Exercise 8.5.9

Question

Use the previous identity to show that
$$\frac{1/2 + D_1(\theta) + D_2(\theta) + \cdots + D_N(\theta)}{N+1} = \frac{1}{2(N+1)} \left[\frac{\sin\left((N + 1) \frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\right]^2$$

Ulisse Mini · Accepted Answer

We need one more trigonometric identity, for any $a, b$:
$$\begin{aligned}
    \sin(a+b)\sin(a-b) &= \left(\sin x \cos b + \sin b \cos a\right) \left(\sin a \cos b - \sin b \cos a\right) \
    &= \sin^2 a \cos^2 b - \sin^b \cos^a \
    &= \sin^2 a \left(1 - \sin^2 b\right) - \sin^2 b \left(1 - \sin^2 a\right) \
    &= \sin^2 a - \sin^2 a \sin^2 b - \sin^2 b + \sin^2 a \sin^2 b\
    \sin(a+b) \sin(a-b) + \sin^2 b &= \sin^2 a
\end{aligned}$$
Assume that $\theta \neq 2n\pi$ for integer $n$; otherwise the right side isn't defined. What follows is a lot of algebra:
\[\begin{aligned}
\frac{1}{2} + \sum_{i=1}^N D_i(\theta)
&= \frac{1}{2} + \frac{1}{2 \sin \frac{\theta}{2}} \sum_{i=1}^N \sin \left(i \theta + \frac{\theta}{2}\right) \
&= \frac{1}{2} + \frac{1}{2 \sin \frac{\theta}{2}} \sum_{i=1}^N \left(\sin i \theta \cos \frac{\theta}{2} + \sin \frac{\theta}{2}\cos i \theta\right) \
&= \frac{1}{2 \sin \frac{\theta}{2}}
    \left[
        \left(\cos \frac{\theta}{2}\sum_{i=1}^N \sin i \theta \right)
        + \left(\sin \frac{\theta}{2} \left( \frac{1}{2} + \sum_{i=1}^N \cos i \theta \right)  \right)
    \right] + \frac{1}{4} \
&= \frac{\cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}}
        \frac{\sin \frac{N \theta}{2} \sin \left((N+1)\frac{\theta}{2}\right)}{\sin \frac{\theta}{2}}
+ \frac{\sin \frac{\theta}{2}}{2 \sin \frac{\theta}{2}}
        \frac{\sin \frac{(2N + 1) \theta}{2}}{2 \sin \frac{\theta}{2}}
+ \frac{1}{4} \
&=\frac{1}{4}  +  \frac{\cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}}
        \frac{\sin \frac{N \theta}{2} \sin \left((N+1)\frac{\theta}{2}\right)}{\sin \frac{\theta}{2}} \
&\quad+ \frac{\sin \frac{\theta}{2}}{2 \sin \frac{\theta}{2}}
        \frac{\sin \frac{N\theta}{2} \cos \frac{(N+1)\theta}{2}
            + \cos \frac{N\theta}{2} \sin \frac{(N+1)\theta}{2}}
        {2 \sin \frac{\theta}{2}}
\
&=\frac{1}{4} + \frac{1}{4 \sin^2 \frac{\theta}{2}} \left[
    \sin \frac{N\theta}{2} \left(
        \sin \frac{(N+1)\theta}{2} \cos \frac{\theta}{2} +
        \sin \frac{\theta}{2} \cos \frac{(N+1)\theta}{2} \right)
\right. \
&\qquad + \left.
    \sin \frac{(N+1)\theta}{2} \left(
        \cos \frac{\theta}{2} \sin \frac{N\theta}{2} +
        \sin \frac{\theta}{2} \cos \frac{N\theta}{2}
    \right)
 \right] \
&=\frac{1}{4} + \frac{1}{4 \sin^2 \frac{\theta}{2}} \left(
\sin \frac{N \theta}{2} \sin \frac{(N+2)\theta}{2} + \sin^2 \frac{(N+1)\theta}{2}
    \right) \
&=\frac{1}{4 \sin^2 \frac{\theta}{2}} \left(
\sin \frac{N \theta}{2} \sin \frac{(N+2)\theta}{2} + \sin^2 \frac{\theta}{2} + \sin^2 \frac{(N+1)\theta}{2}
    \right) \
&=\frac{1}{4 \sin^2 \frac{\theta}{2}} \left(
2\sin^2 \frac{(N+1)\theta}{2}
    \right) \
&=\frac{\sin^2 \frac{(N+1)\theta}{2}}{2 \sin^2 \frac{\theta}{2}} \
&= \frac{1}{2} \left(\frac{\sin \frac{(N+1)\theta}{2}}{\sin \frac{\theta}{2}}\right)^2
\end{aligned} \]
which is the desired identity, except missing a factor of $1/(N+1)$ on both sides.

Exercise 8.5.9

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Exercise 8.5.9

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