prove by induction that Tn(x) is a poly. of deg. n and has highest coefficient of 2^n-1. start with n=1?

Question

Tn(x) is related to the chebyshev polynomials...T0(x)=1, T1(x)=x...

Answer 1

This is a Chebyshev polynomial of the first kind

T_0=1,T_1=x

There is a recurrence relationship for them

T_(n+1) = 2x T_n - T_(n-1)

This holds for n=1,2,...

Assume your theorem holds for n =1,2,...,k-1,k

Then we prove it for n = k+1

The highest coefficient of T_1 is 1 and 2^(1-1) = 1
Degree of T_1 is 1

Hence true for n=1

Let T_(k-1) = 2^(k-2)x^(k-1) + ... + a_0

T_(k) = 2^(k-1) x^k + ...+ b_0

Write the recurrence relationship in terms of k
T_(k+1) = 2x T_k - T_(k-1)

If you substitute these into the recurrence relationship you get

T_(k+1) = 2^(k+1 -1) x^(k+1) + ... - a_0

This shows that it is true for n=k+1

By the second principle of Math Induction since true for n=1,
it is true for n=2,3,....

Answer 2

The Tchebychev polynomials are defined by

cos(nx)=T_n (cosx).

So, T_1(x)=x, T_2 (x)=2x^2 -1, etc.

The true result is that the highest coefficient is 2^(n-1), not 2^n -1.

Answer 3

Please define T_n(x) for us. That will give us
a chance to help you.