These videos are incredible! It is so hard to find videos on such advanced (relatively speaking) topics because Khan Academy only teaches stuff until freshmen year of college. Thank you so much for putting in your time and effort. You are truly making a difference and making the lives of STEM majors around the world so much easier!
An amazing consequence of this logic is that you y_n and y_m can be used as basis functions for expansions of known functions, so for example, the Fourier series of a function is not special. Another cool thing to notice is that the Fourier series can be written using exponentials, and so when you get into the Fourier transform, you can only do said transform if the input functions are "slower" than the exponentials. This places an arbitrary limit on the kinds of functions that are compatible with Fourier analysis. But this is also nice because it means that generating basis functions from other SLPs could easily generate integral transforms that can handle "faster" functions. I thought this was an awesome thing to share, maybe try it out :)
Hey I have a doubt, since both the eigenfunctions are orthogonal, their inner product is going to be zero i.e ym*. yn r(x) dx integrated over interval [a,b] = 0 but here it's written only as ym.yn r(x).dx and not the complex conjugate of ym. Is it because ym is real and thus ym*=ym . Solutions can be complex too right? :/
@@ruchi9917 a good way to think about it is using the Fourier series. We say a periodic function can be represented as a sum of sines and cosines, like f(x) = Sum[A_n*sin(n*x) + b_n*cos(n*x), n=0 to inf] but how do we know what the coefficients a_n and b_n are? In the standard method, we multiply the entire equation by some function, say cos(m*x), and integrate over an appropriate domain. cos(m*x) and cos(n*x) have orthogonality relations like the SLP shown in the video, similar for cos(m*x) and sin(n*x), and this is also true if you change the function you use to multiply the entire function from cos(m*x) to sin(m*x). You have your coefficients now, so you have a Fourier series (congratulations). Now, notice that the domain of integration is finite (he shows this in the video) so what would happen if our Fourier series was defined on an infinite domain? For starters, a periodic function that takes an infinite amount of time to repeat, so at that point we’d be studying an a-periodic function. Secondly, as the domain of integration goes to infinity, the Fourier series converges to an indefinite integral. This is the Fourier transform (assuming you converted the cosine and sine terms to one exponential term using Euler’s magic formula e^( i*w*t) = cos(w*t) + i*sin(w*t), otherwise you’ll get two trig transforms, which is fine, just not as compact to write). So this relied on us identifying two sets of functions that had nice integral orthogonality properties and our guess for sines and cosines worked. But it’s much harder to guess this outcome for something like a Laplace Transform, say. Teachers usually say that you’re just multiplying your input signal by an exponential function to manage the exponential blow-up of unstable functions, and while that’s what happens, you’d really be hard pressed to guess that e^(-s*t) where s E C is the way to go. The SLP is a machine that can find these functions. Set your p(x), q(x), and r(x), and it will generate a two infinite sets of mutually orthogonal functions that you can use to construct powerful, new Integral transforms. I’ve never read about this application of the SLP, but it makes sense, right? I was curious about it last time I watched this video, but didn’t act on it. I think I’ll do some math to see if it works. The real struggle in these things is just solving the 2nd order ODE (I have a non-standard algorithm for this using spectral decomposition and finite element analysis, which is maybe overkill, but it always works - thank you linear algebra!) I hope this explained what I was observing. Edit: sorry if some part of my explanation wasn’t given in the “standard” way, I’m self-taught and only have high-school education so I don’t always say things in the best way. It just makes sense in my head.
@@ozzyfromspace Even I'm self taught and I'm still learning. Please tell me which books you referred, because I want to get on your level of knowledge! Currently I'm halfway through Mathematical methods for Physics and Engineering by Riley, Hobson and Bence. It gets confusing but then these lectures are a save!
@@ruchi9917 For mathematical physics you may also check out the book by Mary L Boas though I think you have completed mathematical physics by now. Also, if you want to know about other good books for other subjects in physics you may refer to any standard Physics honours syllabus of universities like Delhi University , and for M.Sc level, of IITs.
@@ruchi9917 I think it is because the operator is self adjoint and hence its eigenfunction is self adjoint as well i.e. ym is equal to its complex conjugate
Nice proof, i just want to add that the Sturm-Liouville Theorem also includes the fact that one can express any function on the interval [a,b] as an sum of these eigenfunctions. As special cases you get the bessel , legendre , and hermite functions can all be used to express any function interms of them. This gets back bessel , legendre , hermite functions satisfying certain types of Strum-Liouville equations
@@FacultyofKhan Hey I have a doubt, since both the eigenfunctions are orthogonal, their inner product is going to be zero i.e ym*. yn r(x) dx integrated over interval [a,b] = 0 but here it's written only as ym.yn r(x).dx and not the complex conjugate of ym. Is it because ym is real and thus ym*=ym . Solutions can be complex too right? :/
Possibly more than you wanted to know, but check out www.maths.ed.ac.uk/~v1ranick/papers/lutzen.pdf More generally, the SL eqn is the most general form for a second order ODE. These second-order equations are everywhere in physics and engineering. See the many other fine videos on this channel on ODEs and PDEs for examples.
Do any of your videos show to to solve a S-L periodic system? The example in my text skips so many steps, I can't tell how they substituted the BCs into the characteristic equation. I hate my textbook. PS: I signed up on Patreon yesterday. Please don't announce my pledge level. Thanks!
Are you referring to a system with periodic boundary conditions? This video might help, though I don't have a clear idea of the example you're dealing with: th-cam.com/video/heY4cS1v870/w-d-xo.html If you have the example on hand, you could take a picture and post it here and I'll have a look! As for Patreon, I don't know which of my patrons you are (or if you've even pledged yet), but you can look at the description for this video to see if you're on that list: th-cam.com/video/4p83iaUeTro/w-d-xo.html Usually, I ask my patrons beforehand about mentioning them in my pledges. If they don't respond, then I assume they want the full reward (full mention + mention in description). However, if you're on my current patron list and I mentioned your name, then I can take down my video and reupload an edited one (not many people have seen it so hopefully it shouldn't be too bad).
Faculty of Khan Yes, I watched it. I might have to study it more if my professor decides he wants us to derive S-L. Yes, it is a periodic boundary. I am starting the problem over for the third time and trying to match the example from class notes. So far, lamda is less than zero and C1=-C2. I think I am on to something here. :)
Hmm, so the first video I posted was about time dependent boundary conditions. You're referring to periodic BCs, which is a bit different (and it's not something I've covered so far). I can cover it in a later video though; thanks for the suggestion, and good luck with your assignment! Also, you can PM me about the patron thing if you want.
Faculty of Khan Actually, I think I created a second Patreon account when I signed up with you. So, they are fixing it right now. Hopefully by this evening, everything will be fixed. Thanks for the help and support!
Sir, with all due respect, you should check on you pronunciation of "homogeneous." You are saying it like "homogenous." There are two "e's" and neither is silent.
The solution to your equation is y(t)=A*e^(i*sqrt(lambda)*t) + B*e^(-i*sqrt(lambda)*t). You can tell by asking yourself what kind of functions look like themselves when you take 2 derivatives. This matters because if the functions look like themselves, you can set the constants so that everything cancels and you get zero. This allows you to write the solution by inspection. Also, your boundary conditions don’t make sense (certainly not in the context of Sturm-Liouville theory). Try to figure out why :) Edit: you can write the solutions using sines and cosines if you really want to.
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These videos are incredible! It is so hard to find videos on such advanced (relatively speaking) topics because Khan Academy only teaches stuff until freshmen year of college.
Thank you so much for putting in your time and effort. You are truly making a difference and making the lives of STEM majors around the world so much easier!
Thank you for the kind feedback!
I realize I'm kinda off topic but does anybody know of a good place to stream new series online?
@Angel Emmanuel I watch on flixzone. Just google for it =)
@Richard Caspian definitely, have been watching on Flixzone for since april myself :)
@Richard Caspian Thanks, I went there and it seems like they got a lot of movies there :D I really appreciate it!!
An amazing consequence of this logic is that you y_n and y_m can be used as basis functions for expansions of known functions, so for example, the Fourier series of a function is not special. Another cool thing to notice is that the Fourier series can be written using exponentials, and so when you get into the Fourier transform, you can only do said transform if the input functions are "slower" than the exponentials. This places an arbitrary limit on the kinds of functions that are compatible with Fourier analysis. But this is also nice because it means that generating basis functions from other SLPs could easily generate integral transforms that can handle "faster" functions. I thought this was an awesome thing to share, maybe try it out :)
Hey I have a doubt, since both the eigenfunctions are orthogonal, their inner product is going to be zero i.e ym*. yn r(x) dx integrated over interval [a,b] = 0 but here it's written only as ym.yn r(x).dx and not the complex conjugate of ym. Is it because ym is real and thus ym*=ym . Solutions can be complex too right? :/
@@ruchi9917 a good way to think about it is using the Fourier series. We say a periodic function can be represented as a sum of sines and cosines, like f(x) = Sum[A_n*sin(n*x) + b_n*cos(n*x), n=0 to inf] but how do we know what the coefficients a_n and b_n are? In the standard method, we multiply the entire equation by some function, say cos(m*x), and integrate over an appropriate domain. cos(m*x) and cos(n*x) have orthogonality relations like the SLP shown in the video, similar for cos(m*x) and sin(n*x), and this is also true if you change the function you use to multiply the entire function from cos(m*x) to sin(m*x). You have your coefficients now, so you have a Fourier series (congratulations).
Now, notice that the domain of integration is finite (he shows this in the video) so what would happen if our Fourier series was defined on an infinite domain? For starters, a periodic function that takes an infinite amount of time to repeat, so at that point we’d be studying an a-periodic function. Secondly, as the domain of integration goes to infinity, the Fourier series converges to an indefinite integral. This is the Fourier transform (assuming you converted the cosine and sine terms to one exponential term using Euler’s magic formula e^( i*w*t) = cos(w*t) + i*sin(w*t), otherwise you’ll get two trig transforms, which is fine, just not as compact to write). So this relied on us identifying two sets of functions that had nice integral orthogonality properties and our guess for sines and cosines worked.
But it’s much harder to guess this outcome for something like a Laplace Transform, say. Teachers usually say that you’re just multiplying your input signal by an exponential function to manage the exponential blow-up of unstable functions, and while that’s what happens, you’d really be hard pressed to guess that e^(-s*t) where s E C is the way to go. The SLP is a machine that can find these functions. Set your p(x), q(x), and r(x), and it will generate a two infinite sets of mutually orthogonal functions that you can use to construct powerful, new Integral transforms.
I’ve never read about this application of the SLP, but it makes sense, right? I was curious about it last time I watched this video, but didn’t act on it. I think I’ll do some math to see if it works. The real struggle in these things is just solving the 2nd order ODE (I have a non-standard algorithm for this using spectral decomposition and finite element analysis, which is maybe overkill, but it always works - thank you linear algebra!) I hope this explained what I was observing.
Edit: sorry if some part of my explanation wasn’t given in the “standard” way, I’m self-taught and only have high-school education so I don’t always say things in the best way. It just makes sense in my head.
@@ozzyfromspace Even I'm self taught and I'm still learning. Please tell me which books you referred, because I want to get on your level of knowledge! Currently I'm halfway through Mathematical methods for Physics and Engineering by Riley, Hobson and Bence. It gets confusing but then these lectures are a save!
@@ruchi9917 For mathematical physics you may also check out the book by Mary L Boas though I think you have completed mathematical physics by now. Also, if you want to know about other good books for other subjects in physics you may refer to any standard Physics honours syllabus of universities like Delhi University , and for M.Sc level, of IITs.
@@ruchi9917 I think it is because the operator is self adjoint and hence its eigenfunction is self adjoint as well i.e. ym is equal to its complex conjugate
What a fantastic video. Do more proof videos like this. Please
Thank you!
Nice proof, i just want to add that the Sturm-Liouville Theorem also includes the fact that one can express any function on the interval [a,b] as an sum of these eigenfunctions. As special cases you get the bessel , legendre , and hermite functions can all be used to express any function interms of them. This gets back bessel , legendre , hermite functions satisfying certain types of Strum-Liouville equations
Exactly! Sturm-Liouville also includes completeness as you mentioned, but that's quite hard to prove haha
@@FacultyofKhan Hey I have a doubt, since both the eigenfunctions are orthogonal, their inner product is going to be zero i.e ym*. yn r(x) dx integrated over interval [a,b] = 0 but here it's written only as ym.yn r(x).dx and not the complex conjugate of ym. Is it because ym is real and thus ym*=ym . Solutions can be complex too right? :/
Great explanation , I loved it ! Thanks so much .
Supercool precise presentation .. keep it rockin buddy
what is sturm loveulli thery???what is the physcl meaning of this
amazing description 👌👌👌
Hello,
one question please, i searched what was the motivation to create this theory (how or why did Sturm and Liouville started this theory) ?
Possibly more than you wanted to know, but check out www.maths.ed.ac.uk/~v1ranick/papers/lutzen.pdf
More generally, the SL eqn is the most general form for a second order ODE. These second-order equations are everywhere in physics and engineering. See the many other fine videos on this channel on ODEs and PDEs for examples.
Well explained.... Thanks
Great video. I wonder what software you use to generate such video please?
I use Camtasia to record my videos on the computer.
Dude, I love inner product spaces.
Do any of your videos show to to solve a S-L periodic system? The example in my text skips so many steps, I can't tell how they substituted the BCs into the characteristic equation. I hate my textbook. PS: I signed up on Patreon yesterday. Please don't announce my pledge level. Thanks!
Are you referring to a system with periodic boundary conditions? This video might help, though I don't have a clear idea of the example you're dealing with:
th-cam.com/video/heY4cS1v870/w-d-xo.html
If you have the example on hand, you could take a picture and post it here and I'll have a look! As for Patreon, I don't know which of my patrons you are (or if you've even pledged yet), but you can look at the description for this video to see if you're on that list:
th-cam.com/video/4p83iaUeTro/w-d-xo.html
Usually, I ask my patrons beforehand about mentioning them in my pledges. If they don't respond, then I assume they want the full reward (full mention + mention in description). However, if you're on my current patron list and I mentioned your name, then I can take down my video and reupload an edited one (not many people have seen it so hopefully it shouldn't be too bad).
Faculty of Khan Yes, I watched it. I might have to study it more if my professor decides he wants us to derive S-L. Yes, it is a periodic boundary. I am starting the problem over for the third time and trying to match the example from class notes. So far, lamda is less than zero and C1=-C2. I think I am on to something here. :)
Hmm, so the first video I posted was about time dependent boundary conditions. You're referring to periodic BCs, which is a bit different (and it's not something I've covered so far). I can cover it in a later video though; thanks for the suggestion, and good luck with your assignment!
Also, you can PM me about the patron thing if you want.
Faculty of Khan Actually, I think I created a second Patreon account when I signed up with you. So, they are fixing it right now. Hopefully by this evening, everything will be fixed. Thanks for the help and support!
Ah okay. For a second I was worried that I mentioned something in my videos that I shouldn't have. And no problem, glad to help!
Thanks
Can you make video on Hermite ODE and Hermite Polynomials
Sure can! th-cam.com/video/lyqimm6w2kU/w-d-xo.html
@@FacultyofKhan ❤
Thanks 👍 Really appreciate it 💕
Thank you so much
Simple and awesome
Thank you sir!
Is it also true that these operators are diagonisable
Is that an App on a pad you are using to write in the video? What’s the name plzzz
I'm using a Wacom tablet.
Faculty of Khan Thanks :)
Faculty of Khan Thanks :)
what is k1 k2??
That’s my topic for project
my hero
i’m pretty high and have no idea what just happened in this video
😂
next video?
I introduce PDEs in the next video, right here: th-cam.com/video/O3ahEHAX-KU/w-d-xo.html
In the middle of the proof I think I see the product rule pop up again.
I think I’m in the wrong corner of TH-cam lmao
What kind of sick person doesn't want to see the proof?
Sir, with all due respect, you should check on you pronunciation of "homogeneous." You are saying it like "homogenous." There are two "e's" and neither is silent.
Cool
Solve the this Sturm-Liouville problem please. Urgent
y"(t) + λy(t) = 0
y(0) = y'(π) and y'(0) = y(π).
suataltunc@gmail.com
The solution to your equation is y(t)=A*e^(i*sqrt(lambda)*t) + B*e^(-i*sqrt(lambda)*t). You can tell by asking yourself what kind of functions look like themselves when you take 2 derivatives. This matters because if the functions look like themselves, you can set the constants so that everything cancels and you get zero. This allows you to write the solution by inspection.
Also, your boundary conditions don’t make sense (certainly not in the context of Sturm-Liouville theory). Try to figure out why :)
Edit: you can write the solutions using sines and cosines if you really want to.
He’s too fast
you might want to condense the text you are using and not clutterfuck the video with so much text