On Wed, Mar 24, 2010 at 11:55 AM, Ondrej Certik <ond...@certik.cz> wrote:
> On Wed, Mar 24, 2010 at 11:09 AM, Ondrej Certik <ond...@certik.cz> wrote:
>> On Wed, Mar 24, 2010 at 11:01 AM, Freddie Witherden
>> <fred...@witherden.org> wrote:
>>> Hi all,
>>>
>>> After a discussion on IRC I am interested if there is any way to evaluate
>>> sums of the form:
>>>
>>>>>> B = [ Symbols, expressions, ... ]
>>>>>> Sum(n*B[n]*Sqrt(B[n]), (n, 0, len(b)-1))
>>>
>>> I ask because the project I am currently working on is based heavily around
>>> sums of this form:
>>> sum(sum(sum(sum(B[i]*A[i,j]*A[j,k]*A[k,l]*C[l],l,1,s),k,1,s),j,1,s),i,1,s)
>>> = 1/120
>>>
>>> where s, A[i,j] (matrix) and C[i] (vector) are evaluate to Real numbers
>>> with just B[i] being Symbols (so the result of the above expression is an
>>> equation of the form x1*B[1] + x2*B[2] + ... + xn*B[n] with xn being Reals
>>> (of course in Python the indices would be slightly different).
>>>
>>> Can anyone think of a clean way to evaluate these sorts of sums? (Extra
>>> points if you know where such sums arise in practise ;)
>>>
>>> Currently my workflow is:
>>> 1) Use Python to generate the sums;
>>> 2) Evaluate using Maxima;
>>> 3) Paste back into a Python script, parse with regex, solve.
>>
>> Can you paste your whole code so that we can try it? So your problem
>> is that some (complicated?) sums can't be evaluated in sympy, but it
>> works in maxima?
>>
>> Paste the exact expressions, let's see what can be done.
>
> Ok, after chatting at IRC a bit, here are the examples of expressions
> in maxima, that you need to work in sympy:
>
> sum(sum(sum(sum(B[i1]*A[i1,i2]*A[i2,i3]*A[i3,i4]*C[i4]^1, i4, 1, i3),
> i3, 1, i2), i2, 1, i1), i1, 1, s) - 1/120
> sum(sum(sum(B[i1]*A[i1,i2]*A[i2,i3]*C[i3]^2, i3, 1, i2), i2, 1, i1),
> i1, 1, s) - 1/60
> sum(sum(sum(B[i1]*A[i1,i2]*C[i2]^1*A[i2,i3]*C[i3]^1, i3, 1, i2), i2,
> 1, i1), i1, 1, s) - 1/40
> sum(sum(sum(B[i1]*C[i1]^1*A[i1,i2]*A[i2,i3]*C[i3]^1, i3, 1, i2), i2,
> 1, i1), i1, 1, s) - 1/30
> sum(sum(B[i1]*A[i1,i2]*C[i2]^3, i2, 1, i1), i1, 1, s) - 1/20
> sum(sum(B[i1]*C[i1]^1*A[i1,i2]*C[i2]^2, i2, 1, i1), i1, 1, s) - 1/15
> sum(sum(sum(B[i1]*A[i1,i2]*C[i2]^1*A[i1,i3]*C[i3]^1, i3, 1, i1), i2,
> 1, i1), i1, 1, s) - 1/20
> sum(sum(B[i1]*C[i1]^2*A[i1,i2]*C[i2]^1, i2, 1, i1), i1, 1, s) - 1/10
> sum(B[i1]*C[i1]^4, i1, 1, s) - 1/5
>
> e.g.:
>
> (%i1) s : 17;
> (%o1) 17
> (%i2) sum(B[i1]*C[i1]^4, i1, 1, s) - 1/5;
> 4 4 4 4 4 4 4
> (%o2) B C + B C + B C + B C + B C + B C + B C
> 17 17 16 16 15 15 14 14 13 13 12 12 11 11
> 4 4 4 4 4 4 4 4 4
> + B C + B C + B C + B C + B C + B C + B C + B C + B C
> 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2
> 4 1
> + B C - -
> 1 1 5
>
>
> Yes, this should be possible.
Here is one way (thanks again to Fernando for implementing the symarray!):
In [1]: s = 17
In [2]: Bi1 = symarray(s, "B")
In [3]: Ci1 = symarray(s, "C")
In [4]: Add(*(Bi1*Ci1**4)) - S(1)/5
Out[4]:
4 4 4 4 4 4 4
-1/5 + B₀⋅C₀ + B₁⋅C₁ + B₁₀⋅C₁₀ + B₁₁⋅C₁₁ + B₁₂⋅C₁₂ + B₁₃⋅C₁₃ + B₁₄⋅C₁₄
4 4 4 4 4 4 4 4
+ B₁₅⋅C₁₅ + B₁₆⋅C₁₆ + B₂⋅C₂ + B₃⋅C₃ + B₄⋅C₄ + B₅⋅C₅ + B₆⋅C₆ + B₇⋅C₇ +
4 4
B₈⋅C₈ + B₉⋅C₉
Ondrej
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