a): We have the initial-value-problem $y'=-y$, $y(0)=0$ and the interval $[0,1]$. Solve the IVP on the equidistant grid $I_h$, with $h=\frac1N$ with the mid-point-rule and give the difference equation.
b): Calculate a basis $(y^0, y^1)$ of the solution space of the difference equation
c): For the consistend choice of the initial values $y_0=h(y^1)_0$, $y_1=h(y^1)_1$ is the error of approximation of $y(1)$ not processable in a power series.
In the answer I give the solution presented in the tutorials.
The solution given in the tutorials goes as follows:
a) The midpoint rule is given by $y_{k+2}-y_k=2hf(t_{k+1},y_{k+1})=-2hy_{k+1}$
The difference equation is $y_{k+2}+2hy_{k+1}-y_k=0$
b) The characteristic polynomial is $\rho(\lambda)=\lambda^2+2h\lambda-1$
with the two roots $\lambda_{0,1}=-h\pm\sqrt{h^2+1}$
Obviously one root is negative and one is positive, since $h<\sqrt{h^2+1}$. Let $\lambda_0>0$ and $\lambda_1<0$.
Define $(y^1)_h=\lambda_1^k$ this is oscillating since $\lambda_1<0$.
c) The initial value is $y_0=h(y^1)_0, y_1=h(y^1)_1$. The corresponding solution to the multi-step-process is therefor
$y_k=h(y^1_k)=hy_1^k$
The approximation for $y(1)$ is $y_N=h(y^1)_N=h\lambda_1^N$
$\lambda_1=-h-\sqrt{h^2+1}$
$h(y^1)_N=h(-h-\sqrt{h^2+1})^N$
Take $f(x)=\sqrt{x}$, and write
$$h(-h-\sqrt{h^2+1})=h(-h-f(h^2+1))^N$$
$$\stackrel{Taylor pol.}{=}h(-h[f(1)+f'(1)(h^2+1-1)+\mathcal{O}(h^2)^2])$$
$$=h(-h-1+\mathcal{O}(h^2))^N$$
$$=h(-h(\frac1h+1)+\mathcal{O}(h^2))^N$$
$$=h(-h(\frac1h+1))^N\cdot\mathcal{O}(h^2)^0+h\sum_{k=1}^k(-h(\frac1h+1)^N\cdot\mathcal{O}(h^2)^{N-k}\binom{N}{k}$$
$$=h(-(1+h))^N+\mathcal{O}(h^3)$$
$$=(-1)^Nh(1+\frac1N)^N+\mathcal{O}(h^3)\rightarrow h(-1)^\frac1he+\mathcal{O}(h^3)\stackrel{h\to\infty}{\rightarrow} 0$$
This is not differentiable, so you can not develop this into a taylor expansion. But neverless convergent.