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Suppose I have a summation that looks like \begin{align} \sum_{a=0}^{n/2} \sum_{b=0}^a \sum_{c=0}^{n-2b}\sum_{d=0}^c \alpha(a,b,c,d)f(c-2d), \end{align} where $\alpha$ and $f$ are functions of the indices $a,b,c,d$, and I want to change these summations in a particular way. I want the argument of the function $f$ to depend on just one index, and for the summation of that corresponding index to be independent of other indices. That is, I want to rewrite the above summation as \begin{align} \sum_w \sum_x \sum_y \sum_{z=-n/2}^{n/2} \alpha(w,x,y,y/2 - z) f(z), \end{align} where $w,x,y$ have replaced $a,b,c$. Furthermore, we see that $d = y/2 - z$.

I am having trouble figuring out the range of the three summations. I know there is some freedom involved here and I believe that the $w$ summation can be taken to be \begin{align} \sum_{w=0}^{n/2} \end{align} through its identification with the $a$-summation.

How can I systematically figure out what the remaining limits of summation are (for the $x$ and $y$ summations)?

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In this case it is not too hard to transform the index range so that $f$ is dependent on a single index variable only.

We obtain \begin{align*} \sum_{a=0}^{n/2}&\sum_{b=0}^a\color{blue}{\sum_{c=0}^{n-2b}\sum_{d=0}^c\alpha(a,b,c,d)f(c-2d)}\tag{1}\\ &=\sum_{a=0}^{n/2}\sum_{b=0}^a\sum_{0\leq d\leq c\leq n-2b}\alpha(a,b,c,d)f(c-2d)\tag{2}\\ &=\sum_{a=0}^{n/2}\sum_{b=0}^a\sum_{d=0}^{n-2b}\sum_{c=d}^{n-2b}\alpha(a,b,c,d)f(c-2d)\tag{3}\\ &\,\,=\sum_{a=0}^{n/2}\sum_{b=0}^a\color{blue}{\sum_{d=0}^{n-2b}\sum_{c=-d}^{n-2b-2d}\alpha(a,b,c+2d,d)f(c)}\tag{4}\\ \end{align*}

Comment:

  • In (2) we write the index range of the two inner sums in a somewhat more convenient way.

  • In (3) we exchange the order of summation of the two inner sums of (1).

  • In (4) we shift the index $c\to c-2d$ to start from $-d$. This way the argument of $f$ depends on one index variable only.

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