# Prove f(x) = sin(x^2) isn't periodic

1. Oct 7, 2015

### Korisnik

• Member warned about posting homework questions w/o the template
Prove f(x) = sin(x^2) isn't periodic.

f is periodic if f(x + t) = f(x) for all x in f's domain.

I know about 1 way of proving it by derivative and boundedness but is this another way?

If f is periodic then
sin(x^2) = sin((x + t)^2)
Finding t:
(I don't know the proper way of doing this, maybe I ought to add + 2kpi)
x^2=x^2 +2xt + t^2
(I could square root it but it's the same)
t(t + 2x) = 0
t1 = 0 or t2 = -2x

t > 0 so t1 isn't.
t(x) = -2x, but t can't be dependant on x, as it needs to be constant, so t2 isn't a solutio . So I have proved the hipothesis.

2. Oct 7, 2015

### geoffrey159

As you said, your function is periodic iff there is $T>0$ such that $f(x+T) = f(x)$ for all real $x$.
Assume $f$ is periodic. If you set $x = 0$, what constraint does it impose on $T$ ? Now if $x = \sqrt{\frac{\pi}{2}}$, does your function conserve its periodicity with the T you just found ?

3. Oct 7, 2015

### Buzz Bloom

Hi Korisnik:

sin(x^2) = sin((x + t)^2) does not imply that x^2 = (x + t)^2.

I have not found a proof, but I have two suggestions that you might try. However, I am not sure either will be helpful.

1. Use the formula for sin(a+b).

2. Use the formula expressing sin(a) in terms of eia and e-ia.

My apologies. I haven't done proofs for many decades. Geoffrey has the suggested the right approach.

Regards,
Buzz

4. Oct 7, 2015

### Korisnik

Alright, so $T = \sqrt{k\pi}, k\in\mathbb{Z}$, as $T>0$. Now $x = \sqrt{\frac{\pi}{2}}$ so $f\left(\sqrt{\frac{\pi}{2}}\right) = 1$ and $f\left(\sqrt{\frac{\pi}{2}} + \sqrt{k\pi}\right) = \sin\left(\frac{\pi}{2} + k\pi +2\pi\sqrt{\frac{k}{2}}\right)$ which will equal $1$ iff $\frac{\pi}{2} + k\pi +2\pi\sqrt{\frac{k}{2}} = \frac{\pi}{2}+2k\pi$. The solutions are $0$ and $2$. If $k = 0$ $T = 0$. For $k = 2$ $T = \sqrt{2\pi}$. Now $f\left(\sqrt{\frac{\pi}{2}}\right) = 1 = f\left(\sqrt{\frac{\pi}{2}} + \sqrt{2\pi}\right)$ So the function is periodic?

I checked for $x = \sqrt{\frac{\pi}{4}}$ and it didn't work, so I guess it isn't periodic then. Is this now an okay proof or still no?

5. Oct 7, 2015

### Buzz Bloom

Hi Korisnik:

Yes, you have a proof now. But you may want to restructure it a little for clarity.

Step 1 assumes x=0. From this you find all the possible values of T that could possible be the period: T=√kπ, k any integer but 0.
Step 2 assumes x = √π/2. This reduces the possible periods to +/- √2π.
Step 3 assumes x = √π/4. This shows that neither t = +√2π nor t=-√2π are a period.

Regards,
Buzz

6. Oct 7, 2015

### geoffrey159

If $x = 0$, you want to solve $\sin (T^2) = 0$ and $T > 0$. Therefore $T^2 \in \{ k\pi,\ k\in\mathbb{N}-\{0\} \}$. So there is an integer $k \ge 1$ such that $T = \sqrt{k\pi}$.

Now if $x = \sqrt{\frac{\pi}{2}}$ and $f$ is periodic, you must have $1=f(x) = f(x+T) = (-1)^k \cos(\sqrt{2k}\pi )$. Now you see that $\sqrt{2k}$ is odd when $k$ is odd and even when $k$ is even.
What can you say depending on the parity of $k$ ?

7. Oct 7, 2015

### RUber

Essentially, this boils down to the fact that squaring is not linear.
Assume a period exists. Call it T.
Then sin((kT)^2) = 0. Great. So T^2 is a period of the sine function.
For any x, sin((x+kT)^2) = sin(x^2).
But, sin((x+kT)^2) = sin(x^2 + 2kTx +(kT)^2 ). Since T^2 is a period of the sine function, you can say that this is also equal to sin(x^2 + 2kTx).
I am pretty sure it doesn't take too many steps to show that 2kTx can't be a period of sin(x) since it is a function of x.

8. Oct 8, 2015

### Buzz Bloom

Hi Korisnik:

I apologize again. At my age I have become too careless to do proofs, so I will stay out of this thread after this,

Since my quotes of equations get messed up, I have reproduced below an equation from your post #4:

π/2 + kπ +2π√(k/2) = π/2 + 2kπ​

The error you made (which I failed to catch before) is assuming that the k variable on the left hand side is the same as the k variable on the right hand side. These two k's need not be the same value. Therefore your conclusion that k must be 0 or 2 is wrong.

Regards,
Buzz

9. Oct 8, 2015

### Korisnik

Thanks for catching that.
I'm not sure what you mean by that, as decimal numbers (possibly square root of 2k) don't have parity. The obvious is that the sign changes depending on k.

10. Oct 8, 2015

### geoffrey159

Right, the sign changes and $\sqrt{2k}$ must be a positive integer of the same parity than $k$.
Show it can't be odd. This will give you another constraint on the period, assuming it does exist.
Finally, try the case $x=\sqrt{\pi}$, and you will get an absurd result which will prove that the hypothesis of periodicity is false.

11. Oct 8, 2015

### Korisnik

2k = 4t^2, t \in Z, so k is even.
2k = 4t^2 + 4t + 1, so now 2 times k is an odd number, which it can't be, as k is integer: 2k = 2(2t^2 + 2t) + 1 = 2m + 1 (t and m aren't specified, but are integers). And we know that odd squared gives odd, and even squared gives even, not odd, so k can't be odd.

I get this result (plugging in sqrt(k*pi) for T and sqrt(pi) for x):

1 + k + 2*sqrt(k) = m, m is > 0 integer. I'm not sure what you mean by absurd result. How does this yield a solution?

12. Oct 8, 2015

### geoffrey159

Ok, so $k = 2 t^2$. Then the period has an additional constraint : $T = \sqrt{k\pi} = \sqrt{2\pi} t$.
Now try $T$ with $x = \sqrt{\pi}$, with the same kind of reasoning you should be able to conclude.

13. Oct 8, 2015

### Korisnik

Right, I get sqrt(2)*2t + 2t^2 + 1 = m, m is integer. Can't be. Thanks. :)