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I'm studying Chebyshev Interpolation at the moment.
Here n points $$x_1, x_2,..., x_n$$ are chosen within the interval $$-1 \le x \le 1$$ Using the Lagrange formula and minimising, we get: $$\prod_{I=1}^n(x-x_i)=2^{-(n-1)}.T_n(x)$$ Can some explain how they were able to minimise the Lagrange Formula using the Chebyshev Polynomial. Thanks
The main point of Chebyshev polynomials is that they are easily computable (by recursion) and that they form an orthogonal basis through which express any (real) polynomial.
So, the interpolation problem is to minimize:
$$ ||f-p_n||^2_2,$$
where $p_n$ is a polynomial of degree $n$.
Choose $(\phi_n)_{n\in \mathbb{N}}$ to be an orthogonal basis for $P_n$, the space of polynomials with degree $n$ or lower.
Let $p_n=\sum a_i\phi_i $$. Then:
$$||f-p_n||^2_2 =\langle f-\sum a_i\phi_i,f-\sum a_i\phi_i\rangle \\ = ||f||^2_2 -2\langle f, \sum a_i\phi_i \rangle + \langle \sum a_i \phi_i, \sum a_i \phi_i \rangle =: G(a_1,...,a_n)$$
Now, we want to find a minimum of $G$.
Set: $$\frac{\partial G}{\partial a_i}=0.$$
Thus obtain:
$$ 2 \frac{\partial }{\partial a_i}\langle f, \sum a_i \phi_i\rangle= \frac{\partial}{\partial a_i} \langle \sum a_i \phi_i, \sum a_i \phi_i \rangle. $$
It is here that we use the orthogonality of the chosen basis:
$$\dfrac{\partial}{\partial a_i}\langle \sum a_i \phi_i, \sum a_i \phi_i \rangle = \dfrac{\partial}{\partial a_i} a_i^2 \langle \phi_i , \phi_i \rangle.$$ Then:
$$2\langle f, \phi_i\rangle=2a_i\langle \phi_i,\phi_i\rangle.$$
Then:
$$a_i=\langle f, \phi_i\rangle$$.
Moral of the story, Chebyschev polynomials make it easier to compute the coefficients of the interpolating polynomial which solves the minimization problem.
