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I have been trying to find out how the summation is deduced via inspection. However, it is to no avail.

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Could anyone show me a way to quickly deduce this form? Thanks.

user3600725
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  • Why the giant font? – Spencer Mar 08 '15 at 03:08
  • See my answer to http://math.stackexchange.com/questions/986835/finding-the-sum-of-sin0-circ-sin1-circ-sin2-circ-cdots-sin/986860#986860 , a similar method will work here. There isn't a way to just see the result as far as I know. – Spencer Mar 08 '15 at 03:11

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Let $x=\pi ( 2t-0.5 )$

then $\displaystyle \sum_{k=-9}^{9} e^{-ikx} = e^{9ix} \sum_{k=0}^{18} e^{-ikx} $

Now use the formula for the sum of a geometric series ...

$\displaystyle = e^{9ix} \frac{e^{- 19 i x} -1}{e^{ - i x} -1} $

which simplifies to

$\displaystyle = \frac{e^{-10 i x} - e^{9ix}}{e^{ i x} -1}$

$\displaystyle = \frac{ e^{-ix/2} ( e^{-(19/2) i x} - e^{(19/2)ix} ) }{ e^{-i x/2}( e^{ -i x/2} -e^{ i x/2} ) } $

$\displaystyle = \frac{ e^{-(19/2) i x} - e^{(19/2)ix} }{ e^{ -i x/2} -e^{ i x/2} } = \frac{sin(\frac{19x}{2}) }{sin(\frac{x}{2}) }$

WW1
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