Why does the Taylor expansion of $e^x$ satisfy exponential properties?
Sophia Terry 
Suppose I knew nothing about the function $e^x$. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to $1$. Then, the next term should be the antiderivative of the first term, giving me $x$. Doing this again would give me $\frac{x^2}2$. Repeating this process over and over again, I would get $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ Graphing a few terms of this, I might notice that this looks more and more like an exponential the more terms I graph. If I prove that this function satisfies the exponential relationship $f(x+y) = f(x)f(y)$, I would be able to prove that this series is an exponential function. How would I prove this? After this, how would I prove that the base of this exponential function is $e$, which is defined as $\displaystyle{\lim_{n \to \infty}} (1+\frac1n)^n$?
Edit: After expanding $(1+x)(1+y)$ and $(1+x+\frac{x^2}2)(1+y+\frac{y^2}2)$, I can see how extra terms get taken care of when the next degree is added. However, my second question still stands.
$\endgroup$ 95 Answers
$\begingroup$The first answer lies in Cauchy products and the binomial theorem, which show that
$$e^xe^y=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{x^ny^k}{n!k!}=\sum_{n=0}^\infty\frac1{n!}\sum_{k=0}^n\binom nkx^{n-k}y^k=\sum_{n=0}^\infty\frac{(x+y)^n}{n!}=e^{x+y}$$
As per the second question, the proof can be lengthy and various, and many are outlined in
The definition of e by limits of $(1+1/n)^n$ through series expansion.
$\endgroup$ $\begingroup$By the Binomial theorem,
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac{n!}{(n-k)!\,k!\,n^k}=\sum_{k=0}^n\frac1{k!\,n^k}\prod_{j=0}^{k-1}(n-j)=\sum_{k=0}^n\frac1{k!}\prod_{j=0}^{k-1}\left(1-\frac jn\right).$$
But
$$\left(1-\frac kn\right)^k\le\prod_{j=0}^{k-1}\left(1-\frac jn\right)\le1$$ and for any $m\le n$, $$\sum_{k=0}^m\frac1{k!}\left(1-\frac kn\right)^k\le\sum_{k=0}^n\frac1{k!}\left(1-\frac kn\right)^k\le\left(1+\frac1n\right)^n\le\sum_{k=0}^n\frac1{k!}.$$
Taking the limit $n\to\infty$,
$$\sum_{k=0}^m\frac1{k!}\le\lim_{n\to\infty}\left(1+\frac1n\right)^n\le\sum_{k=0}^\infty\frac1{k!}.$$
This shows that $$e^1=\sum_{k=0}^\infty\frac1{k!}=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$
$\endgroup$ 1 $\begingroup$THIS ARTICLE discusses six independent and equivalent characterizations of exponential function, $e^x$.
The OP assumes that the exponential function is characterized as the function $f(x)$ that satisfies the ordinary differential equation
$$f'(x)=f(x)\tag 1$$
subject to the initial condition $f(0)=1$.
Inasmuch as $f'(x)=f(x)$, $\forall x$, then by induction $f\in C^\infty$ with $f^{(n)}(x)=f(x)$. Hence, $f^{(n)}(0)=f(0)=1$ and the Taylor series for $f(x)$ is given by
$$f(x)=\sum_{k=0}^\infty \frac{x^n}{n!}\tag2$$
This establishes the equivalence of the solution of the ODE $(1)$ and the Taylor series representation $(2)$.
That is to say, if we name the function $f(x)$ that satisfies $(1)$, subject to $f(0)=1$, the exponential function, then the exponential function is represented by the Taylor series given in $(2)$. Note that the converse is also true.
Next, the OP tacitly assumes that the exponential function is characterized by the functional equation
$$f(x+y)=f(x)f(y)\tag 3$$
And that if the Taylor series in $(2)$ satisfies $(3)$, then the Taylor series representation is the exponential function.
Other solutions presented on this page have already established (e.g., use of Cauchy Product) that $f(x)$ as given by $(2)$ satisfies the functional equation $(3)$.
However, we have not shown that $f(x)$ as characterized by $(1)$, or $(2)$, or $(3)$ is the function $e^x$, where $e$ is defined as
$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n$$
To do so, one first needs to prove that
$$\lim_{n\to \infty}\left(1+\frac xn\right)^n=\left(\lim_{n\to \infty}\left(1+\frac1n\right)^n\right)^x =e^x\tag4 $$
If $x\in \mathbb{Q}$, then proof of $(4)$ is straightforward. Then, by exploiting the density of the rational numbers, one can prove that $(4)$ is true for $x\in \mathbb{R}$.
$\endgroup$ 0 $\begingroup$Note that $(4)$ provides yet another unique characterization of the exponential function, which can be shown equivalent to $(1)$, $(2)$, and $(3)$.
\begin{eqnarray*} f(x) f(y) = \left( \sum_{i=0}^{\infty} \frac{x^i}{i!} \right) \left( \sum_{j=0}^{\infty} \frac{y^j}{j!} \right) = \sum_{k=0}^{\infty} \sum_{i+j=k} \frac{ x^i y^j }{i! j!} = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{i+j=k} \binom{k}{i,j} x^i y^j = \sum_{k=0}^{\infty} \frac{(x+y)^k}{k!} =f(x+y). \end{eqnarray*}
$\endgroup$ $\begingroup$To prove $f(x+y) = f(x)f(y)$ recall the binomial formula
$$ (x+y)^k = \sum_{i=0}^k \binom{k}{i}x^iy^{k-i} $$
plug this into the formula above for finite $k$, and you'll find that the sums are separable into two terms that multiply (after a simple substitution $(k-i)\mapsto j$). Send $k\to \infty$ and you'll have the final result.
The second part is a similar application of the binomial formula for finite $n$ to the expression $\left(1+\frac{1}{n}\right)^n$, but sending $n\to \infty$ gives you $f(1)$, which could be your definition of $e$.
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