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How to determine Tchebysheff polynomial general expression

  1. Dec 12, 2013 #1
    Hello!!
    Tchebysheff polynomials are often defined with trigonometric functions:

    [itex]T_m (x) =
    \begin{cases} \cos(m \arccos (x)) & -1 \le x \le 1\\
    \mathrm{cosh} (m \mathrm{arccosh} (x)) & x > 1\\(-1)^m \mathrm{cosh} (m \mathrm{arccosh} |x|) & x < 1
    \end{cases}
    [/itex]

    But they are also polynomials, and for m even their definition could be

    [itex]
    T_m (x) = \sum_{n = 0}^{m/2} (-1)^{m/2 - n} \frac{m/2}{m/2 + n} \binom{m/2 + n}{2n}(2x)^{2n}
    [/itex]

    How could one derive the first expression from the latter? That is, how could we pass from a polynomial with powers of x to a [itex]\cos (m \arccos (x))[/itex] function?
    I searched several times in the web for this demonstration, but I never found it. If you can suggest a link or a book instead of the demonstration itself I thank you so much anyway!
    Bye :)

    Emily
     
    Last edited: Dec 12, 2013
  2. jcsd
  3. Dec 12, 2013 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    cos(my) can be written as a polynomial in cos(y) of degree m. If y = arccos(x), then cos(my) is a polynomial in cos(arccos(x)) = x.

    As a simple example the double angle formula, cos(2y) = cos2(y) - sin2(y) = 2cos2(y) - 1.

    Triple angle, quartic angle etc. formulas exist as well. I think the easiest way is to use Euler's formula:

    [tex] cos(my) + i sin(my) = e^{imy} = \left(e^{iy}\right)^n = \left( cos(y) + i sin(y) \right)^{m} [/tex]
    Expanding the right hand side gives cos(my) and sin(my) as polynomials in cos(y) and sin(y) (from which you can write it in terms of only cosine or sine)
     
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