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Frobenius series 
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#1
Apr2509, 07:37 PM

P: 187

Why do we assume that the first term c_{0} in a frobenius series cannot equal 0?
Thanks! 


#2
Apr2509, 08:34 PM

Sci Advisor
P: 1,594

Latex seems to be misbehaving, so I'll write in plain text:
In the Frobenius substitution, the x dependence of the first term is already factored out: y(x) = x^r Sum (a_k x^k) So, the first term in the series is actually a_0 x^r and when we plug the series into the differential equation, the question we are asking is "What is the smallest r for which a_0 does not vanish?" The answer is given by the indicial equation. After solving the indicial equation for r, we are then equipped to ask the next question: "Given that a_0 does not vanish, can I find some sequence a_k such that my formal sum converges and solves the differential equation?" 


#3
Apr2509, 08:56 PM

P: 187

Thanks for the response. Why is it necessary though that a_{0} not vanish?



#4
May509, 03:36 AM

P: 333

Frobenius series
Aa_{0}x^{r} + .... is identically zero. This implies Aa_{0}=0. We may assume a_{0} to be zero or nonzero. But if it is zero then A can be any number. Not an interesting result. 


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