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MATH 128A, SUMMER 2009: HOMEWORK 5 SOLUTIONS
(1) Given a quadrature rule of the form
∫ (^) b a f^ (x)^ dx^ ≈^
∑n i=0 cif^ (xi), let^ f^ (x) be [(x^ −^ x^0 )^...^ (x^ −^ xn)]
a polynomial of degree 2(n + 1). Since f (x) is a continuous function, positive except at its n + 1 roots, the exact value of
∫ (^) b a f^ (x)^ dx must be positive. Since f (x) is zero when x = xi, the estimate for
∫ (^) b a f^ (x)^ dx^ given by the quadrature rule is zero. Since the quadrature can’t integrate f (x) (a polynomial of degree 2(n + 1)) exactly, its degree of precision must be less than 2(n + 1). (2) Use the inner product 〈f, g〉 =
0 e
−xf (x)g(x) dx for functions on [0, ∞). Let f (x) be a polynomial of degree 2n − 1 (or less). use polynomial division to express f (x) in the form q(x)Ln(x)+r(x), where q(x) (the quotient) is a polynomial of degree at most (2n−1)−n = n−1, and r(x) (the remainder) is a polynomial of degree less than n. The quadrature rule estimates
0 e
−xq(x)Ln(x) dx exactly due to the choice of nodes: Since deg q(x) ≤ n − 1, q(x) is orthogonal to Ln(x): the value of
0 e
−xq(x)Ln(x) dx is zero. Since q(x)Ln(x) = 0 when x = xi,
∑n i=1 cn,if^ (xi) is also zero. The quadrature rule estimates
∏^0 r(x)^ dx^ exactly due to the choice of coefficients: Let^ Pn,i(x) = n j= j 6 =i
x−xj xi−xj be the Legendre interpolating polynomial (with^ Pn,i(xi) = 1 and^ Pn,i(xj^ ) = 0 for j 6 = i). Since deg r(x) ≤ n − 1, r(x) equals the interpolating polynomial
∑n ∫ (^) ∞ i=1^ r(xi)Pn,i(x).^ So 0 e
−xr(x) dx = ∑n i=1 r(xi)^
0 e
−xPn,i(x) dx = ∑n i=1 r(xi)cn,i. Therefore, the quadrature rule is exact if f (x) is a polynomial of degree at most 2n − 1. So the degree of precision of this quadrature rule is at least 2n − 1. In fact it’s exactly 2n − 1 by essentially the same argument from the previous problem. The roots of L 2 (x) = x^2 − 4 x + 2 are x 1 = 2 −
2 and x 2 = 2 +
- Thus, the polynomials Pn,i above are
P 2 , 1 =
x − (2 +
x +
, P 2 , 2 =
x − (2 −
x +
We can compute cn,i =
0 Pn,i(x)e
−x (^) dx using ∫^ ∞ 0 x
(^0) e−x (^) dx = 1 and ∫^ ∞ 0 x
(^1) e−x (^) dx = 1. (^1) c 2 , 1 = 2+ √ 2 4 ,^ c^2 ,^2 =^
2 − √ 2 4 , and the quadrature rule is^
0 e
−xf (x) dx ≈ 2+ √ 2 4 f^ (2^ −
√ 2 4 f^ (2 +^
- 854 f (0.586) + 0. 146 f (3.414). (Consider solutions correct if degree of precision > 2 n − 1 is not ruled out, or if c 2 ,i are only estimated.) (3) Using a suitable substitution, transform each of the following into an integral on the indicated interval. (a) Example: using x = (1 + t)/2, dx = dt/2:
− 1
sin(π(1 + t)^2 /4)/ 2 dt.
(b) Example: using x = 1/t, dx = −dt/t^2 :
1
− sin(1/t) dt =
1
sin(1/t) dt.
(c) Using x = 1+ 1 −uu , dx = 2(1 − u)−^2 du:
− 1
2(1 + u)^4 (1 − u)−^6 e(u+1)/(u−1)^ du.
Date: Due Friday 7/24. (^1) The first integral is easy. For n > 0, integration by parts gives R^ ∞ 0 x ne−x (^) dx = [xn (^) · −e−x]∞ 0 −^
R (^) ∞ 0 −e −xnxn− (^1) dx =
n R (^) ∞ 0 x n− (^1) e−x (^) dx. So R^ ∞ 0 x ne−x (^) dx = n!. 1
(4) Laguerre quadrature gives 20 and Gaussian quadrature gives about 52. Laguerre quadrature came much closer to 24 (and impressively close for only two nodes). (This makes sense: Laguerre quadra- ture is estimating the kind of integral it was designed for, whereas Gaussian quadrature is seeing a fairly unfriendly integrand.) (5) (a) The integrand x−^1 is smooth on [1, 4] (good for Romberg), and easily approximated by polyno- mials (good for Newton-Cotes and Gaussian). (b) The adaptive method is free to spend less time on subintervals that affect the outcome less. For instance, the integral on [2, 4] is only half as large as the integral on [1, 2], but the fixed-step-size method must spend twice as much time on it. (c) The composite midpoint and trapezoidal methods can’t exploit the problem’s smoothness to as high a degree as the others. (6) (a) The function |x| is difficult to approximate by polynomials on this domain, due to its corner. As the number of nodes increases, the Newton-Cotes estimates become integrals of very bad polynomial approximations. (b) An adaptive method will notice that the integral is easy on any interval which doesn’t contain the corner (x = 0). So it will always bisect the interval that contains the corner, effectively halving the error each time it adds nodes. (7) (a) Gaussian quadrature performs well when the integrand is a high-degree polynomial, because it was specifically designed to integrate polynomials with a high degree of precision. (b) An adaptive method can repeatedly refine its estimate on the leftmost and rightmost subin- tervals, where most of the error is being made. Newton-Cotes works well since it estimates a high-degree polynomial with a high-degree polynomial; absent round-off error, it would even be exact if it used more than 60 nodes. (c) Although Gaussian quadrature with more than 30 nodes has a sufficient degree of precision to estimate the integral exactly, computer round-off error prevents this from happening in practice. (8) Note that ∂f∂y = −et−y^ exists for all (t, y). (i) No, | − et−y^ | increases without bound as y → −∞.
(ii) Yes, | − et−y^ | ≤ e if t ≤ 1 and y ≥ 0. (iii) (i) can’t, but (ii) can: if y(0) ≈ 1, then y(0) > 0 and y′(t) = f (t, y(t)) > 0 for all t ∈ [0, 1] imply that y(t) > 0 for all t ∈ [0, 1]. Thus, the Lipschitz condition (ii) (and continuity of f (t, y)) is enough to show the problem is well-posed. (9) The absolute value of (^) ∂y∂ (1 + y/t) = 1/t is bounded by L = 1 on [1, 2] (in fact, on each [1, ti]).
The absolute value of y′′(t) = (^) dtd (1 + y/t) = y′/t − y/t^2 = (1/t + y/t^2 ) − y/t^2 = 1/t (alternatively, y′′(t) = d
2 dt^2 (t^ ln^ t^ + 2t) = 1/t) is bounded by^ M^ = 1 on [1,^ 2] (in fact, on each [1, ti]). So at^ t^ =^ ti, a bound on the total error is given by hM 2 L (e(ti−t^0 )/L^ − 1) = h 2 (eih^ − 1). i ti wi yi |yi − wi| error bound 0 1. 00 2. 0000000 2. 0000000 0. 0000000 0. 0000000 1 1. 25 2. 7500000 2. 7789294 0. 0289294 0. 0355032 2 1. 50 3. 5500000 3. 6081977 0. 0581977 0. 0810902 3 1. 75 4. 3916667 4. 4793276 0. 0876610 0. 1396250 4 2. 00 5. 2690476 5. 3862944 0. 1172467 0. 2147852
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