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RSS Feed for Wolfram Community showing any discussions in tag Mathematics sorted by activeVisualizing the 100 first factorials: 100!
https://community.wolfram.com/groups/-/m/t/2296036
![enter image description here][1]
&[Wolfram Notebook][2]
[DONT DELETE: Original Notebook]: https://www.wolframcloud.com/obj/3b2957bd-7371-4fb3-b042-863112b39f31
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=.jpg&userId=20103
[2]: https://www.wolframcloud.com/obj/f97fcd55-358e-4975-8df6-29ddabe7fe7eDaniel Carvalho2021-06-22T19:54:30ZBlank plot of complex polynomial equation solution?
https://community.wolfram.com/groups/-/m/t/2295297
Dear All,
I have a code for a complex polynomial equation which I'm trying to plot solving its real and imaginary parts separately. While I try to plot it gives me a blank plot.
I need your valuable advice which can help me immensely in solving the same.
please advice me.
The code is as attached herewith.
rgdsRahul Chakrabory2021-06-22T11:58:26ZDynamic portraits w/ quasicrystal, waves interference and half-toning
https://community.wolfram.com/groups/-/m/t/2292842
*MODERATORS' NOTE: This post is based on the [tumblr blog found here][1].*
![enter image description here][2]
&[Wolfram Notebook][3]
[1]: https://intothecontinuum.tumblr.com/post/122804795458/mathematica-codehthr-imporththrtzyjpg
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=tumblr_nqffnpsNNO1qfjvexo1_540.gif&userId=20103
[3]: https://www.wolframcloud.com/obj/a66cdce5-bd6f-4882-a109-01a1f0cacd62Sumit Sijher2021-06-17T22:09:28ZHow can I solve this impulsive heat equation please?
https://community.wolfram.com/groups/-/m/t/2295039
I need to solve the following impulsive heat equation:
$$
\left\{\begin{array}{ll}
\partial_{t} \psi(x,t)-\partial_{xx} \psi(x,t)=0, & (x,t)\in (0,1) \times((0, 2) \backslash\{1\}) \\
\psi(0,t)= \psi(1,t)=0, & t \in (0, 2) \\
\psi(x, 0)= x (1-x), & x \in (0,1) \\
\psi(x, 1)=\psi\left(x, 1^{-}\right)+4, & x \in (0,1)
\end{array}\right.
$$
$1^{-}$ denotes the limit to the left!
This is the code I tried in Mathematica, but it's not giving the results
(* problem *)
homogen = If[x = 1, {f[x, t] == Limit[f[x, t], t -> 1, Direction -> "FromBelow"] + 4}, {D[f[x, t], {t, 1}] - D[f[x, t], {x, 2}] == 0}];
(*Initial conditions *)
ic = {f[x, 0] == x*(1 - x)};
(* Dirichlet boundary conditions*)
bc = {f[0, t] == 0, f[1, t] == 0};
(*solution*)
sol = DSolve[{homogen, ic, bc}, f[x, t], {x, 0, 1}, {t, 0, 2}]walid zouhair2021-06-21T16:54:17ZError in using NDSolve
https://community.wolfram.com/groups/-/m/t/2295374
n = 2
sol=NDSolve[{Table[{x_i'[t]==ν_i [t],y_i'[t]==ω_i [t],z_i'[t]==ρ_i [t]},{i,1,n}],
Table[{x_i [0]==x0[i],y_i [0]==y0[i],z_i [0]==z0[i]},{i,1,n}]},
Flatten[Table[{x_i,y_i,z_i},{i,1,n}]],{t,0,T},
MaxSteps→Infinity,Method→{"DiscontinuityProcessing"→False}]
This is the error
NDSolve::ndinnt: Initial condition x0[1.] is not a number or a rectangular array of numbers.
Please someone can help me on how to solve this errorPraviserk Chand2021-06-21T22:50:42ZSlider value change on output
https://community.wolfram.com/groups/-/m/t/2295484
Hello,
As, "d" changes the values of "s" and "t" must also change.
Similar to "c" also : as "c" changes the values of "s" and "t" must also change.
I have tried to do it, but as I move slider "d", the values of "s" and "t" remain same.
Thanks for help to resolve same
Clear[p, t]
sol2 = Table[
Take[Solve[
-10 p - 26 t == 5 &&
-5.3 p + 19 t == (var1 - 50),
{p, t}]], {var1, 10, 20}];
pSol1 = p /. sol2 // Flatten;
Column@Flatten[
var2 = Solve[
# == 0.42 - 2.4 s + 1.52 c &&
d == 15 - 1.4 s + 0.4258 c, {d, s}] & /@ pSol1, 1];
d[c_] = d /. var2 // Flatten;
s[c_] = s /. var2 // Flatten;
var10[c_] = Column@Flatten[Sqrt[c]/(s /. var2 // Flatten)];
Dynamic[Manipulate[
TextGrid[
{{"s", "t"},
{Column[s[c]],
var10[c]}},
Frame -> All,
Background -> {Automatic, 1 -> LightYellow},
ItemStyle -> {Automatic, 1 -> Bold},
Alignment -> {Center, Automatic, {{{2, -1}, {1, -1}} -> "."}}],
{c, 10, 30, 1, Appearance -> {"Labeled", "Open"}},
{d, 2, 12, Appearance -> {"Labeled", "Open"}}],
SaveDefinitions -> True]Ma oj2021-06-22T11:48:05ZParabola, tangents and circumcircle
https://community.wolfram.com/groups/-/m/t/2294788
The circumcircle of a triangle formed by three tangents of a parabola always passes through the focus of this parabola
----------
![enter image description here][1]
&[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=conic.gif&userId=23928
[2]: https://www.wolframcloud.com/obj/a93473a3-ee84-4f11-8f3c-ea0129f64339
[Original]: https://www.wolframcloud.com/obj/eff32e25-7a32-4ca2-87b5-2f7ed9fbdedbShenghui Yang2021-06-21T17:02:19ZMath one-liners from YouTube
https://community.wolfram.com/groups/-/m/t/2285635
There are some daily Youtubers who write out their **symbolic** solutions to a short straight-forward (non-quiz like type) math problem. Oftentimes *Mathematica* can solve those problems too. Maybe we could share here the ones which our software **cannot** solve easily?
Let me go first. In the "[Putnam Exam 2004 | B5][1]" video the youtuber claims that $L = \frac{2}{e}$
$$L=\lim_{x\to 1^-} \prod _{n=0}^{\infty } \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}=\ldots =\frac{2}{e}$$
Imho the *Wolfram L* code for this expression should be:
Limit[Product[((1 + x^(n + 1))/(1 + x^n))^(x^n), {n, 0, Infinity}], x -> 1, Direction -> "FromBelow"]
However, *Mathematica* computes forever without returning a result, i must abort the computation. Maybe my code line is wrong, can you get the desired result?
Obviously [this (hopefully collective) thread][2] (*feel free to memorize the short-URL, if you can't follow/subscribe to the thread for updates*) is nothing urgent. It is **entertaining** to let *Mathematica* have a crack at such symbolic math problems which get ten thousands of views within a week. Thanks for taking an interest!
[1]: https://www.youtube.com/watch?v=2nfW9BTMtMI
[2]: http://bit.ly/mathonelinersRaspi Rascal2021-06-08T08:09:38ZA filled glass or the mathematics of dioptric anamorphism
https://community.wolfram.com/groups/-/m/t/2294643
Back in 1638, Minim Father [Jean François Niceron][1] wrote a ground-breaking book for architects and artists alike: "[La Perspective Curieuse][2]". In this book he describes the mathematical methods to obtain drawings and paintings using perspective, catoptric anamorphism (with cylindrical and conical mirrors) and dioptric anamorphism by means of refraction in crystals.
![enter image description here][3]
The refractive telescope in the above illustration used a polyhedral lens consisting of 8 crystals and was meant to recompose 8 pictures into a single one by means of refraction. Although a [digital reconstruction][4] of the apparatus exists, the actual telescope was lost in a Venice flood. It would be a mathematical tour de force to study the mathematics of the refractive optics through the multi faced lens. To get a feel of what is refractive anamorphism (contrary to [catoptric anamorphism][5]), I made a short study about anamorphism by means of refraction in a cylinder filled with a sugar solution. The crystals used by Niceron had a refractive index of at least 1.55. The highest index I could obtain was 1.448 by using a 65% sucrose solution inside a cylindrical glass.
Let us first see what we can do with Mathematica to emulate refraction through a solution filled cylinder and create a dioptric anamorphic image.
![enter image description here][6]
The above optical ray diagrams show a 3D view (left) and a 2D horizontal {top right) and vertical (bottom right) cross section.
A ray leaving the point R in the x-y plane will meet the viewer's eye at V. This after being refracted at intersection points Q2 between air and cylinder liquid and Q1 between liquid and air. The resulting light ray from V to R is a combination of:
1. refraction in the horizontal plane through Q1 and Q2 (top right) and
2. refraction in the vertical plane through Q1 and Q2 (bottom right).
The result is that the viewer sees the point R in the x-y plane as a virtual point P in a view plane inside the cylinder. R can be defined as the dioptric (refractive) anamorphic of P or, inversely, P can be defined as the refracted image of R.
In both cases, [Snell's law of refraction][7] applies:
![enter image description here][8]
n1 and n2 are the two refractive indices: 1.448 for air -> sucrose solution and 1/1.448 for sucrose solution -> air. Theta 1 and Theta 2 are the incident and refracted angles. Since the relation between anamorphic points R and virtual points P is rather complex, we use an interpolation function ifR to compute the coordinates of R in the x-y plane from the coordinates of P in the view plane. The following code makes a table tblR of R and P coordinates using Snell's law and plain geometry. The interpolation function ifR is derived from the data in tblR:
tblR = With[{c = 1, yv = 10, zv = 10,
nr = 1.448},(*the cylinder radius is 1 and the viewpont is at {10,
0,10)*)
ParallelTable[
Quiet@
Module[{yc, ptVh, ptVv, ptPh, ptPv, ptQ1h, ti1, tr1, ti1v, tr1v,
ptQ2h, ti2, tr2, ti2v, tr2v, ptQ1v, ptQ2v, ptRv, ptR},
(*view plane location and width*)yc = c^2/yv;
wi = Round[c Sqrt[1 - c/yv^2], .001];
ptVh = {yv, 0}; ptPh = {yc, xp}; ptVv = {yv, zv};
ptPv = {yc, zp};
(*HORIZONTAL SECTION*)
(*incoming intersection point Q1*)
ptQ1h =
First@
NSolveValues[{Element[{x, y}, Line[{ptPh, ptVh}]] \[And]
x^2 + y^2 == c^2}, {x, y}];
(*angle of incidence*)ti1 = VectorAngle[ptVh - ptQ1h, ptQ1h];
(*angle of refraction*)tr1 = Sign[xp] ArcSin[Sin[ti1]/nr];
ptQ2h =
First@
SortBy[
NSolveValues[{Element[{x, y},
InfiniteLine[
ptQ1h, -AngleVector[-tr1 + ArcTan @@ ptQ1h]]] \[And]
x^2 + y^2 == c^2}, {x, y}], First];
(*outgoing intersection point Q2*)
(*angle of incidence*)
ti2 = tr1;
(*angle of refraction*)tr2 = ArcSin[Sin[ti2]*nr];
(*VERTICAL SECTION*)
(*incoming ray*)
\
(*intersection point*)
ptQ1v =
First@
NSolveValues[{Element[{x, y}, Line[{ptVv, ptPv}]] \[And]
x == ptQ1h[[1]]}, {x, y}];
(*angle of incidence*)ti1v = VectorAngle[ptVv - ptQ1v, {1, 0}];
(*angle of refraction*)tr1v = -ArcSin[Sin[ti1v]/nr];
(*outgoing ray*)
(*intersection point*)
ptQ2v =
First@
NSolveValues[{Element[{x, y},
HalfLine[ptQ1v, {-1, Tan[tr1v]}]] \[And]
x == ptQ2h[[1]]}, {x, y}];
(*angle of incidence*)ti2v = tr1v;
(*angle of refraction*)tr2v = ArcSin[Sin[ti2v]*nr];
ptRv =
First@
NSolveValues[{Element[{x, y},
HalfLine[ptQ2v, {-1, Tan[tr2v]}]] \[And] y == 0}, {x,
y}];
ptR =
Last@
NSolveValues[{Element[{x, y},
InfiniteLine[ptQ2h,
AngleVector[tr2 + ArcTan @@ ptQ2h]]] \[And]
x == ptRv[[1]]}, {x, y}];
{{xp, zp}, ptR}], {zp, .75, 7.5, .05}, {xp, -.99, .99, .02}]];
ifR = Interpolation[Flatten[tblR, 1]]
![enter image description here][9]
Below, we apply the function ifR to a set of concentric circles in the view plane and compute the corresponding refractive anamorphic mapping in the x-y plane. The hatched disk is the cross section of the cylinder and shows its relative size and position.
![enter image description here][10]
Using the same function ifR, we can make an animation in 3D of a point rotating around a circle in the view plane and see it traveling around the circle's refractive anamorphic image in the x-y plane:
![enter image description here][11]
We are now ready to experiment with some real images. The first trial is with the the "Bugs Bunny" [popular curve][12]. We map the function ifR to the point coordinates and get the refractive anamorphic image:
centerAndScale[g_] :=
Module[{lns, pts, xMin, xMax, yMin, yMax, centeredLns, scaledLines},
lns = Cases[g[[1]], _Line, \[Infinity]];
pts = Flatten[lns[[All, 1]], 1]; {xMin, xMax} =
MinMax[pts[[All, 1]]]; {yMin, yMax} = MinMax[pts[[All, 2]]];
centeredLns = Map[#1 - {(xMax + xMin)/2, +yMin} &, lns, {3}];
Map[1.35` #1/Abs[xMax - xMin] &, centeredLns, {3}]]
bugsbunnyPrimitives =
centerAndScale[
First[
ParametricPlot[
Entity["PopularCurve", "BugsBunnyCurve"]["ParametricEquations"][
t], {t, 0, 40 \[Pi]}]]] /. {x_?NumericQ, y_?NumericQ} :>
1.35 {x, 1.05 y + 1};
refractedPrimitives = MapAt[ifR @@ # &, prims, {All, 1, All}];
Graphics[{{HatchFilling[], Disk[]}, Circle[], AbsoluteThickness[1],
refractedPrimitives}, Axes -> False, Ticks -> None,
AxesOrigin -> {0, 0}]
![enter image description here][13]
Below demonstrates how the printout of the anamorphic image looks like the original after refraction through a glass filled with a sucrose solution:
![enter image description here][14]
The same can be done with any photographic image. We use the function [ImageSquareDivide][15] from the Wolfram Function Repository to divide the image into a set of colored polygons and map the function ifR to the coordinates of polygon's vertices:
mandrill = ImageResize[ExampleData[{"TestImage", "Mandrill"}], 50];
gr = ResourceFunction["ImageSquareDivide"][
mandrill] /. {x_, y_} -> .7 {x, y + 5};
Graphics[gr, Axes -> True, AxesOrigin -> {0, 0}];
grR = Quiet@MapAt[ifR @@ # &, gr, {All, -1, 1, All}];
Graphics[{{HatchFilling[], Disk[]}, Circle[], grR},
AxesOrigin -> {0, 0}]
![enter image description here][16]
Here again, we see the original image after refraction of the anamorphic printout:
![enter image description here][17]
Refraction does not result in the strong deformations obtained by [reflection in cylindrical mirrors][18]. The use of polished glass lenses or crystals as documented by Niceron would undoubtedly result in more spectacular results.
[1]: https://en.wikipedia.org/wiki/Jean_Fran%C3%A7ois_Niceron
[2]: https://bibliotheque-numerique.inha.fr/viewer/11579/?offset=#page=8&viewer=picture&o=bookmark&n=0&q=
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8149niceronbookimages.png&userId=68637
[4]: https://www.mas.bg.ac.rs/_media/istrazivanje/fme/vol45/2/3_aderosa_et_al.pdf
[5]: https://community.wolfram.com/groups/-/m/t/2100897
[6]: https://community.wolfram.com//c/portal/getImageAttachment?filename=3029geometrycombi.png&userId=68637
[7]: https://en.wikipedia.org/wiki/Snell%27s_law
[8]: https://community.wolfram.com//c/portal/getImageAttachment?filename=6198snellslaw.png&userId=68637
[9]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10772interpolationfunction.png&userId=68637
[10]: https://community.wolfram.com//c/portal/getImageAttachment?filename=9820circlescombi.png&userId=68637
[11]: https://community.wolfram.com//c/portal/getImageAttachment?filename=dioptriccircle500.gif&userId=68637
[12]: https://reference.wolfram.com/language/ref/entity/PopularCurve.html
[13]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8432bugsbunnyprintout.png&userId=68637
[14]: https://community.wolfram.com//c/portal/getImageAttachment?filename=bugsbunnycollage.png&userId=68637
[15]: https://resources.wolframcloud.com/FunctionRepository/resources/ImageSquareDivide/
[16]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10239mandrillprintout.png&userId=68637
[17]: https://community.wolfram.com//c/portal/getImageAttachment?filename=mandrillcollage-1.png&userId=68637
[18]: https://community.wolfram.com/groups/-/m/t/2100897
[19]: https://www.researchgate.net/figure/Digital-reconstruction-of-plate-number-23r-after-J-F-Niceron-La-Perspective-Curieuse_fig1_315893906Erik Mahieu2021-06-21T12:02:30ZPlotting NMinimize internal steps
https://community.wolfram.com/groups/-/m/t/2294312
The technique I recently posted for [plotting FindMinimum internal steps][1] also works for NMinimize. The code is attached. ![enter image description here][2]
[1]: https://community.wolfram.com/groups/-/m/t/2274764
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=nminplot.jpg&userId=29126Frank Kampas2021-06-20T18:17:50ZNumerically solving non-linear algebraic equations
https://community.wolfram.com/groups/-/m/t/2293772
Hello, I can not solve the system. Can any one help me with that? Thanks.
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/3f06c6cd-c07e-4236-b25e-c42d290c5587Vedat Erturk2021-06-19T18:05:00ZNDSolve stops working as number of equations/variables increases
https://community.wolfram.com/groups/-/m/t/2293689
When I run the attached code for the case n=11, it works: 22 interpolating functions are outputted. For the case n=12, NDSolve is unable to find any solutions. Can someone help figure out the cause of this error?
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/a3431a26-b58e-4b09-b231-7df98dc234afSanaaya Lakdawala2021-06-19T21:21:49ZSimplifying the conditional expression from a double integral?
https://community.wolfram.com/groups/-/m/t/2293844
I am trying to calculate an integral over y on a function f of two variables f[x,y]. However, the result comes out in the form of a conditional expression. Can anyone please tell how to simplify it? My code goes like this:
w[x_] := 10^-6*Sqrt[1 + (x/(Pi*10^-12))^2]
g[x_, y_] := 1/w[x - y]^2
h[x_] = Integrate[g[x, y], {y, 0, 100*10^-6}]Zainab Chowdhry2021-06-19T14:56:52ZBug in D[ ] in 12.3.0 but not in 12.1.0
https://community.wolfram.com/groups/-/m/t/2293856
Here is the terminal log from 12.1.0, where everything is correct:
$ math12
Mathematica 12.1.0 Kernel for Linux x86 (64-bit)
Copyright 1988-2020 Wolfram Research, Inc.
In[1]:= $Assumptions = Element[x,Reals]
Out[1]= x \[Element] Reals
In[2]:= a=Exp[x]
x
Out[2]= E
In[3]:= D[a*Conjugate[a],x] // ComplexExpand // Simplify
2 x
Out[3]= 2 E
and here is the log from 12.3.0 which shows the problem:
$ math
Mathematica 12.3.0 Kernel for Linux x86 (64-bit)
Copyright 1988-2021 Wolfram Research, Inc.
In[1]:= $Assumptions = Element[x,Reals]
Out[1]= x \[Element] Reals
In[2]:= a=Exp[x]
x
Out[2]= E
In[3]:= D[a*Conjugate[a],x] // ComplexExpand // Simplify
2 x
Out[3]= E (1 + I Im[Conjugate'[x]] + Re[Conjugate'[x]])
In[4]:= D[a*Conjugate[a],x] // ComplexExpand[#, TargetFunctions->Conjugate] & // Simplify
2 x
Out[4]= E (1 + Conjugate'[x])
Of course, if I did "ComplexExpand" and "Simplify" prior to differentiation, then it would work in 12.3.0 also, but this is clearly a bug, i.e. differentiation should be aware of the assumption that 'x' is real-valued. And in 12.1.0 it was aware, as the log above shows.Tigran Aivazian2021-06-19T15:29:34ZHow to write the Meijer g-function in Mathematica?
https://community.wolfram.com/groups/-/m/t/2293265
Hi everyone,
could you please write the code for the following Meijer g-function. I am not familiar with, sorry about it. Thank you very much in advance.
![enter image description here][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=MeijerG.png&userId=2178079Masood Ahmed2021-06-18T21:43:46ZSpeeding up numerical computation of Doble-Egg curves with Clothoids
https://community.wolfram.com/groups/-/m/t/2290697
I kindly ask you if you can give me some useful suggestions to speed up my program written in Wolfram language.
For this purpose, some premises are necessary.
The authors of the article in question use Matlab in their calculations.
I reproduced figures 2, 3, 4, 5, 6, 7,8 and 9 of their article with Mathematica 12.3, and in doing so I noticed some errors in the original text that led the authors to perform a Preprint Rev1. pdf June 8, 2021 (see final quote in the Acknowledgments).
I have found that in these graphs for the tracing of Clothoid the numerical calculation of the function d (Deltafi) = d12 is fundamental, the meaning of which is well described in relations (14) and (15) reported on pages 9 ÷ 10 of the aforementioned article. Figure 5 shows an example of the graphic representation of this function.
Now I correctly reproduce fig. 5 with N = 10,000 but with unacceptable processing time (8.76 h) while the authors take only a few seconds, always with N = 10,000 points. They sent me their Matlab code "Programa que pinta la function d (Delta phi)" see the attachment "Plot function d.rtf".
Can you give me some suggestions / optimization corrections of my Mathematica code in order to speed up this numerical calculation and break it down, if possible, to a few seconds of processing as they do with Matlab?
PS: I am attaching my notebook in fig. 5, with (N = 1,000, t = 178 s) and with (N = 10,000, t = 8.76h)
Sincerely and thanks,
Loris LoriLORIS LORI2021-06-15T07:27:37ZTry to beat these MRB constant records!
https://community.wolfram.com/groups/-/m/t/366628
The MRB constant: ALL ABOARD!
-----------------------------
POSTED BY: Marvin Ray Burns.
Autobiographies and his style: [http://marvinrayburns.com/aboutme.html][1]
On Pi Day, 2021, 2:40 pm EST,
I added a new MRB constant integral.
Map:
----
- First, we have formal identities and theory for **C**<sub>*MRB*</sub>.![CMRB][2]
- Then, at the end of this initial posting, we have world records of the maximum number of digits of **C**<sub>*MRB*</sub>
computations by date.
- Then we have some hints for anyone serious about breaking my record.
- Followed by speed records,
- a program Richard Crandall wrote to check his code for the computing record number of digits
- and a conversation about whether Mathematica uses the same algorithm for computing **C**<sub>*MRB*</sub> by a couple of different methods.
- Then, for a few replies, we compute **C**<sub>*MRB*</sub> from Crandall's eta derivative formulas and see records there.
- There are three replies about "NEW RECORD ATTEMPTS OF 4,000,000 DIGITS!" and the computation is now complete!!!!!.
- We see where I am on a 5,000,000 digit calculation. **(Just recently completed!!!!!!!!!!!!)**
- I describe the MRB supercomputer!!!!!! (faster than some computers with dual platinum Xeon processors) It was used for the 5,000,000 digit calculation.
- Then it comes time for the 6 million digit computation of **C**<sub>*MRB*</sub>. (put on hold, but taken up again at the end)
- We compute **C**<sub>*MRB*</sub> sum via an integral, which certifies the accuracy of **C**<sub>*MRB*</sub> calculations!!!!! (since the sum and integral are vastly different in every way they are computed)
- Then we take up the task of calculating 6,000,000 (6 million) digits of **C**<sub>*MRB*</sub> (for the <s>fourth, fifth time, sixth, seventh time!) (will finish Thu 25 Feb 2021 06:48:06).</s>
- We look for closed forms and find nontrivial, arbitrarily close approximations of **C**<sub>*MRB*</sub>.
- The latest updates on the MRB constant supercomputer 2 (with a GIF of it) and how I'm using it to break new records -- up to 7,500,000 digits -- are in a few replies.
- **On my 8th try I finally report that I successfully computed 6,000,000 digits of **C**<sub>*MRB*** !
- Finally, then I work at proving the accuracy of my computations.
----------
**C**<sub>*MRB*</sub> is defined below. See http://mathworld.wolfram.com/MRBConstant.html.
With over 20 years of research and ideas from users like you, I developed this informal catalog of formulas for the MRB constant.
**C**<sub>*MRB=*</sub>
![eta equals][3]
![enter image description here][4]
That is proven below by an internet scholar going by the moniker "Dark Malthorp:"
![Dark Marthorp's proof][5]
![eta sums][6] 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![(][7]considering the conditionally convergent sum,![enter image description here][8]![)][9] below that. Then the right half is a Taylor expansion of ?(s) around s = 0.
> ![n^(1/n)-1][10]
At
[https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal][11],
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][12]), 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][13], culminating with the Fubini theorem which is essentially the manipulation Helms is using.)"
> ![argrument 1][14] ![argrument 2][15]
----------
----------
----------
We see many more integrals for **C**<sub>*MRB*</sub>.
We can expand
![1/x][16]
into the following.
![xx = 25.656654035][17]
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][18]
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][19]
See
[this notebook from the wolfram cloud][20]
for justification.
----------
----------
Also, in terms of the Euler-Riemann zeta function,
**C**<sub>*MRB*</sub> =![enter image description here][21]
Furthermore, as ![enter image description here][22],
according to [user90369][23] at StackExchange, **C**<sub>*MRB*</sub> can be written as the sum of zeta derivatives similar to the eta derivatives discovered by Crandall.
![zeta hint ][24]Informations about ?<sup>(j)</sup>(k) please see e.g. [here][25], formulas (11)+(16)+(19).![credit][26]
In the light of the parts above, where
**C**<sub>*MRB*</sub>
= ![k^(1/k)-1][27]
= ![eta'(k)][28]
= ![sum from 0][29] ![enter image description here][30]
as well as ![double equals RHS][31]
an internet scholar going by the moniker "Dark Malthorp" wrote:
> ![eta *z^k][32]
----------
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][33]
Which is the same as
![enter image description here][34]
because changing the upper limit to 2N + 1 increases MI by 2i/?.
MKB constant calculations have been moved to their own discussion at [http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH][35] .
![Iimofg->1][36]
![Cauchy's Integral Theorem][37]
![Lim surface h gamma r=0][38]
![Lim surface h beta r=0][39]
![limit to 2n-1][40]
![limit to 2n-][41]
Plugging in equations [5] and [6] into equation [2] gives us:
![left][42]![right][43]
Now take the limit as N?? and apply equations [3] and [4] :
![QED][44]
He went on to note that
![enter image description here][45]
I wondered about the relationship between CMRB and its integrated analog and asked the following.
![enter image description here][46]
So far I came up with
&[Wolfram Notebook][47]
----------
----------
----------
As with any scientific paper, this post contains only reproducible results with methods. These records represent the advancement of consumer-level computers and clever programming over the past 20 years. I see others breaking these records, even after I die!
Here are some record computations. If you know of any others let me know.
- On or about Dec 31, 1998, I computed 1 digit of the (additive inverse of ) **C**<sub>*MRB*</sub> with my TI-92s, by adding 1-sqrt(2)+3^(1/3)-4^(1/4)+... as far as I could. That first digit, by the way, is just 0. Then by using the sum feature, in approximate mode, to compute $\sum _{n=1}^{1000 } (-1)^n \left(n^{1/n}\right),$
I computed 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.
- On Jan 11, 1999, I computed 4 decimals(.1878) of **C**<sub>*MRB*</sub> with the Inverse Symbolic Calculator, with 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).
- In Jan of 1999, I computed 5 correct decimals (rounded to .18786)of **C**<sub>*MRB*</sub> using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95.
- Shortly afterward I tried to compute 9 digits of **C**<sub>*MRB*</sub> 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.
- On Jan 23, 1999, I computed 500 digits of **C**<sub>*MRB*</sub> with an online tool called Sigma. Remarkably the sum in 4. was correct to 6 of the 9 decimal places! See
[http://marvinrayburns.com/Original_MRB_Post.html][48]
if you can read the printed and scanned copy there.
- In September of 1999, I computed the first 5,000 digits of **C**<sub>*MRB*</sub> 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.
- On June 10-11, 2003 over a period, of 10 hours, on a 450 MHz P3 with an available 512 MB RAM, I computed 6,995 accurate digits of **C**<sub>*MRB*</sub>.
- Using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, at 2:04 PM 3/25/2004, I finished computing 8000 digits of **C**<sub>*MRB*</sub>.
- On March 01, 2006, with a 3 GHz PD with 2 GB RAM available, I computed the first 11,000 digits of **C**<sub>*MRB*</sub>.
- On Nov 24, 2006, I computed 40, 000 digits of **C**<sub>*MRB*</sub> in 33 hours and 26 min via my program written in Mathematica 5.2. The computation was run on a 32-bit Windows 3 GHz PD desktop computer using 3.25 GB of Ram.
The program was something like this:
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]]
- Finishing on July 29, 2007, at 11:57 PM EST, I computed 60,000 digits of **C**<sub>*MRB*</sub>. Computed 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.
- Finishing on Aug 3, 2007, at 12:40 AM EST, I computed 65,000 digits of **C**<sub>*MRB*</sub>. Computed in only 50.50 hours on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 5.0 GB of RAM.
- Finishing on Aug 12, 2007, at 8:00 PM EST, I computed 100,000 digits of **C**<sub>*MRB*</sub>. They 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.
- Finishing on Sep 23, 2007, at 11:00 AM EST, I computed 150,000 digits of **C**<sub>*MRB*</sub>. They were 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.
- Finishing on March 16, 2008, at 3:00 PM EST, I computed 200,000 digits of **C**<sub>*MRB*</sub> using Mathematica 5.2. They were computed 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.
- Washed away by Hurricane Ike -- on September 13, 2008 sometime between 2:00 PM - 8:00 PM EST an almost complete computation of 300,000 digits of **C**<sub>*MRB*</sub> was destroyed. 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 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]]
- On September 18, 2008, computation of 225,000 digits of **C**<sub>*MRB*</sub> was started with a 2.66 GHz Core 2 Duo using 64 bit Windows XP. It was completed in 1072 hours. Memory usage is recorded in the attachment pt 225000.xls, near the bottom of this post.
- 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. Like the 300,000 digit computation, this one was almost complete when it failed. The Max memory used was 60.5 GB.
- On Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, I finished computing 250,000 digits of **C**<sub>*MRB*</sub>. 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 in the attached 250000.PD at bottom of this post.
- On Sun 28 Mar 2010 21:44:50 (UTC-0500) EST, I started a computation of 300000 digits of **C**<sub>*MRB*</sub> using an i7 with 8.0 GB of DDR3 RAM onboard, but it failed due to hardware problems.
- I computed 299,998 Digits of **C**<sub>*MRB*</sub>. 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) That 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.
- I computed exactly 300,000 digits to the right of the decimal point
of **C**<sub>*MRB*</sub> 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[]
- Sam Noble of Apple computed 1,000,000 digits of **C**<sub>*MRB*</sub> in 18 days 9 hours 11 minutes 34.253417 seconds.
- Finishing on Dec 11, 2012, Richard Crandall, an Apple scientist, computed 1,048,576 digits
in a lightning-fast 76.4 hours computation time (from the timing command). That's on a 2.93 GHz 8-core Nehalem.
- In Aug of 2018, I computed 1,004,993 digits of **C**<sub>*MRB*</sub> in 53.5 hours 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.
- Sept 21, 2018: I computed 1,004,993 digits of **C**<sub>*MRB*</sub>
in 50.37 hours of absolute time (35.4 hours computation time) with 18
(DDR3 and DDR4) processor cores! Search this post for "50.37 hours"
for documentation.**
- On May 11, 2019, I computed over 1,004,993 digits, using 28 kernels
on 18 DDR4 RAM (of up to 3200 MHz) supported cores overclocked up to
5.1 GHz in 45,5 hours of absolute time and only 32.5 hours of computation time! Search 'Documented in the attached ":3 fastest
computers together 3.nb." ' for the post that has the attached documenting notebook.
- On 10/19/20, using 3/4 of the MRB constant supercomputer 2, I finished an over 1,004,993 digits computation of **C**<sub>*MRB*</sub> in 44 hours of absolute time -- see [https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb][49] for documentation.
- I computed a little over 1,200,000 digits of **C**<sub>*MRB*</sub> in 11
days, 21 hours, 17 minutes, and 41 seconds (finishing on March 31, 2013). I used a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.
- On May 17, 2013, I finished a 2,000,000 or more digit computation of **C**<sub>*MRB*</sub>, 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.
- A previous world record computation of **C**<sub>*MRB*</sub> 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 **C**<sub>*MRB*</sub> 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.
- Finished on Wed 16 Jan 2019 19:55:20, I computed over 4 million digits of **C**<sub>*MRB*</sub>.
It took 4 years of continuous tries. This successful run took 65.13 days computation 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.
See attached [notebook][50].
M R Burns' algorithm:
x = SetPrecision[x, pr];
y = x^n; z = (n - y)/y;
t = 2 n - 1; t2 = t^2;
x =
x*(1 + SetPrecision[4.5, pr] (n - 1)/t2 + (n + 1) z/(2 n t) -
SetPrecision[13.5, pr] n (n - 1) 1/(3 n t2 + t^3 z));
(*N[Exp[Log[n]/n],pr]*)
Example:
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}
R Crandall's algorithm:
While[pc < pr, pc = Min[3 pc, pr];
x = SetPrecision[x, pc];
y = x^n - n;
x = x (1 - 2 y/((n + 1) y + 2 n n));];
(*N[Exp[Log[n]/ n],pr]*)
Example:
ClearSystemCache[]; n = 123456789;
(*n is the n in n^(1/n)*)
x = N[n^(1/n)];
(*x starts out as a machine 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[3 pc, pr];
x = SetPrecision[x, pc];
y = x^n - n;
x = x (1 - 2 y/((n + 1) y + 2 n n));];
(*N[Exp[Log[n]/n],pr]*)
N[n - x^n, 10]](* The error*)]; Print[
t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]]
Gives
{32.1406,0.*10^-9999984}
{104.516,0.*10^-9999984}
More information available upon request.
----------
- Finished on Fri 19 Jul 2019 18:49:02, I computed over 5 million digits of **C**<sub>*MRB*</sub>.
Methods described in the reply below that starts with "Attempts at a 5,000,000 digit calculation ."
For this 5 million calculation of MRB using the 3 node MRB supercomputer:
processor time was 40 days.
and actual time was 64 days.
That is faster than the 4 million digit computation using just one node.
- I finally computed 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 22:02:49 in 160 days.
The MRB constant supercomputer 2 said the following:
Finished on Tue 30 Mar 2021 22:02:49. Processor and actual 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!!!
----------
----------
----------
----------
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 Xeon processor.
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][51]
[1]: http://marvinrayburns.com/aboutme.html
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1ac.JPG&userId=366611
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10514Capture5.JPG&userId=366611
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[10]: https://community.wolfram.com//c/portal/getImageAttachment?filename=10297Capture.JPG&userId=366611
[11]: https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal
[12]: https://en.wikipedia.org/wiki/Riemann_series_theorem
[13]: https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch4.pdf
[14]: https://community.wolfram.com//c/portal/getImageAttachment?filename=7211Capture.JPG&userId=366611
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[20]: https://www.wolframcloud.com/obj/bmmmburns/Published/double%20zeroes%20of%20CMRB.nb
[21]: https://community.wolfram.com//c/portal/getImageAttachment?filename=11i.JPG&userId=366611
[22]: https://community.wolfram.com//c/portal/getImageAttachment?filename=6033d.JPG&userId=366611
[23]: https://math.stackexchange.com/users/332823/user90369
[24]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1558a.JPG&userId=366611
[25]: 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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[34]: https://community.wolfram.com//c/portal/getImageAttachment?filename=580910.JPG&userId=366611
[35]: http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH
[36]: https://community.wolfram.com//c/portal/getImageAttachment?filename=28491.JPG&userId=366611
[37]: https://community.wolfram.com//c/portal/getImageAttachment?filename=76812.JPG&userId=366611
[38]: https://community.wolfram.com//c/portal/getImageAttachment?filename=100173.JPG&userId=366611
[39]: https://community.wolfram.com//c/portal/getImageAttachment?filename=57664.JPG&userId=366611
[40]: https://community.wolfram.com//c/portal/getImageAttachment?filename=74665.JPG&userId=366611
[41]: https://community.wolfram.com//c/portal/getImageAttachment?filename=49236.JPG&userId=366611
[42]: https://community.wolfram.com//c/portal/getImageAttachment?filename=15127.JPG&userId=366611
[43]: https://community.wolfram.com//c/portal/getImageAttachment?filename=92858.JPG&userId=366611
[44]: https://community.wolfram.com//c/portal/getImageAttachment?filename=49309.JPG&userId=366611
[45]: https://community.wolfram.com//c/portal/getImageAttachment?filename=3.PNG&userId=366611
[46]: https://community.wolfram.com//c/portal/getImageAttachment?filename=1021422.JPG&userId=366611
[47]: https://www.wolframcloud.com/obj/b72d7291-a801-496f-977a-34dfd55350c2
[48]: http://marvinrayburns.com/Original_MRB_Post.html
[49]: https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb
[50]: https://community.wolfram.com/groups?p_auth=zWk1Qjoj&p_p_auth=r1gPncLu&p_p_id=19&p_p_lifecycle=1&p_p_state=exclusive&p_p_mode=view&p_p_col_id=column-1&p_p_col_count=6&_19_struts_action=/message_boards/get_message_attachment&_19_messageId=1593151&_19_attachment=4%20million%2011%202018.nb
[51]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ezgif.com-video-to-gif.gif&userId=366611Marvin Ray Burns2014-10-09T18:08:49ZAnimating mathematica surfaces: Wolfram and Blender integration
https://community.wolfram.com/groups/-/m/t/2293492
![enter image description here][1] ![enter image description here][2] ![enter image description here][3]
&[Wolfram Notebook][4]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=cheers_small.gif&userId=20103
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=OneToMany_small.gif&userId=20103
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=pavilion2_small.gif&userId=20103
[4]: https://www.wolframcloud.com/obj/222c3585-5193-448a-bac5-9e3181a2a603Guenther Gsaller2021-06-18T20:26:34ZFormatting a question of solving equation in Wolfram|Alpha?
https://community.wolfram.com/groups/-/m/t/2292994
I want Wolfram|Alpha to solve the diophantine equation 2^100=x^2+y^2+z^2+w^2 for integers. I tried writing "solve 2^100=x^2+y^2+z^2+w^2 for integers" or "Find the integer solutions for 2^100=x^2+y^2+z^2+w^2 ", but it doesn't understand my query.
What format should I write the question in?Devansh Kamra2021-06-18T13:30:15ZPutting the extrema of an NDSolve output into a list?
https://community.wolfram.com/groups/-/m/t/2292592
Hello,
I have a system of ODEs that I have solved using NDSolve.
a = 3/100
c = 1/2
b = d = 1
K = G = 5000
g = 1
solution3 =
Y[t] /. NDSolve[{X'[t] == X[t]*b*H[t]/G - X[t]*d*(X[t] + Y[t])/K,
Y'[t] ==
Y[t]*b*(1 - c*g)*H[t]/G + Y[t]*b*(1 - c*g)*g*(1 - H[t]/G) -
Y[t]*d*(X[t] + Y[t])/K,
H'[t] == Y[t]*b*(1 - c*g)*(1 - H[t]/G)*g - a*H[t], X[0] == K,
Y[0] == 1, H[0] == G}, {X[t], Y[t],
H[t]}, {t, 20,100020}]
I am hoping to find each local maximum for Y[t] and put them into a common list A. Likewise, I am hoping to find each local minimum for Y[t] and put them into a common list B. However, I have no idea how to do this. Any suggestions would be greatly appreciated!
AlexAlex L2021-06-18T00:35:04ZDust loops: artificial chaotic motion of particles
https://community.wolfram.com/groups/-/m/t/2277405
*MODERATORS' NOTE: This post is based on the [tumblr blog found here][1].*
![enter image description here][2]
&[Wolfram Notebook][3]
[1]: https://intothecontinuum.tumblr.com/post/119572912128/maihudson-this-animated-gif-is-20-frames-long
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=dust_loops.gif&userId=20103
[3]: https://www.wolframcloud.com/obj/575b660d-d837-4e75-b0c1-92bee5b0408dSumit Sijher2021-05-27T18:22:56ZSum of first n roots for a function?
https://community.wolfram.com/groups/-/m/t/2292061
I have a function: y = x * e^(-m)
and m is described by this equation: 1/m - m/2 - cot(m) = 0. It has infinite roots
How can I find a sum for the first n roots of equation in my function and plot this?
y = Sum(x * e^(-m)) from 0 to n.
For the first 3 roots: y = x * e^(-m1) + x * e^(-m2) + x * e^(-m3)Nazar Riabenko2021-06-16T22:48:06Z[GIF] Limits (Möbius transformations of the triangular tiling)
https://community.wolfram.com/groups/-/m/t/1179440
![Möbius transformations of the triangular tiling][1]
**Limits**
To make this, I'm applying a family of Möbius transformations $z \mapsto \frac{Z_\infty z - \gamma_1 \gamma_2}{z - z_\infty}$ to the tiling of the plane by equilateral triangles. Specifically, the fixed points of all transformations are $\gamma_1 = 2$ and $\gamma_2 = -2$, and the pole is at $z_\infty = \tan(\pi (t - 1/2))i$ as $t$ varies from 0 to 1. Consequently, the inverse pole (the point to which infinity is sent), is $Z_\infty = \gamma_1 + \gamma_2 - z_\infty = - \tan(\pi (t - 1/2))i$.
In other words, the point infinity gets mapped to just varies along the imaginary axis.
(Actually, there's a slight lie above: I reparametrize $t$ using the `smootheststep` function in order to get it to pause nicely at the beginning/end).
Of course, I can't really tell _Mathematica_ to transform infinitely many triangles, so the animation actually only shows $1001^2$ triangles, which is why there's a hole in the middle. In some ways the hole annoys me, but I also kind of like it: it's a good reminder of the limits of computation (as opposed to imagination).
Of course, transforming a million triangles doesn't really work in a `Manipulate[]` (in fact, the animation took many hours to render), so the code below only shows $21^2$ triangles, which is why it has a much larger hole:
smootheststep[t_] := -20 t^7 + 70 t^6 - 84 t^5 + 35 t^4;
DynamicModule[{?1 = 2, ?2 = -2, z?,
Z?, cols = RGBColor /@ {"#eaeaea", "#0D2C54"}},
Manipulate[
z? = Tan[? (smootheststep[t] - 1/2) ] I;
Z? = ?1 + ?2 - z?;
Graphics[
{cols[[1]],
Table[
If[Abs[3/4 y - Tan[? (smootheststep[t] - 1/2)]] < 1/2 && x == 0, Nothing,
Polygon[
Flatten[
Table[
ReIm[(Z? # - ?1 ?2)/(# - z?)
&[Sqrt[3]/2 (x + 1/4 (-1)^Mod[y, 2]) + 3/4 y I + 1/2 ((1 - s) Exp[I (?)] + s Exp[I (? + 2 ?/3)])]],
{?, ?/2., 2 ?, 2 ?/3}, {s, 0., 1, 1/10}],
1]
]
],
{x, -10, 10}, {y, -10, 10}]},
PlotRange -> {{-((15 Sqrt[3])/8), (15 Sqrt[3])/8}, {-3.25, 3.5}}, ImageSize -> 540, Background -> cols[[-1]]],
{t, .001, 1 - .001}]
]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=triangles5c.gif&userId=610054Clayton Shonkwiler2017-09-08T02:47:08ZThe exploitative segregation of plant roots
https://community.wolfram.com/groups/-/m/t/2235435
- **NEW YORK TIMES**: [How Selfish Are Plants? Let?s Do Some Root Analysis][1]
- **SCIENCE**: [DOI: 10.1126/science.aba9877][2]
![enter image description here][3]
&[Wolfram Notebook][4]
[1]: https://www.nytimes.com/2020/12/14/science/roots-competition-game-theory.html
[2]: https://science.sciencemag.org/content/370/6521/1197
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Roots.jpg&userId=20103
[4]: https://www.wolframcloud.com/obj/3849c6a8-fd41-4929-a677-61d898b8d435Ricardo Martinez Garcia2021-04-01T20:57:00ZImporting a matrix from a txt file?
https://community.wolfram.com/groups/-/m/t/2291836
Hello everyone.
I am trying to read a .mat file created using another language, it's dimensions are 50*10.
The code I am using is
Clear["Global`*"]
SetDirectory[NotebookDirectory[]];
dataText = Import["listOfUVValuesLite.txt", "Table"];
However, once I load this, dataText becomes a vector of dimenson 50, being each element a series of 10 elements. Hoever, mathematica doesn't detect each element of dataText as a row, but as a single element. If I try to write `dataText[[3, 1]]` I get the whole row and `dataText[[3, 2]]`, an error.
Can someone please explain to me how to get a 50X10 or 10X50 matrix form listOfUVValuesLite.txt?
I will attach it so test can be made.
Best regards.
Jaime.Jaime de la Mota2021-06-16T12:49:35Z