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Problem:: X X X X F X X

This document contains the solutions to homework problems 5.4 and 6.2 from an engineering class. For problem 5.4, the student finds the roots of a cubic function graphically and using the bisection and false position methods. For problem 6.2, the student finds the highest real root of another function graphically and using fixed point iteration and Newton's method over 3 iterations starting at x0=3. Matlab code is provided and discussed to show the step-by-step workings.

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nik arif
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5K views22 pages

Problem:: X X X X F X X

This document contains the solutions to homework problems 5.4 and 6.2 from an engineering class. For problem 5.4, the student finds the roots of a cubic function graphically and using the bisection and false position methods. For problem 6.2, the student finds the highest real root of another function graphically and using fixed point iteration and Newton's method over 3 iterations starting at x0=3. Matlab code is provided and discussed to show the step-by-step workings.

Uploaded by

nik arif
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 22

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.4 1 of 22

Problem: (a) determine the roots of f ( x ) = 13 20 x + 19 x 2 3 x 3 graphically. In addition, determine the 1st root of the function using (b) the Bisection Method, (c) the False Position Method. For (b) & (c), assume xl = 1 and xu = 0 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

140 120 100 80 60 40 20


X: -0.4444 Y: -0.09465

0 -20 -2

-1

Using the data curser tool, the apparent roots are: x 0.44 , x 2.0 , x 4.7 .
Ans (a)

hw02soln.docx

HW #02 Due: 02/06/2013 Using the Bisection Method:

Name:

ES 301

Solution Set

5.4 2 of 22

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.

hw02soln.docx

HW #02 Due: 02/06/2013 >> f(xl)*f(xr) ans = -14.3770

Name:

ES 301

Solution Set

5.4 3 of 22

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<5, maxit=50; end %if maxit blank set to 50 if nargin<4, es=0.001; end %if es blank set to 0.001 % bisection iter = 0; xr = xl;

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.4 4 of 22

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 <= es || iter >= maxit, break, end end root = xr; The output from this code is: >> bisection_hw(f,-1,0,1) iter xl xu xr 1 -0.5000 2 -0.5000 3 -0.5000 4 -0.5000 5 -0.4688 6 -0.4531 7 -0.4531 8 -0.4492 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: 0.0000 -0.2500 -0.3750 -0.4375 -0.4375 -0.4375 -0.4453 -0.4453 -0.5000 -0.2500 -0.3750 -0.4375 -0.4688 -0.4531 -0.4453 -0.4492

ea 100.0000 100.0000 33.3333 14.2857 6.6667 3.4483 1.7544 0.8696


Ans (b)

x r = xu

f ( xu )( xl xu ) f ( x l ) f ( xu )

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.4 5 of 22

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

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.4 6 of 22

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)

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.17 7 of 22

Problem: You are designing a spherical water tank for a small village. The volume of liquid (3R h ) where V is in [m3] and h and R have in the tank is computed as: V = h 2 3 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:


30 = h 2

(3(3) h )

3 1 3 f (h ) = h + 3h 2 30 3

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

hw02soln.docx

HW #02 Due: 02/06/2013


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

Name:

ES 301

Solution Set

5.17 8 of 22

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

hr2

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

hr3

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:

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

5.17 9 of 22

>> 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

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

6.2 10 of 22

Problem: Determine the highest real root of f ( x ) = 2 x 3 11.7 x 2 + 17.7 x 5 (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
45 40 35 30 25 20 15 10 5 0 -5
X: 3.586 Y: 0.2433

0.5

1.5

2.5

3.5

4.5

So graphically, the highest positive root is x r = 3.586

Ans (a)

For the Fixed Point Method, we need to solve the function so that we obtain a form x = g (x ) . 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:
x= 2 x 3 + 11.7 x 2 + 5 17.7

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

hw02soln.docx

HW #02 Due: 02/06/2013


xi +1 = 2 xi3 + 11.7 xi2 + 5 17.7

Name:

ES 301

Solution Set

6.2 11 of 22

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

Ans (b)

Ans (b)

Ans (b)

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:

xi +1 = x1

f ( xi ) f ( xi )

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


f ( x ) = 2 x 3 11.7 x 2 + 17.7 x 5 f ( x ) = 6 x 2 23.4 x + 17.7

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

6.2 12 of 22

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)

hw02soln.docx

HW #02 Due: 02/06/2013 >> f(ans) ans = 1.5767e-011

Name:

ES 301

Solution Set

6.2 13 of 22

This is essentially zero. Good results.

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

6.4 14 of 22

Problem: Determine the real roots of f ( x ) = 0.5 x 3 4 x 2 + 5.5 x 1 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

160 140 120 100 80 60 40 20 0 -20

X: 6.263 Y: -0.6259

10

The estimated roots are: x r1 = 0.202

x r 2 = 1.515

x r 3 = 6.263

Ans (a)

To use the Newton-Raphson Method, we need to make three initial guesses (one for each root). The derivative of the function is:
f ( x ) = 1.5 x 2 8 x + 5.5

The following Matlab interactive sessions find the roots:

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

6.4 15 of 22

>> 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

Ans (b)

(4 iterations to meet the stopping criteria.)


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

hw02soln.docx

HW #02 Due: 02/06/2013


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

Name:

ES 301

Solution Set

6.4 16 of 22

Ans (b)

Ans (b)

hw02soln.docx

HW #02 Due: 02/06/2013


ea = 2.3956e-006

Name:

ES 301

Solution Set

6.4 17 of 22

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

6.30 18 of 22

Problem: You are designing a spherical water tank for a small village. The volume of liquid (3R h ) where V is in [m3] and h and R use in the tank is computed as: V = h 2 3 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:

1 f (h ) = h 3 + 3h 2 30 3 f (h ) = h 2 + 6h
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)
hw02soln.docx

HW #02 Due: 02/06/2013


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

Name:

ES 301

Solution Set

6.30 19 of 22

Ans

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

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

7.15 20 of 22

Problem: Use Matlab to determine the roots of (a)

f ( x ) = 0.7 x 3 4 x 2 + 6.2 x 2 (b)

f ( x ) = 3.704 x 3 + 16.3 x 2 21.97 x + 9.34 (c) f ( x ) = x 4 2 x 3 + 6 x 2 2 x + 5 .

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 (b) Ans (a)

Ans (c)

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

7.17 21 of 22

Problem: When trying to find the acidity of a solution of magnesium hydroxide in hydrochloric acid, we obtain the following equation:
A( x ) = x 3 + 3.5 x 2 40

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. accomplishes this:
>> x=linspace(0,5); >> A=x.^3+3.5.*x.^2-40; >> plot(x,A),grid
200

The following command interface session

150

100

50
X: 2.576 Y: 0.3098

-50

0.5

1.5

2.5

3.5

4.5

Results in a concentration of ~ 2.5

Ans

hw02soln.docx

HW #02 Due: 02/06/2013

Name:

ES 301

Solution Set

7.17 22 of 22

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

Ans

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 xterm.

hw02soln.docx

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