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quasar987

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[tex]\frac{1}{2} + \sum_{j=1}^n cos(jx) = \frac{sin([n+\frac{1}{2}]x)}{sin(\frac{x}{2})}[/tex]

?? Thx!

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- #1

quasar987

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[tex]\frac{1}{2} + \sum_{j=1}^n cos(jx) = \frac{sin([n+\frac{1}{2}]x)}{sin(\frac{x}{2})}[/tex]

?? Thx!

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AKG

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No, try n = x = 1.

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I can't tell you off-hand whether or not that identity is true, but at least there is a very similar identity (which might be equal to the one you posted). First, note that

[tex]\cos{jx} = Re(\cos{jx} + i \sin{jx}) = Re(e^{ixj}) = Re(({e^{ix}})^j).[/tex]

The Re function is linear, which means that summing cos(jx) is equivalent to summing (e^(ix))^j and then calculating the real part of that. Hence the problem can be reduced to calculating the sum of a geometric series...

[tex]\cos{jx} = Re(\cos{jx} + i \sin{jx}) = Re(e^{ixj}) = Re(({e^{ix}})^j).[/tex]

The Re function is linear, which means that summing cos(jx) is equivalent to summing (e^(ix))^j and then calculating the real part of that. Hence the problem can be reduced to calculating the sum of a geometric series...

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krab

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- #5

Galileo

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It looks familiar. If I remember correctly, the name is the Dirichletkernel.

[tex]D_n:=\sum_{k=-n}^n e^{ikx}=1+2\sum_{k=1}^n \cos kx[/tex]

Use a geometric expansion to find a closed form. It only works if [itex]e^{ix}\not= 1[/itex]

[tex]D_n(x)=e^{-inx}\sum_{k=0}^{2n}e^{ikx}=\frac{\sin (n+1/2)x}{\sin x/2}[/tex]

If [itex]e^{ix}\not=1[/itex] you can expand the sum geometrically. After some algeblah you'll get the answer. Treat the case [itex]e^{ix}=1[/itex] seperately.

[tex]D_n:=\sum_{k=-n}^n e^{ikx}=1+2\sum_{k=1}^n \cos kx[/tex]

Use a geometric expansion to find a closed form. It only works if [itex]e^{ix}\not= 1[/itex]

[tex]D_n(x)=e^{-inx}\sum_{k=0}^{2n}e^{ikx}=\frac{\sin (n+1/2)x}{\sin x/2}[/tex]

If [itex]e^{ix}\not=1[/itex] you can expand the sum geometrically. After some algeblah you'll get the answer. Treat the case [itex]e^{ix}=1[/itex] seperately.

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- #6

lurflurf

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There was a thread about the correct form a while back

https://www.physicsforums.com/showthread.php?t=86735

- #7

quasar987

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K, thanx, I proved it by induction like the OP.

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