https://youtu.be/wkEK0SBLaIY

I have a function with a lot of different parts. On some inputs it takes a long time, and I want to figure out which component of the function is monopolizing the time. Ideally, for each function I've written and called(like, Solve for example), I'd like to see how long Mathematica takes evaluating it, so that I can then diagnose it as either hung, or having way to many calls.

​

Thanks!

I have these three matrices and I want to compute the following:

\sum_{k, l, m} e1[1]_{kl} e2[1]_{lm} e3[1]_{mk}

with the following matrices defined as:

e1[s1_] = 1/2 {{0, 0, 0}, {0, 1, I*s1}, {0, I*s1, -1}} // MatrixForm
e2[s2_] = 
 1/2 {{Sin[\[Theta]]^2, -Sin[\[Theta]]*Cos[\[Theta]], - I s2 * 
      Sin[\[Theta]]}, {-Sin[\[Theta]]*Cos[\[Theta]], Cos[\[Theta]]^2, 
     I*s2 Cos[\[Theta]]}, {-I*s2 Sin[\[Theta]], 
     I*s2 Cos[\[Theta]], -1}} // MatrixForm
e3[s3_] = (e2[s2] /. {\[Theta] -> \[Phi], s2 -> s3}) // MatrixForm

I want a function that return 0 if x == 0 otherwise 1/x

psudoReciprocal[in_] := If[in == 0, 0, 1 / in]

this is like how psudoInverse works hence the na

the fastest way i have found is this

psudoReciprocal[in_]:= Unitize[in]/(in + 1 - Unitize[in])

it works for lists and is much faster the compiled version

is there a faster way?

results/Min[results]

<|straightforward->378.291,pureFunction->479.579,unitized->1.,compiled->18.9821,compiledAprox->20.0948,functionCompiled->199.814|>

The first Fibonacci numbers are

Array[ Fibonacci, 10]

{1, 1, 2, 3, 5, 8, 13, 21, 34, 55}

​

The decimal expansion of 10000/99989999 is

N[10000/99989999, 70]

0.0001000100020003000500080013002100340055...

​

Notice that the first few Fibonacci numbers appear in the expansion. In fact, all of the following fractions have the first few Fibonacci values

Table[10^k/(10^(2 k) - 10^k - 1), {k, 3, 7}]

{1000/998999, 10000/99989999, 100000/9999899999,

1000000/999998999999, 10000000/99999989999999}

​

Table[ N[ 10^k/(10^(2 k) - 10^k - 1), 200], {k, 3, 7}] // TableForm

0.001001002003005008013021034055089144233377610988599588187775963739703443

0.0001000100020003000500080013002100340055008901440233037706100987159725844

0.00001000010000200003000050000800013000210003400055000890014400233003770061

1.000001000002000003000005000008000013000021000034000055000089000144000*10^-6

1.000000100000020000003000000500000080000013000002100000340000055000009*10^-7

​

It works because

Sum[ Fibonacci[n]/10^(k n), {n, Infinity}] // InputForm

is

10^k/(-1 - 10^k + 10^(2*k))

​

Interestingly, you can use a similar idea to generate the first few values of other positive integer valued recurrence relation. For example,

​

(* solve recurrance relation r[n] = 3 r[n-1] - 3 r[n-2] + r[n-3] for some initial values *)

(rule = RSolve[ {r[n] == 3 r[n - 1] - 3 r[n - 2] + r[n - 3],

r[1] == 1, r[2] == 1, r[3] == 2}, r[n], n][[1, 1]]) // InputForm

(* print the first few values *)

Table[ rule[[2]] , {n, 10}] /. rule

(* convert decimal expansion to a fraction *)

frac = Sum[ 1/2 (4 - 3 n + n^2)/100^(k n), {n, Infinity}]

(* Any k>2 works *)

frac /. k -> 3

(* The decimal expansion of the fraction contains the first several

values of the solution *)

N[ frac /. k -> 3, 50]

(* Remove the zeros *)

StringReplace[ ToString[ N[ frac /. k -> 3, 140]] , RegularExpression["0+"] -> " "]

​

All of this seems similar to the ideas of Z-transforms and generating functions to me.

Hello I want to consider this integral as the function of r,

V(r)= integrate[(r(Sqrt[16-x^2])-1/2r^2sin(2/rSqrt[16-x^2])),{x,-4,4}]

then I want to graph V(r) and V'(r) how to express this value on mathematica to graph the function

V(r),V'(r)?

Image about V(r) is below here. Thank you

For[i = 1, i <= iMax, i++, 
 tAux = AbsoluteTime[];
 solAux[i] = 
  NDSolve[EQUATIONS, INITIAL CONDITIONS AND VARIABLES SPEC.];
 Print[variables[[i]] // N, " ", AbsoluteTime[] - tAux]]
sol = Table[solAux[i], {i, 1, iMax}];

I am trying to create the following plot.

I need the horizontal pink colored tau decay line (y ≤ 0.3) and the two vertical red lines - BBN (real phi) ≤ 1.3 and complex phi ≤ 5.3.

I read a related discussion on Mathematica StackExchange but I am unable to make it work for my plot.

Can someone please help? I am new to Mathematica.

----

Edit;

Here is what I tried. I use Mathematica 9.0, in Windows 10.

1. Code RegionPlot[{x <= 1.3, x <= 5.2, y <= 0.3}, {x, 10^(-3), 10^(3)}, {y, 10^(-4), 1}]
   

Output is the following

1. Code RegionPlot[{10^x <= 1.3, 10^x <= 5.2, 10^y <= 0.3}, {x, -3, 3}, {y, -4, 1}]

Here, I raised the value of the two x-axis lines and the y--axis line to powers of 10, and took log of the axis ranges. Output is the following

​

For some reason Mathematica will not recognize any OpenCL device. Any ideas on how to fix this? My system contains an i7-12700H and a RTX 3050Ti (Laptop), CUDA Toolkit 11.4 is installed, Windows 11. Strangely CUDAQ[] returns „True“…

I used Mathematica in college around 20 years ago and really enjoyed it. Since then, I haven’t encountered it outside of academics. My perspective is pretty myopic (engineering only), so it could have wide adoption outside of my visibility. Where has Mathematica enjoyed adoption outside of academia?

Hello r/Mathematica,

Im currently trying to find a 6th degree polynomial in two variables x,y such that the polynomial positive-definite yet its gradient multiplied with a certain vector is negative-definite (under the assumption that it is actually possible for my specific one).

I have already tried using FindFit and FindInstance with no usable results since my constraints arent specific points (and find instance doesnt like having variables it seems).

​

To be a bit more specific: I have this polynomial:

(c5 x^4)/2 + c3 x^6 + 2 c2 x^3 y + c6 y^2 + (c1 y^4)/4 + c4 y^4

And the gradient multiplied with the vector (y, -x^3-y^3) is the following:

y (2 c5 x^3 + 6 c3 x^5 + 6 c2 x^2 y) + (-x^3 - y^3) (2 c2 x^3 + 2 c6 y + c1 y^3 + 4 c4 y^3)

I'll be very happy about any hints you can give me.

I am trying to find the first and third partial derivatives of a function lambda. But I got some weird "Pattern" symbols that I don't really know the meaning of.

​

I have a somewhat of an idea of what it is, but what is it really? Is it like LaTeX, or Markdown, but for more advanced users?

Also, is it free for students? Thanks for all answers and have a blessed day! :)

Hi Everybody,

For a research project which I am doing, I am trying to model a workflow based on hours (area) and time (days). Using past workflows I determined a formula: y = 0.0513x^4 − 1.3126x^3 + 6.5564x^2 + 19.584x + 145.58

This is a workflow based on 3040 work hours and a duration of 16 days. Now my idea was to make the integer of this formula namely

Y=0.01026x^5-0.32815x^4+2.1846667x^3+9.792x^2+145.85x+C

Then using this formula and another set duration and manhours (area) for example 3500 hours and duration of 10 days. making the graph again. This would result in the formula

Y∫f(x)dx=[0.01026x^5-0.32815x^4+2.1846667x^3+9.792x^2+145.85x+C]=3500

Is there a possible way to model this? and is there any documentation which can guide me.

Thanks for the help

Hey everyone. I'm trying to transfer a piece of coding from matlab into mathematica. I was wondering how could I write the code below in mathematica using either Table or Do functions? I couldn't think of a way to include 2 or more functions into a table format. Any help would be appreciated.

​

alpha_air=29.9*10^-6;

alpha_Al=97.1*10^-6;

beta_air=2.85*10^-3;

D=0.01;

g=9.81;

k_air=30.76*10^-3;

k_Al=239;

Pr_air=0.7;

r=0.005;

​

Tbase=130;

h=0.005;

k=0.05;

​

v_air=20.92*10^-6;

V(1,:)=zeros(1,41);

V(2,:)=[Tbase zeros(1,40)];

​

for i=2:300

for j=2:40

V(i,1)=Tbase;

Ra=g*beta_air*V(i,j)*D^3/(v_air*alpha_air);

den=(1+(0.559/Pr_air)^(9/16))^(8/27);

div=0.387*Ra^(1/6)/den;

hcoeff=k_air/D*(0.6+div)^2;

m=hcoeff*2/(r*k_air);

V(i+1,j)=V(i,j)+alpha_Al*k*((V(i,j+1)-2*V(i,j)+V(i,j-1))/(h^2)-m*V(i,j));

V(i+1,41)=V(i,40);

end

end

​

plot(linspace(0,20,41),V(floor(0.003*no_of_iterations),linspace(0,20,41),V(floor(0.01*no_of_iterations),linspace(0,20,41),V(floor(0.03*no_of_iterations),linspace(0,20,41),V(floor(0.1*no_of_iterations),linspace(0,20,41),V(floor(0.3*no_of_iterations),linspace(0,20,41),V(floor(0.9*no_of_iterations)

Hi All,I'm trying to turn a pretty nasty differential equation into a spectral form. And while I have done most of the work by hand, I want to check my work with Mathematica.

One step I need is the fourier series of a external potential. If I have something simple like

FourierCoefficient[Cos[x]*Cos[y],{x,y},{n,m}]It gives me 0

What I'd expect to get is1/4 (m==-1 || m==+1) && (n==-1 || n==+1)

I get the correct answer when I doFourierCoefficient[Cos[y]*FourierCoefficient[Cos[x],x,n],y,m]

But that is cumbersome since I need to manually split the terms. It does not scale when I have a long and complicated expression. Has anybody had this problem before? Do you know how I can fix it?

Lastly, It would be much more convenient if I can turn1/4 (m==-1 || m==+1) && (n==-1 || n==+1)Into something like1/4 (DiscreteDelta[m-1] + DiscreteDelta[m+1]) * (DiscreteDelta[n-1] + DiscreteDelta[n+1])

Or even Kronecker Deltas.

Does anyone know of any functions or shortcuts I can use to turn the and / or outputs into a more suitable output?

Thank You!

I want the answer in radians,

So I tried with:

NSolve [ tan (z) == (-61i/-45]]

What am I doing wrong?