Summation involving sine and cosine

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SUMMARY

The discussion focuses on deriving the expression for \(\omega^2\) and its derivative with respect to \(K\) from the equation \(\omega^2=(2/M)\sum_{n>0}\frac{A\sin(nk_0a)}{na}(1-\cos(nKa))\). Participants suggest using trigonometric identities, such as \(\sin A \cos B = \frac{1}{2} [ \sin(A-B)+\sin(A+B)]\), and consider employing Euler's formula for a complex approach. The differentiation of \(\omega^2\) with respect to \(K\) is proposed as a direct method, leading to the expression \(\frac{d}{dK}f(K) = \frac{(2/M)}{\sum_{n>0}\frac{A\sin(nk_0a)}{na}(-n\sin(nKa))}\).

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Homework Statement


\omega^2=(2/M)\sum_{n>0}\frac{A\sin(nk_0a)}{na}(1-\cos(nKa))

A, a, and k_0 are constants, n is an integer.

I need to find \omega^2 and \frac{\partial\omega^2}{\partial K}, but I have no idea where to start.

Homework Equations


Not sure, the stuff above.

The Attempt at a Solution


I haven't done something like this in a while. I think I might be able to exploit
\sin A \cos B = \frac{1}{2} [ \sin(A-B)+\sin(A+B)]
but I'm still worried that the \frac{1}{n} term is divergent...

Any ideas on where to start? I'm also considering a complex approach with Euler's formula, but I'm not sure how far I can go with this.Many thanks!
 
Last edited:
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I would just differentiate directly, considering omega squared as a function of K

<br /> f(K) = \omega^2(K)=(2/M)\sum_{n&gt;0}\frac{A\sin(nk_0a)}{na}(1-\cos(nKa))<br />

<br /> \frac{d}{dK}f(K) = \frac{d}{dK}\omega^2(K)=\frac{d}{dK}(2/M)\sum_{n&gt;0}\frac{A\sin(nk_0a)}{na}(1-\cos(nKa))<br />
<br /> <br /> =(2/M)\sum_{n&gt;0}\frac{A\sin(nk_0a)}{na}(\frac{d}{dK}\cos(nKa))<br />
 

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