= Solution
Apply the <exponential-symbol order criterion for a multistep method>, which follows by substituting the <bilateral shift operator> $e^{h\partial_t}$ into the exact-solution residual. Direct expansion gives
$$
\rho(e^z)-z\sigma(e^z)
=-\frac{\alpha+5}{12}z^4-\frac{14\alpha+61}{90}z^5+O(z^6).
$$
All coefficients through degree three vanish. If $\alpha\ne-5$, the fourth-degree coefficient is nonzero, so the normalized <local truncation error> has exact order three. At $\alpha=-5$, the fourth-degree coefficient vanishes and the fifth-degree coefficient is $1/10$, which is nonzero. Thus
$$
\boxed{p=3\text{ if }\alpha\ne-5,\qquad p=4\text{ if }\alpha=-5.}
$$
This is the formal order of the full recurrence. The order-four member is not <zero-stable>, so it is not convergent. All the convergent members from part (a) have order three. At the degenerate value $\alpha=1$, the full recurrence still has formal order three, while cancellation of its common factor gives <Backward Euler method> of order one; that cancellation changes the method rather than its order calculation.
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