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We are given two formal power series $\alpha(x) = \sum_{k=0}^{\infty}a_kx^k$ and $\beta(x) = \sum_{k=0}^{\infty}b_kx^k$ with coefficients in $\mathbb{Q}$ and $a_0=b_0 = 0$. I want to prove that for $\exp(x) = \sum_{k=0}^{\infty}\frac{x^k} {k!}$ we have

$$ \exp(\alpha(x)+\beta(x))=\exp(\alpha(x))\exp(\beta(x)) $$ But I am stuck in the middle of the calculation: $$ \exp(\alpha(x)\beta(x)) = \left(\sum_{k=0}^{\infty}\frac{\alpha(x)^k}{k!}\right)\left(\sum_{k=0}^{\infty}\frac{\beta(x)^k}{k!}\right) = \sum_{k=0}^{\infty}\sum_{n=0}^{k}\frac{\alpha(x)^{k-n}}{(k-n)!}\frac{\beta(x)^n}{n!} $$ And I don't know how to proceed. The notes I am reading, proceeds with $$ \left(\sum_{k=0}^{\infty}\frac{\alpha(x)^k}{k!}\right)\left(\sum_{k=0}^{\infty}\frac{\beta(x)^k}{k!}\right) = \sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n+j=k}\frac{k!}{j!n!}\frac{\alpha(x)^n}{n!}\frac{\beta(x)^j}{j!} = \sum_{k=0}^{\infty}\frac{(\alpha(x)+ \beta(x))^k}{k!} $$ which would conclude the proof but I can't see why the last two equalities are true.

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  • $\begingroup$ Did you mean$$\exp\bigl(\alpha(x)+\beta(x)\bigr)=\exp\bigl(\alpha(x)\bigr)+\exp\bigl(\beta(x)\bigr)?$$ $\endgroup$ – José Carlos Santos Jan 19 '18 at 8:53
  • $\begingroup$ That's a the Newton's binom formula $\endgroup$ – Atmos Jan 19 '18 at 8:55
  • $\begingroup$ Note that unless $a_0 = b_0 = 0$ this is not a formal power series calculation; the sums you write down won't converge $x$-adically. $\endgroup$ – Qiaochu Yuan Jan 19 '18 at 9:03
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By Cauchy's product $$ \left(\sum_{k=0}^{+\infty}\frac{\alpha\left(x\right)^k}{k!}\right)\left(\sum_{k=0}^{+\infty}\frac{\beta\left(x\right)^k}{k!}\right)=\sum_{k=0}^{+\infty}\left(\sum_{p+q=k}^{}\frac{\alpha\left(x\right)^p}{p!}\frac{\beta\left(x\right)^q}{q!}\right) $$ Hence $$ \sum_{k=0}^{+\infty}\left(\sum_{p+q=k}^{}\frac{\alpha\left(x\right)^p}{p!}\frac{\beta\left(x\right)^q}{q!}\right)=\sum_{k=0}^{+\infty}\left(\sum_{p=0}^{k}\frac{\alpha\left(x\right)^p}{p!}\frac{\beta\left(x\right)^{k-p}}{\left(k-p\right)!}\right) $$ That you can write $$ \sum_{k=0}^{+\infty}\left(\sum_{p=0}^{k}\frac{\alpha\left(x\right)^p}{p!}\frac{\beta\left(x\right)^{k-p}}{\left(k-p\right)!}\right)=\sum_{k=0}^{+\infty}\left(\sum_{p=0}^{k}\frac{1}{k!}\binom{k}{p}\alpha\left(x\right)^p\beta\left(x\right)^{k-p}\right) $$ Then you obtain

$$ \left(\sum_{k=0}^{+\infty}\frac{\alpha\left(x\right)^k}{k!}\right)\left(\sum_{k=0}^{+\infty}\frac{\beta\left(x\right)^k}{k!}\right)=\sum_{k=0}^{+\infty}\frac{\left(\alpha(x)+\beta(x)\right)^{k}}{k!}$$

Then the equality holds well.

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  • $\begingroup$ Thanks, this is helpful, although the professor didn't mention anything about Newton's binomial formula and binomial series were introduced after that. $\endgroup$ – laguna Jan 19 '18 at 9:14

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