2

Let $m$ be an odd positive integer.

I showed that

  • $\frac{\sin(mx)}{\sin(x)}$ is a polynomial in $\sin(x)^2$ with degree $\leq \frac{m-1}{2}$ (by induction)

  • $\frac{\sin(mx)}{\sin(x)} = 0$ for $x =\frac{2\pi j}{m}$ and $j=1,..., \frac{m-1}{2}$ (obviously).

Now I need to show that $\frac{\sin(mx)}{\sin(x)}$ is a polynomial in $\sin(x)^2$ with degree $= \frac{m-1}{2}$ and that $C$ so that $\frac{\sin(mx)}{\sin(x)} = C \cdot \prod\limits_{j=1}^{(m-1)/2} (\sin(x)^{2} - \sin(\frac{2\pi j}{m})^{2})$ exists.

$\frac{sin(mx)}{sin(x)}$ has degree $\leq \frac{m-1}{2}$ and if we show that $sin(\frac{2\pi j}{m})^{2}$ is distinct for all $j=1,..., \frac{m-1}{2}$ we should have proved it (?)

I tried to use the unit circle:

If $sin(\frac{2\pi j}{m})^{2}$ shall has to be distinct for all $j$, so has to be $sin(\frac{2\pi j}{m})$. That should make things easier, but I struggle here.

J. W. Tanner
  • 60,406

2 Answers2

1

Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ for $j_1,j_2$

We need $\dfrac{2\pi}m(j_1\pm j_2)=n\pi$ for some integer $n$

$\iff j_1\pm j_2=\dfrac{mn}2$

lab bhattacharjee
  • 274,582
0

Notice that $4\sin(t/2)^2=2-2\cos(t)$ and go from there.

santa
  • 21
  • 2
  • i have absolutely no idea how this could help me...sorry –  Jan 29 '19 at 00:13
  • $\sin(2j\pi/m)^2=\frac{2-2\cos(4j\pi/m)}{4}$ and $j=1,\ldots,\frac{m-1}{2}$,

    $m$ is assumed odd, then $n=(m-1)/2$ even, we will have with $f(t)=\frac{2-2\cos(t)}{4}$, $t_j=\frac{2j\pi}{n+1/2}$, and $j=1,\ldots, n$. Since $f$ is even, this is equivalent to having $t_j=\frac{j\pi}{n+1/2}, j=1,\ldots,n$, and thus distinct $\sin(2j\pi/m)^2$ since $f$ is monotonically increasing in $t_j\in[0,\pi]$.

     – santa Jan 29 '19 at 06:54
  • $m = 7$ so $n = 3$ is not even. –  Jan 29 '19 at 17:40
  • sorry about that. the rest should be ok :) – santa Jan 29 '19 at 18:38
  • but where do you use that $m$ is odd. Because its not true if $m$ is even. –  Jan 29 '19 at 21:05
  • So, I look at it this way. You have the following for $t_j=\frac{2j\pi}{n+1/2}$.

    Assume $n$ defined before is even

    1. Then for $j=1,\ldots,n/2$ you have $t_j$ correspond to the even indices $i=2,4,6,\ldots,n$ in a $t_i=\frac{i\pi}{n+1/2}$
     – santa Jan 29 '19 at 22:43
  • Now consider the indices $j=n/2+1,\ldots,n$

    • First example $j=n/2+1$ you have $t_j=\frac{2(n/2+1)\pi}{n+1/2}=\frac{(n+2)\pi}{n+1/2}=\pi+\frac{3/2\pi}{n+1/2}$. For $j=n/2+\beta$ you will have $\pi+\frac{(2\beta-1)/2\pi}{n+1/2}$.

    Thus for $j=n/2,\ldots,n$ you will have $t_i=\pi+\frac{i\pi}{n+1/2}$ where $i=3/2,7/2,\ldots,(2\beta-1)/2,\ldots,(2n-1)/2$. which are obviously distinct from the $t_i$ in 1.

    You can do the same construction for odd $n$.

    Sorry for a bit sloppy notation.

    – santa Jan 29 '19 at 22:43