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Is there any intuitive argument or visual "proof" that $\exp' = \exp$? Suppose you have defined the Euler number $\mathrm{e}$ as limit of the sequence $(a_n)$ where $a_n = \left (1 + \frac{1}{n} \right)^n \quad \forall n > 0$, and that the $\exp(x)$ is introduced as $\mathrm{e}^x$. The use of the power series of $\exp$ should be avoided.

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    $\begingroup$ question: how do you introduce real powers without knowing the exp function? $\endgroup$ – mookid Oct 15 '14 at 13:53
  • $\begingroup$ If you can use that $\log'(x)=1/x$ and that $\log$ and $\exp$ are inverses, then it follows from the chain rule, as in math.stackexchange.com/a/31392/589. $\endgroup$ – lhf Oct 15 '14 at 14:05
  • $\begingroup$ See also math.stackexchange.com/questions/381397/definition-of-expx. $\endgroup$ – lhf Oct 15 '14 at 14:17
  • $\begingroup$ I agree with @mookid , there is a structural issue regarding properties you can take for granted $\endgroup$ – mvggz Oct 15 '14 at 14:21
  • $\begingroup$ You can prove that $e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ (hint: let $m=\frac nx$). Differentiate that. (P.S. As a bonus, the binomial theorem gives you the Taylor series!) $\endgroup$ – Akiva Weinberger Oct 15 '14 at 14:35
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$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n $$

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    $\begingroup$ Do you intend to commute $\lim$ with $d/dx$ ? $\endgroup$ – lhf Oct 15 '14 at 14:12
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    $\begingroup$ @lhf: Yeah. Rigorously justifying it, but the OP is only asking for an intuitive argument. $\endgroup$ – user14972 Oct 15 '14 at 14:15
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    $\begingroup$ I disagree. The author's definition of $e^x$ would be rather $\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}$, and this definition itself requires to say what the non-integer powers are. $\endgroup$ – Start wearing purple Oct 15 '14 at 14:22
  • $\begingroup$ @O.L.: I opted to leave transforming between the two versions to the reader. $\endgroup$ – user14972 Oct 15 '14 at 14:35
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Once you have defined $exp(x)=e^x$ we can work our way trhough the definition of derivative (which is quite intuitive) for demostrating that $exp'(x)=exp(x)$. The formal definition of derivative is:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

When $h$ tends to $0$.

So, substituting $f(x)$ by $e^x$ we get that:

$$f'(x)=\lim_{h\to0}\frac{exp(x+h)-exp(x)}{h}$$

Working with the expression a bit we can rearrange it to:

$$f'(x)=\lim_{h\to0}\frac{e^x(e^h-1)}{h}$$

Now notice that as $h$ approaches $0$ the denominator of the fraction $h$ and the term $e^h-1$ also approach $0$. Furthermore, when being really close to $0$ they approach it at 'the same speed' and we say they are 'equivalent infinitesimals' which means that we can approximate one to the other. We can substitute then $e^h-1$ with $h$.

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With this in mind we end up with:

$$f'(x)=\lim_{h\to0}\frac{e^x·h}{h}$$

We can cancel the $h$ and get to the final result:

$$exp'(x)=exp(x)$$

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  • $\begingroup$ What would be a explanation that $\lim_{h\to0}a^h-1\not=h$ for $a\not=e$ $\endgroup$ – kingW3 Oct 18 '14 at 12:06
  • $\begingroup$ @kingW3 Note that $\lim_{h\to0}a^h-1$ can also be expressed as $\lim_{h\to0}e^{ln(a^h)}-1=\lim_{h\to0}e^{h·ln(a)}-1$which is the same case as stated above but with $h·ln(a)$ replacing $h$, which means we can aproximate $a^h-1$ to $h·ln(a)$ $\endgroup$ – Ioannes Oct 18 '14 at 17:30
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Provided that you have defined general exponentiation in the first place, you can write

$$\frac{d}{dx} a^x = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}.$$

You now need to intuitively argue that this last limit, as a function of $a$, ranges from $0$ to $\infty$ as $a$ ranges from $1$ to $\infty$. It is also clearly continuous. Therefore by the intermediate value theorem there is some $a$ such that the limit is $1$. We call this number $e$.

Defining general exponentiation in the first place is a huge mess. Indeed, I think that from the perspective of real analysis, it is easier to define $\exp$ by the ODE it satisfies, then prove it has an inverse $\ln$, then prove that $\exp(q \ln x) = x^q$ whenever $q$ is rational. (Provided you can independently prove the power rule for rational powers $q$, this last argument is actually quite simple.)

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Since $$ \frac{e^{x+h}-e^x}{h}=\frac{e^xe^h-e^x}{h}=e^x\frac{e^h-1}{h} $$ you only need to prove that $\exp'(0)=1$.

Perhaps this helps.

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  • $\begingroup$ But how to show that $\exp'(0) = 1$? $\endgroup$ – Julia Oct 15 '14 at 16:00

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