I am working on this problem: Consider αy(t + ah) + βy(t + bh) Find best parameters α,β,a,b that approximates y′(t) at the highest order in h. i.e. Choosing α, β, a, b what is the highest p below that can be achieved? y′(t) − [αy(t + ah) + βy(t + bh)] = O(h^p)
I try to use taylor expansion around y(t) and set coefficient of y(t),y'(t),y''(t) to zero to find a solution for the parameters: α=-β,a∝1/h,b∝1/h. Then consider the orginal eqation αy(t + ah) + βy(t + bh), h vanishes with my solution.I notice something is wrong but I do not understand why.
$$\alpha y(t+ah) + \beta y(t+bh) = \alpha(y(t)+ahy'(t)+a^2h^2/2 y''(t) +\ldots) +$$ $$+\beta (y(t)+bhy'(t)+b^2h^2/2 y''(t) + \ldots).$$ Note that $\alpha + \beta = 0$ (otherwise the values of $y$ won't cancel). We thus get an infinite number of equations on $\alpha, a,b$: $$ \begin{cases} \alpha h (a-b) = 1\\ \alpha h^2 (a^2-b^2)/2 = 0\\ \alpha h^3 (a^3-b^3)/6 = 0\\ \ldots \end{cases} $$ and we need to find the largest $p$ for which the system of first $p-1$ equations will still have a solution. Indeed, using the first equation to rewrite the others, we get $$ \begin{cases} \alpha h (a-b) = 1\\ h (a+b)/2 = 0\\ h^2 (a^2+ab+b^2)/6 = 0\\ \ldots \end{cases} $$ so in particular $a = -b \not = 0$, however then $a^2 + ab + b^2 = a^2$ is still nonzero, and we can't even do better than $O(h^2)$.