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How may I derive the following identity?

$$ \Big( \sum_{n=0}^\infty c_n x^n \Big)^k = \sum_{n=0}^\infty \tilde{c}_n x^{n-1} $$

$$ \text{where } \ \tilde{c}_n = \sum_{\substack{n_1 \dots n_k \\ n_1+ \cdots+n_k +1 = n}} c_{n_1} \dots c_{n_k} \qquad$$

I have no idea how to proceed. Any suggestion is appreciated.

Reference: Page 4 of this document.

Nanashi No Gombe
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  • you meant $$ \tilde{c}^{(k)}n = \sum{\substack{n_1, \dots, n_k \ n_1+ \cdots+n_k = n}} c_{n_1} \dots c_{n_k} = \sum_{m=0}^n \tilde{c}m^{(k-1)}c{n-m}$$ the suggestion is to look at the case $k=2$ then $k=3$ – reuns Jun 22 '17 at 07:40
  • No, I meant what I wrote. That's what's written in the text I am reading. I will add the reference to my question. – Nanashi No Gombe Jun 22 '17 at 07:43
  • You meant $$\Big( \sum_{n=0}^\infty c_n x^n \Big)^k = \sum_{n=0}^\infty \tilde{c}_n x^{n}$$ – reuns Jun 22 '17 at 07:45
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    @user1952009 He meant what he meant. – Andreas Jun 22 '17 at 07:47
  • @Andreas what ? no the OP meant what I and you wrote – reuns Jun 22 '17 at 07:49

1 Answers1

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Just sort for powers of $x$: $$ \Big( \sum_{n=0}^\infty c_n x^n \Big)^k = \\ \sum_{n_1=0}^\infty c_{n_1} x^{n_1} \sum_{n_2=0}^\infty c_{n_2} x^{n_2} \cdots \sum_{n_k=0}^\infty c_{n_k} x^{n_k} \\ = \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_k=0}^\infty c_{n_1} c_{n_2} \cdots c_{n_k} x^{n_1 + n_2 + \cdots + {n_k} }\\ = \sum_{n=0}^\infty \tilde{c}_n x^{n}$$

where the $\tilde{c}_n$ are sums over all products $c_{n_1} c_{n_2} \cdots c_{n_k}$ where ${n_1} + {n_2} + \cdots {n_k}$ matches the power $n$.

Now, for your slightly different notation, you can shift $n$ by $1$ in the sum ${n_1} + {n_2} + \cdots {n_k}$ and in the power of $x$. Then you write the power $x^{n-1}$ and correspondingly you must match ${n_1} + {n_2} + \cdots {n_k} = n-1$ or ${n_1} + {n_2} + \cdots {n_k} +1 = n$.

Andreas
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