Do Chebyshev polynomials form a complete set of independent functions? If yes, what can we say about their shifted versions? E.g. shifted Chebyshev polynomials of the first kind are defined as
$$T_{n}^{*}(x)=T_{n}(2x-1)$$
or other possible shitting like e.g. $T_{n}\left(\frac{x^2}{2}-1\right)$.
Do shifted Chebyshev polynomials form a complete set of independent functions
chebyshev polynomialsorthogonal-polynomials
Best Answer
I assume that by complete you mean a basis for the set of polynomials.
Yes, the Chebyshev polynomials form a complete set of independent functions. That should be clear considering the recurrence relation. For example, for the first kind, that relation is\begin{align} T_0(x) & = 1\\ T_1(x) & = x\\ &\ \ \vdots\\ T_{n+1}(x) & = 2xT_n - T_{n-1}.\end{align} The degree of polynomial $T_n$ is $n$, so any polynomial can be expressed as a linear combination of the $T_n$, and if $c_0 T_0 +\cdots + c_n T_n = 0$, then each $c_i$ is zero. See also the Wikipedia article.
The same holds for the shift $T^*_n(x) = T_n(2x-1)$ because the degree of $T^*_n$ is still $n$ and the argument above applies.
The shift $T^*_n(x^2/2 - 1)$ generates independent polynomials, but they do not form a basis for the polynomials. For example, all the $T^*_n$ in this case are of even degree, so polynomials of odd degree cannot be formed from a linear combination of the $T^*_n$.