Symbolic summation: how to set this index?

I'm trying to write the following symbolic sum in matlab:

However, I can't define the function:

T=(a^(mod(j-3,n))+ a^(n-mod(j-3,n))+a^(mod(j-1-n,n)) + a^(n-mod(j-1-n,n)))*(a^(mod(j-1,n))+ a^(n-mod(j-1,n))+a^(mod(j+1-n,n)) + a^(n-mod(j+1-n,n)))

to compute

symsum(T,j,1,n)

Any help on how to make this calculation?

[John]

The equation appears in my preview. However, once I post the question, it becomes blurred. So I wrote it as a code as well.

[Walter Roberson]

The image file appears to be to a broken image.

[John]

Fixed that!

[madhan ravi]

Nope you didn't!

[John]

[John]

Strange. It's a LaTeX equation that on my end I can see properly (see print screen above)

[madhan ravi]

Have a look about symsum() , that maybe the one you are looking for.

[John]

I know about symsum. The issue is that I haven't been able to define the function to be used as an input in symsum. Anyway, I edited the question to make that point clearer.

Answer 1

[Walter Roberson]

symsum(), like the rest of MATLAB, evaluates the symbolic expression that is passed in as its first argument, before the function itself gets control. This is a problem with your formula because mod() with symbolic inputs is.. weird...

So... Don't.

Instead, construct vectors of powers:

NL = 1:n;
Nm3 = mod(NL-3,n);
Nm1 = mod(NL-1,n);
NNm3 = n - Nm3;
NNm1 = n - Nm1;
Nm1n = mod(NL-1-n, n);

and so on. Then

sum( (a.^Nm3 + a.^NNm3 + a.^Nm1n + a.^NNm1n) .* second_term )

I suggest that you only use symsum() when you are trying to find formula, such as symsum(1/2^x, 1, N) and expecting back the formula 1 - (1/2)^N .

When you have finite limits, it is almost always better to create a vector containing the individual terms, and sum() the vector.

If you were hoping that you could use symsum() to create an expression with generalized n so that you could reason with the formula, or fill in particular n values later, then you should probably give up all hope of that with symsum() or with symfun().

[John]

Must the value for n be set then? Leaving it as a symbolic integer, gives the error message "Unable to compute number of steps from 1 to n by 1."

[Walter Roberson]

Yes, the value for n must be set for use with the colon operator. And mod() with symbolic expressions is nearly useless:

syms n j
mod(j-2, 5)
mod(j-2, n)

ans =
 
j + 3
 
Error using symengine
Invalid second argument.

Error in sym/privBinaryOp (line 1002)
            Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});

Error in sym/mod (line 18)
C = privBinaryOp(A, B, 'symobj::zipWithImplicitExpansion', 'symobj::modp');

You can try rewriting in terms of floor: mod(A,B) -> A - floor(A/B)*B