Solving ODEs by Series Solutions: Legendre's ODE 

Thank you so much!

Any tips on how hard was to get into grad school at Cornell?

OMG a fellow Cornelian!

i guess I am quite off topic but does anybody know of a good website to watch new movies online?

@@finleyahmir7054 😬

Thank you! I'm glad you find my lectures useful!

Thank you! Glad you liked it!

I'll put it on my to-do list.

Hopefully it will also save the lives of all the people who could have been blown up by a nuclear reactor meltdown if this video didn't help you ;)

@@scitwi9164 😂

Glad you found it useful!

Mine's in like an hour

Thank you! Glad you liked it!

No problem, glad you like it!

Glad you like it! Thanks!

No problem, thank you for the kind words! And yeah, that's just how I make my videos (faster writing just allows the lecture to flow better IMO): hopefully it's not too bad but you can still slow down the video or pause.

I don't have to, since the term corresponding to n=0 for the first derivative and those corresponding to n=0,1 for the second derivative are all zero. I could have started them at n=1 and n=2 like you said, but I wanted to keep the starting point consistent.

Haha thank you! Though I'm curious what mm3 is.

mathematical methods 3, its year 2 physics course at ucl (university college london)

That sounds like an interesting course. Good luck!

Danny Best its as if this guy made these videos for us, small world

Thank you Gerardo! When k = n, the coefficient becomes zero, and the series terminates at that coefficient (giving you a polynomial solution). So for example, when you have k = 3, the a5 term becomes zero because of the (3-k) factor. I believe I discuss this in the video. Hope that helps!

Thank you! Glad you liked it!

No problem! I'll definitely put that on my to-do list. My schedule this term is rather packed but whenever I get time, I'll get to that to-do list!

Nope. Graduate student :)

Awesome.. Good luck.. :D

No, it doesn't have to. It doesn't matter, since the term corresponding to n=0 for the first derivative and those corresponding to n=0,1 for the second derivative are all zero. I could have started them at n=1 and n=2, but I wanted to keep things consistent.

@@FacultyofKhan bumping the indexes at least for the second derivative would save us some headache later with matching the indexes ;)

No problem! I'll definitely put that on my to-do list. Thank you for the feedback!

But sir can you have the orthonormality of polynomials also

Sure, I'll try.

but sir as soon as possible please.. because my finals are approaching and I have to prepare

using rodriguez formula: P2(x)= (1/2)(3x^2-1) but in legendre that you presented. a0 is arbitrary constant. that is my question. how to know the exact value of a0 which is -1/2

@@adonisbibat3895 I have this exact same question right now. Did you figure it out?

@@randytudor418 I'm confused about it too. would anyone please enlighten me please?

Sure thing! I'll put that on my to-do list. It shouldn't be too hard considering that I have the material already.

I'll also link it here so you can see when I put it up. Expect another reply in a week or so.

Thanks a lot, that would be great!!! ^.^

No problem! Yes, the power of each sum has to be consistent. Otherwise, I wouldn't be able to combine the terms like I did in 4:24.

ok, and thanks for responding

Thank you very much!

For anyone else who had the same question (I did), thought about this and came to this conclusion… Since y is a linear combo of odd and even, if the odd terminates but the even diverges, then a0 must be =0 in order for the series to converge and describe y. If the even terminates, then the odd won’t and a1 must be =0 to converge to y.

Also, I know Legendre polynomials are orthogonal under the inner product that is the integral of the product from -1 to 1. Is there an easy way to tell this from the differential equation?

Sure! No problem.

thanks a lot u r the best

The even series is still part of the solution to the ODE. It's just that it's not a Legendre polynomial because it continues forever (doesn't terminate). Hope that helps!

Still a Legendre function, though, which is kinda "in between" the two subsequent Legendre polynomials, kinda like a fraction is in between two whole numbers ;)

Yes, you can solve Legendre's equation with the Frobenius method as well. I find the method used in the video to be easier though.

Faculty of Khan thanks.

It can be solved by the (more general) Frobenius's method as well, but it's like using a sledgehammer to crack a nut ;)

@Aryaman Prajin Oh haha how couldn't I see that, thanks!

It's not unsolved; it's just that one of the solutions is a finite polynomial while the other is an infinite series. The finite polynomial is dubbed the Legendre polynomial.

I just used them so I could get expressions for the Legendre polynomials. For those exact values of a0, we have the Legendre polynomials. Hope that helps!

@@FacultyofKhan if we use different value of a0 or a1.. that will derived different answers. it will not be equal anymore using rodriguez formula. please enlighten us

You probably could, but when you look at the indicial equation, your solution would just be r = 0, which is the same as the regular series solution method.

doesnt the initial value of n increase in the differentials. like n=1 for first derivative and n=2 for the second derivative

It doesn't matter, since the term corresponding to n=0 for the first derivative and those corresponding to n=0,1 for the second derivative are all zero. I could have started them at n=1 and n=2 like you said, but I wanted to keep things consistent.

lol well I can't really change my voice now; still, I hope you at least found the lesson useful. You can also turn on captions and put it on mute if the voice is too boring.

yeah thats better idea....thanks

Because the n = 0 term in the derivative (and the n = 0, 1 terms in the second derivative) is zero anyway. It's not necessary to change the index; I didn't do it so I could keep things simple and consistent.

P2(x) corresponds to k = 2, and y_even represents the terms that come from using even indices k on the recursion relation. Also, for a0 = -1/2, y_even becomes the second-order Legendre polynomial, which I've denoted as P2. Hope that helps!

Advanced Engineering Math by Kreyszig (did I spell it right) for this series.

It doesn't matter if the summation is zero for m = 0 and m = 1; I could start it at m = 2 like you said but it doesn't make a difference since the m = 0 and m = 1 terms are zero anyway.

Sure! If you go back to 8:06, you can see the recursion relation at the top (a_n+2 as a function of a_n). All the last part is saying is that when k is an ODD number, the coefficients a_n for ODD values of n continue only up till a certain point, because when n reaches k, you can see that the numerator of the recursion relation immediately becomes zero. Now if one of the coefficients (a_n or a_k) reaches zero, all coefficients that come after that (e.g. a_n+2, a_n+4, etc) for odd values of n will also be zero. This means that the 'odd' index series would terminate, and we end up with a polynomial involving odd powers of x (e.g. P_1, P_3 etc). The same logic applies to EVEN values of k, in which the coefficients a_n for EVEN values of n eventually terminate. If you have any more questions, please ask!

+Professor Khan Thank you Prof!! I really appreciate your respone.. Thank you soo muuchh!!😃

No worries! Glad I could help!

lul

step 1. start lifting, step 2. wait 1-3 years, step 3. learn guitar/piano , step 4. buy a puppy, step 5 walk. step 6. talk, step 7, get her phone nubmer , step 8 sleep, and repeat

so, mechanical engineers are lonely in all parts of the world?

When tf did this turn into a pickup channel?

Or start picking up lonely mechanical engineer girls ;) You can always impress them with solving some crazy ODE :) I wouldn't go with muscles myself, because chick that love muscles are usually empty and boring as hell :P so you don't want to hang out with them anyway, trust me :q

@@3631162 Relax and lighten up cringehere. So just because textbooks show photos of Laplace in formal coat and not laughing does not mean he did not have fun. Think about that.

I believe so! I remember it was used in my Electricity & Magnetism course.

give me name of the video

or else a written solution

Well, I haven't made a video on that right now. But you can look here starting at section 2.2: www.physics.usu.edu/Wheeler/EM3600/Notes08SolvingForPotentials.pdf Hope that helps!

I'm not sure I fully understand your question, but I'll try to explain it according to what I pieced together. I feel like you've asked 2 questions: 1. When you shift the index of the 1st series (around 3:00), why don't you shift the index of all the other series? Answer: The reason is that for now, I just want to make sure the x's are all raised to the same power, so it's easier to change the first series. 2. Why don't you change the starting value of n when you differentiate the series solution y = sum from n = 0 of an*x^n? Answer: I've discussed it in previous responses, but it doesn't matter, since the term corresponding to n=0 for the first derivative and those corresponding to n=0,1 for the second derivative are all zero. I could have started them at n=1 and n=2, but I wanted to keep things consistent.

I'm sure you can also get a chrome extension that allows you to custom adjust your speed (e.g. Video Speed Controller) for next time! Apologies about the speed here though; my videos are slower now FWIW.