The integrals have integrable singularities at

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Special functions and orthogonal polynomials · Numerical and computational mathematics

Error, Chance and the Machine, Ch Integration: error expansions, Gauss rules and adaptivity

Problem

The integrals

\int_{-1}^{1}f(x)/\sqrt{1-x^2}\,\mathrm{d}x

have integrable singularities at

x = \pm1

, which a Gauss-Legendre rule handles badly. A Gauss rule for the weight

\omega(x) = 1/\sqrt{1-x^2}

, with

\int\omega\,\mathrm{d}x = \pi

, takes the singularities into the weight. The orthonormal polynomials

\tilde\psi_k

for this weight obey the recurrence of Section 5.5 with

\alpha_k = 0

for all

k

\beta_1 = 1/\sqrt2

and

\beta_k = \tfrac12

for

k \ge 2

. Use the exact values

\int\omega x^4\,\mathrm{d}x = 3\pi/8

and

\int\omega x^6\,\mathrm{d}x = 5\pi/16

. ((a)) Write the Jacobi matrix

\mathsf{A}

for

n = 3

and find its eigenvalues, the nodes of the three-node rule. ((b)) Find the eigenvectors and hence the weights

w_j

. ((c)) Evaluate the rule on

f = x^4

and

f = x^6

, and state the degree to which it is exact. ((d)) Apply the rule to

\int_{-1}^{1}\dfrac{\mathrm{d}x}{(2-x)\sqrt{1-x^2}} = \dfrac{\pi}{\sqrt3}

and give its error, as value minus exact value. ((e)) Substitute

x = \cos\theta

to show that the rule is a trapezium rule for a periodic integrand with nodes shifted by a constant phase. The function

1/(2-\cos\theta)

has Fourier coefficients

a_k = r^{|k|}/\sqrt3

with

r = 2 - \sqrt3

. Extend Eq. (5.11) to the shifted nodes and use it to predict the error of part (d).

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