Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Thank you very much for your time! –  passenger Nov 14 '12 at 16:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.