Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove that the following definitions of $e^x$ are equivalent, with as simple tools as possible and without any knowledge of $e$ or logarithms ?

$$\sum_{n=0}^{\infty} \frac{x^n}{n!}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$

Also preferably, prove that this is $a^x$ for some real number $a>0$.

share|improve this question
2  
You've tried using the binomial theorem on the right? –  J. M. Oct 4 '11 at 16:35
2  
A simple tool would be to note that both sides ostensibly are their own derivative and that they coincide at $0$, but then showing that these ostensible derivatives are actual derivatives would require somewhat less simple tools. –  joriki Oct 4 '11 at 16:38
    
In your last sentence, how do you want to define $a^x$? –  Chris Eagle Oct 4 '11 at 17:00
add comment

3 Answers 3

up vote 5 down vote accepted

In my answer to Combinatorial proof, I show that for $\displaystyle e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.

To show that $\displaystyle \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$, we could simply note that $$ \left(1+\frac{x}{n}\right)^n=\left(1+\frac{1}{n/x}\right)^{(n/x)x}\tag{1} $$ and just take the limit as $n\to\infty$. However, one might complain that $n/x$ is not an integer. To calm any such complaint, consider the sandwich $$ \left(1+\frac{1}{\lfloor n/x\rfloor+1}\right)^{(\lfloor n/x\rfloor+1)x-x}\le\left(1+\frac{1}{n/x}\right)^{(n/x)x}\le\left(1+\frac{1}{\lfloor n/x\rfloor}\right)^{\lfloor n/x\rfloor x+x}\tag{2} $$ It is easy to see that both $\lfloor n/x\rfloor$ and $\lfloor n/x\rfloor+1$ are integers and that $$ \lim_{n\to\infty}\left(1+\frac{1}{\lfloor n/x\rfloor+1}\right)^{-x}=\lim_{n\to\infty}\left(1+\frac{1}{\lfloor n/x\rfloor}\right)^x=1 $$ Thus, both the left and right sides of $(2)$ tend to $e^x$. Therefore we can use $(1)$ and just take the limit as $n\to\infty$.

share|improve this answer
add comment

Using binomial theorem: $$ \left(1+\frac{x}{n}\right)^n = \sum_{k=0}^n \binom{n}{k} \frac{x^k}{n^k} = \sum_{k=0}^n \frac{x^k}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} = \sum_{k=0}^n \frac{x^k}{k!} \prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right) $$

For each fixed $k$, the limit, as $n\to \infty$, is $\frac{x^k}{k!}$ which establishes equality of formal series.

But these both the series and the limit are known to converge for all $x$.

Also, see if the following $\lim_{n\to \infty} \left( 1 + \frac{x}{n} \right)^n = \left(\lim_{n\to \infty} \left( 1 + \frac{x}{n} \right)^{n/x} \right)^x $ for $x \not= 0$ suggests you the value for the limit.

share|improve this answer
6  
Don't you need to argue with uniform convergence to interchange the limits? –  joriki Oct 4 '11 at 16:49
1  
@joriki Are you talking about the first, or the last limit ? If it is the first one, then my intention was to consider formal series, not worrying about convergence, and justify the equality post-factum, since the lhs is a convergent series. Arguing by uniform convergence would have been more rigorous. –  Sasha Oct 4 '11 at 17:00
add comment

I don't disagree with the other answers but I always found that these kinds of limits and interesting properties came much more naturally from what I thought the definition of $e$ really was: that $e^x$ is its own derivative.

From that starting point, let $y=e^x$, so $dy/dx = e^x$. Now, $x = \ln(y)$. And $dx/dy = 1/e^x = 1/y$. So the derivative of $\ln(x)$ is $1/x$.

By first principles, you get:

\begin{equation} \frac{d\ln(x)}{dx}=\lim_{h->0}\frac{\ln(x+h)-\ln(x)}{h}=\lim_{h->0}\frac{\ln((x+h)/x)}{h}=\lim_{h->0}\ln((1+h/x)^{1/h})=1/x \end{equation}

So swapping $x$ with $1/x$ and $h$ with $1/n$:

\begin{equation} x = \lim_{n->\infty}\ln((1+x/n)^n) \end{equation}

Exponentiating:

\begin{equation} e^x = \lim_{n->\infty}(1+x/n)^n \end{equation}

Obviously you just let $x=1$ to get a limit for $e$. For the other side, it's just the Taylor series. The series is super-differentiable and its derivative is itself. In fact, all of $e^x$'s derivatives are $1$ when $x=0$.

If you think about it, there can only be one function which is its own derivative and which goes through (0, 1). Technically, it's a first order differential equation with one initial condition. You can easily approximate it with arbitrary precision by programming or using a spreadsheet: https://docs.google.com/spreadsheet/ccc?key=0Am_ePpIZW9YMdFI3dFlHOFoxWnpXOTVvWnh5X3FOeGc&hl=en_US

You could also just show that the limit on the right is its own derivative as well.

Alternate method using first principles with $e^x$:

\begin{equation} \frac{de^x}{dx}=\lim_{h->0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h->0}\frac{e^{h}-1}{h} \end{equation}

So:

\begin{equation} \lim_{h->0}\frac{e^h-1}{h}=1 \end{equation}

Rearranging to make $e$ the subject:

\begin{equation} e=\lim_{h->0}(h+1)^{1/h}=\lim_{n->\infty}(1+\frac{1}{n})^n \end{equation}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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