I am studying geophysical engineering and have been learning for the last 6 weeks about Fourier series, transformations, integrals, the wave equation, and now moving into heat transfers. I had no idea why we were learning what we were and how it all relates, but within the first 6 minutes of your video you perfectly explain all of my questions. Great video, thank you!
an Australian teaching in an Australian university.... now that's a first. thank-you so much... its hard to understand my lecture... this makes it so much easier. keep up the good work.
This is really helpful; I come from a small university focused on research and not so much on teaching, and now I see what I'm missing! I'll be following these lectures for sure. Thank you!
Great Video! Exactly what I was looking for. You've explained it in a very easy way to understand. I've seen other videos but your video cleared up a lot for me. Thanks
Great question. We are really only interested in the temperature function on the interval, say, (0,L), that is, the region where the bar lies. Outside the interval we can do whatever we like to the temperature function - it does not matter. Since Fourier series rely on periodic functions, that's how we extend the temp function outside the interval (0,L). Hope this is useful.
Such a great teacher; a Professor. Making things so simple. I wish somebody had provided you with better equipment to do your great work without a blurry screen! Wish you all the best Sir!
The advantage of Fourier series over Taylor series is beautifully explained. You have a way of making mathematics really "cool"! (no reply please - you have enough comments).
Good question. I wanted to make the comparison between Fourier series and Taylor series. What can Fourier series do that Taylor series cannot do? This is an important question that I try to answer. Let's think about the absolute value funtion f(x) := |x| on the interval (-1,1). This function does not have a Taylor series at zero because it is not differentiable, but the function does have a Fourier series. Hope this helps.
Intuitively, I can rephrase my wording as: from -pi to 0, average value is zero, from 0 to +pi, average value is pi/2. Hence, the average value over the interval -pi to +pi is (zero+pi/2)/2 = pi/4 This intuitive method works in example covered during 22:00 because this is not a complicated function. :)
The orthogonality segment suggests application to stream and potential functions in fluid dynamics. Also: At 23:51, you take the integral in two sections (-pi to 0, 0 to pi) but the function is in three (f(x) = x for 0 < x < pi and pi/2 for x = pi). Whither the validity in extending the upper bound, or is it so close as to make no discernible difference?
Great lecture. I am still insure why the heat equation is associated with a method of representing periodic functions (?). I really like the form factor of writing over notes--it makes the lecture dynamic, but keeps a lot of information in view. Helps that you have excellent writing. Thanks
Good Lectures. my two cents: at 22:18, you said consider the a(subknot) as area, wouldn't that be more accurate if I say a(subknot) is the average value of signal over the range 2pi? Because indeed, a(subknot) is called the DC Component in EE.
@syrenissen The videos are based on a course for engineering students where they are applied to solve basic linear partial differential equations (and some ordinary differential equations).
In 15:57, why is it true that Sum (a_m*pi), m from 0 to infinity is always finite? I didn't get why you wrote the infinite sum as a_n where n is some fixed positive integer, as you say.
Hi, the expression at 8:56 is the integral of the product of orthoganal functions. This is demonstrated to equal zero. But the expressions said to be zero at 13:56 are not integral of orthogonal products, but integrals of functions.--they do not appear to be products And so it is unclear to this viewer why they are zero. Thank you for responding.
At 14:30 terms are said to go zero with no explanation. Am I the only who thinks the editing out of significant steps --"leave it for you to do"--makes even intermediate concepts like this hard to follow.
All should be revealed if you compare my integral formula for a_0 with your integral formula for a_0. Try it and see if you can now understand why the two expressions are equivalent.
I've got a question if someone could please answer, why is it at 26:45 that it integrates to 1/(n^2 × π)? i get the n^2 bit but not the π part as you're integrating [1/n × sin(nx)]? As when you plug in π into the equation you get (cos(nπ)-1). If n represented 'nπ' which would make sense as to how he got {1/n^2 × π} then why doesn't the equation resolve to cos[(nπ^2)-1]? hope that makes sense aha, its the first bit that catches me up?
Hi sir, plz provide me answer for this.. why the sine and cosine functions can be used to construct any continuous functions in a Fourier series, while other sets of functions cannot.
They have not appeared yet because I have not taught in a course in which they appear. Virtually all of the topics for my videos are the same as what I teach in class.
Thanks! If you are enjoying my videos then you also might like my new (and free ebook) which is designed to be used together with my videos. The link is on my YT Channel (bookboon.com)
Hello! I have a question. What effect does the function f(x) = Pi/2 when x = -Pi have to do with the function?. You only drew it on the graph and not using it to calculate the Ao, An, and Bn. If it does not have anything to do with the process, why is it there then? Thank you.
sir,u said that every periodic function can be expressed in terms of sines and cosines but u involved a0 which is a constant .And don't say that a0 can be expressed as in terms of sines and cosines
Dear Dr Chris, i am a teacher from Singapore. Been wanting to do a class video that looks like your. Could you share the setup and the equipment used in the the taking of the video? Would like to learn from you so that i could do the same with my students. =)
Great video intro on Fourier series. I did think there is an error in the calculations for An. I think it should be 1(cos n pi - 1)/n^2 instead of the answer you gave as 1(cos n pi - 1)/n^2 pi.
Dr Chris Tisdell at 26min 47sec into the video you are integrating 1/n * (-sin nx) which should give you 1/n * 1/n *(cos nx) but somehow you slipped in an extra pi onto the bottom making your answer 1/n * 1/n pi *(cos nx). Hope that is a better explaination.
you should thank him for yelling, most profs arn't understandable because of the low voice. Plus this loud voice will always keep you up in class and you really do need that in this course.
Calculate the area under the graph and divide by the length of the interval. Or, split in the integral from -\pi to \pi into two separate integrals: from -\pi to zero; and from zero to \pi.
UrbanMedia the integrals from -pi to pi, the function is defined differently for that region so split it into two integrals (-pi to 0 and 0 to pi) and then use the respective f(x) and you should get the same answer.
Hi - the video is available in high definitition, 720p. All you need to do is select it where you see the little gear symbol in the right-hand lower corner.
اليس هناك تنسيقات اخرى غير روسيا و تركيا كن صادقا ومنصفا في تقريراتك بل هناك عدة دول اجنبية فرنسا وايطاليا وامريكا....................................الكل يبحتون على مصالحهم الشخصية.............................
