Answers to Homework 9: Numerical Integration

Transcription

1 Math 8A Spring Handout # Sergey Fomel April 3, Answers to Homework 9: Numerical Integration. a) Suppose that the function f x) = + x ) is known at three points: x =, x =, and x 3 =. Interpolate the function with a natural cubic spline and approximate the integral dx + x = π ) by the integral of the spline. Is the result more accurate than the result of Simpson s rule? Answer: Let the spline on the second interval x [,] be S x) = + b x + c x + d x 3. By the symmetry of the boundary conditions, the corresponding spline on the first interval is S x) = S x) = b x + c x d x 3. The interpolation conditions require S ) = S ) = + b + c + d =. The natural spline boundary conditions require S The continuity conditions require which leads to b =. The condition ) = S ) = c + 6d =. b = S ) = S ) = b, c = S ) = S ) is satisfied automatically. Solving the linear equations for c and d, we obtain c = 3/ and d = /. Therefore The integral approximation is then dx + x S x) = 3 x + x3 S x)dx = + ) = 3 8 = 3.5 This value is more accurate that Simpson s result /3 = )

2 b) Let Sp[ f ] be the approximation of the integral Answer: b a f x)dx 3) by the integral of the natural cubic spline defined at the knots a = x < x <... < x n = b. Prove that h 3 k Sp[ f ] = CT [ f ] c + h3 k k, ) where = x k+ x k, c k are the coefficients in the spline equation k= S k x) = f x k ) + b k x x k ) + c k x x k ) + d k x x k ) 3 for x k x < x k+, 5) and CT [ f ] is the composite trapezoidal rule. The integral of the spline is the sum of the local integrals: Sp[ f ] = b a Sx)dx = x k+ x k S k x)dx. Integrating the natural cubic spline 5) on the interval [x k, x k+ ], we obtain x k+ x k S k x)dx = f k + b k With the help of the recursive relationships and the integral transforms to x k+ x k S k x)dx = f k + b k = f k+ f k d k = c k+ c k 3 h k + c h 3 k k 3 + d h k k. 3 c k + c k+ ), f k+ f k ) h3 k 6 c k + c k+ ) + c k h 3 k 3 + h3 k c k+ c k ) = f k+ + f k ) h3 k c k + c k+ ). The first term in the last expression is exactly the trapezoidal rule on the interval [x k, x k + ]. Therefore, h 3 k Sp[ f ] = CT [ f ] c k + c k+ ). Recalling that the natural cubic spline has c = c n =, we can transform the last sum as follows: h 3 k c k + c k+ ) = h 3 k c k + The final expression takes the form ). h 3 k c k+ = k= h 3 k c k + k= h 3 k c k = k= c k h 3 k + h3 k.

3 . a) Suppose that function ) is known at three points: x =, x =, and x 3 = together with its derivatives. Interpolate the function using Hermite interpolation and approximate integral ) by the integral of the interpolation polynomial. Is the result more accurate than the result of Simpson s rule? Answer: The derivative of f x) is f x) = x + x ). The divided-difference table for Hermite interpolation is then x f [ ] f [, ] f [,, ] f [,, ] f [,,, ] f [,,,, ] f ) = f ) = / f ) = / which leads to the Hermite interpolation polynomial Px) = + x + ) + x + ) x + ) x + x + ) x + x + ) x x ) = 3 x + x Integrating this polynomial, we obtain Px)dx = Px)dx = + ) = 6 5 = 3. This result is more accurate than the result of Simpson s integration /3 = ) b) Let C H[ f ] be the approximation of integral 3) by the integral of the piece-wise cubic polynomial defined by applying Hermite interpolation at each of the intervals [x k, x k+ ], a = x < x <... < x n = b. Prove that h C H[ f ] = CT [ f ] + b b h n n n + k= b k h k h k where = x k+ x k, b k = f x k ), and CT [ f ] is the composite trapezoidal rule., 6) 3

4 Answer: The cubic Hermite interpolation in the interval [x k, x k+ ] follows from the divided-difference table x f [ ] f [, ] f [,, ] f [,, ] x k f k x k f k b k x k+ f k+ f k+ f k fk+ f k b k ) x k+ f k+ b k+ b k+ f k+ f k ) b k+ + b k f k+ f k ) h k which leads to the cubic interpolation polynomial ) fk+ f k x xk ) P k x) = f k + b k x x k ) + b k + b k+ + b k f ) k+ f k x xk ) x x k+ ). Integrating on the interval [x k, x k+ ], we obtain x k+ x k P k x)dx = = f k + b k P k t + x k )dt ) fk+ f k t dt + b k + b k+ + b k f ) k+ f k h t t )dt k h ) k = f k + b k + fk+ f k h b k k 3 b k+ + b k f ) k+ f k h k = f k+ + f k ) + h k b k b k+ ). The first term in the last expression is exactly the trapezoidal rule on the interval [x k, x k + ]. The total integral is, therefore, C H[ f ] = x k+ The last sum transforms as follows: h k b k b k+ ) = x k P k x)dx = CT [ f ] + h k b k h = b b h n n The final expression takes the form 6). h k b k+ = h b + n + k= b k h k h k h k t dt h k b k b k+ ). k=. h k b k n h k b k+ h n b n

5 3. a) The Gauss-Lobatto quadrature rule has the form f x)dx w f ) + w n f ) + w k f x k ) = Lo[ f ], 7) k= where the abscissas x k and weights w k are chosen so that f x) is integrated exactly if it is a polynomial of order n 3 or less. Derive the abscissas and weights for n = 3 and n = and apply the Gauss-Lobatto rule to integral ). Are the results more accurate than the result of Simpson s rule? Hint: Use the symmetry of the interval to constrain the abscissas and weights. Answer: First, let us consider the case n = 3. By the symmetry of the interval, the point x has to be in the center, and we can immediately set x =. Integrating polynomials of the first four powers, we obtain I [] = = w + w + w 3 I [x] = = w + w 3 I [x ] = 3 = w + w 3 I [x 3 ] = = w + w 3 The last two equations define w = w 3 = /3. From the first equation, we get w = w w 3 = /3 The rule is then Lo 3 [ f ] = 3 [ f ) + f ) + f ) ], which is equivalent to Simpson s rule applied on the interval [, ]. In the example problem, the result is equivalent to that of Simpson s rule: Lo 3 [ f ] = 3 = In the case of n =, the symmetry suggests x = x 3 = x and w = w 3 = w. Integrating polynomials of the first six powers, we obtain I [] = = w + w + w I [x] = = w + w I [x ] = 3 = w + w x + w I [x 3 ] = = w + w I [x ] = 5 = w + w x + w I [x 5 ] = = w + w Subtracting the third equation from the first one yields w x ) = 3 = 3 5

6 Subtracting the fifth equation from the third one yields Dividing by the previously equation, we get w x x ) = 3 5 = 5 x = 5 Therefore, x = / 5 and x 3 = / 5. Next, w x ) = 5 w = 3 and w = 5/6. Solving the remaining equations for w and w, we get w = w = /6. The rule is then Lo [ f ] = [ f ) + 5 f ) ) ] f 5 + f ). 6 Applying it to the example problem yields [ ] Lo [ f ] = This is more accurate than Simpson s result. b) The Gauss-Laguerre quadrature rule has the form Answer: e x f x)dx = 8 9 = 3... n w k f x k ) = La[ f ], 8) where the abscissas x k and weights w k are chosen so that f x) is integrated exactly if it is a polynomial of order n or less. Derive the abscissas and weights for n = and n =. Test your formulas by approximating π with π = [ Ɣ3/) ] La[ x] ), 9) where Ɣx) = t x e t dt ) First, let us consider the case n =. Integrating polynomials of the first two powers and applying the general formula I [x k ] = e x x k dx = k! 6

7 we obtain I [] = = w I [x] = = w x The rule is then simply In the example problem, we get La [ f ] = f ) π La [ x] ) =. In the case n =, we need to integrate the first four powers, obtaining I [] = = w + w I [x] = = w x + w x I [x ] = = w x + w x I [x 3 ] = 6 = w x 3 + w x 3 Manipulating these equations by multiplications and subtractions, we can simplify them as follows: w + w )w x 3 + w x 3 ) w x + w x )w x + w x ) = w w x x ) x + x ) = w x + w x )w x 3 + w x 3 ) w x + w x ) = w w x x x x ) = w + w )w x + w x ) w x + w x ) = w w x x ) = Dividing the first and second equations by the last one yields the system of equations with the solution x =, x = +. x + x = x x = An easier way to get this solution is by considering polynomials orthogonal on the interval [, ] with the weight wx) = e x Laguerre polynomials). We constructed a more general case of these polynomials in Homework 7, Problem b). The Laguerre polynomial of the second order is up to scaling by a constant) Px) = x x + and its zeroes are x = and x = +. Equations for the unknowns weights w and w are w + w = w ) + w + ) = with the solution w = + w = 7

8 The rule is then La [ f ] = + In the example problem, we get and La [ x] = + = + f ) π La [ x] ) = + 3. f + ). + + =. Programming) Implement the adaptive quadrature method. Test your program by computing integral ) with the precision of six significant decimal digits. Plot or tabulate the values of x and f x) that were involved in the computation. The algorithm of adaptive integration can be defined either recursively ADAPTIVE RECURSIVE f x),a,b,h, I ) if h < xtol then 3 WARNING did not converge ) return I 5 c a + h/ 6 I R f,a,c) 7 I R f,c,b) 8 D I + I I 9 if D < itol then J I + α D else J ADAPTIVE RECURSIVE f,a,c,h/, I ) + ADAPTIVE RECURSIVE f,c,b,h/, I ) return J 8

9 or sequentially ADAPTIVE NONRECURSIVE f x),a,b,h, I ) J PUSH f,a,b,h, I ) 3 while POP f,a,b,h, I ) do 5 if h < xtol 6 then 7 WARNING did not converge ) 8 return I 9 c a + h/ I R f,a,c) I R f,c,b) D I + I I 3 if D < itol then J J + I + α D 5 else 6 PUSH f,a,c,h/, I ) 7 PUSH f,c,b,h/, I ) 8 return J Both algorithms involve the rule R f,a,b) for computing integral 3), the minimum allowed interval xtol and the requested precision itol. They are initialized with I = R f,a,b) and h = b a. The sequential algorithm operates a queue of intervals using a pair of functions PUSH and POP. Run your program taking R to be the trapezoidal rule. What is the appropriate value of constant α? Answer: An appropriate value for α follows from Richardson s extrapolation. The error of the trapezoidal rule is I [ f ] T [ f ] = A h 3 + Oh 5 ) where A does not depend on h. The composite trapezoidal rule applied on two intervals has the error I [ f ] CT [ f ] = A h/) 3 + Oh 5 ) = A h3 + Oh5 ) Multiplying the second equation by α and the first equation by α and adding them together, we get where I [ f ] R[ f ] = α) A h 3 + α A h3 + Oh5 ), R[ f ] = α) T [ f ] + α CT [ f ] = T [ f ] + α CT [ f ] T [ f ]). The error is minimized if α + α = 9

10 or α = 3 The combination R[ f ] = T [ f ] + CT [ f ] T [ f ]) 3 is equivalent to Simpson s rule Programming) The length of a parametric curve {xt), yt) is given by the integral b a [x t)] + [ y t) ] dt ) The curve that you interpolated in Homework 6 is close to the hypotrochoid defined on the interval t π. xt) = 3 cost + cos3t ; ) yt) = 3 sint sin3t, 3) Estimate the length of this curve to six significant decimal digits applying a numerical method of your choice. Identify the method and plot the {x, y points involved in the computation. Answer:

11 The integrand simplifies as follows: f t) = [x t)] + [ y t) ] 3 = cost cos3t sint sin3t) = 3 5 cost This function is periodic with the period of π/. Therefore, we can integrate it on the interval [,π/] and multiply the result by four: π f t)dt = π/ f t)dt Applying the adaptive integration method of the previous problem leads to the following distribution of points: The estimated length of the curve is.73

12 Solution: C program: #include <math.h> /* for math functions */ #include <stdio.h> /* for output */ #include <stdlib.h> /* for malloc */ #include <assert.h> /* for assertion */ static const double xtol=.e-8; /* x tolerance */ static const double itol=.e-; /* integral computation tolerance */ static const double pi= ; /* Function: adaptive_recursive Adaptive integration by the recursive algorithm. Uses the trapezoidal rule, upgraded to Simpson s by one step of Richardson s extrapolation. fx) - integrand function a,b - integration limits fa = fa) fb = fb) h = b-a I - the trapezoidal estimate of the integral */ double adaptive_recursivedouble *f)double x), double a, double b, double fa, double fb, double h, double I) { double c, fc, I, I, D; h /= ; if h < xtol) { fprintfstderr,"adaptive integration did not converge.\n"); return I; c = a + h; /* midpoint */ fc = fc); I =.5*h*fa+fc); /* trapezoidal from a to c */ I =.5*h*fc+fb); /* trapezoidal from c to b */ D =.*I + I - I)/3.; if fabsd) < itol) { I += D; else { I = adaptive_recursivef,a,c,fa,fc,h,i) + adaptive_recursivef,c,b,fc,fb,h,i); return I; /* Function: adaptive_nonrecursive Adaptive integration by the non-recursive algorithm. Uses the trapezoidal rule, upgraded to Simpson s by one step of Richardson s extrapolation. fx) - integrand function a,b - integration limits

13 fa = fa) fb = fb) h = b-a I - the trapezoidal estimate of the integral */ double adaptive_nonrecursivedouble *f)double x), double a, double b, double fa, double fb, double h, double I) { double c, fc, I, I, D, J; /* Stack is implemented with a linked list */ struct Stack { double a,b,fa,fb,h,i; struct Stack* next; *p, *first; J =.; /* allocate and PUSH the first entry to the stack */ assertfirst = struct Stack*) mallocsizeof*first))); first->a = a; first->fa = fa; first->b = b; first->fb = fb; first->h = h; first->i = I; first->next = NULL; while p = first)!= NULL) { /* POP from the stack */ first = p->next; a = p->a; fa = p->fa; b = p->b; fb = p->fb; h = p->h/; I = p->i; if h < xtol) { fprintfstderr,"adaptive integration did not converge.\n"); return I; c = a+h; fc = fc); I =.5*h*fa+fc); /* trapezoidal from a to c */ I =.5*h*fc+fb); /* trapezoidal from c to b */ D =.*I + I - I)/3.; if fabsd) < itol) { J += I+D); free p); else { p->b = c; p->fb = fc; p->h = h; p->i = I; /* allocate and PUSH the first entry to the stack */ assertfirst = struct Stack*) mallocsizeof*p))); first->a = c; first->fa = fc; first->b = b; first->fb = fb; first->h = h; first->i = I; first->next = p; 3

14 return J; /* example function: problem */ static double funcdouble x) { double y; y =./.+x*x); printf"%f %f\n",x,y); return y; /* example function: problem 5 */ static double func5double t) { double x, y, f; int k; f = sqrt5.-.*cos.*t)); for k=; k < ; k++) { /* account for periodicity */ x =.5*cost) + cos3.*t); y =.5*sint) - sin3.*t); printf"%f %f\n",x,y); t +=.5*pi; return f; /* main program */ int mainvoid) { double a, b, fa, fb, h; double I; a = -.; fa=funca); b =.; fb=funcb); h = b-a; I =.5*h*fa+fb); I = adaptive_nonrecursivefunc, a, b, fa, fb, h, I); fprintfstderr,"integral=%f\n",i); a =.; fa=func5a); b =.5*pi; fb=func5b); /* integrate to p/ due to periodicity */ h = b-a; I =.5*h*fa+fb); I = 6.*adaptive_recursivefunc5, a, b, fa, fb, h, I); fprintfstderr,"integral=%f\n",i); return ;