Why are these ∑cos⁡ and csc⁡ equivalent?Mathematica 'simplifies' this formula∑k=1Rcos⁡2kπxRto this12(csc⁡πxRsin⁡(2R+1)πxR−1)

Assuming you're a bit familiar with complex numbers:eix=cos⁡(x)+isin⁡(x)Let a=2πR then you want to find the sum∑k=1Rcos⁡(akx)Notice that it is way easier to first compute (which we later can relate to∑k=1Reakix=∑k=1R(eaix)kwhich is a geometric series. This evaluates to∑k=0R(eaix)k=eaix(R+1)−1eaix−1But since your series starts at k=1 we have to subtract the first term, so that∑k=1R(eaix)k=eaix(R+1)−1eaix−1−1Now substitute eq. (1) back into (5) to get∑k=1Rcos⁡(akx)+i∑k=1Rsin⁡(akx)=⋯

Answered: "Why are these ∑cos⁡ and csc⁡ equivalent?Mathematica 'simplifies'..."

Adrien Wong

Answered question

2022-03-14

Why are these

\sum \cos

and

\csc

equivalent?
Mathematica 'simplifies' this formula

\sum_{k = 1}^{R} \cos \frac{2 k π x}{R}

to this

\frac{1}{2} (\csc \frac{π x}{R} \sin \frac{(2 R + 1) π x}{R} - 1)

Answer & Explanation

i1a6ldihq

Beginner2022-03-15Added 4 answers

Assuming you're a bit familiar with complex numbers:

e^{i x} = \cos (x) + i \sin (x)

Let

a = \frac{2 π}{R}

then you want to find the sum

\sum_{k = 1}^{R} \cos (a k x)

Notice that it is way easier to first compute (which we later can relate to

\sum_{k = 1}^{R} e^{a k i x} = \sum_{k = 1}^{R} {(e^{a i x})}^{k}

which is a geometric series. This evaluates to

\sum_{k = 0}^{R} {(e^{a i x})}^{k} = \frac{e^{a i x (R + 1)} - 1}{e^{a i x} - 1}

But since your series starts at k=1 we have to subtract the first term, so that

\sum_{k = 1}^{R} {(e^{a i x})}^{k} = \frac{e^{a i x (R + 1)} - 1}{e^{a i x} - 1} - 1

Now substitute eq. (1) back into (5) to get

\sum_{k = 1}^{R} \cos (a k x) + i \sum_{k = 1}^{R} \sin (a k x) = \dots

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