[PDF] [PDF] Solutions for practice problems for the Final, part 3 - Math User

therefore an = 0, and bn can be nonzero (b) Let g(x) = cos(x5)+sin(x2) What coefficients of the Fourier Series of g 



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Solutions for practice problems for the Final, part 3 Note:Practice problems for the Final Exam, part 1 and part 2 are the same as Practice problems for Midterm 1 and Midterm 2.

1. Calculate Fourier Series for the functionf(x), defined on [-2,2], where

We have

f(x) =a0

2+∞

n=1? a ncosπnx2+bnsinπnx2? where a 0=1 2? ?0 -2(-1)dx+? 2 02dx? = 1, a n=1 2? ?0 -2(-1)cosπnx2dx+? 2

02cosπnx2dx?

1 2? (-1)?2πnsinπnx2? 0 -2+ 2?2

πnsinπnx2?

2 0? = 0, n >0, and b n=1 2? ?0 -2(-1)sinπnx2dx+? 2

02sinπnx2dx?

1 2? -(-1)?2πncosπnx2? 0 -2-2?2

πncosπnx2?

2 0? 1 πn(1-cosπn)-21πn(cosπn-1) =3πn(1-(-1)n).

Therefore, we have

f(x) =1

2+∞

n=13πn(1-(-1)n)sinπnx2. An easy way to see that all ofanexcepta0are zero is to note that f(x) =1

2+g(x),

1 whereg(x) is an odd function, g(x) =?3/2, x >0, -3/2, x <0.

2. Calculate Fourier Series for the functionf(x), defined on [-5,5], where

f(x) = 3H(x-2).

By a similar method,

f(x) =9

5+∞

n=1? -3πnsin2πn5cosπnx5+3πn? cos2πn5-(-1)n? sinπnx5?

3. Calculate Fourier Series for the function,f(x), defined as follows:

(a)x?[-4,4], and f(x) = 5. Comparingf(x) with the general Fourier Series expression withL= 4, g(x) =a0

2+∞

n=1? a ncosπnx4+bnsinπnx4? we can see thata0= 10,an=bn= 0 forn >0 will givef(x) =g(x). (b)x?[-π,π], and f(x) = 21 + 2sin5x+ 8cos2x.

Again, forL=π, we have

g(x) =a0

2+∞

n=1(ancosnx+bnsinnx), and settinga0= 42,a2= 8,b5= 2 and the rest of the coefficients zero, we obtainf(x) =g(x). (c)x?[-π,π], and f(x) =8 n=1c nsinnx,withcn= 1/n. 2 zero. (d)x?[-3,3], and f(x) =-4 +6 n=1c n(sin(πnx/3) + 7cos(πnx/3)),withcn= (-1)n.

We have

g(x) =a0

2+∞

n=1? a ncosπnx3+bnsinπnx3?

4. (a) Letf(x) =x+x3forx?[0,π]. What coefficients of the Fourier

Series offare zero? Which ones are non-zero? Why?

f(x) is an odd function. Indeed, f(-x) =-x+ (-x)3=-x-x3=-(x+x3) =-f(x), thereforean= 0, andbncan be nonzero. (b) Letg(x) = cos(x5)+sin(x2). What coefficients of the Fourier Series ofgare zero? Which ones are non-zero? Why? g(x) is an even function. Indeed, g(-x) = cos((-x)5)+sin((-x)2) = cos(-x5)+sin(x2) = cos(x5)+sin(x2) =g(x).

Therefore,bn= 0, andancan be nonzero.

5. Letf(x) = 2x+x4forx?[0,5].

(a) Write down the functionG(x), which is the odd continuation for f(x). Specify what terms will be zero and non-zero in the Fourier expansion forG(x).

We have

G(x) =?2x+x4, x >0,

2x-x4, x <0.

3 Indeed, we can check that ifα >0, thenG(-α) =-2α-(-α)4= -2α-α4=-G(α). In the Fourier expansion forG,an= 0, andbn can be nonzero. (b) Write down the functionV(x), which is the even continuation for f(x). Specify what terms will be zero and non-zero in the Fourier expansion forV(x).

We have

V(x) =?2x+x4, x >0,

-2x+x4, x <0. Indeed, we can check that ifα >0, thenV(-α) =-2(-α) + (-α)4=

2α+α4=V(α). In the Fourier expansion forV,bn= 0, andancan

be nonzero.

6. Supposef(x) is defined forx?[0,7], andf(x) = 2e-4x. Another

function,F(x), is given by the following:

F(x) =∞?

n=0a ncos(πnx/7), where a n=2 7? 7

02e-4xcos?πnx7?

dx. What is the value ofF(3)? What is the value ofF(-2)? The functionF(x) is the cosine Fourier expansion off. On the domain off, that is, forx?[0,7], we haveF(x) =f(x). Therefore, since

3?[0,7], thenF(3) =f(3) = 2e-12.

For the negative values ofx, the cosine series converges to the even extension off(x), which is 2e-4|x|. Therefore,F(-2) =f(2) = 2e-8. Note: a sine Fourier series would give the odd extension, and in this case we would have-f(2) =-2e-8.

7. Let us supposed that both ends of a string of length 25cmare attached

to fixed points at hight 0. Initially, the string is at rest, and has the shape 4sin(2πx/25), wherexis the horizontal coordinate along the string, with zero at the left end. The speed of wave propagation along 4 the string is 3cm/sec. Write down the complete initial and boundary value problem for the shape of the string. We have the following initial and boundary value problem: 2y ∂t2= 9∂2y∂x2, x?[0,25],(1) y(0,t) =y(25,t) = 0,(2) y(x,0) = 4sin(2πx/25),(3) ∂y(x,0) ∂t= 0.(4)

8. Let us suppose that the following boundary value problem is given:

2y ∂t2= 50∂2y∂x2, x?[0,100],(5) y(0,t) =y(100,t) = 0,(6) y(x,0) =x2(100-x),(7) ∂y(x,0) What is the speed of wave propagation along the string? What is the initial displacement of the string at pointx= 20? What is the initial velocity of the string at pointx= 50? At what point of the string is the initial velocity the largest? The speed of wave propagation along the string is⎷

50. The initial

displacement of the string at pointx= 20 is 202(100-20) = 32000. The initial velocity of the string at pointx= 50 is 1/3(100-50) = 50/3. The maximum of the initial velocity is at pointx= 25 (plot the graph of the initial velocity, equation (8), to see this).

9. Let us suppose that the following boundary value problem is given:

2y ∂t2=∂2y∂x2, x?[0,2],(9) y(0,t) =y(π,t) = 0,(10) y(x,0) = 0,(11) ∂y(x,0) ∂t=g(x).(12) 5

Suppose that?2

0g(x)sin?πnx

2? dx=1n3.

Findy(x,t).

For the problem with the zero initial displacement, the solution is given in terms of the initial velocity (herec= 1), y(x,t) =∞ n=1c nsinnπx

2sinnπt2,

with c n=2 nπ? 2

0g(x)sin?πnx2?

dx=2n4π.

10. Let us suppose that the following boundary value problem is given:

2y ∂t2=∂2y∂x2, x?[0,π],(13) y(0,t) =y(π,t) = 0,(14) y(x,0) = 22sin2x+ 8sin6x,(15) ∂y(x,0) ∂t= 0.(16) Findy(x,t), in a closed form (containing no integrals). You will not need to evaluate any integrals.

We look for the solution is the form,

y(x,t) =∞ n=1c nsinnxcosnt. To satisfy initial condition (15), we sett= 0 and obtain, y(x,0) =∞ n=1c nsinnx. To make this equal tof(x) = 22sin2x+8sin6x, we setc2= 22,c6= 8, and the rest of them zero. We obtain, y(x,t) = 22sin2xcos2t+ 8sin6xcos6t. 6quotesdbs_dbs17.pdfusesText_23