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

1. Nov 28, 2017

### jaus tail

1. The problem statement, all variables and given/known data
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]

2. Relevant 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.

3. 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.

2. Nov 28, 2017

### BvU

How ? And: if you are right, and if you can also get the opposite, you have it !

3. Nov 28, 2017

### jaus tail

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. Nov 28, 2017

### jaus tail

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. Nov 28, 2017

### PeroK

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. Nov 28, 2017

### jaus tail

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. Nov 28, 2017

### PeroK

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!

Last edited: Nov 28, 2017
8. Nov 28, 2017

### Gigaz

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. Nov 28, 2017

### Staff: Mentor

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. Nov 28, 2017

### PeroK

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. Nov 28, 2017

### Staff: Mentor

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.

12. Nov 28, 2017

### PeroK

That there might be a continuous function that meets the criteria?

13. Nov 28, 2017

### Staff: Mentor

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. Nov 28, 2017

### Ray Vickson

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, $F(x) = \text{constant}.$

15. Nov 28, 2017

### Delta²

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.

Last edited: Nov 28, 2017
16. Nov 28, 2017

### PeroK

I found that this sequence converges alternatingly to $0$ and $-1$.

17. Nov 28, 2017

### Delta²

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. Nov 28, 2017

### PeroK

Can you prove that? Try starting at 0.5, for example, and see what happens.

19. Nov 28, 2017

### Ray Vickson

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 $(1-\sqrt{5})/2.$

Last edited: Nov 28, 2017
20. Nov 28, 2017

### Delta²

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.

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