[PDF] Solutions to Homework Assignment 2 - Math User Home Pages PDF hw2-solutions.pdf

We can do the integration “by hand” or using Wolfram Alpha (b) One can simply calculate the Fourier series of f ′ For the coefficients Bn we have Bn = 2 π

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Mathematica is the creation of Stephen Wolfram, a theoretical physicist who has made not return an estimate like 0 70710678, as a handheld calculator might As another example, we solve a BVP with a nonconstant coefficient: - ⅆ ⅆx When performing Fourier series calculations, the above command is very useful

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This gives 15 octal digits starting with the coefficient of 8-10 You can tell Mathematica to give you an approximate numerical result, just as a calculator would In Mathematica the Fourier transform of a function fHtL is by default defined to be

[PDF] Solutions to Homework Assignment 2 - Math User Home Pages

We can do the integration “by hand” or using Wolfram Alpha (b) One can simply calculate the Fourier series of f ′ For the coefficients Bn we have Bn = 2 π

1.(a) We haveck=1

2∫

1f(x)eikxdx. We can do the integration \by hand" or using Wolfram Alpha.

The expression we get from the computer in the latter case isck=2kcos(k)+2sin(k)

3k3. For an integer

k̸= 0 this givesck=2(1)k+1

2k2. Fork= 0 we can either calculate directlyc0=1

2 1

1(1x2)dx=2

3 , or obtain the same result by taking the limitk!0 in the above expression we got from Wolfram Alpha.

The formula∫1

1(1x2)2dx= 2∑

kjckj2gives∑1 k=11 k 4=4 90

(b) One can simply calculate the Fourier series off′(x) =2xon the interval (1;1), and check that its

coefficients areikck.

One can also see it without calculation: the Fourier series computed in (a) denes a 2periodic function

onfperon (1;1) which is equal to 1x2forx2(1;1). The functionfperis clearly continuous, smooth away from the points 1 + 2kwherek2Z(the set of integers), and its derivative away from the

points of non-differentiability is a 2periodic extension of the functionf′(x) =2xfrom interval (1;1)

to (1;1)nf1+2k; k2Zg:In particularf′peris piece-wise smooth, and therefore its Fourier series can

be differentiated term by term, see for example p. 114 of the textbook. (c) The extended periodic functionfperis given by the expression 1(x2)2whenx2(1;3):The derivatives from the left (resp. right) of the functionfperatx= 1 are easily calculated to be2 and

2, respectively. Since they are different, the periodically extended function cannot be differentiable at

x= 1. The partial sums∑k=n k=nof Fourier series of the functionf′(x) atx= 1 are easily seen to vanish (note that in this particular exampleckeik+ckeik= 0 for eachk), and hence the Fourier series for f

′(x) gives 0 when evaluated atx= 1. (Note that 0 is the average of the left and right derivative at

x= 1.)

2.We have cos2x=1

2 +1 2 cos2xand this is the cosine series of cos2x. For sin2xwe can similarly write sin 2x=1 2 1 2 cos2x, but this clearlyis notthe sine-Fourier series of sin2x. If we write sin2x=∑1 n=1Bnsinnx, the sum on the right-hand side will be a 2periodic odd function, let us call itfper. We havefper(x) =sin2xforx2(;0) andfper(x) = sin2xforx2(0;). The second derivative f

′′per(x) is easily seen to have the limit 2 asx!0 from the right and2 asx!0 from the left. Hence

′′percannot be continuous at 0 and the functionfpercannot be a nite sum of functions of the form

B ksinkx. For the coefficientsBnwe haveBn=2

0sin2sinnxdx=8

n(n2)(n+2)whennis odd, and B n= 0 whennis even. As we have seen, the second derivative offperis discontinuous atkfor integer

k, and smooth away from those points. Hence the Fourier series off′′perstill converges point-wise. On the

other hand the Foureir series off′′′per(x) cannot converge (point-wise), as itsnth term does not approach

zero: differentiation gives (formally)f′′′per(x) =∑1 n=1n3Bncosnx, andn3Bndoes not approach 0 for n! 1.

3.Our machine can do the Fourier series only for 2-periodic functions, so we change of variables as

follows: Forx2(0;L) we will writeu(x;t) =v(x L ;t), wherev=v(y;t) is an odd 2-peridoc function on the real line. The functionvis dened in three steps: (i) Fory2(0;) we setv(y;t) =u(yL ;t). (ii) Fory2(;0) we letv(y;t) =v(y;t). (iii) we extendvfrom (;) to (1;1) as a 2periodic function. Substituting the expression into the equation foruthe functionu(x;t) =v(x L ;t), we obtain the equation satised byv(y;t), namely 2v @t

2=a2@2v

@y 2 v ; a=c L :(1) We note that the boundary condition forvisv(0;t) =v(;t) = 0, and is satised automatically in view of the requirement thatvbe odd and 2periodic. The functionsu0;u1ate transformed tov0;v1by u i(x;t) =vi(x L ); i= 0;1. We seekv(y;t) as a Fourier series v(y;t) =∑ kc k(t)eikx:(2)

Our task is to determine the coefficientsck(t). Once we have them, the machine can be used to calculate

v(y;t) and thenu(x;t) =v(x L ;t). The equation forck=ck(t) is ck=a2k2ck ckand its general 1 solution is c a 2k2+ :(3) We now determine the values ofAk;Bkfor our particular solution from the conditionsck(0) =Akand

_ck(0) =!kBk. The values ofck(0) and _ck(0) are known from the initial conditions: the Fourier coefficients

ofv0areck(0) and the Fourier coefficients ofv1are _ck(0). Our algorithm can be summarized as follows:

Setvi(y) =ui(Ly

); i= 0;1;and extendvias an odd function of (;). 2. Letck(0) be the Fourier coefficients ofv0and _ck(0) the Fourier coefficients ofv1. (Here we use our machine for the rst time, to calculate Fourier coefficients.) 3.

DetermineAk;Bkby the formulae above.

Sum the Fourier seriesv(y;t) =∑

k(Akcos!kt+Bksin!kt)eiky. (Here we use our machine for the second time, this time to sum a Fourier series.) 5. u(x;t) =v(x L ;t):

4.The general solution of the wave equation in our situation is a sum of terms of the form

B ksin(kx L )sin(!k(ttk)), where!k=kc L T . See, for example, Chapter 4 in the textbook

(formula 4.4.11). Here we are only interested in the \base frequency" of the string, corresponding to

k= 1. Hence we can work with the formula!= L T . The answers can be now easily obtained from the formula. (a) The ratio T has to remain the same, so we have to change the density to 2 . (b) The expression L T has to remain the same, so we have to increaseTto 4T.

5.(a) From the chain rule we have@u

@t =@u ~t@~t @t +@u @x @x ~t=@u ~tv@u @~x. A similar (but easier) calculation gives @u @x =@u @~x. (Here we have a convention which is usual in similar situations: when we take@ ~twe keep ~x constant and when we take @~xwe keep~tconstant, and similarly with thet;xvariables. Hence in the new coordinates the equation becomes ( ~tv@ @~x)2u=c2@2u @~x2, which is the same as@2u ~t22v@2u ~t@~x= (c2v2)@2u @~x2:

If we knowcand can measureu(including its derivatives) in the coordinate frame (~t;~x), we can determine

(b) Consider the motion of the point ~x= 0 watched from the frame (t;x). Setting ~x= 0 in transformation

(6) in the hw2 assignment, we obtaint=~tcoshandx=c~tsinh, which then givesdx dt =csinh cosh= ctanh. This is the velocityvof the origin of the frame (~t;~x) when observed from the frame (t;x). (c) Using the formulae cosh

2sinh2= 1;tanh=sinh

coshand tanh=v c , one obtains cosh=1 1v2 c 2 and sinh=v c 1v2 c

2. This givest=~t

1v2 c 2+v c 2~x 1v2 c

2andx=v~t

1v2 c 2+~x 1v2 c

2, which one can nd in

any textbook of special relativity.

6.(a) Let us rst show thatAA=nI, whereIis the identity matrix. We have

(AA)kl=∑n m=1Akm(A)ml=∑ mw(k1)(m1)w(m1)(l1)=∑ mw(m1)(kl). Whenk=l, the last sum is clearly equal ton. Fork̸=l, let us set=wkl. We note that̸= 1 butn= 1. The last sun can then be written as 1 +++n1=n1 1= 0. (b) One can either say that we have shown in (a) that the matrix 1 p n Ais unitary and this implies the identity 1 n kjfkj2=∑ kjckj2in the hw2 assignment. Alternatively, one can show this identity directly, more or less repeating the calculation in (a): we have∑quotesdbs_dbs20.pdfusesText_26

[PDF] [PDF] Solutions to Homework Assignment 2 - Math User Home Pages

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