Difference between revisions of "HWK10MA527Fall2013" - Rhea

Latest revision as of 17:15, 5 November 2013

Homework 10 collaboration area

From Jake Eppehimer:

I must be doing something wrong on problem 2 of Lesson 31. There doesn't seem to be a way to solve the integral if you use formula (1b) with the answer from problem 1. Any tips?

Also, I don't understand how to do number 5 on Lesson 29. Any help would be appreciated.

Response from Mickey Rhoades Mrhoade

I get 2/pi times a bunch of sinc function integrals which have to be evaluated with the Dirichlet Integral. I can use Table 11.10 relationship #10 and get the answer without chugging out all the integrals.

From Steve Bell: Yes, go ahead and use that table for problem 2 on page 522. Demonstrating that those integrals converge to step functions is rather advanced work and that is not what we are expecting you to do. (Mickey is correct that Dirichlet integral considerations are the key.)

Also, there is a typo in problem 1 on page 517. The integral should be an integral with respect to w, so that dx should be a dw. The point of the problem is to recognize that this integral is an instance of formulas (4) and (5) for the Fourier Integral formulas. Your problem is to figure out what the f(x) is.

From Jake:

What page is this table on? I'm not seeing anything. Thanks.

From Andrew:

Table 1 in 11.10 on Page 534 is a table of Fourier Cosine Transforms.

From Eun Young:

From the lesson 30 lecture note, we know that $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if -1<v<1 and 0 otherwise. Consider only positive v. Then, $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if 0 <v<1 and 0 if v>1.

From this, we can compute $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw . $

Let 2w = t. Then, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw = \int_0^{\infty} \frac{ 2 \sin t }{ \pi \frac t 2} cos (t \frac v 2 ) \frac{dt}{2} = \int_0^{\infty} \frac{2 \sin t}{ \pi t} cos(\frac v 2 t) \ \ dt = 1 \ \ \text{if} \ \ 0 < \frac v 2 < 1 \ \ \ \text{and } 0 \ \ \ \text{if} \ \ \frac v 2 >1 $.

Thus, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw =1 \ \ \text{if} \ \ 0 <v < 2 \ \ \ \text{and } 0 \ \ \ \text{if} \ v >2 $.

Use the second and the last equations. Then, you will get the answer.

From Farhan:

In response to Jake, regarding 11.6.5, I am thinking of a simple proof only with words, like a summation of only even functions will give an even function (I am not 100% sure if I can make this claim). But would like to hear from others if they are thinking of coming up with a mathematical rigorous proof.

From Steve Bell: For p. 509, problem 5, notice that in my lecture, I gave you the fact that the Legendre polynomials P_n for even n are even functions and the P_n for odd n are odd functions. Remember that (even)*(odd) is odd and that the integral of an odd function over a symmetric interval about the origin is zero.

From Craig:

In response to Jake and Farhan, for 11.6 #5 I described a proof with words and then gave examples, as the book asked. I'm not sure if this is exactly what was desired, but it gets the point across.

On a separate note, is anyone else getting a different answer than the book for 11.8 #5? When working it out, I'm getting an extra -2w/w^3 term. My math looks correct, but I still have a bit of doubt.

Response from Mickey Rhoades Mrhoade

I used an explanation with words as well for 11.6 #5. For 11.8 #5 I got the book answer. I initially did substitution with z=wx. dz=w dx and x = z/w. Then you should be have sqrt(2/pi)*(1/w^3) times integral of z^2 cos(z) dz from 0 to w. Use integration by parts with u=z^2 and dv = cos(z). It should be straightforward from there.

From Hzillmer I'm having trouble getting started on 11.5, I must have missed something critical in the notes and the book isn't much help. Can someone point me in the right direction. I wrote down an example but I don't understand how an integral with P(x) in it can be evaluated without knowing P(x).

From Craig:

Hzillmer, in Lecture 29, in the blue text near the middle of the page, Dr. Bell worked out the first few Pn(x) functions. Plug those Pn(x) functions into your integral to get your coefficients. Hope this helps.

From djkees:

Craig: I think you forgot to make x = 0 in that term, it should be 2x*cos(wx)/(w^2), taken when x = 0, = 0, which will kill that term for you and give you the answer in the back of the book.

From Craig:

Ah... yup. Somehow I completely lost that x term from one line to the next. Not sure how I overlooked it. Thanks!

From Jake:

Ok, I got number 2 on Lesson 31. Thank you. For number 5 on Lesson 29, are we supposed to use that crazy formula with all the factorials? I know we were told that we wouldn't need it, but why isn't it in the book? I don't really see any relevance of the chapter to the lecture, but the lecture helped me solve problems 1 and 3. That was just basic algebra. I feel lost with the rest of it.

Craig, I got the answer in the back for that problem.

From Collier:

Jake, in reference to the formula that you are talking about for number 5 in Lesson 29, it is in the book. It should be pg.178, Section 5.2, equation(11). That section explains Legendre's Polynomials a little better.

From Jake:

That did it. Thanks, Collier.

From Collier:

No problem. On another note, for 11.7 Problem 1, I think I am missing something, because I see the similarities to Example 3 on page 516, but it just doesn't seem right that I just show splitting up the given integral equals what they want. [i.e. just using equations (13) and (15)]. Does anyone have any advice on what I might be missing here/what the problem wants us to show?

From Steve Bell: See my remark near the top about p. 517: 1. That dx should be a dw and you will use formulas (4) and (5) from that section to figure out what the f(x) must be.

From Dori: For 11.7 problem 1 I ended up using a Laplace transform to show that they check out. Not sure if that is what they are looking for, but Professor Bell had mentioned it in class. You can show that the A(w) and B(w) work out using the Laplace s-shifting.

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