Solving ODEs by the Power Series Solution Method 

M.I.B I am not fully convinced but I think it doesn’t matter for this particular problem because at the end he releases the first two terms from one of the summations.

Yes exactly . I have seen ALOT of power series solution videos but in almost every one of them even after differentiating they start the series from n=0 . But I think that for y' and y'' , it should start from n=1 and n=2 respectively .

@@darkseid856 it doesn't matter because the first term for y' and first two terms for y'' are zero.

@@shayanmoosavi9139 it slightly matters from a didactic perspective

Yes, had me confused as well.

I think so too, there should be 1/2 on the sinh term.

I think it's a mistake but overall was a good lecture

The exponents would be negative, but the coefficients would be zero, so those terms don't really matter in the end. You can start them at 1 and 2 as you said, but you can also choose to start the summations at zero since the n = 0 for y' and the n = 0,1 terms for y'' are zero. Hope that helps!

Thank you for the kind words! I've already made that correction in the video description, but thank you for pointing it out!

Haha thanks! I've usually heard people using the long form (hyperbolic cosine), but it's good to hear there's an accepted short form for the pronunciation.

you clearly don't get the joke. A math professor undoubtedly knows what's "more typical."

Thank you!

The index you start at for the first and second derivatives 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 as you said, but I wanted to keep things consistent.

Ah, thanks for the clarification. Found your channel through Reddit and I'm really loving it so far. Thanks for the great content and fast response.

No problem and I appreciate the kind words!

You are correct about the powers of x. I'll edit the description to make note of that. Thanks! If you want to convert a power series to a function, you'll pretty much have to use a table. A lot of power series can't even be converted to closed form functions so in a lot of cases, you're better off leaving the series as it is. If you want to convert a function to a power series, you can always use the Taylor expansion. Generally though, converting a power series to a function is just a matter of having the series formula memorized, and as far as I know, there isn't much that can be done beyond that.

Advanced Engineering Mathematics by Erwin Kreyszig is one good option!

No problem, glad you like them!

Thank you so much!

The point of this video was to demonstrate the series method and to show that it's consistent.

I think you've mixed up convergence and equality. The mathematical expression at 6:00 is an equality; what you're talking about is convergence. Convergence just means that if I go far enough in the sequence/series, the coefficient term will eventually approach zero. For example, the series Sum from n = 1 to infinity of (1/n) converges as n approaches infinity, because the lim n -> inf of (1/n) = 0. The series itself is certainly NOT equal to zero because obviously 1/1 + 1/2 + 1/3 + 1/4 + ... =/= 0. However, when I write that a series is zero, I mean that all its terms SUM UP to zero. In this case, since x can take on a large array of real numbers, the only way I can guarantee my series is zero is if I set all its coefficients to zero. Hope that clears things up!

On track! He said differentiable, and that is not correct. Actually, I think it needs to be more than infinitely differentiable. f(x) = e^{-1/x^2} if x not 0 and 0 if x =0 is infinitely differentiable (including at x=0), but not analytic. Meaning, it's Taylor series centered at zero exists, but it does not equal its Taylor series anywhere but the anchor point of zero.

They're both the same thing, but I did mean 2^4 anyway.

Chan Dan no. The n “variables” are the coefficients of the series. All the a_n. They are variable until we know them. As he mentioned, they will be determined uniquely in these types of problems.

Pretty much; that's right. The general power series solution for y would be the same, but the coefficients a_n would be different.

Wow instant reply .Thank you so much for your videos, your work is really appreciated!!

You are most welcome!

Thank you! Glad you liked it!

wow that was quick still watching one of your videos currently !! subbed to you really struggled with the legendre functions for qm until you came

Haha yeah I'm logged in right now so I can instantly respond to my notifications. Also, that's great! I'm glad that my videos helped you!

Great thanks

You can still combine the like terms from p(x)y' and q(x)y to get n terms even when q and p are expressed as series themselves. And when you combine the like terms, you can then solve for the a_n's individually. Hope that helps!

Sorry, but I'm a bit slow. Could you please explain it with an example, e.g. y''+sin(x)y=0. (I happened to find a similar question in Maths stack exchange, which finally gives approximate solutions, as the answerer uses only finite terms in the Taylor expansion. Did you mean the same thing? If so, does it mean that we cannot derive the exact solution using power series solution here?)

There would still be n terms after you simplify and combine everything, because you only have n powers on x (e.g. x^0, x, x^2, x^3, ..., x^n, n is just however far you want to go). In the end, you'll get n equations for the n unknowns (a_n's). For your example, here's what it would look like: i.imgur.com/jQQbcZs.png

Except in class you can't pause, here you can. Your pace doesn't need to match the narrator's, just pause the video