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RSS Feed for Wolfram Community showing any discussions from all groups sorted by activeApril 25th: The Perfect Date?
http://community.wolfram.com/groups/-/m/t/1327147
If you're in touch with iconic comedy films from the year 2000 onward (or just a knowledgeable Sandra Bullock fan), you've undoubtably seen the movie *Miss Congeniality*: the tale of a surly undercover FBI agent in the Miss USA Pageant who ends up making genuine human connections with people she didn't expect to, eventually saving their lives from a depraved former pageant queen.
Anyway, amongst all the amazing quotes scattered throughout the movie (who could forget the iconic "I'm gliding here!" scene?), there's a moment where Miss Rhode Island is asked about her idea of the perfect date:
![Miss Rhode Island, speaking the gospel truth][1]
As satirically funny as it was, she was right! Today was an amazing day, weather-wise, and we can see that with the Wolfram Language:
AirTemperatureData[Here, Today, Mean]
(*Quantity[53.1, "DegreesFahrenheit"]*)
I really did only need a light jacket!
While *I* had a nice day, was this same April 25th being experienced by everyone around the country? Surely the state of Rhode Island was, but what about the rest of the US? I decided to investigate...
##Getting the Data Set###
I started by combining the relevant `Entity` cities. To keep things simple (and to keep cell evaluation under a minute because I'm impatient), I combined my local location here at WRI with the 35 largest cities across the US:
Clear@cities
cities = Join[{Entity[
"City", {"Champaign", "Illinois", "UnitedStates"}]},
CityData[{Large, "UnitedStates"}][[1 ;; 35]]
]
![output of above code][2]
I then got their associated `GeoPosition`:
Clear@positions
positions = GeoPosition /@ cities;
And their temperatures:
Clear@temps
temps = AirTemperatureData[#, Today, Mean] & /@ positions
![enter image description here][3]
You can immediately see I was missing a value. To avoid any issues when plotting, I decided to drop both the value in question, and whatever city it was from:
Clear@tempsClean
tempsClean = Drop[temps, {6}];
Clear@citiesClean
citiesClean = Drop[cities, {6}];
Finally, I combined these into real coordinates for plotting:
Clear@coordinates
coordinates = Partition[
Riffle[citiesClean, tempsClean],
2]
![enter image description here][4]
##Visualizing April 25th Across the US##
To see if April 25th really is the perfect date, we need to set some assumptions beforehand. A priori we know that I indeed had a perfect day. According to `AirTemperatureData`, that temperature was around `53` degrees F. It was also sunny here, so for simplicity, I will assume that's the case for other cities across the country.
Based on this, I feel safe to assume that "not too hot and not too cold" has a floor value of `50` degrees F. Since I don't need a jacket at or above room temperature, I will also make the assumption that `70` degrees F is "too hot" and thus outside the range of "not too hot and not too cold". Since I am assuming sunshine, I would also argue that `65` degrees is also "too hot" for a light jacket, and will use `65` degrees F as my ceiling.
So, with all of that out of the way, let's plot:
ListPlot[coordinates[[All, 2]] -> coordinates[[All, 1]],
Frame -> {False, True},
PlotLabel -> "Mean Temperature of US Cities on April 25th",
FrameLabel -> {"Fahrenheit"}, PlotTheme -> "Detailed",
GridLines -> {{}, {50, 65}}]
![enter image description here][5]
Unsurprisingly, cities located in the south and southwest regions of the US are just above the range of "not too hot and not too cold", and cities in the northern midwest (Milwaukee, Chicago), or in elevated regions (Denver) are just below the specified range...they probably need a bit more than a light jacket to deal with their April 25th.
The only value that seems to be an outlier is Portland, Oregon—which lies just above the specified range. What about Portland today made it noticeably warmer than other cities, given it's location in the Pacific Northwest? I don't really know! And I don't think I would based on this simple analysis either.
##For Next Time##
Next time (i.e.; when I am less constrained by the 24 hr time limit of April 25th), I think I'd like to try creating a formula that not only makes use of `AirTemperatureData` but some other factors that would contribute to what you would actually *feel* when you step outside: humidity, wind speeds, etc. That'd result in a much more varied understanding of the cities: not only would I know which cities fell into the "just a light jacket" range, but *why*. Consequently, I would likely gain insight into why Portland (or any other outlier city for the topic at hand) was a bit warmer than expected.
But, all in all, this was a really fun topical exercise, and a great way to fuse my love of memes and pop culture with some quantitative analysis.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=rhode_island.jpg&userId=515558
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2018-04-25at7.52.10PM.png&userId=515558
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2018-04-25at7.55.22PM.png&userId=515558
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2018-04-25at8.26.44PM.png&userId=515558
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=plot.png&userId=515558Jesse Dohmann2018-04-26T01:21:47ZPrecision Error
http://community.wolfram.com/groups/-/m/t/1327213
While using Mathematica to solve a problem in one of my course, I found a problem which might be related to the precision of calculation. Please see the attached code.The function of L[T] becomes discontinuous at a specific range, and then go negative. However, according to the expression of L[T], it shouldn't go negative at any time, and it should be continuous at all range.I think it might be related to the calculation precision, but I don't know how to fix it.Kaitlyn Frankenfield2018-04-25T17:10:14ZAnti-Symmetric Relation
http://community.wolfram.com/groups/-/m/t/1326965
Is it fair to state that an anti-symmetric relation on set A is expressed by (the reflexive relation on set A) AND (the symmetric relation on set A) [i.e., the intersection of the reflexive relation on set A with the symmetric relation on set A]?Anthony Petosa2018-04-25T18:50:11ZComplete set of values for option "CelestialSystem"?
http://community.wolfram.com/groups/-/m/t/1327013
The documentation mentions "Horizon" (local alt/az) and "Equatorial" (inertial RA/DEC) as typical settings for option "CelestialSystem".
1. I'm assuming setting "Equatorial" specifies a mean equator and equinox at the J2000 epoch, since I see no place to give an epoch for a true-of-date computation.
2. Are there other legal values for this option, such as "ECEF" (earth-fixed: https://en.wikipedia.org/wiki/ECEF)? My guess is No, again, because most such other systems would require some form of time evolution, and there is no obvious place to parameterize the option with a target epoch.
Thank you.Vincent Virgilio2018-04-25T17:27:27ZFraud Analytics Analyst- Enova
http://community.wolfram.com/groups/-/m/t/1326934
Fraud Analytics Overview:
As an Analytics Analyst you will be one of Enova’s most valuable resources. You will develop, enhance and test the company’s rules, models and operational processes for use in determining the appropriate lending criteria and verification procedures, specifically related to preventing fraud. At times, you may be asked to conduct ad hoc analysis using statistical and financial tools to recommend fraud prevention strategies. You will demonstrate the ability to interpret and organize data, and communicate it effectively to cross functional teams to solve business problems, provide requirements and support implementation. You will work with senior management to develop key performance indicators (KPI's) to ensure that our fraud prevention schemes are performing optimally and communicating insights through both presentations and write-ups of results and recommendations.
Qualifications:
Bachelor's degree or equivalent experience, required
Advanced programming skills and the ability to write customized R (or other statistical tool) programs for meaningful data analysis
Working knowledge of SQL
Proficiency in spreadsheet applications and advanced use of statistical applications and databases
Basic knowledge of statistical, econometric or machine learning modeling
Ability to communicate concisely on problems and solutions
Ability to analyze complex business problems and solve them using "divide and conquer" strategies
Analytics Team Overview:
Enova's Analytics team consists of over 30 quantitative professionals dedicated to using the latest cutting-edge techniques to drive business value. We are a shared service for the entire company and operate in four workgroups:
Credit Risk Analytics –we focus on building cutting edge risk, pricing, and underwriting models to optimize our lending decisions.
Fraud Analytics – analysts on the fraud team use advanced data mining techniques to identify and fight online fraud.
Business Analytics – this team uses advanced modeling and simulation techniques to optimize the performance of our loan products and operations.
Marketing Analytics – the marketing analytics team is focused on applying statistical analysis and predictive modeling to help our marketing teams acquire and retain more valuable customers.
At Enova we have a company-wide culture that emphasizes data-driven analysis. That means you spend less time presenting and more time with the fun part, crunching data! We are language agnostic, but primarily use Python, R, SAS, SQL and Mathematica. That means YOU get to pick the tool that works best for you and the analysis at hand.
About Enova:
Enova is a leading provider of online financial services that leverages its advanced technology and analytics to provide access to credit for non-prime consumers and small businesses. Our roots are in Chicago, but we have served over 5 million customers through our six businesses in the U.S. and abroad. We pride ourselves on hiring smart and driven people who bring new and innovative ideas to the table. Our philosophy is, "Life’s short. Work some place awesome."
Many of us consider our people to be Enova’s best perk. But to sweeten the deal, we also have a pretty awesome list of conventional (and less conventional) perks and benefits including competitive salaries, health care benefits, a 401K matching plan, a revamped parental leave program (and brand new nursing rooms for our returning mothers!) summer hours, tuition reimbursement and a sabbatical program. And of course we also have the things you’d expect at a leading tech company in Chicago, such as the snacks, game room, onsite massages/barbers/nail technicians, monthly social events, and sporting sponsorships.
Our goal at Enova is to recruit, hire, develop and maintain a diverse workforce. It is our policy to provide equal employment opportunity for all persons and not discriminate in employment decisions by placing the most qualified person in each job, without regard to any other classification protected by federal, state, or local law.
Click [here][1] to apply!
[1]: https://www.builtinchicago.org/job/data/fraud-analytics-analyst/58512Holly Glenn2018-04-25T17:17:37ZUse NDSolve to solve a set of equations?
http://community.wolfram.com/groups/-/m/t/1323200
While using NDSolve to solve a set of ODEs employing the shooting method, I am not able to get an output.
I need some help regarding what I did wrong with the code
ODEs[\[CapitalOmega]1_, \[CapitalOmega]2_, \[CapitalOmega]3_, \
\[Lambda]_, \[Gamma]_, Mn_, Rd_, Lew_, Nb_, Nt_, Pr_] :=
{f''[\[Eta]] ==
g[\[Eta]]*(g[\[Eta]]^2 + \[Gamma]^2)/(g[\[Eta]]^2 + \[Lambda]*\
\[Gamma]^2),
g'[\[Eta]] == (1/3)*(f'[\[Eta]])^2 - (2/3)*f[\[Eta]]*f''[\[Eta]] +
Mn*f'[\[Eta]],
\[Theta]''[\[Eta]] == -(1/1 + Rd)*(2/3)*Pr*
f[\[Eta]]*\[CapitalTheta]'[\[Eta]] - (Nb/1 +
Rd)*\[CapitalTheta]'[\[Eta]]*\[Phi]'[\[Eta]] - (Nt/1 +
Rd)*(\[CapitalTheta]'[\[Eta]])^2,
\[Phi]''[\[Eta]] == -(2/3)*Lew*
f[\[Eta]]*\[Phi]'[\[Eta]] - (Nt/Nb)*\[CapitalTheta]''[\[Eta]],
f[0] == 0, f'[0] == 1, \[Theta][0] == 1, \[Phi][0] == 1,
g[0] == \[CapitalOmega]1, \[CapitalTheta]'[
0] == \[CapitalOmega]2, \[Phi]'[0] = \[CapitalOmega]3}
Soln[\[CapitalOmega]1_, \[CapitalOmega]2_, \[CapitalOmega]3_, \
\[Lambda]_, \[Gamma]_, Mn_, Rd_, Lew_, Nb_, Nt_, Pr_] :=
NDSolve[ODEs[\[CapitalOmega]1, \[CapitalOmega]2, \[CapitalOmega]3, \
\[Lambda], \[Gamma], Mn, Rd, Lew, Nb, Nt, Pr], {f[\[Eta]],
g[\[Eta]], \[CapitalTheta][\[Eta]], \[Phi][\[Eta]]}, {\[Eta], 0,
10}]
EndCondition[\[CapitalOmega]1_?NumericQ, \[CapitalOmega]2_?
NumericQ, \[CapitalOmega]3_?NumericQ, \[Lambda]_?
NumericQ, \[Gamma]_?NumericQ, Mn_?NumericQ, Rd_?NumericQ,
Lew_?NumericQ, Nb_?NumericQ, Nt_?NumericQ, Pr_?NumericQ] :=
{(First[g[\[Eta]] /.
Soln[\[CapitalOmega]1, \[CapitalOmega]2, \[CapitalOmega]3, \
\[Lambda], \[Gamma], Mn, Rd, Lew, Nb, Nt, Pr]] /. \[Eta] -> 10),
(First[\[CapitalTheta][\[Eta]] /.
Soln[\[CapitalOmega]1, \[CapitalOmega]2, \[CapitalOmega]3, \
\[Lambda], \[Gamma], Mn, Rd, Lew, Nb, Nt, Pr]] /. \[Eta] -> 10),
(First[\[Phi][\[Eta]] /.
Soln[\[CapitalOmega]1, \[CapitalOmega]2, \[CapitalOmega]3, \
\[Lambda], \[Gamma], Mn, Rd, Lew, Nb, Nt, Pr]] /. \[Eta] -> 10)}
EndCondition[0.5, 0.6, 0.8, 0.5, 1, 1, 1, 1, 1, 1, 5]
The equations are:
$g=f''\frac{g^{2}+\lambda\gamma^{2}}{g^{2}+\gamma^{2}}$
$g'=\frac{1}{3}f'^{2}-\frac{2}{3}ff''+Mnf'$
$(1+Rd)\theta''+\frac{2}{3}Prf\theta'+N_{b}\theta'\phi'+N_{t}\theta'^{2}=0$
$\phi''+\frac{2}{3}Lef\phi'+\frac{N_{t}}{N_{b}}\theta''=0$
And the boundary conditions are:
$f=0,\; f'=1,\; \theta=1,\; \phi=1\; at\; \eta=0$
$ f'\rightarrow0,\; \theta\rightarrow0,\; \phi\rightarrow0\; as\; \eta\rightarrow\infty.$Pragyan Sarma2018-04-19T19:37:34ZHow to make filling from vertical epilog line to the y axis in loglogplot?
http://community.wolfram.com/groups/-/m/t/1326846
Hi,
I would like to ask for help on how color everything to the right of a vertical line in a LogLogPlot? From the line to the right y axis.
I made the following:
LogLogPlot[x^2, {x, 0, 2 Pi}, PlotRange -> {{0.001, 5}, {10^-4, 10}},
Epilog -> Line[{{Log[1], Log[10^-4]}, {Log[1], Log[10]}}],
Frame -> True]
I need to exclude all the area from x=1 and to the right, and therefore shade it. I tried with filling but it did not produce anything, and I also tried to make a rectangle using graphics rectangle, but it can not show together with my LogLogPlot.
Thank you in advance,Sophie Møller2018-04-25T12:05:41ZPick factors from Taylor expansion
http://community.wolfram.com/groups/-/m/t/1326924
Hello! I have this code:
genus[Q_, n_Integer] :=
Module[{z, x},
SymmetricReduction[
SeriesCoefficient[
Product[ComposeSeries[Series[Q[z], {z, 0, n}],
Series[x[i] z, {z, 0, n}]], {i, 1, n}], n],
Table[x[i], {i, 1, n}], Table[Subscript[c, i], {i, 1, n}]][[
1]] // FactorTerms];
AgenusTotal[n_Integer] :=
Total[Table[
genus[(Sqrt[#]/2)/Sinh[Sqrt[#]/2] &, i] /. c -> p, {i, 0, n}]];
Which generates certain polynomials. For example for n=3 I get:
$-\frac{p_1}{24}+\frac{7 p_1^2-4 p_2}{5760}+\frac{-31 p_1^3+44 p_2 p_1-16 p_3}{967680}+1$
I need to take the square root of this expression (as a Taylor expansion) and group together terms of similar order (here by order I mean $p_1^3$, $p_1p_2$ and $p_3$ are, for example, of order 3 (each $p_i$ is a polynomial of degree i of another variable), the same way they are grouped in the expression itself. I have this code by now:
Series[Series[
Series[Sqrt[AgenusTotal[3]], {Subscript[p, 1], 0, 5}], {Subscript[p,
2], 0, 5}], {Subscript[p, 3], 0, 5}]
SeriesCoefficient[
SeriesCoefficient[
SeriesCoefficient[
Series[Series[
Series[Sqrt[AgenusTotal[3]], {Subscript[p, 1], 0,
5}], {Subscript[p, 2], 0, 5}], {Subscript[p, 3], 0, 5}], 2],
2], 1];
It works for individual examples, but I would like something more independent and ideally without putting a lot of Series[Series[
Series[ terms or SeriesCoefficient[
SeriesCoefficient[
SeriesCoefficient[ terms by hand (which would be tedious for n large). Also I would like to pick the right terms automatically, without specifying the SeriesCoefficient by hand (as in this way I might miss certain terms). Can someone help me? Thank you!Silviu Udrescu2018-04-25T16:12:58ZCalculate an alternating sum?
http://community.wolfram.com/groups/-/m/t/1326612
I am a beginner, please help me to write a code to obtain the following formula, Thank you very much.
![enter image description here][1]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=%E6%8D%95%E8%8E%B7.PNG&userId=1326375wg Lyu2018-04-25T08:14:02ZImport parameters into System Modeller from Excel?
http://community.wolfram.com/groups/-/m/t/1326247
Hi,
I am have several models of dynamical systems with a significant number of parameters for each. The best way to work with these would be to use an Excel sheet/tabs, and have SystemModeller use this data. With Dymola, I could use ExternData which allows one to specify sheet/row/column of Excel for parameters. Unfortunately, this does not work in SM as it seems to have a dependence on Dymola. There are some workaround for OpenModelica, but I have not had any luck in getting these to work in SM.
Could anyone suggest a way to get ExternData to work in SM, or perhaps some hacks to it?
The alternative is to use Records and Combi-Time and .rtf, but this is really taking a two decade step backwards. ExternData would be great if it could be made to work.
Thanks for the help.Chris Benson2018-04-24T21:03:16ZNest two "For" loops?
http://community.wolfram.com/groups/-/m/t/1326214
Hi All!
I have generated a "For" loop which results random number array called v1, I need to sum it with the same "For" loop (this time this "For" loop generates another number set). And I need to continue summing this for 2000 set of different random number set .
How can I build the code? does it sound like my "For" loop in another for loop?
\[Delta]t = 0.2; tmax = 5
imax = tmax/\[Delta]t
f = {1, 0}
T = 273
th = E^-\[Delta]t
nu = {}
For[v = {0, 0}; i = 1, i <= 7, i++,
If[RandomReal[] < th, v = v + f \[Delta]t,
v = RandomVariate[NormalDistribution[0, Sqrt[T]], {2}]];
v1 = AppendTo[nu, v]]
v1
Thank you very much for your kind help.
Thanks,
LakshithaLakshitha Dinusha Lathpadurage2018-04-24T07:20:25ZIntuitive understanding of the Dataset operation sequence
http://community.wolfram.com/groups/-/m/t/1326689
In <An elementary introduction to Wolfram Language> Chapter 45 example 42 ( In[42] )
planets[All, "Moons", NumberLinePlot[Values[#]] &, Log[#Mass/Quantity[1, "EarthMass"]] &]
This gives a number line plot of the masses of the moons for different planets. I couldn't figured out why it worked and both the documentation and the tutorials weren't very helpful.
Eventually I figured it out myself.
One reads this function forward and then backward, visualizing in ones head not tables but associations. Focus on selections when reading forward, operations when reading backward.
In this case:
Reading forward:
{a list of associations whose keys are planets} -> Select all planets -> {for each planet a list of associations whose keys are its fields("Mass","Radius","Moons") } -> Select "Moons" -> {for field "Moon" a list of associations whose keys are the name of moons} -> Select All (All is equivalant to #& in this case) -> {for each moon a list of associations whose keys are its fields("Mass","Radius")} -> Select "Mass"
Reading backward:
{"Mass" of a specific moon} -> Log["Mass"/EarthMass] -> {Log["Mass"/EarthMass] of a specific moon} -> Step Back -> {a list of associations whose keys are moon name and values are Log["Mass"/EarthMass]} -> Take its Values -> {A list of Log["Mass"/EarthMass]} -> Make a NumberLinePlot -> {A number line plot} -> Step Back -> {a list of associations whose keys are the planets and values are number plots} -> Select All
(reading backwards we only need to step backward twice, since the forward specification has "Moons" and "Mass" these two field specification, which decreased`enter code here` the dimension of the nested associations)
I hope this is helpful for those who just can't wrap their head around the ascending and descending sequence of operations for the dataset.LYU Liuke2018-04-25T14:34:01ZIsolate a fraction from a function?
http://community.wolfram.com/groups/-/m/t/1326756
Hi there
I got the following function:
![enter image description here][1]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Udklip.PNG&userId=1326742
I want to isolate with regards to "a/d", assuming that it is possible.
Any one here who knows how to do this?
Thanks in advanceJoergen Kjaer2018-04-25T11:15:07Z3D matrix transformation illustration
http://community.wolfram.com/groups/-/m/t/1326594
Hi I have a school project and I need to make a dynamic illustration in 3D linear space that shows the effect of a symmetrical degenerate matrix transformation. I have no idea how should I do it can anyone help me with it ?sinkapapa2018-04-25T09:01:12ZGet the right pointing arrow when I type AxesLabel ?
http://community.wolfram.com/groups/-/m/t/1326493
Sometimes, I have found that retyping a long code over again will solve this. I must have typed something wrong somewhere.
But in one case, I triple checked the coding and it seems to be the same as in the text. I'm just starting out with Mathematica, and I'm using the text -
Hands-On Start to Wolfram Mathematica & Programming with the Wolfram Language.
I'm on a mac with High Sierra ver 10.13.4
While I'm here, would you know where I can find errata on Wolfram books?
I also purchased An Elementary Introduction to the Wolfram Language.
Thank youRalph Lyle2018-04-25T02:13:03ZAvoid WordCloud issues in version 11.3 Home Edition?
http://community.wolfram.com/groups/-/m/t/1325903
I am working the exercises in the Elementary Introduction book, and cannot get the examples of the WordCloud function to work. I get the message"The weight specification used is invalid. At least one weight must be a positive real number. I have tried examples from other sources with the same result. Any ideas?CARL CHILDERS2018-04-23T17:17:44ZApply functions to a nested list in parallel?
http://community.wolfram.com/groups/-/m/t/1326196
Hi.
I have a function that generates a nested list containing coordinates, now i want to apply a function to those coordinates.
'CoordinatesGeneration = Function[ {m, n, a, b}, Table[Table[{a /2 (i + j), a (j - i), b /2 (k + h), b (k - h)}, {i, -(1/2) (m - 1), (m - 1)/
2}, {j, -(1/2) (m - 1), (m - 1)/2}], {h, -(1/2) (n - 1), (n - 1)/2}, {k, -(1/2) (n - 1), (n - 1)/2}]];'
What i tried is to flatten the list and then use partition to group the list each 4 elements and then use apply.
Apply[f ,Partition[Flatten[PositionGeneration[10, 10, a, b]], 4], {1}]]]
It worked, but now i want to do the evaluation in parallel. I try to use ParallelMap with Apply inside and ParallelCombine but it only work if the list was less than 4 groups in 5 groups it doesnt do what i wanted.
ParallelCombine[ f @@@ # &, Partition[Flatten[PositionGeneration[3, 3, a, b]], 4], Plus]
I want to parallelize it because when i have big numbers in counters m,n the evalution can take several seconds, so maybe using parallel functions can speed up a bit.
Thank you.Jaime moreno zuleta2018-04-24T16:15:23ZWeb Systems Administrator [Wolfram]
http://community.wolfram.com/groups/-/m/t/1326412
Wolfram, creator of Mathematica, Wolfram|Alpha and the Wolfram Language, has an exciting opportunity available for a Web Systems Administrator. Join the Web Systems Administration team and work in a fast-paced environment where you’ll never lack for new and challenging projects.
The ideal candidate will be good at working independently and in teams with current web standards and initiatives in a flexible, innovative environment, and must have strong Linux skills and experience with web architecture.
**Responsibilities:**
Maintaining production systems
Implementing changes for existing web infrastructure
Designing new tools/utilities to automate processes
Administrating web servers
Building and configuring web servers and databases
**Requirements:**
Strong Linux experience
Strong knowledge of various scripting languages
Strong understanding of networking concepts, technologies, and principles
Strong understanding of virtual machine environments and hypervisors
Understanding of web architecture and the appropriate technology stacks such as Apache and Tomcat
Understanding of Java technologies
Strong documentation skills
Ideal candidate will need to be able to work well in both a team environment and independently
Work well on stressful and fast-paced projects
Ideal candidate will love challenges and enjoy using their own skills and creativity to solve problems
Ideal candidate will be someone who takes pride in their work and achievements
**Preferred Requirements:**
3-5 years of experience
Have already designed/implemented a complex web architecture
Creative and loves to keep on the cutting edge of technology
AWS or hosted cloud services experience
Cloud computing architecture and HPC architecture
Understanding of the Wolfram Language
Bachelor’s degree in computer science, computer engineering, or related field
Location: Champaign, IL
[Apply][1] online at careers.wolfram.com.
Wolfram is an equal opportunity employer and values diversity at its company. Women, people of color, members of the LGBTQ community, individuals with disabilities and veterans are strongly encouraged to apply.
[1]: http://www.wolfram.com/company/careers/opportunities/#op-163230-web-systems-administratorHolly Glenn2018-04-24T16:22:04ZCreate a dynamic object for Graph in Cloud?
http://community.wolfram.com/groups/-/m/t/1325899
I need a dynamic graph in my cloud,but i saw the Graph object in cloud would be a Static image.
A dynamic Graph such as :
![enter image description here][1]
but in cloud(Development Platform ,web CDF) it would be Static graph automatic, such as:
![enter image description here][2]
Have any soluction? thank you.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=QQ%E5%9B%BE%E7%89%8720180424232643.png&userId=1164300
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=QQ%E5%9B%BE%E7%89%8720180424231615.png&userId=1164300PANG KamHing2018-04-24T15:43:01ZCreate a solid volume of revolution using RevolutionPlot3D?
http://community.wolfram.com/groups/-/m/t/1321989
I need to create a solid volume of revolution for f(x) = 2 + square root of x. It is bounded by x = 0, and y = 4. The shaded area between the y axis, y = 4, and the curve needs to be revolved around the y-axis. I don't know how to use the program to do it. If someone could save me the time and create it for me as an SVG or STL file so I can 3d print it as a demonstration for class, it would be much appreciated. Thank you.Grant Pierce2018-04-17T14:36:29ZPerform Integration involving vectors, in 3D?
http://community.wolfram.com/groups/-/m/t/1323495
Since I'm new to Mathematica I was looking for ways to learn vector integration and how to do them efficiently, so I came across this article where they did calculation for field in a point above a "slab" (magnetized bit) using Biot-Savart integration. To calculate following integral which is in vector form; I have problem defining parameters and variables. The integral is as following:
$dH(r)= D \frac{\vec dl \times (\vec r_i' - \vec r)}{|\vec r_i' - \vec r|^3}$
Once problem is defined, they end up with this integral for x direction (I assume one can difine similar for $y$ and $z$).
$H(r)=\sum_{i=1}^{4} C \int_{-l/2}^{l/2} \int_{0}^{1} \frac{\frac{\partial \vec r_i'}{\partial p} \times(\vec r_i'-\vec r)}{|\vec r_i' - \vec r|^3}dt dx$
In my approach I call $C$ and $D$ constants and two vectors:
$\vec r$ (field point) and $\vec r_i'$ (source point), with $\vec r = x \hat x + y \hat y + z \hat z$ and $\vec r_i'$vectors:
$\vec r_1' (x,t) = (x \hat x + \frac{W}{2} \hat y + \frac{T}{2}\hat z)+ (-tW \hat y) = [x \hat x + (\frac{w}{2} - tW) \hat y + \frac{T}{2}\hat z]$
$\vec r_2' (x,t) = (x \hat x - \frac{W}{2} \hat y + \frac{T}{2}\hat z)+ (-tT \hat z) = [x \hat x - \frac{w}{2}\hat y + (+ \frac{T}{2} - tT)\hat z]$
$\vec r_3' (x,t) = (x \hat x - \frac{W}{2} \hat y - \frac{b}{2}\hat z)+(tW \hat y) =[x \hat x + (- \frac{W}{2} + tW) \hat y + \frac{T}{2}\hat z]$
$\vec r_4' (x,t) = (x \hat x + \frac{W}{2} \hat y - \frac{T}{2}\hat z)+(tT \hat z) =[x \hat x + \frac{W}{2}\hat y + (+ \frac{T}{2} + tT)\hat z]$
where $W$ is width of a "slab" in $y$ direction and $T$ is thickness of the slab in $z$ direction. $t$ changes from 0 to 1 and x changes from $-l/2$ to $l/2$
In Mathematica, I don't know how to define variables within vectors and do the summation integration altogether. So my first attempt was to do individual form a individual piece of integral, for example for vector $\vec r_1'$ in $x$ coordinate in following way:
r := {x, y, z};
r1 := {x, W/2 - t W , T/2};
r1 - r = {0, W/2 - W t - y, T/2 - z}
D[r1, t] = {0, -W, 0}
Cross[{0, -W, 0}, {0, W/2 - W t - y, T/2 - z}] = {-((W T)/2) + W z, 0, 0}
Norm[r1 - r]^3 = (Abs[-1 + T/2]^2 + Abs[-1 + W/2 - W t]^2 + Abs[-1 + x]^2)^(3/2)
then when I put pieces to the integrate function:
Integrate[(-((W T)/2) + W z)/(Abs[W/2 - W t - y]^2 + Abs[T/2 - z]^2)^(3/2), {t, 0, 1}]
to calculate only for $t$ variable it takes forever and gives me a conditional result. And I tried using assumption
Integrate[(-((W T)/2) + W z)/((W/2 - W t - y)^2 + (T/2 - z)^2)^(3/2), {t, 0, 1}, Assumptions -> W > 0 && T>0]
or generally
Integrate[Cross[D[r1, t], r1] /Norm[(r1 - r)]^3, {t, 0, 1}, Assumptions -> Im[W] == 0 && Im[T] == 0]
still lengthy integral. And I know doing like this is wrong, because the definitions are not correct and plus integral doesn't even have $x$ component which is basically useless to get the result similar to their. I know there must be a better way, maybe do numerical integration of vectors, or something, so I tried to search similar problems however, I couldn't find anything useful. I would really appreciate if you show me the way you would define such problems to Mathematica and solve it efficiently.
If you want to look at the paper the DOI is provided below.
DOI:10.1007/s00542-011-1245-7
*(this problem was posted in stack exchange too)*Arm Mo2018-04-19T20:09:19ZManipulate histograms and normal distribution code line?
http://community.wolfram.com/groups/-/m/t/1326159
Hello!
My project group and I have been working on a editing a mathematica code to model diffusion. We've got most of it down however, ln21 of the code produces a histogram with a dynamic red line dependent on variable "t" to model the variation in normal distribution with time. The trouble that we ran into here is that only the red line is dynamic while the histogram bars stay static.
We would like to somehow get both the histogram and the line to move simultaneously depending on t.
I've attached the notebook below showing all the progress we've made so far. It would be great if someone could point out what we should do next to in order to make the manipulate command work for both the histogram and the line.Aditi Sharma2018-04-24T13:29:01ZUse CatenateLayer to connect nodes within a NetChain?
http://community.wolfram.com/groups/-/m/t/1326310
**EDIT: Looking closely at my network I have a similar problem of providing inputs to the CropLayer function. The output from the pooling layer (node 30) and node 87 (activation) should feed to the CropLayer. The question is how to feed outputs from one layer to other layers in the network**
I am trying to implement UNET (https://lmb.informatik.uni-freiburg.de/people/ronneber/u-net/) in Mathematica.
I have the following code so far for generating the net partially:
(* encoder *)
encoder = NetEncoder[{"Image", "ImageSize" -> {168, 168}, "ColorSpace" -> "Grayscale"}];
(* decoder *)
decoder = NetDecoder[{"Image", "ColorSpace" -> Automatic}];
(* convolution module *)
Options[convolutionModule] = {"batchNorm" -> True, "downpool" -> False,
"uppool" -> False, "activationType" -> Ramp, "convolution" -> True};
convolutionModule[net_, kernelsize_, padsize_, stride_: {1, 1}, OptionsPattern[]] :=
With[{upPool = OptionValue["uppool"], activationType = OptionValue["activationType"],
convolution = OptionValue["convolution"], batchNorm = OptionValue["batchNorm"],
downpool = OptionValue@"downpool"},
Block[{nnet = net},
If[upPool,
nnet = NetAppend[nnet, DeconvolutionLayer[1, {2, 2}, "PaddingSize" -> {0, 0},
"Stride" -> {2, 2}]];
nnet = NetAppend[nnet, BatchNormalizationLayer[]];
If[activationType === Ramp,
nnet = NetAppend[nnet, ElementwiseLayer[activationType]]
];
];
If[convolution,
nnet = NetAppend[nnet, ConvolutionLayer[1, kernelsize, "Stride" -> stride,
"PaddingSize" -> padsize]]
];
If[batchNorm,
nnet = NetAppend[nnet, BatchNormalizationLayer[]]
];
If[activationType === Ramp,
nnet = NetAppend[nnet, ElementwiseLayer[activationType]]
];
If[downpool,
nnet = NetAppend[nnet, PoolingLayer[{2, 2}, "Function" -> Max, "Stride" -> {2, 2}]]
];
nnet]
]
(* Crop Layer *)
CropLayer[netlayer_] := With[{p = NetExtract[netlayer, "Output"]},
PartLayer[{First@p, 1 ;; p[[2]], 1 ;; Last@p}] ];
(* partial UNET *)
UNET[] :=
Block[{nm, pool1, pool2, pool3, pool4, pool5, kernelsize = {3, 3},
padsize = {1, 1}, stride = {1, 1}},
nm = NetChain@
Join[{ConvolutionLayer[1, {3, 3},
"Input" -> encoder]}, {BatchNormalizationLayer[],
ElementwiseLayer[Ramp],
PoolingLayer[{2, 2}, "Function" -> Max, "Stride" -> {2, 2}]}];
pool1 = nm[[-1]];
nm = convolutionModule[nm, kernelsize, padsize, stride,"downpool" -> True];
pool2 = nm[[-1]];
nm = convolutionModule[nm, kernelsize, padsize, stride,"downpool" -> True];
pool3 = nm[[-1]];
nm = convolutionModule[nm, kernelsize, padsize, stride,"downpool" -> True];
pool4 = nm[[-1]];
nm = NetAppend[nm, DropoutLayer[]];
nm = convolutionModule[nm, kernelsize, padsize, stride, "downpool" -> True];
pool5 = nm[[-1]];
nm = convolutionModule[nm, kernelsize, padsize, stride, "uppool" -> True];
nm = convolutionModule[nm, kernelsize, padsize + 1, stride, "uppool" -> True];
nm = NetAppend[nm, CropLayer@pool3];
with `NetInformation` I can generate the net plot below:
NetInformation[(nm = UNET[]), "MXNetNodeGraphPlot"]
![enter image description here][1]
My problem: how do I catenate the output from the pooling layer i.e. node 30 with the output from node 91.
I tried using `NetGraph` with `CatenateLayer` but could not find a way to connect node 30 within the `NetChain` with the second input of `CatenateLayer`.
![enter image description here][2]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=2xfEG.png&userId=942204
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=6MvH1.png&userId=942204Ali Hashmi2018-04-24T09:00:06Z[CALL] for package submissions to a new package manager
http://community.wolfram.com/groups/-/m/t/1325863
I recently set up a package server in GitHub and need some people to submit packages to see how this can work. The server is here: https://paclets.github.io/PacletServer/
[ ![pm pick](https://i.stack.imgur.com/JIqln.png) ](https://paclets.github.io/PacletServer/)
It can handle one-line installation, gives a log of who changed which paclets, and can deal with submissions from anyone.
I'm looking for people to test this system out. Szabolcs kindly submitted a few of his and gave me feedback, but obviously he really knows what he's doing and so the system didn't bump into any real issues. That is likely not the case for most of us (not for me at least).
I'd like people to submit their packages via the instructions [on the GitHub repo](https://github.com/paclets/PacletServer#contributing-a-paclet) so I can see where things can be improved.b3m2a1 2018-04-24T06:36:25ZFind missing elements in a List?
http://community.wolfram.com/groups/-/m/t/1324489
Hi,
Is there some easy methods that could be used here for filling missing elements in List A based on elements of List B?
List A = { 190.5904`, 190.7829`, 190.9854`, 191.0015`, 191.0078`, 191.0233`, "", "", "", "", "", "", "", "", "", ""};
List B = { 9.4413`, 10.2572`, 7.8034`, 6.7098`, 6.3064`, 8.1171`, 6.618`, 6.618`, 6.378`, 7.944`, 7.8`, 6.294`, 7.356`, 6.936`, 6.252`, 7.344`};
Thanks in advance to everyone who respond.M.A. Ghorbani2018-04-20T17:57:15ZA 5-chromatic unit distance graph
http://community.wolfram.com/groups/-/m/t/1320004
On April 8, Aubrey de Grey posted the paper "[The chromatic number of the plane is at least 5](https://arxiv.org/abs/1804.02385)" on arxiv.org.
This tackles the long unsolved [Hadwiger-Nelson Problem](http://mathworld.wolfram.com/Hadwiger-NelsonProblem.html), where the plane is colored with the minimum colors so that no two points are a unit distance apart. There are many unit distance 4-chromatic graphs known, such as the [Golomb Graph](http://mathworld.wolfram.com/GolombGraph.html) (below), [Moser Spindle](http://mathworld.wolfram.com/MoserSpindle.html), and [Braced square](http://mathworld.wolfram.com/BracedPolygon.html). With these graphs you need at least 4 colors to avoid vertices of the same color being joined by an edge. With these, the *chromatic number of the plane* is at least 4.
![Golomb graph][1]
The hexagonal grid can color the plane with 7 colors so that no two points are distance 1 apart. For decades, the chromatic number has been known to be one of the values 4, 5, 6, or 7. A few months ago, Aubrey contacted me due to my Demonstration [Moser Spindles, Golomb Graphs and Root 33](http://demonstrations.wolfram.com/MoserSpindlesGolombGraphsAndRoot33/), where I built up a large dense graph filled with many overlapping 4-chromatic graphs.
I've also found a new graphs on my own. Here's one I call the Nested-7 graph, which might be the smallest unit-distance 4-chromatic graph without a 4-cycle.
points=Flatten[Table[RootReduce[#.RotationMatrix[{0,-1,1}[[n]]Pi/21]&/@(CirclePoints[7]/ EuclideanDistance[{0,1},CirclePoints[7][[4+n]]])],{n,1,3}],1];
unitlines=Select[Subsets[Range[21],{2}],Chop[N[1-Quiet@EuclideanDistance@@points[[#]]]]==0&];
Graphics[{EdgeForm[{Black,Thick}],Line[points[[#]]]&/@unitlines,White, Disk[#,.05]&/@points,Black, Table[Style[Text[n,points[[n]]],10],{n,1,Length[points]}]}, ImageSize-> 500]
![Nested 7 graph][2]
Aubrey had a Mathematica program for coloring a graph, and wondered if I could help him in his search. A few weeks ago, he found a unit-distance graph that seemed to need 5 colors. See the attached notebook.
Length /@ {v, e}
The graph has 1585 vertices and 7909 edges. It's a big one. Here are a few set up items, and then a graphic. We could definitely just draw the graph, but that many points and lines is hard to visually analyze.
hexgrids = {{1, 2, 3, 10, 45, 74, 75, 111, 140, 141, 148, 206, 207, 214, 249, 278, 279, 338, 339},
{1, 398, 399, 406, 441, 470, 471, 507, 536, 537, 544, 602, 603, 610, 645, 674, 675, 734, 735},
{3, 794, 795, 802, 837, 866, 867, 903, 932, 933, 940, 998, 999, 1006, 1041, 1070, 1071, 1130, 1131},
{794, 1190, 1191, 1198, 1233,1262, 1263, 1299, 1328, 1329, 1336, 1394, 1395, 1402, 1437, 1466,1467, 1526, 1527}};
importantedges = {{3, 399}, {3, 1191}, {75, 471}, {141, 537}, {207, 603}, {207, 999}, {279, 675}, {339, 735}, {867, 1263}, {933, 1329}, {999, 1395}, {1071, 1467}, {1131, 1527}};
importantvertices = {1, 3, 794};
colors = {Red, Green, Blue, Cyan, Gray};
Graphics[{Point[v],EdgeForm[Black],Thick,
Table[{colors[[n]],Line[v[[#]]]&/@Select[e,Length[Intersection[hexgrids[[n]],#]]==2&], Disk[v[[#]],.02]&/@hexgrids[[n]]},{n,1,4}],
colors[[5]],Line[v[[#]]]&/@importantedges,Black,Disk[v[[#]],.04]&/@importantvertices}]
![Aubrey graph][3]
Basically, there are four triangle grids, which I've colored red, blue, green, and cyan. There are 13 important edges, which I've colored in gray, and three important vertices where the triangle grids overlap. Of the important edges, six connect cyan and blue, six connect red and green, and one at the bottom connects blue and red.
If these 73 vertices are removed, the remaining 1512 vertices split into four 378-vertex graphs that connect solely to one of the four triangular grids. The purpose of all these other vertices is to restrict the number of colorings
If vertex 3 and edge {207, 999} are removed, the graph splits in half. Add vertex 3 back in and the graph looks like the following.
Graphics[Line[v[[#]]] & /@ Intersection[e, Subsets[Sort[Append[First[ConnectedComponents[
Graph[#[[1]] \[UndirectedEdge] #[[2]] & /@ Complement[Select[e, Not[MemberQ[#, 3]] &], {{207, 999}}]]]], 3]], {2}]]]
![Half of Aubrey graph][4]
In this half graph, the 12 outer vertices have the property that any 4-coloring forces the opposing vertices to have the same color.
If the graph had a 4-coloring, the two bottom vertices would have the same color as the vertex at the top. But these two vertices are connected, so they cannot have the same color. Therefore, a 4-coloring doesn't exist.
There is now a polymath project for finding the [smallest 5-chromatic unit-distance graph](https://polymathprojects.org/2018/04/10/polymath-proposal-finding-simpler-unit-distance-graphs-of-chromatic-number-5/).
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=GolombGraph_800.gif&userId=21530
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=nested7.jpg&userId=21530
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=AubreyGraph.jpg&userId=21530
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Aubreyhalf.jpg&userId=21530Ed Pegg2018-04-13T18:12:49ZAvoid problem with inverse Laplace transform?
http://community.wolfram.com/groups/-/m/t/1207437
Hi,
Can anybody explain to me please why MATHEMATICA 10 returns the correct answer to the command:
LaplaceTransform[Exp[t]*Erfc[Sqrt[t]],t,s]
(the answer is 1/(Sqrt[s] +s) )
but it refuses to answer the command:
InverseLaplaceTransform[1/(Sqrt[s] +s),s,t]
I would appreciate any suggestions how to obtain the inverse transform in a possibly alternative way.
LeslawLeslaw Bieniasz2017-10-24T08:46:05ZCreate a framed legend at the third plot with Show?
http://community.wolfram.com/groups/-/m/t/1325015
Anyone who can help me use legend with Show command. I have three plots, and I need to frame the legend in the final graph that gathers them.
I have two problems, first, for the last plot(fig) the vertical axes has a different name, when I used Show to gather the three plots, the output depends on the first element to name the axes and I can't use the third one because its range is smaller than the others. Secondly, I don't know how to frame the legend in the plot that gathers them.
Please see my code that is attachedGhady Almufleh2018-04-21T21:04:17ZImplement a GAN (generative adversarial network)?
http://community.wolfram.com/groups/-/m/t/1311452
Hello everyone,
I am trying to code a generative adversarial network using Mathematica's neural networks but I have a few questions still.
1. Since I am no computer scientist (yet) I intend on taking the structure of a GAN and implementing it into Mathematica. However I have trouble finding Fractionally-Strided Convolution layers. Is that the same as Deconvolution layers in Mathematica or should I use some other layers ?
2. Is there a specific encoder for text ? Since my input will be a piece of text describing the output image, a built in encoder might help greatly.
3. And finally, since GANs are composed of a generator and a discriminator battling each other, how do I, in Mathematica, implement this back and forth battle between the two nets ?
I really am a beginner in using neural networks so any help is greatly appreciated. Because of that, I am fully aware that entire swathes of computer science might have to been explained to me, however a quick direction towards the important topics can help me direct my research. Thank you for your answers !Luca Montanelli2018-03-29T18:38:23ZTry to beat these MRB constant records!
http://community.wolfram.com/groups/-/m/t/366628
POSTED BY: Marvin Ray Burns .
**MKB constant calculations have been moved to their own discussion at**
[Calculating the digits of the MKB constant][1] .
I think this important point got buried near the end.
When it comes to mine and a few more educated people's passion to calculate many digits and the dislike possessed by a few more educated people; it is all a matter telling us that the human mind is multi faceted in giving passion, to person a, for one task and to person b for another task!
The MRB constant is defined below. See http://mathworld.wolfram.com/MRBConstant.html
> ![enter image description here][2]
Here are some record computations. If you know of any others let me know..
1. On or about Dec 31, 1998 I computed 1 digit of the (additive inverse of the) MRB constant with my TI-92's, 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.
2. On Jan 11, 1999 I computed 3 digits of the MRB constant with the Inverse Symbolic Calculator.
3. In Jan of 1999 I computed 4 correct digits of the MRB constant using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95.
4. Shortly afterwards I computed 9 correct digits of the MRB constant using Mathcad 7 professional on the Pentium II mentioned below.
5. On Jan 23, 1999 I computed 500 digits of the MRB constant with the online tool called Sigma.
6. In September of 1999, I computed the first 5,000 digits of the MRB Constant 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.
7. On June 10-11, 2003 over a period, of 10 hours, on a 450mh P3 with an available 512mb RAM: I computed 6,995 accurate digits of the MRB constant.
8. Using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, on 2:04 PM 3/25/2004, I finished computing 8000 digits of the MRB constant.
9. On March 01, 2006 with a 3GH PD with 2GBRAM available, I computed the first 11,000 digits of the MRB Constant.
10. On Nov 24, 2006 I computed 40, 000 digits of the MRB Constant in 33hours and 26min via my own program in written in Mathematica 5.2. The computation was run on a 32-bit Windows 3GH PD desktop computer using 3.25 GB of Ram.
11. Finishing on July 29, 2007 at 11:57 PM EST, I computed 60,000 digits of MRB Constant. Computed in 50.51 hours on a 2.6 GH AMD Athlon with 64 bit Windows XP. Max memory used was 4.0 GB of RAM.
12. Finishing on Aug 3 , 2007 at 12:40 AM EST, I computed 65,000 digits of MRB Constant. Computed in only 50.50 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 5.0 GB of RAM.
13. Finishing on Aug 12, 2007 at 8:00 PM EST, I computed 100,000 digits of MRB Constant. They were computed in 170 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 11.3 GB of RAM. Median (typical) daily record of memory used was 8.5 GB of RAM.
14. Finishing on Sep 23, 2007 at 11:00 AM EST, I computed 150,000 digits of MRB Constant. They were computed in 330 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 22 GB of RAM. Median (typical) daily record of memory used was 17 GB of RAM.
15. Finishing on March 16, 2008 at 3:00 PM EST, I computed 200,000 digits of MRB Constant using Mathematica 5.2. They were computed in 845 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 47 GB of RAM. Median (typical) daily record of memory used was 28 GB of RAM.
16. Washed away by Hurricane Ike -- on September 13, 2008 sometime between 2:00PM - 8:00PM EST an almost complete computation of 300,000 digits of the MRB Constant was destroyed. Computed for a long 4015. Hours (23.899 weeks or 1.4454*10^7 seconds) on a 2.66GH Core2Duo 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]]
17. On September 18, 2008 a computation of 225,000 digits of MRB Constant was started with a 2.66GH Core2Duo 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. .
18. 250,000 digits was attempted but failed to be completed to a serious internal error which 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.
19. On Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, I finished computing 250,000 digits of the MRB constant. with a multiple step Mathematica command running on a dedicated 64bit XP using 4Gb DDR2 Ram on board 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.
20. On Sun 28 Mar 2010 21:44:50 (UTC-0500) EST, I started a computation of 300000 digits of the MRB constant using an i7 with 8.0 GB of DDR3 Ram on board.; But it failed due to hardware problems.
21. I computed 299,998 Digits of the MRB constant. 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 GH 2.80 GH
with 8GB physical DDR3 RAM. Windows 7 reserved an additional 48.929
GB virtual Ram.
22. I computed exactly 300,000 digits to the right of the decimal point
of the MRB constant 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 16GB 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 at 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[]
23. 314159 digits of the constant took 3 tries do to hardware failure. Finishing on September 18, 2012 I computed 314159 digits, taking 59 GB of RAM. The digits are came from the Mathematica 8.0.4 code
DateString[]
NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]},
WorkingPrecision -> 314169, Method -> "AlternatingSigns"] // Timing
DateString[]
Where I have 10 digits to round off. (The command NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]},
WorkingPrecision -> big number, Method -> "AlternatingSigns"] tends to give about 3 digits of error to the right.)
**The following records are due to the work of Richard Crandall found [here][3]. **
24. Sam Noble of Apple computed 1,000,000 digits of the MRB constant in 18 days 9 hours 11 minutes 34.253417 seconds
25. Finishing on Dec 11, 2012 Ricard Crandall, an Apple scientist, computed 1,048,576 digits
in a lighting fast 76.4 hours. That's on a 2.93 Ghz 8-core Nehalem
26. I computed a little over 1,200,000 digits of the MRB constant in 11
days, 21 hours, 17 minutes, and 41 seconds,( finishing on on March 31 2013). I used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.
27. On May 17, 2013 I finished a 2,000,000 or more digit computation of the MRB constant, using only around 10GB of RAM. It took 37 days 5 hours 6 minutes 47.1870579 seconds. I used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.
28. Finally, I would like to announce a new unofficial world record computation of the MRB constant that was finished on Sun 21 Sep 2014 18:35:06. It took 1 month 27 days 2 hours 45 minutes 15 seconds. I computed 3,014,991 digits of the MRB constant with Mathematica 10.0. I Used my new version of Richard Crandall's code, below, optimized for my platform and large computations. I also used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 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 actually accurate to 2,009,993 digits. (They were used as MRBtest2M.)
(**Fastest (at MRB's end) as of 25 Jul 2014*.*)
DateString[]
prec = 3000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] :=
Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 12,
tsize = 2^7, chunksize, start = 1, ll, ctab,
pr = Floor[1.005 pre]}, chunksize = cores*tsize;
n = Floor[1.32 pr];
end = Ceiling[n/chunksize];
Print["Iterations required: ", n];
Print["end ", end];
Print[end*chunksize]; d = ChebyshevT[n, 3];
{b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
iprec = Ceiling[pr/27];
Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
x = N[E^(Log[ll]/(ll)), iprec];
pc = iprec;
While[pc < pr, pc = Min[3 pc, pr];
x = SetPrecision[x, pc];
y = x^ll - ll;
x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(**N[Exp[Log[ll]/ll], pr]**)x, {l, 0, tsize - 1}], {j, 0, cores - 1},
Method -> "EvaluationsPerKernel" -> 4]];
ctab = ParallelTable[Table[c = b - c;
ll = start + l - 2;
b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 2];
s += ctab.(xvals - 1);
start += chunksize;
Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0,
end - 1}];
N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; DateString[]
Print[MRBtest2]
MRBtest2 - MRBtest2M
t2 From the computation was {1.961004112059*10^6, Null}.
Here are a couple of graphs of my record computations in max digits/ year:
![enter image description here][4]![enter image description here][5]
[1]: http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=68115.JPG&userId=366611
[3]: http://www.marvinrayburns.com/UniversalTOC25.pdf
[4]: /c/portal/getImageAttachment?filename=7559mrbrecord1.JPG&userId=366611
[5]: /c/portal/getImageAttachment?filename=mrbrecord3.JPG&userId=366611Marvin Ray Burns2014-10-09T18:08:49ZAvoid trouble with NDSolve and dependent variables?
http://community.wolfram.com/groups/-/m/t/1325048
My goal here is to MPF and Wee1 using NDsolve (and eventually with different values of DDS), but for some reason I keep getting that my dependent variables are more numerous than my equations. I'm getting very frustrated because I thought I was making sure that it would be the other way around as well as trying many other methods, but to no avail.
I know this is way more code than is probably necessary... but I couldn't figure it out! Is there anyone who can help?
Here's my code:
(These are just constants)
k1 = 1.5; k2 = 0.001; k3 = 10.0; k4 = 0.02; k5 = 6.0; k6 = 0.04; k7 = \
0.005; k8 = 0.00000001; k9 = 1; k10 = 1.0; k11 = 1.0; k12 = 0.0005; \
k13 = 1.0; k14 = 0.01; k15 = 1.0; k16 = 0.01; k17 = 1; k18 = 1; k19 \
= 0.1; k20 = 0.01; k21 = 0.1; k22 = 1.0; k23 = 0.01; k24 = 0.01; k25 \
= 1.0; k26 = 0.01; k27 = 1.0; k28 = 100.0; k29 = 1.0; k30 = 0.01; k31 \
= 0.01; k32 = 0.0001; k33 = 1.0; k34 = 0.1; k35 = 1.0; k36 = 1.0; \
kp = 0.0001; kM = 0.00094; kW = 0.00054; kj = 0.04; jW = 1.8; kd = \
0.01; kdeg = 0.772; kdamp = 0.02; kA = 0.2; ki = 0.01; vm = 0.00005; \
kWee1 = 0.0002; kin = 0.0013; k1d = 0.026; k2d = 0.0013; kca = 0.004; \
kcm = 0.005; kwip11 = 0.00054; kwip12 = 0.04; jwip1 = 1.8; kwip13 = \
0.001; k1p21 = 0.0001; k2p21 = 0.135; jp21 = 2; kdin1 = 0.000054; \
kdin2 = 0.0027; jdin1 = 0.4; kdin3 = 0.135; jdin2 = 0.5; kdin4 = \
0.00135; kAIP1 = 0.0011; kAIP2 = 0.027; jAIP1 = 0.3; kAIP3 = 0.01;
Manipulate[
sol = NDSolve[{
MPF'[t] ==
k17 * (Cdc25a[t] + Cdc25ps216a[t]) * preMPF[t] +
k18 * P21MPF[t] - k14 * MPF[t] * Wee1[t] -
k19 * MPF[t] * p21[t] - k20 * MPF[t]^2,
Cdc25a'[t] ==
k15 * MPF[t] * Cdc25[t] + k30 * Cdc25ps216a[t] - ki * Cdc25[t] -
k30 * Chk1p[t] * Cdc25a[t] - k32 * Cdc25a[t],
Chk1p'[t] == k9 * Chk1[t] * 1 - k10 * Chk1p[t],
(* EQ for ATR'[t] is not given*)
Chk1'[t] == k9 * Chk1p[t] - k10 * Chk1[t],
Cdc25'[t] ==
ki * Cdc25[t] + vm - k15 * MPF[t] * Cdc25[t] -
k23 * Chk1p[t] * Cdc25[t],
Cdc25ps216a'[t] ==
k31 * Chk1p[t] * Cdc25a[t] + k25 * MPF[t] * Cdc25ps216[t] -
k30 * Cdc25ps216a[t] - k24 * Cdc25ps216a[t],
preMPF'[t] == (k12)/(1 + k13 * P53[t]) + k14 * k15 * k16 -
k17*(Cdc25a[t] + Cdc25ps216a[t]) * preMPF[t],
P21MPF'[t] == k19*p21[t] - k18 * P21MPF[t],
p21'[t] ==
k1p21 + k2p21*(p53a[t]^3)/(jp21^3 + p53a[t]^3) +
k18 * P21MPF[t] + k16 - k22 * p21[t] - k19 * MPF[t] * p21[t],
p53a'[t] == k1d * p53[t] - kin * p53a[t] - k2d * p53a[t],
p53'[t] ==
ks + k1 * (DDS *
Exp[-k8 *
t]) - (((Dego - kdeg * (DDS * Exp[-k8 * t]) -
DDS * Exp[-kdamp * DDS * t])) * p53[t] * Mdm2 [t])/(ka +
p53[t]) + kM * p53a[t] - k1d * p53[t] - k2 * p53[t],
Cdc25ps216'[t] ==
k23 * Chk1p[t] * Cdc25[t] - k25 * MPF[t] * Cdc25ps216[t] +
k24 * Cdc25ps216a[t] - k28 * F1433[t] * Cdc25ps216[t],
(*14-3- = F1433*)
F1433'[t] ==
k26 * p53[t] + k27 - k28 * Cdc25ps216[t] * F1433[t] -
k29 * F1433[t],
(**)
Wee1'[t] == kWee1 + k33 * Wee1p[t] - k34 * MPF[t] * Wee1[t],
(**)
MPF[0] == MPFo, Cdc25a[0] == Cdc25ao, Chk1p[0] == Chk1po,
Chk1[0] == Chk1o, Cdc25[0] == Cdc25o,
Cdc25ps216a[0] == Cdc25ps216ao, preMPF[0] == preMPFo,
P21MPF[0] == P21MPFo, p21[0] == p21o, p53a[0] == p53ao,
p53[0] == p53o, Cdc25ps216[0] == Cdc25ps216o, F1433[0] == F1433o,
Wee1[0] == Wee1o}, {MPF, Cdc25a, Chk1p, Chk1, Cdc25,
Cdc25ps216a, preMPF, P21MPF, p21, Cdc25ps216, F1433, Wee1}, {t,
8}];
Plot[Evaluate[{MPF[t], Wee1[t]} /. sol], {t, 0, TFinal},
PlotRange -> All, PlotLegends -> {"[MPF], [Wee1]"},
PlotStyle -> {Green, Cyan}],
{{TFinal, 5, "Time"}, 0.2, 8},
{{DDS, 0, "DDS"}, 0, 0.008},
{{MPFo, 1, "[MPF] Initial"}, 0, 10},
{{Cdc25ao, 1, "[Cdc25a] Initial"}, 0, 10},
{{Chk1po, 1, "[Chk1p] Initial"}, 0, 10},
{{Chk1o, 1, "[Chk1] Initial"}, 0, 10},
{{Cdc25o, 1, "[Cdc25] Initial"}, 0, 10},
{{Cdc25ps216o, 1, "[Cdc25ps21a] Initial"}, 0, 10},
{{Cdc25ps216ao, 1, "[Cdc25ps216a] Initial"}, 0, 10},
{{preMPFo, 1, "[preMPF] Initial"}, 0, 10},
{{P21MPFo, 1, "[P21MPF] Initial"}, 0, 10},
{{p53ao, 1, "[P53a] Initial"}, 0, 10},
{{p53o, 1, "[P53] Initial"}, 0, 10},
{{F1433o, 1, "[F1433] Initial"}, 0, 10},
{{Wee1o, 1, "[Wee1] Initial"}, 0, 8},
{{ks, 1, "ks"}, 0, 8},
{{K, 1, "K"}, 0, 8},
{{Dego, 1, "Dego"}, 0, 8}
]
Out[33]= Manipulate[
sol = NDSolve[{Derivative[1][MPF][t] == k17*(Cdc25a[t] + \
Cdc25ps216a[t])*
preMPF[t] + k18*P21MPF[t] - k14*MPF[t]*Wee1[t] - \
k19*MPF[t]*p21[t] -
k20*MPF[t]^2, Derivative[1][Cdc25a][t] == k15*MPF[t]*Cdc25[t] \
+
k30*Cdc25ps216a[t] - ki*Cdc25[t] - k30*Chk1p[t]*Cdc25a[t] -
k32*Cdc25a[t], Derivative[1][Chk1p][t] == k9*Chk1[t]*1 -
k10*Chk1p[t], Derivative[1][Chk1][t] == k9*Chk1p[t] - \
k10*Chk1[t],
Derivative[1][Cdc25][t] == ki*Cdc25[t] + vm - \
k15*MPF[t]*Cdc25[t] -
k23*Chk1p[t]*Cdc25[t], Derivative[1][Cdc25ps216a][t] ==
k31*Chk1p[t]*Cdc25a[t] + k25*MPF[t]*Cdc25ps216[t] -
k30*Cdc25ps216a[t] - k24*Cdc25ps216a[t], \
Derivative[1][preMPF][t] ==
k12/(1 + k13*P53[t]) + k14*k15*k16 - k17*(Cdc25a[t] + \
Cdc25ps216a[t])*
preMPF[t], Derivative[1][P21MPF][t] == k19*p21[t] - \
k18*P21MPF[t],
Derivative[1][p21][t] == k1p21 + k2p21*(p53a[t]^3/
(jp21^3 + p53a[t]^3)) + k18*P21MPF[t] + k16 - k22*p21[t] -
k19*MPF[t]*p21[t], Derivative[1][p53a][t] ==
k1d*p53[t] - kin*p53a[t] - k2d*p53a[t], Derivative[1][p53][t] \
==
ks + k1*(DDS*Exp[(-k8)*t]) - ((Dego - kdeg*(DDS*Exp[(-k8)*t]) -
DDS*Exp[(-kdamp)*DDS*t])*p53[t]*Mdm2[t])/(ka + p53[t]) +
kM*p53a[t] - k1d*p53[t] - k2*p53[t], \
Derivative[1][Cdc25ps216][t] ==
k23*Chk1p[t]*Cdc25[t] - k25*MPF[t]*Cdc25ps216[t] +
k24*Cdc25ps216a[t] - k28*F1433[t]*Cdc25ps216[t],
Derivative[1][F1433][t] == k26*p53[t] + k27 - k28*Cdc25ps216[t]*
F1433[t] - k29*F1433[t], Derivative[1][Wee1][t] ==
kWee1 + k33*Wee1p[t] - k34*MPF[t]*Wee1[t], MPF[0] == MPFo,
Cdc25a[0] == Cdc25ao, Chk1p[0] == Chk1po, Chk1[0] == Chk1o,
Cdc25[0] == Cdc25o, Cdc25ps216a[0] == Cdc25ps216ao,
preMPF[0] == preMPFo, P21MPF[0] == P21MPFo, p21[0] == p21o,
p53a[0] == p53ao, p53[0] == p53o, Cdc25ps216[0] == Cdc25ps216o,
F1433[0] == F1433o, Wee1[0] == Wee1o}, {MPF, Cdc25a, Chk1p, \
Chk1,
Cdc25, Cdc25ps216a, preMPF, P21MPF, p21, Cdc25ps216, F1433, \
Wee1},
{t, 8}]; Plot[Evaluate[{MPF[t], Wee1[t]} /. sol], {t, 0, TFinal},
PlotRange -> All, PlotLegends -> {"[MPF], [Wee1]"},
PlotStyle -> {Green, Cyan}], {{TFinal, 5, "Time"}, 0.2, 8},
{{DDS, 0, "DDS"}, 0, 0.008}, {{MPFo, 1, "[MPF] Initial"}, 0, 10},
{{Cdc25ao, 1, "[Cdc25a] Initial"}, 0, 10}, {{Chk1po, 1, "[Chk1p] \
Initial"},
0, 10}, {{Chk1o, 1, "[Chk1] Initial"}, 0, 10},
{{Cdc25o, 1, "[Cdc25] Initial"}, 0, 10},
{{Cdc25ps216o, 1, "[Cdc25ps21a] Initial"}, 0, 10},
{{Cdc25ps216ao, 1, "[Cdc25ps216a] Initial"}, 0, 10},
{{preMPFo, 1, "[preMPF] Initial"}, 0, 10},
{{P21MPFo, 1, "[P21MPF] Initial"}, 0, 10}, {{p53ao, 1, "[P53a] \
Initial"},
0, 10}, {{p53o, 1, "[P53] Initial"}, 0, 10},
{{F1433o, 1, "[F1433] Initial"}, 0, 10}, {{Wee1o, 1, "[Wee1] \
Initial"}, 0,
8}, {{ks, 1, "ks"}, 0, 8}, {{Global`K, 1, "K"}, 0, 8},
{{Dego, 1, "Dego"}, 0, 8}]
NDSolve::underdet: There are more dependent variables, {Cdc25[t],Cdc25a[t],Cdc25ps216[t],Cdc25ps216a[t],Chk1[t],Chk1p[t],F1433[t],Mdm2[t],MPF[t],p21[t],P21MPF[t],p53[t],P53[t],p53a[t],preMPF[t],Wee1[t],Wee1p[t]}, than equations, so the system is underdetermined.
ReplaceAll::reps: {NDSolve[{(MPF^\[Prime])[t]==-0.01 MPF[<<1>>]^2-0.1 MPF[t] p21[t]+P21MPF[t]+(Cdc25a[<<1>>]+Cdc25ps216a[<<1>>]) preMPF[t]-0.01 MPF[t] Wee1[t],(Cdc25a^\[Prime])[t]==-0.01 Cdc25[t]-0.0001 Cdc25a[t]+0.01 Cdc25ps216a[t]-0.01 Cdc25a[t] Chk1p[t]+1. Cdc25[t] MPF[t],(Chk1p^\[Prime])[t]==Chk1[t]-1. Chk1p[t],(Chk1^\[Prime])[t]==-1. Chk1[t]+Chk1p[t],(Cdc25^\[Prime])[t]==0.00005 +0.01 Cdc25[t]-0.01 Cdc25[t] Chk1p[t]-1. Cdc25[t] MPF[t],(Cdc25ps216a^\[Prime])[t]==-0.02 Cdc25ps216a[t]+0.01 Cdc25a[t] Chk1p[t]+1. Cdc25ps216[t] MPF[t],(preMPF^\[Prime])[t]==0.0001 +0.0005/Plus[<<2>>]-(Cdc25a[<<1>>]+Cdc25ps216a[<<1>>]) preMPF[t],(P21MPF^\[Prime])[t]==0.1 p21[t]-P21MPF[t],<<12>>,preMPF[0]==1,P21MPF[0]==1,p21[0]==p21o,p53a[0]==1,p53[0]==1,Cdc25ps216[0]==1,F1433[0]==1,Wee1[0]==1},<<1>>,{t,8}]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing.
NDSolve::dsvar: 0.00010214285714285715` cannot be used as a variable.
ReplaceAll::reps: {NDSolve[{(MPF^\[Prime])[0.000102143]==-0.01 MPF[<<1>>]^2-0.1 MPF[0.000102143] p21[0.000102143]+P21MPF[0.000102143]+(Cdc25a[<<1>>]+Cdc25ps216a[<<1>>]) preMPF[0.000102143]-0.01 MPF[0.000102143] Wee1[0.000102143],(Cdc25a^\[Prime])[0.000102143]==-0.01 Cdc25[0.000102143]-0.0001 Cdc25a[0.000102143]+0.01 Cdc25ps216a[0.000102143]-0.01 Cdc25a[0.000102143] Chk1p[0.000102143]+1. Cdc25[0.000102143] MPF[0.000102143],<<24>>,F1433[0]==1,Wee1[0]==1},{MPF,Cdc25a,Chk1p,Chk1,<<4>>,p21,Cdc25ps216,F1433,Wee1},{0.000102143,8}]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing.
NDSolve::dsvar: 0.00010214285714285715` cannot be used as a variable.
ReplaceAll::reps: {NDSolve[{(MPF^\[Prime])[0.000102143]==-0.01 MPF[<<1>>]^2-0.1 MPF[0.000102143] p21[0.000102143]+P21MPF[0.000102143]+(Cdc25a[<<1>>]+Cdc25ps216a[<<1>>]) preMPF[0.000102143]-0.01 MPF[0.000102143] Wee1[0.000102143],(Cdc25a^\[Prime])[0.000102143]==-0.01 Cdc25[0.000102143]-0.0001 Cdc25a[0.000102143]+0.01 Cdc25ps216a[0.000102143]-0.01 Cdc25a[0.000102143] Chk1p[0.000102143]+1. Cdc25[0.000102143] MPF[0.000102143],<<24>>,F1433[0.]==1.,Wee1[0.]==1.},{MPF,Cdc25a,Chk1p,<<6>>,Cdc25ps216,F1433,Wee1},{0.000102143,8.}]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing.
General::stop: Further output of ReplaceAll::reps will be suppressed during this calculation.
NDSolve::dsvar: 0.10214295918367347` cannot be used as a variable.
General::stop: Further output of NDSolve::dsvar will be suppressed during this calculation.eIsaac Lee2018-04-22T01:37:16ZMake vector/mutlivariate clustering/machine learning of a dataset?
http://community.wolfram.com/groups/-/m/t/1324841
Hi,
Does anyone know how to perform a vector level clustering of a data set ?
**The problem to solve:** I have a data set of 1450 samples. Each sample is a vector with 10 scalar data (numbers).
The data is structured in a matrix, i.e. a list of lists of numbers. {{1,2,1,...},{3,1,7,...}...}
When I use the function Find Clusters, it returns a **classification of the scalars themselves**, i.e. each number, but not of the vectors.
I want to be able to **classify the vectors** {1,2,1,...} as single objects, as **opposed to each scalar component**, which is what Mathematica does wenn I call the function Findclusters on the matrix itself.
Does anyone know how to proceed to do this ?
Thanks a lot for the answer.
Best,
Emmanuel.Emmanuel Daugeras2018-04-21T10:45:55ZDownload manager won't start downloading and displays errors
http://community.wolfram.com/groups/-/m/t/1211541
Hello everyone!
Could anyone please help me with this situation? I am trying to download the latest version of Mathematica, but I am having this problem with download manager utility and I can't find any solution. The download process nevers starts and I only see those error messages...
![enter image description here][1]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=helpmeifyoucan.png&userId=1211525
I would really appreciate some explanation for fixing this, thanks a lot :)
(2 hours after creating this thread Update) - **Timeout in I/O** has just been replaced by **Socket error** :(Samuel Čech2017-10-31T18:31:36ZRearrange a matrix of data to Map LinearModelFit?
http://community.wolfram.com/groups/-/m/t/1325340
Hello, Community
I have built the following matrix (10,10) of data
xdata = Table[RandomVariate[NormalDistribution[], n], 10];
\[Epsilon]data = Table[RandomVariate[NormalDistribution[0, RandomChoice[{1/3, 1, 3}]], n], 10];
\[Beta]data = Table[RandomChoice[{1/3, 1, 3}, n], 10];
ydata = xdata \[Beta]data + \[Epsilon]data;
I rearranged the data according to
data = Table[Transpose[{ydata[[k]], xdata[[k]]}], {k, 10}];
But when I map *LinearModelFit* to estimate 10 equations like below,
lsFunc = LinearModelFit[#, x, x] & /@ data
It doesn't work {:^(
Can someone please let me know my mistake and how to fix it?
Also, how can I Map properties to the LinearModelFit results? For example, I want to extract the list of *"BestFitParameters"*.Thadeu Freitas Filho2018-04-23T01:33:56Z[WSS16] Quantum Computing with the Wolfram Language
http://community.wolfram.com/groups/-/m/t/897811
**Introduction to the Problem**
While a gate-based quantum computer has yet to be implemented at the level of more than a handful of qubits, and some worry that the decoherence problem will remain an obstacle to real-world use of these machines; the field of theoretical quantum computing has its own virtue apart from these problems of construction and implementation. The theory of quantum computation and quantum algorithms have been used as powerful tools to tackle long-standing problems in classical computation such as proving the security of certain encryption schemes and refining complexity classifications for
some approaches to the Traveling Salesman problem. Moreover, learning how to apply quantum effects like superposition, interference, and entanglement in a useful, computational, manner can help students gain a better understanding of how the quantum world really works. These educational and research advantages of quantum computing, along with the ever-present goal of designing new quantum algorithms that can provide us with speedups over their classical counterparts, furnish ample reason to make the field as accessible as possible. The goal of this project was to do just that by using the Wolfram language to design functionality that allows for researchers and students alike to engage with quantum computing in a meaningful way.
**Getting it Done**
This project involved the design and development of a suite of functions that allows for the simulation of quantum computing algorithms. The overarching goal was a framework that allows for easy implementation of quantum circuits, with minimal work done by the user. The specific design challenges were to have a tool simple enough to be used as an educational aide and powerful enough for researchers. To this end circuits can be built iteratively, allowing students, and those new to quantum computing, to build a working knowledge of the field as they increase the complexity of the algorithms. The system has a universal set of gates allowing it to carry out any operation possible for a quantum computer (up to limits on the number of qubits due to the size of the register).
----------
*Short note on this: I have not rigorously tested the system yet, but unless you want to wait several hours for your computation to complete, I suggest not attempting computations with more than ~20 qubits. To classically simulate an N-qubit register, requires a state vector of length 2<sup>N</sup>. Interestingly, it is this insight into the computational difficulty of simulating a quantum state that led Feynman to realize the power that quantum computing could have.*
----------
The project has functionality for the following gates: Hadamard, X, Y, Z, Rotation (any angle, about any axis), C-NOT, C-anything, SWAP, and QFT. It takes input in standard quantum circuit notation, and can output circuit diagrams, and the corresponding unitary transformation matrix as well as return the probabilities for results of measurements on a given qubit. Moreover, there is built in functionality for easy circuit addition, allowing one to stitch together large circuits from smaller ones, a boon for comprehension and testing.
**A Simple Example**
We initialize some random circuit by specifying it's corresponding circuit notation. For sake of brevity, we start with a medium-sized circuit that is already formed, and perform operations on it, but one can easily build a circuit up qubit-by-qubit and gate-by-gate with the applyQ and circuitQ functions. Below we name some variable `quantumCircuit` using the function `circuitQ` to which we pass some circuit notation. This notation is just a matrix representing the quantum logic circuit, with the gates and qubits arranged schematically.
quantumCircuit= circuitQ[{{"H", "R[1,Pi/2]", "N", "SWAP1"}, {"H", 1, "C",
"SWAP2"}, {"X", 1, "C", 1}}];
`circuitQ` outputs the circuit diagram corresponding to the notation given:
![enter image description here][1]
But, say I wish to alter the circuit. We can add in as many layers of gates or extra qubits as we wish, without having to deal with the pesky notation matrix. Here I add a Hadamard gate to the second qubit after the SWAP using the function `applyQ`:
applyQ[quantumCircuit, "H", 2]
the output of which is:
![enter image description here][2]
One can also use `Append`, `Join`,`Nest` and a variety of other Wolfram language functions to build up highly complex circuits. However, the `circuitQ` function is overloaded, and one can also perform computations with it. We will now build the actual unitary transformation matrix that corresponds to the circuit diagram:
unitar=matrixBuild@quantumCircuit
which, for our circuit, produces:
![enter image description here][3].
Now we can easily perform operations with circuit. Let's specify some random 3 qubit initial state (in the computational basis):
initalState = {1, 0, 0, 1, 0, 0, 1, 0} // Normalize
![enter image description here][4]
We can pass this initial state to the circuit easily with:
premeasure=unitar.initialState
which gives back the state of the quantum register (in this case our 3 qubits) after they have been operated on by the circuit, but pre-measurement:
![enter image description here][5]
We can now sample our state using the `projection` function. Here we will calculate the probability of getting state |0> when measuring qubit #3:
projection[2,0,premeasure]
which, for our case, gives back a probability of 2/3.
**Wrap Up**
This was only a very simple example. Using `applyQ` and `circuitQ` one can build and modify highly complex quantum circuits easily. `matrixBuild` does all the math of calculating the corresponding unitary transformation matrix for you. All that is left is for the user to pass an initial state and see the output. A good learning technique is to start with a very simple circuit and initial state, and slowly build up in complexity, performing measurements at each step, to build an intuition and working knowledge of any given quantum circuit.
An obvious next step for the project would be to add functionality that allows for the easy implementation of a general quantum oracle. I would also like to add more gates to the gate library, including: $\sqrt{SWAP}$, Tofolli, and QFT<sup>-1</sup> which were left out due to lack of time and are trivial to implement. These tools would make it significantly easier for researchers to model any given quantum circuit.
**Where is the NKS?**
Finding quantum algorithms that perform useful tasks faster than their classical counterparts is an open area of research. However, it is often quite difficult to design these algorithms to take advantage of interference, as well as the structure in a given computational problem that may be useful to exploit. As such, there are only a small number of important quantum algorithms that are currently known. Hopefully this tool will allow for NKS-style search experiments for interesting behavior in quantum circuits. Similar searches have been carried out for classical circuits, and the tools I built will make it easy to generate vast sets of random quantum circuits that follow certain rules. What remains is to build useful analytic tools for combing the space.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=circuit.jpeg&userId=896802
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=circuit2.jpeg&userId=896802
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=unitar.jpeg&userId=896802
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=3114initialstate.jpeg&userId=896802
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=finalstate.jpeg&userId=896802Aaron Tohuvavohu2016-08-02T02:35:38ZMy five-part introduction to Wolfram Mathematica
http://community.wolfram.com/groups/-/m/t/1325468
Over the past week, I have read through and updated the five-part introductory tutorial on Wolfram Mathematica and computer science in general that I originally created back in 2014 when I first fell in love with this language. Intended for motivated beginners, the PDF files are available on DropBox, as linked on [my Wolfram Mathematica page][1]. Hopefully other people will find these tutorials useful, but at least I had a good time writing them.
[1]: https://docs.google.com/document/d/1SnbBfzAMmn6m8VvnW1RxMOXfvhxkjbquXjcIKYR9lbE/edit?usp=sharingIlkka Kokkarinen2018-04-23T02:23:26ZEstimate the best fit function for a time series?
http://community.wolfram.com/groups/-/m/t/1325097
Hi!
I'm working on some financial data, and got stuck once computing the sample autocorrelation function for a time series.
First I import the data, then turn it into a Time Series using TimeSeries[data,Automatic]. Then I run CorrelationFunction and it returns a proper autocorrelation function. But if I then try to fit a function to this sample autocorrelation, the FindFit says something like
FindFit::fitd: First argument in FindFit is not a list or a rectangular array.
Any ideas. My objective is to estimate a nonlinear function out of the sample autocorrelation function
thanks,
Jussijussi lindgren2018-04-22T16:36:06ZGet a symbolic Lyapunov solver?
http://community.wolfram.com/groups/-/m/t/1325427
Hi,
Is there any symbolically Lyapunov Solver in Mathematica/Wolfram? I already searched on internet and I couldn`t found anything promising.
Thank you in advance!Ilie Denis2018-04-22T20:46:03ZGenerate a rectangular Code using the CodingTheory Package?
http://community.wolfram.com/groups/-/m/t/1324698
Hello, I'm new in the community.
Does anyone know how to generate a rectangular Code of length between 80 and 100 and then get the parameters of that Code?
I only know that I have yo use the CodingTheory Package.
Thanks.Juan Marcos Díaz2018-04-22T20:41:49ZInstall MathToolBox package?
http://community.wolfram.com/groups/-/m/t/1324339
The MathToolBox package, listed at http://packagedata.net and available at https://yadi.sk/d/oC5lXLWa3PVEhi, contains a single `.m` file, namely, `MathToolBox.m`. But when I look at the contents of that file, I find a bunch of individual`BeginPackage` expressions there, none seeming to be named `MathToolBox`.
How should this be installed so that one can readily load it?
(I do wish that authors who publicly post packages would always include installation instructions! Today, the standard way seems to be to have a directory `ThePackageName` that contains the actual package `ThePackageName.m`, whose code begins with ``BeginPackage["ThePackageName`"]``; and a subdirectory `Kernel` that contains a simple `init.m` file of the form ``Needs[["ThePackageName`ThePackageName`"]``, or perhaps the same thing but with `Get` instead of `Needs`.)Murray Eisenberg2018-04-20T20:23:55ZUse GPUs to compute faster RandomFunction and ParallelTable?
http://community.wolfram.com/groups/-/m/t/1324662
Hi, I have a code where I use RandomFunction and ParallelTable. Could somebody guide me how to make use of my GPUs for accelerating this computation? I have looked through the CUDA Programming (http://reference.wolfram.com/language/CUDALink/tutorial/Programming.html#135446596) but I do not understand the codes which are later linked through CUDAFunctionLoad.Rafael Petrosian2018-04-22T17:25:40ZCalculate Resultant of an equation?
http://community.wolfram.com/groups/-/m/t/1324461
code file is attached.Deepak Singh2018-04-20T14:50:27ZPreserve SystemModeler generated source files?
http://community.wolfram.com/groups/-/m/t/1322648
Is it possible to configure SystemModeler to preserve the output source files? By default it generates the source and then purges the files after compilation.natehollis2018-04-18T12:51:23ZAdd the ["BestFitParameters"] property to a function from a package?
http://community.wolfram.com/groups/-/m/t/1323091
I am building a MonteCarlo simulation to compare the performance of Least Squares and Quantile Regressions. To automate the notebook, I need to be able to extract the parameters from the regressions. For the Least Squares method, I can run a *LinearModelFit* and use the property *["BestFitParameters"]*.
For Quantile Regressions, I run the *QuantileRegressionFit* from the package below (written by Anton Antonov - Quantile regression Mathematica package, source code at GitHub, [MathematicaForPrediction][1], package QuantileRegression.m, (2013)).
However, the function doesn't come with the same property, *["BestFitParameters"]*, so I don't know how to extract the parameters.
Can anyone help me with extracting the optimal parameters for the *QuantileRegressionFit* function?
Thanks in advance,Thadeu Freitas Filho2018-04-19T09:47:54Z[✓] Solve a system of coupled ODE's?
http://community.wolfram.com/groups/-/m/t/1324899
Hello!
I am currently working on neutrino's oscillations and I need to solve a system of coupled ODE's.
The problem is that I have never coded in Mathematica so my code is probably awful and... doesn't work at all.
The system to solve is the following:![enter image description here][1] (at the end, one must take the modulus of the amplitude)
with everything a constant except V_e a function of time.
This is my code which does not return the right solution (see below)
Gf = 0;
Xmax = 100000;
O13 = N[Degree, 8.54];
O23 = N[Degree, 47.2];
O12 = N[Degree, 33.62];
dcp = N[Degree, 234];
dm21 = 7.40*10^(-5);
dm31 = 2.494*10^(-3);
p = 1;
M = {{0, 0, 0}, {0, dm21, 0}, {0, 0, dm31}};
R1 = {{1, 0, 0}, {0, Cos[O23], Sin[O23]}, {0, -Sin[O23], Cos[O23]}};
R2 = {{Cos[O13], 0, Sin[O13]*Exp[-I*dcp]}, {0, 1,
0}, {-Sin[O13]*Exp[I*dcp], 0, Cos[O13]}};
R3 = {{Cos[O12], Sin[O12], 0}, {-Sin[O12], Cos[O12] , 0}, {0, 0, 1}};
U = R1*R2*R3;
Ve[x_] = Sqrt[2]*Gf*Exp[-2*x/Xmax];
P[x_] = 1/(2 p) * U*M*
ConjugateTranspose[U] + {{Ve[x], 0, 0}, {0, 0, 0}, {0, 0, 0}};
A[x_] = Table[ai[i, x], {i, 1, 3}];
A0 = Table[0, {3}];
A0[[1]] = 1;
diffEq = Table[
A'[x][[i]] == -I (P[x].A[x])[[i]], {i, 1, 3}]; initialCond =
Table[A[0][[i]] == A0[[i]], {i, 1, 3}];
sol = NDSolve[{Join[diffEq, initialCond]}, A[x], {x, 0, Xmax}];
Plot[Evaluate[
Table[(ai[i, x] /. sol)*Conjugate[ai[i, x] /. sol], {i, 1, 3}]], {x,
0, Xmax}]
The solution is supposed to be ![enter image description here][2]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture.JPG&userId=1324885
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=untitled.jpg&userId=1324885
Thank you very much for your help !Loic Sablon2018-04-21T21:07:39ZPeriodic piecewise functions?
http://community.wolfram.com/groups/-/m/t/490391
How would one create a piecewise periodic function (without using recursions)?
For example I'd like to make this function Piecewise[{{x, 0 <= x <= Pi}, {x - 2 Pi, Pi < x < 2 Pi}}] repeat itself to infinity...Nev Nev2015-05-03T15:46:31ZUse manipulate to create a slider showing its current value?
http://community.wolfram.com/groups/-/m/t/1324753
i have been using manipulate to create myself a slider, but i have my result at certain point when i move the slider, but i am not able to find the value at which my slider is at currently, kindly replymourishwaran j2018-04-21T04:45:53Z[✓] Differentiate a function defined via Piecewise[]?
http://community.wolfram.com/groups/-/m/t/1323015
Using the "default value" syntax of Piecewise[] one can define the function equal to x*sin(1/x) for non-zero x and equal to 0 for x=0 in the following compact form:
y[x_] := Piecewise[{{x*Sin[1/x], x != 0}}]
Now, if we try to calculate the value of its derivative at x=0, then Mathematica assumes that it depends only on the value of y[x] at x=0:
y'[0]
0
But of course this is not true --- this function is not differentiable at x=0, because for x!=0 we have:
y'[x] = Sin[1/x] - Cos[1/x]/x
And the above expression has no limit as x approaches 0 and Mathematica knows this very well:
Limit[Sin[1/x] - Cos[1/x]/x, x -> 0]
Indeterminate
Limit[y'[x],x->0]
Indeterminate
Also, just taking the definition of derivative (as a limit) at x=0 we would end up with Limit[Sin[1/h],h->0] which of course doesn't exist.
So, I think there is a bug here: when one applies the differentiation operator D to something which has the head of Piecewise it shouldn't differentiate the expression for each condition independently, because the value of a derivative of a function at some point depends not only on the value of the function at that point, but also on all the values of the function in the infinitesimal neighbourhood of that point.Tigran Aivazian2018-04-18T21:26:56ZGet absolute value of a complex number?
http://community.wolfram.com/groups/-/m/t/1320443
Hi!
I'll start right away.
I have a question about *Complex Numbers*.
I have a complex number, or a bunch of complex numbers, like you can see below. Here I need to know the absolute value of the number B squared, but in the result there is still the imaginery unit in it. I said that all the variables are real numbers greater than zero, so it should be possible to give me a correct term for the absolute value of B.
Or what is my mistake here? How do I use complex numbers with variables in it? How do I make sure that the variables are part of the reals numbers an grater than zero?
Input
\[Alpha] \[Element] Reals; \[Lambda] \[Element] Reals; L \[Element] Reals; \[Alpha] > 0; \[Lambda] > 0; L > 0;
\[CapitalOmega] = (\[Alpha] - I*\[Lambda])/(\[Alpha] + I*\[Lambda])*Exp[\[Alpha]*L];
c = (2*I*\[Lambda])/(I*\[Lambda] + \[Alpha])*Exp[(\[Alpha] - I*\[Lambda])*L/2]/(1 + \[CapitalOmega]^2);
d = c*\[CapitalOmega];
B = c*Exp[(-\[Alpha] + I*\[Lambda])*L/2] + d*Exp[(\[Alpha] + I*\[Lambda])*L/2] - 1;
Abs[B]^2
Output
Abs[-1 + (
2 I E^(L \[Alpha] + 1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (\[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda])^2) + (
2 I E^(1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (-\[Alpha] + I \[Lambda])) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2
In[7]:= Simplify[
Abs[-1 + (
C E^(L \[Alpha] +
1/2 L (\[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]))/(\[Alpha] + I \[Lambda]) + (
2 I E^(1/2 L (\[Alpha] - I \[Lambda]) +
1/2 L (-\[Alpha] + I \[Lambda])) \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2]
Out[7]= Abs[-1 + (
C E^(1/2 L (3 \[Alpha] + I \[Lambda])) (\[Alpha] -
I \[Lambda]))/(\[Alpha] + I \[Lambda]) + (
2 I \[Lambda])/((1 + (
E^(2 L \[Alpha]) (\[Alpha] - I \[Lambda])^2)/(\[Alpha] +
I \[Lambda])^2) (\[Alpha] + I \[Lambda]))]^2
I hope you can help me. Thanks in advance! :)
Best regards,
ThomasThomas Vergeiner2018-04-15T09:10:20Z[✓] Calculate a definite integral?
http://community.wolfram.com/groups/-/m/t/1323897
Why the output of a definite integral is in between braces key![What is the meaning of this?, it suppose to show an algebraic expression in function of the integers limits][1]
What is the meaning of this?, it suppose to show an algebraic expression in function of the integers limits
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Braces.png&userId=1323881Oscar Moreno2018-04-20T03:42:57Z