Prove $\frac{1}{\exp(z)}=\exp(-z)$ by Taylor series

Question

Prove $$\frac{1}{\exp(z)}=\exp(-z)$$ by Taylor series.

Since $$\exp(z)=\sum_{k \geq 0}\frac{z^k}{k!}$$ What can we do with $\frac{1}{\exp(z)}$?

Answer 1

So you will need to make use of discrete convolution, i.e. a way to take the product of two series and end up with a series with two indices (think of it as an extension of polynomial multiplication, or in parallel with its continuous counter part, integral convolution).

The formula for convolution gives $$ \exp(z)\exp(-z)=\left(\sum_{i=0}^\infty\frac{z^k}{k!}\right)\left(\sum_{i=0}^\infty\frac{(-z)^j}{j!}\right)\\ =\sum_{k=0}^\infty\sum_{l=0}^k\frac{z^l(-z)^{k-l}}{l!(k-l)!} $$ now notice that the summand looks a lot like the binomial expansion of $$ (z-z)^k=0^k $$ but off by a $k!$. So we divide by $k!$ to compensate and find that the product is equal to $$ \sum_{k=0}^\infty\frac{0^k}{k!}=1 $$ $e^0=1$ if you like, or since $0^k=0$ for every $k>1$ but is $1$ for $k=0$.

Answer 2

Theorem $$e^{ax}.e^{bx}=e^{(a+b)x} $$

(We will use the series definition and the cauchy product of two series, it works on formal power series also)

Proof: We will take the product of the two series $e^{ax}=\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k}$ and $e^{bx}=\sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k}$

$$\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k} \sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k} =\sum\limits^{\infty}_{k=0}c_{k}x^{k} $$ where $c_{k}=\sum\limits^{k}_{p=0}\frac{a^{p}}{p!} \frac{b^{k-p}}{(k-p)!}$ multiplying by $k!$ we have $c_{k}k!=\sum\limits^{k}_{p=0}k!\frac{a^{p}}{p!} \frac{b^{k-p}}{(k-p)!}=\sum\limits^{k}_{p=0} {k \choose p}a^{p}b^{k-p}$ by the binomial theorem $c_{k}k!=(a+b)^{k}$ so $c_{k}=\frac{(a+b)^{k}}{k!}$, then we have

$$\sum\limits^{\infty}_{k=0}\frac{a^{k}}{k!}x^{k} \sum\limits^{\infty}_{k=0}\frac{b^{k}}{k!}x^{k} =\sum\limits^{\infty}_{k=0}\frac{(a+b)^{k}}{k!}x^{k} $$

Corollary

Take $a=1$, $b=-1$, then the result is $e^{0}=1$ by the series. Then $e^{-x}=\frac{1}{e^x}.$