6.1 The function can be set up for fixed-point iteration by solving it for x

Transcription

1 1 CHAPTER The function can be set up for fied-point iteration by solving it for 1 sin i i Using an initial guess of 0 = 0.5, the first iteration yields 1 sin a % 3% Second iteration: sin a % 9.96% The process can be continued as tabulated below: i i a E t E t,i / E t,i % % % % % % % % % Thus, after nine iterations, the root is estimated to be with an approimate error of %. To confirm that the scheme is linearly convergent, according to the book, the ratio of the errors between iterations should be E g '( ) E i 1 1 i cos Substituting the root for gives a value of which is close to the values in the last column of the table. 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 1 i Using an initial guess of 5, the first iteration yields

2 0.9(5) a 100% 57.5% Second iteration: 0.9(11.76) a 100% 83.6% 71.8 Clearly, this solution is diverging. An alternative is to solve for the second-order, i i 0.9 Using an initial guess of 5, the first iteration yields i 1 a 1.7(5) % 43.0% Second iteration: i 1 a 1.7(3.496) % 14.14% This version is converging. All the iterations can be tabulated as iteration i a % % % % % % % % % Thus, after 9 iterations, the root estimate is with an approimate error of %. The result can be checked by substituting it back into the original function, f ( ) 0.9( ) 1.7( )

3 3 (b) The formula for Newton-Raphson is i1 i i.5 i i Using an initial guess of 5, the first iteration yields 0.9(5) 1.7(5).5 i (5) 1.7 a % 46.0% Second iteration: 0.9( ) 1.7( ).5 i ( ) a 100% 17.1% The process can be continued as tabulated below: iteration i f( i ) f'( i ) a % % % E % E % After 5 iterations, the root estimate is with an approimate error of %. The result can be checked by substituting it back into the original function, 14 f ( ) 0.9( ) 1.7( ) (a) >> = linspace(0,4); >> y =.^3-6*.^+11*-6.1; >> plot(,y) >> grid

4 4 Estimates are approimately 1.05, 1.9 and (b) The formula for Newton-Raphson is 3 i i i i1 i i i Using an initial guess of 3.5, the first iteration yields 3 (3.5) 6(3.5) 11(3.5) (3.5) 1(3.5) 11 a % 9.673% Second iteration: 3 ( ) 6( ) 11( ) ( ) 1( ) 11 a % 3.995% Third iteration: 3 ( ) 6( ) 11( ) ( ) 1( ) % 0.70% a (c) For the secant method, the first iteration: 1 =.5 f( 1 ) = = 3.5 f( 0 ) = 1.775

5 (.5 3.5) a 100% 9.098% Second iteration: 0 = 3.5 f( 0 ) = = f( 1 ) = a ( ) ( ) % 5.57% Third iteration: 1 = f( 1 ) = = f( ) = a ( ) ( ) % % (d) For the modified secant method, the first iteration: 0 = 3.5 f( 0 ) = = f( ) = (3.5) a 100% 9.389% Second iteration: 1 = f( 1 ) = = f( ) = ( ) a 100% 4.041% Third iteration: = f( ) = = f( + ) =

6 6 0.01( ) a 100% 0.869% (e) >> a = [ ] a = >> roots(a) ans = (a) >> = linspace(0,4); >> y = 7*sin().*ep(-)-1; >> plot(,y) >> grid The lowest positive root seems to be at approimately 0.. (b) The formula for Newton-Raphson is i1 i i i 7sin( i ) e 1 i 7 e (cos( ) sin( )) Using an initial guess of 0.3, the first iteration yields i sin(0.3) e e (cos(0.3) sin(0.3)) a % 107.8%

7 7 Second iteration: sin( ) e e (cos( ) sin( )) a 100% % Third iteration: sin( ) e e (cos( ) sin( )) a 100% 0.453% (c) For the secant method, the first iteration: 1 = 0.5 f( 1 ) = = 0.4 f( 0 ) = ( ) a % 14,78% Second iteration: 0 = 0.4 f( 0 ) = = f( 1 ) = ( ) ( ) a % 98.75% Third iteration: 1 = f( 1 ) = = f( ) = ( ) a % 1.97% (d) For the modified secant method: First iteration: 0 = 0.3 f( 0 ) = = f( ) = (0.3)

8 8 a % 108.8% Second iteration: 1 = f( 1 ) = = f( ) = ( )( ) ( ) a 100% 15.18% Third iteration: = f( ) = = f( + ) = ( )( ) ( ) a 100% 0.45% Errata: In the first printing, the problem specified five iterations. Fourth iteration: 3 = f( 3 ) = = f( ) = ( )( ) a 100% 0.001% Fifth iteration: 3 = f( 3 ) = = f( ) = ( )( ) ( ) a 100% 0.000% (a) The formula for Newton-Raphson is i i i i i i1 i 4 3 i i i i Using an initial guess of 0.585, the first iteration yields

9 9 a % % Second iteration a 100% % 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 ) a % E+09.84E % E E % E % E % % % % % % % % % % % % % % % % % E % (b) For the modified secant method: First iteration: 0 = f( 0 ) = = f( ) = (0.585) a % % Second iteration: 1 = f( 1 ) = =.3034 f( ) = ( )( ) ( )

10 10 a % % Again, the result seems to be diverging. However, the computation eventually settles down and converges on a root at = 0.. The iterations can be summarized as iteration i i + i f( i ) f( i + i ) a % % % % % % % % % % % E % Eplanation of results: The results are eplained by looking at a plot of the function. The guess of is located at a point where the function is relatively flat. Therefore, the first iteration results in a prediction of.3 for Newton-Raphson and.193 for the secant method. At these points the function is very flat and hence, the Newton-Raphson results in a very high value (90.075), 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 (%) % mait = (optional) maimum allowable iterations % output: % root = real root % if necessary, assign default values

11 11 if nargin<5, mait=50; end %if mait blank set to 50 if nargin<4, es=0.001; end %if es blank set to % Secant method iter = 0; while (1) rn = r - func(r)*(rold - r)/(func(rold) - func(r)); iter = iter + 1; if rn ~= 0, ea = abs((rn - r)/rn) * 100; end if ea <= es iter >= mait, break, end rold = r; r = rn; end root = rn; Test by solving Prob. 6.3: format long f=@() ^3-6*^+11*-6.1; secant(f,.5,3.5) ans = function root = modsec(func,r,delta,es,mait) % modsec(func,r,delta,es,mait): % uses 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, mait=50; end %if mait blank set to 50 if nargin<4, es=0.001; end %if es blank set to if nargin<3, delta=1e-5; end %if delta blank set to % Secant method iter = 0; while (1) rold = r; r = r - delta*r*func(r)/(func(r+delta*r)-func(r)); iter = iter + 1; if r ~= 0, ea = abs((r - rold)/r) * 100; end if ea <= es iter >= mait, break, end end root = r; Test by solving Prob. 6.3: format long f=@() ^3-6*^+11*-6.1; modsec(f,3.5,0.0) ans = The equation to be differentiated is

12 1 gm f ( m) tanh c d gc d t v m Note that d tanh u sech u d u d d Therefore, the derivative can be evaluated as df ( m) gm gc d 1 m cdg gc d 1 cd g sech t t tanh t dm c d m cdg m m gm cd The two terms can be reordered df ( m) 1 cd g gc d 1 gm m cdg gc d tanh t t sech dm gm c t d m cd cdg m m The terms premultiplying the tanh and sech can be simplified to yield the final result df ( m) 1 g gc d g gc tanh t t sech dm mc d m m m d t 6.9 (a) The formula for Newton-Raphson is 3 6i 4i 0.5 i i1 i 68i 1.5i Using an initial guess of 4.5, the iterations proceed as iteration i f( i ) f'( i ) a % % % % % % % % % E % Thus, after an initial jump, the computation eventually settles down and converges on a root at = (b) Using an initial guess of 4.43, the iterations proceed as

13 13 iteration i f( i ) f'( i ) a 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 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), i1 i a i which can be epressed as i

14 14 i 1 i a/ i 6.11 (a) The formula for Newton-Raphson is i i1 i i i tanh 9 sech 9 Using an initial guess of 3., the iterations proceed as iteration i f( i ) f'( i ) a % % % Note that on the fourth iteration, the computation should go unstable. (b) The solution diverges from its real root of = 3. 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 4 3 i i i i i1 i 3 i i i Using an initial guess of 16.15, the iterations proceed as iteration i f( i ) f'( i ) a % % % % % % E % E %

15 15 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 The solution can be formulated as ( ) T 4 f T T T T A MATLAB script can be used to generate the plot and determine all the roots of this polynomial, clear,clc,clf,format long g cp=[1.95e e e e ]; T=[0:100]; cp_plot=polyval(cp,t); plot(t,cp_plot),grid =[1.95e e e e ]; roots() ans = i i The only realistic value is This value can be checked using the polyval function, >> polyval(,544.09) ans = e The solution involves determining the root of

16 16 6 f( ) MATLAB can be used to develop a plot that indicates that a root occurs in the vicinity of = f=@()./(1-).*sqrt(6./(+))-0.05; = linspace(0,.); y = f(); plot(,y),grid The fzero function can then be used to find the root format long fzero(f,0.03) ans = The coefficient, a and b, can be evaluated as >> format long >> R = 0.518;pc = 4600;Tc = 191; >> a = 0.47*R^*Tc^.5/pc a = >> b = *R*Tc/pc b = The solution, therefore, involves determining the root of 0.518(33.15) f( v) 65,000 v vv ( ) 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 = 0.518;pc = 4600;Tc = 191; a = 0.47*R^*Tc^.5/pc; b = *R*Tc/pc; T = ;p = 65000;

17 17 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(0.00,0.004); >> fv = fvol(v); >> plot(v,fv) >> grid Thus, a root is located at about The fzero function can be used to refine this estimate, >> vroot = fzero('fvol',0.008) vroot = The mass of methane contained in the tank can be computed as 3 mass V m v The function to be evaluated is 3 1rh f ( h) V r cos ( rh) rhh L r 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',0.5,[],r,l,v) h =

18 (a) The function to be evaluated is TA 500 TA f( TA) 10 cosh 10 TA 10 The solution can be obtained with the fzero function as >> format long >> TA = fzero(inline('10-/10*cosh(500/)+/10'),1000) TA = e+003 (b) A plot of the cable can be generated as >> = linspace(-50,100); >> w = 10;y0 = 5; >> y = TA/w*cosh(w*/TA) + y0 - TA/w; >> plot(,y),grid 6.18 The function to be evaluated is t f() t 9e sin( t) 3.5 A plot can be generated with MATLAB, >> t = linspace(0,); >> ft 9*ep(-t).* sin(*pi*t) 3.5; >> y=ft(t); >> plot(t,y),grid

19 19 Thus, there appear to be two roots at approimately 0.1 and 0.4. The fzero function can be used to obtain refined estimates, >> t = fzero(ft,[0 0.]) t = >> t = fzero(ft,[0. 0.8]) t = The function to be evaluated is 1 1 f( ) C 1 Z R L Substituting the parameter values yields 1 6 f ( ) The fzero function can be used to determine the root as >> fzero('0.01-sqrt(1/5065+(0.6e-6*-./).^)',[1 1000]) ans = The following script uses the fzero function can be used to determine the root as format long k1=40000;k=40;m=95;g=9.81;h=0.43; fd=@(d) *k*d^(5/)/5+0.5*k1*d^-m*g*d-m*g*h; fzero(fd,1) ans =

20 0 6.1 If the height at which the throw leaves the right fielders arm is defined as y = 0, the y at 90 m will be 0.8. Therefore, the function to be evaluated is f ( ) tan cos ( 0 /180) Note that the angle is epressed in degrees. First, MATLAB can be used to plot this function versus various angles: format long g=9.81;v0=30;y0=1.8; fth=@(th) *tan(pi*th/180)-44.1./cos(pi*th/180).^; thplot=[0:70];fplot=fth(thplot); plot(thplot,fplot),grid Roots seem to occur at about 40 o and 50 o. These estimates can be refined with the fzero function, theta1 = fzero(fth,[30 45]) theta = fzero(fth,[45 60]) with the results theta1 = theta = Therefore, two angles result in the desired outcome. We can develop plots of the two trajectories: yth=@(,th) tan(th*pi/180)*-g//v0^/cos(th*pi/180)^*.^+y0; plot=[0:dist];yplot=yth(plot,theta1);yplot=yth(plot,theta); plot(plot,yplot,plot,yplot,'--') grid;legend(['angle=' numstr(theta1)],['angle=' numstr(theta)])

21 angle= angle= Note that the lower angle would probably be preferred as the ball would arrive at the catcher sooner. 6. The equation to be solved is 3 f ( h) Rh h V 3 Because this equation is easy to differentiate, the Newton-Raphson is the best choice to achieve results efficiently. It can be formulated as 3 Ri i V 3 i1 i Ri i or substituting the parameter values, 3 (3) i i 30 3 i1 i (3) i i The iterations can be summarized as iteration i f( i ) f'( i ) a % % E % Thus, after only three iterations, the root is determined to be with an approimate relative error of 0.007%. 6.3 >> format short g >> r = [ ]; >> a = poly(r) a = >> polyval(a,1) ans = -1890

22 >> b = poly([- -5]) b = >> [q,r] = deconv(a,b) q = r = >> = roots(q) = >> a = conv(q,b) a = >> = roots(a) = >> a = poly() a = >> a = [ ]; >> r = roots(a) r = >> a = [ ]; >> r = roots(a) r = Therefore, the transfer function is ( s4)( s3)( s) Gs () ( s6)( s5)( s3)( s1) 6.5 The equation can be rearranged so it can be solved with fied-point iteration as H 3/5 /5 ( Qn) ( B Hi ) i1 3/10 BS Substituting the parameters gives, Hi (0 Hi) /5

23 3 This formula can be applied iteratively to solve for H. For eample, using and initial guess of H 0 = 0, the first iteration gives /5 H (0 (0)) Subsequent iterations yield i H a % % % % Thus, the process converges on a depth of We can prove that the scheme converges for all initial guesses greater than or equal to zero by differentiating the equation to give g ' H 3/5 This function will always be less than one for H 0. For eample, if H = 0, g = Because H is in the denominator, all values greater than zero yield even smaller values. Thus, the convergence criterion that g < 1 always holds. 6.6 This problem can be solved in a number of ways. One approach involves using the modified secant method. This approach is feasible because the Swamee-Jain equation provides a sufficiently good initial guess that the method is always convergent for the specified parameter bounds. The following functions implement the approach: function ffact = prob066(ed,ren) % prob066: friction factor with Colebrook equation % ffact = prob066(ed,ren): % uses modified secant equation to determine the friction factor % with the Colebrook equation % input: % ed = e/d % ReN = Reynolds number % output: % ffact = friction factor mait=100;es=1e-8;delta=1e-5; iter = 0; % Swamee-Jain equation: r = 1.35 / (log(ed / / ReN ^ 0.9)) ^ ; % modified secant method while (1) rold = r; r = r - delta*r*func(r,ed,ren)/(func(r+delta*r,ed,ren)... -func(r,ed,ren)); iter = iter + 1; if r ~= 0, ea = abs((r - rold)/r) * 100; end if ea <= es iter >= mait, break, end end ffact = r; function ff=func(f,ed,ren) ff = 1/sqrt(f) + *log10(ed/ /ReN/sqrt(f));

24 4 Here are implementations for the etremes of the parameter range: >> prob066( ,4000) ans = >> prob066(0.05,4000) ans = >> prob066( ,1e7) ans = >> prob066(0.05,1e7) ans = The Newton-Raphson method can be set up as 0.5i e i1 i 0.5i (4 i ) e (3 0.5 ) (a) i f() f'() a % % % E % E % % i (b) The case does not work because the derivative is zero at 0 = 6. (c) i f() f'() E-5 7.1E+4-0 This guess breaks down because, as depicted in the following plot, the near zero, positive slope sends the method away from the root The optimization problem involves determining the root of the derivative of the function. The derivative is the following function, 5 3 f '( )

25 5 The Newton-Raphson method is a good choice for this problem because The function is easy to differentiate It converges very rapidly The Newton-Raphson method can be set up as 5 3 i i i1 i 4 60i 18i First iteration: (1) 6(1) (1) 18(1) a % 11.43% Second iteration: 5 3 1( ) 6( ) ( ) 18( ) a %.84% Since a < 5%, the solution can be terminated. 6.9 Newton-Raphson is the best choice because: You know that the solution will converge. Thus, divergence is not an issue. Newton-Raphson is generally considered the fastest method You only require one guess The function is easily differentiable To set up the Newton-Raphson first formulate the function as a roots problem and then differentiate it 0.5 f ( ) e f '( ) 0.5e 5 These can be substituted into the Newton-Raphson formula 0.5i e 55i 0.5e 5 i1 i 0.5i First iteration: 0.5(0.7) e 5 5(0.7) (0.7) 0.5e a 100% 1.98%

26 6 Therefore, only one iteration is required (a) function [b,fb] = fzeronew(f,l,u,varargin) % fzeronew: Brent root location zeroes % [b,fb] = fzeronew(f,l,u,p1,p,...): % uses Brent s method to find the root of f % input: % f = name of function % l, u = lower and upper guesses % p1,p,... = additional parameters used by f % output: % b = real root % fb = function value at root if nargin<3,error('at least 3 input arguments required'),end a = l; b = u; fa = f(a,varargin{:}); fb = f(b,varargin{:}); c = a; fc = fa; d = b - c; e = d; while (1) if fb == 0, break, end if sign(fa) == sign(fb) %If necessary, rearrange points a = c; fa = fc; d = b - c; e = d; end if abs(fa) < abs(fb) c = b; b = a; a = c; fc = fb; fb = fa; fa = fc; end m = 0.5 * (a - b); %Termination test and possible eit tol = * eps * ma(abs(b), 1); if abs(m) <= tol fb == 0. break end %Choose open methods or bisection if abs(e) >= tol & abs(fc) > abs(fb) s = fb / fc; if a == c %Secant method p = * m * s; q = 1 - s; else %Inverse quadratic interpolation q = fc / fa; r = fb / fa; p = s * ( * m * q * (q - r) - (b - c) * (r - 1)); q = (q - 1) * (r - 1) * (s - 1); end if p > 0, q = -q; else p = -p; end; if * p < 3 * m * q - abs(tol * q) & p < abs(0.5 * e * q) e = d; d = p / q; else d = m; e = m; end else %Bisection d = m; e = m; end c = b; fc = fb; if abs(d) > tol, b = b + d; else b = b - sign(b - a) * tol; end fb = f(b,varargin{:}); end (b) >> [,f] = fzeronew(@(,n) ^n-1,0,1.3,10) = 1 f = 0