- #1
EmilyRuck
- 136
- 6
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
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
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