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

given the integral

[tex]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[/tex]

where [tex]T_n(x)[/tex] is the chebyshev polynomial of the first kind:

[tex]T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)[/tex]

[tex]T_0(x)=1[/tex]

[tex]T_1(x)=x[/tex]

Important detail -> [tex]0 < z < r < 1[/tex]

prove that:

[tex]I_{n+1} = I_{n-1}[/tex]

and

[tex]I_0 = I_1 = \frac{\pi}{2}[/tex]

3. The attempt at a solution

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

The other proof if giving me a hard time.

Substituting the [tex]T_n(x) = 2 x T_{n-1}(x)- T_{n-2}(x)[/tex], I got:

[tex]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 - [/tex]

[tex]- \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 -[/tex]

[tex]- \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)[/tex]

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:

[tex]xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))[/tex]

Any help will be very welcome!

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# Homework Help: Help with one small demonstration

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