Orthogonal Polynomials: an Illustrated Guide

Home , Chebyshev polynomials, Gegenbauer polynomials, Legendre polynomials, Orthogonal polynomials, Weight function

Orthogonal Polynomials: An Illustrated Guide

Avaneesh Narla

December 10, 2018

Contents

1 Deﬁnitions 1

2 Example 1 2

3 Three-term Recurrence Relation 3

4 Christoﬀel-Darboux Formula 5

5 Zeros 6

6 Gauss Quadrature 8 6.1 Lagrange Interpolation ...... 8 6.2 Gauss quadrature formula ...... 8

7 Classical Orthogonal Polynomials 11 7.1 Hermite Polynomials ...... 11 7.2 Laguerre Polynomials ...... 12 7.3 Legendre Polynomials ...... 14 7.4 Jacobi Polynomials ...... 16 7.5 Chebyshev Polynomials of the First Kind ...... 17 7.6 Chebyshev Polynomials of the Second Kind ...... 19 7.7 Gegenbauer polynomials ...... 20

1 Deﬁnitions

Orthogonal polynomials are orthogonal with respect to a certain function, known as the weight function w(x), and a defined interval. The weight function must be continuous and positive such that its moments (µn) exist. Z b n µn := w(x)x dx, n = 0, 1, 2, ... a The interval may be infinite. We now define the inner product of two polynomials as follows Z ∞ hf, giw(x) := w(x)f(x)g(x) dx −∞

1 We will drop the subscript indicating the weight function in future cases. Thus, as always, a ∞ sequence of polynomials {pn(x)}n=0 with deg(pn(x)) = n are called orthogonal polynomials for a weight function w if hpm, pni = hnδmn Above, the delta function is the Kronecker Delta Function

There are two possible normalisations: If hn = 1 ∀n ∈ {0, 1, 2...}, the sequence is orthonormal. If the coeﬃcient of highest degree term is 1 for all elements in the sequence, the sequence is monic. We note that orthogonal polynomials are thus linearly independent and generate bases for all polynomials of any arbitrary degree.

Important Immediate Properties: 1. hQ, pniw(x) = ckpn and hQ, pniw(x) 6= 0 =⇒ Q has a term of degree n. Question 1. Can you have two distinct sets of orthogonal polynomials corresponding to the same weight function but diﬀerent intervals? What about the same interval? Is there any relation between the two distinct sets? Also, do all weight functions have corresponding sequence of orthogonal polynomials? If not, what are the deﬁning properties of the weight functions that do?

What additional restrictions do you need on w(x) to make the inner product space complete?

Answers: For two sets of orthogonal polynomials corresponding to the same weight function but diﬀerent intervals, look at the example set and the Legendre Polynomials. For two sets of orthogonal polynomials corresponding to the same weight function but the same interval, look at the two kinds of Chebyshev Polynomials.

2 Example 1

We choose the weight function to be a constant (say c), and choose the interval (0,a). We can construct the sequence through the Gram-Schmidt process (by moving from the basis {1, x, x2, x3...} 2 to the orthogonal basis). We start with the sequence {1, x, x , ...}. We choose p0(x) = 1 for now, we can later normalise the sequence to be orthonormal. Thus by the Gram-Schmidt Process,

Proof. hx, p0(x)i hx, 1i a p1(x) = x − projp0(x)x = x − p0(x) = x − = x − hp0(x), p0(x)i h1, 1i 2 R a ca2 R a Since, hx, 1i = 0 c · x dx = 2 and h1, 1i = 0 cdx = ca

2 2 2 2 2 2 hx , p0(x)i hx , p1(x)i =⇒ p2(x) = x − projp0(x)x − projp1(x)x = x − p0(x) − p1(x) = hp0(x), p0(x)i hp1(x), p1(x)i

2 2 a 3 4 2 2 2 2 hx , 1i hx , x − 2 i a 2 ca 1 ca 12 a 2 a a 2 a x − − a a x − = x − · − · 3 x − = x − −ax+ = x −ax+ h1, 1i hx − 2 , x − 2 i 2 3 ca 12 ca 2 3 2 6 as, Z a ca3 hx2, 1i = c · x2 dx = 0 3

2 Z a 4 3 a 4 4 4 2 2 cx cax ca ca ca hx , (x − a/2)i = cx · (x − a/2) dx = − = − = 0 4 6 0 4 6 12 Z a 3 2 2 a 3 3 3 3 2 cx cax ca x ca ca ca ca h(x − a/2), (x − a/2)i = c(x − a/2) dx = − + = − + = 0 3 2 4 0 3 2 4 12 And similarly by continuing the Gram-Schmidt process, we get that

3a 3a2 a3 9a2 2a3 a4 p (x) = x3 − x2 + x − , p (x) = x4 − 2ax3 + x2 − x + 3 2 5 20 4 7 7 70 5a 20a2 5a3 5a4 a5 p (x) = x5 − x4 + x3 − x2 + x − 5 2 9 6 42 252

√ R a 2 The orthonormal polynomials would be qn(x) = pn(x)/ hn, where hn = 0 cpn(x) dx

Question 2. What is the general expression for pn(x) in the above example? What about hn(x)? Is this a complete inner product space?

3 Three-term Recurrence Relation

∞ Theorem 1. A sequence of orthogonal polynomials {pn(x)}n=0 satisﬁes

pn+1(x) = (Anx + Bn)pn(x) + Cnpn−1(x), n = 1, 2, 3...

3 where, kn+1 An hn An = , n = 0, 1, 2, 3... and Cn = − · , n = 1, 2, 3... kn An−1 hn−1 where kn is the leading coeﬃcient of pn(x)

Proof. We first note that the degree of pn+1(x) is the same as the degree of xpn(x). Thus, by defining the coefficients of the leading terms of each polynomial, we note that the leading term of p (x) is k xn+1, which is equal to the leading term of kn+1 xp := A xp . n+1 n+1 kn n n n

=⇒ hpn+1, xpn(x)i = An (1)

Now we consider the lower order terms. pn+1 − Anxpn is a polynomial of degree ≤ n and n X pn+1 − Anxpn = ckpk(x) (2) k=0 This is because orthogonal polynomials are the basis for all polynomials. The orthogonality property now gives us that, n X hpn+1(x) − Anxpn(x), pk(x)i = cmhpm(x), pk(x)i = ckhpk(x), pk(x)i = hkck (3) m=0

= hpn+1(x), pk(x)i − hAnxpn(x), pk(x)i = −Anhpn(x), xpk(x)i (4)

−An =⇒ ck = hpn(x), xpk(x)i hk But, hpn(x), xpk(x)i = 0 for k < n − 1. Thus, ∀k < n − 1, ck = 0 Therefore, from (2) 1 1 =⇒ pn+1 − Anxpn = −An hpn(x), xpn(x)i · pn(x) + hpn(x), xpn−1(x)i · pn−1(x) hn hn−1 Or, equivalently

pn+1(x) − Anxpn(x) = cnpn(x) + cn−1pn−1(x), n = 1, 2, 3... Further, by using the same argument as (1), we have that

kn−1 An hn hn−1cn−1 = −Anhpn(x), xpn−1(x)i = −An hn =⇒ cn−1 = − · kn An−1 hn−1

The term B in Th. 1 can be represented as −An hp (x), xp (x)i but the inner product cannot n hn n n be put in terms of coeﬃcients as simply as the other terms. Also, we note that if we represent the polynomials as monic polynomials, we get the simpler representation,

hn pn+1(x) = xpn(x) + Bnpn(x) + Cnpn−1(x) where Cn = − , n = 1, 2, 3... hn−1 If they are orthogonal polynomials, we get the representation,

hn+1 An pn+1(x) = ((Anx + Bn)pn(x) + Cnpn−1(x)) where Cn = − , n = 1, 2, 3... hn An−1 Question 3. Which is more computationally viable, the Gram-Schmidt Process or the Three-Term Recurrence Relation? Is there any other algorithm to do it better (like FFT)?

4 4 Christoﬀel-Darboux Formula

As a consequence of the Three-Term Recurrence Relation, we have that

∞ Theorem 2. A sequence of orthogonal polynomials {pn(x)}n=0 satisﬁes n X pk(x)pk(y) kn pn+1(x)pn(y) − pn+1(y)pn(x) = · , n = 0, 1, 2... hk hnkn+1 x − y k=0 and as y → x

n 2 X pk(x) kn 0 0 = · (pn+1(x)pn(x) − pn+1(x)pn(x)), n = 0, 1, 2... hk hnkn+1 k=0 Proof. The three term recurrence relation implies that

pn+1(x)pn(y) = (Anx + Bn)pn(x)pn(y) + Cnpn−1(x)pn(y), n = 1, 2, 3... and pn+1(y)pn(x) = (Anx + Bn)pn(y)pn(x) + Cnpn−1(y)pn(x), n = 1, 2, 3... Subtracting the two equations above, we have

pn+1(x)pn(y) − pn+1(y)pn(x) = An(x − y)pn(x)pn(y) + Cn[pn−1(x)pn(y) − pn−1(y)pn(x)]

=⇒ An(x − y)pn(x)pn(y) = Cnpn−1(x)pn(y) − Cnpn−1(y)pn(x) − pn+1(x)pn(y) + pn+1(y)pn(x)

Dividing all through by Anhn, we get (x − y)p (x)p (y) −p (y)p (x) + p (x)p (y) −p (y)p (x) + p (x)p (y) =⇒ n n = n−1 n n−1 n + n+1 n n+1 n hn An−1hn−1 Anhn n n n X (x − y)pk(x)pk(y) X −pk−1(y)pk(x) + pk−1(x)pk(y) X −pk+1(y)pk(x) + pk+1(x)pk(y) =⇒ = + hk Ak−1hk−1 Akhk k=1 k=1 k=1 By reordering, n n n X pk(x)pk(y) X pk+1(x)pk(y) − pk(x)pk+1(y) X pk(x)pk−1(y) − pk−1(x)pk(y) =⇒ (x − y) = − hk Akhk Ak−1hk−1 k=1 k=1 k=1 By eliminating all the common terms in the two sums on the RHS, n X pk(x)pk(y) pn+1(x)pn(y) − pn(x)pn+1(y) p1(x)p0(y) − p0(x)p1(y) =⇒ (x − y) = − hk Anhn A0h0 k=1 p (x)p (y) − p (x)p (y) k (p (x)k − k p (y)) = n+1 n n n+1 − 0 1 0 0 1 Anhn k1h0 as A = k1 and p (x) = p (y) = k 0 k0 0 0 0 k2(x−y) Further, p (x) − p (y) = k (x − y) =⇒ k0(p1(x)k0−k0p1(y)) = 0 = (x − y) p0(x)p0(y) 1 1 1 k1h0 h0 h0 n X pk(x)pk(y) pn+1(x)pn(y) − pn(x)pn+1(y) =⇒ (x − y) = hk Anhn k=0

5 n X pk(x)pk(y) kn =⇒ = (pn+1(x)pn(y) − pn(x)pn+1(y)) hk (x − y)kn+1hn k=0 Thus, we prove the Christoﬀel-Darboux formula. To prove the conﬂuent form, we take the limit as y → x p (x)p (y) − p (x)p (y) p (x)(p (x) − p (y)) − p (x)(p (x) − p (y)) lim n+1 n n n+1 = lim n n+1 n+1 n+1 n n y→x x − y x→y x − y

0 0 = pn(x)p n+1 − pn+1(x)p n(x)

5 Zeros

∞ Theorem 3. If {pn(x)x=0 is a sequence of orthogonal polynomials on the interval (a,b) with respect to the weight function w(x), then the polynomial pn(x) has exactly n real simple zeros on the interval (a,b)

Proof. Since the degree of pn(x) is n, the number of distinct real roots is at most n. Suppose that pn(x) has m ≤ n distinct real roots x1, x2...xm in (a,b) of odd multiplicity (we ignore the roots of even multiplicity). Then, the polynomial

pn(x)(x − x1)(x − x2)...(x − xm) does not change sign on the interval (a,b) as each root occurs an even number of times in the interval (a,b). Thus,as w(x) > 0∀x ∈ (a, b)

Z b w(x)pn(x)(x − x1)(x − x2)...(x − xm)dx 6= 0 a

But, by orthogonality this integral equals 0 if m

This can be seen in the roots of the previous example. We can also see in that example that the roots of the polynomials are interlacing.

∞ Theorem 4. If {pn(x)x=0 is a sequence of orthogonal polynomials on the interval (a,b) with respect to the weight function w(x), then the zeros of pn(x) and pn+1(x) separate each other. Proof. This follows from the conﬂuent form of the Christoﬀel-Darboux formula, stated as:

n 2 X pk(x) kn 0 0 = · (pn+1(x)pn(y) − pn+1(y)pn(x)), n = 0, 1, 2... hk hnkn+1 k=0 We note that Z b 2 hn = w(x){pn(x)} dx > 0, n = 0, 1, 2... a as w(x) > 0 ∀x.

n 2 kn 0 0 X pk(x) =⇒ · (pn+1(x)pn(x) − pn+1(x)pn(x)) = > 0 hnkn+1 hk k=0

6 Now, suppose xn,k and xn,k+1 are two consecutive zeros of pn(x) with xn,k < xn,k+1. Since all n zeros 0 0 of pn(x) are real and distinct, pn(xn,k) and pn(xn,k+1) should have opposite signs (as pn(xn,k) = 0 and pn(xn,k+1) = 0, but there are no zeros in between xn,k and xn,k+1 by deﬁnition, hence the curve must be increasing in one direction and decreasing in the other). Hence we have that

0 0 pn(xn,k) = 0 = pn(xn,k+1) and pn(xn,k) · pn(xn,k+1) < 0

Then, as per the formula above,

kn 0 kn 0 · (−pn+1(xn,k)pn(xn,k)) > 0 and · (−pn+1(xn,k+1)pn(xn,k+1)) > 0 kn+1 kn+1 2 kn 0 0 =⇒ · (−pn+1(xn,k)pn(xn,k)) · (−pn+1(xn,k+1)pn(xn,k+1)) > 0 kn+1 0 0 =⇒ (pn+1(xn,k)pn(xn,k)) · (pn+1(xn,k+1)pn(xn,k+1)) > 0 0 0 =⇒ pn+1(xn,k) · pn+1(xn,k+1) < 0 as pn(xn,k) · pn(xn,k+1) < 0

Further, the continuity of pn+1(x) implies that there should be an odd number of zeros of pn+1(x) between xn,k and xn,k+1 as the sign must change between them. However, this holds for any two consecutive zeros of pn. Thus, the only way this is possible is if there is one zero of pn+1 between any two consecutive zeros of pn. Therefore, a < xn,k < xn+1,k+1 < xn,k+1 < b ∀n > 0, k < n where xn,k is the kth root of pn and so on.

This can be seen in the following graph of roots plotted for the ﬁrst ﬁve non-trivial Orthonormal Polynomials of the constant function as in the above example:

Question 4. Is there any relationship between the Cauchy Interlacing Theorem and Orthogonal Polynomials that lends to this property?

7 6 Gauss Quadrature

6.1 Lagrange Interpolation

If f is a continuous function on (a,b) and x1 < x2 < x3 < ... < xn are n distinct points in (a,b), then there exists exactly one polynomial P with degree ≤ n − 1 such that P (xi) = f(xi) for all i = 1, 2, 3..., n (there are inﬁnite polynomials of degree n or greater). This is because if there existed another polynomial, say Q(x) with such a property, we have that (P − Q)(x) has n distinct zeros, for each xi. But this is not possible as (P − Q)(x) has only n zeros. The existence of the unique polynomial P can be easily demonstrated (and the polynomial found) by using Lagrange interpolation as follows. Deﬁne

p(x) = (x − x1)(x − x2)...(x − xn)

n X p(x) =⇒ p0(x) = (x − x ) i=1 i

0 p(x) =⇒ p (xi) = = (xi − x1)...(xi − xi−1)(xi − xi+1)...(xi − xn) := qi(xi) x − x i xi where q (x) := p(x) . And consider i x−xi

n n n X p(x) X (x − x1)...(x − xi−1)(x − xi+1)...(x − xn) X qi(x) P (x) = f(x ) = f(x ) = f(x ) i (x − x )p0(x ) i (x − x )...(x − x )(x − x )...(x − x ) i q (x ) i=1 i i i=1 i 1 i i−1 i i+1 i n i=1 i i

We see that thus, P (xi) = f(xi) for all i = 1, 2, 3..., n.

6.2 Gauss quadrature formula

∞ Let {pn(x)}n=0 be a sequence of orthogonal polynomials on the interval (a,b) with respect to the n weight function w(x). Then we take the n distinct roots of pn(x) = {xi}i=1 such that x1 < x2 < ... < xn. We now consider a f(x) of degree ≤ 2n − 1, and P(x) as the polynomial of degree

f(x) = P (x) + r(x)pn(x) where r(x) is a polynomial of degree ≤ n − 1. This can also be written as

n X p(x) f(x) = f(x ) + r(x)p (x) i (x − x )p0(x ) n i=1 i i

Z b n Z b Z b X w(x)pn(x) =⇒ w(x)f(x) dx = f(x ) + w(x)r(x)p (x) dx i (x − x )p0 (x)dx n a i=1 a i n a Since r(x) is of a degree

Z b n Z b X w(x)pn(x) =⇒ w(x)f(x) dx = f(x )λ such that λ := dx i n,i n,i (x − x )p0 (x) a i=1 a i n

8 This is known as the Gauss quadrature formula, and it gives us the value of the integral in the case that f is a polynomial of degree <2n and if the values of f(xi) is known for the n zeros x1 < x2 < x3...xn of the polynomial pn(x). If f is not a polynomial of degree <2n, this leads to an approximation of the integral:

Z b n Z b X w(x)pn(x) w(x)f(x) dx ≈= f(x )λ such that λ := dx i n,i n,i (x − x )p0 (x) a i=1 a i n

n The coefficients {λn,i}i=1 are called Christoffel numbers. We note that these numbers do not depend on the function f. We will now try to approximate sin(x) using this formula and the set of orthogonal polynomials used above. We found the roots of the polynomial, and found the respective Christoffel numbers.

This graph shows the five Christoffel numbers for the 5th orthonormal polynomial for the weight function described above. Further, I used the fifth orthonormal polynomial of the constant weight function to approximate some functions, and found that for a random polynomial of degree 9, and for other functions such as sin(x) and exp(x), the Gauss Quadrature was able to approximate the function with a near 100% precision (the error is thought to be due to the errors in numerical integration). This is incredible, as after the initial steps of computing the zeros of the orthonormal polynomial, and computing its Christoffel numbers, the integration of any function will only take 2n steps(one for computing the function at each point, and then to add it perform summation), where n is the order of the orthonormal polynomial. And we have seen that we achieve near 100% accuracy even with just a degree 5 polynomial.

1 % Christoffel Approximation 2 pol=[1;-5/2;+20/9;-5/6;+5/42;-1/252];% The orthonormal polynomial is defined by its coefficents 3 xi=roots(pol);%MATLAB finds the roots of the polynomial.(This is the first source of numerical error) 4 p5=@(x) x.^5-5/2*x.^4+20/9*x.^3-5/6*x.^2+5/42*x-1/252;% The orthonormal polynomial defined asa function inMATLAB 5 dp5=@(x) 5*x.^4-4*5/2*x.^3+3*20/9*x.^2-2*5/6*x+5/42;% The derivative of the orthonormal polynomial defined asa function inMATLAB

9 6 func =@(x) exp(x);% The function to be approximated 7 w=@(x) x.^0;% The weight function 8 9 approx =0; 10 for i=1:length(xi) 11 fu=@(x) w(x).*p5(x)./((x-xi(i,1)).*dp5(xi(i)));% The integrand 12 lambda(i,1)=integral(fu,0,1);% The lambdas are the Christoffel numbers 13 approx=approx+lambda(i)*func(xi(i,1)); 14 end 15 16 actual=integral(func,0,1);% Computes the actual value of the integral 17 error=abs((actual-approx)/actual) These Christoﬀel numbers are all positive, as can be seen below:

Z b pn(x) λn,i = w(x)ln,i(x)dx with ln,i(x) := 0 a (x − xi)pn(x)

2 2 pn(x) pn(x) We consider l (x) − ln,i(x) = 2 0 2 − 0 which is a polynomial of degree < 2n − 2 n,i (x−xi) p (x) (x−xi)p (x) n n n which vanishes at the zeros {xn,k}k=1 of pn(x). Hence,

2 ln,i(x) − ln,i(x) = pn(x)s(x)

pn(x) 1 for some polynomial q of degree ≤ n − 2. Note: q(x) is actually equal to 2 0 2 − 0 (x−xi) pn(x) (x−xi)pn(x) Z b Z b 2 =⇒ w(x)(ln,i(x) − ln,i(x))dx = w(x)pn(x)q(x)dx = 0 a a by orthogonality. Hence we have,

Z b Z b 2 λn,i = w(x)ln,i(x)dx = w(x){ln,i(x)} dx > 0 a a Now, we can also prove that

∞ Theorem 5. If {pn(x)x=0 is a sequence of orthogonal polynomials on the interval (a,b) with respect to the weight function w(x), then we have that between any two zeros of pm(x) there is at least one zero of pn(x), for any m < n.

Proof. Suppose that xm,k and xm,k+1 are two consecutive zeros of pm(x) and that there is no zero of pn(x) in (xm,k, xm,k+1). Then consider the polynomial

p (x) g(x) = m (x − xm,k)(x − xm,k+1) Then we have g(x)pm(x) ≥ 0 for x∈ / (xm,k, xm,k+1)

as then (x − xm,k) and (x − xm,k+1) have the same sign and their product is positive. Now the Gauss quadrature formula gives

n Z b X w(x)g(x)pm(x) dx = λn,ig(xn,ipm(xn,i) a i=1

10 n where {xn,i}i=1 are the zeros of pn(x). Since we assumed that there are no zeros of pn(x) in (xm,k, xm,k+1) we conclude that g(xn,ipm(xn,i) > 0 for all i=1,2...n. Further, we have λn,i > 0 for all i=1,2...n. Which implies that the sum at the RHS cannot vanish. However, the integral at the left-hand side is zero by orthogonality (as we are only considering m < n). This contradiction proves that there should be at least one zero of pn(x) between any two consecutive zeros of pm(x). This can again be seen in the graph of the roots of the Orthogonal Polynomials.

7 Classical Orthogonal Polynomials

7.1 Hermite Polynomials The weight function for this set of orthonormal polynomials is e−x2 . The interval on which the inner product for this set of orthonormal polynomials is defined is (−∞, ∞). The first five polynomials of this set are:

2 x 4 x2 − 2 8 x3 − 12 x 16 x4 − 48 x2 + 12 32 x5 − 160 x3 + 120 x We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph:

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

11 We also plotted the Christoffel numbers for the first five polynomials of this set.

7.2 Laguerre Polynomials The weight function for this set of orthonormal polynomials is e−xxα. The interval on which the inner product for this set of orthonormal polynomials is defined is (0, ∞). The first five polynomials of this set are:

α − x + 1 3 α α2 x2 − x (α + 2) + + + 1 2 2 2 11 α α2 5 α α 3 α3 x3 − x + + 3 + x2 + + α2 + − + 1 6 2 2 2 2 6 6 25 α α3 3 α2 13 α α 2 α2 7 α 35 α2 5 α3 α4 x4 −x + + + 4 −x3 + +x2 + + 3 + + + + +1 12 6 2 3 6 3 4 4 24 12 24 24

12 137 α α3 47 α α 5 α4 7 α3 71 α2 77 α + x2 + α2 + + 5 + x4 + − x + + + + 5 − ... 60 12 12 24 24 24 12 24 12 α2 3 α 5 15 α2 17 α3 α4 α5 x5 x3 + + + + + + − + 1 12 4 3 8 24 8 120 120 We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph (for a random parameter, α = 0.6324

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

We also plotted the Christoffel numbers for the first five polynomials of this set.

13 7.3 Legendre Polynomials The weight function for this set of orthonormal polynomials is 1. The interval on which the inner product for this set of orthonormal polynomials is defined is (-1,1). The first five polynomials of this set are:

x 3 x2 1 − 2 2 5 x3 3 x − 2 2 35 x4 15 x2 3 − + 8 4 8 63 x5 35 x3 15 x − + 8 4 8

14 We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph:

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

We also plotted the Christoffel numbers for the first five polynomials of this set.

15 7.4 Jacobi Polynomials The weight function for this set of orthonormal polynomials is (1 − x)α(1 + x)β. The interval on which the inner product for this set of orthonormal polynomials is deﬁned is (-1,1). The ﬁrst two polynomials of this set are:

α β α β − + x + + 1 2 2 2 2 α2 α β 7 α β2 7 β 3 β α β α α2 β2 α2 3 α β2 3 β 1 x2 + + + + + − − − + + + x + − − − 8 4 8 8 8 2 8 4 8 8 8 4 4 4 4 2 We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph:

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set (interestingly, some of them had imaginary parts).

16 We also plotted the Christoffel numbers for the first five polynomials of this set.

7.5 Chebyshev Polynomials of the First Kind

The weight function for this set of orthonormal polynomials is √ 1 . 1−x2 The interval on which the inner product for this set of orthonormal polynomials is defined is (-1,1). The first five polynomials of this set are:

x 2 x2 − 1 4 x3 − 3 x 8 x4 − 8 x2 + 1 16 x5 − 20 x3 + 5 x

17 We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph:

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

18 7.6 Chebyshev Polynomials of the Second Kind

2 x 4 x2 − 1 8 x3 − 4 x 16 x4 − 12 x2 + 1 32 x5 − 32 x3 + 6 x We plotted the ﬁrst ﬁve polynomials on the interval and obtained the following graph:

19 We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

We also plotted the Christoffel numbers for the first five polynomials of this set.

7.7 Gegenbauer polynomials The weight function for this set of orthonormal polynomials is (1 − x2)α–1/2. The interval on which the inner product for this set of orthonormal polynomials is defined is (-1,1). The first five polynomials of this set are:

2 α x x2 2 α2 + 2 α − α 4 α3 8 α x3 + 4 α2 + − x 2 α2 + 2 α 3 3 α 2 α4 22 α2 α2 − x2 2 α3 + 6 α2 + 4 α + x4 + 4 α3 + + 4 α + 2 3 3 2

20 4 α5 8 α4 28 α3 40 α2 32 α 4 α4 44 α2 + + + + x5+ − − 8 α3 − − 8 α x3+α3 + 3 α2 + 2 α x 15 3 3 3 5 3 3 We plotted the ﬁrst ﬁve polynomials (for a random α on the interval and obtained the following graph:

We then plotted the zeros of the ﬁrst ﬁve polynomials of this set.

We also plotted the Christoffel numbers for the first five polynomials of this set.

21 22