Why is A0=1 in the Frobenius Method?

[asdf1]

for the frobenius method, why do you suppose that a0=1?

 

[HallsofIvy]

You don't. You assume that it is non-zero which is just a choice of "c".

The idea of Frobenius method is to write a solution in the form
\Sigma_{n=0}{\infnty}a_n x^{n+c}
where c, while a constant, is not necessarily a positive integer.
If I, choosing c to be, say, c_{0[/sup], I find that a0= 0 but a1 is not, I could just take c= c0+ 1 instead.}

 

[asdf1]

thank you very much! :)