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Help with one small demonstration

  1. Jan 9, 2008 #1
    Now in LaTeX hope it's ok (the preview sucks):

    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]

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

    prove that:
    [tex]I_{n+1} = I_{n-1}[/tex]
    [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!
    Last edited: Jan 9, 2008
  2. jcsd
  3. Jan 9, 2008 #2


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    Homework Helper

    This is extremely hard to read. Could you write in Latex?
  4. Jan 9, 2008 #3
    I could, but I need to know how to add latex code here.
  5. Jan 9, 2008 #4
    There is a [tex]\sum[/tex] symbol in the tool bar above the reply box. Click on this and a latex reference should appear :)
  6. Jan 9, 2008 #5
    I just noticed that [tex]T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)[/tex] is just the same as
    [tex]xT_n(x) = \frac{1}{2}(T_{n+1}(x) + T_{n-1}(x))[/tex]

    Last edited: Jan 9, 2008
  7. Jan 9, 2008 #6
    thanks man!
  8. Jan 14, 2008 #7
    no ideas? :(
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