NLA MT25, Chebyshev polynomials
Flashcards
@Define the Chebyshev polynomials $T _ k(x)$ in terms of trigonometric functions and derive a three term recurrence.
Let $z = \exp(i\theta)$ and set $x = \frac 1 2(z + z^{-1}) = \cos \theta$. Then
\[T _ k(x) = \frac 1 2(z^k + z^{-k}) = \cos(k\theta)\]Then we have
\[\frac 1 2 (z + z^{-1})(z^k + z^{-k}) = \frac 1 2 (z^{k+1} + z^{-(k+1)}) + \frac 1 2 (z^{k-1} + z^{-(k-1)})\]Hence
\[2x T _ k(x) = T _ {k+1}(x) + T _ {k-1}(x)\]@State a result about the behaviour of Chebyshev polynomials inside and outside $[-1, 1]$.
Inside $[-1, 1]$:
For all $x \in [-1, 1]$,
\[\vert T _ k(x) \vert \le 1\]Outside $[-1, 1]$:
\[\vert T _ k(x) \vert \gg 1\]grows rapidly with $ \vert x \vert $ and $k$, and has the fastest growth among $p \in \mathcal P _ k$, with $ \vert p(x) \vert \le 1$ on $x \in [-1, 1]$.
@State a result about the symmetry of Chebyshev polynomials.
\[\vert T _ k(x) = \vert T _ k(-x) \vert\]
for all $x \in \mathbb R$.