Exercises for numerical integration. Øyvind Ryan

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1 Exercises for numerical integration Øyvind Ryan February 25, 21

2 1. Vi ar r(t) = (t cos t, t sin t, t) Solution: Koden blir % Oppgave.1.11 b) quad(@(x)sqrt(2+t.^2),,2*pi) a. Finn astigeten, farten og akselerasjonen. Answer: v(t) = (cos t t sin t,sin t + t cos t,1) v(t) = 2 + t 2 b. Vis at buelengden fra t = til t = 2π er 2π 2 + t 2 dt. Bruk numerisk integrasjon til å beregne dette integralet. Answer: 22.4 c. Løs integralet i b) ved regning. Bruk substitusjonen t = eu e u 2. Answer: π 2 + 4π 2 + ln( 2π + 2π 2 + 1) 2. Mark eac of te following statements as true or false. a. Numerical integration metods are usually constructed by dividing te interval of integration into many subintervals and using some sort of approximation to te area under te function on eac subinterval.. In tis exercise we are going to study te definition of te integral for te function f (x) = e x on te interval [,1]. a. Determine lower and upper sums for a uniform partition consisting of 1 subintervals. Answer: I 1.678, I b. Determine te absolute and relative errors of te sums in (a) compared to te exact value e 1 = of te integral. Answer: I I I I.85, I = I I I I.87, I =.51. c. Write a program for calculating te lower and upper sums in tis example. How many subintervals are needed to acieve an absolute error less tan 1? 4. Mark eac of te following statements as true or false. 2

3 a. Wen we use te midpoint rule for numerical integration, round-off errors due to subtraction of two similar numbers is a major source of errors. b. Te midpoint rule gives te exact result for polynomials of degree 1. c. Te midpoint rule gives te exact result for polynomials of degree 2. d. Te global error in te midpoint metod is one order lower tan te local error. e. Wen we decrease te step lengt in te midpoint rule by a factor of, te error is reduced by rougly a factor of We use te midpoint rule to approximate te integral x 2 dx using te midpoint rule wit 2 subintervals. Wat is te result? 5/16 1/4 4/9 2/5 Answer: 5/16. Solution: We get tat I mid = 1 2 f (1/4) f (/4) = = = 5 16, so tat te first alternative is correct. 6. Calculate an approximation to te integral π/2 sin x dx = x2 wit te midpoint rule. Split te interval into 6 subintervals. Answer: Approximation:.5624 (wit 6 digits). 7. Mark eac of te following statements as true or false. a. Te trapezoidal rule is usually more accurate tan te midpoint rule.

4 b. Because every point of measurement in te trapezoidal rule is used in two different subintervals, we must evaluate te function we want to integrate twice at every point. 8. We use te trapezoidal rule to approximate te integral x 2 dx using te trapezoidal rule wit 2 subintervals. Wat is te result? 1/2 /8 5/9 /5 Answer: /8 Solution: We ave tat I tr ap = 1 ( ( )) f () + f (1) 1 + f = 1 ( ) = 4 8, so tat te second alternative is correct. 9. Calculate an approximation to te integral π/2 sin x dx = x2 wit te trapezoidal rule. Split te interval into 6 subintervals. Answer: Approximation: (wit 6 digits). 1. In tis exercise you are going to program an algoritm like algoritm?? for te trapezoidal rule. If you cannot program, use te trapezoidal rule manually wit 1 subintervals, ceck te error, and skip te second part of (b). a. Write a program tat implements te midpoint rule as in algoritm?? and test it on te integral e x dx = e 1. Answer: Approximation wit 1 subintervals: (wit 6 digits). b. Determine a value of tat guarantees tat te absolute error is smaller tan 1 1. Run your program and ceck wat te actual error is for tis value of. (You may ave to adjust algoritm?? sligtly and print te absolute error.) Answer:

5 11. Wen is alved in te trapezoidal rule, some of te function values used wit step lengt /2 are te same as tose used for step lengt. Derive a formula for te trapezoidal rule wit step lengt /2 tat makes it easy to avoid recomputing te function values tat were computed on te previous level. 12. Mark eac of te following statements as true or false. a. Simpson s rule requires tat we use an odd number of measurement points. b. Simpson s rule is exact for polynomials of degree or lower. 1. a. (Exam 21) Wic of te integration metod (trapezoidal, midpoint and Simpson s) will be most accurate for a polynomial of degree 1? Just te trapezoidal rule. Just Simpson s rule. Just te midpoint rule. All will be equally accurate. b. (Continuation exam 29) We use Simpson s metod to calculate approximations to b a f (x)dx (We do not take round off errors into account). Ten te result will always be correct if f (x) is a trigonometric function. a logaritmic function. a polynomial of degree 2 on te form g (x)/(x) were f and g are polynomials of degree 2. c. (Exam 28) Te midpoint rule evaluates te integral of f on te interval [a,b] by te approximation b a f (x)dx (b a)f ((a + b)/2). Wic of te following statements are true (we do not take round off errors into account)? Te midpoint rule is more accurate tan Simpson s rule. Te midpoint rule and te trapezoidal rule always give te exact same error. 5

6 Te midpoint rule only gives error if f (x) = c for som arbitrary constant c. Te midpoint rule gives error if f (x) is an arbitrary straigt line in te x, y-plane. 14. Calculate an approximation to te integral π/2 sin x dx = x2 wit Simpson s rule. Split te interval into 6 subintervals. Answer: Approximation: (wit 6 digits). 15. a. How many function evaluations do you need to calculate te integral dx 1 + 2x wit te trapezoidal rule to make sure tat te error is smaller tan 1 1. Answer: evaluations. Solution: Since te error in te trapezoidal rule is bounded by (b a) 2 6 max x [a,b] f (x), and since f 8 (x) = (wic attains maximum absolute value for x = (1+2x) ), we need to coose so tat (b a) 2 6 max x [a,b] f (x) = , wic gives tat 1 5 /2. Te number of function evaluations is ten 1/ = / , wic means tat at least evaluations are needed. b. How many function evaluations are necessary to acieve te same accuracy wit te midpoint rule? Answer: evaluations. Solution: Since te error in te midpoint rule is bounded by (b a) 2 24 max x [a,b] f (x), we obtain as in (a) tat (b a) 2 24 max x [a,b] f (x) = , wic gives tat 1 5. Te number of function evaluations is ten 1/ = 1 5 / 5775., wic means tat at least 5776 evaluations are needed. 6

7 c. How many function evaluations are necessary to acieve te same accuracy wit Simpson s rule? Answer: 8 evaluations. Solution: Since te error in Simpson s rule is bounded by (b a) 4 18 max x [a,b] f (i v) (x), and since f (i v) (x) = 84 (wic attains maximum absolute value for (1+2x) 5 x = ), we need to coose so tat (b a) 4 18 max x [a,b] f (i v) (x) = , wic gives tat ( /84) 1/4. Te number of function evaluations is ten 1/ = ( /84) 1/ , wic means tat at least 8 evaluations are needed. 16. In tis exercise you are going to program an algoritm like algoritm?? for Simpson s rule. If you cannot program, use Simpson s rule manually wit 1 subintervals, ceck te error, and skip te second part of (b). a. Write a program tat implements Simpson s rule as in algoritm?? and test it on te integral e x dx = e 1. Approximation wit 1 subintervals: (wit 1 dig- Answer: its). b. Determine a value of tat guarantees tat te absolute error is smaller tan 1 1. Run your program and ceck wat te actual error is for tis value of. (You may ave to adjust algoritm?? sligtly and print te absolute error.) Answer: a. Verify tat Simpson s rule is exact wen f (x) = x i for i =, 1, 2,. Solution: It is enoug to verify Simpson s rule on eac interval. Te integrals become x dx = 1 4 ((a + )4 (a ) 4 ) = 1 4 (8a + 8a ) = 2a + 2a x 2 dx = 1 ((a + ) (a ) ) = 1 (6a2 + 2 ) = (6a ) xdx = 1 2 ((a + )2 (a ) 2 ) = 1 4a = 2a 2 dx = 2. 7

8 We also get tat te rule itself becomes ((a ) + 4a + (a + ) ) = (6a + 6a 2 ) = 2a + 2a ((a )2 + 4a 2 + (a + ) 2 ) = (6a ) ((a ) + 4a + (a + )) = 2a ( ) = 2. Tis sows tat Simpsons rule is exact for te given polynomials. b. Use (a) to sow tat Simpson s rule is exact for any cubic polynomial. Solution: For f (x) = bx + cx 2 + dx + e te integral is f (x)dx = b Te rule itself now gives x dx + c (f (a ) + 4f (a) + f (a + )) x 2 dx + d xdx + e = b ((a ) + 4a + (a + ) ) + c ((a )2 + 4a 2 + (a + ) 2 ) + d ((a ) + 4a + (a + )) + e ( ) We see tat equality follows from tat we ave equality for x, x 2, x,1. 1dx. c. Could you reac te same conclusion as in (b) by just considering te error estimate (??)? Solution: Since f (4) (x) = for all x for every tird degree polynomial, it follows directly from te error estimate tat te metod is exact for suc functions. 18. We want to design a numerical integration metod b a f (x)dx w 1 f (a) + w 2 f (a 1/2 ) + w f (b). Determine te unknown coefficients w 1, w 2, and w by demanding tat te integration metod sould be exact for te tree polynomials f (x) = x i for i =, 1, 2. Do you recognise te metod? Answer: w 1 = w = (b a)/6, w 2 = 2(b a)/. 8

9 19. Find te distance covered by Jupiter, Mars, and Eart, for te orbital data files used in Example??. Solution: It is straigtforward to write a function wic computes te arc lengt for te orbit data in a given file. Te following code can be used for tis: function dist=finddistances(filename) coords=findcoords(filename); N=size(coords,2); lengts=zeros(1,n-1); for k=1:(n-1) lengts(k)=norm(coords(:,k)-coords(:,k+1)); end dist=sum(lengts); If you run tis code for te tree different orbits, we get 1.9 for te eart, for Mars, and for Jupiter. Te planet closest to te sun tus covers te biggest distance witin te same period, as promised by Kepler s laws. 9

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