I'm trying to show that for the k monic chebyshev polynomial, T*_k, that 1/2^k
1 = max_[-1,1] |T*_k| <= max_[-1,1] |p_k| where p_k is any monic polynomial. The easiest way to proceed is by induction. So for k=1, p_1(x) = x+a_0. max_[-1,1] |x+a_0| = 1 +|a_0| >= 1 = 2^-1+1 = 2^0. So it's true for the base case. Here is where I am running into trouble. For ease of reading, the following maxes are over the interval [-1,1].
Assume max|T*_k| <= max|p_k(x)|. We need to show max|p_(k+1)| >= max|T*_(k+1)|. There is a recursion for monic Chebyshev polynomials: T*_(k+1) = 2x T*_k - T*_k-1. So if we start with max|T*_(k+1)| = max |2xT*_k - T*_(k-1)|. But I don't really know where to take it from here. If you play with the recursion formula, you are left with a |T*_(k-1)| which doesn't go away. I would appreciate some help from you ree rees.
1 = max_[-1,1] |T*_k| <= max_[-1,1] |p_k| where p_k is any monic polynomial. The easiest way to proceed is by induction. So for k=1, p_1(x) = x+a_0. max_[-1,1] |x+a_0| = 1 +|a_0| >= 1 = 2^-1+1 = 2^0. So it's true for the base case. Here is where I am running into trouble. For ease of reading, the following maxes are over the interval [-1,1].
Assume max|T*_k| <= max|p_k(x)|. We need to show max|p_(k+1)| >= max|T*_(k+1)|. There is a recursion for monic Chebyshev polynomials: T*_(k+1) = 2x T*_k - T*_k-1. So if we start with max|T*_(k+1)| = max |2xT*_k - T*_(k-1)|. But I don't really know where to take it from here. If you play with the recursion formula, you are left with a |T*_(k-1)| which doesn't go away. I would appreciate some help from you ree rees.
