development of $\sum^{M-1}_{m=-(M-1)}e^{-imh\xi}$

Question

$$ \sum^{M-1}_{m=-(M-1)}e^{-imh\xi} = \frac{\sin (M-\frac12)h\xi }{\sin \frac12 h\xi}$$ Can you show the process of the development?

Thank you for advance.

Answer 1

Set $m=-r,$

$$\sum_{m=-(M-1)}^{M-1}(e^{-i2x})^m =\sum_{r=-(M-1)}^{M-1}(e^{i2x})^r$$

$$=e^{-2ix(M-1)}\cdot\dfrac{(e^{2ix})^{2M-1}-1}{e^{i2x}-1}$$

$$=\dfrac{e^{ix}}{e^{ix}}\cdot\dfrac{(e^{ix})^{2M-1}-(e^{-ix})^{2M-1}}{e^{ix}-e^{-ix}}$$

Now $e^{iy}-e^{-iy}=2i\sin y$

Answer 2

Set $z=h\xi$ for the sake of brevity, and assume that $z\neq 2k\pi$ for an integer $k$. Then using the formula for the sum of a geometric series, we get $$ \sum_{m=0}^{M-1}e^{-imz}=\frac{e^{-iMz}-1}{e^{-iz}-1}=\frac{e^{-i(M-\frac{1}{2})z}-e^{\frac{iz}{2}}}{e^{-\frac{iz}{2}}-e^{\frac{iz}{2}}}$$ and $$ \sum_{m=-(M-1)}^{-1}e^{-imz}=\sum_{m=1}^{M-1}e^{imz}=e^{iz}\frac{e^{i(M-1)z}-1}{e^{iz}-1}=\frac{e^{i(M-\frac{1}{2})z}-e^{\frac{iz}{2}}}{e^{\frac{iz}{2}}-e^{-\frac{iz}{2}}}$$ Adding these two expressions yields $$ \frac{e^{i(M-\frac{1}{2})z}-e^{-i(M-\frac{1}{2})z}}{e^{\frac{iz}{2}}-e^{-\frac{iz}{2}}}=\frac{\sin((M-\frac{1}{2})z)}{\sin(\frac{z}{2})}$$ using the fact that $$ \sin w=\frac{e^{iw}-e^{-iw}}{2i}$$ for all $w\in\mathbb{C}$.

Answer 3

Um, that's classic tricks.

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Let $$S=\sum_{m=-(M-1)}^{M-1}e^{imh\epsilon}$$ Then $$e^{ih\epsilon}S=\sum_{m=-(M-2)}^{M}e^{imh\epsilon}$$ $$(e^{ih\epsilon}-1)S=e^{iMh\epsilon}-e^{-i(M-1)h\epsilon}$$ $$S=\frac{e^{iMh\epsilon}-e^{-i(M-1)h\epsilon}}{e^{ih\epsilon}-1}$$ Because, $$e^{ih\epsilon}-1=cosh\epsilon-1+isinh\epsilon=-2sin(\epsilon/2)\cdot(cos(\epsilon/2)-sin(\epsilon/2))$$

and $$e^{iMh\epsilon}-e^{-i(M-1)h\epsilon}=cosM\epsilon-cos(M-1)\epsilon+i(sinM\epsilon+sin(M-1)\epsilon)\\ =2sin(M-0.5)\epsilon \cdot sin(\epsilon/2)+2sin(M-0.5)\epsilon \cdot cos(\epsilon/2)$$

Therefore $$S=\frac{sin(M-0.5)\epsilon}{sin(\epsilon/2)}$$

Answer 4

As compactly as possible: Set $r:=e^{-ix}$. Then $\sum_{n=-N}^Ne^{-inx}$ is equal to $$\sum_{n=-N}^Nr^n \stackrel{(a)}=\frac{r^{-N}-r^{N+1}}{1-r} \stackrel{(b)}=\frac{r^{-(N+\frac12)}-r^{N+\frac12}}{r^{-\frac12}-r^{\frac12}} \stackrel{(c)}=\frac{e^{i(N+\frac12)x}-e^{-i(N+\frac12)x}}{e^{i\frac12x}-e^{-i\frac12x}} \stackrel{(d)}=\frac{\sin(N+\frac12)x}{\sin\frac12x} $$ In step (a) use the formula for the sum of a geometric sequence; in (b) multiply top and bottom by $r^{-\frac12}$; in (c) substitute $r=e^{-ix}$; in (d) use $e^{it}-e^{-it}=2i\sin t$.

Replace $x$ by $h\xi$ and $N$ by $M-1$ and you've got your formula.