I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$

So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$

But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$

Any idea ?

  • 1
    $\begingroup$ try the other way $\lim_{n \to \infty} (1+\frac{x}{n})^n(1+\frac{y}{n})^n$. Then ignore the higher order terms $\endgroup$ – user265328 Sep 25 '15 at 16:50
  • $\begingroup$ I got it, but I would like to conclude like this if it's possible :-) $\endgroup$ – idm Sep 25 '15 at 16:54
  • $\begingroup$ I think it does not matter. Anyways I will try to explain it to you without writing it mathematically. While going in the direction you want, we have to add and substract the terms we neglected to make $\lim_{n\to \infty} (1 + \frac{x}{n})^n(1+\frac{y}{n})^n = \lim_{n\to \infty} (1+\frac{x+y}{n})^n$. Then we will be left with two terms $(1+\frac{x+y}{n})^n - (something)$. This something has powers of $n$ in the denominator and will go to $0$ after taking the limit. I personally do not like this approach. $\endgroup$ – user265328 Sep 25 '15 at 17:11

My favorite approach to these problems is to first prove:

Lemma: If $g$ is a real-valued function on the natural numbers such that $$\lim_{n\to\infty}n(g(n)-1)=c$$ then: $$\lim_{n\to\infty} g(n)^{n} = e^c$$


Case c=0:

Restrict to $n$ so that $n|g(n)-1|<1$. Then: $$\begin{align} g(n)^n &= (1+(g(n)-1))^n\\ & = 1+ \sum_{k=1}^{n}\binom{n}{k}(g(n)-1)^k\end{align}$$ Since $\binom{n}{k}\leq \frac{n^k}{k!}$ and $\frac{1}{k!}\leq \frac{1}{2^{k-1}}$:

$$\begin{align} |g(n)^n-1| &\leq \sum_{k=1}^{n}\frac{1}{k!}\left(n|g(n)-1|\right)^k\\ &\leq n|g(n)-1|\sum_{k=1}^\infty \frac{1}{2^{k-1}}\\ &=2n|g(n)-1| \end{align}$$

So $\left|g(n)^n-1\right|\to 0$.

General case:

Let $h(n)=\frac{g(n)}{1+c/n}$. Then $$n(h(n)-1)=\frac{n(g(n)-1)-c}{1+c/n}\to 0.$$ So, by the case $c=0$, we have that $h(n)^n\to 1$. But we know that $(1+c/n)^n\to e^c$, so we are done.

The case $c=0$ used only binomial theorem and basic inequalities. The general case requires knowledge that $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n\to e^c.$

The lemma can be rewritten in little-$o$ notation as:

If $g(n)=1+\frac{c}{n}+o\left(\frac1n\right)$ then $g(n)^n\to e^c$.

The problem at hand

Next, show that, for fixed $x,y$ that: $$\begin{align}G(n)&=\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) \end{align}$$

satisfies the property that $n(G(n)-1)\to x+y$ since $$n(G(n)-1)=x+y+\frac{xy}{n}.$$

This shows what you wanted.

Another application: Euler's formula

If $h(x)$ is a function such that $h(x)=1+cx+o(x)$ then $h(x/n)^n\to e^{cx}$. This condition $h(x)=1+cx+o(x)$ is equivalent to $h(0)=1,h'(0)=c$.

One interesting case is $h(x)=\cos x+i\sin x$. We can show purely geometrically that $\cos x+i\sin x= 1+ix+o(x)$, and by induction prove that $h(x)^n=h(nx)$, so we have that $$\cos x+i\sin x =h(x)= \lim_{n\to\infty} h(x/n)^n=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$

The geometric proof is roughly, for all $x\in(0,\pi/2)$:

  1. $\sin x<x$. This is because $\frac{1}{2}\sin x$ is the area of a triangle contained in the circle wedge of angle $x$ which as area $x/2$.

  2. $0\leq 1-\cos x<\frac{1}{2}x^2$. This is true because $1-\cos x = 2\sin^2(x/2)<\frac{x^2}{2}$ by step (1).

  3. $x<\tan x$. This is because the wedge of angle $x$ is contained in the triangle $(0,0)$, $(1,0)$, $(1,\tan x)$ of area $\frac{1}{2}\tan x$.

  4. From (3) and (1), we get $x\cos x< \sin x<x$. But $\cos x =1+o(x)$ so $x\cos (x)=x+o(x^2)$ and thus $\sin(x)=x+o(x^2)$.

This shows that $\cos x +i\sin x = 1+ix + o(x)$.

  • $\begingroup$ what is $g(n)$ ? $\endgroup$ – idm Sep 25 '15 at 17:03
  • $\begingroup$ Any function with the property that I've stated. Edited the answer to make that clear. $\endgroup$ – Thomas Andrews Sep 25 '15 at 17:07
  • 1
    $\begingroup$ Well done! ... +1 $\endgroup$ – Mark Viola Sep 25 '15 at 21:01
  • $\begingroup$ Beautiful proof! $\endgroup$ – Hasan Saad Sep 25 '15 at 23:47
  • $\begingroup$ In past I had shown that $(1 + x/n)^{n}(1 + y/n)^{n} - (1 + (x + y)/n)^{n}$ tends to $0$ (see math.stackexchange.com/a/541330/72031), but it appears that showing $a/b \to 1$ is easier here than showing $a - b \to 0$. Plus derivation of the final result involving $\cos x + i\sin x$ is truly a gem. +1 $\endgroup$ – Paramanand Singh Sep 27 '15 at 10:37

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

  • 1
    $\begingroup$ It would facilitate to use $$1+\frac1n\le \left(1+\frac1{n^2}\right)^n\le 1+\frac2n$$and squeeze the second term in Lines $3-5$. $\endgroup$ – Mark Viola Sep 26 '15 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.