# [✓] Build a matrix with the help of loop structures?

GROUPS:
 Please, friends, I ask your help. Based on my old minded and averaged skilled Fortran programmer, I would know how to build a matrix based on the Loop structure, as presented in the attached file. There is no need to having the same structure or similarity in commands, but only doing the same task. I thank you very much! Scabral!Fortran example
1 year ago
7 Replies
 Michael Helmle 1 Vote Hi Sergio, this should work. Please check carefully against the results of the old code xm = Array[x, {10, 10}]; For[i = 1, i <= 10, i++, xi = (N[i] - 0.5) h; x[i, i] = const * Log[0.44705*h*xi*2.]; For[j = 1, j <= 10, j++, If[i == j, Break[]]; xj = (N[j] - 0.5) h; x[i, j] = const* Log[Abs[xi - xj]* Abs[xi + xj]]; x[j, i] = x[i, j]; ] ] Here you can see the structure of the matrix before putting in the numbers MatrixForm[xm] Finally the variables are replaced by the numeric values prms = {g -> 1.*10^8 / 2.83, freq -> 377., h -> 5.*10 (-3), perm -> 4.* \[Pi]* 10^(-7), eo -> 1.}; MatrixForm[xm] /. const -> (freq * perm* g * h^2) / (2. \[Pi]) /. prms 
1 year ago
 Thank you so much , Michael ! Absolutely great from you!
1 year ago
1 year ago
 Hi, dears ! Unfortunately, it didn´t work. In fact, I need a Wolfram version to this programof building a matrix X . The key point seems to be the presence of j as the argument of logarithm. Thank you !Mo= 0,4PiE-6 Sigma = 35335689,05 w = 120*Pi h = 0.005 const = (w * Mo * sigma * h^2 )/(2*Pi)Define x(10,10) Do i =1 to 10, step = 1 X(i,i) = const LN(0.44705h) Do j = 1 to 10, step =1 If (i= j ) Continue X(I,j) = const* LN( j * h) X(j,i) =X (i,j) Continue
 Hi Sergio,I propose you to verify the following code with Table[]:  mo = 4/10 \[Pi]; \[Sigma] = 3533568905/100; w = 120 \[Pi]; h = 5/1000; const = (w mo \[Sigma] h^2)/(2. \[Pi]); x = Table[ Which[i < j, const Log[j h], i == j, const Log[0.44705 h], i > j, const Log[i h]], {i, 10}, {j, 10}]; x//MatrixForm Your matrix is defined as a list $x$. Element $(i,j)$ of the matrix $x$ can be addressed by applying the syntax $x[[i,j]]$.
 Murray Eisenberg 1 Vote I haven't checked for efficiency of evaluation, but for sheer simplicity of code, why not just the following? x[i_, j_] := const Log[Max[i, j] h] x[i_, i_] := const Log[0.44705 h] Table[x[i, j], {i, 1, 10}, {j, 1, 10}] // MatrixForm That uses the symmetry of the condition for different i and j, and it exploits the fact that a specific rule, here x[i_,i_]:= always overrides a general rule, here:  x[i_,j_]:=