Numerical Integration: Orthonormal Basis of Polynomials and Trapezoidal Method, Study notes of Calculus

Download Numerical Integration: Orthonormal Basis of Polynomials and Trapezoidal Method and more Study notes Calculus in PDF only on Docsity! Introduction to Numerical Integration Adrian Down March 16, 2006 1 Orthonormal basis of polynomials 1.1 Review Recall that we constructed on orthonormal basis on the space of quadratic functions on [−1, 1], a0(x) = 1√ 2 a1(x) = x a2(x) = 3x2 − 1 2 √ 2 5 1.2 Laplace’s equation These polynomials can be continued using a recursion process. Each resulting polynomial an is of degree n−1 and is orthonormal to the other polynomials in the set, am · an = ∫ 1 −1 amandx = δmn The recursion relation is given in problem 7. It can also be derived from the Laplace partial differential equation describing the spherical harmonics in R3, ∂11u + ∂22u + ∂33u = 0 where ∂11 = ∂2 ∂x2 and so on for x2 = y and x3 = z. 1 Write the equation in spherical polar coordinates. We seek solutions that are dependent only on r and θ, u = rnp(x) where x = cos θ. Note. The x in our assumed form of the solution of Laplace’s differential equation is not the same as the x coordinate in R3. The differential equation for p(x) becomes,{ (1− x2)p′ } + n(n + 1)p = 0 ∈ (−1, 1) (1) Note. x = 1 corresponds to cos θ = 1 ⇒ θ = 0, i.e. the north pole of the coordinate system. Likewise x = −1 corresponds to the south pole. 1.3 Orthogonality The functions that satisfy 1 are the Legendre polynomials. pn defined in this way is an nth degree polynomial which is bounded at x = ±1. These polynomials can be normalized such that∫ 1 −1 a2ndx = 1 We found the first three polynomials above in the previous lecture. We showed these functions are orthogonal. Now we show that all other pm con- structed in this was are orthogonal. Proof. Consider the differential equation 1 applied to two particular Legendre polynomials pn and pm. Multiply the equation for pm by pn and the equation for pn by pm, pn { (1− x2)p′m } + n(n + 1)pmpn = 0 pm { (1− x2)p′n } + n(n + 1)pnpm = 0 Subtract these two equations and integrate from −1 to 1,∫ 1 −1 {( 1− x2 ) (pnp ′ m − p′npm) }′ dx = {n(n + 1)−m(m + 1)} ∫ 1 −1 pmpndx 2 The total error is then, (h) = T (h)− ∫ b a f(x)dx = n−1∑ j=0 j(h) = h2 12 n−1∑ j=0 f ′′(xj)h + O(h 3) Note. In our original expression for the error, we had a term of the form O(h4). When this term is summed over n, it becomes O(nh4). Since n = b−a h , O(nh3) reduces to O(h3). In the limit h → 0, the summation becomes an integral, lim h→0 n−1∑ j=0 f ′′(xj)h → ∫ b a f ′′(x)dx The integral of a derivative can be evaluated using the Fundamental Theorem of Calculus, (h) = h2 12 {f ′(b)− f ′(a)}+ O(h3) In general, (h) is a power series in h. However, it can be shown using advanced analysis that only even powers of h appear in this expansion. Theorem (Deep result). Let f(x) be analytic on an open interval [a, b]. Then, T (h)− ∫ b a f(x)dx = c2h 2 + c4h 4 + . . .︸ ︷︷ ︸ even powers only 2.3 Example Consider a simple integral for which we know the value exactly,∫ 1 0 exdx = e− 1 5 The formula for the trapezoidal approximation is, T (h) = n∑ 0 ( ejh ) h− 1 2 (1 + e) (2) The summation over ejh = (ej) h can be expressed as a geometric series, n∑ 0 ( ej )h = 1− ( eh )n+1 1− eh With this substitution, (2) becomes, T (h) = h 1− ( eh )n+1 1− eh − h 2 (1 + e) (3) This result can be simplified,( eh )n+1 = e(n+1)h = enhe Also, nh = n b− a n = b− a = 1− 0 = 1 Hence, ( eh )n+1 = eeh Using this substitution, (3) becomes, T (h) = h ehe− 1 eh − 1 − 1 2 (1 + e)h = h 2 eh + 1 eh − 1 (e− 1) = h 2 e h 2 + e− h 2 e h 2 − e−h2 (e− 1) Note. • We factored a negative from the numerator and denominator of the first term. • The exponential term was written in this way so that it corresponds to the hyperbolic tangent function. 6 The final expression for the trapezoidal approximation is, T (h) = h 2 tanh ( h 2 )(e− 1) The error term is, (h) = T (h)− ∫ 1 0 exdx = T (h)− (e− 1) = ( h 2 tanh ( h 2 ) − 1) (e− 1) This expression for (h) can be Taylor expansion in even powers of h, starting with h2. The Taylor series expansion for (h) in terms of h is in even powers of h, as claimed above. 3 Midterm 2 There will be four problems. One or two will have multiple sections. 1. The first problem will concern polynomial interpolation and error esti- mation thereof. Be aware that it is sometimes possible to estimate the error of polynomial approximation without writing an explicit expres- sion for the polynomial itself. f(x)− p(x) = f (n+1)(ζ) (n + 1)! n∏ j=0 (x− xj) • If the interval is of length L, we can make a maximum bound on the product as, ∣∣∣∣∣ n∏ j=0 (x− xj) ∣∣∣∣∣ ≤ Ln+1 A much better bound can be found by actually maximizing the argument of the product. 7