# Proof of homomorphism property of the exponential function for formal power series

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.

• Did you mean$$\exp\bigl(\alpha(x)+\beta(x)\bigr)=\exp\bigl(\alpha(x)\bigr)+\exp\bigl(\beta(x)\bigr)?$$ – José Carlos Santos Jan 19 '18 at 8:53
• That's a the Newton's binom formula – Atmos Jan 19 '18 at 8:55
• 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. – Qiaochu Yuan Jan 19 '18 at 9:03

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