Symsum multidimensionalfunction error sym/colon

A = [0 1 2 3; -1 0 -2 -3; 4 0 8 9; -6 -7 8 0]; t= 2; E = [0 0 0 0; 0 0 0 0; 0 1 0 0; -0 0 0 0];

function f = partko(k)

t=evalin('base','t'); % t is number in {1,2,...k,}

A=evalin('base','A'); % A is real 4x4-Matrix

E=evalin('base','E'); % E is real 4x4-Matrix, where all entries are 0, entry (i,j) is 1

f=((t^(k+1))/factorial(k+1))*kommpo2(E,A,k);

________

innerfuntkion %% commutator potential

function f = kommpo2(E,A,k)

a=E;

for i=1:1:k

a=komm(a,A);

end

f=a;

__

%% innerfunciton matrix commutator regular

function f = komm(E,A)

f=E*A-A*E;

__

I am running the symsum function in order to receive a sum > syms k; symsum(partko(k),0,Inf)

I seems, the inner function is to complicated for matlab to evaluate with symsum, since the following erroer message appears, even if sum is finite!:

EDU>> symsum(partko(k),0,10) Error using sym/colon (line 27) Cannot compute the number of steps from 1 to k by 1.

Error in kommpo2 (line 4) for i=1:1:k

Error in partko (line 6) f=((t^(k+1))/factorial(k+1))*kommpo2(E,A,k);

How can I modify my application? Anny ideas? Thank you very much.

Answer 1

[Walter Roberson]

You have the line

   for i=1:1:k

In order for that to be calculated, k has to have a definite value. But it doesn't: it is the "syms k".

Please remember that when you have the line

symsum(partko(k),0,10)

that that means that the expression partko(k) should be evaluated, and that the result should be passed to symsum().

[Michael S]

Of course. At the end I need a sum over the function partko(k), where kommpo2(k) is evaluated.

I expext symsum to send k from (0:inf). What would you suggest to do?

[Walter Roberson]

?? Commutating an infinite number of times is going to require an infinite time to process ?? Did you mean from (0:10) ?

It appears to me that what you want is probably to add those terms rather than use symsum(). symsum() does not add the terms: instead it finds the generalized summation expression for an indefinite variable value followed by substituting in the given limits -- similar in spirit to how one forms a definite integral. And just like how an indefinite integral has a "+ C" term that is expected to cancel out when subtracting the values at the two limits, so it is with symsum() -- that there might be an error in the lower terms (compared to just adding up the terms) that cancels out when one goes high enough. So if you have a particular number of terms (e.g., 0 to 10) then evaluate them individually and sum the result, rather than using the asymptotic symsum()

[Michael S]

I don't know, which term is equal to 0. Thats why I want to use summation till Inf. If symsum is not intented for this, fair enough. Is there a summation function which can do this, without individually/manually choosing the last term of my sum?

[Walter Roberson]

It sounds as if you do not need to work symbolically at all, and instead just need a "while" loop.

[Michael S]

Ok, I made a while-loop and it seems to work out:

A00=zeros(4,4);
  l=0;
  while (norm(partko(l))~= 0)
  A00 = A00+partko(l);
  l=l+1;
end