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

1. Nov 28, 2017

### WWGD

But, will the terms on polys remain within $[-1,1]$?

2. Nov 28, 2017

### Staff: Mentor

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?

3. Nov 28, 2017

### Delta²

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 $i>k(x)$.

Last edited: Nov 29, 2017
4. Nov 28, 2017

### Staff: Mentor

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 $1$.

5. Nov 28, 2017

### Delta²

$q_n(x)$ alternates but $f(q_n(x))$ converges because f(-1)=f(0)=0 .

6. Nov 28, 2017

### Staff: Mentor

I haven't checked, but what happens at $\xi = \dfrac{1}{2}(1-\sqrt{5})$ with $q_n(\xi)$ and $p_n(\xi)\,$?

7. Nov 28, 2017

### Delta²

$q_n(\xi)=\xi$ , $p_n(\xi)=\xi^n$. No problem here either because $f(\xi)=0$

8. Nov 28, 2017

### Staff: Mentor

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.

9. Nov 28, 2017

### Delta²

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.

10. Nov 28, 2017

### Staff: Mentor

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.

11. Nov 28, 2017

### Delta²

even for x neighbouring to $\xi$ , $2^np_n(x)$ is totally different than $2^n(\xi \pm \epsilon)^n$.

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

### Staff: Mentor

I'm not saying it won't work, only that it requires more care. It's not obvious, at least to me, why it will work. I assume that some topological argument might be needed to make the step from iterations to the continuum. I would prefer a theorem before cases, but that's a personal view.

13. Nov 29, 2017

### WWGD

If you are using that to approximate $f$ in $[0,1]$ then the limit should be finite, as continuous function in compact is bounded. But then you have a limit of the type $\infty . \epsilon$ which should be dominated by $\epsilon$ in order for the limit to be finite.

14. Nov 29, 2017

### WWGD

Part of the problem is that there are continuous functions on compact intervals with uncountably many zeros and even with the added condition that there are non-zero values between any two zeros, i.e., without f being zero in (a,b) , e.g. the distance function d(x,C) , for C a Cantor set. So it is difficult to discount many choices on these grounds alone. Basically, any closed subspace can be the zero set of a continuous function. This seems to allow a great amount of functions with uncountably-many zeros in a compact subspace; I thought such functions would be rare.

15. Nov 29, 2017

### StoneTemplePython

It seems like a lot more brain power has gone into this than perhaps the question wanted...

Another approach would be to show that for any any $x \in (-1,0)$, after a large enough finite number of iterations, the resulting x becomes 'trapped' in $\{(-1, 0.99], [0.01, 0)\}$ and with respect to the magnitude of the function the effect of the x scaling over two iterations is a multiple less than one fourth, which overwhelms the scaling of $2^2 = 4$, and hence the infinite product decays toward zero. This can be shown for 3 iterations at a time as well. It's sufficient to show this for 10 iterations for starting x = 0.5.

With respect to the $2x$ factor. Another way to think about it is how it relates to $GM \leq AM$ where the inequality is strict unless all values are the same. If we wanted to have some fun, look at the magnitude of the scalars $\big \vert \approx 0\big \vert, \big \vert \approx -1\big \vert$, and see that the arithmetic mean is $\frac{1}{2}$ which is strictly larger than the geometric mean. Hence the effect of scaling by $x$ in our infinite product must overwhelm the scaling by 2. The infinite product decays toward zero.

16. Nov 29, 2017

### jaus tail

We've only been taught of limits and derivatives and integration.
I tried using derivatives.
f(x) = 2x(f(x2-a))
Let f(x^2 - 1) = g(x)
So
f(x) = 2x g(x)
f'(x) = 2xg'(x) + 2g(x)
f'(x) = 2xf'(x)(2x) + 2f(x^2 - 1)
f'(x) (1-4x^2) = 2f(x^2 - 1)
So we get f'(x) = 2f(x^2 - 1)/(1-4x^2)

Now we got that f(-1) = f(1) = 0.
So if f(x) is non zero at any point between 1 and -1, then f'(x) must be positive some place and negative some place.
But above we see that f'(x) is an even function.
So f'(x) is not satisfying condition of positive and negative. It's either positive or negative and if it's one of them then it can have only 1 zero.
And so we can say that f(x) is constant at zero.

Aren't all continuous functions derivative? I'm not sure what's the difference. Sine and consine are continous and differentiable. So i guess it should be true.

17. Nov 30, 2017

### PeroK

Not all continuous functions are differentiable. One example is the modulus $|x|$, which is not differentiable at $x = 0$.

There is, in fact, the Weierstrass function, which is continuous but nowhere differentiable!

https://en.wikipedia.org/wiki/Weierstrass_function

18. Nov 30, 2017

### jaus tail

So I guess we can take two conditions.
1) f(x) is differentiable then we can go with my solution.
and 2) f(x) is not differentiable. Any idea how to prove in this case?

19. Nov 30, 2017

### PeroK

First:

$g'(x) = 2xf'(x^2-1) \ne 2xf'(x)$

But also, I don't understand how you get that an even function can't take positive and negative values.

I wonder where you got this question, because it seems to be a little beyond your level at the moment.

20. Nov 30, 2017

### PeroK

See @Ray Vickson's solution in post #14.

21. Nov 30, 2017

### Delta²

If you are allowed to use integral and derivatives and all that, follow Ray Vickson solution at post #14. It is mostly correct, and it uses the theorem that $F(x)=\int_0^x{f(y)dy}$ is differentiable and it is $F'(x)=f(x)$ (that is the fundamental theorem of calculus). it doesn't need the assumption that f(x) is differentiable.

If you are not allowed to use integral and derivatives then you can follow my approach explained in post #15, #20 and #28 which I believe is also correct.

22. Nov 30, 2017

### Delta²

$2^np_n(x)$ is dominated by $p_n(x)$ for any $x\neq\xi$ its $\lim_{n->\infty}{p_n(x)}=0$ and the convergence of $p_n(x)$ to 0 is such that it overcomes $2^n$. I believe I prove this at post #28.

23. Nov 30, 2017

### PeroK

If you take an interval close enough to -1 then you can generate a sequence of function values of increasing magnitude without bound in that interval. Which contradicts continuity, unless the function is zero on the interval. After that your argument simplifies for other points.