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Power series when to use Frobenius method

by John777
Tags: frobenius, method, power, series
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John777
#1
Oct28-10, 10:25 AM
P: 27
Hi, I'm new to the forum and need some help regarding my calc class. Any help you could provide would be greatly appreciated.

In doing a power series series solution when should I use the frobenius method and when should I use the simple power series method. The simple method seems a little faster, but I know there is a certain type of problem where you must use frobenius.

Frobenius being y=[tex]\Sigma[/tex]AnXn+s

Regular method being y =[tex]\Sigma[/tex]AnXn
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kof9595995
#2
Oct28-10, 11:44 PM
P: 679
These 2 are equivalent
John777
#3
Oct29-10, 09:06 AM
P: 27
Quote Quote by kof9595995 View Post
These 2 are equivalent
Don't take this the wrong way as I'm just trying to learn, but why do they teach both methods? There is no difference between them?

Mute
#4
Oct29-10, 10:11 AM
HW Helper
P: 1,391
Power series when to use Frobenius method

Quote Quote by John777 View Post
Don't take this the wrong way as I'm just trying to learn, but why do they teach both methods? There is no difference between them?
There is a difference between them, but for differential equations without a singularity at some value of x the difference disappears because you will be forced to conclude s = 0.

When you have a differential equation with a singularity at some value of x, you will find a non-trivial value of s when you do a power series around the singular point.

i.e., if you have a singularity at a point x = c, you would plug in a series

[tex]y = \sum_{n=0}^\infty A_n(x-c)^{n+s}[/tex]

and you would get s = some non-zero number. If there were no singularity at x = c, you would find s = 0.
Trifis
#5
Dec25-12, 03:02 PM
P: 148
Can you explain what correction does the xs factor contribute exactly? I don't see why the Frobenius method improves the failing ordinary power series method.
HallsofIvy
#6
Dec28-12, 06:45 AM
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Emeritus
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Thanks
PF Gold
P: 39,361
You use "Frobenius" method when the point about which you are exanding (the "[itex]x_0[/itex]" in [itex]\sum a_n(x-x_0)^n[/itex]) is a "regular singular point". That means that the leading coefficient has a singularity there, but not "too bad" a singularity: essentially that is acts like [itex](x- x_0)^{-n}[/itex] for nth order equations but no worse. Every DE text I have seen explains all that.


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