# Symbolic summation: how to set this index?

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John on 23 Jan 2019
Commented: Walter Roberson on 25 Jan 2019
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 on 24 Jan 2019
Strange. It's a LaTeX equation that on my end I can see properly (see print screen above)
madhan ravi on 24 Jan 2019
Have a look about symsum() , that maybe the one you are looking for.
John on 24 Jan 2019
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.

Walter Roberson on 25 Jan 2019
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.
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 on 25 Jan 2019
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 on 25 Jan 2019
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)