Stability region for multiderivative multistep methods My question concerns the ODE $y^\prime = f(y)$, where $y$ is a function of time, $t$. Consider the following numerical scheme to approximate $y(nh)$:
$$
15y_{n+2} - 7hf(y_{n+2}) + h^2f^\prime(y_{n+2}) = 15y_{n} + h\left(16f(y_{n+1})+7f(y_{n})\right) + h^2f^\prime(y_{n}), \quad n\in \mathbb{Z}^+.
$$
According to my old notes, this solution should have a bounded domain of linear stability. I set out trying to show this, by applying the method to the usual test problem $y^\prime = \lambda y$, $y(0)=1$. Doing so, I rearrange to obtain:
$$15y_{n+2} - 7h\lambda y_{n+2} + h^2\lambda^2y_{n+2} - 15y_n - 16h \lambda y_{n+1} - 7h \lambda y_n - h^2 \lambda^2 y_n = 0, \quad n\in\mathbb{Z}^+.$$
Now, this is a linear recurrence relation with characteristic equation 
$$(15 - 7q+ q^2)w^2 - 16q w - (15 + 7q + q^2) = 0,$$
where for brevity I have written $q=h\lambda$. The stability domain is then the set of all $q\in\mathbb{C}$ for which the roots of this quadratic lie in the unit disc. The Schur-Cohn criterion provides necessary and sufficient conditions for this to happen, and the full statement of the criterion can be found here.
However, when I apply the criterion to this problem and work through the algebra, I obtain that:
(1) $$\Re q \leq 0 $$
(2) $$-14\Re q (15 + \vert q\vert^2) \geq -16 \Re q (15 + \vert q\vert^2)$$
Clearly, something has gone wrong here. Can anyone identify and clear up this problem? I would very much appreciate it. 
 A: Computing the truncation error
Insert the Taylor series around $t=t_{n+1}$ to compute the order of the method
\begin{align}
15[y_{+1}&-y_{-1}]-7h[y'_{+1}+y'_{-1}]+h^2[y''_{+1}-y''_{-1}]\\
&=30[hy'+\tfrac16h^3y'''+\tfrac1{120}h^5y^{(5)}+\tfrac1{7!}h^7y^{(7)}]
\\&\quad-14h[y'+\tfrac12h^2y'''+\tfrac1{24}h^4y^{(5)}+\tfrac1{6!}h^6y^{(7)}]
\\&\quad+2h^2[hy'''+\tfrac16h^3y^{(5)}+\tfrac1{5!}h^5y^{(7)}]
\\
&=16hy'+\underbrace{\frac{2(15-49+42)}{7!}}_{=\frac{16}{7!}}h^7y^{(7)}+O(h^9)
\end{align}
Recapitulate the Schur-Cohn test
The Schur-Cohn test posits that if the roots of $p(w)=a_2w^2+a_1w+a_0=0$ are inside (and not on) the unit circle, then the roots of the reverse and conjugate polynomial $p^*(w)=\bar a_0w^2+\bar a_1w+\bar a_2=0$ are outside (and not on) the unit circle. Note that for $|w|=1$ on the unit circle, $|p(w)|=| p^*(w)|$.

*

*Now if $|a_2|>|a_0|$, it would follow that $|\bar a_0 p(w)|<|a_2 p^*(w)|$, so that $p^*(w)$ and
$$
T[p](w)=a_2 p^*(w)-\bar a_0p(w)=(a_2\bar a_1-\bar a_0a_1)w+(|a_2|^2-|a_0|^2)]
$$
have the same number zero of roots inside the unit circle. This is possible if and only if the root of the linear function $T[p]$ is not inside the unit circle,  $|a_2\bar a_1-\bar a_0a_1|\le |a_2|^2-|a_0|^2$.


*The contrary case $|a_2|<|a_0|$ is not possible, as a linear function can not have $2$ roots inside the unit circle (or at all).


*If $|a_2|=|a_0|$, then $T[p](0)=0$, which is not compatible with either case.
Application of the test
$|a_2|>|a_0|$ means that $|15-7q+q^2|>|15+7q+q^2|$, which is equivalent to $Re(\frac{7q}{15+q^2})<0\iff 0>Re(q(15+\bar q^2))=(15+|q|^2)Re(q).$ That is, that $Re(q)$ is negative. Allowing roots on the unit circle in the limit would include the case $Re(q)=0$.
The second condition gives
$$
16|(15-7q+q^2)\bar q+(15+7\bar q+\bar q^2)q|\le |15+q^2-7q|^2-|15+q^2+7q|^2
\\\iff
32|(15+|q|^2)Re(q)|\le -4Re(7\bar q(15+q^2))=-28(15+|q|^2)Re(q)
\\\iff
8|Re(q)|\le -7Re(q).
$$
This is indeed a condition that is never satisfied, except for $Re(q)=0$, meaning that the method is only weakly stable.
