# What is the intuition/meaning behind the first characteristic polynomial of a linear multistep method?

Given a linear multistep method $$y_{n+s} + \sum_{k = 0}^{s-1} a_k y_{n + k} = h \sum_{k=0}^{s} b_k f_{n + k}$$ the first characteristic polynomial is defined as $$\rho(z) = z^s + \sum_{k=0}^{s-1} a_k z^k.$$

A lot can be said about the method just by looking at the properties of the polynomial $$\rho$$, but why?

For example, the method cannot be consistent if $$\rho(1)\neq0$$. This is the only relation that make sense to me since the new value $$y_{n+s}$$ has to be a "complete" linear combination (i.e. $$\sum_{k=0}^{s-1}\alpha_k=-1$$) of the previous values plus a displacement given by the right hand side of the method's expression.

But why, for example, the method is stable iff $$\rho$$ meets the criterion of Dahlquist? I understand each step of the proof, but I still don't get the intuition.

(The criterion of Dahlquist is met if the polynomial only has roots with $$|z| \leq 1$$ and the roots with $$|z|=1$$ are simple).

• Thinking of it as the characteristic polynomial of a recursive sequence may help. – David Rubio Jan 16 at 19:34

The same way that $$\rho(1)=0$$ for the method to be consistent because the combination of the previous values $$y_{n+k}$$ must be "complete", the evaluations of $$\rho$$ in other points other than $$1$$ speak about how the coefficients are distributed.
If $$\rho$$ where defined like this: $$\rho(z)=1+\sum_{k=0}^{s-1} \alpha_k z^{s-k}$$ the Dahlquist condition would require $$|z| \geq 1$$.
It still seems incredible how much can be said about the method from $$\rho$$.