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### p1038F07

Course: EE 103, Fall 2007
School: UCLA
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Fall EE103 2007 Lecture Notes (SEJ) Section 8 SECTION 8: INTRODUCTION TO ORTHOGONAL POLYNOMIALS............. 122 Orthogonal Polynomials........................................................................................... 124 Trigonometric Polynomials...................................................................................... 125 Background:...

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Fall EE103 2007 Lecture Notes (SEJ) Section 8 SECTION 8: INTRODUCTION TO ORTHOGONAL POLYNOMIALS............. 122 Orthogonal Polynomials........................................................................................... 124 Trigonometric Polynomials...................................................................................... 125 Background: .......................................................................................................... 125 Trigonometric Interpolation ................................................................................ 127 Examples of Trigonometric Interpolation:......................................................... 129 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 SECTION 8: INTRODUCTION TO ORTHOGONAL POLYNOMIALS Ex: Consider the following data for which we wish to compute a least-squares approximating polynomial. [x y] 10.0000 10.2000 10.4000 10.6000 10.8000 11.0000 0 0.0040 0.0160 0.0360 0.0640 0.1000 The X matrix, for the approximating quadratic, is X= 100.0000 104.0400 108.1600 112.3600 116.6400 121.0000 10.0000 10.2000 10.4000 10.6000 10.8000 11.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 The condition number of X T X is given by MatLab to be cond(X'*X) = 1.5598e+010. Keep in mind that the "basis polynomials" that we've been using are, for a quadratic leastsquares approximating polynomial, 2 ( x) = x2 1 ( x ) = x 0 ( x) 1 However, suppose we choose other "basis polynomials"; e.g., 2 ( x ) = ( x - 10.5) 2 - 0.1166667 1 ( x ) = x - 10.5 0 ( x) 1 Do not, at this point, be concerned with the derivation of these polynomials; however, as before, the X matrix becomes 122 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 2 ( x1 ) 1 ( x1 ) ( x ) ( x ) 2 2 1 2 X = 2 ( x6 ) 1 ( x6 ) 1 1 1 and we wish to compute c2 , c1 , c0 so that P2 ( x ) = c2 2 ( x ) + c11 ( x ) + c0 0 ( x ) is the least-squares approximating polynomial. In this case, the X matrix is 0.1333 -0.0267 -0.1067 -0.1067 -0.0267 0.1333 -0.5000 -0.3000 -0.1000 0.1000 0.3000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 and the condition number is cond(X'*X) = 100.4464, a remarkable reduction. However, this is not all: The matrix X'X = 0.0597 0 0.0000 0 0.7000 0 0.0000 0 6.0000 a diagonal matrix! Therefore, the system of equations, X T Xc = X T y can be trivially solved. Of course, we need to see why this has occurred. The functions x m , x m -1 , , x, 1 comprise a set of "basis polynomials" for the vector space of m th degree polynomials. That is, any m th degree polynomial is a linear combination of these "basis polynomials"; e.g., am x m + am-1 x m -1 + a1 x + a0 123 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 Orthogonal Polynomials Suppose we can find, or construct, another set of "basis polynomials" for the vector space of m th degree polynomials; e.g., m ( x ), m -1 ( x ), and, suppose further, that , 1 ( x ), 0 ( x ) ( x ) ( x ) = 0 k =1 i k j k n for all i , j with i j Definition: When this latter condition holds, we say the "basis polynomials" are orthogonal. [Note: The definition depends upon the x values.] HW: Show that the "basis polynomials" x 2 , x , 1 are not orthogonal with respect to the data of the above example; then show that the functions 2 ( x ) = ( x - 10.5) 2 - 0.1166667 1 ( x ) = x - 10.5 0 ( x) 1 are orthogonal. Recall the following about "orthogonal" vectors. The vectors { v1 , v 2 , orthogonal if for all i , j (i j ) v i v j = 0 Therefore, the matrix V TV is a diagonal matrix. To see this, consider the (i , j )th , i j , element of the matrix V TV . That element is the inner product of the i th row of V T with the j th column of V . But the i th row of V T is the i th column of V . But we've assumed orthogonality and, therefore, the (i , j )th , i j , element is zero. If the "basis functions" m ( x ), m -1 ( x ), data set ( xk , yk ), k = 1, , 1 ( x ), 0 ( x ) are orthogonal (with respect to the T , v k } are said to be , n ) then the X T X matrix is a diagonal matrix, where 124 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 m ( x1 ) m ( x2 ) X = m ( x3 ) ( x ) m n m -1 ( x1 ) m-2 ( x1 ) 1 ( x1 ) 0 ( x1 ) m -1 ( x2 ) m-2 ( x2 ) 1 ( x2 ) 0 ( x2 ) m -1 ( x3 ) m -2 ( x3 ) 1 ( x3 ) 0 ( x3 ) m-1 ( xn ) m -2 ( xn ) ( X T X )c = X T y 1 ( xn ) 0 ( x n ) and, hence, the solution for the coefficients, via, is simply computed. To see this, note that the definition of "orthogonal basis functions" is ( x ) k =1 i k n j ( xk ) = 0 for all i , j with i j and, further note, this is precisely the "inner product" of the i th and j th columns of X . There is a computational implementation of the Gram-Schmidt process for computing a set of orthogonal "basis polynomials", given the data ( xk , yk ), k = 1, However, we will not develop that method in these notes. , n for the problem. Rather, we introduce approximation/interpolation based upon "trigonometric polynomials", indeed, orthogonal trigonometric polynomials. This topic is at the heart of Fourier analysis and engineering applications. Trigonometric Polynomials Perhaps the most important engineering applications of polynomial interpolation and approximation involve the use of trigonometric polynomials. Background: Let f be a function of period 2 ; i.e., f ( x + 2 ) = f ( x ) , x . Fourier's great insight is that virtually any such function can be expressed as f ( x) = a0 2 + [a j cos( jx ) + bk sin( jx )] j =0 (1) Of course, cos( jx ), sin( jx ) , both have period 2 . If one is concerned with a function f of period T, then (1) can be rewritten as 125 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 f ( x) = a0 2 + [a j cos( jw0 x ) + b j sin( jw0 x )] j =0 (2) where w0 = 2 / T , the so-called "fundamental frequency". Ex: Show that cos( jw0 x ) and sin( jw0 x ) have periods equal to T . The leading coefficient, a0 2 , is generally included in various discussions merely to enable closed-form solutions for the a j , b j coefficients. Those solutions are: 1 aj = - f ( x ) cos( jx )dx , j = 0,1, 2, j = 0,1, 2, (3) bj = 1 f ( x)sin( jx)dx , - These solutions are the consequence of the fact that the "trigonometric basis functions" are orthogonal, in the following sense: cos( kx ) sin( rx )dx = 0 - cos( kx ) cos( rx )dx = 0 , k r - sin( kx ) sin( rx )dx = 0 , k r - HW: Show that typical examples of the above are true; you may use Matlab's symbolic toolbox, if you wish. Note: One could also choose the interval of integration to be [0, 2 ] , or [0, T ] or [ - T , T ] when using the fundamental frequency w0 = 2 / T . 2 2 The solutions given in (3) can be derived, although a good deal of somewhat technical mathematics is involved (of course, deriving the values of the coefficients depends upon the creativity of Fourier: i.e., equation (1) ). The simplest derivation that is for a0 . Integrating both sides of equation (1) yields - a f ( x )dx = 0 + [a j cos( jx ) + b j sin( jx )] dx - 2 j =0 = a0 2 - dx + a j cos( jx )dx + b j sin( jx )dx j =0 - j =0 - 126 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 assuming the interchanges of integration and summation are valid. summations are equal to zero and, therefore, we have a 0 = -1 - The last two f ( x )dx Note: We'd have exactly the same result if the interval of integration is [0, 2 ] . The usefulness of Equations (1) or (2) arises by utilizing a modification when given a finite sample for f (e.g., we've sampled the generally unknown function f , assumed to be of period T ) T xk = k N , k = 0,1, 2, ,N f ( xk ) , f ( x0 ) = f ( xN ) on the interval [0, T ] . Equation (2) suggests we use a finite sum f ( x ) Fm ( x ) a0 2 + [a j cos( jw0 x ) + b j sin( jw0 x )] j =0 m (4) where m should be chosen, relative to N , so that 2m + 1 N (note: b0 doesn't appear since sin(0) = 0 ; therefore m should be chosen so that N is large enough to provide enough information to determine the 2m + 1 coefficients a0 , a1 , Trigonometric Interpolation1 , am ; b1 , b2 , , bm ). Rather than use (4), we "equivalently" use in its place, dropping the leading term a0 / 2 , f ( x ) Fm ( x ) [a j =0 m j cos( jw0 x ) + b j sin( jw0 x )] (5) Assume f to be a T periodic function; we have "sampled" the function at N + 1 points, equally spaced on the interval [0, T ] , and we have the following data at hand: ( x0 , f 0 ), ( xN , f N ) , f N = f 0 ,N xk = kT / N , k = 0, 1, 1 This discussion follows that of Charles Van Loan, Introduction to Scientific Computing. 127 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 where, of course, f k = f ( xk ) . Assume that N is even and that m = N / 2 . We seek coefficients a0 , points ( x0 , f 0 ), , am , b0 , , bm so that the function in (5), Fm ( x ) , interpolates the data , am , b0 , , bm so that , ( x N , f N ) . That is, we seek coefficients a0 , f ( xk ) = Fm ( xk ) , k = 0, , N (6) As noted above, b0 does not actually appear in Fm ( x ) because sin( 0) = 0 . It's also the case that bm does not appear because sin( jw0 x N ) = sin( jw0T ) = sin( 2 j ) = 0 . Also, the first and last equations of (6) are identical, due to periodicity of f , i.e., f n = f 0 . Therefore, we delete the last equation. We now have an NxN system of linear equations to determine the coefficients of the interpolating trigonometric polynomial: f k = a0 + [a j cos( jw0 xk ) + b j sin( jw0 xk )] + am cos( mw0 xk ) , k = 0, j =1 m -1 , N -1 (7) and we see that we seek m + 1 " a " coefficients and m - 1 " b " coefficients. The a ' s are associated with the cosine function and the b ' s are associated with the sine function. Since w0 = 2 / T , m = N / 2 , and each xk = kT / N , we have that each jw0 xk = jk / m . Hence, (7) becomes f k = a0 + [a j cos( jk / m ) + b j sin( jk / m )] + am cos( k ) , k = 0, j =1 m -1 , N -1 (8) 128 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 The following, as an example, is a resulting 6x6 matrix ( N = 6, m = 3 ): 1 1 1 1 1 1 1 1 1 1 1 1 cos 0 cos 3 cos 23 cos 33 cos 43 cos 53 1.0000 0.5000 -0.5000 -1.0000 -0.5000 0.5000 cos 0 cos 23 cos 43 cos 63 cos 83 cos 10 3 1.0000 -0.5000 -0.5000 1.0000 -0.5000 -0.5000 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 sin 0 sin 3 sin 23 sin 33 sin 43 sin 53 0 0.8660 0.8660 0.0000 -0.8660 -0.8660 sin 0 a0 sin 23 a1 sin 43 a2 = sin 63 a3 sin 83 b1 sin 10 b2 3 0 0.8660 -0.8660 -0.0000 0.8660 -0.8660 f0 f1 f2 f3 f4 f5 Examples of Trigonometric Interpolation: Figure 1: function with a jump discontinuity at 1 129 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 Figure 2: 12 point trigonometric polynomial interpolation Figure 3: 24 point trigonometric polynomial interpolation 130 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 Figure 4: 48 point trigonometric polynomial interpolation Figure 5: Bell shaped function 131 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 Figure 6: 8 point trigonometric polynomial interpolation Figure 7: 16 point trigonometric polynomial interpolation 132 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 The following MatLab function, due to Charles Van Loan, produces the solution for the interpolation equations (8): the coefficients, a j ' s for the cosine functions, and the b j ' s for the sine functions, and, of course, the corresponding orthogonal matrix. function [F,P] = CSInterp(f) % F = CSInterp(f): Adopted from Charles Van Loan % f is a column n vector and n = 2m. % F.a is a column m+1 vector and F.b is a column m-1 % vector so that if % tau = (0:n-1)'*pi/m, then % % f = F.a(1)*cos(0*tau) +...+ F.a(m+1)*cos(m*tau) + % F.b(1)*sin(tau) +...+ F.b(m-1)*sin((m-1)*tau) n = length(f); m = n/2; tau = (pi/m)*(0:n-1)'; P = []; for j=0:m, P = [P cos(j*tau)]; end for j=1:m-1, P = [P sin(j*tau)]; end y = P\f; %note: this code makes no use of P'P, diagonal. F = struct('a',y(1:m+1),'b',y(m+2:n)); %F.a are the weights on the cosines; F.b for the sines. Chart 1: Matlab Function for the solution of the Trigonometric Interpolation Eqs Of course, we're generally interested in evaluating the interpolating trigonometric polynomial at points other than x0 , x1 , interpolating polynomial. , xN , and especially if we desire to plot the The following is a succinct MatLab function, also due to Charles Van Loan, that produces the values of the trigonometric interpolating polynomial at each of the points in tvals (a selection of domain values chosen by the user); the other two inputs are (1) the output "structure" F from CSInterp.m, and (2) the period T. 133 EE103 Fall 2007 Lecture Notes (SEJ) Section 8 function Fvals = CSeval(F,T,tvals) % Due to Charles Van Loan % F.a is a length m+1 column vector, F.b is a length m-1 column vector, % T is a positive scalar, and tvals is a column vector. % If % F(t) = F.a(1) + F.a(2)*cos((2*pi/T)*t) +...+ F.a(m+1)*cos((2*m*pi/T)*t) + % F.b(1)*sin((2*pi/T)*t) +...+ F.b(m-1)*sin((2*m*pi/T)*t) % % then Fvals = F(tvals). Fvals = zeros(length(tvals),1); tau = (2*pi/T)*tvals; for j=0:length(F.a)-1, Fvals = Fvals + F.a(j+1)*cos(j*tau); end for j=1:length(F.b), Fvals = Fvals + F.b(j)*sin(j*tau); end Chart 2: Matlab Function for Evaluating the Trigonometric Interpolation Polynomial at a selection of domain values, tvals. %clear all;clc; n=16; x=linspace(-1,1,n+1)'; f=1./(1+25*x.^2); [F,P]=CSinterp(f(1:n)); xd=linspace(0,2,100)'; Fvals=CSeval(F,2,xd); plot(xd-1,Fvals,'r',x(1:n),f(1:n),'bd'); hold on; f1=inline('1./(1+25*x.^2)'); fplot(f1,[-1,1]); hold on; label=strcat({'familiar (?) function 16 sample points ,'},{' f=1./(1+25*x.^2) '}); title(label);grid on; hold off; Chart 3 (Tutorial): Matlab Script for Solving for a,b coefficients, and Plotting the sampled points, the actual sampled function, and the Trig Interpolating Polynomial (note selection of xd and xd-1 (why?)) 134
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MCDB 101A Summer 2008 Exam 3 7/31/08 (100 points total)Name_Key_ Perm #_Circle the best answer for each of the multiple choice questions (3 points each) 1. Which of the following processes requires RecA? a. Formation of a cointegrant during replicative
UCSB - MCDB - 101A
MCDB 101A Summer 2008 Exam 3 7/31/08 (100 points total)Name _ Perm #_Circle the best answer for each of the multiple choice questions (3 points each) 1. Which of the following processes requires RecA? a. Formation of a cointegrant during replicative tra
UCSB - MCDB - 101A
MCDB 101A Summer 2008 Exam 2 7/21/08 (100 points total)Name_Key_ Perm #_Circle the best answer for each of the multiple choice questions (3 points each) 1. Which of the following evidence supports the Holliday model? a. Mismatches that are generated as
UCSB - MCDB - 101A
MCDB 101A Summer 2008 Exam 2 7/21/08 (100 points total)Name_ Perm #_Circle the best answer for each of the multiple choice questions (3 points each) 1. Which of the following evidence supports the Holliday model? a. Mismatches that are generated as a co
UCSB - MCDB - 101A
MCDB 101A M2008 Exam 1 Name_Key_ 7/7/08 (100 points total) Perm #_ Multiple Choice: Circle the letter corresponding to the best answer (3 points for each correct answer) 1. A mutant organism that requires the amino acid tryptophan in order to grow is know
UCSB - MCDB - 101A
MCDB 101A M2008 Exam 1 Name_ 7/7/08 (100 points total) Perm #_ Multiple Choice: Circle the letter corresponding to the best answer (3 points for each correct answer) 1. A mutant organism that requires the amino acid tryptophan in order to grow is known as
UCSB - MCDB - 101A
MCDB 101A Summer 2009 2nd Midterm 7/20/09 (100 points total)Name_Key_ Perm #_Multiple Choice: Circle the letter corresponding to the best answer (3 points for each correct answer) 1. Which is true of RecA protein? No answer was correct: 3 free points! a
UCSB - MCDB - 101A
MCDB 101A S09 Midterm 1 7/6/09 (100 points total)Name_Key_ Perm #_Multiple Choice: Circle the letter corresponding to the best answer (3 points for each correct answer) 1. If a DNA molecule consists of 20% A, 30% T and 20% G, which of the following is c
UCSB - MCDB - 101A
MCDB 101A Final Exam July 31, 2009 200 points totalName_KEY_ Perm#_Choose the best answer to each of the following multiple choice questions (3 points each) 1. The Chi sequence is GCTGGTGG. What is the probability of finding this exact series of nucleot
UCSB - MCDB - 101A
Transcriptional Regulation Regulation of Gene ExpressionTranscriptional inhibition/enhancement mRNA degradation/stabilization Translational inhibition/enhancement Protein degradation/stabilization Inhibition/stimulation of protein function12Lactose
UCSB - MCDB - 101A
mRNATranscription and Translation Colinear with gene sequence Codes for one or more proteins One DNA strand serves as coding strand for a given mRNA (has same sequence as mRNA) Other strand is template strand (has a sequence complementary to that of the
UCSB - MCDB - 101A
DNA Replication&quot;It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material&quot;-Watson and Crick1DNA Replication ModelsFig 3.12Meselson and Stahl Experiment Gro
UCSB - MCDB - 101A
DNA: The Genetic MaterialGriffith and Bacterial Transformation Streptococcus pneumoniae polysaccharide capsule = virulence factor S-strain (encapsulated)-lethal to mice R-strain-nonpathogenic12The Griffith Mixed Culture ExperimentNature of the Trans
UCSB - MCDB - 101A
Clicker QuizzesMolecular Genetics of ProkaryotesIntroduction12Prokaryotes vs. Eukaryotes What is Genetics? Griffiths &quot;.the study of genes&quot; Snyder &quot;The manipulation of DNA to study cellular and organismal functions&quot; Russell &quot;[a study of]. the biologi
American College of Gastroenterology - BUS - 500
Cash Outflow Working Cap. Oper. Exp. Overhaul Cost Savings Salvage Value Cash Flows Discount 10% PV Amount Sum of PV NPVBeg. End Yr. 1 Yr. 1 (300,000) (30,000) (60,000) 135,000 (330,000) 1 (330,000) 374322 \$44,322 75,000 0.91 68175End Yr. 2End Yr. 3En