Applied Numerical Methods With MATLAB for Engineers and Scientists

Transcription

1 Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists Steven C. Chapra Tufts University

2 CHAPTER. You are given the following differential equation with the initial condition, v(t ), dv dt cd g m v Multiply both sides by m/c d m c d dv dt m c d g v Define a mg / c d m cd dv dt a v Integrate by separation of variables, dv a v cd dt m A table of integrals can be consulted to find that d tanh a a a Therefore, the integration yields tanh a v a cd t + C m If v at t, then because tanh (), the constant of integration C and the solution is tanh a v a cd t m This result can then be rearranged to yield gm gcd v tanh t c d m. This is a transient computation. For the period from ending June :

3 Balance Previous Balance + Deposits Withdrawals Balance The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date Deposit Withdrawal Balance -May $ 5. $. $ 7.6 -Jun $ 45. $ 6.8 $ Jul $ 4.9 $ 5.5 $ 6.8 -Aug $ $ 7. $ Sep $ At t s, the analytical solution is (Eample.). The numerical results are: step v() absolute relative error % 5.8.5% % where the relative error is calculated with analytical numerical absolute relative error % analytical The error versus step size can be plotted as.%.%.% relative error Thus, halving the step size approimately halves the error..4 (a) The force balance is

4 v m c g dt dv ' Applying Laplace transforms, V m c s g v sv ' ) ( Solve for m c s v m c s s g V / ' () ) / ' ( () The first term to the right of the equal sign can be evaluated by a partial fraction epansion, m c s B s A m c s s g / ' ) / ' ( () ) / ' ( ) / ' ( ) / ' ( m c s s Bs m c s A m c s s g Equating like terms in the numerators yields A m c g B A ' + Therefore, ' c' mg B c mg A These results can be substituted into Eq. (), and the result can be substituted back into Eq. () to give m c s v m c s c mg s c mg V / ' () / ' ' / ' / Applying inverse Laplace transforms yields t m c t m c e v e c mg c mg v ) '/ ( ) '/ ( () ' ' + or

5 v v() e ( c'/ m) t mg + c' ( c'/ m) t ( e ) where the first term to the right of the equal sign is the general solution and the second is the particular solution. For our case, v(), so the final solution is mg v c' ( c'/ m) t ( e ) (b) The numerical solution can be implemented as.5 v ( ) () v ( 4) (9.6) The computation can be continued and the results summarized and plotted as: t v dv/dt Note that the analytical solution is included on the plot for comparison. 4

6 .5 (a) The first two steps are c (.).(). 9.8 Bq/L c (.) 9.8.(9.8) Bq/L The process can be continued to yield t c dc/dt (b) The results when plotted on a semi-log plot yields a straight line The slope of this line can be estimated as ln(8.77) ln(). Thus, the slope is approimately equal to the negative of the decay rate..6 The first two steps yield y (.5) + 4 y () sin () [ sin (.5).] [.]

7 The process can be continued to give t y y c m gm t.7 v( t) ( e ) c jumper #:.5 9.8(68.) 68. v ( t) ( e ) m / s.5 4 t 9.8(75) 75 jumper #: ( e ) e.45 e.8666 t ln.45 ln e.8666 t.8666 t t. sec.8 Q in Q out Q Q + Q 6

8 + va 5 A A m.9 M M in out [ + + MP + 5] [ ] Metabolic production grams. % body weight IW 6 % Intracellular water body weight % IW 6 % body water TW % Transcellular water of body water.5 % 7

9 CHAPTER. >> q ;R 5;L 5;C e-4; >> t linspace(,.5); >> q q*ep(-r*t/(*l)).*cos(sqrt(/(l*c)-(r/(*l))^)*t); >> plot(t,q). >> z linspace(-,); >> f /sqrt(*pi)*ep(-z.^/); >> plot(z,f) >> label('z') >> ylabel('frequency'). (a) >> t linspace(5,,6) 8

10 t (b) >> linspace(-,,7) (a) >> v -:.75: v (b) >> r 6:-: r >> F [ 5 9 6]; >> [ ]; >> k F./ k.e+ * >> U.5*k.*.^ U >> ma(u) ans. >> TF :.6:9.; >> TC 5/9*(TF-); >> rho 5.589e-8*TC.^-8.56e-6*TC.^+6.56e-5*TC ; >> plot(tc,rho) 9

11 .7 >> A [.5. ;.. 8 ;.5..5;..7 4 ;.. 5.5] A >> U sqrt(a(:,))./a(:,).*(a(:,).*a(:,4)./(a(:,)+*a(:,4))).^(/) U >> t ::6; >> c [ ]; >> tf :7; >> cf 4.84*ep(-.4*tf); >> plot(t,c,'s',tf,cf,'--')

12 .9 >> t ::6; >> c [ ]; >> tf :7; >> cf 4.84*ep(-.4*tf); >> semilogy(t,c,'s',tf,cf,'--'). >> v ::8; >> F [ ]; >> vf :; >> Ff.74*vf.^.984; >> plot(v,f,'d',vf,ff,':')

13 . >> v ::8; >> F [ ]; >> vf :; >> Ff.74*vf.^.984; >> loglog(v,f,'d',vf,ff,':'). >> linspace(,*pi/); >> c cos(); >> cf -.^/+.^4/factorial(4)-.^6/factorial(6); >> plot(,c,,cf,'--')

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15 CHAPTER. The M-file can be written as function sincomp(,n) i ; tru sin(); ser ; fprintf('\n'); fprintf('order true value approimation error\n'); while () if i > n, break, end ser ser + (-)^(i - ) * ^(*i-) / factorial(*i-); er (tru - ser) / tru * ; fprintf('%d %4.f %4.f %.8f\n',i,tru,ser,er); i i + ; end This function can be used to evaluate the test case, >> sincomp(.5,8) order true value approimation error The M-file can be written as function futureworth(p, i, n) nn :n; F P*(+i).^nn; y [nn;f]; fprintf('\n year future worth\n'); fprintf('%5d %4.f\n',y); This function can be used to evaluate the test case, >> futureworth(,.8,8) year future worth

16 . The M-file can be written as function annualpayment(p, i, n) nn :n; A P*i*(+i).^nn./((+i).^nn-); y [nn;a]; fprintf('\n year annualpayment\n'); fprintf('%5d %4.f\n',y); This function can be used to evaluate the test case, >> annualpayment(5,.76,5) year annualpayment The M-file can be written as function Tavg avgtemp(tmean, Tpeak, tstart, tend) omega *pi/65; t tstart:tend; Te Tmean + (Tpeak-Tmean)*cos(omega*(t-5)); Tavg mean(te); This function can be used to evaluate the test cases, >> avgtemp(5.,.,,59) ans >> avgtemp(.,.6,8,4) ans.98.5 The M-file can be written as function vol tankvol(r, d) if d < R vol pi * d ^ / ; elseif d < * R v pi * R ^ / ; v pi * R ^ * (d - R); vol v + v; else error('overtop') end This function can be used to evaluate the test cases, 5

17 >> tankvol(,.5) ans.9 >> tankvol(,.) ans.6755 >> tankvol(,.) ans 7.4 >> tankvol(,.)??? Error using > tankvol overtop.6 The M-file can be written as function [r, th] polar(, y) r sqrt(.^ + y.^ ); if < if y > th atan(y / ) + pi; elseif y < th atan(y / ) - pi; else th pi; end else if y > th pi / ; elseif y < th -pi / ; else th ; end end th th * 8 / pi; This function can be used to evaluate the test cases. For eample, for the first case, >> [r,th]polar(,) r.44 th 9 The remaining cases are 6

18 y r θ The M-file can be written as function polar(, y) r sqrt(.^ + y.^ ); n length(); for i :n if (i) < if y(i) > th(i) atan(y(i) / (i)) + pi; elseif y(i) < th(i) atan(y(i) / (i)) - pi; else th(i) pi; end else if y(i) > th(i) pi / ; elseif y(i) < th(i) -pi / ; else th(i) ; end end th(i) th(i) * 8 / pi; end ou [;y;r;th]; fprintf('\n y radius angle\n'); fprintf('%8.f %8.f %.4f %.4f\n',ou); This function can be used to evaluate the test cases and display the results in tabular form, >> polar(,y) y radius angle

19 .8 The M-file can be written as function grade lettergrade(score) if score > 9 grade 'A'; elseif score > 8 grade 'B'; elseif score > 7 grade 'C'; elseif score > 6 grade 'D'; else grade 'F'; end This function can be tested with a few cases, >> lettergrade(95) ans A >> lettergrade(45) ans F >> lettergrade(8) ans B.9 The M-file can be written as function Manning(A) A(:,5) sqrt(a(:,))./a(:,).*(a(:,).*a(:,4)./(a(:,)+*a(:,4))).^(/); fprintf('\n n S B H U\n'); fprintf('%8.f %8.4f %.f %.f %.4f\n',A'); This function can be run to create the table, >> Manning(A) n S B H U

20 . The M-file can be written as function beam() linspace(,); nlength(); for i:n uy(i) -5/6.*(sing((i),,4)-sing((i),5,4)); uy(i) uy(i) + 5/6.*sing((i),8,) + 75*sing((i),7,); uy(i) uy(i) + 57/6.*(i)^ *(i); end plot(,uy) function s sing(,a,n) if > a s ( - a).^n; else s; end This function can be run to create the plot, >> beam(). The M-file can be written as function cylinder(r, L) h linspace(,*r); V (r^*acos((r-h)./r)-(r-h).*sqrt(*r*h-h.^))*l; plot(h, V) This function can be run to the plot, >> cylinder(,5) 9

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22 CHAPTER 4 4. The true value can be computed as f '(.) 6(.577) (.577 ),5,9 Using -digits with chopping chopping 6 6(.577) chopping f '(.577).46 (.996) ,5 This represents a percent relative error of ε t,5,9 6,5,5,9 9.8% Using 4-digits with chopping chopping 6 6(.577) chopping f '(.577).46 (.9987).46.,48,5 This represents a percent relative error of ε t,5,9,48,5,5,9.9% Although using more significant digits improves the estimate, the error is still considerable. The problem stems primarily from the fact that we are subtracting two nearly equal numbers in the denominator. Such subtractive cancellation is worsened by the fact that the denominator is squared. 4. First, the correct result can be calculated as y.7 7(.7) + 8(.7).5.45

23 (a) Using -digits with chopping (.7) 7(.87). 8(.7) This represents an error of ε t % (b) Using -digits with chopping y ((.7 7).7 + 8).7.5 y ( ).7.5 y ( ).7.5 y y.97.5 y.47 This represents an error of ε t % Hence, the second form is superior because it tends to minimize round-off error. 4. (a) For this case, i and h. Thus, the Taylor series is f "() f () f ( ) f () + f '() + + +!! For the eponential function, f () f '() f "() f () () Substituting these values yields, ()

24 f ( ) !! which is the Maclaurin series epansion. (b) The true value is e and the step size is h i+ i The complete Taylor series to the third-order term is h i i i f ( i+ ) e e h + e e i h! Zero-order approimation: f ().5 e.7788 ε t %.7% First-order approimation: f ( ) (.75).947 ε t % 47.% Second-order approimation:.75 f () (.75) ε t %.5% Third-order approimation:.75 f () (.75) ε t %.4% Use ε s.5.5%. The true value cos(π/4).777 zero-order:

25 π cos 4 ε t.777 % 4.4%.777 first-order: π ( π / 4) cos ε t %.9%.777 ε a % 44.6% second-order: π ( π / 4) cos ε t %.456%.777 ε a %.4%.7749 third-order: π ( π / 4) cos ε t %.5%.777 ε a %.46%.77 Because ε a <.5%, we can terminate the computation. 4.5 Use ε s.5.5%. The true value sin(π/4).777 zero-order: 4

26 π sin ε t %.%.777 first-order: π ( π / 4) sin ε t %.47%.777 ε a %.46%.7465 second-order: π ( π / 4) sin ε t %.5%.777 ε a %.5%.774 Because ε a <.5%, we can terminate the computation. 4.6 The true value is f(). zero order: ( 6) f ( ) f () 6 ε t % 6.8% first order: f '() 75() () f () 6 + 7() 8 8 ε t % 9.% second order: 5

27 f "() 5() 8 8 f () 8 + () third order: f () () ε t % 4.5% 5 f () 77 + () 6 ε t %.% Because we are working with a third-order polynomial, the error is zero. This is due to the fact that cubics have zero fourth and higher derivatives. 4.7 The true value is ln().986 zero order:.986 f ( ) f () ε t % %.986 first order: f '( ) f () + () f '().986 ε t % 8.5%.986 second order: f "( ) f () third order: f () ( ) f () + 6 fourth order: f "() f "() ε t % % ε t % 4.7%.986 6

28 f (4) 6 ( ) 4 f (4) () 6 f () (.) ε t %.4%.986 The series is diverging. A smaller step size is required to obtain convergence. 4.8 The first derivative of the function at can be evaluated as f '() 75() () The points needed to form the finite divided differences can be computed as i.75 f( i ) i. f( i ) i+.5 f( i+ ) 8.46 forward: 8.46 f '().565 Et backward: f '() Et centered: f '() Et Both the forward and backward differences should have errors approimately equal to E t f "( i ) h The second derivative can be evaluated as f "() 5() 88 Therefore, E t which is similar in magnitude to the computed errors. 7

29 For the central difference, E t f () ( i ) h 6 The third derivative of the function is 5 and E t 5 (.5) which is eact. This occurs because the underlying function is a cubic equation that has zero fourth and higher derivatives. 4.9 The second derivative of the function at can be evaluated as f '() 5() 88 For h., () f "() 88 (.) For h.,.765 () f "() 88 (.) Both are eact because the errors are a function of fourth and higher derivatives which are zero for a rd -order polynomial. 4. Use ε s.5.5%. The true value /(.). zero-order:. ε t. % %. first-order:

30 .. ε t % %. ε a. % 9.99%. second-order: ε t %.%. ε a.. %.99%. third-order: ε t %.%. ε a.. %.99%. The approimate error has fallen below.5% so the computation can be terminated. 4. Here are the function and its derivatives f ( ) sin f f f '( ) f "( ) sin () (4) cos ( ) cos ( ) sin 9

31 Using the Taylor Series epansion, we obtain the following st, nd, rd, and 4 th order Taylor Series functions shown below in the MATLAB program f, f, and f4. Note the nd and rd order Taylor Series functions are the same. From the plots below, we see that the answer is the 4 th Order Taylor Series epansion. :.:.; f--.5*sin(); subplot(,,); plot(,f);grid;title('f()--.5*sin()');hold on f-.5; eabs(f-f); %Calculates the absolute value of the difference/error subplot(,,); plot(,e);grid;title('st Order Taylor Series Error'); f *((-.5*pi).^); eabs(f-f); subplot(,,); plot(,e);grid;title('nd/rd Order Taylor Series Error'); f *((-.5*pi).^)-(/48)*((-.5*pi).^4); e4abs(f4-f); subplot(,,4); plot(,e4);grid;title('4th Order Taylor Series Error');hold off f()--.5*sin().8 st Order Taylor Series Error nd/rd Order Taylor Series Error..5 4th Order Taylor Series Error

32 4. f() f(-) f(+) f'()-theory f'()-back f'()-cent f'()-forw First Derivative Approimations Compared to Theoretical f'() Theoretical Backward Centered Forward values f() f(-) f(+) f(-) f(+) f''()- Theory f''()- Back f''()-cent f''()- Forw

33 Approimations of the nd Derivative f''() f''()-theory f''()-backward f''()-centered f''()-forward values 4. function eps macheps % determines the machine epsilon e ; while e+> e e/; end eps *e; >> macheps ans.4e-6 >> eps ans.4e-6

34 CHAPTER 5 5. The function to evaluate is gm gc d f ( cd ) tanh t v( t) c d m or substituting the given values 9.8(8) 9.8 ( ) tanh cd f c 4 d 6 c 8 d The first iteration is r f (.) f (.5).869(.456) Therefore, the root is in the first interval and the upper guess is redefined as u.5. The second iteration is r ε a.5.5 % %.5 f (.) f (.5).869(.847).79 Therefore, the root is in the second interval and the lower guess is redefined as u.5. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % % % % Thus, after si iterations, we obtain a root estimate of.965 with an approimate error of.%. 5. function root bisectnew(func,l,u,ead) % bisectnew(l,u,es,mait):

35 % uses bisection method to find the root of a function % with a fied number of iterations to attain % a prespecified tolerance % input: % func name of function % l, u lower and upper guesses % Ead (optional) desired tolerance (default.) % output: % root real root if func(l)*func(u)> %if guesses do not bracket a sign change error('no bracket') %display an error message and terminate end % if necessary, assign default values if nargin<4, Ead.; end %if Ead blank set to. % bisection r l; % compute n and round up to net highest integer n round( + log((u - l)/ead) +.5); for i :n rold r; r (l + u)/; if r ~, ea abs((r - rold)/r) * ; end test func(l)*func(r); if test < u r; elseif test > l r; else ea ; end end root r; The following is a MATLAB session that uses the function to solve Prob. 5. with E a,d.. >> fcd inline('sqrt(9.8*8/cd)*tanh(sqrt(9.8*cd/8)*4)-6','cd') fcd Inline function: fcd(cd) sqrt(9.8*8/cd)*tanh(sqrt(9.8*cd/8)*4)-6 >> format long >> bisectnew(fcd,.,.,.) ans The function to evaluate is 9.8(8) 9.8 ( ) tanh cd f c 4 d 6 c 8 d 4

36 The first iteration is r.978(..) (.978) f (.) f (.489).869(.5).9 Therefore, the root is in the first interval and the upper guess is redefined as u.489. The second iteration is r.5(..489) (.5) ε a %.7%.465 Therefore, after only two iterations we obtain a root estimate of.465 with an approimate error of.7% which is below the stopping criterion of %. 5.4 function root falsepos(func,l,u,es,mait) % falsepos(l,u,es,mait): % uses the false position method to find the root % of the function func % input: % func name of function % l, u lower and upper guesses % es (optional) stopping criterion (%) (default.) % mait (optional) maimum allowable iterations (default 5) % output: % root real root if func(l)*func(u)> %if guesses do not bracket a sign change error('no bracket') %display an error message and terminate end % default values if nargin<5, mait5; end if nargin<4, es.; end % false position iter ; r l; while () rold r; r u - func(u)*(l - u)/(func(l) - func(u)); iter iter + ; if r ~, ea abs((r - rold)/r) * ; end test func(l)*func(r); if test < u r; elseif test > l r; else ea ; 5

37 end if ea < es iter > mait, break, end end root r; The following is a MATLAB session that uses the function to solve Prob. 5.: >> fcd inline('sqrt(9.8*8/cd)*tanh(sqrt(9.8*cd/8)*4)-6','cd') fcd Inline function: fcd(cd) sqrt(9.8*8/cd)*tanh(sqrt(9.8*cd/8)*4)-6 >> format long >> falsepos(fcd,.,.,) ans Solve for the reactions: R 65 lbs. R 85 lbs. Write beam equations: << <<6 6<< << M + (6.667 ) 65 () M M + ( )( ) + 5( () M ()) M 5( ()) + ( 4.5) 65 () M M + ( ) (4) M 65 Combining Equations: Because the curve crosses the ais between 6 and, use (). () M Set 6 ; L U 6

38 M ( L ) 54 M ( ) U M ( ) 7 replaces r R L L + U 8 M ( L ) 7 M ( ) U M ( ) 5 r replaces R U M ( L ) 7 M ( ) 5 U M ( ) r replaces R L M ( L ) 77.5 M ( ) 5 U M ( ). 5 r replaces R L M ( L ).5 M ( ) 5 U M ( ) 8. 5 r replaces R L M ( L ) 8.5 M ( ) 5 U M ( ). 475 r replaces R U M ( L ) 8.5 M ( ).475 U M ( ). 475 r replaces R L M ( L ).475 M ( ).475 U M ( ) r R U replaces M ( L ).475 M ( ) U r M ( R ).8984 Therefore, 8. 9 feet 5.6 (a) The graph can be generated with MATLAB >> [-:.:6]; 7

39 >> f--*+8*.^-.75*.^; >> plot(,f) >> grid This plot indicates that roots are located at about.4,.5 and 4.7. (b) Using bisection, the first iteration is r +.5 f ( ) f (.5) 9.75(.475) Therefore, the root is in the second interval and the lower guess is redefined as l.5. The second iteration is r ε a.5 (.5) % %.5 f (.5) f (.5).475( 5.58) Therefore, the root is in the first interval and the upper guess is redefined as u.5. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % 8

40 % % % % % Thus, after eight iterations, we obtain a root estimate of with an approimate error of.9%, which is below the stopping criterion of %. (c) Using false position, the first iteration is r ( ) ( ) f ( ) f (.8745) 9.75( ).49 Therefore, the root is in the first interval and the upper guess is redefined as u The second iteration is r ( (.8745)) ( ) ε a (.8745) % 4.5% f ( ) f ( ) 9.75(.89669) Therefore, the root is in the first interval and the upper guess is redefined as u The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % % % Therefore, after five iterations we obtain a root estimate of.44 with an approimate error of.45%, which is below the stopping criterion of %. 5.7 A graph of the function can be generated with MATLAB >> [-.5:.:.5]; >> fsin()-.^; >> plot(,f) >> grid 9

41 This plot indicates that a nontrivial root (i.e., nonzero) is located at about.85. Using bisection, the first iteration is r f (.5) f (.75).946(.988).7 Therefore, the root is in the second interval and the lower guess is redefined as l.75. The second iteration is r ε a % 4.9%.875 f (.75) f (.875).99(.985).9 Because the product is positive, the root is in the second interval and the lower guess is redefined as l.875. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % % % 4

42 Therefore, after five iterations we obtain a root estimate of.8965 with an approimate error of.75%, which is below the stopping criterion of %. 5.8 (a) A graph of the function indicates a positive real root at approimately (b) Using bisection, the first iteration is r f (.5) f (.5).869(.579).59 Therefore, the root is in the second interval and the lower guess is redefined as l.5. The second iteration is r ε a.65.5 %.8%.65 f (.5) f (.65).57(.756).6876 Therefore, the root is in the first interval and the upper guess is redefined as u.65. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % 4

43 Thus, after three iterations, we obtain a root estimate of.475 with an approimate error of.4%. (c) Using false position, the first iteration is r (.5 ) f (.5) f (.6877).8694(.75574) Therefore, the root is in the first interval and the upper guess is redefined as u The second iteration is r.4974( ) ε a % 8.8%.4974 f (.5) f (.4974).8694(.6945).9 Therefore, the root is in the first interval and the upper guess is redefined as u The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % Therefore, after three iterations we obtain a root estimate of with an approimate error of.6%. 5.9 (a) Equation (5.6) can be used to determine the number of iterations 5 log n + + log E a, d.5 which can be rounded up to iterations..45 (b) Here is an M-file that evaluates the temperature in o C using iterations of bisection based on a given value of the oygen saturation concentration in freshwater: function TC TempEval(osf) % function to evaluate the temperature in degrees C based % on the oygen saturation concentration in freshwater (osf). l + 7.5; u ; if fta(l,osf)*fta(u,osf)> %if guesses do not bracket error('no bracket') %display an error message and terminate 4

44 end r l; for i : rold r; r (l + u)/; if r ~, ea abs((r - rold)/r) * ; end test fta(l,osf)*fta(r,osf); if test < u r; elseif test > l r; else ea ; end end TC r - 7.5; end function f fta(ta, osf) f e5/Ta e7/Ta^; f f +.48e/Ta^ e/Ta^4; f f - log(osf); The function can be used to evaluate the test cases: >> TempEval(8) ans >> TempEval() ans >> TempEval(4) ans (a) The function to be evaluated is 4 f ( y) ( + y) 9.8( y + y / ) A graph of the function indicates a positive real root at approimately.5. 4

45 (b) Using bisection, the first iteration is r f (.5) f (.5).58(.946).9986 Therefore, the root is in the second interval and the lower guess is redefined as l.5. The second iteration is r ε a.5 % 5% f (.5) f ().946(.689).864 Therefore, the root is in the first interval and the upper guess is redefined as u. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % % % % % % 44

46 After eight iterations, we obtain a root estimate of.5785 with an approimate error of.5%. (c) Using false position, the first iteration is r.8(.5.5) f (.5) f (.458).58(.79987) Therefore, the root is in the first interval and the upper guess is redefined as u.458. The second iteration is r.79987(.5.458) ε a %.96%.46 f (.5) f (.46).58(.786) The root is in the first interval and the upper guess is redefined as u.46. The remainder of the iterations are displayed in the following table: i l f( l ) u f( u ) r f( r ) ε a % % % % % % % % % After ten iterations we obtain a root estimate of.977 with an approimate error of.59%. Thus, after ten iterations, the false position method is converging at a very slow pace and is still far from the root in the vicinity of.5 that we detected graphically. Discussion: This is a classic eample of a case where false position performs poorly and is inferior to bisection. Insight into these results can be gained by eamining the plot that was developed in part (a). This function violates the premise upon which false position was based that is, if f( u ) is much closer to zero than f( l ), then the root is closer to u than to l (recall Fig. 5.8). Because of the shape of the present function, the opposite is true. 45

47 CHAPTER 6 6. The function can be set up for fied-point iteration by solving it for ( ) i+ sin i Using an initial guess of.5, the first iteration yields (.5) sin ε a % % Second iteration: ( ). 754 sin ε a % 9.96% The process can be continued as tabulated below: iteration i ε a % %.759.9% % % % % % % Thus, after nine iterations, the root is estimated to be with an approimate error of.97%. 6. (a) The function can be set up for fied-point iteration by solving it for in two different ways. First, it can be solved for the linear, i+.9i.5.7 Using an initial guess of 5, the first iteration yields 46

48 .9(5) ε a % 57.5% Second iteration:.9(.76) ε a % 8.6% Clearly, this solution is diverging. An alternative is to solve for the second-order, i+.7 i Using an initial guess of 5, the first iteration yields.7(5) i+.496 ε a % 4.% Second iteration:.7(.496) i+.69 ε a % 4.4% This version is converging. All the iterations can be tabulated as iteration i ε a % % % % % 47

49 % % % % Thus, after 9 iterations, the root estimate is.869 with an approimate error of.6%. The result can be checked by substituting it back into the original function, f (.869).9(.869) +.7(.869) (b) The formula for Newton-Raphson is i+ i.9i +.7i i Using an initial guess of 5, the first iteration yields.9(5) +.7(5) (5) +.7 i ε a % 46.% Second iteration:.9(.44658) +.7(.44658) (.44658) +.7 i ε a % 7.% The process can be continued as tabulated below: iteration i f( i ) f'( i ) ε a % % % E % E % After 5 iterations, the root estimate is.864 with an approimate error of.%. The result can be checked by substituting it back into the original function, f (.864).9(.864) +.7(.864)

50 6. (a) >> linspace(,4); >> y.^-6*.^+*-6.; >> plot(,y) >> grid Estimates are approimately.5,.9 and.5. (b) The formula for Newton-Raphson is i+ i i 6 i i + i i 6. + Using an initial guess of.5, the first iteration yields (.5) 6(.5) + (.5) 6..5 (.5) (.5) +.94 ε a % 9.67% Second iteration: (.94) 6(.94) + (.94) (.94) (.94) ε a %.995% 49

51 Third iteration: (.68699) 6(.68699) + (.68699) (.68699) (.68699) ε a %.7% (c) For the secant method, the first iteration:.5 f( ) f( ) (.5.5) ε a % 9.98% Second iteration:.5 f( ) f( ) (.5.7) (.455).879 ε a % 5.57%.879 Third iteration:.7 f( ) f( )..(.7.879) (.).9 ε a %.889%.9 (d) For the modified secant method, the first iteration:.5 f( ) δ.57 f( + δ )

52 .(.5) ε a % 9.7% Second iteration:.757 f( ) δ.775 f( + δ ).6856.(.757) ε a % 4.7%.84 Third iteration:.84 f( ) δ.4675 f( + δ ).54.(.84) ε a %.% (e) >> a [ -6-6.] a (a) >> roots(a) ans >> linspace(,4); >> y 7*sin().*ep(-)-; >> plot(,y) >> grid 5

53 The lowest positive root seems to be at approimately.. (b) The formula for Newton-Raphson is i+ i 7e 7 sin( ) e i i (cos( ) sin( )) i i i Using an initial guess of.5, the first iteration yields. 7e. 7 sin(.) e (cos(.) sin(.)) ε a % 7.8%.4476 Second iteration: e sin(.4476) e (cos(.4476) sin(.4476)) ε a % 4.776% Third iteration: e sin(.6949) e (cos(.6949) sin(.6949))

54 ε a %.45% (c) For the secant method, the first iteration:.4 f( ) f( ) (.4.) ε a.947. % 5.4%.947 Second iteration:. f( ) f( ).6.6(..947) (.6) ε a %.% Third iteration:.947 f( ) f( ) ( ) ε a % 4.68% (d) For the modified secant method, the first iteration:. f( ) δ. f( + δ ).5478.(.) ε a % 8.8%

55 Second iteration:.4698 f( ).75 + δ.455 f( + δ ).49.(.4698)(.75) (.75).694 ε a % 5.8% Third iteration:.694 f( ).7 + δ.76 f( + δ ).4456.(.694)(.7) (.7).78 ε a %.45% 6.5 (a) The formula for Newton-Raphson is i+ i 5 i i 4 i i i i i i i Using an initial guess of.585, the first iteration yields ε a % % Second iteration ε a % % 54

56 Thus, the result seems to be diverging. However, the computation eventually settles down and converges (at a very slow rate) on a root at 6.5. The iterations can be summarized as iteration i f( i ) f'( i ) ε a % E+9.84E % E+9.6E+8.874% E % E % % % % % % % % % % % % % % % % % E % (b) For the modified secant method, the first iteration:.585 f( ) δ.665 f( + δ ) (.585) ε a % 7.447% Second iteration:.975 f( ) δ.4 f( + δ ) (.975)(.969) (.969)

57 ε a % 48.87% Again, the result seems to be diverging. However, the computation eventually settles down and converges on a root at.. The iterations can be summarized as iteration i i +δ i f( i ) f( i + δ i ) ε a % % % % % % % % % % %...4E % Eplanation of results: The results are eplained by looking at a plot of the function. The guess of.585 is located at a point where the function is relatively flat. Therefore, the first iteration results in a prediction of. for Newton-Raphson and.9 for the secant method. At these points the function is very flat and hence, the Newton-Raphson results in a very high value (9.75), whereas the modified false position goes in the opposite direction to a negative value (-4.49). Thereafter, the methods slowly converge on the nearest roots Modified Newton secant Raphson function root secant(func,rold,r,es,mait) % secant(func,rold,r,es,mait): % uses secant method to find the root of a function % input: % func name of function % rold, r initial guesses % es (optional) stopping criterion (%) 56

58 % mait (optional) maimum allowable iterations % output: % root real root % if necessary, assign default values if nargin<5, mait5; end %if mait blank set to 5 if nargin<4, es.; end %if es blank set to. % Secant method iter ; while () rn r - func(r)*(rold - r)/(func(rold) - func(r)); iter iter + ; if rn ~, ea abs((rn - r)/rn) * ; end if ea < es iter > mait, break, end rold r; r rn; end root rn; Test by solving Prob. 6.: >> secant(inline('^-6*^+*-6.'),.5,.5) ans function root modsec(func,r,delta,es,mait) % secant(func,rold,r,es,mait): % uses the modified secant method % to find the root of a function % input: % func name of function % r initial guess % delta perturbation fraction % es (optional) stopping criterion (%) % mait (optional) maimum allowable iterations % output: % root real root % if necessary, assign default values if nargin<5, mait5; end %if mait blank set to 5 if nargin<4, es.; end %if es blank set to. if nargin<, deltae-5; end %if delta blank set to. % Secant method iter ; while () rold r; r r - delta*r*func(r)/(func(r+delta*r)-func(r)); iter iter + ; if r ~, ea abs((r - rold)/r) * ; end if ea < es iter > mait, break, end end root r; 57

59 Test by solving Prob. 6.: >> modsec(inline('^-6*^+*-6.'),.5,.) ans The equation to be differentiated is gm f ( m) tanh c d gc d t v m Note that d tanh u sech d u du d Therefore, the derivative can be evaluated as df ( m) dm gm sech c d gc d t m m cd g t cd g m + tanh gc d t m cd gm g c d The two terms can be reordered df ( m) c d g gcd gm m cd g gcd tanh t t sech t dm gm c d m cd cd g m m The terms premultiplying the tanh and sech can be simplified to yield the final result df ( m) g gcd g gcd tanh t t sech t dm mc d m m m 6.9 (a) The formula for Newton-Raphson is i+ i i + 6i i i i Using an initial guess of 4.5, the iterations proceed as iteration i f( i ) f'( i ) ε a % % % % 58

60 % % % % % E % Thus, after an initial jump, the computation eventually settles down and converges on a root at (b) Using an initial guess of 4.4, the iterations proceed as iteration i f( i ) f'( i ) ε a E+ 669.% E % E % E % E % % % % E % E % This time the solution jumps to an etremely large negative value The computation eventually converges at a very slow rate on a root at Eplanation of results: The results are eplained by looking at a plot of the function. Both guesses are in a region where the function is relatively flat. Because the two guesses are on opposite sides of a minimum, both are sent to different regions that are far from the initial guesses. Thereafter, the methods slowly converge on the nearest roots. 59

61 The function to be evaluated is a This equation can be squared and epressed as a roots problem, f ( ) a The derivative of this function is f '( ) These functions can be substituted into the Newton-Raphson equation (Eq. 6.6), i+ i i a i which can be epressed as i+ i + a / i 6. (a) The formula for Newton-Raphson is i+ i tanh sech i ( i 9) ( 9) i Using an initial guess of., the iterations proceed as iteration i f( i ) f'( i ) ε a % 6

62 % % (b) The solution diverges from its real root of. Due to the concavity of the slope, the net iteration will always diverge. The following graph illustrates how the divergence evolves The formula for Newton-Raphson is i+ i 4.74i.84i +.55i.8i i i i Using an initial guess of 6.5, the iterations proceed as iteration i f( i ) f'( i ) ε a % % % % % % E % E 9.45.% As depicted below, the iterations involve regions of the curve that have flat slopes. Hence, the solution is cast far from the roots in the vicinity of the original guess. 6

63 The solution involves determining the root of 6 f ( ).5 + MATLAB can be used to develop a plot that indicates that a root occurs in the vicinity of.. >> f inline('./(-).*sqrt(6./(+))-.5') f Inline function: f()./(-).*sqrt(6./(+))-.5 >> linspace(,.); >> y f(); >> plot(,y) The fzero function can then be used to find the root 6

64 >> format long >> fzero(f,.) ans The coefficient, a and b, can be evaluated as >> format long >> R.58;pc 46;Tc 9; >> a.47*r^*tc^.5/pc a >> b.866*r*tc/pc b The solution, therefore, involves determining the root of.58(.5) f ( v) 65, + v.866 v( v +.866).5 MATLAB can be used to generate a plot of the function and to solve for the root. One way to do this is to develop an M-file for the function, function y fvol(v) R.58;pc 46;Tc 9; a.47*r^*tc^.5/pc; b.866*r*tc/pc; T 7.5-4;p 65; y p - R*T./(v-b)+a./(v.*(v+b)*sqrt(T)); This function is saved as fvol.m. It can then be used to generate a plot >> v linspace(.,.4); >> fv fvol(v); >> plot(v,fv) >> grid 6

65 Thus, a root is located at about.8. The fzero function can be used to refine this estimate, >> vroot fzero('fvol',.8) vroot The mass of methane contained in the tank can be computed as V mass 68.7 m v The function to be evaluated is f ( h) V r cos r h ( r h) r rh h L To use MATLAB to obtain a solution, the function can be written as an M-file function y fh(h,r,l,v) y V - (r^*acos((r-h)/r)-(r-h)*sqrt(*r*h-h^))*l; The fzero function can be used to determine the root as >> format long >> r ;L 5;V 8; >> h fzero('fh',.5,[],r,l,v) h

66 6.6 (a) The function to be evaluated is TA 5 TA f ( TA ) cosh + T A The solution can be obtained with the fzero function as >> format long >> TA fzero(inline('-/*cosh(5/)+/'),) TA e+ (b) A plot of the cable can be generated as >> linspace(-5,); >> w ;y 5; >> y TA/w*cosh(w*/TA) + y - TA/w; >> plot(,y),grid 6.7 The function to be evaluated is t f ( t) 9e sin(π t).5 A plot can be generated with MATLAB, >> t linspace(,); >> y 9*ep(-t).* sin(*pi*t) -.5; >> plot(t,y),grid 65

67 Thus, there appear to be two roots at approimately. and.4. The fzero function can be used to obtain refined estimates, >> t fzero('9*ep(-)*sin(*pi*)-.5',[.]) t >> t fzero('9*ep(-)*sin(*pi*)-.5',[..8]) t The function to be evaluated is f ( ω ) + ωc Z R ωl Substituting the parameter values yields f ( ω). +.6 ω 565 ω 6 The fzero function can be used to determine the root as >> fzero('.-sqrt(/565+(.6e-6*-./).^)',[ ]) ans. 6.9 The fzero function can be used to determine the root as 66

68 >> format long >> fzero('*4*^(5/)/5+.5*4*^-95*9.8*-95*9.8*.4',) ans If the height at which the throw leaves the right fielders arm is defined as y, the y at 9 m will be.8. Therefore, the function to be evaluated is π 44. f ( θ ) tan θ 8 cos ( πθ /8) Note that the angle is epressed in degrees. First, MATLAB can be used to plot this function versus various angles. Roots seem to occur at about 4 o and 5 o. These estimates can be refined with the fzero function, >> theta fzero('.8+9*tan(pi*/8)-44../cos(pi*/8).^',) theta 7.88 >> theta fzero('.8+9*tan(pi*/8)-44../cos(pi*/8).^',[4 6]) theta Thus, the right fielder can throw at two different angles to attain the same result. 6. The equation to be solved is 67

69 f ( h) π Rh π h V Because this equation is easy to differentiate, the Newton-Raphson is the best choice to achieve results efficiently. It can be formulated as i+ i π πri πr π i i i V or substituting the parameter values, i+ i π π () i i π () π i i The iterations can be summarized as iteration i f( i ) f'( i ) ε a % % E % Thus, after only three iterations, the root is determined to be with an approimate relative error of.7%. 6. >> r [ ]; >> a poly(r) a >> polyval(a,) ans >> b poly([- 6]) b -4 - >> [q,r] deconv(a,b) q r

70 >> roots(q) >> a conv(q,b) a >> roots(a) >> a poly() a >> a [ 9 6 4]; >> r roots(a) r >> a [ ]; >> r roots(a) r Therefore, the transfer function is ( s + 4)( s + )( s + ) G ( s) ( s + 6)( s + 5)( s + )( s + ) 69

71 CHAPTER 7 7. >> Aug [A eye(size(a))] Here s an eample session of how it can be employed. >> A rand() A >> Aug [A eye(size(a))] Aug (a) [A]: [B]: {C}: [D]: 4 [E]: [F]: G : (b) square: [B], [E]; column: {C}, row: G (c) a 5, b 6, d undefined, e, f, g 6 (d) MATLAB can be used to perform the operations 5 8 () [ E ] + [ B] 8 9 () [ E ] [ B] () [A] + [F] undefined (4) 5[ F ] (5) [A] [B] undefined (6) [ B ] [ A] T (7) [G] [C] 56 (8) [ C ] 6 5 (9) 4 7 T [ D ] () I [B] The terms can be collected to give 7

72 Here is the MATLAB session: >> A [-7 ; 4 7;-4-7]; >> b [;-;4]; >> A\b >> AT A' AT >> AI inv(a) AI [ X ] [ Y ] [ X ] [ Z] 5 [ Y ] [ Z] [ Z ] [ Y ] Terms can be combined to yield 7

73 k k k + k k k + k m g m m Substituting the parameter values g g A MATLAB session can be used to obtain the solution for the displacements >> K[ - ;- -; - ]; >> m[;;.5]; >> mgm*9.8; >> K\mg The mass balances can be written as ( Q 5 + Q Q Q ) c 5 c c + ( Q + Q 4 + Q 5 Q Q Q 4 ) c 5 c c c + ( Q Q + Q c 4 Q 4 ) c c + Q 44 c 4 + ( Q 54 Q + Q The parameters can be substituted and the result written in matri form as Q c 5 ) c 5 Q c c c c c c 4 c MATLAB can then be used to solve for the concentrations >> Q [6 - ; - ; - 9 ; ; - - 4]; >> Qc [5;;6;;]; 7

74 >> c Q\Qc c The problem can be written in matri form as F F F H V V MATLAB can then be used to solve for the forces and reactions, >> A [ ; ; ; ;.5 ; ] >> b [ - ]'; >> F A\b F Therefore, F 5 F 4 F 866 H V 5 V The problem can be written in matri form as 5 5 i i i i 5 i i MATLAB can then be used to solve for the currents, 7

75 >> A [ ; - - ; - ; -; ; ]; >> b [ ]'; >> i A\b i >> k ;k 4;k 4;k4 ; >> m ;m ;m ; >> km [(/m)*(k+k), -(k/m),; -(k/m), (/m)*(k+k), -(k/m);, -(k/m),(/m)*(k+k4)]; >> [.5;.4;.]; >> km km* km Therefore, & &.9, & &, and & &. m/s. 74

76 CHAPTER 8 8. The flop counts for the tridiagonal algorithm in Fig. 8.6 can be summarized as Mult/Div Add/Subtr Total Forward elimination (n ) (n ) 5(n ) Back substitution n n n Total 5n 4 n 8n 7 Thus, as n increases, the effort is much, much less than for a full matri solved with Gauss elimination which is proportional to n. 8. The equations can be epressed in a format that is compatible with graphing versus : which can be plotted as Thus, the solution is 4, 5. The solution can be checked by substituting it back into the equations to give 4(4) 8(5) (5) (a) The equations can be epressed in a format that is compatible with graphing versus :

77 which can be plotted as Thus, the solution is approimately 4, 6. The solution can be checked by substituting it back into the equations to give.(4) + (6) 6 (4) + 7.4(6) Therefore, the graphical solution is not very good. (b) Because the lines have very similar slopes, you would epect that the system would be ill-conditioned (c) The determinant can be computed as..(7.) ( ) This result is relatively low suggesting that the solution is ill-conditioned. 8.4 (a) The determinant can be evaluated as D ( ) D ( ) + (5) + 7( ) 69 (b) Cramer s rule 76

78 (c) Pivoting is necessary, so switch the first and third rows, Multiply pivot row by /5 and subtract the result from the second row to eliminate the a term Pivoting is necessary so switch the second and third row, Multiply pivot row by.4/( ) and subtract the result from the third row to eliminate the a term The solution can then be obtained by back substitution (.94)

79 (d) + (.46768) (.46768) + 7(.94) (.46768) (.94) 5(.98557) (.46768) 8.5 Prob. 8.: >> A[-. ;- 7.4]; >> det(a) ans.86 Prob. 8.4: >> A[ - 7; -;5 - ]; >> det(a) ans (a) The equations can be epressed in a format that is compatible with graphing versus : The resulting plot indicates that the intersection of the lines is difficult to detect: Only when the plot is zoomed is it at all possible to discern that solution seems to lie at about 4.5 and. 78

80 (b) The determinant can be computed as.5..5( ) ( )(.). which is close to zero. (c) Because the lines have very similar slopes and the determinant is so small, you would epect that the system would be ill-conditioned (d) Multiply the first equation by./.5 and subtract the result from the second equation to eliminate the term from the second equation, The second equation can be solved for This result can be substituted into the first equation which can be solved for (e) Multiply the first equation by./.5 and subtract the result from the second equation to eliminate the term from the second equation,

81 The second equation can be solved for This result can be substituted into the first equation which can be solved for Interpretation: The fact that a slight change in one of the coefficients results in a radically different solution illustrates that this system is very ill-conditioned. 8.7 (a) Multiply the first equation by / and subtract the result from the second equation to eliminate the term from the second equation. Then, multiply the first equation by / and subtract the result from the third equation to eliminate the term from the third equation Multiply the second equation by.8/( 5.4) and subtract the result from the third equation to eliminate the term from the third equation, Back substitution can then be used to determine the unknowns ( 5.4.7( 6)) (7 6 (8)).5 (b) Check: 8

82 (.5) + (8) ( 6) 7 (.5) 6(8) + ( 6) ( 6) (a) Pivoting is necessary, so switch the first and third rows, Multiply the first equation by /( 8) and subtract the result from the second equation to eliminate the a term from the second equation. Then, multiply the first equation by /( 8) and subtract the result from the third equation to eliminate the a term from the third equation Pivoting is necessary so switch the second and third row, Multiply pivot row by.75/( 5.75) and subtract the result from the third row to eliminate the a term The solution can then be obtained by back substitution ( )