How can I prove that $e^x \cdot e^{-x}=1$ using Taylor series?

Question

When proving $e^x.e^{-x}=1$ by using Taylor series, there are infinite many terms of $e^x$ and $e^{-x}$.

Is there any fancy way to combine terms by terms to show that eventually it is equal to $1$?

Answer 1

$$\begin{align} e^x\,e^{-x}&=\sum_{m=0}^\infty\frac{x^m}{m!}\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}\\\\ &=\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(-1)^nx^{m+n}}{m!\,n!}\end{align}$$

Now, let $p=m+n$. Then, we can write

$$\begin{align} e^x\,e^{-x}&=\sum_{m=0}^\infty \sum_{p=m}^\infty\frac{(-1)^{p-m}x^{p}}{m!\,(p-m)!}\\\\ &=\sum_{p=0}^\infty \sum_{m=0}^p\frac{(-1)^{p-m}x^{p}}{m!\,(p-m)!}\\\\ &=\sum_{p=0}^\infty (-1)^px^{p}\sum_{m=0}^p\frac{(-1)^{m}}{m!\,(p-m)!}\\\\ &=\sum_{p=0}^\infty \frac{(-1)^px^{p}}{p!}\sum_{m=0}^p\binom{p}{m}\,(-1)^m\\\\ &=1+\sum_{p=1}^\infty \frac{(-1)^px^{p}}{p!}\left(\sum_{m=0}^p\binom{p}{m}\,(-1)^m\right) \tag 1\\\\ \end{align}$$

Noting that the binomial expansion of

$$\begin{align} 0&=(1-1)^p\\\\ &=\sum_{m=0}^p\binom{p}{m}(-1)^m \tag 2 \end{align}$$

Using $(2)$ in $(1)$ reveals $e^x\,e^{-x}=1$. And we are done!

Answer 2

Set $e^x=\sum_{n\ge 0}a_nx^n$, where $a_n=\frac{1}{n!}$. Similarly, set $e^{-x}=\sum_{n\ge 0} b_n x^n$, where $b_n=\frac{(-1)^n}{n!}$.

By the usual formula for multiplying power series, we have $$e^x e^{-x}=\sum_{i\ge 0}c_i x^i$$ where $$c_i=\sum_{j=0}^i a_jb_{i-j}=\sum_{j=0}^i \frac{1}{j!}\frac{(-1)^{i-j}}{(i-j)!}=\frac{(-1)^i}{i!}\sum_{j=0}^i(-1)^j{i\choose j}=\frac{(-1)^i}{i!}(1-1)^i=\begin{cases}0&i\ge 1\\1 & i=0\end{cases}$$ Note that $(1-1)^i=\sum_{j=0}^i(-1)^j{i\choose j}$ by the binomial theorem. In the special case $i=0$, there is just one summand, and that is $1$.

Answer 3

EDIT: As per the comments discussion my answer is perhaps not in the spirit of the exercise and my argument can be made cleaner, but I'll keep it as is.

Here is a method which avoids the headache of formulas for multiplying infinte sums.

You could naively compute the Taylor series by computing the derivatives $$ \frac{d^k}{dx^k} e^{x} e^{-x} $$ so for $k=1$ you'd get $$ e^x e^{-x} - e^x e^{-x} =0 $$ for all $x$ and so all the derivatives vanish. Thus the taylor series for the function has only a constant term and thus $e^x e^{-x}$ is constant. If you were then to multiply the Taylor series for each product you'd quickly see that the only way for their product to have a constant term, i.e. a term with no $x^n$ is if you multiplied the constant terms of the two taylor series. Given that both taylor series have constant term $1$ you get the desired result.