COT4501 Spring Homework VII

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1 COT451 Spring 1 Homework VII The ssignment is due in clss on Thursdy, April 19, 1. There re five regulr problems nd one computer problem (using MATLAB). For written problems, you need to show your work nd it is insufficient to just give the results or nswers. For the computer problems, turn in your results (e.g., grphs, plots, simple nlysis nd so on) nd lso printout of your (MATLAB) code. Problem 1 (3pts) 1. True or flse: Evluting definite integrl is lwys well-conditioned problem.. True of flse: The degree of qudrture rule is the degree of the interpolting polynomil on which the rule is bsed. 3. How cn you estimte the error in qudrture rule without computing the derivtives of the integrnd function tht would be required by Tylor series expnsion? 4. Newtwon-Cotes nd Gussin qudrture rules re both bsed on polynomil interpoltion. Wht specific property chrcterizes Newton-Cotes qudrture rule for given number of nodes? Wht specific property chrcterizes Gussin qudrture rule for given number of nodes? 5. Suppose tht the qudrture rule f(x)dx w i f(x i ) is exct for ll constnt functions. Wht does this imply bout the weights w i or the nodes x i?. For ech of the following properties, stte which type of qudrture, Newton-Cotes or Gussin, more ccurtely fits the description: Esier to compute nodes nd weights Esier to pply for generl intervl [, b] More ccurte for the sme number of nodes Hs mximl degree for the number of nodes Nodes esy to reuse s order of rule chnges 1

2 Solution 1. True. True 3. by using the E(f) T (f) m(f) 3 4. () it chooses eqully spced points in the intervl [, b] (b) weights nd nodes chosen to mximize degree of resulting rule. 5. Since integrtion of the constnt function (f(x i ) = c) is just (b )c the summtion of the weights is equl to (b ). The weights could be replced by single weight nd we just need vlue on single node.. () Newton-Cotes (b) Newton-Cotes (c) Gussin (d) Gussin (e) Newton-Cotes Problem (15pts) 1. Compute the pproximte vlue of the integrl 1 x3 dx, first by the midpoint rule nd then by the trpezoid rule.. Use the difference between these two results to estimte the error in ech of them. 3. Combine the two results to obtin the Simpson s rule pproximtion to the integrl. 4. Would you expect the ltter to be exct for this problem? Why? Solution 1. M f = (1 )f(1/) = 1(1/) 3 = 1/8; T f = (1 )(f()+f(1))/ = (+1)/ = 1/. E (T M)/3 = (1/ 1/8)/3 = 1/8 so E M 1/8 nd E T 1/4 3. S = M/3 + T/3 = (1/8)/3 + (1/) = 3 = 1/1 + 1/ = 1/4 4. Yes, becuse Simpson s rule integrtes ll polynomils of degree 3 exctly.

3 Problem 3 (pts) Suppose tht Lgrnge interpoltion t given set of nodes x 1,, x n is used to derive qudrture rule. Prove tht the corresponding weights re given by the integrls of the Lgrnge bsis functions for i = 1,..., n. w i = l i (x)dx, Solution Given set of nodes x i, i = 1,..., n the Lgrnge polynomil interpolting the corresponding vlues of the integrnd function f is we therefore hve p(x)dx = p(x) = f(x i )l i (x) f(x i )l i (x)dx = compring it with the qudrture rule we cn drive tht w i = l i (x)dx, f(x i ) l i (x)dx Problem 4 (pts) Given sufficiently smooth function f : R R, use Tylor series to derive second-order ccurte, one-sided difference pproximtion to f (x) in terms of the vlues of f(x), f(x + h), nd f(x + h). Solution In the Tylor series expnsion solving for the f (x) f(x + h) = f(x) + f (x)h + f (x) f (x) = similrly using Tylor series nd solving for the f (x) f(x + h) f(x) h f (x) f(x + h) = f(x) + f (x)h + f (x) f (x) = f(x + h) f(x) h h + f (x) h , h f (x) h +..., (h) + f (x) (h) , f (x)h f (x) h +..., 3 3

4 If we now subtrct the second of these series forf (x) from twice the first nd ignoring higher powers f 3f(x) + 4f(x + h) f(x + h) (x) h it s second order ccurte nd one sided difference pproximtion to f (x) Problem 5 (pts) Archimedes pproximted the vlue of π by computing the perimeter of regulr polygon inscribing or circumscribing circle of dimeter 1. The perimeter of n inscribed polygon with n sides is given by nd tht of circumscribed polygon by p n = n sin(π/n), q n = n tn(π/n), nd these vlues provide lower nd upper bounds, respectively on the vlue of π. 1. Using the power series expnsions for the sine nd tngent functions, show tht p n n q n cn be expressed in the form nd p n = + 1 h + h q n = b + b 1 h + b h where h = 1/n. Wht re the true vlues of nd b?. Given the vlues p = 3. nd p 1 = 3.158, use Richrdson extrpoltion to produce better estimte for π. similrly, given the vlues q = nd q 1 = 3.154, use Richrdson extrpoltion to produce better estimte for π. Solution so sin(x) = x x3 3! + x5 5! x7 7! +... p n = nsin( π n ) p n = π π3 3! h + π5 5! h4 + π7 7! h +... where h = 1/n, hence = π Similrly for the tngent function tn(x) = x + x3 3 + x x

5 therefore q n = π + π3 3 h + π5 15 h4 + 17π7 315 h +... therefore b = π Using the Richrdson extrpoltion with p = nd q = = F (1/) + F (1/) F (1/1) 1 = = b = F (1/) + F (1/) F (1/1) 1 = = Computer Problem 1 (3pts) Use numericl integrtion to verify or refute ech of the following conjectures 1. 1 x3 dx = dx =.5 π(5x + 1) 1 dx = x e 9x + e 14(x 1 ) π dx =. 9 1 x dx =. 1 5e 5x dx = log(x)dx = 1 Solution 5

6 1 function cp8_4 % numericl integrtion to test conjectures fs = {'sqrt(x.ˆ3)' '1./(1+1*x.ˆ)'... 3 '(exp(-9*x.ˆ)+exp(-14*(x-.5).ˆ))/sqrt(pi)'... '5./(pi*(5*x.ˆ+1))'... 4 '1./sqrt(bs(x))' '5*exp(-5*x)' 'log(x)'}; c = [ ]; 5 = [ -9 1e-1]; b = [ ]; disp('truth or flsity of conjecture for given tolernce using qudl:'); 7 disp(' tol bcdefg'); 8 for k = 1:7 9 tol = 1ˆ(-k); 1 for i = 1:7 11 f = inline(fs{i},'x'); 1 if bs(qudl(f,(i),b(i),tol)-c(i)) tol, torf(i) = 'T'; 13 else torf(i) = 'F'; end 14 end; fprintf('%8.1e %s\n', tol, torf); 15 end; disp(' ');

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