- #1

- 155

- 11

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!

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!

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