Simple Calculus view on Fermat's Last Theorem
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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 $ale ble c$ such that,
$$int_0^{a} f(x)dx = int_b^{c} f(x)dxtag1$$
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.
real-analysis number-theory conjectures
edited Apr 5 '18 at 13:59
PJTraill
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
$endgroup$
|
show 7 more comments
$begingroup$
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 $ale ble c$ such that,
$$int_0^{a} f(x)dx = int_b^{c} f(x)dxtag1$$
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.
real-analysis number-theory conjectures
edited Apr 5 '18 at 13:59
PJTraill
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
$endgroup$
1
$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
1
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
8
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
1
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14
|
show 7 more comments
$begingroup$
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 $ale ble c$ such that,
$$int_0^{a} f(x)dx = int_b^{c} f(x)dxtag1$$
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.
real-analysis number-theory conjectures
edited Apr 5 '18 at 13:59
PJTraill
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
$endgroup$
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 $ale ble c$ such that,
$$int_0^{a} f(x)dx = int_b^{c} f(x)dxtag1$$
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.
real-analysis number-theory conjectures
real-analysis number-theory conjectures
edited Apr 5 '18 at 13:59
PJTraill
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
edited Apr 5 '18 at 13:59
PJTraill
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
edited Apr 5 '18 at 13:59
PJTraill
688519
edited Apr 5 '18 at 13:59
PJTraill
688519
edited Apr 5 '18 at 13:59
PJTraill
688519
688519
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
asked Apr 1 '18 at 14:29
Doctorwho2311Doctorwho2311
373210
373210
1
$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
1
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
8
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
1
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14
|
show 7 more comments
1
$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
1
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
8
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
1
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14
1
1
$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 at 14:50
$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 at 14:50
5
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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
$begingroup$
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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
1
1
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
8
8
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
1
1
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Interesting. But not true. If you define $f(x)=dfrac1{x+1}$, then$$int_0^1f(x),mathrm dx=int_1^3f(x),mathrm dx,$$because$$int_alpha^beta f(x),mathrm dx=logleft(frac{beta+1}{alpha+1}right).$$
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
add a comment |
$begingroup$
(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!
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
$endgroup$
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 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 :)
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– Pedro A
Apr 2 '18 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 '18 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 '18 at 15:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Interesting. But not true. If you define $f(x)=dfrac1{x+1}$, then$$int_0^1f(x),mathrm dx=int_1^3f(x),mathrm dx,$$because$$int_alpha^beta f(x),mathrm dx=logleft(frac{beta+1}{alpha+1}right).$$
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
add a comment |
$begingroup$
Interesting. But not true. If you define $f(x)=dfrac1{x+1}$, then$$int_0^1f(x),mathrm dx=int_1^3f(x),mathrm dx,$$because$$int_alpha^beta f(x),mathrm dx=logleft(frac{beta+1}{alpha+1}right).$$
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
add a comment |
$begingroup$
Interesting. But not true. If you define $f(x)=dfrac1{x+1}$, then$$int_0^1f(x),mathrm dx=int_1^3f(x),mathrm dx,$$because$$int_alpha^beta f(x),mathrm dx=logleft(frac{beta+1}{alpha+1}right).$$
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
Interesting. But not true. If you define $f(x)=dfrac1{x+1}$, then$$int_0^1f(x),mathrm dx=int_1^3f(x),mathrm dx,$$because$$int_alpha^beta f(x),mathrm dx=logleft(frac{beta+1}{alpha+1}right).$$
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
edited Dec 29 '18 at 19:20
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
answered Apr 1 '18 at 14:41
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
add a comment |
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
$begingroup$
what about when a,b,c are distinct integers, and not 0
$endgroup$
– Doctorwho2311
Apr 1 '18 at 16:53
8
8
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 at 16:55
$begingroup$
@Doctorwho2311 What about $a=1$, $b=3$, and $c=7$ then?
$endgroup$
– José Carlos Santos
Apr 1 '18 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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 at 17:25
$begingroup$
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 $f(infty)=infty$
$endgroup$
– Doctorwho2311
Apr 1 '18 at 17:25
5
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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
$begingroup$
@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 $lim_{xto+infty}f(x)=+infty$, you should have stated it from the start.
$endgroup$
– José Carlos Santos
Apr 1 '18 at 17:31
add a comment |
$begingroup$
(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!
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
$endgroup$
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 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 '18 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 '18 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 '18 at 15:45
add a comment |
$begingroup$
(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!
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
$endgroup$
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 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 '18 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 '18 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 '18 at 15:45
add a comment |
$begingroup$
(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!
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
$endgroup$
(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!
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
answered Apr 1 '18 at 22:08
Pedro APedro A
2,0661827
2,0661827
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 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 '18 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 '18 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 '18 at 15:45
add a comment |
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 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 '18 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 '18 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 '18 at 15:45
1
1
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 at 3:18
$begingroup$
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?
$endgroup$
– Doctorwho2311
Apr 2 '18 at 3:18
1
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 '18 at 10:22
$begingroup$
@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 '18 at 10:22
1
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 '18 at 14:21
$begingroup$
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 '18 at 14:21
1
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 '18 at 15:45
$begingroup$
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 '18 at 15:45
add a comment |
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$begingroup$
For $a=b=c=0$, we have $0ge0ge0$ and $0$ certainly is integer, but independent of $f$, $(1)$ reduces to $0=0$ which obviously is true.
$endgroup$
– celtschk
Apr 1 '18 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.
$endgroup$
– Steven Stadnicki
Apr 1 '18 at 14:57
1
$begingroup$
It is far too general. There are many ways to produce counter-examples.
$endgroup$
– DanielWainfleet
Apr 1 '18 at 21:04
8
$begingroup$
Bummer that your teachers won't discuss this. Can't imagine turning this question away.
$endgroup$
– Randall
Apr 2 '18 at 2:40
1
$begingroup$
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?
$endgroup$
– Acccumulation
Apr 2 '18 at 16:14