# Proving $f(x+y)=f(x)f(y)$ for exponential function $f$

I'm sorry if the title was a bit misleading but I don't know how else to phrase it.

I'm having trouble understanding some lecture notes and I couldn't find extra information by Googling.

Anyways, the exponential function is represented as

$$\lim_{n}\left(1+\frac{x}{n}\right)^n$$

Then, this inequality follows:

$$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)=\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)\ge 1+\frac{x+y}{n}$$

Ok, following so far. Then there's this thing:

\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}\left[\sum\limits_{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\right] \end{align*}

That last one amost broke me. After figuring out it's just the formula for

$$a^n-b^n$$

I finally get to the part I just don't understand. It says "Because of the last inequality follows:"

\begin{align*} \frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1}&\le\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n-\left(1+\frac{x+y}{n}\right)^n\\ &\le\frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1} \end{align*}

I'm supposed to study for an exam on Monday and I still have a ton of stuff to go through but I've been staring at the page for the last 2 hours. I don't know if this is supposed to be one of the harder proofs or maybe I'm just missing something (in my brain).

Help is much appreciated.

P.S Sorry if my Latex is bad, this is the second time I'm using it.

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Your $\LaTeX$ was pretty good under the circumstances; if you want to see how I changed it, just right-click on any expression and select Show Math As and TeX Commands. (I inserted a missing exponent that I’m pretty sure was just a typo.) – Brian M. Scott Jan 12 '13 at 22:27

## 1 Answer

We have

\begin{align*} &\frac{xy}{n^2}\left[\sum\limits_{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\right]\\ &\qquad=\frac{xy}n\cdot\frac1n\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*}

Your first inequality indicates that you’re assuming that $x,y\ge 0$, so

$$\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\ge\left(1+\frac{x+y}n\right)^{n-1}$$

for $k=0,\dots,n-1$. There are $n$ terms, so

$$\frac1n\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\ge\left(1+\frac{x+y}n\right)^{n-1}\;,$$

and hence

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

The rest comes from your first inequality.

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Well, yoyo answered first with the same solution but you answered with a more detailed explanation. Which answer should I rate as the accepted one? – Luka Horvat Jan 12 '13 at 22:45
@Darwin: That’s entirely up to you. – Brian M. Scott Jan 12 '13 at 22:46