Trinity College Dublin

Questions, comments and answers.

4 October 2010
BookI am in the 22S3 Senior Freshman course on Fourier Analysis with you and I was just wondering is there a book that we will follow or any suggestions of books so that I can get extra examples to practise from during the course?
Reply Lots of books cover this material, I am suggesting Kreyszig, it's on the website

5 October
Tutorials How do you know if you are in group A, B or C for the tutorials? Or should we just go to any one we want? I haven't seen anything up on the noticeboards. I know you said today that the tutorials won't be starting til next week, it's just all a bit confusing, sorry.
Reply The tutorial lists will be posted on the maths department notice boards when they are available, these notice boards are on the ground floor of the Hamilton bldg, opposite the Salmon and Synge. If your tutorial time is bad, or clashes, there is no problem with you changing to another group.

Further message Apologies, I found the answer on the TCD portal.
Reply You now know more than me!

5 October
Tutorials Just checking what tutorial do you take for sf fourier analysis so I can switch into it?
Reply I'll announce these details when I know them, you can then switch in to, or out of, my tutorial, as suits.

24 October 2010
Integration period I know you have to integrate over one period when calculating a_n and b_n for example, but does it matter what interval? I was trying a problem where f(t)=t^2, and it only seemed to work when I integrated from 2pi to 0, rather than from pi to minus pi. Was I making a mistake or is there some way of picking a suitable integral? Thanks :)
Reply Well if you have a periodic function it shouldn't matter what period you integrate over, so if f(t_+2\pi)=f(t) then
\int_{-\pi}^\pi f(t)dt= \int_0^{2\pi}f(t)dt
However, you have to be careful if $f(t)$ is defined on a particular interval in a piecewise continuous way you need to careful. Say, for example you are told f(t)=t^2 from -pi to pi and is periodic, then the Fourier integral from -pi to pi will just be
\int_{-\pi}^\pi f(t)e^{-int}dt=\int_{-\pi}^\pi t^2 e^{-int}dt
but the integral from 0 to pi would be much more complicated because f(t)=(t-2pi)^2 for pi \int_0^{2\pi}f(t)e^{-int}dt=\int_0^\pi t^2 e^{-int}dt+\int_\pi^{2\pi} (t-2\pi)^2 e^{-int}
You can do a change of variables to show that this is the same as the other thing. This a bit of a confusing issue so let me know if I haven't explained it.

24 October 2010
Examples in lectures I'm find the course very interesting, but I just wanted to say that doing more examples on the topic discussed would help a lot, including some examples from past papers. Some parts of the course are a little bit confusing, and examples would help. Although I fully understand the method and the material so far, I still don't fully understand the motivation behind some calculations.
Reply Thanks for your comments. I am sorry there hasn't been enough examples, it is actually hard to find useful and informative Fourier examples, the second half of the course, ODEs, is much better in this regard. I am now intending to make a homework sheet of examples with solutions available, hopefully this will help.

1 November 2010
T-shirts Where do you get the awesome tshirts?
Reply Thanks re the awesome bit; I used to only wear black tshirts till a student feedback form criticized my "macarbre vampire-like sense of style"; it was kind of a hobby for about a year finding fun tshirts on the web, makes me sad now though since I stopped buying them two years ago because of the recession.

22 November 2010
Schols Hi Conor. I'm a 3rd year doing schols again, I'm just wondering whether you will provide formulas once more i.e. the Fourier Formulas in particular
Reply This will be the same as before, the format of the paper won't change.

8 December 2010
CSS I couldn't help but notice you like CSS the band. Seen as no one else i know has ever heard of them, i wanted to tell you of an awesome mashup of one of their songs with some hiphop. It's called "Stop that booty" by Super Mash Bros. Super Mash Bros are two college students in california and are becomming an underground phenomenom in America. They mix hiphop songs over rock (in a way you cant imagine!). The css song features midway through this track with some 'out there' hip hop lyrics over it (to warn you). Sorry if i have wasted your time, I just wanted to share the music. Honoured to be a student of yours.
Reply That's very cool, thanks and thanks for your kind remarks and for everyone else: http://youtu.be/s24vcToyrBs. Music suggestions always welcomed!

8 December 2010
Exam tutorial I just wanted to ask, will there be some sort of exam-preparation tutorial coming up? At the end of the year or after xmas? I thought it would be quite helpful to do a revision course with past exam questions ans stuff. Like a exam-related tutorial. Thanks!
Reply I hadn't thought about it, I won't do a schol tutorial, I feel that makes too much of a fuss about schols in a way that ultimately puts some people off doing it, they feel shy about involving themselves. I would be happy to do an exam preparation tutorial, it seems a good idea to have one at the end of the next semester. Maybe if someone from the class could email me next semester we could try to arrange something. In the meantime, if you are stuck on anything, do contact me, I have no problem with helping students who come to my office.

8 December 2010
Epic Quotes Mathsoc quotes thing
Reply If my father is done for toad smuggling I will be very annoyed.

8 December 2010
Laplace Hi,I was wondering if we need to know about about Laplace transforms for schols? I saw Q5 on last years paper but I dont remember doing m/any Laplace things this year..... Thanks,
Reply They hadn't done Laplace either, it is supposed to be a self contained question, all you need to know about laplace is given.

20 December 2010
Thanks Thanks for the awesome course! :) Hope you liked the tshirt. :)
Reply Loved the tshirt, many thanks!

22 December 2010
Fourier Series (Odd and even functions) Dear Dr Houghton, I'm having a slight difficulty with the fourier series of a block wave in the lecture you gave on the 8th of October. Here, the component a_n goes to zero, and I'm a little bit perplexed. In the online notes, I you say that the integrand is odd, but I'm a little confused as to why. Any help with this would be greatly appreciated! (Oh, and merry christmas/happy new year/pleasant any other time period)
. Reply The a_0 goes to zero when the function is odd, ie when f(-t)=-f(t). The key point is that if you multiply an even and an odd function you get an odd function, say g(-t)=g(t) then if h(t)=g(t)f(t) then h(-t)=f(-t)g(-t)=-f(t)g(t)=-h(t), odd. Now cos (stuff X t) is even so in the calculation of a_n the integrand, the thing inside the integral, is odd if f(t) is odd. Finally, integrating an odd function, h(t) say, over an interval of the form [-A,A] always gives zero, for any A:
int_{-A}^A h(t)dt = int_{-A}^0 h(t) dt + int_0^A h(t)dt
but, doing a change of variables in the first integral t'=-t you get dt'=-dt h(t)=h(-t')=-h(t') by oddness of h and t=-A means t'=A, so
int_{-A}^0 h(t)dt = int_A^0 h(t')dt' = int_0^A h(t')dt'
which is minus the second term, hence, the integral is zero. Let me know if this hasn't helped.

22 December 2010
PS scripts. I was just wondering are the problem sheets corrected yet, and are they available to be collected?
Reply I am sorry but no, I'm still catching stuff up, tomorrow at lunch time. If you aren't still around then, I cam scan yours and email it to you, email me

22 December 2010
PS2 solnI was wondering about the very last line in the solution for tutorial 2. I don't know how you lost the (1/2pi), as you have it so it just becomes (1/pi).
The solution was wrong, the error you mentioned and there was a sign wrong as well, corrected now. Thanks!

31 December 2010
PS3 Q2 In tutorial 3, question 2, where did you get the second line of the solution: [ie^-ik(pi/2)+ie^ik(pi/2)]/i(1 - k) + [-ie^-ik(pi/2)-ie^ik(pi/2)]/i(-1 - k) from?
Reply The limits pi/2 and -pi/2 are substituted and then the result is simplified using, for example, e^i(1-k)pi/2=e^{i pi/2} e^{-ik pi/2}=i e^{-ik pi/2} since e^{i pi/2}=i, you can see this by expanding e^{i pi/2}=cos pi/2+ i sin pi/2.

31 December 2010
Complex Fourier series Is c_n for a fxn that has period other than 2pi 1/L[int(-L/2,L/2) f(t)e^-int*pi dt? If so, why? A bit later Further to the question "Is c_n for a fxn that has period other than 2pi 1/L[int(-L/2,L/2) f(t)e^-int*pi dt? If so, why?" I've worked out c_n from a_n and b_n, and got 1/L[int(-L/2,L/2) f(t)e^-int*2pi/L dt, is this correct?
Reply Your second answer is correct, you can see this the way you did it, by doing the calculation starting with the real series, or directly. To do it directly, notice that the series has to have Fourier modes e^{int 2pi/L} to have the correct periodicity. Now to calculate the c_n you need something orthogonal to that and
int_0^L e^{int 2pi/L}e^{-imt 2pi/L}dt = L\delta_{mn}
so if you multply
f(t)=sum_{n=-\infty}^\infty c_n e^{int 2pi/L}
across by e^{-imt 2pi/L} and integrate both sides.

31 December 2010
Fourier PS Q2 I'm a bit confused as to why in question 2 the solution is for a_n is -6/(n*pi)^2 for n odd. I'm probably overlooking something, but I'm not sure what. Sorry to bother you over Christmas.
Reply The solution looks right to me;

int_1^2 3 cos n\pi t dt = 3/n\pi sin n\pi t ]_1^2 = 0 since sin n\pi and sin 2 n \pi are both zero for all n.

that just leaves the term with the t in it. By parts, with u=3t dv=cos n\pi t dt,

int_0^1 3t cos n\pi t dt = \frac{3t}{n\pi} sin n \pi t ]_0^1 - \frac{3}{n\pi} int_0^1 sin n\pi t dt

The boundary term, the first term, is zero since sin n \pi =0 for all n, the second term integrates to give

\frac{3}{n^2\pi^2} [cos n\pi -1] and use the fact that cos n\pi =(-1)^n so the term in square brackets gives -2 for n odd and 0 for n even.

31 December 2010
Fourier PS Q3 Hi, I was looking at the solutions you posted up for the Fourier Analysis worksheet for Senior Freshman maths for scientists. In line 14 of the solution for question 3, should it not be 16/3 is equal to all that? If not, can I email you my answer to see where I'm going wrong? Thanks, and hope you had a good Christmas.
Reply You're right, I'm wrong. Corrected now. Thanks!

31 December 2010
Fourier PS Q3 Is Q3, why is the series starting at n!=0? Why does |c_n|^2 become 4*pi^2/n^2, should it not be 4/(pi^2*n^2)?
Reply So the n=0 case is done seperately, the integral works different in that case because it has no exponential, that's what gives the two. The pi^2 should be on the bottow, you're right: I have corrected the solution sheet. Thanks.

8 February 2011
Other Maths Worksheet Hi, you mentioned putting up another worksheet as preparation for exams, I was just wondering if it'd be up soon? Thanks :)
Reply Thanks for reminding me, I will try to get to it this week, sorry!

7 April 2011
Pre-exam tutorial Will you be having a pre-exam tutorial?
Reply I can organize tutorials on an ad hoc basis, the easiest thing would be for you to email me directly I will arrange something with you.

15 May 2011
Corrections I think I've spotted a few mistakes in the solutions to that sheet. For q6 on the solutions (7 on the sheet), shouldn't the 25/6 disappear in the derivative since it's a constant? In q8 in the solutions (9 on the sheet, shouldn't the t*exp(3t) term go to zero for y'(0)? Also in q9 in the solutions (10 on the sheet), I think the -18C term should actually be -18Ct^2 since you'll be left with an 18Ct^2 term left over then and you won't be able to solve for C. Also, for future reference, I think a lot of people were confused by your usage of != instead of equals with a line through it, leading to questions about how n! could be = 0. To make up for the bother I'm causing, here's some ducklings and a yoyo.
Reply Thanks a million for the corrections, I have made two of your three corrections, the middle one I amn't so sure about; the one in the expression for \dot{y}(0) comes from differentiating the te^{3t}
d/dt (te^{3t}) = e^{3t} + 3t e^{3t}
so it doesn't go to zero at t=0.
I am sorry about the misnumbering of the solutions, I don't want to risk changing it now since it would mean renaming all the files, also, I have added a note to explain "!=", but might change it if I get chance.
Thanks for the yo-yo duck video!

15 May 2011
How can question 11 on your ODE work sheet be solved? Is it just a case of letting sin(4t) = (e^(4it) - e^(-4it))/2i?
Reply Yes in short, the solution is posted now.

15 May 2011
Fourier analysis 1.In tutorial 2, Question 2 I can get to the answer that (sint)^3 = 3/4sint - 1/4sin3t but I just don’t understand how this is a Fourier series or how we would know that this was the answer when we came to this point? Since you don’t need to use the Fourier equation at all for it? 2. Is f~(k) what the Fourier Transform is, or is the Fourier Transform the f(t) that we get when we sub f~(k) back into the equation for f(t)? I ask because in the tutorial answers there is an exmple of subbing f~(k) back in to find f(t) but this f(t) is in a different form from the original f(t) . . . I thought the Fourier Transform of f(t) was f~(k) and vice-versa or is that wrong? 3. I'm really stuck on tutorial 3, Q2. In the solution sheet, going from the 2nd line of working to the 3rd line, it looks like you’ve said e^ix = ie^y . . . . . (where x= t-tk & y= -tk) is that true? I don’t understand why that is, how can you move the 'it' out of the exponents power and put the 'i' in front of the e and the 't' has disappeared? I know it must be right I just can’t understand why that is.
Reply So for 1) it is a trick, you could do it using the Fourier formula but the integrals would be really hard, it turns out you can do it much more simply by spotting that you can just use trignometric rules to express sin^3 t in terms of sin nt and cos nt, therefore giving the Fourier series without having to do any integrals.
So it is sin^t = sin t(sin^2) = sint t (1-cos 2t)/2
and then use sin t cos 2t =(sin 3t - sin t)/2
2) \tilde{f(k)} is the Fourier transform, when you substitute that back in to get an integral formula for f(t)
f(t)= \frac{1}{2\pi} int_{-\infty}^\infty \tilde{f(k)} e^{-ikt} dk
that's called the Fourier integral of f(t), if you could do that integral it would give back the original f(t), if f(t) is continuous, and if f(t) is discontinuous it would be the same except at the discontinuities.
3) So going from the 2nd to the 3rd line I have am evaluating the limits, that is, I have put in that t=\pi/2 and t=-\pi/2, line 2 come from evaluating the integrals and there is a limit sign ]_{\pi/2}^\{pi/2} on it! Next use that e^{i\pi/2}=i and so on, you can see where these come from using the Euler formula e^{i\theta}=cos{\theta}+i\sin{\theta}

15 May 2011
ODE solns q10 Just looking at the first one now, should equation (3) not be -16 + 20i +6 = -i, if y = C*exp(4it)?
Reply Yes, thanks, hopefully it is corrected now.
Further message hat's great thanks. As a thank you, here's a capoeira cat.

15 May 2011
Series with constant term on RHS Sorry to bother you again, but I'm having a lot of trouble figuring out what to do for series solutions whenever you have something equal to a constant. This threw me in the schols exam as well. I was doing 2010 q6, and I ended up with a set of series on the left hand side, each starting at the same n and with same power of t, but they were equal to a constant on the right hand side. This could be wrong. I understand that the method is probably simple enough, but I thought I'd ask to get something definite. Thanks!
Results So you just do the constant term separate to the non-constant terms. This isn't an example you'd do using a series but say you had \dot{y}=5 so the series gives
\sum_{n=0} ^\infty (n+1)a_{n+1} t^n=5
then take out the n=0 term to give
a_1 + \sum_{n=1}^\infty a_{n+1}t^n=5
so a_1=5 is the only constant terms, for t^n n>0 we have a_n=0 and so the solution is y=a_0+5t.

15 May 2011
Fourier series and transforms Sorry for bothering you again...
I got the ODEs sorted but I dont fully understand when to use which formula. During the lectures, we did the expansion
f(t)=
{1 0 {-1 -Pi using both Ao, An, Bn formula as well with complex series formula Cn... Did we do this just to show that these formulas are identical(the latter being for complex fxns)
I don't fully understand which one when to use. For on/off pulse we always must use Fourier transform one, right? Thanks!
Reply You use the series for periodic functions and transforms for functions that decay to zero for t goes to plus and minus infinity. If you are asked for complex you must use c_n, if you are asked for the trigonometric series, use a_n and b_n, otherwise you have a choice between the two, if the function is even or odd, use the a_n and b_n since the even or odd allows you to set one or the other to zero.

15 May 2011
Series with a constant term in the schol paper. On the schols paper you asked for the series solution of t*dy/dt - 3y = k, where k is a constant. How is this done, I'm not sure what to do when it's not equal to zero. Am I right in saying that a_0 + \sum_{n=1}^\infty a_n(n-3)t^n=k, so a_0 = -k/3?
Reply That was a slightly tricky question, now you have a_0=-k/3, but where is the arbitrary constant to come from since the recursion relation seems to just say ? The answer is, when n=3 where is no equation so
y=a_3t^3-k/3