What is the approximation used here?

[Unconscious]

I'm reading a paper (Beamwidth and directivity of large scanning arrays, R. S. Elliott, Appendix A) in which the author starts from this expression:

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

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?

 

[mathman]

The first line looks funny. -1+1 at end?? I went no further.

 

[Charles Link]

I didn't work out every single detail, but I got a term that is $K/(K+p)$. When the negative $p's$ are summed, this will change that to a $K/(K-p)$. (Notice the final sum is over positive $p$). The common denominator becomes $K^2-p^2$. It's a lot of algebra, but if the author was careful, he might have got it right.
Edit: Notice $2(1-\frac{p^2}{p^2-K^2})=2 \frac{K^2}{K^2-p^2}=\frac{K}{K+p}+\frac{K}{K-p}$.
I think the author got it right.

 

[Unconscious]

Thanks for your answer, tomorrow I'll try to do calculations again.
Thank you again :)