Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 39B (Numerical Analysis)

Interpolate through the $s+1$ values at $t_n,\dots ,t_{n+s}$, and differentiate the interpolation polynomial at its newest node. In the backward coordinate $\tau =(t-t_{n+s})/h$, the Newton form iswith the $j=0$ factor interpreted as one and $\nabla y_m=y_m-y_{m-1}$. For $j\ge 1$, the derivative of the factor at zero is $(j-1)!/j!=1/j$. Thus the backward differentiation formula isIf $w$ denotes the forward-shift variable, ${\nabla }^j{y}_{n+s}$ corresponds to $w^{s-j}(w-1{)}^j$. Consequently the requested polynomial isThe first form is a polynomial valid also at $w=0$. The second form is convenient for relating it to the truncated series of $\log w$. Differentiation of the degree-$s$ interpolant has local derivative error $O(h^s)$, giving order $s$ for smooth solutions.

For $s=2$, part (a) gives $\rho (w)=\frac{3}{2}{w}^2-2w+\frac{1}{2}$ and the right-side polynomial is $w^2$. On the scalar test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. Its characteristic equation isThe linear stability domain contains every $z$ for which both roots have modulus strictly less than one. For real $-1/2\le z<0$, the roots are positive; their larger numerator is less than three while their denominator exceeds three, so both are less than one. At $z=-1/2$ the repeated root is $1/2$, also strictly inside the unit circle. For $z<-1/2$, the roots are complex conjugates and their product isThus in every case both roots lie strictly inside the unit disk. Repeated roots there are allowed, since polynomial factors times their powers still tend to zero. Therefore