Cs 201 Final 2010 Solution

8/10/2019 Cs 201 Final 2010 Solution

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Solution to CSCI E-201 Numerical Analysis Final

Name:

November 10, 2012

1) (25 points) Gram-Schmidt

Use the Gram-Schmidt algorithm or Housholder Reflectors to derive a QR decomposition ofthe following matrix. The first two columns are orthogonal.

2 1 02 0 32

1 1

=

13

1 013

0 313

1 1

2

3 0 00 1 00 0 1

=

13

12

013

0 313

12

1

2

3 0 0

0

2 00 0 1

=

13

12 1

613

0 26

13 1

2 1

6

2

3 0 43

0

2 12

0 0 56

=

0.5774 0.7071 0.40820.5774 0.0000 0.8165

0.5774 0.7071 0.4082

3.4641 0 2.30940 1.4142 0.707

0 0 2.0412

The hard work is in reworking the last column. In slow motion:

03

1

13

13

13

03

1

131313

12

0 12

03

1

12

0

12

=

03

1

4

3

11

1

+1

2

10

1

=5

6

2

2) (30 points) Numeric Integration

We wish to devise an integration method for the interval [1, 1] that samples the functionf(x) which we wish to evaluate at the points x0 = 12 , x1 = 0, x2= 12 .

Our approximation will be

1

1

f(x)dx

A0f(x0) +A1f(x1) +A2f(x2) =A0f(

1

2

) +A1f(0) +A2f(1

2

)

Use the method of undetermined coefficients to find values A0, A1, A2, so that the ap-proximation is exact for functions f(x) that are constant, linear, or quadratic.

(a) Equations Provide the system of equations for{Ai}.

A0+A1+A2 =

1

1dx= 2

1


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Solution to CSCI E-201 Numerical Analysis Final Name:

A0x0+A1x1+A2x2 =

1

1xdx= 0

A0x2

0+A1x2

1+A2x2

2 =

1

1x2dx=

2

3

(b) Solution Solve the system of equations and find{Ai}.We take the second and third equations, and see that

(A0 A2) = 01

4(A0+A2) =

2

3

A0 = A2=4

3

A1= 2 2 43

= 23

(c) Degree of Precision What is the degree of precision of this method?This works up to cubics, as A0 and A2 will cancel.

A0x3

0+A1x3

1+A2x3

2=

1

1x3dx= 0

However, it does not work for x4, as we see below:

1

12=

1

16(A0+A2) =A0x

4

0+A1x4

1+A2x4

2=

1

1x4dx=

2

5

(d) Interpolation Sketch (draw freehand) the three components of the Lagrange Inter-polating polynomials that can be used to determine the weights {Ai}. For example, the firstpolynomial runs through the points{(1/2, 1), (0, 0), (1/2, 0)}.

The key insight is that A1 is parabola opening downward with a negative area. If wemove x0 andx2 out, we will get a parabola with area of 0. I have uploaded an image of theareas to the Exam page.

(e) Revising If we keep x0 =x2 but modify x2 and repeat this exercise, we can finda value ofx2 where the weight A1 changes sign. What is the value ofx2 when A1 = 0?

If we want A1 = 0, thenA0 = A2= 1 andA0x2

0= 1

3 , so x0 = 13 .

(f) Degree of Precision What is the degree of precision of the method using these newvalues ofx0 andx2 that make A1= 0? Is this a method we have seen before?

This is the two point Gaussian Integration we studied, which also has precision 3.

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3) (10 points) Runga-Kutta

The Butcher tableau below describes a third-order Runga-Kutta method.

Write down MATLAB code or Pseudo Code that could be used to take h, f(t, y) and

(ti, wi), and compute (ti+1, wi+1) using the method described by the tableau.

01/2 1/2

1 -1 21/6 4/6 1/6

k1 = f(ti, wi)

k2= f(ti+h

2, wi+

h

2k1)

k3= f(ti+h, wi+h(2k2 k1))

wi+1= wi+1

6(k1+ 4k2+k3)

4) (35 points) Initial Value Problem

Given the differential equation y = 15(2 y), and initial condition y(0) = 1, or w0 = 1,compute an estimate of (t, y) (0.1, w1) after one step with h = 0.1 using each of thefollowing methods. Show your work.

(e) Analytic Solution Find closed form solution to this problem.

y

(2 y) = 15 dy

y 2=

15 dt

ln(y 2) = 15t+Ky 2 =Ce15t

Whent= 0,y= 1, so C= 1y= 2 e15t

Whent= 0.1, we havey(0.1) = 2 e1.5 = 1.777

Note the natural solution y(t) = 2. Compare to a) below.

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(a) Euler method

wi+1= 1 +hf(t, y) = 1 + 0.1 15(2 1) = 2.5 We have overshot the solution y(t) = 2.

Error is 2.5

1.777 = 0.7231

(b) Backwards Euler method

wi+1= 1 +hf(t, wi+1) = 1 + 1.5(2 wi+1) = (1 + 3) 1.5wi+1= 4 1.5wi+1

wi+1= 4

2.5= 8

5= 1.6

Error is0.1769

(c) Trapezoid Method

wi+1= wi+ h

2(f(ti, wi) +f(ti+h, wi+f(ti, wi))) =wi+

h

2(f(0, 1) +f(0.1, 2.5))

wi+1= 1 + h

2(15 15

2) = 1 +h 15

4 = 1.375

Error = 1.375 1.777 = 0.4018

(d) Classic 4-th order Runge-Kutta

Enter your intermediate computations for the Runge-Kutta solution in the table below.

k1 f(ti, wi) f(1) 15

k2 f(ti+ h

2 , wi+ h

2k1) f(1.75) 3.75k3 f(ti+

h

2, wi+

h

2k2) f(1.1875) 12.18750

k4 f(ti+h, wi+hk3) f(2.218750) -3.28125w1 w0+

h

6(k1+k2+k3+k4) 1 + (43.59375)/60 1.7265625

Error is 1.7265 1.7768 = 0.0503

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Cs 201 Final 2010 Solution