Is it possible to work out the derivative of $e^x$ using the summation definition of $e = \sum_n 1/n!$?

Question

So I know this question is a bit obtuse because usually we define $e$ in terms of the $\lim_{n \to \infty} (1 + 1/n)^n$ definition, and then compute derivatives of $e^x$ from there appealing to the limit definition of $e$, and then appeal to Taylor series of $e^x$ to finally arrive at $e = \sum_n 1/n!$. However, how hard would it be to do the other way around, to compute the derivative of $e^x$? I.e., is it possible to show the derivative of $(\sum_n 1/n!)^x$ with respect to $x$ is equal to itself, somewhat "directly", without showing the summation formulation of $e$ is equivalent to the typical limit definition of $e$?

Answer 1

Check the following: $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$

Now you can take the derivative term by term (you actually need to justify why one can do so), and you obtain: $$\sum_{n=0}^\infty \frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}=\sum_{k=0}^\infty \frac{x^{k}}{k!}=e^x$$

Answer 2

Note that if we define $$ f(x)=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{1} $$ we get $$ \begin{align} f(x)f(y) &=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{j=0}^\infty\frac{y^j}{j!}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k\frac{x^{k-j}}{(k-j)!}\frac{y^j}{j!}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k\frac1{k!}\binom{k}{j}x^{k-j}y^j\\ &=\sum_{k=0}^\infty\frac{(x+y)^k}{k!}\\[6pt] &=f(x+y)\tag{2} \end{align} $$ Induction shows that for $p,q\in\mathbb{Z}$, $f(p/q)^q=f(p)=f(1)^p$, therefore, $f(p/q)=f(1)^{p/q}$.

Thus, for all $x\in\mathbb{Q}$, $$ f(x)=f(1)^x\tag{3} $$ Since $f$ is continuous, $(3)$ holds for all $x\in\mathbb{R}$.

Since we've set $f(1)=e$, we get that $$ e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{4} $$ Thus, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}e^x &=\sum_{k=0}^\infty\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^k}{k!}\\ &=\sum_{k=0}^\infty k\frac{x^{k-1}}{k!}\\ &=\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}\\ &=\sum_{k=0}^\infty\frac{x^k}{k!}\\[6pt] &=e^x\tag{5} \end{align} $$

Answer 3

For any base, $$\frac{b^{x+h}-b^x}{h}=b^x\frac{b^h-1}{h},$$ so that it suffices to compute the derivative at $x=0$.

If we define $L_b:=\left(b^x\right)'\Big|_{x=0}=\lim_{h\to0}\dfrac{b^h-1}{h}$, then

$$\left(b^x\right)'=L_bb^x.$$

The power series that satisfies this differential equation with the initial condition $b^0=1$ is found by identification of the coefficients, which yields the recurrence $nc_n=L_bc_{n-1}$ and

$$b^x=\sum_{n=0}^\infty\frac{L^n_b x^n}{n!}.$$ As the power series is strictly increasing, the equation $$\sum_{n=0}^\infty\frac{L^n_b x^n}{n!}=b$$ has at most one solution in $x$.

With $$b=e=\sum_{n=0}^\infty\frac1{n!},$$ this solution must be $$L_ex=1.$$

Then $$e^{1/L_e}=e$$ implies $$L_e=1$$ and $$\left(e^x\right)'=e^x.$$

Answer 4

I have given some links to my blog posts in the comments to the question and I would like to give some remarks on the way you want to define $e^{x}$ and show that derivative of $e^{x}$ is $e^{x}$ itself.

First note that the most frequent definition of $e$ is given by $$e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}\tag{1}$$ Using this definition it is possible to prove that $$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots\tag{2}$$ so that both these definitions are equivalent. For a simple proof see this answer. Thus you see that we can easily go from $(1)$ to $(2)$ without defining $e^{x}$ and finding its Taylor series (like you mention in your question). In fact it is possible to prove more, namely $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{x^{2}}{2!} + \cdots\tag{3}$$ without using any properties of $e$ or $\log$ function. See this post for the same.

Now coming to your brutal definition of $e^{x}$ as limit of $e^{x_{n}}$ where $x_{n}$ is a sequence of rationals tending to $x$ (this is what we mean by "extending definition of a function $f(x)$ from rational values of $x$ to irrational values of $x$ by continuity"). This is possible if we establish the existence of such a limit and also the fact that it is independent of specific sequence $x_{n}$ so that if another sequence $y_{n} \to x$ then both $e^{x_{n}}$ and $e^{y_{n}}$ tend to same limit.

When we proceed in this manner we can easily see that the properties of exponentiation hold i.e. $e^{x + y} = e^{x}e^{y}$ for all $x, y$. The derivative of $e^{x}$ is then easily seen to be $$\lim_{h \to 0}\frac{e^{x + h} - e^{x}}{h} = e^{x}\lim_{h \to 0}\frac{e^{h} - 1}{h}$$ and thus our goal is achieved if we show that $$\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1\tag{4}$$ This is another difficult story. In my blog I have used another technique to establish $(4)$. Using some tricky but not so difficult inequalities I have shown that the limit $$\lim_{h \to 0}\frac{a^{h} - 1}{h}$$ exists for all $a > 0$ (where $a^{h}$ has been defined brutally). And hence this limit can be used to define a function of $a$ for $a > 0$. Lets call it $L(a)$ so that $$L(a) = \lim_{h \to 0}\frac{a^{h} - 1}{h}\tag{5}$$ for all $a > 0$. Then we can easily establish that $$L(ab) = L(a) + L(b), L(1/a) = -L(a), L(1) = 0\tag{6}$$ and that $L(a)$ is a strictly increasing continuous and differentiable function of $a$ and the range of this function is whole of $\mathbb{R}$ and its derivative is $L'(a) = 1/a$ (so that $L(a)$ is what we call $\log a$). Hence there is a unique number $\xi > 0$ such that $L(\xi) = 1$. Then I show that this unique $\xi$ is same as the number $e$ defined by limit $(1)$ (this requires us to use the fact that $L'(1) = 1$). This way I establish $(4)$. Note that proof of $(4)$ in this manner is independent of the proof of equation $(3)$ which deals with power series for $e^{x}$. More details available in this post.