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$$e^x= 1+x/1!+x^2/2!+x^3/3!+x^4/4!\cdots$$ $$\frac{d}{dx}e^x= \frac{d}{dx}1+\frac{d}{dx}x+\frac{d}{dx}x^2/2!+\frac{d}{dx}x^3/3!+\frac{d}{dx}x^4/4!+\cdots$$ $$\frac{d}{dx}e^x=0+1+2x/2!+3x^2/3!+4x^3/4!\cdots$$ $$\frac{d}{dx}e^x= 1+x/1!+x^2/2!+\cdots=e^x$$

Of course this proof is circular because in order to find the expansion requires knowing the derivative, but ignoring that fact, is this proof valid?

Teoc
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  • It is not necessarily a circular proof: some authors treat the series expansion of $e^z$ as a definition, then later establish other equivalent formulations and properties of the function. It depends on what you take as the definition of the exponential function. – heropup Jan 27 '15 at 02:26
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    Its valid only if you have a theorem about applying the derivative to an infinite sum of functions, or, what is the same thing, to a sequence of function. You can get such a theorem at http://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability – user24142 Jan 27 '15 at 02:28

3 Answers3

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Firstly, in order to get the expansion one only must know the values of the derivatives at a single point (namely, the origin).

Secondly, to make that argument hold, you need some results regarding term-by-term differentiation.

user1337
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  • That's not necessarily true. If you define the exponential using the series (which is a perfectly valid definition), then you don't need to know the derivatives beforehand. – hasnohat Jan 27 '15 at 02:30
  • I did not pose this question, but in an attempt to learn more as I find it intriguing, how else could one prove that the derivative of $e^x$ is indeed $e^x$? By definition, the derivative of any transcendental function in my basic Calc. I course is shown to be something like $$\frac{d}{dx}\left(a^x\right)=a^{x}\log\left|a\right|,$$ which in the case of $e^x$, $\log\left|e\right|$ would of course just be 1. Perhaps an attack such as actually solving in terms of the definition of derivative would be acceptable? – bjd2385 Jan 27 '15 at 02:32
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    @bd1251252 The classic proof is: $$\lim_{h\to 0} \frac{e^xe^h-e^x}{h}=e^x\lim_{h\to 0}(\frac{e^h-1}{h})$$, applying L'Hopital we get that the limit is 1, and therefore $\frac {d}{dx} e^x=e^x$ Of course, the derivative of an arbitary base can be proved by $a^x=e^{x\log a}$ and using the chain rule. – Teoc Jan 27 '15 at 02:37
  • @MathNoob Thanks for taking the time to show me! I will study hard and get through this discrete math textbook so I may move on to more difficult analysis like this. – bjd2385 Jan 27 '15 at 03:13
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It's fine as long as you know that power series converge uniformly on bounded sets. The important point is that the series of derivatives converges locally uniformly.

user208259
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It is valid after you have shown (or defined) that $e^x$ is indeed equal to that series and that deriving the series is the same as deriving each term individually (under suitable conditions, of course). It is also not circular and works also for complex numbers.

Zero
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