We search the highest order for the multi-step process
$y_{k+2}-(1+\alpha)y_{k+1}+\alpha y_k = h(\frac{3-\alpha}{2}f_{k+1}-\frac{1+\alpha}{2}f_k)$.
It is $f\in C^3$. We find $\alpha$ with the Taylor expansion.
I have a question on the following calculation:
$$\tau_h(t)=\frac1h(y(t+2h)-(1+\alpha)y(t+h)+\alpha y(t))-(\frac{3-\alpha}{2}y'(t+h)-\frac{1+\alpha}{2}y'(t))$$
$$=y'(t)\underbrace{(2-(1+\alpha)-\frac{3-\alpha}{2}+\frac{1+\alpha}{2})}_{=0}$$
$$+y''(t)\underbrace{(\frac{4h}{2}-(1+\alpha)\frac{h}{2}-\frac{3-\alpha}{2}h)}_ {=0}$$
$$+y'''(t)\underbrace{(\frac86h^2-(1+\alpha)\frac{h^2}{6}-\frac{3-\alpha}{2}\frac{h^2}{2})}_{=\dfrac{h^2}{12}(5+\alpha)}$$
$$+\mathcal{O}(h^3)$$
So for $\alpha=-5$ the order would be 3, which is the highest achievable for $f\in C^3$.
But I do not understand, where this Taylor expansion comes from. Is the Taylor expansion done, just for the $y(t)$, or do you have to do it for every $y(t+2h), y(t+h), y'(t+h), y'(t)$?
Could you elaborate more details of the calculation, or explain how we get to it?
Thanks in advance.
All values and derivatives are Taylor expanded at $t$ so they can be compared.
You could just as well start with a polynomial or power series, $y(t+s)=\sum a_ks^k$, $y'(t+s)=\sum ka_ks^{k-1}$ and then start comparing coefficients starting at the lowest degree.
Note that the resulting method is extremely unstable, as local errors, truncation or floating point, are magnified by a factor $5^{k}=\exp(\ln5\cdot \frac{t-t_0}h)$, which thus grows the faster the smaller the step size $h$.