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Unformatted text preview: 18.06 Problem Set 5 Due Wednesday, 19 March 2008 at 4 pm in 2106. Problem 1: Do problem 12 from section 4.3 (pg. 217) in the book. There is a typo in part c. It should read something like this: c) Lets analyze what happens when b = (1 , 2 , 6). In this case b x = 3 and the projection onto the line is p = (3 , 3 , 3). Check that p is perpendicular to e . Also find the projection matrix P . Solution (4+3+3 points) a) We know the vectors a and b , so we can just write this down: a T a = a a = m , a T b = a b = i b i , so b x = 1 m i b i . This is the average of the b i . b) Ill continue to use b x to denote the average of the b i . The error vector is e = ( b 1 b x,b 2 b x,... ). The variance is then k e k 2 = X i ( b i b x ) 2 = X i b 2 i 2 b x X i b i + m b x 2 = X i b 2 i m b x 2 = X i ( b 2 i b x 2 ) which is the usual expression. The standard deviation is the square root of this quantity. c) We have b = (1 , 2 , 6) and p = (3 , 3 , 3). The error is e = ( 2 , 1 , 3), and p e = 0 as it should. The matrix P is P = a ( a T a ) 1 a T = 1 3 1 1 1 1 1 1 1 1 1 Problem 2: Do problem 17 from section 4.3 (pg. 217) in the book. Solution (10 points) 1 The points are ( 1 , 7), (1 , 7), and (2 , 21). So, our matrices are A = 1 1 1 1 1 2 b = 7 7 21 We find the coeffiecients ( C,D ) by setting C D = b x = ( A T A ) 1 A T b = 3 2 2 6 1 35 42 = 1 14 6 2 2 3 35 42 = 9 4 So the best fit line is b = 9+4 t . When you draw it, it will look about right; Im not going to draw it here. Problem 3: Find a function of the form f ( t ) = C sin( t )+ D cos( t ) that approximates the three points (0 , 0), ( / 2 , 2), and ( , 1). As explained in the book, the method is the same as for fitting a line using leastsquares! (See pg. 212 for a quadratic example.) The difference is that the matrix A we use will no longer have columns with entries 1 and t i but rather sin( t i ) and cos( t i ). Solution (10 points) Our matrix A will have columns given by sin( t i ) and cos( t i ), but everything else is the same. A = 1 1 1 B = 2 1 2 We find b x in the usual way. b x = ( A T A ) 1 A T b = 1 0 0 2 1 2 1 = 1 0 1 2 2 1 = 2 1 2 Our best fit equation is f ( t ) = 2 sin( t ) 1 2 cos( t ). Again, if you draw the graph, youll see that it looks about right. Problem 4: a) Show that if Q is orthogonal (i.e. Q is square with orthonormal columns) then so is Q T . Use the criteria Q T Q = I . b) If Q 1 and Q 2 are orthogonal, show that their product Q 1 Q 2 is as well. c) If Q has orthonormal columns but is not square, does Q T have the same property? What can you say about QQ T and Q T Q in this case? Give an example for a 3 by 2 matrix Q ....
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.
 Spring '08
 CHANILLO

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