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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).

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  • $\begingroup$ Thinking of it as the characteristic polynomial of a recursive sequence may help. $\endgroup$ – David Rubio Jan 16 at 19:34
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I've came up with something that may point in the right track:

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 $\rho$.

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