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We have the following expression: $$ A(x) = f'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j_1} \Phi(k,c,j) f^{k-c-j}g^{c-j} \sum_{l=1}^{n-1} {n \choose l}\rho(f,n-l,c-1)\rho(g,l,k-c) + g'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j_2} \Phi(k,c,j) f^{k-c-j}g^{c-j} \sum_{l=1}^{n-1} {n \choose l} \rho(f,n-l,c)\rho(g,l,k-c-1) $$ The sums $j_1$ and $j_2$ are at the moment the same summation (so I will represent there bounds as the variable $j$). They are defined by the following inequality: $$ k-c \ge j \ge 0 \quad \quad \quad (k-c \le c) $$ and $$ c \ge j \ge 0 \quad\quad\quad (c \le k-c) $$ Now i am trying to get the $l$ summation into the following form: $$ \sigma(f,g,n,k,c) = \sum_{l=1}^{n-1} {n \choose l} \rho(f,n-l,c)\rho(g,l,k-c) $$ Now i will show you how I accomplished this: \begin{align*} f'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j_1} \Phi(k,c,j) f^{k-c-j}g^{c-j} \sum_{l=1}^{n-1} {n \choose l}\rho(f,n-l,c-1)\rho(g,l,k-c) \\ = f'\sum_{c=0}^{n-2} \sum_{k=c+2}^n \sum_{j_1} \Phi(k,c+1,j) f^{k-c-j-1}g^{c-j+1} \sum_{l=1}^{n-1} {n \choose l}\rho(f,n-l,c)\rho(g,l,k-c-1) \\ = f'g\sum_{c=0}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j_1} \Phi(k+1,c+1,j) f^{k-c-j}g^{c-j} \sigma(f,g,n,k,c) \end{align*} Now for the other summation: \begin{align*} g'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j_2} \Phi(k,c,j) f^{k-c-j}g^{c-j} \sum_{l=1}^{n-1} {n \choose l} \rho(f,n-l,c)\rho(g,l,k-c-1) \\ =g'f\sum_{c=1}^{n-1} \sum_{k=c}^{n-1} \sum_{j_2}\Phi(k+1,c,j) f^{k-c-j}g^{c-j} \sigma(f,g,n,k,c) \end{align*} This is swell and all, but there is one problem with what I have done. I have no idea how to implement these boundary changed on my $j_1$ and $j_2$ summations. Can someone please explain how this can be done?

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  • $\begingroup$ Is it wrong to just have $$\sum_{j_i=0}^a$$ for each $j_i$ where $a=\min\{c,k-c\}$? $\endgroup$ – Adina Goldberg Sep 26 '15 at 20:07
  • $\begingroup$ @AdinaGoldberg Well this is new for me, Is this method of expressing the bounds of $j$ easy to manipulate with what I have done? $\endgroup$ – Eric L Sep 26 '15 at 20:10
  • $\begingroup$ It seems that the only summation you move is the summation over $l$, which has no relationship to the bounds on the $j_i$ summations. So it seems nothing should change about the bounds on the $j_i$. $\endgroup$ – Adina Goldberg Sep 26 '15 at 20:14
  • $\begingroup$ @AdinaGoldberg Well since the sum $j$ is relying on the values of $c$ and $k$ and since i shifted them over one, wouldn't that result in the some sort of difference with the $j$ bounds? $\endgroup$ – Eric L Sep 26 '15 at 20:17
  • $\begingroup$ Just note that $a$ is a function of $c$ and $k$, so it will change as $c$ and $k$ vary. Perhaps it would be good to write it as $a(c,k)$ instead to explicitly note the dependence. $\endgroup$ – Adina Goldberg Sep 26 '15 at 20:17
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It looks like for the $j_1$ sum you want the upper bound to be $a(c+1,k+1)$ and for the $j_2$ sum you want the upper bound to be $a(c,k+1)$, where $a(c,k) = \min\{c,k-c\}$.

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