20

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 $a(b^x)$?

jwodder
  • 1,497
P.B.G.
  • 223

4 Answers4

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 $c$.

florence
  • 12,819
  • Thanks, is there a way to do it without defining the exponential function? – P.B.G. May 05 '17 at 10:54
  • 1
    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 $F$. – Travis Willse May 05 '17 at 10:58
  • 13
    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!} $. – Martin Argerami May 05 '17 at 13:14
  • 6
    Why are $\ln x, e^x$ differentiable? That is worth mentioning at this level. – zhw. May 05 '17 at 15:20
  • 2
    @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". – Carsten S May 05 '17 at 19:11
  • @CarstenS: proving that basic fact about power series is certainly easier than proving both the Fundamental Theorem of Calculus and the Inverse Function Theorem, which are required for the "log" approach. – Martin Argerami May 05 '17 at 19:35
  • @MartinArgerami, I think without the Fundamental Theorem you have to actually work to show that uniform convergence implies that you limit and derivative commute. I would not enjoy it. I am not sure about the inverse function theorem, you may be right about that one. Well, it all depends on what one already knows. – Carsten S May 05 '17 at 19:40
  • Unfolding the proof of Picard's theorem, you get a construction which isn't too bad: iterate the construction $f \mapsto \lambda x . 1 + \int_0^x f(t),dt$, and then prove that for instance on the interval $[-\frac{1}{2}, \frac{1}{2}]$ the sequence of iterates is uniformly Cauchy and therefore has a uniform limit. – Daniel Schepler May 05 '17 at 23:29
  • Oh, and incidentally, it just so happens if you start the iteration with the 0 function, the iterates are the partial sums of the Taylor series for $e^x$. – Daniel Schepler May 05 '17 at 23:41
  • 1
    @P.B.G. So according to youe question you wan t to prove that the solutions have the form $y=ab^x$, but according to your comment you do not want to introduce the exponential form? That may really be ambitious ... – Hagen von Eitzen May 06 '17 at 07:16
  • 4
    @P.B.G.: without defining the exponential function? What does "$b^x$" mean to you? – Martin Argerami May 06 '17 at 21:59
42

$f(x) = 0$ 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.

user2357112
  • 2,401
  • Good example, producing an example is sometimes the easiest way to prove this kind of thing. $e^x$ would also work, obviously. – EJoshuaS - Stand with Ukraine May 05 '17 at 18:44
  • 18
    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. – guest May 06 '17 at 03:54
  • 14
    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. – user2357112 May 06 '17 at 06:58
  • 2
    It does not solve the OP's problem, @user2357112; you know that, so don't pretend you don't. You are smart enough to know what the OP is looking for, so you can simply make a note that they should exclude the trivial solution by positing y(0)=1 in a comment. – guest May 06 '17 at 07:52
  • 9
    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. – user2357112 May 06 '17 at 08:15
  • 5
    This answer is incomplete even for the question as written (as usually understood). If you answer this on a test you may wake up to a bad surprise. "this function" refers to a generic solution; granted it would be better if phrased differently. – quid May 06 '17 at 13:55
  • 7
    @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. – Todd Wilcox May 06 '17 at 21:35
  • 8
    @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: http://www.smbc-comics.com/?id=2208 – Todd Wilcox May 06 '17 at 21:40
  • 2
    @ToddWilcox it seems amply clear that "this function" is intended as "such a function" or to be still more clear "every such function." Or differently the "is" is a "must be." That this is not clear in OP is likely an error of translating and/or paraphrasing the original source. For the context of this site, this is not answer. If the wording was on an exam, I'd say it is ambiguous enough that the correct course of action would be to ask the instructor what was intended. Put differently there is no exact wording of the requirements to begin with. – quid May 06 '17 at 21:52
  • 1
    @ToddWilcox further, it is not very clear to me what's your authority when asserting "what math is really about" or which "students are mathematician." In my opinion it is a quality of a mathematician to grasp the essence of a situation and to deal with accordingly, as opposed to focusing solely on a formality. Differently, in your comic I think the roles are rather reverted. – quid May 06 '17 at 22:02
  • 1
    @quid Why do you interpret the question as if it was a poem? – Kamil Maciorowski May 06 '17 at 22:39
  • @KamilMaciorowski because it is ill-formed as written, and thus needs interpretation. There is no (particular) function that could be referred to at the point where "this function" appears in the text. It's more than a bit ironic that those insisting on the importance of taking things as written, organized system of thoughts, etc., seem not to care about this issue. – quid May 06 '17 at 22:42
  • 2
    @guest "You are smart enough to know what the OP is looking for". Knowing better what somebody else wants is patronizing. The OP had all means to tell us. It is rude to assume he or she isn't smart enough to do so. People are not born with the knowledge of trivial solutions so they may want them also, we don't know it beforehand. Even if we all agreed the trivial solution is enough so it was the only answer here, and the OP realized it's not what he or she really wants, then the OP could ask another question with additional condition to get non-trivial solution or every possible one. – Kamil Maciorowski May 07 '17 at 00:27
  • @KamilMaciorowski once again, the original post is ambiguous as written. If you disagree, please explain what "this function" would refer to. (Note that the post does not say "Give an example of a function [...]; prove that this function [...]". A proof of existence may well not yield any example at all.) The reasonable course of action is thus to assume the more general/interesting version, and include mention of the special case/other interpretation as a side remark. – quid May 07 '17 at 14:26
  • @quid "A proof of existence may well not yield any example at all". True. But what it does is proving that at least one such-and-such function exists. This function (whatever it is, known to us or not) is "this function" the OP refers to. And you're right: the post does not say "give an example of a function". It also does not forbid this angle of attack. Giving a valid example is an excellent proof of existence, isn't it? Sometimes giving an example is hard and other methods may be easier (including ones that don't yield any example), sometimes it's easy. In this case it was easy. – Kamil Maciorowski May 07 '17 at 16:16
  • @KamilMaciorowski I have nothing against the proof of existence given in this answer. Please read carefully my objections. The problem comes later. It is that, as you say, "this function" can only refer to some unspecified function that is it's own derivative. Therefore one needs to establish the claim on the form of such a function for an unspecified function, that is, every function that is its own derivative. Not only one particular one. The objection would also apply to an answer that says $e^x$ has the property and is of that form. – quid May 07 '17 at 16:36
  • 2
    @quid I would easily take your course of thought in the first place, if only the OP wrote "every such a function" instead of "this function". However I got your point when I asked myself the same question ending with "…of the form $ax+b$". – Kamil Maciorowski May 07 '17 at 17:20
  • @quid The question as stated is "How can I prove that a function which is its own derivative exists? And how can I prove that this function is in the form of $a(b^x)$?" For questions that merely ask whether such a function exists, demonstrating even a single function that fits those criteria is proof that such a function exists. By trying to psychoanalyze the questioner, you are making the solution more complicated than it needs to be. – Abion47 May 07 '17 at 18:24
  • 1
    @Abion47 'For questions that merely ask whether such a function exists, demonstrating even a single function that fits those criteria is proof that such a function exists.' Indeed. This is was not my objection. You seem to miss the point. It is: OP is not clearly formulated (concerning the second part of the question). It is odd to chose a trivial interpretation over more interesting ones (mentioning the trivial one as an aside). – quid May 07 '17 at 18:48
  • @quid Whether or not you approve of the trivial answer does not make the answer any less valid. Also, you are making assumptions as to what it is that OP wants. If OP wanted a proof that, say, shows all functions which are their own derivatives were in the form $a(b^x)$, then he would've asked for that, or in the very least he would've clarified. The fact that he chose an answer that is just a more thorough version of this answer exists makes your objection largely unfounded. – Abion47 May 07 '17 at 18:54
  • @Abion47 "The fact that he chose an answer that is just a more thorough version of this answer exists makes your objection largely unfounded." I have no idea what you are referring to. Anyway, the accepted answer shows precisely that every solution must be of the form $ce^x$. Note 'And do show that every function which is its own derivative is [...]' On the first part of your comment. OP is ambiguous; it is not coherent strictly as written (for details see above). The question is just which interpretation one chooses, a meaningful one or a trivial one. – quid May 07 '17 at 19:04
  • @quid: "Anyway, the accepted answer shows precisely that every solution must be of the form $ce^x$" - actually, it doesn't. It skips over any justification for saying that the inverse of the integral used is of the form $a(b^x)$, instead jumping straight to using $e^x$ notation, and it doesn't do anything to prove that solutions not of the form $a(b^x)$ can't exist. – user2357112 May 07 '17 at 21:23
  • 1
    I'll skip replying to the first part, let me focus on: "and it doesn't do anything to prove that solutions not of the form a(bx) can't exist" since this is my main issue. I can only assume you did not read that answer to the end. It takes any solution $f$, considers the derivative of $f/e^x$. Shows that it is $0$ and thus concludes that the quotient $f/e^x$ is constant and therefore $f$ is of the form $ce^x$. To reiterate, the $f$ is an arbitrary solution and nothing but $f' = f$ is used, and it follows it is of the form $ce^x$ What is not clear about that? – quid May 07 '17 at 22:49
  • If your issue is about the "precisely." I did not mean to insist on the fact that that answer was particularly precise or detailed (though I find it alright). The "precisely" was intended as a counter-point to the mysterious allegation that "he chose an answer that is just a more thorough version of this answer exists makes your objection largely unfounded", when the accepted answer does contain a paragraph whose sole purpose is to establish (i.e., is precisely about) what I find missing in this answer. I feel that part is also rather precise, but that wasn't the point I meant to make. – quid May 07 '17 at 22:57
  • 1
    @quid: Ah, whoops, it does indeed show that all solutions must be a constant multiple of the solution given. I didn't read the part in the middle carefully enough where it introduces $f$. My mistake. – user2357112 May 07 '17 at 23:07
  • The phrase "this function" should indeed be understood as "every such a function" to avoid a subtle trap. Consider "this function returns $0$ for $x=0$". The negation is "this function doesn't return $0$ for $x=0$". You can find an example to "prove" the first, and another example to "prove" the second. Now it's easy to say "Proven! This function returns $0$ for $x=0$ and this function doesn't return $0$ for $x=0$". With "every such a function" approach you can prove none of the two statements and there's no contradiction. The root problem is "this function" isn't an exact phrase. – Kamil Maciorowski May 08 '17 at 06:30
21

An intuitive answer:

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

$$f'(x)\approx\frac{f(x+h)-f(x)}h.$$

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.

  • 3
    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. – Martin Argerami May 05 '17 at 16:57
  • 1
    @MartinArgerami: the answer covers any $nh$, but this is unimportant. The goal was to show the appearance of $e$. –  May 05 '17 at 17:19
  • 2
    This can be seen as solving the differential equation $y' = y$ using Euler's method with $n$ steps of size $1/n$. – Daniel Schepler May 05 '17 at 23:34
  • @DanielSchepler: of course. –  May 06 '17 at 17:12
15

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 $$ y(x)=a_0\,e^x. $$

Martin Argerami
  • 205,756
  • 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. – Paul May 05 '17 at 13:55
  • Yes, good points. – Martin Argerami May 05 '17 at 16:13
  • 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 – dafinguzman May 05 '17 at 20:14