- #1

- 74

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##\frac{\sin\left [ \left ( 2N+1 \right ) u_0\right ]}{\sin(u_0)}\sum_{p=-P}^Pa_p\cos(p\pi)\left [\frac{\sin(u_0)}{\sin(u_p)} -1+1 \right ]##

where ##a_p=a_{-p}## and ##u_p=\frac{\pi L}{(N+1)\lambda}\left(\cos \theta'-\cos\theta_0+\frac{p\lambda}{L}\right)##, and says that we can approximate it in the following way:

##( 2N+1)\frac{\sin\left [ \left ( 2N+1 \right ) u_0\right ]}{( 2N+1)u_0}\left \{ a_0+\sum_{p=1}^P2a_p\cos(p\pi) \right \} - ( 2N+1)\frac{\sin\left [ \left ( 2N+1 \right ) u_0\right ]}{( 2N+1)u_0}\sum_{p=1}^P2a_p\cos(p\pi)\frac{p^2}{p^2-K^2}##

where ##K=\frac{L}{\lambda}\left(\cos\theta'-\cos\theta_0\right)##, because "##u_0## and ##u_p## are small and ##P## is a small integer".

I simply can't understand why. My attempt, using ##\sin u_p\approx u_p##, gives the same approximate expression, but with the difference that I have a factor ##\frac{p}{p+K}## instead of the factor ##\frac{p^2}{p^2-K^2}##.

What am I missing?