Product of Chebyshev polynomials

Chebyshev polynomials satisfy a lot of identities, much like trig functions do. This point will look briefly at just one such identity.

Chebyshev polynomials T_n are defined for n = 0 and 1 by

T₀(x) = 1
T₁(x) = x

and for larger n using the recurrence relation

T_n+1(x) = 2xT_n(x) - T_n-1(x)

This implies

T₂(x) = 2xT₁(x) - T₀(x) = 2x² - 1
T₃(x) = 2xT₂(x) - T₁(x) = 4x³ - 3x
T₄(x) = 2xT₃(x) - T₂(x) = 8x⁴ - 8x² + 1

and so forth.

Now for the identity for this post. If m a n, then

2 T_mT_n = T_m+n + T_m-n.

In other words, the product of the mth and nth Chebyshev polynomials is the average of the (m + n)th and (m - n)th Chebyshev polynomials. For example,

2 T₃(x) T₁(x) = 2 (4x³ - 3x) x = T₄(x) + T₂(x)

The identity above is not at all apparent from the recursive definition of Chebyshev polynomials, but it follows quickly from the fact that

T_n(cos I) = cos nI.

Proof: Let I = arccos x. Then

2 T_m(x) T_n(x)
= 2 T_m(cos I) T_n(cos I)
= 2 cos mI cos nI
= cos (m+n)I + cos (m-n)I
= T_m+n(cos I) + T_m-n(cos I)
= T_m+n(x) + T_m-n(x)

You might object that this only shows that the first and last line are equal for values of x that are cosines of some angle, i.e. values of x in [-1, 1]. But if two polynomials agree on an interval, they agree everywhere. In fact, you don't need an entire interval. For polynomials of degree m+n, as above, it is enough that they agree on m + n + 1 points. (Along those lines, see Binomial coefficient trick.)

The close association between Chebyshev polynomials and cosines means you can often prove Chebyshev identities via trig identities as we did above.

Along those lines, we could have taken

T_n(cos I) = cos nI

as the definition of Chebyshev polynomials and then proved the recurrence relation above as a theorem, using trig identities in the proof.

Forman Acton suggested in this book Numerical Methods that Work that you should think of Chebyshev polynomials as "cosine curves with a somewhat disturbed horizontal scale."