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L is in this problem? | L is in this problem? | ||
| − | Answer: | + | Answer: |
| + | |||
| + | L=pi in this problem. The given Fourier series for f(x) tells | ||
you what the coefficients A_n need to be. Plug them back into the | you what the coefficients A_n need to be. Plug them back into the | ||
formula to get u(x,t). The relevant formula is on page 543 of the book. | formula to get u(x,t). The relevant formula is on page 543 of the book. | ||
| Line 121: | Line 123: | ||
four times the sum of 1/n^2. | four times the sum of 1/n^2. | ||
| − | Follow up question: | + | Follow up question: |
| + | |||
| + | Shouldn't we pull the 2 out of the Fourier series before squaring? This way the RHS is divided by 2. | ||
Answer: | Answer: | ||
Since the 2 is part of the Fourier coefficient, it is important to | Since the 2 is part of the Fourier coefficient, it is important to | ||
| − | square it when you sum up the squares. | + | square it when you sum up the squares. (It doesn't apply in this |
| − | term has a special coefficient in Parseval's identity. | + | problem, butdDon't forget that the a_0 |
| + | term has a special coefficient in Parseval's identity.) | ||
Question: | Question: | ||
| Line 134: | Line 139: | ||
Answer: Your e^-2t term should be e^-2(t-1). | Answer: Your e^-2t term should be e^-2(t-1). | ||
| − | |||
Recall L(u(t-a)f(t-a))=e^-as*F(s) | Recall L(u(t-a)f(t-a))=e^-as*F(s) | ||
| Line 143: | Line 147: | ||
I can't really see what I am doing wrong and maybe it is when I try to evaluate this for y(2). For the Heaviside, isn't it off if t<1 and on for t>1 for u(t-1)? Does this mean equal to 1 or 0? Maybe I am totally wrong on something before this. | I can't really see what I am doing wrong and maybe it is when I try to evaluate this for y(2). For the Heaviside, isn't it off if t<1 and on for t>1 for u(t-1)? Does this mean equal to 1 or 0? Maybe I am totally wrong on something before this. | ||
| + | |||
| + | Answer: | ||
| + | |||
| + | It took me a while to get this one too, probably because studying for 24 total hours this weekend took it out of me. I think there are two places you are going wrong. First, check your equation for Y. | ||
Question: 21 | Question: 21 | ||
| − | In the new solutions set, Professor Bell states that the key to this problem is knowing that the Fourier Cosine Transform, taken twice, is the function. I knew that already, but I still don't see how to execute this problem. Can someone please explain a little better? (I know there are other ways to get the answer, but I want to understand how to apply the Fourier cosine transform here. | + | In the new solutions set, Professor Bell states that the key to this problem is knowing that the Fourier Cosine Transform, taken twice, is the function. I knew that already, but I still don't see how to execute this problem. Can someone please explain a little better? (I know there are other ways to get the answer, but I want to understand how to apply the Fourier cosine transform here.) |
| + | Answer: | ||
| + | |||
| + | The problem gives that | ||
| + | |||
| + | <math>\mathcal{F}_c(e^{-x^2})=\sqrt{\frac{2}{\pi}}\ \frac{1}{1+w^2}.</math> | ||
| + | |||
| + | If you apply the Fourier Cosine Transform again, you get the function back. | ||
| + | That means that | ||
| + | |||
| + | <math>\sqrt{\frac{2}{\pi}} | ||
| + | \int_0^\infty \sqrt{\frac{2}{\pi}}\ \frac{1}{1+w^2} \cos(wx)dw</math> | ||
| + | |||
| + | is equal to | ||
| + | |||
| + | <math>e^{-x^2}.</math> | ||
| + | |||
| + | Now ask yourself what that 2 is doing there in the cosine term inside the integral. | ||
| + | |||
| + | Reply: Thanks. I don't know why none of us (0/4) came up with that. I'm a little bothered by the dx transforming into a dw, after we perform the second fourier cosine transform. (Either that or if you call the function resulting for the first fourier transform a f(w), then why is there a cos(wx) term resulting from a second fourier cosine transform?) Do you know why that's allowed? | ||
| + | |||
| + | Answer: | ||
| + | |||
| + | The w's and x's are just dummy variables. As a function on the real | ||
| + | line, exp(-x^2) is the same thing as exp(-t^2). To keep track of | ||
| + | the variables, we write the formulas so that the Fourier Cosine | ||
| + | Transform takes a function of x to a function of w. The inverse | ||
| + | Fourier Cosine Transform does the reverse, and the inverse Fourier | ||
| + | Transform is itself. | ||
Question, Problem 24: | Question, Problem 24: | ||
Can someone explain the end of the solution that Prof. Bell posted. I follow all the way to the end, but then it seems like the signs get changed and the solution should be 1/4 instead of 3/4. I guess I am confused by which terms to plug the 0 and Pi/2 in. | Can someone explain the end of the solution that Prof. Bell posted. I follow all the way to the end, but then it seems like the signs get changed and the solution should be 1/4 instead of 3/4. I guess I am confused by which terms to plug the 0 and Pi/2 in. | ||
| + | |||
| + | Reply: | ||
| + | |||
| + | What Prof. Bell posted is good. Do you understand how to get | ||
| + | |||
| + | phi = 1/2sinx + 1/8cos2x and | ||
| + | |||
| + | psi = 1/2sinx - 1/8cos2x ? | ||
| + | |||
| + | If you do, then calculate phi(x-2t) and psi(x+2t) if x = pi/4 and t = pi/8. E.g. you will have for phi: | ||
| + | |||
| + | phi(x-2t) = phi(pi/4-2*pi/8) = phi(pi/4-pi/4) = phi(0) = 1/2sin(0) + 1/8cos(2*0) = 1/8. | ||
| + | |||
| + | Then for psi(x-2t) = psi(pi/2) = 1/2 + 1/8. | ||
| + | |||
| + | 1/8 + 1/2 + 1/8 = 3/4 | ||
| + | --[[User:Bpavlov|Bpavlov]] 20:43, 13 December 2010 (UTC) | ||
| + | |||
| + | |||
| + | Problem 27 solution: In the posted solution to this problem why are the limits of integration 0 to 1 for the A(p) term? It seams from the function they should be -1 to 1. | ||
| + | |||
| + | Answer: | ||
| + | |||
| + | Since the function f(x) was even, you could simplify A(p) by | ||
| + | making it twice the integral from zero to infinity, but even | ||
| + | if you don't do that, you'll get the same answer. | ||
Latest revision as of 01:59, 14 December 2010
Work area for Practice Final Exam questions
Question:
For problem 1 on the practice problems, is the reason the answer is D is because det(A) has infinitely many solutions?
Answer:
No, that's not the reason. 3x3 matrices form a vector space, so the question as to whether or not the set of 3x3 matrices A such that det(A)=0 is a vector space is asking if this set is a subspace. There are only two things to check for the subspace condtion:
1. If v_1 and v_2 are in the set, is their sum in there too?
2. If v is in the set, is cv also in for any constant c?
Matrices with zero determinant satisfy 2, but not 1. It is pretty easy to come up with two matrices that fail. Let's see ...
1 0 0 0 1 0 0 0 0
and
0 0 0 0 0 0 0 0 1
for example.
Follow up question: Then why is iii) a vector space. I can think of some symmetric 3x3 matrices that have determinant = 0. For example:
1 0 0 0 0 0 0 0 1
I don't understand how the set of ALL symmetric 3x3 matrices could be considered vector spaces?
Answer: Parts i) and iii) are independent. It is not assumed that the determinant of the matrices are zero in part iii. The set of 3x3 symmetric matrices is a vector space because, if you add two of them together, you get a symmetric matrix. If you multiply one by a constant, you get one. They satisfy the two subspace conditions.
Question:
Can anyone fill in the blanks on the last problem (23) Professor Bell worked in class today?
I follow up to the u(x,t) = 1/2(sin2x)(cos2t) + ......
Where did the 1/2(sin2x)(cos2t) come from?
Answer:
When we did the method of separation of variables to solve the string problem, we got solutions of the form
X_n(x) T_n(t).
Then we took linear combinations and realized what the coefficients had to be from plugging in the initial conditions.
The cos 2t term is the T(t) part of the solution that goes with sin 2x. (The B_n coefficients are zero because there is zero initial velocity.)
The part of this problem that makes the final answer so short and sweet is that
cos pi/2 =0, cos 3 pi/2 =0,... , cos (odd) pi/2 =0
makes all those terms in the infinite sum go away.
Follow up Question: I understand the separation of variables, the X_n(x) T_n(t) solutions, and the linear combination. I don't understand how the f(x)is used to get the u(x,t)=1/2 sin2x cos2t. Also, how do we know what L is in this problem?
Answer:
L=pi in this problem. The given Fourier series for f(x) tells you what the coefficients A_n need to be. Plug them back into the formula to get u(x,t). The relevant formula is on page 543 of the book.
Question:
Problem 23: Can this problem be solved using D'Alembert method?
U(x,t)=(1/2)f(x+t)+(1/2)f(x-t)?
I tried but the result is different.
Answer:
Yes, it can be done via D'Albert's solution. Be sure to take into account that f(x) is sin(2x) only up to pi/2 and equal to zero from pi/2 to pi. D'Alembert's solution should work out the same.
Question:
Can someone explain the purpose of the infinite sum 1/n^2 in problem 30? I understand how to use the Parseval's identity, but that last term in the problem statement is really confusing me.
Answer:
When you square the coefficients of the Fourier series, you get four times the sum of 1/n^2.
Follow up question:
Shouldn't we pull the 2 out of the Fourier series before squaring? This way the RHS is divided by 2.
Answer:
Since the 2 is part of the Fourier coefficient, it is important to square it when you sum up the squares. (It doesn't apply in this problem, butdDon't forget that the a_0 term has a special coefficient in Parseval's identity.)
Question:
Has anyone had any success with Problem 15? I keep solving this on and getting the solution A. I know I'm not doing it correctly. Any hints?
Answer: Your e^-2t term should be e^-2(t-1).
Recall L(u(t-a)f(t-a))=e^-as*F(s)
Question: 16
I guess I am having some trouble somewhere in this because I can't seem to come up with the right answer. It seems straight forward with take the Laplace, solve for Y and then inverse Laplace. When I Laplace, I get sY-1=exp(-s)/(s+2). Y=1/s+exp(-s)/(s(s+2)). Then partial fractions to get Y=1/s+exp(-s)*[1/(2s)-1/(2*(s+2))]. I think the inverse Laplace gives y(t)=1+1/2*u(t-1)*[1-exp(-2*(t-1))].
I can't really see what I am doing wrong and maybe it is when I try to evaluate this for y(2). For the Heaviside, isn't it off if t<1 and on for t>1 for u(t-1)? Does this mean equal to 1 or 0? Maybe I am totally wrong on something before this.
Answer:
It took me a while to get this one too, probably because studying for 24 total hours this weekend took it out of me. I think there are two places you are going wrong. First, check your equation for Y.
Question: 21
In the new solutions set, Professor Bell states that the key to this problem is knowing that the Fourier Cosine Transform, taken twice, is the function. I knew that already, but I still don't see how to execute this problem. Can someone please explain a little better? (I know there are other ways to get the answer, but I want to understand how to apply the Fourier cosine transform here.)
Answer:
The problem gives that
$ \mathcal{F}_c(e^{-x^2})=\sqrt{\frac{2}{\pi}}\ \frac{1}{1+w^2}. $
If you apply the Fourier Cosine Transform again, you get the function back. That means that
$ \sqrt{\frac{2}{\pi}} \int_0^\infty \sqrt{\frac{2}{\pi}}\ \frac{1}{1+w^2} \cos(wx)dw $
is equal to
$ e^{-x^2}. $
Now ask yourself what that 2 is doing there in the cosine term inside the integral.
Reply: Thanks. I don't know why none of us (0/4) came up with that. I'm a little bothered by the dx transforming into a dw, after we perform the second fourier cosine transform. (Either that or if you call the function resulting for the first fourier transform a f(w), then why is there a cos(wx) term resulting from a second fourier cosine transform?) Do you know why that's allowed?
Answer:
The w's and x's are just dummy variables. As a function on the real line, exp(-x^2) is the same thing as exp(-t^2). To keep track of the variables, we write the formulas so that the Fourier Cosine Transform takes a function of x to a function of w. The inverse Fourier Cosine Transform does the reverse, and the inverse Fourier Transform is itself.
Question, Problem 24:
Can someone explain the end of the solution that Prof. Bell posted. I follow all the way to the end, but then it seems like the signs get changed and the solution should be 1/4 instead of 3/4. I guess I am confused by which terms to plug the 0 and Pi/2 in.
Reply:
What Prof. Bell posted is good. Do you understand how to get
phi = 1/2sinx + 1/8cos2x and
psi = 1/2sinx - 1/8cos2x ?
If you do, then calculate phi(x-2t) and psi(x+2t) if x = pi/4 and t = pi/8. E.g. you will have for phi:
phi(x-2t) = phi(pi/4-2*pi/8) = phi(pi/4-pi/4) = phi(0) = 1/2sin(0) + 1/8cos(2*0) = 1/8.
Then for psi(x-2t) = psi(pi/2) = 1/2 + 1/8.
1/8 + 1/2 + 1/8 = 3/4 --Bpavlov 20:43, 13 December 2010 (UTC)
Problem 27 solution: In the posted solution to this problem why are the limits of integration 0 to 1 for the A(p) term? It seams from the function they should be -1 to 1.
Answer:
Since the function f(x) was even, you could simplify A(p) by making it twice the integral from zero to infinity, but even if you don't do that, you'll get the same answer.
