9

How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ using the fact that $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$ ?

So, $$e^x=\lim_{n\to\infty }\sum_{k=0}^n \binom{n}{k}\left(\frac{x}{n}\right)^k=\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$

I think that I have to prove that $\frac{n!}{(n-k)!n^k}=1$ but I didn't have success.

Ennar
  • 24,364
idm
  • 12,049

4 Answers4

26

1. Solution using $\,e^{\,x} = \dfrac{d}{dx}\,e^{\,x}\,$ property of exponent

Assume that the exponent function can be represented as a series with unknown coefficients: $$ e^{\,x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots = \sum_{n=0}^{\infty} a_n x^n $$

Recall the fundamental property of exponent $\, \dfrac{d}{dx} \big(e^{\,x} \big) = e^{\,x}$. Applying this property to the series for of exponent, we get \begin{align} \dfrac{d}{dx} \,e^{\,x} = e^{\,x} & \implies \dfrac{d}{dx} \left(\sum_{n=0}^{\infty} a_n x^n \right) = \sum_{n=0}^{\infty} a_n x^n \\ & \implies \dfrac{d}{dx} \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \\ & \implies 0 + a_1 + 2 \, a_2\, x + 3 \, a_3\, x^2 + \ldots = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \\ & \implies \sum_{n=1}^{\infty}n \,a_n\,x^{n-1} = \sum_{n=0}^{\infty} a_n x^n \iff \sum_{n=0}^{\infty}\left(n+1\right) a_{n+1}\,x^{n} = \sum_{n=0}^{\infty} a_n x^n \\ & \implies \left(n+1\right)a_{n+1} = a_n \\ & \implies a_{n+1} = \frac{a_n}{n+1} \end{align} The last equation can be rewritten as $\,a_{n} = \dfrac{1}{n}a_{n-1}, \,$ so that $$ a_{n} = \frac{1}{n}\,a_{n-1} = \frac{1}{n}\,\frac{1}{n-1} \,a_{n-2}= \frac{1}{n}\,\frac{1}{n-1}\,\frac{1}{n-2}\,a_{n-3} = \ldots = \frac{1}{ n!}\,a_0\tag{1.1} $$ Observer that $\,e^0 = 1,\,$ so we can write $$ e^{\,x}\Big\rvert_{x=0} = \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big)\Big\rvert_{x=0} = a_0 = 1 $$ This fact combined with equation $(1.1)$ gives us the explicit expression for coefficient $\,a_n = \dfrac{1}{n!}.\,$ Therefore we finally write $$ \bbox[4pt, border:2.5pt solid #FF0000]{\ \ e^{\,x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\ \,} $$ Q.E.D.

EDIT: As requested in comments, here I provide solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ expression.

2. Solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ property of exponent

I do not believe that it is possible (at least within reasonable timespan) express the exponentsas $\,e^{\,x}=\sum_{k=0}^\infty \frac{x^k}{k!}\,$ using only algebraic operations and the expression formula $\, e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ as starting point. However, it is possible to show the equivalence of these two definitions of $\,e^{\,x}\,$ as $\,n\to \infty.\,$

Indeed, for any $x\ge 0$ let us define $$ S_n = \sum_{k=0}^n \frac{x^k}{k!}, \qquad L_n = \left(1+\frac{x}{n}\right)^n. $$ Then, by Newton's binomial $$ \begin{aligned} L_n & = \sum_{k=0}^n {n\choose k} \,\frac{x^k}{n^k} = 1 + x + \sum_{k=2}^{n} \frac{n\cdot \left(n-1\right)\cdot \left(n-2\right)\cdot \ldots\cdot \left(n-(k-1)\right)}{k! \,n^k}= \\ & = 1 + x + \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) + \frac{x^3}{3!}\,\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) + \ldots + \frac{x^n}{n!}\,\left(1 - \frac{1}{n} \right) \cdots \left(1 - \frac{n-1}{n} \right) \leq S_n \end{aligned} $$ Therefore $$ \limsup_{n\to\infty}L_n \leq \limsup_{n\to\infty}S_n = e^{\,x}.\tag{2.1} $$

On the other hand, for any positive integer $\, m\,$ such that $\,2\le m \le n\,$ we have $$ 1 + x + \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right) + \ldots + \frac{x^m}{m!}\,\left(1 - \frac{1}{n} \right)\left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right) \le L_n $$ If we fix $\,m\,$ and let $\,n\to\infty,\,$ then we get $$ S_m = 1 + x + \frac{x^2}{2!} + \ldots + \frac{x^m}{m!} \leq \liminf_{n\to\infty}L_n\tag{2.2} $$ Letting $\,m\to\infty\,$ in inequality $(2.2)$ and combining it with inequality $(2.1)$, we get $$ e^{\,x} = \limsup_{n\to\infty}L_n\leq \lim_{n\to\infty} S_n \leq \liminf_{n\to\infty}L_n = e^{\,x} $$ and thus $$ \bbox[5pt, border:2.5pt solid #FF0000]{\lim_{n\to \infty}S_n = \sum_{n=0}^\infty \frac{x^n}{n!}=e^{\,x}} $$

Vlad
  • 6,938
  • How would you prove that the exponential function is indeed analytic? – Gabriel Romon Aug 23 '15 at 18:41
  • @LeGrandDODOM That depends on which of several equivalent definitions of exponential function do you choose, and which domain do you consider. In this specific case $e^x$ is defined on reals, and we could simply show that the function is infinitely differentiable and that its Taylor series converges to the function. The former is obvious, the latter follows directly from the way we constructed our series. Our series happened to coincide with Taylor series for Exponent, so the function is indeed analytic. If we were unable to construct such series, we would not have analyticity of $e^x$. – Vlad Aug 23 '15 at 19:05
  • 1
    I want to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ from $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$, so you don't answer to my question. – idm Aug 23 '15 at 19:49
  • @idm I am sorry my solution is unsatisfactory for you as it doesn't use $e^x=\lim_{n\to\infty}(1+x/n)^n$. However, I have to point out that you have not specified this requirement in your original post. The only explicit question you are asking is "How to prove that $e^x=\sum{k=0}^\infty x^k/k!$?"_ The rest of your post looks like attempted solution; it is not clearly stated that the proof has to rely on a specific expression. Please edit your post in order to make clear all the requirement for the solution. Otherwise people may keep waisting time typing invalid solutions for you. – Vlad Aug 23 '15 at 20:51
  • @Vlad: Thank you for your comment, I corrected it :-) – idm Aug 23 '15 at 21:23
  • @idm No problem. I have added the second version of the solution in my post. Is that what you was looking for? – Vlad Aug 23 '15 at 22:59
  • Nice ;-) you miss a limit in the last formula though :-) – Ant Aug 23 '15 at 23:49
  • @Ant Thank you for pointing it out! – Vlad Aug 24 '15 at 00:01
  • @Vlad Part 2 was really very well done!! +1! And I hope that you don't mind that I took the liberty of correcting a typo - the former definition of $S_n$ had no dependence on $n$ since the upper on the sum was $\infty$. – Mark Viola Aug 24 '15 at 03:21
  • @Dr.MV Thank you for fixing it! – Vlad Aug 24 '15 at 03:29
  • In your first solution you need to prove that derivative of $e^{x}$ is $e^{x}$ from the definition $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$ and it is not that difficult to do so but also not obvious to omit. Moreover if we could show that $(e^{x})' = e^{x}$ then its series expansion follows from Taylor's theorem and there is no need for what you have done. – Paramanand Singh Aug 24 '15 at 03:32
  • @ParamanandSingh I don't believe that Vlad is starting with that definition of $e^x$, but rather one that asserts that it is a function with a power series and one that is its own derivative. That said, Part 2 answers the OP's question exceptionally well without the hand-waving argument for taking the limit of the binomial sum. – Mark Viola Aug 24 '15 at 03:36
  • @Dr.MV: Fully agree! that's why I did not comment about anything in part 2. I think he should clarify his assumptions in part 1. Rest is fine. – Paramanand Singh Aug 24 '15 at 03:38
  • @ParamanandSingh Vlad does state these assumptions clearly and upfront. – Mark Viola Aug 24 '15 at 03:39
  • @Dr.MV: Ok sir! my +1 for Vlad. But once you have $(e^{x})' = e^{x}$ I prefer to say that the series for $e^{x}$ exists by Taylor's theorem and I don't think there is a need to assume the existence of series and calculate the coefficients. Only when you don't have a suitable/easy mechanism for evaluation of derivatives at $x = 0$ you will want to resort to calculate the coefficients in this manner. – Paramanand Singh Aug 24 '15 at 03:43
  • @ParamanandSingh That is perfectly reasonable. – Mark Viola Aug 24 '15 at 03:45
  • @Vlad: Why do you take the $\limsup$ and the $\liminf$ and not directly the limit ? Because I think that $\lim_{n\to \infty }S_n$ and $\lim_{n\to\infty }L_n$ both exist. Thanks, – idm Sep 11 '15 at 21:22
  • @idm In section 2 technically we do not know from the beginning that $, \lim_{n\to \infty }S_n,$ exists. – Vlad Sep 12 '15 at 00:34
10

To complete your method, that is, without assuming the derivative of $e^x$, you just have to write the coefficient of $\frac{x^k}{k!}$ as $$\frac{n(n-1)(n-2)...(n-k+1)(n-k)!}{(n-k)!n^k}$$ $$=(1)(1-\frac 1n)(1-\frac 2n)...(1-\frac{k-1}{n})\rightarrow1$$ as $n\rightarrow\infty$

David Quinn
  • 35,087
  • 1
    The sum of interest extends from $k=0$ to $k=n$. So, what happens to the limit (as $n\to \infty$) of the term in the post herein for $k=n$, $k=n-1$, $k=n-N$ for any fixed $N$? – Mark Viola Aug 23 '15 at 21:27
  • As Euler put it, when $j$ is infinite, $\frac{j-1}j=1$. Euler played a lot with infinities. He literally meant that $j$ was an infinite number. I'm pretty sure he knew that that wasn't rigorous; I'm not sure he cared. This is the same man that wrote $\ln p=\dfrac{p^0-1}0$, an equation I find delightful. – Akiva Weinberger Aug 24 '15 at 03:13
  • (The "true" interpretation of that equality is $\displaystyle\ln p=\lim_{\epsilon\to0}\frac{p^\epsilon-1}\epsilon$, by the way.) – Akiva Weinberger Aug 24 '15 at 03:14
9

Assume that we know nothing about the exponential function and the behavior of the limit of $(1+x/n)^n$ as $n\to\infty$. How would we establish the convergence and its equivalence to the power series $f(x)$ given by

$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} \ ? $$

A rough idea may be to expand using binomial theorem and take limit termwise, but such manipulation needs some justification since interchangng limitig operators may be invalid in some cases. (If you know the dominated convergence theorem, then you can utilize it in a straightforward way. But this is like nuking a mosquito.) So here is one possible solution to ths technical issue:

To this end, we utilize the binomial theorem to expand

$$\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^{\infty} \frac{n(n-1)\cdots(n-k+1)}{n^k} \frac{x^k}{k!}. $$

The last equality holds because $ n(n-1)\cdots(n-k+1) = 0$ for $k > n$. Now fix $N$, and for $n > N$ we decompose the difference as

$$ \left| \left(1+\frac{x}{n}\right)^n - f(x) \right| \leq \sum_{k=0}^{N} \left| \frac{n(n-1)\cdots(n-k+1)}{n^k} - 1 \right| \frac{|x|^k}{k!} + 2 \sum_{k=N+1}^{\infty} \frac{|x|^k}{k!}. $$

Taking limsup as $n \to \infty$, we see that

$$ \limsup_{n\to\infty} \left| \left(1+\frac{x}{n}\right)^n - f(x) \right| \leq 2 \sum_{k=N+1}^{\infty} \frac{|x|^k}{k!}. \tag{1} $$

But the left-hand side does not depend on $N,$ so by taking $N\to\infty$, we find that the left-hand side of (1) is actually 0. This proves that

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

as desired. ////

Sangchul Lee
  • 181,930
1

My first thought would be to list defining properties of $f(x) = e ^{x}$ and prove that $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ shares these properties.

$$ f \left( 0 \right) = 1 \\ \frac{d}{dx} f = f $$

If $\frac{d}{dx} y = y$ , then $y = Ce^x$. If $y(0) = Ce^0$ = $1$, then $C = 1$. Therefore $e^x$ is the only function with both these properties.

Now it's simple enough to show that $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ has these same defining features.

crb233
  • 1,021
  • The trouble is to prove those properties from the definition as $\lim_{n \to \infty} (1 + x / n)^n$... – vonbrand Aug 23 '15 at 23:15
  • @vonbrand Well, evaluating $\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right) ^{n}$ at $x = 0$ is trivial. And, I believe it may be possible to say that $\frac{d}{dx} \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right) ^{n} = \lim_{n \to \infty} \frac{d}{dx} \left( 1 + \frac{x}{n} \right) ^{n}$. It is easy to show that this second expression is equal to the original limit. Is this correct? – crb233 Aug 24 '15 at 01:07