Show identity for $e^x$

Question

I want to show that, using the definition $e^x := \lim_{n \to \infty} (1+\frac{x}{n})^n$, the following identity holds:

$$ e^{x+y}=e^xe^y$$

This is what I tried:

\begin{align} e^xe^y &=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n \\ &= \lim_{n \to \infty}\left(\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)\right)^n \\ &=\lim_{n \to \infty}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n \end{align}

However, this seems to be more than $e^{x+y}=\lim_{n \to \infty}(1+\frac{x+y}{n})^n$. How can I resolve this seeming contradiction to the upper identity?

@SeanRoberson But how could I get rid of the remaining terms that are too much? — MinecraftShamrock, Sep 05 '17 at 19:34
@MinecraftShamrock Use $(a+b)^n = a^n + n a^{n-1} b + \frac{n(n-1)}{2} a^{n-2} b^2 + \cdots$ for $a = 1 + \frac{x+y}{n}$ and $b = \frac{xy}{n^2}$. — Math Lover, Sep 05 '17 at 19:41
Use the fact (proved easily using binomial theorem) that if $n(a_{n} - 1)\to 0$ then $a_{n} ^{n} \to 1$ and set $a_{n} =(1+x/n) (1+y/n)/(1+(x+y)/n)$. — Paramanand Singh, Sep 05 '17 at 19:49

mechanodroid · Accepted Answer · 2017-09-05T21:21:17.440

For $x, y \geq 0$ we have: $$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) = 1 + \frac{x+y}{n} + \frac{xy}{n^2}\geq 1+\frac{x+y}{n}$$

For $n \in \mathbb{N}$, using the identity $u^n-v^n = (u-v)\sum_{k=0}^{n-1}u^{n-1-k}v^k$ we have:

\begin{align}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1 + \frac{x+y}{n}\right)^n &= \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^n - \left(1+\frac{x+y}{n}\right)^n \\ &= \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k \end{align}

Using the first inequality we get:

$$\frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} \leq \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1+\frac{x+y}{n}\right)^n \leq \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}$$

The left inequality is obtained by noticing that $\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right) = \left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) \geq \left(1+\frac{x+y}{n}\right)$ so by replacing the larger term with the smaller one in every product, we get a lower bound for the sum:

\begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\geq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1+\frac{x+y}{n}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k \\ &= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}\right)^{n-1} \\ &= \frac{xy}{n^2} \cdot n\,\left(1+\frac{x+y}{n}\right)^{n-1} \\ &= \frac{xy}{n} \left(1+\frac{x+y}{n}\right)^{n-1} \end{align}

Similarly, for the right inequality we replace the smaller term with the larger one: \begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\leq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^k \\ &= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\ &= \frac{xy}{n^2} \cdot n\,\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\ &= \frac{xy}{n} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\ &= \frac{xy}{n} \left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1} \end{align}

Letting $n\to\infty$ both sides converge to $0$ so by the squeeze theorem we have: $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1+\frac{x+y}{n}\right)^n = 0$$ So $$e^xe^y = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{x+y}{n}\right)^n = e^{x+y}$$

Just for clarification, the limits of the lower and upper bound are obtained using the limit of product and quotient theorem:

$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} = \lim_{n\to\infty} \frac{xy}{n} \frac{\left(1+\frac{x+y}{n}\right)^n}{\left(1+\frac{x+y}{n}\right)} = 0 \cdot\frac{e^{x+y}}{1} = 0$$

$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1} = \lim_{n\to\infty}\frac{xy}{n}\frac{\left(1+\frac{x}{n}\right)^{n-1}}{\left(1+\frac{x}{n}\right)}\frac{\left(1+\frac{y}{n}\right)^{n-1}}{\left(1+\frac{y}{n}\right)} = 0\cdot \frac{e^x}{1}\cdot\frac{e^y}{1} = 0$$

I don't understand by which reason the second formula is true. From the beginning, it is clear that $(1+\frac{x}{n})^n(1+\frac{y}{n})^n=(1+\frac{x+y}{n}+\frac{xy}{n^2})^n$. But as soon as you subtract something, the equality sign does no longer hold. — MinecraftShamrock, Sep 05 '17 at 20:27
@MinecraftShamrock You are right, of course. A term was missing. — mechanodroid, Sep 05 '17 at 20:30
Thank you. Also, how are your inequalities deduced from the first one? — MinecraftShamrock, Sep 05 '17 at 20:39
@MinecraftShamrock I have added an explanation, fixed a typo in the upper bound, and added clarification for the limits. — mechanodroid, Sep 05 '17 at 21:02
Aah, thanks a lot! I think the the two $\geq \leq$ signs are swapped. But I understand now. — MinecraftShamrock, Sep 05 '17 at 21:19

Show identity for $e^x$

1 Answers1