x)^(1/(1 + x)));
CMRB = NSum[f[n], {n, 0, Infinity}, WorkingPrecision -> 30,
Method -> "AlternatingSigns"];
M2 = NIntegrate[f[t], {t, 0, Infinity I}, WorkingPrecision -> 50];
part = NIntegrate[(Im[2 f[(-t)]] + (f[(-t)] - f[(t)]))/(-1 +
E^(-2 I \[Pi] t)), {t, 0, Infinity I}, WorkingPrecision -> 50];
CMRB (1 - I) - (M2 - part)
gives
> 6.103779*10^-23 - 6.103779*10^-23 I.
Where the integral does not converge, but Mathematica can give it a value:
![enter image description here][198]
----------
----------
----------
----------
----------
Here is my mini cluster of the fastest 3 computers (the MRB constant supercomputer 0) mentioned below:
The one to the left is my custom-built extreme edition 6 core and later with an 8 core 3.4 GHz Xeon processor with 64 GB 1666 MHz RAM..
The one in the center is my fast little 4 core Asus with 2400 MHz RAM.
Then the one on the right is my fastest -- a Digital Storm 6 core overclocked to 4.7 GHz on all cores and with 3000 MHz RAM.
![first 3 way cluster][199]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1ac.JPG&userId=366611
[2]: https://en-academic.com/dic.nsf/enwiki/11755
[3]: https://www.amazon.com/exec/obidos/ASIN/193563819X/ref=nosim/ericstreasuretro
[4]: http://www.marvinrayburns.com/UniversalTOC25.pdf
[5]: http://web.archive.org/web/20081121134611/http://www.irancivilcenter.com/en/tools/units/math_const.php
[6]: https://www.amazon.com/exec/obidos/ASIN/0521818052/ref=nosim/ericstreasuretro
[7]: https://web.archive.org/web/20010616211903/http://pauillac.inria.fr/algo/bsolve/constant/itrexp/itrexp.html
[8]: https://pdfs.semanticscholar.org/f3dc/9f828442123015c5d52535fdefc0ab450bf5.pdf
[9]: https://web.archive.org/web/20210812083640/http://www.bitman.name/math/article/962
[10]: https://arxiv.org/pdf/0912.3844v3.pdf
[11]: http://ebyte.it/library/educards/constants/MathConstants.pdf
[12]: https://web.archive.org/web/20090702134146/http://www.turkbiyofizik.com/sabitler.html
[13]: https://mathworld.wolfram.com/MRBConstant.html
[14]: http://oeis.org/wiki/MRB_constant
[15]: https://web.archive.org/web/20030415202103/http://pi.lacim.uqam.ca/eng/table_en.html
[16]: https://web.archive.org/web/20001210231700/http://www.lacim.uqam.ca/piDATA/mrburns.txt
[17]: https://oeis.org/A037077
[18]: https://en.wikipedia.org/wiki/MRB_constant
[19]: https://community.wolfram.com//c/portal/getImageAttachment?filename=106291.jpg&userId=366611
[20]: https://community.wolfram.com//c/portal/getImageAttachment?filename=62082.jpg&userId=366611
[21]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-15100048.jpg&userId=366611
[22]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-08043948.png&userId=366611
[23]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-19060404.jpg&userId=366611
[24]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-15005626.jpg&userId=366611
[25]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-15010434.jpg&userId=366611
[26]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-17062423.jpg&userId=366611
[27]: https://mathworld.wolfram.com/GelfondsConstant.html
[28]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-19070749.jpg&userId=366611
[29]: https://www.wolframcloud.com/obj/bmmmburns/Published/5_11_2022.nb
[30]: https://community.wolfram.com//c/portal/getImageAttachment?filename=44623.jpg&userId=366611
[31]: https://community.wolfram.com//c/portal/getImageAttachment?filename=104034.jpg&userId=366611
[32]: https://community.wolfram.com//c/portal/getImageAttachment?filename=101513.PNG&userId=366611
[33]: https://en.wikipedia.org/wiki/Oscillatory_integral
[34]: https://scholar.google.com/scholar?hl=en&as_sdt=0,15&q=%22MKB%20constant%22&btnG=
[35]: https://arxiv.org/abs/0912.3844
[36]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capturemkb2.JPG&userId=366611
[37]: https://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=zHVSqCM8
[38]: https://www.researchgate.net/publication/309187705_Gelfond%27_s_Constant_using_MKB_constant_like_integrals
[39]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-03083042.jpg&userId=366611
[40]: https://community.wolfram.com//c/portal/getImageAttachment?filename=73892.JPG&userId=366611
[41]: https://web.archive.org/web/20120310223600/http://pi.lacim.uqam.ca/eng/records_en.html
[42]: https://community.wolfram.com//c/portal/getImageAttachment?filename=P1.JPG&userId=366611
[43]: https://community.wolfram.com//c/portal/getImageAttachment?filename=P2.JPG&userId=366611
[44]: https://community.wolfram.com//c/portal/getImageAttachment?filename=P3.JPG&userId=366611
[45]: https://community.wolfram.com//c/portal/getImageAttachment?filename=P5.JPG&userId=366611
[46]: http://marvinrayburns.com/Original_MRB_Post.html
[47]: https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb
[48]: https://en.wikipedia.org/wiki/MRB_constant
[49]: https://en.wikipedia.org/wiki/MRB_constant
[50]: https://community.wolfram.com//c/portal/getImageAttachment?filename=w1.jpg&userId=366611
[51]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-24021607.jpg&userId=366611
[52]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-28212004.jpg&userId=366611
[53]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-27102212.jpg&userId=366611
[54]: http://mathworld.wolfram.com/MRBConstant.html
[55]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Capture30.JPG&userId=366611
[56]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2021-11-14113747.png&userId=366611
[57]: https://mathworld.wolfram.com/SteinersProblem.html
[58]: https://community.wolfram.com//c/portal/getImageAttachment?filename=limit_nto1on_eq1.jpg&userId=366611
[59]: https://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1
[60]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-04014455.jpg&userId=366611
[61]: https://www.amazon.com/exec/obidos/ASIN/0521818052/ref=nosim/ericstreasuretro
[62]: https://mathworld.wolfram.com/Closed-FormSolution.html#:~:text=An%20equation%20is%20said%20to,not%20be%20considered%20closed-form.
[63]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-08065337.jpg&userId=366611
[64]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-06204151.jpg&userId=366611
[65]: https://community.wolfram.com//c/portal/getImageAttachment?filename=88723.jpg&userId=366611
[66]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-02-01115012.png&userId=366611
[67]: https://community.wolfram.com//c/portal/getImageAttachment?filename=24492.jpg&userId=366611
[68]: https://community.wolfram.com//c/portal/getImageAttachment?filename=iest.gif&userId=366611
[69]: https://community.wolfram.com//c/portal/getImageAttachment?filename=7284test2.gif&userId=366611
[70]: https://community.wolfram.com//c/portal/getImageAttachment?filename=88723.jpg&userId=366611
[71]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-02-12161446.jpg&userId=366611
[72]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-20174334.jpg&userId=366611
[73]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-20175444.jpg&userId=366611
[74]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-02-13031818.jpg&userId=366611
[75]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-12032123.jpg&userId=366611
[76]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-12201237.jpg&userId=366611
[77]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-25051449.jpg&userId=366611
[78]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-25052650.jpg&userId=366611
[79]: https://web.archive.org/web/20010616211903/http://pauillac.inria.fr/algo/bsolve/constant/itrexp/itrexp.html
[80]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Mc1.jpg&userId=366611
[81]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Mc3.jpg&userId=366611
[82]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-08050441.jpg&userId=366611
[83]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-08211605.jpg&userId=366611
[84]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-08211514.jpg&userId=366611
[85]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-27065833.jpg&userId=366611
[86]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-04-02091305.jpg&userId=366611
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[88]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-29143511.jpg&userId=366611
[89]: https://en.wikipedia.org/wiki/Nth_root#Geometric_constructibility
[90]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-29040006.jpg&userId=366611
[91]: https://mathblog.com/the-geometry-of-the-mrb-constant/
[92]: https://en.wikipedia.org/wiki/Minkowski_space
[93]: https://www.vixra.org/abs/1609.0082
[94]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-01-31045436.png&userId=366611
[95]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-01-31050317.png&userId=366611
[96]: https://scholar.google.com/scholar?hl=en&as_sdt=0,15&q=%22MRB%20constant%22&btnG=
[97]: https://en.wikipedia.org/wiki/Richard_Crandall
[98]: https://scholar.google.com/scholar?hl=en&as_sdt=0,15&q=key%20fundamental%20constant%20zeta&btnG=
[99]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-05-02113359.jpg&userId=366611
[100]: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.5959&rep=rep1&type=pdf
[101]: https://www2.mpia-hd.mpg.de/~mathar/
[102]: https://arxiv.org/abs/0912.3844
[103]: https://pdfs.semanticscholar.org/f3dc/9f828442123015c5d52535fdefc0ab450bf5.pdf
[104]: https://www.academia.edu/search?q=%22MRB%20constant%22
[105]: https://physics.stackexchange.com/questions/5207/number-of-dimensions-in-string-theory-and-possible-link-with-number-theory
[106]: https://scholar.google.com/citations?hl=it&user=iDJ9f5gAAAAJ&view_op=list_works&sortby=pubdate
[107]: https://reference.wolfram.com/language/ref/MeijerG.html
[108]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8236Capture20.JPG&userId=366611
[109]: https://www.wolframcloud.com/obj/bmmmburns/Published/Oct_11_2021.nb
[110]: https://reference.wolfram.com/language/ref/MeijerG.html
[111]: https://www.wolframcloud.com/env/bd78bab5-4bb6-445c-977d-7d7b0abb0f05
[112]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Captureb.PNG&userId=366611
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[129]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8348Capture.JPG&userId=366611
[130]: https://arxiv.org/abs/0912.3844
[131]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2021-11-14085836.png&userId=366611
[132]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2021-11-14085338.png&userId=366611
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[136]: https://www.wolframcloud.com/obj/bmmmburns/Published/Nov_9_2021.nb
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[155]: https://www.wolframcloud.com/obj/bmmmburns/Published/double%20zeroes%20of%20CMRB.nb
[156]: https://community.wolfram.com//c/portal/getImageAttachment?filename=11i.JPG&userId=366611
[157]: https://community.wolfram.com//c/portal/getImageAttachment?filename=6033d.JPG&userId=366611
[158]: https://math.stackexchange.com/users/332823/user90369
[159]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1558a.JPG&userId=366611
[160]: https://digitalcommons.wku.edu/cgi/viewcontent.cgi?referer=http://www.google.de/url?sa=t&rct=j&q=&esrc=s&source=web&cd=27&ved=2ahUKEwjMx5SuxbnjAhVLLpoKHcBPBWo4FBAWMAZ6BAgAEAI&url=http://digitalcommons.wku.edu/cgi/viewcontent.cgi?article=2093&context=theses&usg=AOvVaw0gQx0dl_Nw4esC2IQc0LEo&httpsredir=1&article=2093&context=theses
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[173]: https://community.wolfram.com//c/portal/getImageAttachment?filename=left.gif&userId=366611
[174]: https://community.wolfram.com//c/portal/getImageAttachment?filename=3959Capture7.JPG&userId=366611
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[176]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10297Capture.JPG&userId=366611
[177]: https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal
[178]: https://en.wikipedia.org/wiki/Riemann_series_theorem
[179]: https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch4.pdf
[180]: https://community.wolfram.com//c/portal/getImageAttachment?filename=7211Capture.JPG&userId=366611
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[183]: https://community.wolfram.com//c/portal/getImageAttachment?filename=580910.JPG&userId=366611
[184]: http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH
[185]: https://community.wolfram.com//c/portal/getImageAttachment?filename=28491.JPG&userId=366611
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[194]: https://community.wolfram.com//c/portal/getImageAttachment?filename=3.PNG&userId=366611
[195]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1021422.JPG&userId=366611
[196]: https://www.wolframcloud.com/obj/d0181107-938a-4119-a2ed-a75b72dd285d
[197]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8a.JPG&userId=366611
[198]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2022-03-05002204.jpg&userId=366611
[199]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ezgif.com-video-to-gif.gif&userId=366611Marvin Ray Burns2014-10-09T18:08:49ZDirectly emulate Ridge Regression and LASSO with a neural network?
https://community.wolfram.com/groups/-/m/t/2179969
I would like for pedagogic purposes to show how Ridge Regression and LASSO (and presumably other forms of regularized regression such as ElasticNet) can be emulated using a neural network in the same way that one can emulate conventional regression. But for the life of me I can't figure out how to get the coefficients out of the LinearLayer and feed them into some sort of penalty function. I suspect I am just being dense. Surely there is some sort of NetGraph and some sort of Layer that can do this. Can anyone help? (Yes, I know there are many other ways of preventing overfitting, but this is the one I want to use).Seth Chandler2021-01-31T14:33:40ZFormulaData doesn't solve equation
https://community.wolfram.com/groups/-/m/t/2532075
Hello everyone,
I tried to solve an equation with the FormulaData command.
I copied an example from the documentation for FormulaData and ran it. It outputted a few times: Interpreting unit "AmpereConstant". And after a few seconds, it runs into a recursion limit.
&[Wolfram Notebook][1]
This happens with every formula I try to solve, even those that have nothing to do with electricity.
Can someone help me with this or has the same problem?
Thanks for your help.
Jesse
[1]: https://www.wolframcloud.com/obj/d070bccb-92b5-4725-aa3b-4bddaa81890aJesse Palarus2022-05-15T09:49:51ZCan't retrieve a Key in an Association, but the key is listed in it
https://community.wolfram.com/groups/-/m/t/2531767
Hello,
I've encountered a strange behavior in Mathematica, and it is quickly reproducible
Table[q, {q, 0.01, 0.99, 0.01}]
Table[-Sqrt[2] InverseErfc[2 q],{q, 0.01, 0.99, 0.01}]
ndQuantile = AssociationThread[%% -> %]
Calling ndQuantile[0.35] prints that this Key is missing, but I most definitely can see it listed in the Association dataset. Calling every other keys seem to work fine (0.4, 0.45 etc).
Using Table to make a list of every values in the Association lets me retrieve the desired value, so I'm lost.
Table[ndQuantile[x],{x,0.01,0.99,0.01}]
%[[35]]
Am I doing something wrong ?
Thank you.Clarisse Wagner2022-05-14T17:49:29ZInitial cell is evaluating and outputting twice in Wolfram Cloud notebook
https://community.wolfram.com/groups/-/m/t/2531545
I have a notebook set up in Wolfram Cloud with an initialization cell that _should_ run whenever it is opened. Two issues:
a) The initialization cell is not running at open;
b) If I choose to Evaluate the notebook, the initialization cell runs twice (if I change the cell to _not_ be an initialization cell and run the notebook, it correctly outputs one time).
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/a6ecbbfd-dea2-485d-8a4c-1f0acd16c2e4Steven Buehler2022-05-14T18:00:17ZRe-exploring the structure of Chinese character images
https://community.wolfram.com/groups/-/m/t/2516662
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/afc17dcc-20ac-4990-a933-02fed606f0afAnton Antonov2022-04-22T17:14:10ZExpress arbitrary number in complex filed as a cyclotomic polynomial.
https://community.wolfram.com/groups/-/m/t/2531019
GAP system uses [Cyclotomic Numbers](https://www.gap-system.org/Manuals/doc/ref/chap18.html#X7DFC03C187DE4841) for Trace (character table) related computation. So, I want to know if I can convert the express of arbitrary number in complex filed, e.g., `3^(1/7)`, into a cyclotomic polynomial with the help of Wolfram language, so that I can use it conveniently in GAP.
Regards,
HZHongyi Zhao2022-05-13T13:56:10ZArea in multidimensional space using Wolfram|Alpha?
https://community.wolfram.com/groups/-/m/t/2529301
Can someone please tell me how to use Wolfram|Alpha to calculate area in a multidimensional space, given the coordinates of the vertices? E.g., the area of a triangle in 5D with vertices (0,0,0,1,2), (1,2,1,3,2), (2,0,1,1,1) ?Tim Roberts2022-05-11T03:38:05ZHow to force Dynamic Clock in a Button to wait for mouse click to start?
https://community.wolfram.com/groups/-/m/t/2361954
Abstract
--------
My task sounds simple and straight forward but not all is as it first appears. I just need a `Button` that waits to be `Clicked` before displaying a `countdown` timer in the label. `Button` will eventually do a lot more which I have coded before including evaluation control however everything I tried to stall `countdown` just doesn't work. The `Clock` in `countdown` always starts when the `Button` is created even if the current value of label isn't displayed as demonstrated in `Button` samples given below.
I even tried wrapping and embedding `Button` with multiple layers of mouse `EventHandlers` which worked with styles like `Background` but failed completely on `countdown` (did not update label). I'm almost positive I did something wrong. Unfortunately that code was lost. *Please help me find a solution. I have been trying to solve this off and on for months*.
`countdown` (Clock)
-----
Here is my latest iteration of `countdown`. As you can see it's wrapped in an anonymous `Function` which has attributes `HoldAll` (should not `Evaluate` until all arguments are provided) in an attempt to force the `Clock` in `countdown` to wait on a mouse event. A few of the various `Buttons` attempted are shown afterwards.
countdown =
Function[{s, n}, Dynamic[s - Clock[{0, s, 1}, s, n]]];
(* demo countdown timer *)
Row[{"almost bare naked countdown \[Rule] ", countdown[10, 1]}]
Button 0: Global
-------
This is the only `Button` not wrapped in a `DynamicModule` and in which label `x` is set to an intermediary variable named `timer` (this variable is only used here) so that `Clock` in `countdown` can be observed to start without a `Click` event (`Evaluate Cells` may need to be executed more than once). Immediate `Evaluation` (before mouse event) is not how a `Button` is supposed to behave.
Quiet@ClearAll[x];
Row[{"Global label (x) set to intermediate variable (timer) \[Rule] ",
Button[Dynamic[x = timer], (timer := countdown[60, 1]) &,
ImageSize -> {75, 30}]}]
Button 1: Click
--------
After momentary wait click `Button`. If less than 60 `Clock` in `countdown` did not wait for `Click` event. Also note that `Clock` continued to run dynamically despite not displaying updated value in label `x` before `Click` event. This even continues to run if `Cell` containing `Button` is scrolled out of view. This is not how `Dynamic` is supposed to work. Although `AutoAction -> False` is the default setting it is explicitly set for reasons which will become clear in *Button 2: (Hover)*.
DynamicModule[{x},
Row[{"DynamicModule Button with regular Click \[Rule] ",
Button[Dynamic[x, (#) &,
Initialization :> (x = "wait 5s then Click")], (x :=
countdown[60, 1]) &, ImageSize -> {150, 30},
AutoAction -> False]}]]
Button 2: Hover
---------------
After momentary wait hover `Button`. If less than 60 `Clock` in `countdown` did not wait for `Hover` event. Again note that this *Button 2 (Hover)* exhibits all the same dynamics as *Button 1 (Click)*. That is because `AutoAction` does not supercharge `Evaluation`. The only thing it does is emulate a `Click` event with `Hover`.
DynamicModule[{x},
Row[{"DynamicModule Button with Hover \[Rule] ",
Button[Dynamic[x, (#) &,
Initialization :> (x = "wait 5s then Hover")], (x :=
countdown[60, 1]) &, ImageSize -> {150, 30},
AutoAction -> True]}]]
Why Stress Click vs. Hover?
---------------------------
I stress the role of option `AutoAction` because I posted similar questions before at mathematica.stackexchange.com...
1. https://mathematica.stackexchange.com/questions/253172/please-help-with-dynamic-autosave-button-that-only-halfway-works
2. https://mathematica.stackexchange.com/questions/253544/my-dynamic-button-evaluates-as-soon-as-its-created-how-do-i-stop-this-behavior
but the few well-meaning good folks who tried to help only provided halfway good but mostly unsatisfactory answers. They were under the mistaken impression that `AutoAction` would control `Clock` evaluation or that there was nothing wrong with my code. I believe they didn't wait long enough after `Button` was rendered to observe the immediate invocation of `countdown` without a mouse event. It was mostly *my bad* for not properly setting up demos. That is what I tried to avoid here.
Closing Thoughts
----------------
The official Wolfram Documentation isn't very clear on how `Clock` evaluation is supposed to work. I don't believe there is anything wrong with `Clock` but it could have been designed with better evaluation control. However it is absolute on how `Dynamic` and `Button` are designed to work. `Dynamic` objects should never update when not in view (unless contained in `DynamicWrapper`) and `Button` should never permit evaluation of expression unless invoked by a mouse or similar event. I believe these are *bugs* with both and `Button` and `Dynamic`. However I still have faith that there is a simple work-around (without injecting code with spaghettis-like *low-level* procedural loops and routines) using some of the deeper `Evaluation` controls provided by *Mathematica*.Jules Manson2021-09-08T12:31:33ZQuantum random walk: a drunken quantum sailor
https://community.wolfram.com/groups/-/m/t/2531168
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/afa3a8a9-3261-49be-bfe8-54bc222cbec3Mads Bahrami2022-05-13T19:35:25ZAn entropic lens on stabilizer states
https://community.wolfram.com/groups/-/m/t/2531074
*WOLFRAM MATERIALS for the ARTICLE:*
> *An Entropic Lens on Stabilizer States*, Cynthia Keeler, William Munizzi, Jason Pollack (2022).
> arXiv:2204.07593 [quant-ph].
> https://doi.org/10.48550/arXiv.2204.07593
> [Full article in PDF][1]
> GitHub: https://github.com/WMunizzi/StabilizerStateData
&[Wolfram Notebook][2]
[1]: https://arxiv.org/pdf/2204.07593
[2]: https://www.wolframcloud.com/obj/e7e83c2e-9164-4c15-8c8f-43bae69c720fWilliam Munizzi2022-05-13T18:31:48ZPerformance measures for FindClusters?
https://community.wolfram.com/groups/-/m/t/2528464
I have tried different combinations of parameters (Method, CriterionFunction ,and DistanceFunction) in FindClusters.<br />
But I still don't know what the best combination.<br />
What is the performance index of FindClusters? How to get it?Tsai Ming-Chou2022-05-10T13:13:46ZHow to write a function with a conditional expression on itself?
https://community.wolfram.com/groups/-/m/t/2529718
How to write such a function, since it giving recursion error problem
y=33x^(3/2) UnitStep[y-1]+x^(-1)*5*E^x UnitStep[1-y]John Wick2022-05-12T07:03:57ZNSolve gives an empty set as a solution
https://community.wolfram.com/groups/-/m/t/2530639
This gives a result
NSolve[{7463^x*2070^y*15^w == 3, 7060^x*2035^y*15^w == 5,
8874^x*1945^y*15^w == 23}, {w, x, y}, Reals]
This doesn't
NSolve[{7463^x*2070^y*15^w*10^z == 3, 7060^x*2035^y*15^w*10^z == 5,
8874^x*1945^y*15^w*10^z == 23, 9918^x*1949^y*15^w*10^z == 17}, {w,
x, y, z}, Reals]S Sarkar2022-05-13T04:24:15ZMultiple errors while using MintNFT[ ]?
https://community.wolfram.com/groups/-/m/t/2351041
Has anyone been able to get the MintNFT function to work? I'm pretty much using exactly the code from
https://resources.wolframcloud.com/FunctionRepository/resources/MintNFT
but I get this when I call the actual minting function:
> Select::normal: Nonatomic expression expected at position 1 in
Select[ServiceConnections`Private`$authenticatedservices,ServiceConnections`Private`serviceName[#1]===IPFS&].
and also
> ExternalStorageUpload::multser: One service was chosen from multiple name services.
Any advice?William Doyle2021-08-25T01:56:03Z[WSS20] Investigating quantum interference effects in Wolfram Models
https://community.wolfram.com/groups/-/m/t/2028586
![enter image description here][1]
&[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=WM.png&userId=2026505
[2]: https://www.wolframcloud.com/obj/dcafbe94-019c-40b9-a1f9-c6bb4a390461Hatem Elshatlawy2020-07-14T15:45:04ZSolving two equations using Wolfram|Alpha?
https://community.wolfram.com/groups/-/m/t/2530442
Hello,
I am trying to Solve these two equations on WOLFRAM for x and y, but I couldn't
2 ((L x^3)/12 + L x (W/2 - x/2)^2) + ((W - 2 x)^3 y)/6 == B
2 (((W - 2 x) y^3)/12 + (W - 2 x) y (L/2 - y/2)^2) + (x L^3)/6 == A
Can anyone guide me on how to do that?
Thanks in Advance.Raghed Arnouk2022-05-12T14:58:51ZStars orbiting Sagittarius A*: evidence for a black hole
https://community.wolfram.com/groups/-/m/t/2530346
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/5a4a012d-db7a-4fbf-97a4-d83f03573ba9Jeff Bryant2022-05-12T19:16:56ZCodeParser and CodeInspector
https://community.wolfram.com/groups/-/m/t/1931315
[![enter image description here][2]][1]
&[Wolfram Notebook][3]
[1]: https://youtu.be/rOa5IntICFA
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2020-04-09at12.54.33PM.png&userId=11733
[3]: https://www.wolframcloud.com/obj/afe2a2fb-ee55-4df5-a6fb-9bc16dd08af7Brenton Bostick2020-04-09T15:04:38ZA function to close and open sections in a notebook?
https://community.wolfram.com/groups/-/m/t/2528182
Hi, All!
I want to write a function in Mathematica 9 that opens and closes all sections in my notebook. I do it as follows:
openCloseAll[nb_, target_String, to : (Open | Closed)] :=
Do[SelectionMove[cell, All, CellGroup, AutoScroll -> False];
With[{content =
Block[{$Context = "FrontEnd`", $ContextPath = {"System`"}},
NotebookRead[nb]], from = to /. {Closed -> Open, Open -> Closed}},
If[MatchQ[content, Cell[CellGroupData[{Cell[_, target, ___], __}, from]]],
NotebookWrite[nb, Cell[CellGroupData[content[[1, 1]], to]],
AutoScroll -> False]]];, {cell, Cells[CellStyle -> target]}]
Then I create two buttons to deal with sections:
Button["Close sections", openCloseAll[EvaluationNotebook[], "Section", Closed]]
and
Button["Open sections", openCloseAll[EvaluationNotebook[], "Section", Open]]
At last, I create a section with a single cell in it (a comment between two expressions - it is important!):
something1
(*a comment*)
something2
The resulting file attached below is the first notebook. When I push successively "Close sections" and "Open sections" buttons, I see a garbage in the section instead of the first two lines of the cell (see the second attached file). What happens?Vladimir Ivanov2022-05-09T21:23:58ZExport to PNG gives error on Raspberry Pi?
https://community.wolfram.com/groups/-/m/t/2465013
Dear Wolfram Community,
for some time now I observe the following issue under `Wolfram` installed within a docker container on a Raspberry Pi.
When I try to export something to png, e.g.,
Export["test.png", Plot[x, {x, 0, 1}]]
I get
Import::fmterr: Cannot import data as PNG format.
On the other hand I can export to, e.g., jpg just fine. Moreover, I can import png images without problems.
**Some information on my environment:**
The Wolfram engine is installed in a Docker container running on a Raspberry Pi 4 with Ubuntu 20.04.
I tried different base images which both showed the same above described behavior:
- the official `python:3.10-bullseye` container, where I added the deb
```
http://archive.raspberrypi.org/debian bullseye main
```
and added the `armhf` architecture to `dpkg` before installing the `wolfram-engine` package.
(This setup did already work in the past, but at some point a rebuild of the image must have broken things, which I didn't realize as I did not export to png for some time.)
- the image `navikey/raspbian-bullseye` where I just added the `armhf` architecture to `dpkg` and then installed the `wolfram-engine` package.
Can somebody else reproduce this problem or has a hint on what I'm doing wrong.
Thanks and Best Regards,
DominikDominik V.2022-02-06T17:44:43ZResearch opportunity for out of the box type of grad student
https://community.wolfram.com/groups/-/m/t/2530004
I was hoping that I could find a grad student that would be willing to investigate my analysis of the MRB constant, as well as my procedural and numerical methods found in the Wolfram Community post, [Try to beat these MRB constant records][1].
I've spent 23 years developing them, and I'd enjoy seeing someone bringing some of them into academia!
I'd like to reward you for your time to the best of my ability, so perhaps you might find something that you can perfect to add to your CV!
Let's keep our first contact on this board as a monument for both of ours's sake.
[1]: https://community.wolfram.com/groups/-/m/t/366628?p_p_auth=t2LiUcqWMarvin Ray Burns2022-05-12T09:47:58Z[GiF] Tangents (Tangent lines to a Bessel function)
https://community.wolfram.com/groups/-/m/t/1105345
![Tangent lines to a Bessel function][1]
**Tangents**
This shows 900 tangent lines to the Bessel function $J_4$. The Bessel function itself is not explicitly shown at all, but, as in [_Bessel_][2], the graph of the function emerges quite clearly from the approximations.
One note: the GIF is made up out of 1801 frames, which seems to make the Javascript on the post preview page very unhappy, and the original file output by Mathematica was 275 MB(!). Using [gifsicle][3] to reduce the color palette and only record diffs between consecutive frames, I was able to get this down to 2.9 MB, or about 100 times smaller (and thereby actually reasonable to post online).
Here's the code, which makes use of one simple function that returns the tangent line to a function at a given point:
TangentLine[f_, x_] := InfiniteLine[{x, f[x]}, {1, f'[x]}];
DynamicModule[{f, b = N[BesselJZero[4, 6]], n = 450,
cols = RGBColor /@ {"#FF5200", "#003355"}},
f[x_] := BesselJ[4, x];
Manipulate[
Graphics[
{Opacity[.1], cols[[1]],
Table[TangentLine[f, x], {x, Max[-b, s - 2 b], Min[s, b], b/n}]},
PlotRange -> {{-b, b}, {-2, 2}}, AspectRatio -> 2/3,
ImageSize -> 540, Background -> cols[[-1]]],
{s, -b, 3 b}]
]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=lines9c.gif&userId=610054
[2]: http://community.wolfram.com/groups/-/m/t/1052218
[3]: http://lcdf.org/gifsicle/Clayton Shonkwiler2017-05-24T05:00:14ZCalculations with 7th root of unity
https://community.wolfram.com/groups/-/m/t/2529431
Hi, I'm trying to do some simple computations with 7th roots of unity that fail. I define
m=Exp[2Pi I/7]
and then, as a sanity check, I ask Mathematica to check that 1+m+...+m^6=0, but I don't see how to get this done.
Simplify[Sum[m^k,{k,0,6}]]
does not work.
For the record, I also tried the analogous thing when m=Exp[2Pi I/5], and it also failed. But it did work when I defined
m=Cos[2Pi/5]+I*Sin[2pi/5]Dimitris Koukoulopoulos2022-05-11T18:33:13Z