HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

1 of 22

hw02soln.docx

Problem: (a) determine the roots of ( ) 32 3192013 xxxxf += graphically. In addition, determine the 1st root of the function using (b) the Bisection Method, (c) the False Position Method. For (b) & (c), assume 1=lx and 0=ux and use a stopping criteria of 1%.

Solution:

The following interactive Matlab session results in the plot shown:

>> x=linspace(-2,5); >> f=-3.*x.^3+19.*x.^2-20.*x-13; >> plot(x,f),grid

Using the data curser tool, the apparent roots are: 44.0x , 0.2x , 7.4x .

-2 -1 0 1 2 3 4 5-20

0

20

40

60

80

100

120

140

X: -0.4444Y: -0.09465

Ans (a)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

2 of 22

hw02soln.docx

Using the Bisection Method:

Ill use Matlab in interactive mode to do the calculations and show/explain the results here.

>> f=inline('-3*x^3+19*x^2-20*x-13','x') f = Inline function: f(x) = -3*x^3+19*x^2-20*x-13

>> xl=-1; xu=0; >> f(xl) ans = 29 >> f(xu) ans = -13

A sign change occurs so a root exists between these values. Bisect the range.

>> xr=(xl+xu)/2 xr = -0.5000

>> f(xl)*f(xr) ans = 61.6250

This is positive, so the root is in the upper interval Therefore, xr becomes xl for the next iteration.

>> xl=xr; >> xr=(xl+xu)/2 xr =

-0.2500

Determine the approximate error.

>> ea=abs((xr-xl)/xr)*100 ea = 100

This is greater than the stopping criteria, so we continue.

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

3 of 22

hw02soln.docx

>> f(xl)*f(xr) ans = -14.3770

This is negative, so the root is in the lower interval. xr becomes xu for the next iteration.

>> xu=xr; >> xr=(xl+xu)/2 xr = -0.3750

Determine the approximate error.

>> ea=abs((xr-xu)/xr)*100 ea = 33.3333

This is still greater than the stopping criteria of 1%, so we continue. I modified the bisection function written in class to report the summary of results as seen below:

function root = bisection_hw(func,xl,xu,es,maxit) % root = bisection_hw(func,xl,xu,es,maxit): % uses bisection method to find the root of a function % This is a modification of the generic function to report % the results as request by HW Pr. 5.4(b). % input: % func = name of function % xl, xu = lower and upper guesses % es = (optional) stopping criterion (%) % maxit = (optional) maximum allowable iterations % output: % root = real root

if func(xl)*func(xu)>0 %if guesses do not bracket a sign change disp('no bracket') %display an error message return %and terminate end % if necessary, assign default values if nargin

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

4 of 22

hw02soln.docx

disp('iter xl xu xr ea') while (1) xrold = xr; xr = (xl + xu)/2; iter = iter + 1; if xr ~= 0, ea = abs((xr - xrold)/xr) * 100; end test = func(xl)*func(xr); if test < 0 xu = xr; elseif test > 0 xl = xr; else ea = 0; end fprintf('\n%3d %8.4f %8.4f %8.4f %8.4f', iter,xl,xu,xr,ea) if ea = maxit, break, end end root = xr;

The output from this code is:

>> bisection_hw(f,-1,0,1) iter xl xu xr ea

1 -0.5000 0.0000 -0.5000 100.0000 2 -0.5000 -0.2500 -0.2500 100.0000 3 -0.5000 -0.3750 -0.3750 33.3333 4 -0.5000 -0.4375 -0.4375 14.2857 5 -0.4688 -0.4375 -0.4688 6.6667 6 -0.4531 -0.4375 -0.4531 3.4483 7 -0.4531 -0.4453 -0.4453 1.7544 8 -0.4492 -0.4453 -0.4492 0.8696 ans =

-0.4492

Note that the first three lines agree with the manual (by-hand) iterations shown above.

Finally, look at the same problem using the False Position Method.

The general form of this method is:

( )( )( ) ( )ul

uluur

xfxfxxxf

xx

=

Ans (b)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

5 of 22

hw02soln.docx

Ill again use an interactive Matlab session to perform the calculations.

>> xu=0;xl=-1; >> xr=xu-f(xu)*(xl-xu)/(f(xl)-f(xu)) xr = -0.3095 >> f(xl)*f(xr) ans = -142.1078

This is negative, so the root is in the lower interval and xr becomes xu for the next iteration.

>> xu=xr; >> xr=xu-f(xu)*(xl-xu)/(f(xl)-f(xu)) xr = -0.4093

Calculate the approximate error.

>> ea=abs((xr-xu)/xr)*100 ea = 24.3832

This is greater than the stopping criteria, so we iterate.

>> f(xl)*f(xr) ans = -41.2992

This is negative, so again, the root is in the lower interval and xr becomes xu for the next iteration.

>> xu=xr; >> xr=xu-f(xu)*(xl-xu)/(f(xl)-f(xu)) xr = -0.4370 >> ea=abs((xr-xu)/xr)*100 ea = 6.3271 >> f(xl)*f(xr) ans = -11.0776

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.4

6 of 22

hw02soln.docx

The error seems to be dropping rapidly, so Ill just continue with the hand calculations. >> xu=xr; >> xr=xu-f(xu)*(xl-xu)/(f(xl)-f(xu)) xr = -0.4443 >> ea=abs((xr-xu)/xr)*100 ea = 1.6475 >> f(xl)*f(xr) ans = -2.9070

>> xu=xr; >> xr=xu-f(xu)*(xl-xu)/(f(xl)-f(xu)) xr = -0.4462 >> ea=abs((xr-xu)/xr)*100 ea = 0.4290

This is less that my 1% stopping criteria, so we stop.

The estimate of the root is -0.4462 after 5 iterations.

Ans (c)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.17

7 of 22

hw02soln.docx

Problem: You are designing a spherical water tank for a small village. The volume of liquid in the tank is computed as: ( )

332 hRhV = pi where V is in [m3] and h and R have

units of [m]. If R = 3m, to what depth must the tank be filled so that it holds 30 m3? Use three iterations of the Bisection Method to determine your answer. Use initial guesses of 0 and R and determine your approximate error after each iteration. Check your answer using the bisection function developed for Pr. 5.4.

Solution:

Plugging in the given information, the function becomes:

( )( )

( ) 303313

3330

23

2

+=

=

hhhf

hh

pipi

pi

Ill use a command session in Matlab to perform the manual calculations required for the 1st three iterations to find h. As the problem suggests, Ill start with initial guess of hl=0 & hu=3.

>> f=inline('-1/3*pi*h^3+3*pi*h^2-30','h') f = Inline function: f(h) = -1/3*pi*h^3+3*pi*h^2-30 >> hl=0;hu=3; >> hr=(hl+hu)/2 hr = 1.5000 hr1

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.17

8 of 22

hw02soln.docx

>> f(hl)*f(hr) ans = 369.8562

This is positive, so the root is in the upper interval and we set hl = hr for the next iteration.

>> hl=hr; >> hr=(hl+hu)/2 hr = 2.2500 >> ea=abs((hr-hl)/hr)*100 ea = 33.3333

This is a fairly large error so well repeat.

>> f(hl)*f(hr) ans = -71.3170

This is negative, so the root is in the lower interval and well set hu = hr for the next iteration.

>> hu=hr; >> hr=(hl+hu)/2 hr = 1.8750 >> ea=abs((hr-hl)/hr)*100 ea = 20

Still a large error. However, the problem only asked for the 1st three iterations. I ran this function to a stopping criteria of 1% using the edited function from the last problem and the table summarizing the results is shown below:

hr2

hr3

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 5.17

9 of 22

hw02soln.docx

>> bisection_hw(f,0,3,1) iter xl xu xr ea 1 1.5000 3.0000 1.5000 100.0000 2 1.5000 2.2500 2.2500 33.3333 3 1.8750 2.2500 1.8750 20.0000 4 1.8750 2.0625 2.0625 9.0909 5 1.9688 2.0625 1.9688 4.7619 6 2.0156 2.0625 2.0156 2.3256 7 2.0156 2.0391 2.0391 1.1494 8 2.0156 2.0273 2.0273 0.5780 ans =

2.0273

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.2

10 of 22

hw02soln.docx

Problem: Determine the highest real root of ( ) 57.177.112 23 += xxxxf (a) graphically, (b) using fixed point iteration (3 iterations w/x0=3), (c) The Newton-Raphson Method (3 iterations w/x0=3).

Solution:

Using Matlab, the following plot was prepared:

>> x=linspace(0,5); >> f=2.*x.^3-11.7.*x.^2+17.7.*x-5; >> plot(x,f),grid

So graphically, the highest positive root is 586.3=rx

For the Fixed Point Method, we need to solve the function so that we obtain a form ( )xgx = . There are many ways to do this. But the simplest way is to move the 17.7x term to the left of the equal sign and then divide thru by 17.7 to obtain:

7.1757.112 23 ++

=

xxx

This form of the equation is then converted to an iterative form as follows:

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-5

0

5

10

15

20

25

30

35

40

45

X: 3.586Y: 0.2433

Ans (a)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.2

11 of 22

hw02soln.docx

7.1757.112 23

1++

=+ii

ixx

x

Ill use Matlab in interactive mode to perform the iterations starting with x = 3 @ i = 0.

>> x=3; >> xn=(-2*x^3+11.7*x^2+5)/17.7 xn =

3.1808

>> ea=abs((xn-x)/xn)*100 ea = 5.6838 >> x=xn; >> xn=(-2*x^3+11.7*x^2+5)/17.7 xn =

3.3340

>> ea=abs((xn-x)/xn)*100 ea = 4.5942 >> x=xn; >> xn=(-2*x^3+11.7*x^2+5)/17.7 xn =

3.4425

>> ea=abs((xn-x)/xn)*100 ea = 3.1542

This is still more error than would normally be acceptable. However, the answer is converging on the highest root.

The Newton-Raphson Method calls for us to solve the iterative equation:

( )( )i

ii

xfxf

xx

=+ 11

For this we need to determine the derivative of our function.

( ) 57.177.112 23 += xxxxf ( ) 7.174.236 2 += xxxf

Ans (b)

Ans (b)

Ans (b)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.2

12 of 22

hw02soln.docx

An interactive Matlab session to perform the calculations shows:

>> f=inline('2*x^3-11.7*x^2+17.7*x-5','x') f = Inline function: f(x) = 2*x^3-11.7*x^2+17.7*x-5

>> df=inline('6*x^2-23.4*x+17.7','x') df = Inline function: df(x) = 6*x^2-23.4*x+17.7

>> x=3; >> xn=x-f(x)/df(x) xn = 5.1333

>> ea=abs((xn-x)/xn)*100 ea = 41.5584 >> x=xn; >> xn=x-f(x)/df(x) xn = 4.2698

>> ea=abs((xn-x)/xn)*100 ea = 20.2256 >> x=xn; >> xn=x-f(x)/df(x) xn = 3.7929

>> ea=abs((xn-x)/xn)*100 ea = 12.5712

This is still a large error. Running this problem though the newtraph function developed in class reveals the following result at a stopping criteria of 0.001 error.

>> newtraph(f,df,3) ans = 3.5632

Ans (c)

Ans (c)

Ans (c)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.2

13 of 22

hw02soln.docx

>> f(ans) ans = 1.5767e-011

This is essentially zero. Good results.

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.4

14 of 22

hw02soln.docx

Problem: Determine the real roots of ( ) 15.545.0 23 += xxxxf to within 0.01% (a) graphically, (b) using the Newton-Raphson Method.

Solution:

Using Matlab, the graphical solution is:

>> x=linspace(0,10); >> f=0.5.*x.^3-4.*x.^2+5.5.*x-1; >> plot(x,f),grid

The estimated roots are: 202.01 =rx 515.12 =rx 263.63 =rx

To use the Newton-Raphson Method, we need to make three initial guesses (one for each root). The derivative of the function is:

( ) 5.585.1 2 += xxxf

The following Matlab interactive sessions find the roots:

0 1 2 3 4 5 6 7 8 9 10-20

0

20

40

60

80

100

120

140

160

X: 6.263Y: -0.6259

Ans (a)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.4

15 of 22

hw02soln.docx

>> f=inline('0.5*x^3-4*x^2+5.5*x-1','x') f = Inline function: f(x) = 0.5*x^3-4*x^2+5.5*x-1 >> df=inline('1.5*x^2-8*x+5.5','x') df = Inline function: df(x) = 1.5*x^2-8*x+5.5

>> x=0; >> xn=x-f(x)/df(x) xn =

0.1818 >> ea=abs((xn-x)/xn)*100 ea = 100

>> x=xn; >> xn=x-f(x)/df(x) xn =

0.2134 >> ea=abs((xn-x)/xn)*100 ea = 14.7893

>> x=xn; >> xn=x-f(x)/df(x) xn =

0.2143 >> ea=abs((xn-x)/xn)*100 ea = 0.4466

>> x=xn; >> xn=x-f(x)/df(x) xn =

0.2143 >> ea=abs((xn-x)/xn)*100 ea = 4.0809e-004

(4 iterations to meet the stopping criteria.)

>> x=1.5; >> xn=x-f(x)/df(x)

Ans (b)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.4

16 of 22

hw02soln.docx

xn =

1.4800 >> ea=abs((xn-x)/xn)*100 ea = 1.3514 >> x=xn; >> xn=x-f(x)/df(x) xn =

1.4798 >> ea=abs((xn-x)/xn)*100 ea = 0.0156 >> x=xn; >> xn=x-f(x)/df(x) xn =

1.4798 >> ea=abs((xn-x)/xn)*100 ea = 2.0929e-006

>> x=6; >> xn=x-f(x)/df(x) xn =

6.3478 >> ea=abs((xn-x)/xn)*100 ea = 5.4795 >> x=xn; >> xn=x-f(x)/df(x) xn =

6.3065 >> ea=abs((xn-x)/xn)*100 ea = 0.6547 >> x=xn; >> xn=x-f(x)/df(x) xn =

6.3059 >> ea=abs((xn-x)/xn)*100 ea = 0.0101 >> x=xn; >> xn=x-f(x)/df(x) xn =

6.3059 >> ea=abs((xn-x)/xn)*100

Ans (b)

Ans (b)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.4

17 of 22

hw02soln.docx

ea = 2.3956e-006

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.30

18 of 22

hw02soln.docx

Problem: You are designing a spherical water tank for a small village. The volume of liquid in the tank is computed as: ( )

332 hRhV = pi where V is in [m3] and h and R use

units of [m]. If R = 3m, to what depth must the tank be filled so that it holds 30 m3? Use three iterations of the Newton-Raphson Method to determine your answer. An initial guess of R will always converge. Check your answer using the newtraph function developed in class.

Solution:

This is a repeat of Prob. 5.17. As we found then, the function reduces to:

( )( ) hhhf

hhhfpipi

pipi

6

30331

2

23

+=

+=

A Matlab interactive session for three iterations follows:

>> f=inline('-1/3*pi*h^3+3*pi*h^2-30','h') f = Inline function: f(h) = -1/3*pi*h^3+3*pi*h^2-30 >> df=inline('-pi*h^2+6*pi*h','h') df = Inline function: df(h) = -pi*h^2+6*pi*h

>> h=3; >> hn=h-f(h)/df(h)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 6.30

19 of 22

hw02soln.docx

hn = 2.0610 >> ea=abs((hn-h)/hn)*100 ea = 45.5581 >> h=hn; >> hn=h-f(h)/df(h) hn = 2.0270 >> ea=abs((hn-h)/hn)*100 ea = 1.6769 >> h=hn; >> hn=h-f(h)/df(h) hn = 2.0269 >> ea=abs((hn-h)/hn)*100 ea = 0.0067

This is fairly good agreement with problem 5.17. Using the newtraph function from class to the default (0.001) stopping criteria yields:

>> newtraph(f,df,3)

ans =

2.0269

Ans

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 7.15

20 of 22

hw02soln.docx

Problem: Use Matlab to determine the roots of (a) ( ) 22.647.0 23 += xxxxf (b)

( ) 34.997.213.16704.3 23 ++= xxxxf (c) ( ) 5262 234 ++= xxxxxf .

Solution:

Ill use the built-in function roots to find these roots. The following interactive session shows the results.

>> a=[0.7 -4 6.2 -2]; >> roots(a)

ans =

3.2786 2.0000 0.4357

>> a=[-3.704 16.3 -21.97 9.34]; >> roots(a)

ans =

2.2947 1.1525 0.9535

>> a=[1 -2 6 -2 5]; >> roots(a)

ans =

1.0000 + 2.0000i 1.0000 - 2.0000i -0.0000 + 1.0000i -0.0000 - 1.0000i

Ans (a)

Ans (b)

Ans (c)

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 7.17

21 of 22

hw02soln.docx

Problem: When trying to find the acidity of a solution of magnesium hydroxide in hydrochloric acid, we obtain the following equation:

( ) 405.3 23 += xxxA

where x is the hydronium ion concentration. Find the hydronium ion concentration for a saturated solution (acidity equals zero) using two different methods in Matlab.

Solution:

1st use the graphical method. The following command interface session accomplishes this:

>> x=linspace(0,5); >> A=x.^3+3.5.*x.^2-40; >> plot(x,A),grid

Results in a concentration of ~ 2.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-50

0

50

100

150

200

X: 2.576Y: 0.3098

Ans

HW #02 Due: 02/06/2013

ES 301 Name:

Solution Set 7.17

22 of 22

hw02soln.docx

For the other method, Ill use Matlabs built-in roots function.

>> A=[1 3.5 0 -40]; >> roots(A)

ans =

-3.0338 + 2.5249i -3.0338 - 2.5249i 2.5676

Note that the only real root is 2.5676. Also, the coefficients vector needed to be entered in highest order to lowest order and padded with zero for the missing x-term.

Ans