3
$\begingroup$

It was shown in here that $$\left\lvert \frac{\sin(nx)}{n\sin(x)}\right\rvert \le1\,\,\forall x\in\mathbb{R}-\{\pi k: k\in\mathbb{Z}\}$$ iff $n$ is a non-zero integer.

Using the similar argument in the same post, we are able to show that $$\left\lvert \frac{\cos(nx)}{n\cos(x)}\right\rvert \le1\,\,\forall x\in\mathbb{R}-\{\frac{(2k+1)\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer (Please alert me if this is wrong).

Now, by some graph sketching, it seems that $$\left\lvert \frac{\sin(nx)}{n\sin(x)} + \frac{\cos(nx)}{n\cos(x)} \right\rvert \le\left\lvert \frac{n+1}{n}\right\rvert \,\,\forall x\in\mathbb{R}-\{\frac{k\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer.

I am not sure if the above inquality is true. Please enlighten me!

$\endgroup$
  • $\begingroup$ If it is true, we may write $$\left\lvert \frac{\sin(nx)}{\sin(x)} + \frac{\cos(nx)}{\cos(x)} \right\rvert \le\left\lvert n+1\right\rvert \,\,\forall x\in\mathbb{R}-\{\frac{k\pi}{2} : k\in\mathbb{Z}\}$$ iff $n$ is an odd non-zero integer, instead. $\endgroup$ – pipi Oct 22 '12 at 5:49
3
$\begingroup$

You could make progress as follows:$$\frac{\sin(nx)}{\sin(x)} + \frac{\cos(nx)}{\cos(x)}=2\frac{\sin(nx)\cos(x)+\cos(nx)\sin(x)}{2\cos(x)\sin(x)}=\frac{2\sin((n+1)x)}{\sin(2x)}$$

Then with $y=2x$ you find that this is equal to $\cfrac {2\sin(\frac{(n+1)y}{2})}{\sin(y)}$ and you just need to jiggle with the factors to use the original result [odd $n$ gives an integer]

$\endgroup$
  • $\begingroup$ @ Mark Bennet I manage to get it after I posted the question! Anyway, it is a nice result to share! It may be an extension of the previous question... $\endgroup$ – pipi Oct 22 '12 at 6:21
2
$\begingroup$

Multiplying everything by $n\sin(x)\cos(x)$ and using the addition formula for sines, one sees that the task is to prove $|\sin((n+1)x)|\leqslant|n+1|\cdot|\sin(x)\cos(x)|$. Since $n$ is odd, $n=2k-1$ and, replacing $x$ by $\frac12x$, one is left with $|\sin(kx)|\leqslant |k|\cdot|\sin(x)|$.

The last inequality depends only on $|k|$ and it obviously holds for $k=0$ and $k=1$ hence a recursion over $k\geqslant1$ shows that it holds for every integer $k$.

$\endgroup$
0
$\begingroup$

$$\frac{2\sin((n+1)x)}{n\sin(2x)}\frac{2\sin((n+1)x)}{n\sin(2x)}\frac{\sin(nx)\cos(x) + \cos(nx)\sin(x)}{n\sin(x)\cos(x)} =\frac{2\sin((n+1)x)}{n\sin(2x)}=\frac{2(n+1))}{n} \frac{\sin((n+1)x)}{\frac{n+1}{2}\sin(2x)}$$

If $n+1$ is even, this inequality follows immediately from the mentioned one, while if $n+1$ is odd you should be able to prove it exactly the same way as the stated one. Actually, since you need to prove it anyhow, show directly that

$$\left| \frac{\sin((n+1)x)}{\frac{n+1}{2}\sin(2x)} \right| \leq 1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.