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by: Vivien Bradtke V


Marketplace > Texas A&M University > Mathematics (M) > MATH 414 > FOURIER SERIES WAVELET
Vivien Bradtke V
Texas A&M
GPA 3.6


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This 7 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 414 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/226037/math-414-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Math 414 Solutions to Chapter 4 Al Boggess Spring 1999 1 The given function 71 0 xlt14 7 4 14zlt12 99 2 12 xlt34 73 34 lt1 can be written as 7 gt4x 717 4 gt4z 7 1 717 2 gt4z 7 2 7 3 gt4z 7 3 Using the equations M490 7 W290 290 2 M496 71 7 296 7 44296 2 M496 7 2 WW 712 794290 712 gt2 7122 M496 7 3 WW 7 12 7 7 296 7 12 7 44290 7122 into the terms for x and collecting terms7 we have ag 32 gt2x712 gt2x 717 32 2m 52112x 71 The terms involving 1amp2m and 112z 7 1 belong to W1 We need to further decompose the terms involving gt2z and gt2z 7 1 The key equations are 290 7 4495 Mam2 295 71 7 44 7 Mam2 Substituting these equations and collecting terms gives 1 7 7 gt 7 12Wm 7 3MWQE 52 290 71 1 to The term involving gtz is the V0 component of f The term involving sz is the W0 component of f The sum of the terms involving diam and 112z 7 1 is the W1 component of f The dimension of a vector space is the number of basis elements So suppose the dimension of A is n and the dimension of B is m Let a1 an is a basis for A and 131 bm is a basis for B We would like to show that the collection a17amb1bm is a basis for A 63 B First we show this collection spans An arbitrary element of A 63 B is of the form a b with a E A and b E B Since 117 an is a basis for A7 there are constants r1 776 with arla1rnan Likewise7 there are constants 31 3m with b slbl smbm Adding these two equations yields abna1 rnan81b1 smbm which shows this collection spans To show linear independence7 suppose that 11rnanslb1smbm 0 We wish to show that all the coefficients the 7s and the ss are zero Let ar1a1rnan EAand let I 31b1smbm E B Thus the above equation becomes a b 0 Taking the inner product of both sides with this equation with I gives a bbgt 01 0 Since 071 0 by hypothesis7 we conclude that ltbbgt 07 which means that 31131 smbm b 0 Since b17bm is a basis for B7 they are linearly independent So all the s coe icients must be zero In the same manner taking the inner product with a7 we can show that a 0 and hence all the r coe icients must be zero If A and B are not orthogonal7 they may have a nonempty intersection In this case7 the dimensions are related by dimA B dimA dimB 7 dimA O B OJ ln fact7 a basis for A B can be constructed as follows Start with a basis for A O B7 say 01 ck assuming the dimension of A O B is Expand this collection to a basis for A as 01 cak1 an Also expand the collection of cs to a basis for B as 01ck7bk17bm Then a basis for A B can be shown to be bm 3177Ck7ak177ambk17 which is a collection of n m 7 k elements V0 is spanned only by gtz since any translate of j is zero on the interval 0 g x g 1 Thus7 V0 has dimension 1 Likewise7 V1 is spanned by gt2m and gt2z 7 1 All other translates of gt2m are zero on the interval 0 g x g 1 ln general7 the collection gt2 xik k0122 71 is a basis for V All other translates of gt2 m are zero on the interval 0 g x g 1 Therefore the dimension of Vn is 2 The same argument with 1 replaced by 1 shows that the dimension of WW is also 2 For the second part7 we use the decomposition Vn Wn1 Wn2 WOVO The dimension of the right side using the previous problem is 2 12 22020 From the formula for the geometric series Til 172 2TH 2 220 Therefore 2 12 220202 7112 which agrees with the dimension of Vn computed in the rst part of the problem 4 Suppose f 2k ak gt2z 7 k is orthogonal to V0 which is spanned by the gtz 71 We wish to show a0 7a1 a2 7a3 Now 1 is orthogonal to gtz 7 l for all 1 Suppose l 0 then since gtz is one for 0 g x g 1 and zero otherwise we have 1 0 7 lt ltzgtgt 7 2mm 7 k k The only values of k which contribute to this sum are k 0 and 1 Any other translate of gt2z is zero on the interval 0 g x g 1 Therefore the above equation becomes 1 1 0 lo0 110 gt2m 71 Since the graphs of gt2m and gt2z 7 1 are boxes of width 12 and height 1 both integrals are 12 Therefore the above equation becomes 0 002 717 012 which implies that LG 7011 as desired In general 1 is orthogonal to gtz 7 l for any 1 and so the above arguments can be repeated over the interval 1 g x g l 1 and we conclude that am 7012111 01 First we claim that if lf ml M then 11 7 MM S M195 7 241 This follows from the Mean Value Theorem which states He 7 y f c 7 y for some 0 between z and y Since lf cl M the previous inequal ity now follows So on the interval k2 x g k 12 which has width 12 the function f varies by no more than M2 The function fnz 2k ak gt2nm 7 k with ak fk2 has constant value fk2k on the interval k2 x g k 12 Thus fn and f differ by no more than M2 To make this quantity less than 6 we must require MQ e Me 2 a 10g2ME g n as claimed Since Vj1 63 W717 the function gt27z 7 k E can be written as 2735 i k Zal j71l l j71l 1 leZ for some choice of constants a and 31 Our goal is to compute 31 Since g1 and L j11 are orthonormal7 we have lt gt2jm 7 k1Jj71lgtL2 2171gt2 0 mix 7 10de 51 Now we use the de ning equation for 11 1W 271k1317k 2 7 k k with z replaced by 274 7 Z We obtain 00 31 2lt71gt2 gt27z 7 k 271k p1k 2iz 7 217 k dz 00 k Since 272 gt27m 7 kk E Z is an orthonormal set7 the only con tributing term on the right occurs when 21 k k or k k 7 21 and we obtain 1 Bi 7 2 lt71gt271kp17k21 2 So lmjilvz larlm 71 by de nition of 15 27171kpl7k21 2j71 71 by 9 the projection of 271 7 k onto Wfl is l j71ix Thus from 17 Z 5115711 Z 271P17k211M2j71 1 leZ leZ 8 The orthogonality condition can be stated as m 7 mm 71W 7 6k 7 0 By replacing z 71 by z in the above integral and relabeling n k 71 this orthogonality condition can be restated as m 7 mm dz 7 0 lt3 Recall that Plancherel s identity for the Fourier transform states 00 7 00 A 7 2 fzgzdz f5ggd5 for f 9 7L 00 We apply this identity with x zMz 7 n and gz As shown in class7 zMz 7 w e m So the orthogonality condi tion 3 can be restated as 00 A A7 wlt5gt gtlt5gt m d5 7 0 00 By diViding up the real line into the intervals I 2717 7 27rj 1 for j E Z this equation can be written as 27W A if 7mg 2 memos d5 7 jez 27 Now replace 5 by E 27rj The limits of integration change to 0 and 27139 27r A A Z gtlt5 27rj 5 27rjgte mlt quot7gt d5 7 0 0 JEZ Since 527 17 for j E Z7 this equation becomes A2 2 3 whme m 15 039 4 jEZ H5 7 2w 2 3lt5 2mgtwlt5 2m jeZ The orthogonality condition 4 becomes 27r 1270 new 7 0 lt5 The function F is 27139 periodic because F527r 2wZ mmj1gtgtzz3lt52wltj1gt jEZ 2w 2345 2wj gtzz35 M j where the last equality uses the change of index 3 j 1 Since F is periodic it has a Fourier series Zanemm where the Fourier coefficients are given by an 127139 027 F e mg 15 Thus the or thonormality condition 5 is equivalent to an O which in turn is equivalent to the statement FE O This completes the proof


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