I'm trying to prove that $e^{-x}=1/(e^{x})$ using power series. Here's what I have thus far.
I want to show that
$\sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = (\sum_{n=0}^{\infty} \frac{(x)^n}{n!})^{-1}$
The left hand side is clearly: $1 – x + x^2/2! – x^3/3! + … – … +…$
The right hand side is the inverse of $(\sum_{n=0}^{\infty} \frac{(x)^n}{n!})$, meaning $(\sum_{n=0}^{\infty} \frac{(x)^n}{n!})*(\sum_{n=0}^{\infty} \frac{(x)^n}{n!})^{-1} = 1$.
Therefore, it must be true that (if $a_0+ a_1 + a_2…$ is the sequence for e^x which is known, and $b_0 + b_1 + b_2…$is the sequence for $1/e^{x}$ which we don't know)
$a_0b_0 = 1$
$a_1b_0 + a_0b_1 = 0$
$a_2b_0 + a_1b_1 + a_0b_2 = 0$
…
Therefore, $b_0 = 1$.
Then we can see from the second line, using what we found out about $b_0$, that $b_1= -x$. Next we can use that information to see that $b_2 = x^{2}/2$. In this way it seems clear that the sequence for $e^{-x}$ is exactly what we found for the left hand side of the generating function equation. However, saying that it looks like the same thing is not proof, and I'm not sure how to prove this.
Best Answer
Define the function $$\exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!},$$ and consider the product $$\exp(z) \exp(-z) = \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^m z^{k+m}}{k! m!}.$$ If we collect like powers of $z$; i.e., letting $n = k+m$, this double sum is equivalent to $$\exp(z) \exp(-z) = \sum_{n=0}^\infty \sum_{m=0}^n \frac{(-1)^m}{m! (n-m)!} z^n = \sum_{n = 0}^\infty c_n \frac{z^n}{n!},$$ where $$c_n = \sum_{m=0}^n (-1)^m \binom{n}{m}.$$ Therefore, it suffices to show that $c_0 = 1$ and $c_n = 0$ for all $n \in \mathbb Z^+$. The zero case is trivial. The positive case is a direct corollary of the binomial theorem: $$(x+y)^n = \sum_{m=0}^n \binom{n}{m}x^{n-m} y^m,$$ with the choice $x = 1, y = -1$: $$0^n = \sum_{m=0}^n \binom{n}{m} (-1)^m = c_n.$$ Therefore, $$\exp(z)\exp(-z) = 1.$$
It is worth noting that nowhere in the above proof do we make any claims that $\exp(z)$ is a function whose value is some constant raised to the power of $z$; i.e., we don't assume $\exp(z) = e^z$, and thus we do not use the multiplicativity of $\exp$. All that is being done here is to show that this particular power series satisfies the desired property.