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I have the following:

$$\sum_{i=0}^n \sum_{k=0}^i a_i {i \choose k} (-b)^{i-k} x^k$$

I want to find a way to express $x^k$ in terms of the outer summation so it would be $x^i$. Is there any way to do this?

Feltie
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  • Is that what you mean? $$ \sum_{i=0}^n \sum_{k=0}^i a_i {i \choose k} (-b)^{i-k} x^k = \sum_{k=0}^n \sum_{i=k}^n a_i {i \choose k} (-b)^{i-k} x^k = \sum_{k=0}^n x^k \sum_{i=k}^n a_i {i \choose k} (-b)^{i-k} $$ – Clement C. Dec 04 '17 at 02:56
  • Ya, this is exactly what I needed. Could you explain why you can do $\sum_{k=0}^n \sum_{i=k}^n$ – Feltie Dec 04 '17 at 03:07
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    Rewrite the first double sum $$\sum_{i=0}^n \sum_{k=0}^n \mathbf{1}{k\leq i} [...]$$ are you OK with this step? If so, then you can switch the summations:$$ \sum{k=0}^n \sum_{i=0}^n \mathbf{1}{k\leq i} [...]$$ and /then/ finally get $$ \sum{k=0}^n \sum_{i=k}^n [...]$$ (where I write "[...]" for the actual summands, that I'm too lazy to type in the comment). I write $\mathbf{1}_A$ for the indicator of $A$, i.e. the function that is $1$ if $A$ is satisfied, and $0$ otherwise. – Clement C. Dec 04 '17 at 03:10
  • Ok I got it now. Thanks! – Feltie Dec 04 '17 at 03:12

1 Answers1

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I think the closest one can get, is by simplyfying the inner summation. Rearranging like this, $$\sum_{i=0}^n a_i \sum_{k=0}^i {i \choose k} (-b)^{i-k} x^k$$ Then using the binomial theroem, we can get $$\sum_{k=0}^i {i \choose k} (-b)^{i-k} x^k = (x-b)^{i}$$ $$\sum_{i=0}^n a_i (x-b)^{i}$$ Since we have no other information on $a_i$ we can not further manipulate this expression to yield a "nicer" form.

Andrew Kovács
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  • I was actually working backwards from $\sum_{i=0}^n a_i (x-b)^{i}$ trying to express it as something like $\sum_{i=0}^n c x^i$ so I would need to find a c – Feltie Dec 04 '17 at 03:04