You probably know the Pascal triangle, of the binomial coefficients.
I draw it here for later reference. Let *B(i,j)* denote the j-th element of the
i-th row; the index i runs from 0 on, and j goes from 0 to i.
For example *B(3,2)* = 3.

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ...

Let's draw another triangle, with 1 on the left side and 2 on the right side,

2 1 2 1 3 2 1 4 5 2 1 5 9 7 2 1 6 14 16 9 2 ...If

Let's define

1 1 -1 2 -3 1 4 -8 5 -1 8 -20 18 -7 1 ...Prove the following:

For example,

cos(3x) = T(3,0) cosThe claim explains why i called the triangle after Chebyshev. The polynomial expansions of cos(nx) in terms of powers of cos(x) are called Chebyshev polynomials of first kind, cos(n x)=T^{3}(x) + T(2,1) cos(x) = 4 cos^{3}(x) -3 cos(x) cos(4x) = T(4,0) cos^{4}(x) + T(3,1) cos^{2}(x) + T(2,2) = 8 cos^{4}(x) - 8 cos^{2}(x) + 1

What about the Chebyshev polynomials of second kind?
Let's go back to Pascal triangle and define
*D(i,j)* = (-)^{i+j} 2^{j} *B(i,j)*.

1 -1 2 1 -4 4 -1 6 -12 8 1 -8 24 -32 16 ...Next consider the polynomials (the sum runs over the k such that n+k is even)

UFor example,_{n}(x) = Sum_{k}D((n+k)/2, k) x^{k}

UProve that sin((n+1)x) = sin(x) U_{1}(x) = D(1,1) x^{1}= 2 x U_{2}(x) = D(1,0) x^{0}+ D(2,2) x^{2}= -1 + 4 x^{2}. U_{3}(x) = D(2,1) x^{1}+ D(3,3) x^{3}= -4 x + 8 x^{3}.

By the way, Pascal triangle is related to Fibonacci numbers. Write the triangle left-aligned,

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ...The sum of the diagonals are the Fibonacci numbers: 1, 1, 2, 3, 5, ... If you do the same for the Chebyshev triangle you get the Lucas numbers: 2, 1, 3, 4, 7, ...

Marco Corvi - 2005