use power series expansion to prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$

Question

prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$

hint: If $\sum_{n=0}^{\infty} a_n \sum_{n=0}^{\infty} b_n = \sum_{n=0}^{\infty} c_n$ then

$ c_n=\sum_{k=0}^{n} a_k b_{n-k} $ provided that $ \sum_{n=0}^{\infty} a_n , \sum_{n=0}^{\infty} b_n$ are convergent

using power series expansion: $e^{z}= \sum_{n=0}^{\infty} {z^n\over n!}$

$e^{z_1}= \sum_{n=0}^{\infty} {z_1^n\over n!}$ and $e^{z_2}= \sum_{n=0}^{\infty} {z_2^n\over n!}$

$e^{z_1}e^{z_2}= \sum_{n=0}^{\infty} {z_1^n\over n!} \sum_{n=0}^{\infty} {z_2^n\over n!}=\sum_{k=0}^{n} {z_1^k\over k!} {z_2^{n-k}\over {n-k}!} = e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}$ [Using binomial thm.]

Now, $e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}=e^{z_1}e^{z_2}$ but iam not sure of the binomial thm part

Answer 1

Let $S_n(x)=\sum_{j=0}^n x^j/j!.$

Then $S_n(x)S_n(y)-S_n(x+y)=R_n(x,y)$ where $$R_n(x,y)=\sum_{j\le n\land k\le n\land j+k>n}x^jy^k/(j!k!).$$ The number of terms in the RHS above cannot be more than $(n+1)^2.$ And each term cannot have a modulus greater than $r^{2n}/m!$ where $r=\max (1,|x|,|y|),$ and (i) $m=1+n/2$ if $n$ is even, and (ii) $m=(n+1)/2$ if $n$ is odd.

So $|R_n(x,y)|\le (n+1)^2r^{2n}/m!,$ which $\to 0$ as $n\to \infty$ because $ m> n/2.$

Answer 2

We obtain \begin{align*} \color{blue}{e^{z_1}e^{z_2}}&=\sum_{k=0}^\infty\frac{z_1^k}{k!}\sum_{l=0}^{\infty}\frac{z_2^l}{l!}\\ &=\sum_{n=0}^\infty\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{z_1^kz_2^l}{k!l!}\tag{1}\\ &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{z_1^kz_2^{n-k}}{k!(n-k)!}\tag{2}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{n!}{k!(n-k)!}z_1^kz_2^{n-k}\right)\frac{1}{n!}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}z_1^kz_2^{n-k}\right)\frac{1}{n!}\\ &=\sum_{n=0}^\infty\frac{(z_1+z_2)^{n}}{n!}\tag{3}\\ &\,\,\color{blue}{=e^{z_1+z_2}} \end{align*} and the claim follows.

Comment:

• In (1) we change the order of summation by introducing $n$ and summing up according to increasing $n$.

• In (2) we set $l=n-k$.

• In (3) we apply the binomial theorem.