# recurrence relation of integral

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 $I_n( \phi)$

..................................................................................................................................................

Can someone help with (a)?

-
Note: $I_{n+2} ( \phi) - 2\cos \phi I_{n+1}( \phi)+ I_n( \phi)$ is not an equation. Do you mean this is equal to zero? – Thomas Andrews Nov 14 '12 at 13:27
Also, you are missing an $x$ in the characteristic quadratic, $x^2-2(\cos \phi )x +1$ – Thomas Andrews Nov 14 '12 at 13:28
Note that $c_1$ and $c_2$ are not constants, they are functions of $\phi$. (Think of it as solving a recurrence for each $\phi$, getting different $c_i$). And the $c_i$ are not necessarily real, so while it is true that $c_1(\phi)+c_2(\phi)=0$ it is not true that $(c_1(\phi)+c_2(\phi))\cos \phi = \pi$ – Thomas Andrews Nov 14 '12 at 13:31
@ThomasAndrews: What is the other equation? – passenger Nov 14 '12 at 13:42
It's $c_1(\phi)x_1 + c_2(\phi)x_2 = \pi$. What else would it be? – Thomas Andrews Nov 14 '12 at 13:55

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.

-
Thank you very much for your time! – passenger Nov 14 '12 at 16:44