How can I evaluate a Chebishev polynomial in python?

[confused_engineer]

TL;DR

I can't find a python function which provides me with the evaluation of a Chebishev polynomial of a concrete order at a concrete point

Hello everyone. I need to construct in python a function which returns the evaluation of a Chebishev polynomial of order k evaluated in x. I have tested the function chebval form these documents, but it doesn't provide what I look for, since I have tested the third one, 4t^3-3t and

import numpy as np
import numpy.polynomial.chebyshev as cheb

gfg = cheb.chebval((3), (3))

does not return 4*(3)^3-3*3, but instead it returns 3. I have a code which does this but for Legendre polynomials, but I cannot reproduce it whith these because the recurrence relationship uses the last two terms, not the first two as Legendre's. The code is as follow.

import numpy as np
import numpy.polynomial.chebyshev as cheb

gfg = cheb.chebval((3), (3))
 print(gfg)
x=5
K=2

pn2 = 2*(x)**K+3print(pn2)

So, if someone could tell me how to do this but with Chebishev's polynomials I would be most grateful.

Thanks for reading.

 

[tnich]

The recurrence relation for Chebyshev polynomials of the first kind is $T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x)$. So to calculate $T_{n}(x)$, you just need a loop. Start with $T_0=1$, and $T_1=x$ and iterate until you get $T_n$.

 

[confused_engineer]

The recurrence relation for Chebyshev polynomials of the first kind is $T_{n}(x) = 2xT_{n-1}(x) - T_{n-2}(x)$. So to calculate $T_{n}(x)$, you just need a loop. Start with $T_0=1$, and $T_1=x$ and iterate until you get $T_n$.

Thanks for the answer, but I am afraid that is exactly the problem. I need a function that returns the evaluation of a polynomial of order n evaluated at x. For that, I need to calculate the polynomial of order n based on the polynomials of order (n-1) and (n-2). Currently, I am using mt.cos(n*mt.acos(x)) to evaluate the polynomial, but a numerical error arrises.

Is there a way to calculate the Chevishev polynomial of order n based on the first two?

 

[Ibix]

Currently, I am using mt.cos(n*mt.acos(x)) to evaluate the polynomial, but a numerical error arrises.

Your example was $n=3$, $x=3$. At least according to Wikipedia, $T_n(x)=\cosh(n\cosh^{-1}(x))$ for $x>1$, rather than $\cos$ and $\cos^{-1}$ as it is for $-1\leq x\leq 1$. Does that fix your numerical error?