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Power series when to use Frobenius method 
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#1
Oct2810, 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]A_{n}X^{n+s} Regular method being y =[tex]\Sigma[/tex]A_{n}X^{n} 


#2
Oct2810, 11:44 PM

P: 679

These 2 are equivalent



#3
Oct2910, 09:06 AM

P: 27




#4
Oct2910, 10:11 AM

HW Helper
P: 1,391

Power series when to use Frobenius method
When you have a differential equation with a singularity at some value of x, you will find a nontrivial 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(xc)^{n+s}[/tex] and you would get s = some nonzero number. If there were no singularity at x = c, you would find s = 0. 


#5
Dec2512, 03:02 PM

P: 148

Can you explain what correction does the x^{s} factor contribute exactly? I don't see why the Frobenius method improves the failing ordinary power series method.



#6
Dec2812, 06:45 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

You use "Frobenius" method when the point about which you are exanding (the "[itex]x_0[/itex]" in [itex]\sum a_n(xx_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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