You're probably one on the clearest communicators I have ever come across. You're not teaching, my friend. What you're doing is more than that......... Thank you for these awesome videos! I'm not formally educated so your content makes these ideas accessible to me in ways that a math textbook just can't. Keep up the great work, Faculty of Khan 🎊🔥
fabulous! you just explained Power Series solution in 11 minutes ...something that took my professors a whole month to do and they finally failed! Thank you ,teacher
Thank you for this video, i really don't get the concept of power series solution for ODE when reading the textbook. But after watching this video only in 11 minutes i really get the concept.
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 .
Epic video! Minor thing to point out: around 9:50 when you expanded the y(x) power series, you forgot to tack on the x^(2*n) and x^(2*n+1) pieces onto the nth terms. I got confused for a brief second lol. Again, wonderful presentation! I'm watching your entire series. :)
Wonderful video! This was a really helpful explanation, and I'll be pointing others to this. One minor error, though, early on you say that if y is assumed to be continuous and differentiable, then it is equal to its Taylor Series. I think a function like y = 0 if x=0 and e^(-1/x^2) otherwise, contradicts this. We can show that y is C^infinity, but it does not equal its MacLaurin Series (except at x=0). Really, we need to assume y is analytic, which isn't quite the same. Regardless, I will be referring to this and your other videos often! Many thanks!
Isn't the correct answer to the y''-4y=0 example y(x)=(a_0)cosh(2x)+(1/2)(a_1)sinh(2x), since the coefficients of the odd terms in the Taylor expansion are of form (2^2k)(a_1)/(2k+1)! rather than (2^(2k+1))(a_1)/(2k+1)! ? And does this even matter?
Dear Sir, Why the summations for y'(x) and y''(x) start in zero? I think they should start in 1 and 2 respectively, as in other case the exponents would be negative!
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!
Quick question, but when you differentiate power series don't you have to increase the value that n starts at each time? E.g. for the first derivative it goes from 1 to infinity, then 2 to infinity for the second derivative.
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.
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!
09:56 You forgot the powers of `x` for the last two coefficients. Anyway, in general, how can I know what function is encoded by a particular power series? How can I decode the power series back into a function? Is it just a table lookup? Or is there some more "algorithmic" method?
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.
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.
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.
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.
Em... In the example, the p(x), q(x) and r(x) are all constant and thus we don't need to take the Taylor expansion of them, but what if the coefficients are more complicated and we have to express them in series? In that case, the number of terms that corresponds to p(x)y' or q(x)y is n^2 rather than n. I'm curious what technique shall be used to solve for a_n then.
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
Just one complaint. Slow down your teaching at times I felt like I was right back in class furiously taking notes while completely losing what the prof is saying........ Khan is suppose to be different, just slow down, no one is timing you.......
You're probably one on the clearest communicators I have ever come across. You're not teaching, my friend. What you're doing is more than that......... Thank you for these awesome videos! I'm not formally educated so your content makes these ideas accessible to me in ways that a math textbook just can't. Keep up the great work, Faculty of Khan 🎊🔥
fabulous! you just explained Power Series solution in 11 minutes ...something that took my professors a whole month to do and they finally failed! Thank you ,teacher
these videos are timeless, thank you for the clear and fun explanations which my professors were never able to do
Thank you for this video, i really don't get the concept of power series solution for ODE when reading the textbook. But after watching this video only in 11 minutes i really get the concept.
In 10:47 shouldn't it be (a1)/2*sinh(2*x) in order to be consistent with the series? Or was (a1)/2 implicitly relabeled as a1?
I think so too, there should be 1/2 on the sinh term.
Confused. Shouldn't the indices for y' and y" start from n=1 and n=2 respectively?
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
Epic video! Minor thing to point out: around 9:50 when you expanded the y(x) power series, you forgot to tack on the x^(2*n) and x^(2*n+1) pieces onto the nth terms. I got confused for a brief second lol. Again, wonderful presentation! I'm watching your entire series. :)
Thank you for the kind words! I've already made that correction in the video description, but thank you for pointing it out!
Wonderful video! This was a really helpful explanation, and I'll be pointing others to this.
One minor error, though, early on you say that if y is assumed to be continuous and differentiable, then it is equal to its Taylor Series. I think a function like y = 0 if x=0 and e^(-1/x^2) otherwise, contradicts this. We can show that y is C^infinity, but it does not equal its MacLaurin Series (except at x=0). Really, we need to assume y is analytic, which isn't quite the same.
Regardless, I will be referring to this and your other videos often! Many thanks!
I wish you were my teacher... love your content!!
Thank you!
Brilliant! Whoever thought of this is genius!
Isn't the correct answer to the y''-4y=0 example
y(x)=(a_0)cosh(2x)+(1/2)(a_1)sinh(2x),
since the coefficients of the odd terms in the Taylor expansion are of form (2^2k)(a_1)/(2k+1)! rather than (2^(2k+1))(a_1)/(2k+1)! ?
And does this even matter?
Yes, had me confused as well.
Dear Sir, Why the summations for y'(x) and y''(x) start in zero? I think they should start in 1 and 2 respectively, as in other case the exponents would be negative!
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!
Quick question, but when you differentiate power series don't you have to increase the value that n starts at each time? E.g. for the first derivative it goes from 1 to infinity, then 2 to infinity for the second derivative.
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!
Can some somebody explain 6:00? Wouldn't it be possible to have a series where all coefficients are non zero but the series converges to zero?
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!
You have done a lot of help!
09:56 You forgot the powers of `x` for the last two coefficients.
Anyway, in general, how can I know what function is encoded by a particular power series? How can I decode the power series back into a function? Is it just a table lookup? Or is there some more "algorithmic" method?
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.
One recommendation: I think it's more typical to call cosh "kawsh" instead of "coshine."
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."
Can you recommend a textbook to follow for “homework” problems
Advanced Engineering Mathematics by Erwin Kreyszig is one good option!
Thank you for these videos, they are very helpful.
No problem, glad you like them!
2:27 do you mean that in system of equation, those 'n' variables are (x-x_0),(x-x_0)^2...,(x-x_0)^n ?
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.
At 7:48, did you mean to write 2^4 or 4^2?
They're both the same thing, but I did mean 2^4 anyway.
Can a possible reoccurence relation be: a_(n+2) = (4 / (n + 2)(n + 1)) * a_n ? where instead of going 2 terms back we go 2 terms forward?
Amazing explanation 🤩🤩🤩🤩🤩
what if the expanded terms are not zero? what to do then
You forgot your x^(2k) and x^(2k+1) when expanding out your series at around 9:46
Man, you are a genius, thank you for this vid
Thank you so much!
so if the equation was y''+p(x)y'-4y=0 the only thing that would change is the recursion relation ?
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!!
Thank you so much for your explanation, one question, which device &app you used for digital handwriting
Wouldnt it be easier just to change change it to e"(iwS) instead of doing the whole taylor series? Solves the equation same way... Just a question...
The point of this video was to demonstrate the series method and to show that it's consistent.
Thank you Sir ,,, you are a saviour
You are most welcome!
excellent video keep up the good work!!!!!!!!
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
In order to use Taylor series, the function must be infinitely (real) differentiable. Nice videos though. I've watched (and liked) quite a few.
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.
Em... In the example, the p(x), q(x) and r(x) are all constant and thus we don't need to take the Taylor expansion of them, but what if the coefficients are more complicated and we have to express them in series? In that case, the number of terms that corresponds to p(x)y' or q(x)y is n^2 rather than n. I'm curious what technique shall be used to solve for a_n then.
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
Hermano muchas gracias.
Magical!
👍👍
Just one complaint. Slow down your teaching at times I felt like I was right back in class furiously taking notes while completely losing what the prof is saying........ Khan is suppose to be different, just slow down, no one is timing you.......
Except in class you can't pause, here you can. Your pace doesn't need to match the narrator's, just pause the video
An android indeed
Quraan ko v dikhaye
You should try and sound more excited. Maybe get some female energy going. :)