43
$\begingroup$

My question (more of a hypothesis really) is basically this: If a function $f(x)$ is defined such that $f'(x)$ is not constant and never the same for any 2 values of $x$. Then there do not exist positive integers $a,b,c$ and $a\le b\le c$ such that,

$$\int_0^{a} f(x)dx = \int_b^{c} f(x)dx\tag1$$

Taking $f(x)=x^n$ then

  • for $n \gt 1$, $(1)$ becomes Fermat's Last Theorem, while
  • for $n=1$, $(1)$ becomes $a^2 + b^2 = c^2$

Maybe this has something to do with the "curviness" (I am 16 so please forgive me for non-technical language) of the graphs of $x^n$ since for $n=1$ the graph is linear and for $n>1$ the graph is curvy.

My school teachers are not talking to me about this because "it is out of syllabus", so I thought maybe this community could help.

$\endgroup$
11
  • 1
    $\begingroup$ For $a=b=c=0$, we have $0\ge0\ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true. $\endgroup$
    – celtschk
    Apr 1, 2018 at 14:50
  • 5
    $\begingroup$ By the fundamental theorem of calculus, integration doesn't really get you anything here; what you're saying here is essentially equivalent to the statement that for any (differentiable) $F$ with $F(0)=0$, there are no integers $a,b,c$ with $F(a)+F(b)=F(c)$. But 'functions' are so arbitrary that this can't be true. $\endgroup$ Apr 1, 2018 at 14:57
  • 2
    $\begingroup$ It is far too general. There are many ways to produce counter-examples. $\endgroup$ Apr 1, 2018 at 21:04
  • 14
    $\begingroup$ Bummer that your teachers won't discuss this. Can't imagine turning this question away. $\endgroup$
    – Randall
    Apr 2, 2018 at 2:40
  • 2
    $\begingroup$ Have you considered the fact that for n>2, Fermat's Theorem has arbitrarily "close" counterexamples (i.e. for any epsilon, there are a,b,c such that (c^n-a^n-b^n)/c^n < epsilon)? How does this fact fit in with your hypothesis? $\endgroup$ Apr 2, 2018 at 16:14

2 Answers 2

42
$\begingroup$

Interesting. But not true. If you define $f(x)=\dfrac1{x+1}$, then$$\int_0^1f(x)\,\mathrm dx=\int_1^3f(x)\,\mathrm dx,$$since$$\int_\alpha^\beta f(x)\,\mathrm dx=\log\left(\frac{\beta+1}{\alpha+1}\right).$$

$\endgroup$
7
  • $\begingroup$ what about when a,b,c are distinct integers, and not 0 $\endgroup$ Apr 1, 2018 at 16:53
  • 8
    $\begingroup$ @Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then? $\endgroup$ Apr 1, 2018 at 16:55
  • $\begingroup$ are there infinitely many such functions like the one you gave which are not consistent with the hypothesis. also what if there is one more condition that binds $f(x)$ that $f(\infty)=\infty$ $\endgroup$ Apr 1, 2018 at 17:25
  • 5
    $\begingroup$ @Doctorwho2311 I suppose that there are mor functions like this one. The one I used was this first one that I was able to think of. And if you wanted to impose that $\lim_{x\to+\infty}f(x)=+\infty$, you should have stated it from the start. $\endgroup$ Apr 1, 2018 at 17:31
  • 1
    $\begingroup$ @BCLC With $f(x)=x^n$, the equality$$\int_0^af(x)\,\mathrm dx=\int_b^cf(x)\,\mathrm dx$$becomes $\frac{a^{n+1}}{n+1}=\frac{c^{n+1}}{n+1}-\frac{b^{n+1}}{n+1}$, which is equivalent to $a^{n+1}+b^{n+1}=c^{n+1}$. $\endgroup$ Jun 3, 2021 at 15:38
30
$\begingroup$

(Too long for a comment, therefore adding as an answer)

Creating conjectures is a great way to challenge your understanding and improve your skills. Seeing that you're only 16 and you're already creating conjectures involving integrals that your teachers are avoiding to tackle, I thought I'd give my 2 cents (reminds me of myself).

As for your specific question, there are many counterexamples, as others pointed out. I'm going to talk about some general ideas of creating conjectures as a way to improve your skills.

What I'd like to say is basically that you should give a lot of thought in your choice of hypothesis. This is not silly nitpicking, this is a serious advice - good conjectures always have meaningful and logical hypothesis. By questioning yourself why you're using certain hypothesis often makes it clearer if your conjecture makes sense or not.

For example:

a function $f(x)$ is defined such that $f'(x)$ is not constant and never the same for any 2 values of $x$

I know you didn't say this, but assume for a moment that $f'$ is continuous. If this is the case, it is clear that from your hypothesis, $f'$ must be strictly monotonic. So, question yourself if you have good reasons to include in your conjecture functions that are differentiable but not $C^1$. If yes, why? (I don't see any reason). If not, then you should choose a less convoluted statement (such as "$f'$ is strictly monotonic").

(Also, note that "is not constant" is redundant anyway)

Also, I know it's cool that your conjecture becomes Fermat's Last Theorem in a certain case, but you should also question yourself what fundamental difference does it make that your integration limits are integers. Since your hypotheses are incredibly broad, I don't see why integer limits would have anything special - again, this is a suggestion for you to rethink your hypothesis. Why integers are special here? Why not allow any reals? By asking yourself this kind of question, you have the means to assess the value of your own conjecture yourself.

(Observe, for example, that if your conjecture was true, it would be trivial to prove a broader version for rational limits, by shrinking/enlarging $f$ by a constant...)

Keep up the good work!

$\endgroup$
4
  • 2
    $\begingroup$ I absolutely welcome your mentorship and love your advice. I think my queries have been answered by @Hamsterrific as well as @JoseCarlosSantos(by providing a counterexample proving the conjecture incorrect), is there anyway i can "check" both your answers? $\endgroup$ Apr 2, 2018 at 3:18
  • 1
    $\begingroup$ @Doctorwho2311 thank you :) there's no way to check more than one answer... I don't mind if you accept the other one, since that's the one that really answers your specific question :) $\endgroup$
    – Pedro A
    Apr 2, 2018 at 10:22
  • 1
    $\begingroup$ Also, if the conjecture were true, it would be trivial to prove a broader version for irrational limits. But $a^n+b^n=c^n$ is trivial to solve if $a$, $b$, and $c$ are allowed to be irrational. This would be another indication that, while OP's statement is very interesting, it's not likely to be true. $\endgroup$
    – Teepeemm
    Apr 2, 2018 at 14:21
  • 1
    $\begingroup$ As a side note: the hypotheses in the original question imply that $f'$ is strictly monotone and continuous since a derivative always has the intermediate value property. $\endgroup$
    – Adayah
    Apr 2, 2018 at 15:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .