How can I prove that a function that is its own derivative exists? And how can I prove that this function is of the form $a(b^x)$?


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    $\begingroup$ There are probably already a few posts about this problem on this site. For example, using Approach0 I was able to find this question: Proof that $C\exp(x)$ is the only set of functions for which $f(x) = f'(x)$. $\endgroup$ – Martin Sleziak May 5 '17 at 10:45
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    $\begingroup$ 3Blue1Brown actually goes through this in his "Essence of Calculus" videos that are posting on YouTube this week (sorry, but I'm not able to go searching for the link right now). $\endgroup$ – Paul Sinclair May 5 '17 at 18:30
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    $\begingroup$ The first part's easy - can't you just write down $e^x$? Seriously, though, you don't need to do anything special - all you need to do to prove that "some $x$ exists" is produce an example of $x$. $\endgroup$ – EJoshuaS May 5 '17 at 18:39
  • $\begingroup$ See this answer math.stackexchange.com/a/1292586/72031 $\endgroup$ – Paramanand Singh May 5 '17 at 18:46
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    $\begingroup$ One way to prove there exists such a function is to prove that the derivative of the zero function, $f(x) = 0$ for all $x,$ is the zero function. $\endgroup$ – Dave L. Renfro May 5 '17 at 18:47

There are two ways you could show it. The harder route would be to prove the existence and uniqueness theorem for ordinary differential equations, thus showing there exists solutions to $y'=y$.

The more direct way would be to just construct the function $e^x$ and show that it's its own derivative. You would start by defining $$\ln(x) = \int_1^x \frac{1}{t}\, dt$$ and prove that it's a strictly increasing function on $(0,\infty)$ with range $(-\infty, \infty)$. It follows that $\ln(x)$ has an inverse, which we should dub $e^x$. As for finding the derivative of this new and mysterious function: $$y=e^x$$ $$\ln(y)=x$$ Taking the $x$ derivative of both sides, $$\frac{y'}{y} = 1$$ $$\implies y'=y$$ And do show that every function which is its own derivative is a constant multiple of $e^x$, suppose that $f'=f$. Then, noting that $e^x$ is nowhere zero, $$\frac{d}{dx} \frac{f(x)}{e^x} = \frac{f'(x)e^x-f(x)e^x}{(e^x)^2} = \frac{f(x)e^x-f(x)e^x}{(e^x)^2} = 0$$ Therefore, $$\frac{f(x)}{e^x}$$ is constant since it has a connected domain, and so $f(x) = ce^x$ for some $c$.

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    $\begingroup$ Like @florence says in the first paragraph of the answer, this follows from the existence of (short-time) solutions to ordinary differential equations, which shows just as well that there is a function $f$ that satisfies the equation $F(x, y, y', \ldots, y^{(m)}) = 0$ for any reasonable function $F$. $\endgroup$ – Travis May 5 '17 at 10:58
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    $\begingroup$ I think it would be fair to first write the function as $\exp (x) $, then find $e=\exp (1) $, and show that the properties of the logarithm imply that $exp (x)=e^x $ for $x $ rational, and then extend to irrational. Also, you omitted the easiest way, which is to define $e^x=\sum_{k=0}^\infty\frac {x^k}{k!} $. $\endgroup$ – Martin Argerami May 5 '17 at 13:14
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    $\begingroup$ Why are $\ln x, e^x$ differentiable? That is worth mentioning at this level. $\endgroup$ – zhw. May 5 '17 at 15:20
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    $\begingroup$ @MartinArgerami, the power series approach does not seem easier to me. You would at least have to say that the power series converges uniformly on each compactum and that that implies that you can differentiate (or easier: integrate, then use the fundamental theorem) "under the sum". $\endgroup$ – Carsten S May 5 '17 at 19:11
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    $\begingroup$ @P.B.G.: without defining the exponential function? What does "$b^x$" mean to you? $\endgroup$ – Martin Argerami May 6 '17 at 21:59

$f(x) = 0$ is trivially its own derivative, and is of the form $a(b^x)$ for $a=0$ and any positive $b$. That's all we need to solve the problem posed.

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    $\begingroup$ Eh, some questioners may find it charming to point out trivialities coming from minor deficiencies in the question posed. Others might not, and may get frustrated and disappointed. Especially if it is in an answer (highly upvoted!) and not a comment. $\endgroup$ – guest May 6 '17 at 3:54
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    $\begingroup$ The point isn't to be charming; the point is to solve the problem. If this problem showed up on a test, I would give this answer and use the time I saved to check my work, and I would be frustrated and disappointed if I didn't answer this way and then realized later that I could have. $\endgroup$ – user2357112 May 6 '17 at 6:58
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    $\begingroup$ But it does solve the problem. Sometimes problems are easy. Easy problems arise naturally in a wide variety of contexts, on their own or as part of the solution to harder problems, and recognizing when a problem is easy and not being blind to the easy answer is a crucial mathematical skill. $\endgroup$ – user2357112 May 6 '17 at 8:15
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    $\begingroup$ @quid Any math teacher who marks as incorrect a succinct and valid proof based on the exact wording of the requirement (such as this answer) was erroneously given a degree of any sort in mathematics, IMHO. While it is helpful to keep the spirit of mathematics in ones mind while practicing it (by spirit, I mean the idea that the technical details are not the only important parts of mathematics), what math is really about is assiduously holding oneself to obeying the structures of an organized system of thought. A student finding such an elegant solution to one of my problems earns my respect. $\endgroup$ – Todd Wilcox May 6 '17 at 21:35
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    $\begingroup$ @guest I myself have posed problems such as the one in this question, knowing that there is the "expected" answer as well as one or more "trivial" answers with the express purpose of understanding when a student is a good student or a mathematician. The former will show that $e^x$ is its own derivative. The latter will answer with the zero function. See: smbc-comics.com/?id=2208 $\endgroup$ – Todd Wilcox May 6 '17 at 21:40

An intuitive answer:

For smooth functions and "small" $h$, we have


Then $f'(x)=f(x)$ yields

$$f(x+h)\approx(1+h)f(x),$$ and $$f(x+2h)\approx(1+h)f(x+h)\approx(1+h)^2f(x),$$ $$\cdots$$ $$f(x+nh)\approx(1+h)^nf(x).$$

Now with $nh=1$,

$$f(x+1)\approx \left(1+\frac1n\right)^n f(x),$$ which should ring a bell.

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    $\begingroup$ I guess you want $f(x+y)=f(x)e^y$, not just the case $y=1$. On another note, I have some distrust of these "intuitive" reasonings because when I was studying physics in another life these things crept everywhere; thing is, whenever I would try to do the same, my results were always wrong. So I always got the feeling that one does these kind of computations because they already work by other reasons, but that it is extremely easy to stray from the "correct" path. $\endgroup$ – Martin Argerami May 5 '17 at 16:57
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    $\begingroup$ @MartinArgerami: the answer covers any $nh$, but this is unimportant. The goal was to show the appearance of $e$. $\endgroup$ – Yves Daoust May 5 '17 at 17:19
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    $\begingroup$ This can be seen as solving the differential equation $y' = y$ using Euler's method with $n$ steps of size $1/n$. $\endgroup$ – Daniel Schepler May 5 '17 at 23:34
  • $\begingroup$ @DanielSchepler: of course. $\endgroup$ – Yves Daoust May 6 '17 at 17:12

If you postulate a solution to $y=y'$ of the form $g(x)=\sum_{k=0}^\infty a_kx^k$, by the equality of the power series on both sides of the equality one gets $$ a_{k+1}=(k+1)a_k,$$and then one deduces that $$a_k=\frac{a_0}{k!},\ \ \ \ k=1,2,\ldots.$$So $$y(x)=a_0\,\sum_{k=0}^\infty\frac{x^k}{k!}.$$One can then focus on the case where $a_0=1$, say $g(x)=\sum_{k=0}^\infty\frac{x^k}{k!}$. Define $e=g(1)$. Using the series one can show that $$ g(x+y)=g(x)g(y).$$ It follows that $$\tag{1}g(x)=e^x$$ for $x$ rational. As $g$ is continuous (infinitely differentiable, even), it has to be $e^x=g(x)$ for irrational $x$, too. Thus $$ y(x)=a_0\,e^x. $$

  • $\begingroup$ Possibly a new learner needs the recurrence relationship on a[k] explicitly written out to see why a[k]=1/k! follows from the assumption that the function being expanded is its own derivative. Similarly, for exposition I'd write f(x)=... instead of exp(x). The danger I suppose, is that then you'd just be doing a bunch of future visitors homework for them as they could copy/paste the entirety and turn it in. $\endgroup$ – Paul May 5 '17 at 13:55
  • $\begingroup$ Yes, good points. $\endgroup$ – Martin Argerami May 5 '17 at 16:13
  • $\begingroup$ This is an excellent answer! It doesn't require knowledge about $e$ or the natural logarithm, and even helps explain what they are. It just needs some power series machinery and the binomial theorem. +1 $\endgroup$ – dafinguzman May 5 '17 at 20:14

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