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Consider the integral defined by

$\displaystyle{ I_k( \phi) = \int_0^{\pi} \frac{ \cos(k\theta) - \cos( k \phi) }{ \cos \theta - \cos\phi} d \theta} $

(a) Show that $I_k( \phi) $ satisfies the difference equation

$\displaystyle { I_{n+2} ( \phi) - 2\cos \phi I_{n+1}( \phi)+ I_n( \phi)=0, \quad I_0 (\phi)=0 , \quad I_1( \phi) = \pi }$

(b) Solve the difference equation in part (a) to find $I_n( \phi)$

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Can someone help with (a)?

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    I've added more detail to my solution for (a).2012-11-14

1 Answers 1

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For $(a)$, first show that $\cos (n+2)x - 2\cos x\cos(n+1)x + \cos nx = 0$ for all $x$. This follows by almost direct application of the sum rules for $\cos$. Indeed, it might be easier to show if you write it as $\cos(m+1)x + \cos(m-1)x = 2\cos x\cos mx$ where $m=n+1$. The rest of $(a)$ follows with some manipulation. (It's not quite as easy as it looks.)

For $(b)$, you've assumed $c_1$ and $c_2$ are real values. They are not. They are possibly complex functions of $\phi$.

The actual resulting formula should be $I_n(\phi)=\frac{\pi\sin n\phi}{\sin \phi}$

One other thing to note is that if $\sin\phi = 0$ then $x_1=x_2$, so you have to adjust your general formula for the recurrence relationship to the case where your recurrence polynomial has repeated roots. Then $x_1=x_2=x=\pm 1$. If $x=+1$ then $I_n = c_0+nc_1$ and we get that $I_n = n\pi$. If $x=-1$, then $c_0=0$ and $c_1=-\pi$ and $I_n=(-1)^{n+1}\pi n$. This is actually just the limit - it is the value which makes $I_n(\phi)$ continuous at these values.

In the calculation for $(a)$, when you do the substitution listed at the top in the expression $\frac{\cos(n+2)\theta - \cos(n+2)\phi}{\cos\theta-\cos\phi}$ you get:

$\frac{2\cos \theta \cos(n+1)\theta - \cos n\theta - (2\cos\phi\cos(n+1)\phi -\cos n\phi)}{\cos\theta-\cos\phi}$

The trick is to write $\cos \theta = (\cos\theta - \cos\phi) + \cos\phi$. Substituting, we get:

$2\cos(n+1)\theta + 2\cos\phi\frac{\cos(n+1)\theta - \cos(n+1)\phi}{\cos\theta-\cos\phi} - \frac{\cos n\theta -\cos n\phi}{\cos\theta -\cos\phi}$

Then integrating, you get $I_{n+2}(\phi)=\int_{0}^\pi 2\cos(n+1)\theta\ d\theta + 2\cos\phi I_{n+1}(\phi) - I_n(\phi)$

But $\int_{0}^\pi 2\cos(n+1)\theta\ d\theta=0$.

So you are done.

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    Thank you very much for your time!2012-11-14