Just want to confirm: Indicial equation is obtained by subbing in the y (mentioned in 7:45 of video) into your initial equation. Then looking at the coefficient of a_0? Can we find other indicial equations in similar initial equations by looking at the coefficient of a_0?
Yes, indicial equation is obtained by setting the coefficient of a_0 equal to zero. So if you don't want to remember the formula, this would be an alternate method.
after 8mins of the video. professor write frobenius general equation i.e y = x^r * sigma(An*X^(n+r)), is there a small mistake here, I think it should be sigma(An*X^n) not X^(n+r)
I had a question why x had to be 0 in P(x) and Q(x) of the indicial equation. Now I think I have work it out. Because P(x) and Q(x) can be witten as power series: P(x)=p0+p1*x+p2*x^2+... So items related to P(x)-p0 will make the power of x higher so that it will not exist on the very first item, the coefficient of the lowest power of x (or what we called indicial equation). But only P(0) will exist on it.
The 0 is your singularity. So if you chose a different center x₀, let's say 7, around which you expand your power series, then you would need P(7) and Q(7) instead. Unfortunately, Daniel didn't show any examples as of yet what should you do when the singularity is not at x₀=0, but somewhere else. But these examples can be easily generalized, from x and its powers, to (x-x₀) and its powers. Just substitute (x-x₀) everywhere you see x in your power series. So p(x) is in fact (x-x₀)·P(x), and q(x) is in fact (x-x₀)²·Q(x), and you evaluate them at your particular x₀ instead of 0.
You need to find if the singularity at x=0 is a regular singularity or irregular singularity ( ruclips.net/video/iEQKwwvfk2s/видео.html ) Let me state what happens if your DE is second order. 1. If x=0 is a regular singularity and the indicial equation has two solutions that differ by a non-integer, then you have two Frobenius solutions. 2. If x=0 is a regular singularity and the indicial equation has two solutions that differ by an integer, then you have one Frobenius solution at least. The other one maybe also Frobenius type, or it could be the Frobenius solution with a logarithmic singularity. 3. If x=0 is an ordinary point (not singular), then only power series solutions exist (no need for Frobenius). 4. If x=0 is an irregular singularity, then most likely it won't have any Frobenius type solution, but in rare cases it has just 1 Frobenius solution (but not two). For higher order DE with regular singularity, same principle applies. You have the same number of Frobenius solution as the order, but some of them may have logarithmic singularity if any roots differ by an integer.
@@daniel_an So what is it about "differing by an integer" that messes this up and why logarithms help? My guess is that it has something to do with the fact that the powers in a power series also differ by whole numbers, so it might be that the powers from one series coincide with the powers of the other, so they're not linearly independent. (Because, for example, it works when one series contains only even powers and the other contains only odd powers.) But I'm not sure why logarithms fix it.
@@bonbonpony You got the idea right about differing in n. I don't have a good insight why log fixes it, other than that log function satisfies the diff eq xy'-1=0 which has a singularity at 0. This d.e. indicates that some d.e. with singularities must have log as solutions, and it happens that regular singularities with integer differences in the indicial values sometimes have it.
Golden content! Thanks so much for the playlist on Frobenius method. Please sir, dividing through by (2x^2) assumes that x cannot be equal to zero, right?
Thanks! To answer your question, equation sign in differential equations is equality of functions, meaning that the two sides should be equal for ALL values of x. So we can safely divide by 2x^2 - just think of x being a non-zero value. Now, even after dividing by 2x^2, the two sides can still be considered to be equal for x=0 as well (This is called the identity theorem in complex analysis. It's a property of analytic functions.)
Yes you can. You can solve it using the Frobenius method. When you get the solution you will realize that the solution is in the form of a power series.
It can be applied, but it would be an overkill, considering that a regular power series would do the job equally well, just with less work needed to find it.
He explained it in one of the previous videos: because if it is (or in fact a bunch of first coefficients are 0 until the first non-zero one), you can just "shift" the series by multiplying it by a power of x, which is precisely what the Frobenius coefficient in front of it does. So you will have an equivalent solution, just with a different value for your "shift" r.
Silent for one minute, who never found this channel. Such clear explanation. Thank you very much.
The cleanest explanation i could find on this topic.. good job
I saw all three videos and I feel much confident about the Frobenius method. Thank you!
omg....analytic gibberish tripped me up soo much for no reason. Thanks for keeping it simple.
This guy knows what he is talking about. Thank you! Better than just throwing equations at us
I was so confused until I found this video, a very good explanation thanks!
I'm a second year chemistry student struggling with quantum mechanics, these videos have been a life saver. Thank you so much!
Just want to confirm: Indicial equation is obtained by subbing in the y (mentioned in 7:45 of video) into your initial equation. Then looking at the coefficient of a_0? Can we find other indicial equations in similar initial equations by looking at the coefficient of a_0?
Yes, indicial equation is obtained by setting the coefficient of a_0 equal to zero. So if you don't want to remember the formula, this would be an alternate method.
I actually understand this now, thank you.
Only if I was able to give an award ...
Thank u so much sir.
after 8mins of the video. professor write frobenius general equation i.e
y = x^r * sigma(An*X^(n+r)), is there a small mistake here, I think it should be sigma(An*X^n) not X^(n+r)
Indeed, that's a mistake. Thanks!
I had a question why x had to be 0 in P(x) and Q(x) of the indicial equation. Now I think I have work it out. Because P(x) and Q(x) can be witten as power series: P(x)=p0+p1*x+p2*x^2+... So items related to P(x)-p0 will make the power of x higher so that it will not exist on the very first item, the coefficient of the lowest power of x (or what we called indicial equation). But only P(0) will exist on it.
The 0 is your singularity. So if you chose a different center x₀, let's say 7, around which you expand your power series, then you would need P(7) and Q(7) instead. Unfortunately, Daniel didn't show any examples as of yet what should you do when the singularity is not at x₀=0, but somewhere else. But these examples can be easily generalized, from x and its powers, to (x-x₀) and its powers. Just substitute (x-x₀) everywhere you see x in your power series. So p(x) is in fact (x-x₀)·P(x), and q(x) is in fact (x-x₀)²·Q(x), and you evaluate them at your particular x₀ instead of 0.
Thank you for the video, sir. How do we find how many Frobenius series solutions exist?
You need to find if the singularity at x=0 is a regular singularity or irregular singularity ( ruclips.net/video/iEQKwwvfk2s/видео.html ) Let me state what happens if your DE is second order.
1. If x=0 is a regular singularity and the indicial equation has two solutions that differ by a non-integer, then you have two Frobenius solutions. 2. If x=0 is a regular singularity and the indicial equation has two solutions that differ by an integer, then you have one Frobenius solution at least. The other one maybe also Frobenius type, or it could be the Frobenius solution with a logarithmic singularity. 3. If x=0 is an ordinary point (not singular), then only power series solutions exist (no need for Frobenius). 4. If x=0 is an irregular singularity, then most likely it won't have any Frobenius type solution, but in rare cases it has just 1 Frobenius solution (but not two).
For higher order DE with regular singularity, same principle applies. You have the same number of Frobenius solution as the order, but some of them may have logarithmic singularity if any roots differ by an integer.
@@daniel_an Thank you so much for the explanation, sir. 🙏 I now have a better understanding of the Frobenius method.
@@daniel_an So what is it about "differing by an integer" that messes this up and why logarithms help?
My guess is that it has something to do with the fact that the powers in a power series also differ by whole numbers, so it might be that the powers from one series coincide with the powers of the other, so they're not linearly independent. (Because, for example, it works when one series contains only even powers and the other contains only odd powers.) But I'm not sure why logarithms fix it.
@@bonbonpony You got the idea right about differing in n. I don't have a good insight why log fixes it, other than that log function satisfies the diff eq xy'-1=0 which has a singularity at 0. This d.e. indicates that some d.e. with singularities must have log as solutions, and it happens that regular singularities with integer differences in the indicial values sometimes have it.
Golden content! Thanks so much for the playlist on Frobenius method.
Please sir, dividing through by (2x^2) assumes that x cannot be equal to zero, right?
Thanks! To answer your question, equation sign in differential equations is equality of functions, meaning that the two sides should be equal for ALL values of x. So we can safely divide by 2x^2 - just think of x being a non-zero value. Now, even after dividing by 2x^2, the two sides can still be considered to be equal for x=0 as well (This is called the identity theorem in complex analysis. It's a property of analytic functions.)
@@daniel_an Thank you so much sir for your speedy reply.
Sir, if x=0 is an ordinary point, can Frobenius method be applied?
Yes you can. You can solve it using the Frobenius method. When you get the solution you will realize that the solution is in the form of a power series.
Tunji Ade Thank you sir... I have got that!
Duk Man thank you...
No then we use legendary differential equation to solve it. If x=0 the double derivative Will be vanished which can't give the solution
It can be applied, but it would be an overkill, considering that a regular power series would do the job equally well, just with less work needed to find it.
Sir, why the constant term a_0 is non zero?
He explained it in one of the previous videos: because if it is (or in fact a bunch of first coefficients are 0 until the first non-zero one), you can just "shift" the series by multiplying it by a power of x, which is precisely what the Frobenius coefficient in front of it does. So you will have an equivalent solution, just with a different value for your "shift" r.
Finally get it!!! Thank u ❤️
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Amazing explanation, thank you!
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Yea saludos desde la facultad de quimica UNAM México
Thank you sir 🤍🤍
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Great professor, dead class.
Stupid-ass students😅