sin(nx)sin(x)\frac{\sin(nx)}{\sin(x)}

\(\begin{align}

\frac{\sin(nx)}{\sin(x)} &= \frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}} = \frac{e^{inx}}{e^{ix}} \cdot \frac{e^{-2inx}-1}{e^{-2ix}-1} & {\color{red} \sin(\theta) = \frac{e^{i\theta}-e^{-i\theta}}{2i} }\\

&= \frac{e^{inx}}{e^{ix}} \cdot \sum_{k=0}^{n-1} \left (e^{-2ix} \right )^k = e^{ix(n-1)} \cdot \sum_{k=0}^{n-1} e^{-2ixk}=\sum_{k=0}^{n-1} e^{ix(n-1-2k)} & {\color{red} \sum_{k=0}^{n-1} x^k = \frac{x^n - 1}{x-1}} \\

&= \underbrace{e^{ix(n-1)}+e^{ix(n-3)} +... {\color{blue} + e^{-ix(n-3)} + e^{-ix(n-1)}}}_{n \ term} \\

&= \left\{\begin{matrix} e^{ix(n-1)}+e^{ix(n-3)}+ \dots+e^{4ix}+e^{2ix}+\underbrace{e^0}_{k=\frac{n-1}{2}} + {\color{blue} e^{-2ix}+e^{-4ix} \dots + e^{-ix(n-3)} + e^{-ix(n-1)}} & n \ is \ odd\\

e^{ix(n-1)}+e^{ix(n-3)}+ \dots+e^{3ix}+e^{ix} + {\color{blue} e^{-ix}+e^{-3ix}\dots+ e^{-ix(n-3)} + e^{-ix(n-1)}} & n \ is \ even \end{matrix}\right.\\

&= \left\{\begin{matrix} 1+\sum_{k=1}^\frac{n-1}{2} {\color{red} 2} \cdot \frac{e^{2k ix}+e^{-2k ix}}{\color{red} 2} & n \ is \ odd\\

\sum_{k=1}^\frac{n}{2} {\color{red} 2} \cdot \frac{e^{(2k-1) ix}+e^{-(2k-1) ix}}{\color{red} 2} & n \ is \ even \end{matrix}\right.\\

&= \left\{\begin{matrix} 1+2 \sum_{k=1}^\frac{n-1}{2} \cos(2kx) & n \ is \ odd\\

2 \sum_{k=1}^\frac{n}{2} \cos\left ( (2k-1)x \right ) & n \ is \ even \end{matrix}\right.

\end{align} \)

#Calculus

Note by Hassan Abdulla
11 months, 4 weeks ago

No vote yet
1 vote

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Comments

Relevant: Dirichlet kernel.

Pi Han Goh - 11 months, 4 weeks ago

Can we generalise this to non integral values of nn ? What if we wanted to find 0πsin(xy)sinxdx\int_{0}^{\pi}\frac{\sin\left(xy\right)}{\sin x}dx when yy varies over the reals greater than 22 ? @Pi Han Goh @Hassan Abdulla

Aaghaz Mahajan - 11 months, 4 weeks ago

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If yy is not an integer, then the integrand (might) diverge at x=πx=\pi.

Pi Han Goh - 11 months, 4 weeks ago
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