# Frobenius series

1. Apr 25, 2009

### IniquiTrance

Why do we assume that the first term c0 in a frobenius series cannot equal 0?

Thanks!

2. Apr 25, 2009

### Ben Niehoff

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. Apr 25, 2009

### IniquiTrance

Thanks for the response. Why is it necessary though that a0 not vanish?

4. May 5, 2009

### matematikawan

Let say the first term that we obtained on substituting the Frobenius series into the DE as

Aa0xr + .... is identically zero.

This implies Aa0=0.
We may assume a0 to be zero or nonzero. But if it is zero then A can be any number. Not an interesting result.

Last edited: May 5, 2009