Prove f(x) is zero in range [-1,1]

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  • #1
jaus tail
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Homework Statement


A function f(x) is continuous in the range [-1,1]
f(x) = 2xf(x2 - 1)

Prove that f(x) = 0 everywhere in range [-1,1]

Homework Equations


I don't know how to proceed.
By putting values as x = 0, I got f(0) = 2*0 = 0.
f(1) = 2*1*f(0) = 0.

And I also get f(negative values) = f(positive values). So it's like a cosine wave.

But how to prove that the values are equal to zero.

The Attempt at a Solution


How to proceed? Continuous means value of f(x) when I approach from left side = value of f(x) when I approach from right side. Not sure how to imply it here.
 

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  • #2
BvU
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And I also get f(negative values) = f(positive values)
How ? And: if you are right, and if you can also get the opposite, you have it !
 
  • #3
jaus tail
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No, I think i wrote wrong.
Putting x = -1,
I get f(-1) = 2(-1)*f((-1)2 -1 )
which is f(-1) = -2f(0)
and f(1) = 2f(0)

f(0) = 0*f(-1) = 0.
So f(0) = 0
So f(-1) = 0 and so is f(1) from underlined parts.

So f(x) = 0 for x = -1, 0, 1. But how to prove for other values of x?

when x = 0.5. I get f(0.5) = 2*0.5 f(0.25 - 1) = f(0.75)
and when I put x = 0.75 I get
f(0.75) = 2*0.75 f(0.5625-1) = 1.5f(-0.4375)

I don't know how to prove that all these f(x) are equal to zero.
 
  • #4
jaus tail
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How ? And: if you are right, and if you can also get the opposite, you have it !
Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)
 
  • #5
PeroK
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Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)

This looks tricky. What happens if you start from any point in ##x_1 \in [-1, 1]## and generate a recursive sequence by ##x_{n+1} = x_n^2 - 1##?
 
  • #6
jaus tail
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What's a recursive sequence?

I tried with mean value theorem.
a = -1, b = 1
there is a point c between a and b such that f'(c) = [ f(b) - f(a) ] / [b - a]
this is zero since f(-1) = f(1) = 0
So since f'(c) is zero, we get a constant value and since one constant value is zero we can say that f(x) = constant = 0.
 
  • #7
PeroK
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What's a recursive sequence?

I tried with mean value theorem.
a = -1, b = 1
there is a point c between a and b such that f'(c) = [ f(b) - f(a) ] / [b - a]
this is zero since f(-1) = f(1) = 0
So since f'(c) is zero, we get a constant value and since one constant value is zero we can say that f(x) = constant = 0.

It doesn't say that ##f## is differentiable. But, another idea is to use the fact that a continuous function attains its max and min on a closed interval.

Regarding my previous idea. If you start with a number. E.g. ##x_1 = 1/2##, then ##x_2 = x_1^2 - 1 = -3/4## and ##x_3 = x_2^2 - 1 = -7/16## and so on. This sequence may converge to a point ##x## where ##x^2 - 1 = x## (if there is such a point).

These are just ideas. I haven't spotted the solution yet, so they may not help!
 
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  • #8
Gigaz
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Simple proof:

Go to the point where x²-1=x and call it P12. (Should be -0.6180339887498949 and 1.6180339887498949)
f(P12)=2P12 f(P12)
1/2=P12 except if f is zero. And because P12 is clearly not 1/2, f must be zero.

Edit: Wait im stupid. That says only that f(P12) must be zero...
 
  • #9
fresh_42
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It doesn't say that ##f## is differentiable.
So maybe integration is an idea. What about ##F(x):= \int_{-1}^x f(y)dy\,##? With ##g(x)=x^2-1## we get easier conditions plus differentiability. However, I haven't found the trick, yet.
 
  • #10
PeroK
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So maybe integration is an idea. What about ##F(x):= \int_{-1}^x f(y)dy\,##? With ##g(x)=x^2-1## we get easier conditions plus differentiability. However, I haven't found the trick, yet.

Do you think it might be false?

The condition on the interval ##[0, 1]## is irrelevant. If you can satisfy the condition on ##[-1, 0]## that would do it.
 
  • #11
fresh_42
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Do you think it might be false?
What could be false? I only thought continuity invites to be integrated to ##F'(x)=f(x)## with ##F(0)=F(-1)=F(1)=0## and ##F=F\circ g## if I didn't make a mistake. These are easier to handle.
The condition on the interval ##[0, 1]## is irrelevant. If you can satisfy the condition on ##[-1, 0]## that would do it.
Sure, but do we have it on one of them? I'm looking for an argument without iterations.
 
  • #12
PeroK
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What could be false?

That there might be a continuous function that meets the criteria?
 
  • #13
fresh_42
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That there might be a continuous function that meets the criteria?
I don't get it, sorry. Since it is supposed to result in ##f(x)=0## it'll be both, continuous and differentiable. I think that this is the only solution, I just don't see the killer argument. I simply thought, that we can get differentiability for free by an integration.
 
  • #14
Ray Vickson
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Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)

What could be false? I only thought continuity invites to be integrated to ##F'(x)=f(x)## with ##F(0)=F(-1)=F(1)=0## and ##F=F\circ g## if I didn't make a mistake. These are easier to handle.

Sure, but do we have it on one of them? I'm looking for an argument without iterations.

Iteration plus integration will do the job. Let
$$ F(x) = \int_0^x f(t) \, dt,$$
so that we get
$$F(x) = F(x^2-1) - F(-1).$$
Since ##f## is an odd function, ##F## is even; thus ##F(-1) = F(1)##, so we have ##F(x) = F(x^2-1) - F(0)##. Using ##F(0)=0## we have ##F(1) = -F(1)##, so ##F(1) =0.## Therefore, we have
$$F(x) = F(x^2 - 1).$$

Now, for any ##x_0 \in (-1,1)## the iterative scheme ##x_1 = x_0^2-1, x_2 = x_1^2 - 1, \ldots## converges to the negative solution of ##x = x^2-1##, which is ##\tau \equiv (1-\sqrt{r})/2 \doteq -0.61803##. (We can see this by looking at the "cobweb diagram" of the iteration scheme ##x_{n+1} = x_n^2 - 1.##). Since ##F(x_{n+1}) = F(x_n)## for all ##n \geq 0## we have ##F(x_0) = \lim_{n \to \infty} F(x_n ) = F(\tau)## for any ##x_0##. In other words, ##F(x) = \text{constant}.##
 
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  • #15
Delta2
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My idea is that ##f(x)=2xf(x^2-1)=2x2(x^2-1)f((x^2-1)^2-1)=...=2^np_n(x)f(q_n(x))## where ##p_n## and ##q_n## polynomials of degree ##2^n-1## and ##2^n##. If we can prove that ##\lim_{n->\infty}q_n(x)=0## then I think we got it.
 
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  • #16
PeroK
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Now, for any ##x_0 \in (-1,1)## the iterative scheme ##x_1 = x_0^2-1, x_2 = x_1^2 - 1, \ldots## converges to the negative solution of ##x = x^2-1##, which is ##\tau \equiv (1-\sqrt{r})/2 \doteq -0.61803##.

I found that this sequence converges alternatingly to ##0## and ##-1##.
 
  • #17
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I found that this sequence converges alternatingly to ##0## and ##-1##.
That's only if you start with 0, 1 or -1. for any other x in (-1,1) it converges to -0.618...=##\frac{1-\sqrt{5}}{2}##.

Now I see clearly that the polynomial ##q_n(x)## in post #15 converges to -0.618.. for any x in (-1,0) or (0,1) and f(-0.618...)=0.
 
  • #18
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That's only if you start with 0, 1 or -1. for any other x in (-1,1) it converges to -0.618...=##\frac{1-\sqrt{5}}{2}##

Can you prove that? Try starting at 0.5, for example, and see what happens.
 
  • #19
Ray Vickson
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Can you prove that? Try starting at 0.5, for example, and see what happens.

Yes, you are right. However, the basic limiting argument still holds: ##F(x_0) = F(x_n) ## for all ##n##, and for very large ##n## we have that ##x_n## is either ##0## or ##-1## (alternating). But since ##F(0) = F(-1) = 0## we have ##F(x_0) = 0##. That is for any ##x_0 \in (-1,1)## different from ##(1-\sqrt{5})/2.##
 
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  • #20
Delta2
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Can you prove that? Try starting at 0.5, for example, and see what happens.

Well I ve got to eat my hat, seems you are right. But still the sequence ##f(q_n(x))## converges alternatingly to f(0) or f(-1) and since they are equal to 0, converges to zero.
 
  • #22
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I think he is right but still your proof is mostly correct because ##F(x_{2n})## converges to ##F(0) ## and ##F(x_{2n+1})## converges to ##F(-1)## but ##F(0)=F(-1)=0##.
 
  • #23
Ray Vickson
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And, yet:

0.5, -0.75, -0.4375, -0.8086, -0.34618, -0.88016 etc.

I edited the message to remove those incorrect statements, but you responded before I made the changes. Apparently, the version you responded to remains unchanged. As the edited version makes plain, the basic argument still goes through.
 
  • #24
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Well, by repeatedly applying the IVT we have that f is continuous with infinitely-many zeros. Only non-zero function I know satisfies this is the distance function d(x,S) for ##S## a closed subspace ( of [-1,1] ). Can we exclude this ? (Yes) EDIT: Does anyone know of EDIT any other non-zero continuous function that can have infinitely-many zeros ( in a compact subset of Real line)? Obviously not a poly, but that does not narrow it too much. Maybe we can show the 0's are dense , then extending by continuity we can show the function is identically 0?

EDIT2: My bad, should have been obvious that a "contraction" of sinx or cosx would have produced an example:https://math.stackexchange.com/questions/48746/existence-of-non-constant-continuous-functions-with-infinitely-many-zeros
STILL, this gives us countably-many. Could we have uncountably-many zeros? This would imply the zero set would have a limit point in ##[-1,1]##, which excludes Complex-analytic functions ( if the function was analytic to start with ). Seems like a Baire-Cat -type argument could show it.
 
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  • #25
Delta2
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Guys do you find my idea in post #15, further explained in post #20 as correct? My approach doesn't use integrals or derivatives or the fundamental theorem of calculus or IVT which I believe probably they aren't yet being taught to the OP...
 
  • #26
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My idea is that ##f(x)=2xf(x^2-1)=2x2(x^2-1)f((x^2-1)^2-1)=...=2^np_n(x)f(q_n(x))## where ##p_n## and ##q_n## polynomials of degree ##2^n-1## and ##2^n##. If we can prove that ##\lim_{n->\infty}q_n(x)=0## then I think we got it.
But, will the terms on polys remain within ##[-1,1]##?
 
  • #27
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I still think that ##F = F \circ g## with ##g(x)=x^2-1\, , \,F'(x)=f(x)## and ##F(-1)=F(0)=F(-1)## is the key. To write it this way is too invitingly. Why should there be the factor two otherwise?
 
  • #28
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I still think that ##F = F \circ g## with ##g(x)=x^2-1\, , \,F'(x)=f(x)## and ##F(-1)=F(0)=F(-1)## is the key. To write it this way is too invitingly. Why should there be the factor two otherwise?

yes you got a point here, the 2 factor "screams" that 2x is the derivative of ##x^2-1##.

Still I believe it can be done my way and without the use of integrals and derivatives. My main difficulty is proving that ##2^np_n(x)## is bounded but I believe it can be done.

It is ##p_n(x)=q_1(x)q_2(x)...q_{n-1}(x)##
For ##q_n(x)## we have that ##\lim_{n->\infty}{q_{2n}(x)}=0##, hence we can find a ##k(x)## such that ##q_{2n}(x)>-\frac{1}{4}## for ##n>k(x)##.

So it will be ##p_n(x)=q_1(x)...q_{2k(x)}(x)q_{2k(x)+1}(x)...q_n(x)=P_{2k(x)}(x)q_{2k(x)+1}(x)...q_n(x)##.

to prove that ##2^np_n(x)## is bounded we break it down to
##2^{2k(x)}P_{2k(x)}2^{n-2k(x)}q_{2k(x)+1}(x)...q_n(x)##

##2^{2k(x)}P_{2k(x)}## is bounded cause its finite. The rest part of product I believe it is also bounded for any n because ##q_{2i}(x)>\frac{-1}{4}## for any ##i>k(x)##.
 
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  • #29
fresh_42
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yes you got a point here, the 2 factor "screams" that 2x is the derivative of ##x^2-1##. Still I believe it can be done my way and without the use of integrals and derivatives. My main difficulty is proving that ##2^np_n(x)## is bounded but I believe it can be done.
But this still leaves the same convergence problem as with Ray's idea: ##q_n(x)## alternates between ##-1## and ##0## and I'm not sure that the boundaries won't cause problems. The compactness of ##[-1,1]## might play a role, but it is simultaneously an obstacle to get a "norm" below ##1##.
 
  • #30
Delta2
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But this still leaves the same convergence problem as with Ray's idea: ##q_n(x)## alternates between ##-1## and ##0## and I'm not sure that the boundaries won't cause problems. The compactness of ##[-1,1]## might play a role, but it is simultaneously an obstacle to get a "norm" below ##1##.
##q_n(x)## alternates but ##f(q_n(x))## converges because f(-1)=f(0)=0 .
 
  • #31
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##q_n(x)## alternates but ##f(q_n(x))## converges because f(-1)=f(0)=0 .
I haven't checked, but what happens at ##\xi = \dfrac{1}{2}(1-\sqrt{5})## with ##q_n(\xi)## and ##p_n(\xi)\,##?
 
  • #32
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##q_n(\xi)=\xi## , ##p_n(\xi)=\xi^n##. No problem here either because ##f(\xi)=0##
 
  • #33
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Yes, but ##2^n\xi^n## explodes and as we don't know much about ##f##, it cannot be compensated by ##f(q_n(x))##, at least we don't know.
 
  • #34
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you can prove another way ##f(\xi)=0## check post #8. We must exclude ##\xi## from this treatment and treat it as a special case.
 
  • #35
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you can prove another way ##f(\xi)=0## check post #8. We must exclude ##\xi## from this treatment and treat it as a special case.
That's not my point. ##f(\xi)=0## directly follows from the defining equation. But the exclusion of ##\xi## might not be sufficient, because ##2^n (\xi \pm \varepsilon)^n## is also divergent and polynomials are continuous, so at first glance the construction looks unstable.
 

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