How to Bound a Fraction Involving Sine Functions?

[Peter_Newman]

Hi,

I have given the following, which I would like to show that this estimation is correct, where $|\theta| \leq \frac{\pi}{^2}$ and $M \geq 1$:

$$\frac{1}{M^2}\frac{\sin^2(M\theta)}{\sin^2(\theta)} \geq \frac{4}{\pi^2}$$

I would approach an estimation of the denominator via $\sin(x) \leq x$ for $x \geq 0$, then I would have the following:

$$\frac{1}{M^2}\frac{\sin^2(M\theta)}{\sin^2(\theta)} \geq \frac{1}{M^2}\frac{\sin^2(M\theta)}{(\theta)^2}$$

However, I can't come up with a reasonable estimate for the numerator (to achieve the required limit of $4/\pi^2$). Maybe my first thought is already wrong. But I simply do not come to this estimation.

I would be very grateful for helpful tips/hints!

 

[fresh_42]

We can get rid of the squares if we only consider positive parameters. That makes it easier to use e.g. the series expansion of the sine function, or the definition by the exponential function.

I would first try the latter.

Edit: What about $\theta = \pi/3\, , \,M=3$?

 

[Peter_Newman]

Hello,

I also thought about the Taylor series of $\sin^2(M\theta)$ and $\sin(\theta)$, but here the shortening/rewriting is similarly difficult. Likewise with the exponential notation.

I have put both together once, with no possibility of shortening or estimating the limit.

Series:
$\sin^2(M\theta) = M^2x^2 - \frac{M^4x^4}{3} + \frac{2M^6x^6}{45}$
$\sin^2(\theta) = x^2 - \frac{x^4}{3} + \frac{2x^6}{45}$

$$\frac{1}{M^2}\frac{\sin^2(M\theta)}{\sin^2(\theta)} = \frac{1}{M^2}\frac{M^2x^2 - \frac{M^4x^4}{3} + \frac{2M^6x^6}{45} + ... }{x^2 - \frac{x^4}{3} + \frac{2x^6}{45} + ...}$$

Exponential notation:

$$\frac{1}{M^2}\frac{\sin^2(M\theta)}{\sin^2(\theta)} = \frac{1}{M^2}\frac{-\frac{1}{4}e^{-2iM\theta} - \frac{1}{4}e^{2iM\theta} + \frac{1}{2}}{-\frac{1}{4}e^{-2i\theta} - \frac{1}{4}e^{2i\theta} + \frac{1}{2}}$$

 

[PeroK]

Hi,

I have given the following, which I would like to show that this estimation is correct, where $|\theta| \leq \frac{\pi}{^2}$ and $M \geq 1$:

$$\frac{1}{M^2}\frac{\sin^2(M\theta)}{\sin^2(\theta)} \geq \frac{4}{\pi^2}$$

That's roughly equivalent to $$\sin(M\theta) \ge \frac{2M}{\pi} \sin \theta$$
That's true for small $\theta$, as $\frac 2 \pi < 1$, but can't hold generally.

Although, it does hold for $M\theta \le \frac \pi 2$.

 

[Peter_Newman]

Hi @PeroK,

I think this $\sin(M\theta) \ge \frac{2M}{\pi} \sin \theta$ helps. But how came you up with this? It looks like something I have already seen, but I don't remember what it was called...

 

[PeroK]

Hi @PeroK,

I think this $\sin(M\theta) \ge \frac{2M}{\pi} \sin \theta$ helps. But how came you up with this? It looks like something I have already seen, but I don't remember what it was called...

If $M\theta$ is small, then $\sin(M\theta) \approx M\theta$.

But, as $M\theta$ increases beyond $\frac \pi 2$ it decreases and gets to zero at $\pi$. So, that can't work.

You need to change your bound to $|M\theta| \le \frac \pi 2$. But, even then if $M$ is large it's not going to work. Or, perhaps it does?

 

[PeroK]

Yes, try to prove it for $|M\theta| \le \frac \pi 2$.

 

[Peter_Newman]

Hello

and thanks for your help! I have not yet managed an analytical proof, I do not know where exactly to start. Is it possible to prove this using the gradient of the function or an intermediate value theorem? But I had it plotted for me:

Any further help on an analytical proof of $\sin(M\theta) \ge \frac{2M}{\pi} \sin \theta$ would be greatly appreciated!

 

[PeroK]

Hello

and thanks for your help! I have not yet managed an analytical proof, I do not know where exactly to start. Is it possible to prove this using the gradient of the function or an intermediate value theorem? But I had it plotted for me:

Any further help on an analytical proof of $\sin(M\theta) \ge \frac{2M}{\pi} \sin \theta$ would be greatly appreciated!

Okay, it's not too hard. Let $f(\theta) = \sin(M\theta) - \frac{2M}{\pi}\sin(\theta)$.

If we differentiate $f(\theta)$ we get a single local maximum somewhere in the range $(0, \frac \pi {2M})$. The minimum value, therefore, occurs at one of the endpoints.

We have $f(0) = 0$ and $f(\frac \pi {2M}) = 1 - \frac{2M}{\pi}\sin(\frac \pi {2M}) > 1 - 1 = 0$.

The last inequality follows from $\sin(\theta) < \theta$, for $\theta > 0$.

 

[Peter_Newman]

If I understand this correct your reasoning was $h(x) = f(x) - g(x) \geq 0$, than it follows that $f(x)$ is greater than $g(x)$? Can one say this?

Another question, how could you say so quickly that the maximum is in the interval $(0, \frac \pi {2M})$? For $M = 1$ you can quickly show that there is a maximum between $0$ and $\pi/2$ (via the zero of the first derivative...). But that's my underlying question, how did you then make the jump from there to the interval $(0, \frac \pi {2M})$?

Thanks for the help!

 

[PeroK]

If I understand this correct your reasoning was $h(x) = f(x) - g(x) \geq 0$, than it follows that $f(x)$ is greater than $g(x)$? Can one say this?

That's what greater than means!

Another question, how could you say so quickly that the maximum is in the interval $(0, \frac \pi {2M})$? For $M = 1$ you can quickly show that there is a maximum between $0$ and $\pi/2$ (via the zero of the first derivative...). But that's my underlying question, how did you then make the jump from there to the interval $(0, \frac \pi {2M})$?

It's just calculus (first and second derivatives etc.) to get maxima/minima on a closed interval.

 

[Peter_Newman]

I agree with you in essence, @PeroK! But I would like to take another quick look here.

So $f'(\theta) = M\cos(M\theta) - \frac{2M}{\pi}\cos(\theta)$ is the first derivative we are interested here in the zeros, because we can conclude with these then on a possible maximum. So we have:
$$f'(\theta) = M\cos(M\theta) - \frac{2M}{\pi}\cos(\theta) = 0$$
Now I would be very interested in how you proceeded here to find the zeros.

---

So my thought was and can be described approximately as follows, I picked out the individual terms and looked when these are zero, so that one has in the ideal case an identity, something like this:
$$f'(\theta) = M\cos(M\theta) - \frac{2M}{\pi}\cos(\theta) = 0$$
With that I now look at the individual cosine terms and check when these are 0.
$\cos(M\theta) = 0, \text{ if } M\theta = \frac{\pi}{^2} \Rightarrow \theta = \frac{\pi}{2M}$
$cos(\theta) = 0, \text{ if } \theta = \frac{\pi}{^2}$
$\cos(M\theta) = cos(\theta) = 0, \text{ if } \theta = 0$

I would now conclude that there is a maximum between $0$ and $\frac{\pi}{2M}$, is it ok to do it that way?

 

[PeroK]

To see that there is a single zero of $f'(\theta)$ I took out the factor of $M$ and looked at the graphs of the two components. This can be made rigorous with a little more calculus.

I can't see that looking for zeroes of each component achieves anything.

 

[Peter_Newman]

Ah ok, so you made an estimate with the plot then. However, I cannot come up with an analytical solution of $f'(\theta) = \cos(M\theta) - \frac{2}{\pi}\cos(\theta) = 0$ (I took out the factor of $M$...). I once used the CAS of my choice (WolframAlpha), which unfortunately could not give me any further hints here...

 

[PeroK]

Ah ok, so you made an estimate with the plot then. However, I cannot come up with an analytical solution of $f'(\theta) = \cos(M\theta) - \frac{2}{\pi}\cos(\theta) = 0$ (I took out the factor of $M$...). I once used the CAS of my choice (WolframAlpha), which unfortunately could not give me any further hints here...

You're not going to get a precise value for the zero. It's enough to show that there is a zero.

Just sketch the graphs.

 

[Peter_Newman]

YES, I already thought that you can't get an exact zero. But that a zero exists could be found out by a sign change criterion...

I would be interested, how this "This can be made rigorous with a little more calculus" looks exactly

 

[PeroK]

There's nothing mysterious. We have two monotonic functions that must intersect once and only once on the interval $(0, \frac{\pi}{2M})$.

 

[Peter_Newman]

You had said before that you pulled out the $M$. So we look at these two functions $f_1(\theta) = \cos(M\theta)$ and $f_2(\theta) = \frac{2}{\pi}\cos(\theta)$ and find that they intersect somewhere in the interval from $0$ to $\frac{\pi}{2M}$.

 

[PeroK]

You had said before that you pulled out the $M$. So we look at these two functions $f_1(\theta) = \cos(M\theta)$ and $f_2(\theta) = \frac{2}{\pi}\cos(\theta)$ and find that they intersect somewhere in the interval from $0$ to $\frac{\pi}{2M}$.

Yes, exactly. Looking at the graphs: The first starts at $1$ and decreases to $0$, the second starts at $\frac 2 {\pi}$, which is less than $1$ and decreases to some positive value at $\theta = \frac{\pi}{2M}$. That gives one point of intersection.

 

[Peter_Newman]

If so I have no further questions! Thank you @PeroK ! This (#19) is what I wanted to express with my post #12

 

[Peter_Newman]

One question has come up for me after all. So we now have two claims, one $|M\theta| \leq \frac{\pi}{2}$ and $|\theta| \leq \frac{\pi}{2}$, right? Originally, only this was given $|\theta| \leq \frac{\pi}{2}$ and thus alone the estimation (post #1) was given...

 

[PeroK]

One question has come up for me after all. So we now have two claims, one $|M\theta| \leq \frac{\pi}{2}$ and $|\theta| \leq \frac{\pi}{2}$, right? Originally, only this was given $|\theta| \leq \frac{\pi}{2}$ and thus alone the estimation (post #1) was given...

You need the the bound on $M\theta$.

 

[Peter_Newman]

Good morning @PeroK. Yes, that's right, we discussed that and it's understandable . But what I don't quite understand, how then in the original it just says that the estimation is for $|\theta| \leq \frac{\pi}{2}$, I find that a bit strange (Is this a typo?!).

 

[PeroK]

Good morning @PeroK. Yes, that's right, we discussed that and it's understandable . But what I don't quite understand, how then in the original it just says that the estimation is for $|\theta| \leq \frac{\pi}{2}$, I find that a bit strange.

It's a typo, I guess.

 

[Peter_Newman]

Yes, I accept that then so :) Because mathematically we had already discussed this well and understandably, that was so far great. But if one looks then into the book, one wonders however somewhat...