Intereting Posts

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

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- Prove that there exist a constant $c$ such that all the roots of $P(x)+c.T_n(x)$ are real

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;
n=45;
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*-')
```

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