The Tschebychev polynomials (of the 1st kind) T_{n}(x) are defined through the recurrence relation

T_{0}(x) = 1

T_{1}(x) = x

T_{n+2}(x) = 2x · T_{n+1}(x) − T_{n}(x)

for n ≥ 0.

Theorem:

T_{n}(x) satisfies the differential equation (x^{2} − 1) · Tn^{′′}(x) + x · Tn^{′}(x) − n^{2} · T_{n}(x) = 0 for all n ≥ 0.

Proof:

Let F := ∑_{n=0}^{∞} T_{n}(x)z^{n} be the generating function of the sequence of polynomials. The recurrence is equivalent to the equation

(1 − 2x · z + z^{2}) · F = 1 − x · z

Proof 1:

F is a rational function in z, F = \frac{1−xz}{1−2xz+z^{2}} . From the theory of recursions with constant coefficients, we know that we have to perform a partial fraction decomposition. So let p(z) = z^{2} − 2x · z + 1 be the denominator and α = x + \sqrt{x^{2} − 1} and α^{−1} its zeroes. The partial fraction decomposition reads

F = \frac{1 − xz}{1 − 2xz + z^{2}} = \frac{1}{2} (\frac{1}{1 − αz} + \frac{1}{1 − α^{−1}z} )

hence T_{n}(x) = \frac{1}{2}(α^{n} + α^{−n}). Note that the field Q(x)(α), being a finite dimensional extension field of Q(x) in characteristic 0, has a unique derivation extending \frac{d}{dx} on Q(x). We can therefore try to construct an annihilating differential operator for T_{n}(x) by combination of annihilating differential operators for α^{n} and α^{−n}. In fact, L_{1} := (α − x)\frac{d}{dx} − n satisfies L_{1}[α^{n}] = 0, and L_{2} := (α − x)\frac{d}{dx} + n satisfies L_{2}[α^{−n}] = 0. A common multiple of L_{1} and L_{2} is easily found by solving an appropriate system of linear equations:

L = (x^{2}−1) (\frac{d}{dx})^{2}+x\frac{d}{dx} −n^{2} = ((α − x)\frac{d}{dx} + n)·L_{1} = ((α − x)\frac{d}{dx} − n)·L_{2} It follows that both L[α^{n}] = 0 and L[α^{−n}] = 0, hence L[T_{n}(x)] = 0.

Proof 2:

Starting from the above equation, we compute a linear relation for the partial derivatives of F. Write ∂_{x} = \frac{d}{dx} and Δ_{z} = z \frac{d}{dz}. One computes

(1 − 2xz + z^{2}) · F = 1 − xz

(1 − 2xz + z^{2})^{2} · ∂_{x}F = z − z^{3}

(1 − 2xz + z^{2})^{3} · ∂_{x}^{2}F = 4z^{2} − 4z^{4}

(1 − 2xz + z^{2})^{2} · Δ_{z}F = xz − 2z^{2} + xz^{3}

(1 − 2xz + z^{2})^{3} · ∂_{x}Δ_{z}F = z + 2xz^{2} − 6z^{3} + 2xz^{4} + z^{5}

(1 − 2xz + z^{2})^{3} · Δ_{z}^{2}F = xz + (2x^{2} − 4)z^{2} − (2x^{2} − 4)z^{4} − xz^{5}

Solve a 6 × 6 system of linear equations over Q(x) to get

x · ∂_{x}F + (x^{2} − 1) · ∂_{x}^{2}F − Δ_{z}^{2}F = 0

This is equivalent to the claimed equation (x^{2} − 1) · Tn^{′′}(x) + x · Tn^{′}(x) − n^{2} · T_{n}(x) = 0.

References

1 Bruno Haible: D-finite power series in several variables. Diploma thesis, University of Karlsruhe, June 1989. Sections 2.12 and 2.15.