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I am trying to prove that $(1 + f(x))^a(1 + f(x))^b = (1 + f(x))^{a+b}$ in the world of formal power series. At a certain point in the prove I get

\begin{align*} (1 + f(x))^a(1+f(x))^b& = \ldots \\ & = \lim\limits_{N \to \infty} \left(\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right]\right) \\ & = \lim\limits_{N \to \infty} \left(\left[\binom{a}{0} \binom{b}{0} \right] f(x)^0 + \left[\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1}\right] f(x)^1 + \ldots \right) \\ & = \lim\limits_{N \to \infty} \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n \end{align*}

I tried $\binom{a}{0} \binom{b}{0} = \frac{1}{0!} \cdot \frac{1}{0!} = \frac{(a+b)^{\underline{0}}}{0!}$, $\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1} = \frac{a}{1!} \cdot \frac{1}{0!} + \frac{1}{0!} \frac{b}{1!} = \frac{(a+b)^{\underline{1}}}{1!}$, $\ldots$, but I don't seem to be able to generalize and prove this.

Could anybody get me in the right direction of how to prove that $\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right] = \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n$?
Or am I approaching this problem the wrong way?

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    $\begingroup$ What do "this" and "that" refer to? $\endgroup$ – joriki Jun 13 '16 at 12:00
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The development chain goes like this: $$ \eqalign{ & \left( {1 + f(x)} \right)^a \left( {1 + f(x)} \right)^b = \sum\limits_k {\left( \matrix{ a \cr k \cr} \right)f(x)^k } \sum\limits_j {\left( \matrix{ b \cr j \cr} \right)f(x)^j } \cr & = \sum\limits_j {\sum\limits_k {\left( \matrix{ a \cr k \cr} \right)\left( \matrix{ b \cr j \cr} \right)f(x)^k } } f(x)^j = \cr & = \sum\limits_n {\left( {\sum\limits_k {\left( \matrix{ a \cr k \cr} \right)\left( \matrix{ b \cr n - k \cr} \right)} } \right)\;} f(x)^n = \cr} $$ here the Chu-Vandermonde identity comes into play,(https://en.wikipedia.org/wiki/Vandermonde's_identity) giving: $$ \eqalign{ & = \sum\limits_n {\left( \matrix{ a + b \cr n \cr} \right)\;} f(x)^n = \cr & \left( {1 + f(x)} \right)^{a + b} \cr} $$

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  • $\begingroup$ Any hints on why the Cauchy product is applicable with those power series $f(x)$? $\endgroup$ – msBlümchen Jun 14 '16 at 8:15
  • $\begingroup$ @msBlümchen sorry, I do not catch the actual content of your question, pls. reword more explicitely $\endgroup$ – G Cab Jun 14 '16 at 16:12
  • $\begingroup$ could you provide any insights on why the Cauchy product of limits of power series has such a nice form? $\endgroup$ – msBlümchen Jun 18 '16 at 8:04
  • $\begingroup$ I think you can find enough explanations in the two Wikipedia's articles on "Cauchy product" and "Vandermonde" plus "Chu-Vandermonde" convolution. If still you have some doubts, since space here is limited, you shall focus on which pass you have doubts: product -> convolution, convolution of binomial coefficients, ...? $\endgroup$ – G Cab Jun 18 '16 at 10:12

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