I searched the site and found that someone said FindRoot is equivalent to fsolve, however, without proper options, FindRoot is going to run forever. In my case, I am trying to do the numerical approximations, but it seems that Mathematica is doing a lot of symbolic calculations.

Here is the original Matlab code I want to translate:

x0 = 0.06*ones(14,1);

options = optimoptions(‘fsolve’,’Display’,’iter’,’TolFun’,1e-60,’TolX’,1e-60);

[x1,fval14] = fsolve(@myfun14,x0,options);

And my attempt is:

x1=FindRoot[myfun14[Array[x, 14]] == 0, Transpose[{Array[x, 14], ConstantArray[0.06, 14]}]]

Thanks!

The full original matlab code and the mathematica code I translated are as follows for reference:

Matlab code:

function F14 = myfun14(x)

b = [1/2;1/4;0;1/8;0;0;0;1/16;0;0;0;0;0;0];

A = [0,0,0,0,0,0,0,0,0,0,0,0,0,x(1);

1,0,0,0,0,0,0,0,0,0,0,0,0,x(2);

0,1,0,0,0,0,0,0,0,0,0,0,0,x(3);

0,0,1,0,0,0,0,0,0,0,0,0,0,x(4);

0,0,0,1,0,0,0,0,0,0,0,0,0,x(5);

0,0,0,0,1,0,0,0,0,0,0,0,0,x(6);

0,0,0,0,0,1,0,0,0,0,0,0,0,x(7);

0,0,0,0,0,0,1,0,0,0,0,0,0,x(8);

0,0,0,0,0,0,0,1,0,0,0,0,0,x(9);

0,0,0,0,0,0,0,0,1,0,0,0,0,x(10);

0,0,0,0,0,0,0,0,0,1,0,0,0,x(11);

0,0,0,0,0,0,0,0,0,0,1,0,0,x(12);

0,0,0,0,0,0,0,0,0,0,0,1,0,x(13);

0,0,0,0,0,0,0,0,0,0,0,0,1,x(14)];

B = A^(16-15);

for n=5:50

B = B + A^(2^n-15)/2^(n-4);

end

F14 = (x – b – 1/32 * B * x);

x0 = 0.06*ones(14,1);

options = optimoptions(‘fsolve’,’Display’,’iter’,’TolFun’,1e-60,’TolX’,1e-60);

[x1,fval14] = fsolve(@myfun14,x0,options);

[x2,fval14] = fsolve(@myfun14,x1,options);

[x3,fval14] = fsolve(@myfun14,x2,options);

[x4,fval14] = fsolve(@myfun14,x3,options);

[x5,fval14] = fsolve(@myfun14,x4,options);

[x,fval14] = fsolve(@myfun14,x5,options);

for t=1:200

[x,fval14] = fsolve(@myfun14,x,options);

end

A = [0,0,0,0,0,0,0,0,0,0,0,0,0,x(1);

1,0,0,0,0,0,0,0,0,0,0,0,0,x(2);

0,1,0,0,0,0,0,0,0,0,0,0,0,x(3);

0,0,1,0,0,0,0,0,0,0,0,0,0,x(4);

0,0,0,1,0,0,0,0,0,0,0,0,0,x(5);

0,0,0,0,1,0,0,0,0,0,0,0,0,x(6);

0,0,0,0,0,1,0,0,0,0,0,0,0,x(7);

0,0,0,0,0,0,1,0,0,0,0,0,0,x(8);

0,0,0,0,0,0,0,1,0,0,0,0,0,x(9);

0,0,0,0,0,0,0,0,1,0,0,0,0,x(10);

0,0,0,0,0,0,0,0,0,1,0,0,0,x(11);

0,0,0,0,0,0,0,0,0,0,1,0,0,x(12);

0,0,0,0,0,0,0,0,0,0,0,1,0,x(13);

0,0,0,0,0,0,0,0,0,0,0,0,1,x(14)];

xfinal = (A)^((10^9)-15)*x;

Mathematica code for the function definition part:

myfun14[list_] := Module[{b, A, B},

b = Transpose@{{1/2., 1/4., 0., 1/8., 0, 0, 0, 1/16., 0, 0, 0, 0, 0,

0}};

A = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, list[[1]]},

{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, list[[2]]},

{0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, list[[3]]},

{0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, list[[4]]},

{0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, list[[5]]},

{0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, list[[6]]},

{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, list[[7]]},

{0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, list[[8]]},

{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, list[[9]]},

{0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, list[[10]]},

{0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, list[[11]]},

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, list[[12]]},

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, list[[13]]},

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, list[[14]]}};

B = A + Sum[MatrixPower[A, (2^n – 15)]/2.^(n – 4), {n, 5, 50}];

(list – b – 1/32.*B*list)]

=================

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– bbgodfrey

Jun 30 ’15 at 16:02

@bbgodfrey myfun14is included in the code parts

– Nick

Jun 30 ’15 at 16:04

What exactly are you trying to do to the companion matrix of a polynomial?

– J. M.♦

Jun 30 ’15 at 16:07

@Guesswhoitis. The description of the original problem and code is here, jingplusplus.com/2015/06/12/mathpuzzleslotteryproblem

– Nick

Jun 30 ’15 at 16:10

=================

2 Answers

2

=================

Changing B*list to B.list seems to solve the problem. Also, I recommend dropping unnecessary decimal points from myfun14. Doing so gives

(* {x[1] -> 0.5, x[2] -> 0.273438, x[3] -> 0.0128174, x[4] -> 0.125601,

x[5] -> 0.00588754, x[6] -> 0.000275978, x[7] -> 0.0000129365,

x[8] -> 0.0625006, x[9] -> 0.00292972, x[10] -> 0.00013733,

x[11] -> 6.43736*10^-6, x[12] -> 3.01751*10^-7,

x[13] -> 1.41446*10^-8, x[14] -> 6.63028*10^-10} *)

Addendum

The recently updated code in the Question truly takes forever to run, because FindRoot first attempts to evaluate symbolically the gigantic expression created by myfun14[Array[x, 14]]. This can be circumvented by using the option, Evaluated -> False.

FindRoot[myfun14[Array[x, 14]], Transpose[{Array[x, 14], ConstantArray[6/100, 14]}],

Evaluated -> False]

(* {x[1] -> 0.500634, x[2] -> 0.266567, x[3] -> 0.00914348,

x[4] -> 0.130162, x[5] -> 0.00916464, x[6] -> 0.00235785,

x[7] -> 0.00297105, x[8] -> 0.065152, x[9] -> 0.00401983,

x[10] -> 0.00206912, x[11] -> 0.00273191, x[12] -> 0.00231037,

x[13] -> 0.00135524, x[14] -> 0.00136153} *)

Runtime on my PC is just less than one second.

Thank you, your method worked. I still have two things not clear : 1. I added those decimal points to let mathematica do numerical computation instead of keeping the rational numbers, but I have made things worse; 2. Although the program seems to be identical with the matlab one, this step gives different and wrong answers compared to matlab, where the correct answer should be approximately {0.5006, 0.2666, 0.0091, 0.1302, 0.0092, 0.0024, 0.0030, 0.0652, 0.0040, 0.0021, 0.0027, 0.0023, 0.0014, 0.0014}. Can you explain a little bit?

– Nick

Jun 30 ’15 at 17:02

@Nick 1) The problem appears to have been MatrixPower[A, (16 – 15.)]. MatrixPower expects an integer for its second argument, and 15. is a real number. 2} I tried different initial guesses and a much higher WorkingPrecision for FindRoot, but the answer did not change. My guess is that there is a typo somewhere.

– bbgodfrey

Jun 30 ’15 at 17:57

Thanks! I will do some check.

– Nick

Jul 1 ’15 at 3:06

I have located the typo, I forgot to change the ^ into MatrixPower in the definition of myFun14. But after changing this, the question returned to its original state: the function again becomes evaluating forever:( I have updated the mathematica code.

– Nick

Jul 1 ’15 at 3:16

@Nick Just a guess: replace 2.^(n – 4) by 2^(n – 4). In general, use rationale numbers, unless there is a good reason not to.

– bbgodfrey

Jul 1 ’15 at 3:55

An alternative solution would be changing the myfun14[list_] to myfun14[list_ /; VectorQ[list, NumberQ]], the result can be obtained after a few seconds. Setting a high AccuracyGoal is also necessary to get the correct answer for the problem.