8
$\begingroup$

This question already has an answer here:

How do you differentiate $e^x$?

I looked on many sites, including similar questions here but most answers seemed circular.

The only known definition of $e$ to be used in this proof is $$ e=\lim_{n \to\infty} \left(1+\frac{1}n \right)^n $$

What I did is:

$$ \begin{align*} (e^x)' &=\lim_{h\to0}\frac{e^{x+h}-e^x}h \\ &= e^x\lim_{h\to0}\frac{e^{h}-1}h \end{align*} $$

But I don't know how to go on, I know $\lim_{h\to0}\frac{e^{h}-1}h=1$ but I don't know how to prove it, I can't use the $e^x$ taylor expansion as that would imply diferentiating $e^x$.

Edit: I also can't use the derivative of $\ln(x)$.

$\endgroup$

marked as duplicate by Claude Leibovici, N. F. Taussig, kingW3, Emily, Cookie Apr 5 '15 at 17:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Are you allowed to use Newton's binomial theorem? If so then you just have to expand in binomial series $(1 + 1/n)^{hn}$, substract 1 and find that the only term surviving the limiting process is $h/h =1$, from whence the limit follows. $\endgroup$ – Rogelio Molina Apr 4 '15 at 0:25
  • $\begingroup$ @RogelioMolina Yes, I'm allowed to use the binomial theorem, could you elaborate on that? I couldn't follow you. $\endgroup$ – YoTengoUnLCD Apr 4 '15 at 0:26
  • $\begingroup$ See characterizations of the exponential function, especially the paragraph about their equivalence. $\endgroup$ – Lucian Apr 4 '15 at 0:29
  • $\begingroup$ Another point of interest: You've defined $e$ as a limit. How is $e^h$ defined for all $h\in \mathbb {R}$? $\endgroup$ – zhw. Apr 4 '15 at 1:42
7
$\begingroup$

We must calculate the limit

$$ \lim_{h \to 0} \frac{e^h -1}{h} $$ using the definition we have for $e$ we have

$$ \lim_{h \to 0}\lim_{n \to \infty}\frac{(1+1/n)^{hn} -1}{h} $$ we expand using the binomial theorem:

$$ (1+ 1/n)^{hn} = \sum_{k=0}^{\infty}\binom{hn}{k}(1/n)^k = 1 + h + h^2 \cdots $$ Then the tricky part (this needs to be justified carefully), we exchange the limits to get:

$$ \lim_{n \to \infty} \lim_{h\to 0}\frac{1 + h + h^2 \cdots -1}{h} = \lim_{n \to \infty} \lim_{h\to 0}(1 + h\cdots) = \lim_{n \to \infty}1 =1 $$

$\endgroup$
  • $\begingroup$ I thought the binomial theorem was only applicable if we are raising to an integer power. As $h$ grows small, $hn$ may no longer be an integer, right? $\endgroup$ – layman Apr 4 '15 at 1:03
  • $\begingroup$ The binomial theorem (or a version of it) is valid for all complex powers. See en.wikipedia.org/wiki/Binomial_series $\endgroup$ – Moya Apr 4 '15 at 1:40
  • $\begingroup$ Even if you don't accept the general binomial theorem, for $h$ rational, there will be a subsequence of $(1+1/n)^{hn}$ where the exponent is an integer, so provided you accept that the original sequence at least converges, there is no loss of generality in assuming $hn$ is integral. $\endgroup$ – Strants Apr 4 '15 at 1:46
  • $\begingroup$ I'm having trouble seeing why $$ \sum_{k=0}^{\infty}\binom{hn}{k}(1/n)^k = 1 + h + h^2 \cdots $$ for the second term I obtain $\frac{h^{2}}{2}-\frac{h}{2n}$ $\endgroup$ – user135520 Feb 7 '16 at 16:36
6
$\begingroup$

Bernoulli's inequality gives that for any $n\geq 2$ we have: $$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n-1}\right)^n\tag{1} $$ hence $n\left(e^{\frac{1}{n}}-1\right)$ is between: $$ 1 < n\left(e^{\frac{1}{n}}-1\right) < 1+\frac{1}{n-1}\tag{2}$$ hence, by squeezing: $$ \lim_{n\to +\infty}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}=1 \tag{3}$$ that, together with $(2)$, gives: $$ \lim_{r \to 0}\frac{e^r-1}{r}=1\tag{4}$$ that is enough to grant: $$ \lim_{h\to 0}\frac{e^{x+h}-e^{x}}{h}=e^x \tag{5} $$ as wanted.

$\endgroup$
  • 1
    $\begingroup$ (+1) I was going to go this way before I saw your answer. I am just too slow in my old age! Bernoulli's Inequality is so very useful. $\endgroup$ – robjohn Apr 4 '15 at 1:57
  • $\begingroup$ I hope you don't mind. I've added an answer to add a bit more detail about how $(1)$ is derived from Bernoulli's Inequality. $\endgroup$ – robjohn Apr 4 '15 at 2:14
5
$\begingroup$

This is really a comment on Jack D'Aurizio's answer, but it is too long for a comment. Let me show how Bernoulli's Inequality is used to prove $(1)$ from Jack D'Aurizio's answer.

In this answer, it is shown, using Bernoulli's Inequality, that $$ \left(1+\frac1n\right)^n\tag{1} $$ is an increasing sequence and that $$ \left(1+\frac1n\right)^{n+1}\tag{2} $$ is a decreasing sequence. This means that for all $n$ $$ \left(1+\frac1n\right)^n\le\overbrace{\lim_{k\to\infty}\left(1+\frac1k\right)^k}^{\large e}\le\left(1+\frac1n\right)^{n+1}\tag{3} $$ Since $(3)$ is true for all $n$, we can substitute $n\mapsto n-1$ in the right-side inequality to get the inequality from Jack's answer: $$ \left(1+\frac1n\right)^n\le e\le\left(1+\frac1{n-1}\right)^n\tag{4} $$

$\endgroup$
3
$\begingroup$

\begin{equation*} e^x=\lim_{n\rightarrow \infty}\left(1+\frac {x} {n}\right)^n =\lim_{n\rightarrow \infty}\sum_{k=0}^n\binom n k\frac {x^k} {n^k} =\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {k!}\frac {x^k} {n^k}\end{equation*} $$ =\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n(n-1)...(n-k+1)} {n^k}\frac {x^k} {k!}=\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {n} {n}\frac {n-1}{n}...\frac{n-k+1} {n}\frac {x^k} {k!}$$ $$\lim_{n\rightarrow \infty}\sum_{k=0}^n\frac {1} {1}\left(1-\frac {1}{n}\right)...\left(1-\frac{k-1} {n}\right)\frac {x^k} {k!}=\sum_{k=0}^\infty\frac {x^k} {k!}$$ Hence we have that $$e^x=\sum_{k=0}^\infty\frac {x^k} {k!}\implies(e^x)'=\sum_{k=1}^\infty\frac{kx^{k-1}} {k!}=\sum_{k=1}^\infty\frac{x^{k-1}} {(k-1)!}=\sum_{k=0}^\infty\frac{x^{k}} {k!}$$ It follows that $e^x=(e^x)'$

$\endgroup$
2
$\begingroup$

Define the log function as $\log x=\int_1^x \frac{dt}{t}$. The derivative is obviously $1/x$. The exponential function is the inverse of the log function. So, we may write

$$\begin{align} x&=\log (e^x)\\ &=\int_1^{e^x} \frac{dt}{t} \end{align}$$

Taking the derivative of both sides and applying the chain rule reveals

$$\begin{align} \frac{dx}{dx}&=1\\ &=\frac{d\log(e^x)}{dx}\\ &=\frac{1}{e^x}\frac{de^x}{dx} \end{align}$$

whereupon solving for $\frac{de^x}{dx}$ shows that

$$\frac{de^x}{dx}=e^x$$

So, the derivative of the exponential function is itself!

$\endgroup$
-1
$\begingroup$

$$e^x=\sum_{n\ge 0} \frac{x^n}{n!}$$ $$\frac{de^x}{dx}=\frac{d}{dx}[1+x+x^2/2!+x^3/3!+\cdots]=0+1+x+x^2/2+\cdots=\sum_{n\ge 0} \frac{x^n}{n!}=e^x$$

This uses a theorem justifying interchanging the derivative with the summation.

$\endgroup$
  • $\begingroup$ From the original post: but I don't know how to prove it, I can't use the ex Taylor expansion as that would imply differentiating $e^x$. $\endgroup$ – MathMajor Apr 4 '15 at 1:34
  • $\begingroup$ @GabrielH The Taylor expansion only requires knowing the derivative at a single point, and that can be easily found by approximation with secant lines. $\endgroup$ – Cyclohexanol. Apr 4 '15 at 1:35
  • 1
    $\begingroup$ The OP has likely not even learned that term by term differentiation of an infinite series is allowed ... $\endgroup$ – MathMajor Apr 4 '15 at 1:36
  • 1
    $\begingroup$ It is the OP's problem for not including enough context. $\endgroup$ – Cyclohexanol. Apr 4 '15 at 1:37
  • 1
    $\begingroup$ Okay, that may be the case. It wasn't me who down voted. I just pointed out that answer is most likely not helpful to the OP. $\endgroup$ – MathMajor Apr 4 '15 at 1:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.