# Why does the Taylor expansion of $e^x$ satisfy exponential properties?

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.

• Use log rules and show that the properties for the homomorphism hold. Aug 22, 2017 at 19:36
• I would prove the exponential property one degree at a time. Aug 22, 2017 at 19:38
• This is one of many ways to define the exponential function; then you prove the other definitions as properties. I think it's the approach Gleason takes in his Abstract Analysis text. amazon.com/Fundamentals-Abstract-Analysis-Andrew-Gleason/dp/… Aug 22, 2017 at 19:46
• If you knew nothing about the exponential function, then how would you be able to develop its Taylor series representation? What properties are you actually assuming to begin? Aug 22, 2017 at 20:09
• @MarkViola: I am not sure but it seems OP is just defining a function by a power series. And in the interior of region of convergence a power series is actually the Taylor series of its sum (this is not an obvious result btw). Aug 23, 2017 at 7:48

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

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 $$e^1=\sum_{k=0}^\infty\frac1{k!}=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$

• This was nice. +1 Aug 23, 2017 at 8:05

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 $$x\in \mathbb{R}$$.

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*}

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 $e$.