Help with one small demonstration

1. Jan 9, 2008

pocaracas

Now in LaTeX hope it's ok (the preview sucks):

1. The problem statement, all variables and given/known data

given the integral

$$I_n(r,z)= \int_z^r \frac{T_n(\frac{p}{z})T_n(\frac{p}{r})}{p \sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp$$

where $$T_n(x)$$ is the chebyshev polynomial of the first kind:
$$T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)$$
$$T_0(x)=1$$
$$T_1(x)=x$$

Important detail -> $$0 < z < r < 1$$

prove that:
$$I_{n+1} = I_{n-1}$$
and
$$I_0 = I_1 = \frac{\pi}{2}$$

3. The attempt at a solution

$$I_0 = I_1 = \frac{\pi}{2}$$ is relatively simple. I just changed variables $$y= p^2$$ and let Mathematica do the rest.

The other proof if giving me a hard time.

Substituting the $$T_n(x) = 2 x T_{n-1}(x)- T_{n-2}(x)$$, I got:

$$I_{n+1}(r,z) = \int_z^r \frac{4p T_n(\frac{p}{z}) T_n(\frac{p}{r})} {\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp -$$

$$- \int_z^r \frac{2r T_n(\frac{p}{z})T_{n-1}(\frac{p}{r})}{\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp -$$

$$- \int_z^r\frac{2z T_{n-1}(\frac{p}{z}) T_n(\frac{p}{r})}{\sqrt{r^2-p^2} \sqrt{p^2-z^2}} dp + I_{n-1}(r,z)$$

Now, the remainder integrals should be zero, but I can't figure out how to get there.
If it helps, I've found this relation:
$$xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))$$

Any help will be very welcome!

Last edited: Jan 9, 2008
2. Jan 9, 2008

Defennder

This is extremely hard to read. Could you write in Latex?

3. Jan 9, 2008

pocaracas

I could, but I need to know how to add latex code here.

4. Jan 9, 2008

Pomico

There is a $$\sum$$ symbol in the tool bar above the reply box. Click on this and a latex reference should appear :)

5. Jan 9, 2008

pocaracas

I just noticed that $$T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)$$ is just the same as
$$xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))$$

:(

Last edited: Jan 9, 2008
6. Jan 9, 2008

pocaracas

thanks man!

7. Jan 14, 2008

no ideas? :(