Exponential function of negative argument [closed]

Right, so what are we allowed to know?

Let's say we know that $$e^x=\sum_{i=0}^\infty\frac{x^i}{i!}$$ and $$e^{-x}=\sum_{i=0}^\infty\frac{(-1)^ix^i}{i!}$$

Multiply these together, and we get

$$e^xe^{-x}=\left(\sum_{i=0}^\infty\frac{x^i}{i!}\right)\left(\sum_{i=0}^\infty\frac{(-1)^ix^i}{i!}\right)$$

which equals

$$ \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{x^i}{i!}\frac{(-1)^jx^j}{j!} =\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^jx^{i+j}}{i!j!} $$

writing $k=i+j$, and we get

$$\sum_{k=0}^\infty\sum_{j=0}^k\frac{(-1)^jx^{k}}{(k-j)!j!}=\sum_{k=0}^\infty\sum_{j=0}^k\frac{x^k}{k!}\frac{(-1)^jk!}{(k-j)!j} =\sum_{k=0}^\infty\left[\sum_{j=0}^k\frac{(-1)^jk!}{(k-j)!j}\right]\frac{x^k}{k!}$$

The binomial theorem helps us simplify the square bracket. We know that

$$\sum_{j=0}^k\frac{a^jk!}{(k-j)!j}=(1+a)^k$$

with the special case $a=-1$ and $k=0$ yielding $1$. That is,

$$ \sum_{j=0}^k \frac{(-1)^j k!}{(k-j)! j} = \begin{cases} 1, & \text{if } k=0, \\ 0, & \text{otherwise}. \end{cases} $$

So $$\sum_{k=0}^\infty\left[\sum_{j=0}^k\frac{(-1)^jk!}{(k-j)!j}\right]\frac{x^k}{k!}=1{\cdot}\frac{x^0}{0!}+\sum_{k=1}^\infty0{\cdot}\frac{x^k}{k!}=1+0=1$$

That is, $e^xe^{-x}=1$, so we can conclude $e^{-x}=\frac1{e^x}$.

This property can be naturally generated from $\exp(x+y)=\exp(x)\exp(y)$, by taking $y=-x$ to show $\exp(x)\exp(-x)=\exp(0)=1$.

As for $\exp(x+y)=\exp(x)\exp(y)$, it is natural enough “matching resulting series between each other”. Or you can try to show that $f(y)=\dfrac{\exp(x+y)}{\exp(x)}$ satisfies the definition of $\exp(y)$.

You know that $u:=\exp$ is the unique solution of $f'-f = 0$ and $f(0) = 1$. Hence $v:x \mapsto \exp(-x)$ is the unique solution of $f'+f = 0$ and $f(0) = 1$ (both can be shown by differentiating the power series). Hence $uv: x\mapsto \exp(x)\exp(-x)$ is the unique solution of $f' = 0$ and $f(0) = 1$ since $(uv)' = u'v+uv' = uv - uv = 0$. But a solution is obviously also the constant function $1$. Hence $uv = 1$, i.e., $e^{-x} = \dfrac{1}{e^x}$ for all $x \in \mathbb{R}$.

Let us start with the definition of the $\exp$ function: \begin{align*} \exp(x) = \sum_{n=0}^{\infty}\frac{x^{n}}{n!} \end{align*} which converges for every $x\in\mathbb{R}$. Now we shall prove that $\exp(x + y) = \exp(x)\exp(y)$. Indeed, \begin{align*} \exp(x)\exp(y) & = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{y^{n - k}}{(n - k)!}\\ & = \sum_{n = 0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{k!(n - k)!}x^{k}y^{n - k}\\ & = \sum_{n = 0}^{\infty}\frac{(x + y)^{n}}{n!}\\ & = \exp(x + y) \end{align*} Thus, $1 = \exp(0) = \exp(x + (-x)) = \exp(x)\exp(-x)$, from whence the desired relation holds.

Hopefully this helps!

You can make an ansatz: invert the formal power series $\sum_{n \geq 0} a_nX^n$. So find $\sum_{n \geq 0} b_n X^n$ with $1 = (\sum_{n \geq 0} a_n X^n)(\sum_{n \geq 0} b_n X^n) = \sum_{n \geq 0} \sum_{k = 0}^n a_kb_{n-k} X^n$, i.e., $a_0 b_0 = 1$ and $\sum_{k=0}^n a_kb_{n-k} = 0$ for $n \geq 1$. Hence $b_0 = 1/a_{0}$ and $b_n = -\frac{1}{a_0}\sum_{k=1}^n a_k b_{n-k}$ for $n \geq 1$.

For $\exp$ it is $a_n = \frac{1}{n!}$. Hence $b_0 = 1/a_0 = 1$ and $b_n = -\sum_{k = 1}^n \frac{b_{n-k}}{k!}$. Hence $b_0 = 1$, $b_1 = -b_0/1! = -1$, $b_2 = -b_1/1! - b_0/2! = 1-1/2 = 1/2 = 1/2!$, $b_3 = -b_2/1!-b_1/2!-b_0/3! = -1/2+1/2-1/3! = -1/3!$, $\dots$ So claim that $b_{n} = (-1)^{n}/n!$ and show it by induction.