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I've been taught that the definition of the exponential function is the following power series: $$\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ Here's my question: how do we know that this series is equal to $e^x$? That is to say, how do we know that the function defined by this power series can be expressed in the form $c^x$ where $c$ is a constant and that $c=e$? If we look at another power series, say, $\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{2k!}$, there is obviously no constant $c$ for which $c^x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{2k!}$, so what tells us that there exists such a constant for the first series?

David Zhang
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  • Using Taylor's inequality, we can prove that the partial sum $\sum_{k=0}^n \frac{x^k}{k!}$ converges pointwise, and even uniformly over any compact subset of $\mathbb{R}$, towards $e^x$. – zuggg Jul 12 '13 at 06:45
  • This might help – Pedro Jul 12 '13 at 06:53
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    What's your definition of $c^x$? Some people define it to mean $e^{x \log c}$! – Qiaochu Yuan Jul 12 '13 at 07:34
  • As Qiaochu Yuan has pointed out, it all depends upon definition of exponentiation (and logarithms too). Using any standard definition like for example $$y = f(x) = \sum_{n = 0}^{\infty}\frac{x^{n}}{n!}$$ and inverse $ x = g(y)$ we can show that the definition $a^{b} = f(b\cdot g(a))$ satisfies all the properties which we associate with exponents. Under this definition we can see that $f(x) = {f(1)}^{x} = e^{x}$ if we define $e = f(1)$. – Paramanand Singh Jul 12 '13 at 08:11
  • If we don't want to use logarithms (function $g$ in previous comment) we can directly show that $f(a + b) = f(a) \cdot f(b)$ and thus $f(n) = {f(1)}^{n}$ for any integer $n$ and by a little more effort we can show that $f(r) = {f(1)}^{r}$ for any rational $r$. Defining $e = f(1)$ we can argue that the defining $e^{x}$ as $f(x)$ is a natural choice when $x$ is irrational. – Paramanand Singh Jul 12 '13 at 08:17

2 Answers2

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It is only after some manipulation that it becomes something familiar. Your other example of course is $\cos x$.
A common approach is to find what differential equation it satisfies. Let $y(x) = \sum_{k=0}^{\infty}\frac{x^k}{k!}$.
Differentiate it term by term, and you find $\frac{dy}{dx}=y(x)$.
That is a differential equation. It is separable, and you get $x=c+\log y$. You can tell $c=0$ because $y=1$ when $x=0$. Lastly, the inverse function of $\log x$ is $e^x$.
Your other series is a solution to $\frac{d^2y}{dx^2}+y=0$, and so are $\cos x$ and $\sin x$, so it must be one of those, or a combination of those.

Empy2
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It is equally well recognised that: $$\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x$$ and you can get from this to the power series by going a binomial expansion: $$\left(1+\frac xn\right)^n=1+n\frac{x}{n}+\frac{n(n-1)}{2!}\left(\frac xn\right)^2...$$ by now taking the limit as $n\to\infty$ and then converting to sigma notation you should get this same series.

You can also get this from the taylor series formula, which is easy since: $$f^{(n)}(x)=f(x)\,\,\,n\in\mathbb{N}$$ where $f(x)=e^x=\exp(x)$

Henry Lee
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