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

[Korisnik]

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.

 

[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 ?

 

[Buzz Bloom]

Hi Korisnik:

Unfortunately your proof is flawed.
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 e^ia and e^-ia.

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

Regards,
Buzz

 

[Korisnik]

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 ?

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?

 

[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

 

[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 $k$ ?

 

[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.

 

[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

 

[Korisnik]

Thanks for catching that.

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 $k$ ?

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.

 

[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.

 

[Korisnik]

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.

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?

 

[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.

 

[Korisnik]

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.

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