Therefore, the root is in the second interval and the lower guess is redefined as x l = The second iteration is

Transcription

1 1 CHAPTER The function to evaluate is gm gc d f ( cd ) tanh t v( t) c d m or substituting the given values 9.81(65) 9.81 ( ) tanh cd f c 4.5 d 35 c 65 d The first iteration is f (.) f (.5) ( ) Therefore, the root is in the second interval and the lower guess is redefined as x l.5. The second iteration is ε a % 9.9%.75 f (.5) f (.75) ( ).633 Therefore, the root is in the first interval and the upper guess is redefined as x u.75. The remainder of the iterations are displayed in the following table: i x l f(x l ) x u f(x u ) f( ) ε a % % % % Thus, after five iterations, we obtain a root estimate of with an approximate error of 1.15%. 5. function root bisectnew(func,xl,xu,ead)

2 % bisectnew(xl,xu,es,maxit): % uses bisection method to find the root of a function % with a fixed number of iterations to attain % a prespecified tolerance % input: % func name of function % xl, xu lower and upper guesses % Ead (optional) desired tolerance (default.1) % output: % root real root if func(xl)*func(xu)> %if guesses do not bracket a sign change error('no bracket') %display an error message and terminate % if necessary, assign default values if nargin<4, Ead.1; %if Ead blank set to.1 % bisection xr xl; % compute n and round up to next highest integer n round(1 + log((xu - xl)/ead) +.5); for i 1:n xrold xr; xr (xl + xu)/; if xr ~, ea abs((xr - xrold)/xr) * 1; test func(xl)*func(xr); if test < xu xr; elseif test > xl xr; else ea ; root xr; The following is a MATLAB session that uses the function to solve Prob. 5.1 with E a,d.1. >> format long >> fcd@(cd) sqrt(9.81*65/cd)*tanh(sqrt(9.81*cd/65)*4.5)-35; >> bisectnew(fcd,.,.3,.1) ans The function to evaluate is 9.81(65) 9.81 ( ) tanh cd f c 4.5 d 35 c 65 d The first iteration is

3 (..3) ( ) f (.) f (.71833) ( ).946 Therefore, the root is in the first interval and the upper guess is redefined as x u The second iteration is (..7183) ( ) ε a %.66%.7457 Therefore, after only two iterations we obtain a root estimate of.7457 with an approximate error of.66% which is below the stopping criterion of %. 5.4 function root falsepos(func,xl,xu,es,maxit) % falsepos(func,xl,xu,es,maxit): % uses the false position method to find the root % of the function func % input: % func name of function % xl, xu lower and upper guesses % es (optional) stopping criterion (%) (default.1) % maxit (optional) maximum allowable iterations (default 5) % output: % root real root if func(xl)*func(xu)> %if guesses do not bracket a sign change error('no bracket') %display an error message and terminate % default values if nargin<5, maxit5; if nargin<4, es.1; % false position iter ; xr xl; while (1) xrold xr; xr xu - func(xu)*(xl - xu)/(func(xl) - func(xu)); iter iter + 1; if xr ~, ea abs((xr - xrold)/xr) * 1; test func(xl)*func(xr); if test < xu xr; elseif test > xl xr; else ea ;

4 4 if ea < es iter > maxit, break, root xr; The following is a MATLAB session that uses the function to solve Prob. 5.1: >> format long >> fcd@(cd) sqrt(9.81*65/cd)*tanh(sqrt(9.81*cd/65)*4.5)-35; >> falsepos(fcd,.,.3,) ans Solve for the reactions: R 1 65 lbs. R 85 lbs. Write beam equations: <x<3 3<x<6 6<x<1 1<x<1 x M + (16.667x ) 65x 3 (1) M 65x x x 3 M + 1( x 3)( ) + 15( x () M 5x + 415x 15 3 (3)) 3 M 15( x (3)) + 3( x 4.5) 65x 3 (3) M 185x M + 1(1 x) (4) M 1x 1 65x A plot of these equations can be generated: <x<3 3<x<6 6<x<1 1<x<1 Combining Equations: Because the curve crosses the axis between 6 and 1, use (3).

5 5 (3) M 185 x Set x L 6 ; x 1 U L U ) 54 ) ) 17 x r replaces x R x L L + x U 8 L U ) x 9 ) r ) 15 replaces R x U L U ) x 8.5 ) 15 r ) replaces R x L L U ) x 8.75 ) 15 r ) replaces R x L L U ) x ) 15 r ) replaces R x L L U ) x ) 15 r ) replaces R x U L U ) 8.15 ) ) replaces R x L L U ) ) ) replaces R x U ) L x U ) r M ( ).8984 Therefore, x feet x R 5.6 (a) The graph can be generated with MATLAB >> x[-1:.1:6]; >> f-14-*x+19*x.^-3*x.^3;

6 6 >> plot(x,f) >> grid This plot indicates that roots are located at about.5, and 4.7. (b) Using bisection, the first iteration is f ( 1) f (.5) 8(1.15) 31.5 Therefore, the root is in the second interval and the lower guess is redefined as x l.5. The second iteration is ε a.5 (.5) 1% 1%.5 f (.5) f (.5) 1.15( ) Therefore, the root is in the first interval and the upper guess is redefined as x u.5. The remainder of the iterations are displayed in the following table: i x l x u f(x l ) f( ) f(x l ) f( ) ε a % % %

7 % % % % % Thus, after eight iterations, we obtain a root estimate of.4766 with an approximate error of.83%, which is below the stopping criterion of 1%. (c) Using false position, the first iteration is 14( 1 ) ( 14) f ( 1) f (.33333) 8( ) Therefore, the root is in the first interval and the upper guess is redefined as x u The second iteration is ( 1 (.33333)) ( ) ε a.4364 (.33333) 1% 3.59%.4364 f ( 1) f (.4364) 8( 1.418) Therefore, the root is in the first interval and the upper guess is redefined as x u The remainder of the iterations are displayed in the following table: i x l f(x l ) x u f(x u ) f( ) f(x l ) f( ) ε a % % % % Therefore, after five iterations we obtain a root estimate of with an approximate error of.357%, which is below the stopping criterion of 1%. 5.7 A graph of the function can be generated with MATLAB >> x[-.5:.1:1.5]; >> fsin(x)-x.^3; >> plot(x,f);grid This plot indicates that a nontrivial root (i.e., nonzero) is located at about.9.

8 8 Using bisection, the first iteration is f (.5) f (.75).35446( ).967 Therefore, the root is in the second interval and the lower guess is redefined as x l.75. The second iteration is ε a 1% 14.9%.875 f (.75) f (.875).59764(.97616).5359 Because the product is positive, the root is in the second interval and the lower guess is redefined as x l.875. All the iterations are displayed in the following table: i x l f(x l ) x u f(x u ) f( ) ε a % % % % Consequently, after five iterations we obtain a root estimate of with an approximate error of 1.69%, which is below the stopping criterion of %. The result can be checked by substituting it into the original equation to verify that it is close to zero.

9 9 f ( x) sin( x) x 3 sin(.91875) (a) A graph of the function indicates a positive real root at approximately x (b) Using bisection, the first iteration is ε a 1% 6% +.5 f (.5) f (1.5) (.1957) Therefore, the root is in the first interval and the upper guess is redefined as x u 1.5. The second iteration is ε a 1% 4.86%.875 f (.5) f (.875) ( ) Consequently, the root is in the second interval and the lower guess is redefined as x l.875. All the iterations are displayed in the following table: i x l x u f(x l ) f( ) f(x l ) f( ) ε a % % Thus, after three iterations, we obtain a root estimate of 1.65 with an approximate error of 17.65%. (c) Using false position, the first iteration is.759(.5 ) f (.5) f ( ) (.75678).6797 Therefore, the root is in the first interval and the upper guess is redefined as x u The second iteration is

10 1 ε a.75678( ) % 13.% f (.5) f (1.717) (.67).931 Consequently, the root is in the first interval and the upper guess is redefined as x u All the iterations are displayed in the following table: iteration x l x u f(x l ) f(x u ) f( ) f(x l ) f( ) ε a % % After three iterations we obtain a root estimate of with an approximate error of 4.414%. 5.9 (a) Equation (5.6) can be used to determine the number of iterations 4 log Δx n log E a, d which can be rounded up to 1 iterations. (b) Here is an M-file that evaluates the temperature in o C using 1 iterations of bisection based on a given value of the oxygen saturation concentration in freshwater: function TC TempEval(osf) % function to evaluate the temperature in degrees C based % on the oxygen saturation concentration in freshwater (osf). xl ; xu ; if fta(xl,osf)*fta(xu,osf)> %if guesses do not bracket error('no bracket') %display an error message and terminate xr xl; for i 1:1 xrold xr; xr (xl + xu)/; if xr ~, ea abs((xr - xrold)/xr) * 1; test fta(xl,osf)*fta(xr,osf); if test < xu xr; elseif test > xl xr; else ea ;

11 11 TC xr ; function f fta(ta, osf) f e5/Ta e7/Ta^; f f e1/Ta^ e11/Ta^4; f f - log(osf); The function can be used to evaluate the test cases: >> TempEval(8) ans >> TempEval(1) ans >> TempEval(1) ans Note that these values can be compared with the true values to verify that the errors are less than.5: o sf Approximation Exact Error (a) The function to be evaluated is 4 f ( y) 1 (3 + y) (3 y + y / ) A graph of the function indicates a positive real root at approximately (b) Using bisection, the first iteration is

12 f (.5) f (1.5) 3.58(.3946) Therefore, the root is in the second interval and the lower guess is redefined as x l 1.5. The second iteration is ε a 1% 5% f ( 1.5) f ().3946(.6189).1864 Therefore, the root is in the first interval and the upper guess is redefined as x u. All the iterations are displayed in the following table: i x l f(x l ) x u f(x u ) f( ) ε a % % % % % % % After eight iterations, we obtain a root estimate of with an approximate error of.5%. (c) Using false position, the first iteration is.8133(.5.5) f (.5) f (.4583) 3.581(.79987) Therefore, the root is in the first interval and the upper guess is redefined as x u The second iteration is.79987( ) ε a 1% 1.96% f (.5) f (.4363) 3.58(.7861) The root is in the first interval and the upper guess is redefined as x u All the iterations are displayed in the following table:

13 13 i x l f(x l ) x u f(x u ) f( ) ε a % % % % % % % % % After ten iterations we obtain a root estimate of.977 with an approximate error of 1.59%. 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 1.5 that we detected graphically. Discussion: This is a classic example of a case where false position performs poorly and is inferior to bisection. Insight into these results can be gained by examining the plot that was developed in part (a). This function violates the premise upon which false position was based that is, if f(x u ) is much closer to zero than f(x l ), then the root is closer to x u than to x l (recall Figs. 5.8 and 5.9). Because of the shape of the present function, the opposite is true Errata: In the first printing, the solution to the differential equation should have been: S S vm t + k s ln( S / S) Subsequent printings should show the correct solution. The problem amounts to determining the root of f ( m s S) S v t + k ln( S / S) S The following script calls the bisect function (Fig. 5.7) for various values of t in order to generate the solution. S1;vm.5;ks; S - vm * t + ks * log(s / S) - S; t:5;s:5; for i 1:n S(i)bisect(f,.1,1.1,1e-6,1,t(i)); plot(t,s)

14 The function to be solved is (4 + x) f ( x).16 (4 x) (8 x) (a) A plot of the function indicates a root at about x (b) The shape of the function indicates that false position would be a poor choice (recall Fig. 5.9). Bisection with initial guesses of and can be used to determine a root of after 8 iterations with ε a.493%. Note that false position would have required 68 iterations to attain comparable accuracy. i x l x u f(x l ) f( ) f(x l ) f( ) ε a % x % % x % x1-5 4.% x % x1-6.99% x %

15 This problem can be solved by determining the root of the derivative of the elastic curve dy dx w 1EIL 4 4 ( 5x + 6L x ) L Therefore, after substituting the parameter values, we must determine the root of 4 f ( x) 5x +,16,x A plot of the function indicates a root at about x 7. E+11 1E+11-1E+11 -E Bisection can be used to determine the root. Here are the first few iterations: i x l x u f(x l ) f( ) f(x l ) f( ) ε a x x x x x x % x x x1.% x x x % x x1 9.95x % After iterations, the root is determined as x This value can be substituted into Eq. (P5.13) to compute the maximum deflection as y.5 1(5,)3,(6) ( (68.38) 5.14 The solution can be formulated as i(1 + i) f ( i) 5, 5,5 6 (1 + i) 1 6 A plot of this function suggests a root at about.86: + 7,(68.38) (68.38)).51519

16 A numerical method can be used to determine that the root is (a) The solution can be formulated as f ( t) 1. 75,.45t 3, ( e + 1,).8t 1 + 9e A plot of this function suggests a root at about 4: (b) The false-position method can be implemented with the results summarized as i t l t u f(t l ) f(t u ) t r f(t r ) f(t l ) f(t r ) ε a x x % x % x1 4.9% x1-3.1% (c) The modified secant method (with δ.1) can be implemented with the results summarized as i t i f(t i ) δt i t i +δt i f(t i +δt i ) f (t i ) ε a % % %

17 17 For both parts (b) and (c), the root is determined to be t At this time, the ratio of the suburban to the urban population is 135,14.5/11, The solution can be formulated as f ( N) q N + N + 4 where n i ρ μ 1 μ Substituting this value along with the other parameters gives f ( N) N + N A plot of this function indicates a root at about N E+6 4.E+6.E+6.E+.E+ -.E+6 1.E+1.E+1-4.E+6 (b) The bisection method can be implemented with the results for the first 5 iterations summarized as i N l N u N r f(n l ) f(n r ) f(n l ) f(n r ) ε a 1 5.x x1 1 1.x1 1.3x x1 5-7x x1 9 1.x x1 9.3x x x % 3 7.5x1 9 1.x x x1 5.6x x % x1 9 1.x x1 9.6x x x % x x x1 9.6x x x % After 15 iterations, the root is with a relative error of.3%. (c) The modified secant method (with δ.1) can be implemented with the results summarized as i N i f(n i ) δn i N i +δn i f(n i +δn i ) f (N i ) ε a

18 18 9.x x1 4 9.x x x x x1 9.6x x x % 9.8x x1 9.8x x x % 3 9.8x x1-9.8x x x % 5.17 Using the given values, the roots problem to be solved is f ( x) ( x x +.81) 3 / A plot indicates roots at about.3 and A numerical method can be used to determine that the roots are.9537 and The solution can be formulated as ( Re f ) f ( f ) 4 log1.4 1 f We want our program to work for Reynolds numbers between,5 and 1,,. Therefore, we must determine the friction factors corresponding to these limits. This can be done with any root location method to yield.1155 and.913. Therefore, we can set our initial guesses as x l.8 and x u.1. Equation (5.6) can be used to determine the number of bisection iterations required to attain an absolute error less than.5,.1.8 log Δx n log E a, d which can be rounded up to 11 iterations. Here is a MATLAB function that is set up to implement 11 iterations of bisection to solve the problem. function ffanning(func,xl,xu,varargin) test func(xl,varargin{:})*func(xu,varargin{:}); if test>,error('no sign change'), for i 1:11 xr (xl + xu)/; test func(xl,varargin{:})*func(xr,varargin{:}); if test <

19 19 xu xr; elseif test > xl xr; else break fxr; This can be implemented in MATLAB. For example, >> 4*log1(Re*sqrt(f))-.4-1/sqrt(f); >> format long >> ffanning(vk,.8,.1,5) f Here are additional results for a number of values within the desired range. We have included the true value and the resulting error to verify that the results are within the desired error criterion of E a < Re Root Truth E t x x x x x x x The solution can be formulated as 4 f ( T ) T T T T MATLAB can be used to determine all the roots of this polynomial, >> format long >> x[1.95e e e e ]; >> roots(x) ans 1.e+3 * i i The only realistic value is This value can be checked using the polyval function,

20 >> polyval(x, ) ans e The solution can be formulated as m f ( t) u ln gt v m qt Substituting the parameter values gives 15, f ( t), ln 9.81t 75 15,,7t A plot of this function indicates a root at about t Because two initial guesses are given, a bracketing method like bisection can be used to determine the root, i t l t u t r f(t l ) f(t r ) f(t l ) f(t r ) ε a % % % % % % % Thus, after 8 iterations, the approximate error falls below 1% with a result of t Note that if the computation is continued, the root can be determined as