Symbolic summation: how to set this index?

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John
John 2019-1-23
评论： Walter Roberson 2019-1-25
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?
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madhan ravi
madhan ravi 2019-1-24
Have a look about symsum() , that maybe the one you are looking for.
John
John 2019-1-24
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.

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Walter Roberson
Walter Roberson 2019-1-25
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().
  2 个评论
John
John 2019-1-25
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
Walter Roberson 2019-1-25
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

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