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I have a problem calculating/verifying the partial derivative below:

$$ \frac{\partial}{\partial{\phi_k}} \sum_{i=1 } ^{N} \sum_{j=1}^{N}\cos{(\phi_i-\phi_j})$$

My result, after doing the expansion for specific values of $i,j=1,2,3$ e.g., and generalizing , is the following:

$$-2\sum_{j=1}^{N}\sin{(\phi_k-\phi_j})$$

Could someone verify my result? Thank you in advance.

noir
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1 Answers1

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You have: $$\frac{\partial}{\partial{\phi_k}} \sum_{i=1 } ^{N} \sum_{j=1}^{N}\cos{(\phi_i-\phi_j})$$ Which equals the following, tossing out all terms that have neither $i$ nor $j$ equal to $k$ and then accounting for the $(i,j)=(k,k)$ term which would otherwise be counted twice. $$ \begin{align} &\frac{\partial}{\partial{\phi_k}} \sum_{j=1}^{N}\cos{(\phi_k-\phi_j})+\frac{\partial}{\partial{\phi_k}} \sum_{i=1}^{N}\cos{(\phi_i-\phi_k})-\frac{\partial}{\partial{\phi_k}}\cos{(\phi_k-\phi_k})\\ &=\sum_{j=1}^{N}-\sin{(\phi_k-\phi_j})+ \sum_{i=1}^{N}\sin{(\phi_i-\phi_k})-0\\ &=\sum_{j=1}^{N}\sin{(\phi_j-\phi_k})+ \sum_{i=1}^{N}\sin{(\phi_i-\phi_k})\\ &=\sum_{i=1}^{N}\sin{(\phi_i-\phi_k})+ \sum_{i=1}^{N}\sin{(\phi_i-\phi_k})\\ &=2\sum_{i=1}^{N}\sin{(\phi_i-\phi_k}) \end{align}$$

2'5 9'2
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  • And since the sine function is odd, you can interchange the variables inside, and the result is the same right? And for index variable in the final result, j is also acceptable. – noir Jan 27 '20 at 20:48