 Solution

Apply the exponential-symbol order criterion for a multistep method. The exact-solution residual of the linear multistep method is encoded by $E(z)=\rho (e^z)-z\sigma (e^z)$; vanishing through degree $p$ is equivalent to the order conditions for a linear multistep method through order $p$. Here $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)=1-\alpha$, and direct Taylor expansion givesFor example, for $n\ge 2$ the coefficient of $z^n$ can be calculated asFor $n=2,3$ this is zero, verifying the lower conditions explicitly. The constant and linear coefficients also vanish for every $\alpha$. Therefore the formal order is at least three for every $\alpha$.

The fourth-degree term vanishes exactly when $\alpha =-1$. For that value the fifth-degree coefficient is $-1/90\ne 0$, soAt $\alpha =1$ these are formal residual conditions only: the characteristic polynomial has a double root at one, so the method is not convergent. Canceling its common polynomial factor would change the recurrence and give a different order calculation. This distinction is relevant to part (b).