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I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done?

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This is usually done by restricting $x$ to the natural numbers and expanding the right hand side using the binomial theorem and some rather straightforward estimates. Any decent text on real analysis contains this. –  t.b. May 15 '11 at 8:02
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This identity, in isolation, is certainly confusing. It becomes less so once you know that it is just a special case of a more general identity

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

Why does this hold? Well, let's call the function on the left $e^x$. Then by inspection, $\frac{d}{dx} e^x = e^x$. General theory tells you that the solution to $y' = y$ is unique provided you fix the value of $y(0)$ (and this is also physically intuitive), and here $e^0 = 1$ by inspection, so that tells you that $e^x$ is the unique function satisfying $e^0 = 1$ and $\frac{d}{dx} e^x = e^x$.


Chandru1's deleted answer reminded me that the proof of this is actually completely elementary. Suppose $f$ is a function which satisfies $f'(x) = f(x)$. Then

$$\frac{d}{dx} \left( e^{-x} f(x) \right) = e^{-x} f'(x) - e^{-x} f(x) = 0$$

hence $e^{-x} f(x) = C_f$ for some constant $C_f$. Substituting $f(x) = e^x$ gives $e^{-x} e^x = C$, and substituting $x = 0$ gives $e^{-x} e^x = 1$. Now it follows that $f(x) = C_f e^x$.


Why does the function on the RHS also satisfy this property? Let's call this function $\exp(x)$. Then

$$\exp(cx) = \lim_{n \to \infty} \left( 1 + \frac{cx}{n} \right)^n = \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^{mc} = \exp(x)^c$$

where $m = \frac{n}{c}$. Setting $c = 1 + \frac{y}{x}$, this gives

$$\exp(x + y) = \exp(x)^{1 + \frac{y}{x}} = \exp(x) \exp(x)^{ \frac{y}{x} } = \exp(x) \exp(y)$$

from which it follows that

$$\frac{d}{dx} \exp(x) = \lim_{h \to 0} \frac{\exp(x) \exp(h) - \exp(x)}{h} = \exp(x) \lim_{h \to 0} \frac{\exp(h) - 1}{h}$$

so it follows that $\exp(x)$ satisfies $\exp(0) = 1$ and $\frac{d}{dx} \exp(x) = \exp(x)$, which means it must equal $e^x$ - at least, as soon as we know that

$$\lim_{h \to 0} \frac{\exp(h) - 1}{h} = 1.$$

But

$$\exp(h) = \lim_{n \to \infty} \left( 1 + \frac{h}{n} \right)^n = \lim_{n \to \infty} \left( 1 + {n \choose 1} \frac{h}{n} + {n \choose 2} \frac{h^2}{n^2} + ... \right) = \lim_{n \to \infty} \left( 1 + h + R(h, n) \right)$$

where $|R(h, n)| \le \sum_{k \ge 2} |h|^k = O(|h|^2)$ by the ratio test, from which the conclusion follows.


That last step, by the way, is an echo of the standard proof, where one expands

$$\left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$

and uses the fact that $\frac{1}{n^k} {n \choose k} \to \frac{1}{k!}$ as $n \to \infty$. While this is fairly hands-on, in my opinion it obscures an underlying principle behind this problem, which is that $e^x$ is a very special function and all of the equivalent ways of defining it are based on one of its special properties.

One really nice way to interpret the above identity is that it is precisely what you get by applying Euler's method to the ODE $y' = y$ with initial conditions $y(0) = 1$.

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At the 1st step, you have to force $0^0 = 0$; that's probably obtained with some limit argument. –  Luke May 15 '11 at 17:46
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No, my convention here is that $x^0 = 1$ for all $x$. That is standard for talking about polynomials and power series, and is justifiable in several other ways; I am never varying the exponent in a neighborhood of $0$ so arguments about indeterminate forms aren't relevant. –  Qiaochu Yuan May 15 '11 at 17:55
    
My bad. $0^0 \triangleq 1$, because of that very reasonable convention, so that every term, except for the first one, in $\sum_{n \ge 0} \frac{0^n}{n!}$ equals zero. Now everything makes sense. –  Luke May 15 '11 at 19:48
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Another way of viewing $e^{x}$, is if suppose $f : \mathbb{R} \to \mathbb{R}$ is a differentiable function which satisfies $f'(x) = f(x)$ for all $x \in \mathbb{R}$, then $f(x) = e^{x} + C$. To see this we have $f'(x) = f(x) \Longrightarrow e^{-x}f'(x) = e^{-x} f(x) \Longrightarrow e^{x} \cdot [f'(x) - f(x)]= 0 \Longrightarrow \frac{\rm d}{\rm dx } \Bigl[e^{-x}f(x)\Bigr]=0$ which says that $e^{-x} f(x) = C$. Hence $f(x) = Ce^{x}$.

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You need some words to justify going from the second line to the third line. It is not generally true that if $a_{n,m}$ is a sequence of sequences such that $\lim_{n \to \infty} a_{n,m} = a_m$, then $\lim_{n \to \infty} \sum_m a_{n,m} = \sum a_m$. –  Qiaochu Yuan May 15 '11 at 8:18
    
@Qiaochu: I have deleted that answer and added another one. Could you please elaborate as to what was wrong. –  user9413 May 15 '11 at 11:23
    
@Chandru1: exactly what I said. You implicitly assumed a statement that, in full generality, is false (take $a_{n,m} = 1$ if $n = m$ and $0$ otherwise). –  Qiaochu Yuan May 15 '11 at 12:10
    
@Qiaochu: But I really am not able to understand where to use your result. There is no $m$ and the quantity depends on $n$ only. –  user9413 May 15 '11 at 12:12
    
@Chandru1: in your application you had $a_{n,m} = \frac{1}{n^m} {n \choose m} x^m$. You wanted to claim that since $\lim_{n \to \infty} a_{n,m} = \frac{x^m}{m!}$, this implies that $\lim_{n \to \infty} \sum_m a_{n,m} = \sum_m \frac{x^m}{m!}$. You cannot deduce this without more information about how $a_{n,m}$ converges because, as I said, this result is false in full generality. –  Qiaochu Yuan May 15 '11 at 12:15
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