Express First kind Chebyshev polynomial in terms of monomials

First kind Chebyshev polynomial of order n ($T_n$) is defined in terms of cosine function as follow:

1) $T_n(\cos x)=\cos n x$

Instead of working with cosine functions, I want to express Chebyshev polynomial in terms of monomials (powers of x) using the following recursion formula: $T_0(x)=1; T_1(x)=x; T_{n+1}(x)=2xT_n(x)−T_{n−1}(x)$ for $x \in [-1, 1]$.

For example, first kind Chebyshev polynomial of order 2 can be expressed as:

Usig formula 1) $T_2(cosx)=cos(2x)$,
Usig recursive formula) $T_2(x)=2xT_1(x)-T_0(x)=2x^2-1$

If we plot $cos2x$ vesrus $cosx, cosx\in [-1,1]$ and also $2x^2-1$ versus $x, x\in [-1, 1]$, the two plot should match.

However as "n" increases ($n\ge45$), the $T_n$ obtained as a polynomial by the recursive formula doesn't match the Chebyshev polynomial of formula (1).

Why is that????? The recursion formula is not valid for higher orders????

here is my Matlab code that compares two formulas:

clc;clear;close all;
syms x;


Tr(1,1)=1+0*x; Tr(2,1)=x;
for i=2:n; Tr(i+1,1)= expand(2*x*Tr(i,1)-Tr(i-1,1));end;
hold on;for x=-1:0.01:1; plot(x,eval(Tr(n)),'--bo'); end; 

t=[0:0.01:2*pi]; plot(cos(t),cos((n-1)*t),'k*-')
  • 1
    $\begingroup$ The Chebyshev polynomials do satisfy the recurrence you state. $\endgroup$ Sep 2, 2017 at 6:26
  • $\begingroup$ You have not articulated the descrepancy between the two ways of getting the Chebyshev polynomials. It would improve your Question to point out an actual disagreement. $\endgroup$
    – hardmath
    Sep 2, 2017 at 14:49
  • $\begingroup$ The sampled points in your code (i.e., t=0:0.1:2*pi [followed by cos(t)] vs. x=-1:0.1:1) are different. Change, e.g., "x=-1:0.01:1; plot(x,eval(Tr(n)),'--bo');" to " t=0:0.01:2*pi; x=cost(t); plot(x,eval(Tr(n)),'--bo');". $\endgroup$
    – Kasra
    Sep 3, 2017 at 3:14


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