Solutions to Homework 3

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1 Solutions to Homework Section. d) I wrote code, but that was not necessary. For MatLab output, see the appendix. The free cubic spline with six digit rounding of the coefficients is (x + ) (x + ) for x,.5] p(x) = (x +.5) (x +.5) +.088(x +.5) for x.5, 0] x x.099x for x 0,.5] 8d) I wrote code, but that was not necessary. For MatLab output, see the appendix. The clamped cubic spline with six digit rounding of the coefficients is (x + ) (x + ) (x + ) for x,.5] p(x) = (x +.5) (x +.5) (x +.5) for x.5, 0] x +.90x x for x 0,.5] Section.5 For a and c, let (x 0, y 0 ) = (0, 0) and (x, y ) = (5, ). Let (ˆx 0, ŷ 0 ) and (ˆx, ŷ ) denote the left and right guidepoints, respectively. Then α 0 = ˆx 0 x 0, α = x ˆx, β 0 = ŷ 0 y 0, and β = y ŷ. MATLAB output and figures are in the appendix. a) (ˆx 0, ŷ 0 ) = (, ) and (ˆx, ŷ ) = (6, ). So, α 0 = β 0 = β = = α. Therefore, (α 0 + α ) = 0, (α + α 0 ) =, (β 0 + β ) =, and (β + β 0 ) =. Using Equations (.) and (.5) from page 6, we have x(t) = ( 5) + (0)] t + (5) ()] t + ()t + 0 = 0t + t + t y(t) = ( ) + ()] t + () ()] t + ()t + 0 = t t + t. c) (ˆx 0, ŷ 0 ) = (, ) and (ˆx, ŷ ) = (6, ). So, ˆx 0, ˆx, α 0, and α are unchanged from problem a and therefore so is x(t). Now β 0 = = β, so (β 0 + β ) = 0, and (β + β 0 ) =. From Equation (.5) from page 6, we have y(t) = ( ) + (0)] t + () ()] t + ()t + 0 = t + t + t.

2 Section. 6b) Use the most accurate -point formulat to fill in the missing data from the following table: x f(x) f (x) in the first three cases h =., and in the final case h =.. For the first and last data point, we must use the forward -point formula (where h < 0 actually makes it a backward formula) and the two middle cases will use the central -point formula. f (7.) ( 68.9) + ( 7.698) ( )] = f (7.6) ( 68.9)] = f (7.8) ( 7.698)] = f (8.0) ( ) + ( )] = b) Now we now the data was generated from f(x) = ln(x+) (x+) so that f (x) = x+ (x+) and f () (ξ) = (x+). Therefore, our table should actually be x f(x) f (x) Thus, for the final three values our approximations are exact. The absolute error for f (7.) is.000. Let E(x) denote the theoretical error bound for f (x). The theoretical bound for the forward -point formula is f () (ξ) h and for the central -point formula it is half of that. For x = 7. and x = 7.6, ξ 7., 7.8] and for x = 7.8 and x = 8.0 ξ 7.6, 8.0]. Since f () (x) is decreasing on both of these intervals, its maximum is achieved at the left endpoint of the appropriate interval. Hence, ( ).0 E(7.) (ξ + ) ( ).0 E(7.6) (ξ + ) 6 ( ).0 E(7.8) (ξ + ) 6 ( ).0 E(8.0) (ξ + ) (9.) (9.) (9.6) (9.6) ( ).0 =.0000 ( ).0 = ( ).0 = ( ).0 =.00000

3 Obviously, we meet this bound with the last three approximations. The first approximation however does not achieve this bound. This is almost certainly due to the precision of the initial data as the original data is less precise than the bounds. G8) To derive an approximation technique for f () (x 0 ) we expand f(x) as a Taylor series centered at x 0 and evaluate at the points x 0 ± h. f(x 0 + h) = f(x 0 ) + f (x 0 )h + f (x 0 )h + 6 f () (x 0 )h + f () (x 0 )h + 0 f (5) (x 0 )h 5 + f(x 0 h) = f(x 0 ) f (x 0 )h + f (x 0 )h 6 f () (x 0 )h + f () (x 0 )h 0 f (5) (x 0 )h 5 + so f(x 0 + h) f(x 0 h) = f (x 0 )h + f () (x 0 )h + 60 f (5) (x 0 )h 5 +. () Then solving () for f () (x 0 ) results in f () (x 0 ) = h f(x 0 + h) f(x 0 h)] 6 h f (x 0 ) 0 f (5) (x 0 )h +. () Substituting h for h in () and multiplying the new equation by provides f () (x 0 ) = h f(x 0 + h) f(x 0 h)] 6 h f (x 0 ) 8 0 f (5) (x 0 )h +. () Subtracting () from () and dividing the resulting equation by produces a centered approximation for f () (x 0 ) with error of order O(h ). f () (x 0 ) = h f(x 0 + h) f(x 0 + h) + f(x 0 h) f(x 0 h)] 7 60 f (5) (x 0 )h +. () 9) Of course we want to minimize error. Finding the minimum of the error e(h) is basic calculus. e (h) = ɛ h + h ɛ M must equal zero. Thus h = h M or h = ( ɛ M derivative is positive and therefore e(h) has a minimum at h = ( ɛ M ). Since h is positive, the second ).

4 Section. 8) Let N (h) = h f(x 0 + h) f(x 0 )]. Then f (x 0 ) = N (h) h f (x 0 ) h 6 f (x 0 ) + O(h ) (5) ( ) h so f (x 0 ) = N h f (x 0 ) h f (x 0 ) + O(h ). (6) So two times equation (6) minus equation (5) gives us f (x 0 ) = Let N (h) = N ( h ) N (h). Then ( ) ] h N N (h) + h f (x 0 ) + O(h ). (7) f (x 0 ) = N ( h Thus four times equation (8) minus equation (7) gives us ( ) ] h f (x 0 ) = N N (h) ) + h 8 f (x 0 ) + O(h ). (8) + O(h ). (9) Now we solve equation (9) for f (x 0 ) and unravel the definition of N (h) in terms of the original function f. f (x 0 ) = ( ) h N N (h) + O(h ) = ( ) ( )] h h N N ( ) ] h N N (h) + O(h ) = 8 ( ) h N 6 ( ) h N + N (h) + O(h ) = 8 ( f x 0 + h ) ] f(x 0 ) 6 ( f x 0 + h ) ] f(x 0 ) + h h h f(x 0 + h) f(x 0 )] + O(h ) = ( f x 0 + h ) ( f x 0 + h ) ] + f (x 0 + h) f(x 0 ) + O(h ) (0) h Substituting h for h in equation (0) and arranging the terms in an aesthetically pleasing order gives us a O(h ) approximation of the derivative of f at x 0 : f (x 0 ) = h f (x 0 + h) f(x 0 + h) + f(x 0 + h) f(x 0 )] + O(h ).

5 Section. a) With the trapezoid rule, we have h =.5 and.5.5 cos x dx.5 cos (.5) + cos (.5) ] = cos (.5) a) With Simpson s method, we have h =.5 and.5.5 cos x dx.5 cos (.5) + cos (0) + cos (.5) ] = 6 cos (.5) + ] (In the calculations we used that cos is an even function. The absolute errors are a).007 and 6a) ) 6) Here we have h = b a so h = (b a). Then our quadrature method can be written as b a f(x)dx = (b a) f(a + h) + f(b).] The degree of accuracy of this quadrature method is. We verify that it produces exact results for polynomials of degree 0,, and but fails to do so for a polynomial of degree. Suppose f(x) = x 0 =. Then b a f(x)dx = b a. The quadrature approximation is (b a) () + ] = b a so the quadrature is exact for constant functions. Now suppose f(x) = x = x so that b a f(x)dx = (b a ) = (b a)(b + a). The approximation is (b a) (a + h) + b] = (b a) a + (b a) + b] = (b a)(b + a) so the quadrature is exact for polynomials of degree. Suppose f(x) = x. Then b a f(x)dx = (b a ) = (b a)(b + ab + a ). Then b b a a x dx = (b + ab + a ). Applying the quadrature method, we have b a b a x dx (a + h) + b ] = (a + ah + h ) + b ] = ] a (b a) + a(b a) + + b = a + ab a + b + ] (b ab + a ) = (a + ab + b ) + ] (b + ab + a ) = b + ab + a. So the method is exact for polynomials of degree as well. Finally, suppose f(x) = x. Then b a f(x)dx = (b a ) = (b a)(b +ab +a b+b ). So we need only check that f(a+h)+f(b)] = (b + ab + a b + b ). However f(a + h) + f(b) = (a + h) + b = (a + a h + ah + h ) + b. We simply verify that the coefficient for b is incorrect. b appears only in the terms b and h = 9 (b ab + a b a ) so the coefficient for b is 0 9 in the quadrature formula instead of. Therefore, this method does not exactly recover the integral for a cubic. 5

6 Section. ) f(x) = x ln x so f (x) = ln x + and f (x) = x. Let E T (h), E M (h), and E S (h) denote the ABSOLUTE error for the trapezoidal, midpoint, and Simpson s methods, respectively, when approximating f(x)dx. In this case, b a = and f (x) for all x, ]. Part (c) is done prior to part (b). (a) By Theorem.5, E T (h) = f (µ) h h. Since we want E T (h) 0 5, we have h.000 or h.0095 (truncating rather than rounding to ensure the bound). Since h = b a n = n, then n 9 will suffice. (c) By Theorem.6, E M (h) = f (µ) 6 h = f (µ) h = E T (h). Then E M (h) 0 5 E T (h) or h So we need h.0077 and n 0. (b) By Theorem., E S (h) = f () (µ) 80 h. Since f () (x) is decreasing on, ], then f () (µ). Thus E S (h) 0 5 h 90 ( 0 5) h (.0009).70. Then n = h Therefore, n 6 will suffice. The approximations are (a. Trapezoid).66086, (c. Midpoint).66877, and (b. Simpson) which are all within 0 5 with Simpson s absolute error about half the other two absolute errors. ) Let T o be the set of all times m6 for m =,,..., and S o the sum of the respective speeds. Let T e be the set of times n6 for n =,,..., and S e the sum of the respective speeds. Then, since h = the length of the track in feet is approximately ( + S o + S e + ) = ) Let g(t) = π e t /. It is moderatley helpful to recognize that this is the pdf of the normal distribution with mean 0 and standard deviation. So G(x) = x f(t)dt is the cummulative distribution function. We know that the normal distribution is symmetric about the mean (here 0). Additionally, loosely speaking, the probability you are within standard deviations of the mean is approximately.95 for any distribution. Also, for the continuous random variable X, P(X x) = G(x) =.5 + x 0 g(t)dt. So x = is a reasonable initial guess since we translate this problem to one asking for what value x do we have P(X x) =.95. First I want to ensure that my function f(x) = x 0 g(t)dt is accurate to within 0 5. By Theorem., the error can be determined using f () (x) = π (x x )e x /. By graphing (this is fine as we don t need to be exact, but just get some bound) the absolute value of this function, we see that f () (x).5 on 0, ). So E S (x).5 x 5 80 since b = x and a = 0. Now we know that it is very n unlikely that x so let s use x =. We want E S (x) 0 5 so or n 5. So in my n composite Simpson method, I use n = 6. Finally, if F (x) = f(x).5, the F (x) = f (x) = g(x) and I run Newton s method with tolerance 0 5. From above, I used an initial approximation of. My algorithms return the approximation.685 which, using Simpsons method, gives an approximation to the integral of The error of 0 8 is certainly within our limits. For the Trapezoid method, I need E T (x).5 x 0 5. Again, using x =, then n = 8 n will suffice. I obtain.6858 as the approximate root to F T (x) (same as F above but uses the Trapezoid method). This returns a Trapezoid approximation to the integral of with the same absolute error. One could change the stopping criteria in Newton s method to a function value tolerance for this problem. 6

7 Section.6 f) We want to use adaptive quadrature (painfully by hand to understand exactly how the method works) to approximate the integral.6 x x dx with tolerance 0. We use the book s notation S(a, b). S(,.6) =. () (.). (.6) ] S(,.) =.05 () (.5).6775 (.) ].6. S(.,.6) =.05 (.). (.5).8975 (.6) ] Now S(,.6) (S(,.) + S(.,.6)) = <.05 = 5(0 ). Therefore the theoretical development of adaptive quadrature says we should stop with this second level of approximation as our approximation of the error meets our tolerance. Thus.6 x x dx S(,.) + S(.,.6) = (The actual value is so the absolute error is and we achieve the desired accuracy.) Section.7 f) By a double substitution,.6 have.6 ( x x dx = x x dx =. u du =.78.78w.dw. Then for n = 5, we ).78( ). +.78( ). ( ) (.58690). +.78(.58690). ]. Evaluating this in Matlab returns The actual value of the integral is and our absolute error is Alternatively, without the first the substitution and with f(x) = x, we have x.6 f(x)dx =. f(.x +.)dx with absolute error (In both Gaussian Quadratures above, we use 5 nodes with error less than However, our adaptive quadrature uses 6 nodes and has error greater than.9 0. Optimal nodes certainly improve our accuracy.) 7

8 APPENDIX: MATLAB OUTPUT. d) >> x= ]; >> a= ]; >> cubicfree(x,a) ans = >> cubicclamp(x,a,.5560,.58676) ans = a) and c) This function also returns the correct values for the polynomial coefficents in matrix form. The first four terms are the coefficients for t^0 through t^ in x(t) while the last four terms are the coefficients for t^0 through t^ in y(t). The plots of the curves are in this same order on the next page. >> aa=0 0;5 ]; >> bb= ]; >> cc=6 ]; >> bezierplot(aa,bb,cc) ans = >> dd=6,]; >> bezierplot(aa,bb,dd) ans =

9 a c