Talk:Derivative of Exponential Function

$\frac{dy}{dx}=\frac{e^x(e^h-1)}{h}$

$\lim_{h \to 0}{(e^h-1)}=h$

$\frac{dy}{dx}=\frac{e^xh}{h}$

Is the second line above correct? It seems to me that as h → 0, e^h→ 1 and (e^h --1)→ 0

Definition of e

According to Stewart's Calculus: Early Transcendentals, the definition of the number e is the number such that its derivative at x=0 is 1.

This is equivalent to stating $~\lim_{h \to 0} \frac{e^h - 1}{h} = 1 $.

Why? This is simply because

$f(x)=e^x\rightarrow f^'(x)=\lim_{h \to 0}\frac{e^{x+h}-e^x}{h} = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$

Defining $f^'(0)=1$, we get

$ 1 = f^'(0) = \lim_{h \to 0}\frac{e^0(e^h - 1)}{h} = \lim_{h \to 0}\frac{1\cdot(e^h-1)}{h}=\lim_{h \to 0}\frac{e^h-1}{h}$

There ya go.

Infinite Series

Take a common definition of the exponential function: $e^x = \sum_{n = 0}^{\infty} {\frac{x^n}{n!}}$

Evaluating the limit based on this definition: $\lim_{h \to 0} \frac{e^h - 1}{h}=\lim_{h \to 0} \frac{\sum_{n = 0}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \frac{\frac{h^0}{0!}+\sum_{n = 1}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \frac{1+\sum_{n = 1}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \sum_{n = 1}^{\infty} {\frac{h^{n-1}}{n!}}$

$=\lim_{h \to 0} \sum_{n = 0}^{\infty} {\frac{h^{n}}{(n+1)!}}$

$=\sum_{n = 0}^{\infty} {\frac{0^n}{(n+1)!}}$

$=\frac{0^0}{(0+1)!}+\sum_{n = 1}^{\infty} {\frac{0^n}{(n+1)!}}$

$=\frac{1}{1}+0$

$=1$

Question

This might be a little too picky and it's not even on the main proof page but "definition of the number e is the number such that its derivative at x=0 is 1" clearly can't be correct. Do you mean the function $ a^x $ with base $e$?

One assumes, but you've caught me without my copy of Stewart's Calculus: Early Transcendentals, so I can't check. That does seem to be a fair definition of the exponential function, though. See above for a proof based on an alternate definition of $e^x$. Note that this talk page is getting kind of long (longer than many of the proof pages[yeah, largely my fault :D]). We might want to put some of it on the page itself. --Cynic 00:50, 25 June 2008 (UTC)

i don't like the assertion about "by definition... (e^h-1)/h -->1". that isn't the canonical definition of e. certainly it is a consequence of that definition, but consequences require substantiation.

e^h has a limit definition, whose argument can be expanded out by the binomial theorem. the expanded version clearly validates the desired limit.

sorry for providing a text heuristic instead of a print-worthy rendition of the proof. i have no experience with this interface, and frankly no interest in sending time figuring it out. --Misanthropope‎

I don't think anyone asserted that was the definition, Matt just showed how it was a consequence.

However, we definitely need to pic a definition for $e$ and $e^x$. Wikipedia gives three possible definitions for the exponential function (of which I think the first one is probably the best):

• $e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots$

• Less commonly, e^x is defined as the solution y to the equation: $x = \int_{1}^y {dt \over t}.$

• It is also the following limit: $e^x = \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^{n}.$

And defines $e$ as the unique real number such that $\frac{d}{dx}e^x=e^x$ --Cynic 18:51, 14 December 2008 (UTC)

"... Matt just showed how it was a consequence."

Did I? Where? (Or are you talking about a different Matt?)

I prefer, as a definition from first principles, the one that goes:

$e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^{n}.$

It gives an ultra-convenient "real world" definition for it: the limiting case of interest earned on a bank account, for example.

The definition $e^x = \sum_{n = 0}^{\infty} {\frac{x^n}{n!}}$ is a consequence of $\frac{d}{dx}e^x=e^x$ and Taylor's Theorem.

The book I'm currently plundering has it defined (in a roundabout way) from the solution y to the equation: $x = \int_{1}^y {dt \over t}.$

Point is you can only define it by saying what it is, and then sort-of saying "... and it has these properties" - each of which can be derived from each other, as all definitions are ultimately equivalent.

Ultimately the most generally useful way of defining it in most applications tends to be $e^x = \sum_{n = 0}^{\infty} {\frac{x^n}{n!}}$ but in order to reach the fact that it's a valid definition (i.e. that it converges for all $x$, etc.) one needs to thrash through a lot of undergrowth: limits, differentiation, integration, the whole ball of wax - and until we have that in place it sort of hangs in the middle of nowhere and all "definitions" need to be taken on trust.

Bear with me, I'll get to it - I'm still fumbling around in the basics at the moment. --Matt Westwood 19:22, 14 December 2008 (UTC)

My bad, it was Grambottle. Wow, I completely forgot about him (no offense) since he hasn't made any edits since May. --Cynic-----(talk) 21:32, 14 December 2008 (UTC)

Hi. --Grambottle 07:10, 17 December 2008 (UTC)

o.. kay. I have defined $e$ as $e = \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$.

From that I have proved that $e = {1 \over 0!} + {1 \over 1!} + {1 \over 2!} + {1 \over 3!} + {1 \over 4!} \cdots$

which is a start.

This at least gives us a consistent definition of $e$. --Matt Westwood 22:48, 17 December 2008 (UTC)

Umm.... That denominator is $e$. Your fraction represents $e^{-1}$. You might wanna double-check your calculations. --Grambottle 03:42, 18 December 2008 (UTC)

Yeah I know, it was late last night when I put that bit together. I thought Cynic had fixed it. Seems not.--Matt Westwood 06:16, 18 December 2008 (UTC)

BTW: given that $e = \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$, I believe we can then prove that $e^x = \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$ directly, but first we need to show that the binomial theorem continues to hold for non-integral (and indeed, real) values of $x$. I'm working towards it! --Matt Westwood 07:23, 18 December 2008 (UTC)

Notation

I find that $e^x$ is a lot easier to read than $\exp x$. Of course, that could just be me, or it could be that the exp notation is more common in higher mathematics. --Cynic (talk) 22:21, 22 January 2009 (UTC)

The operation of associating $2.718281828...^x \ $ is a multifunction in $\C \ $, whereas the map $\text{ exp }:\C \to \C$ is a well-defined transformation of the complex plane that corresponds to the principal branch of $2.718282828...^x \ $. Frequently, the exponential function is simply abbreviated $e^x \ $, with the understanding that it refers only to exponential function and not to $e=2.718281828...., \ $ raised to the $x \ $ power. I haven't contributed to anything regarding exponentials too heavily, but that's a necessary distinction when discussing the exponential as a complex function, though unnecessary and cumbersome for real analysis or any calculus at an elementary level. Zelmerszoetrop 22:29, 22 January 2009 (UTC)

As you say, the distinction is necessary at complex analysis level (which I'm working towards, I'll start as soon as I've finished up defining the trig functions in terms of $D_{xx} f \left({x}\right) = -A f \left({x}\right)$.

While I'm about it, note that a lot of the above discussion is rendered irrelevant by the definition of natural logarithm and exponential and proving all those messy old sequences and series directly from them via various calculus results. There are three given definitions, and all are in the process of being proved equivalent. It should tighten it up a little. Matt Westwood

First proof

Hello, the argument in the first proof for the claim that (e^h - 1)/h --> 1 as h --> 0 was circular (it had used l'Hopital's rule, which already assumed that the derivative of e^x is e^x). I changed it to a new one; I hope someone will point out if I've made an error. Mag487 23:59, 22 August 2009 (UTC)

What part of l'hopital's rule depended on d(e^x)/dx = e^x? I'm probably being dense here... --Cynic (talk) 02:59, 23 August 2009 (UTC)

Sorry, I was unclear. The proof relied (it has referred the reader to this page) on using l'Hopital's rule on the quantity $\lim_{h\to 0} \frac{\exp h - 1}{h}$ to show that it's equal to one. However, in order to apply the rule here, we have to already know the derivative of the function e^x in the numerator. Mag487 05:07, 23 August 2009 (UTC)

Ah, I see. Thanks a lot for fixing it! --Cynic (talk) 00:40, 24 August 2009 (UTC)