family of linear one-step methods These questions are not homework it is recommended exercises to prepare for the final exam. 
Consider the family of linear one-step methods defined by $$y_n = y_{n-1} + h(\theta f_n + (1 - \theta)f_{n-1})$$ where $0\leq \theta \leq 1$.
2) For what values of $\theta$ in the interval $0\leq \theta \leq 1$ is the associated method convergent?
3) Show that the values $\theta$ in the interval $0\leq \theta < 1$ where the associated method is convergent, the order $p$ can be at most $2$.
Attempted proof for 2):
Let's first prove convergence of forward Euler i.e. when $\theta = 0$, the forward Euler method says that $$u_{n+1} = u_n + hf(t_n,u_n)$$ and Taylor expansion with remainders of each component gives $$ y_j(t_{n+1}) = y_j(t_n) + hf_j(t_n,y(t_n)) + \frac{1}{2}h^2y_j''(t_n + \theta_j h)$$ $$= y_j(t_n) + hf_j(t_n,y(t_n)) + h\tau_{nj}$$ for some unknown collection of numbers $\theta_j$ between $0$ and $1$. This defines the local truncation error $\tau_n$, the amount by which the exact solution $y$ fails to satisfy the method after one step. Subtraction gives a difference equation for the error $$e_{n+1} = u_{n+1} - y(t_{n+1}) = e_n + h[f(t_n,u_n) - f(t_n,y(t_n))] + h\tau_n$$ Assume $f$ is Lipschitz constant $L$ and the local truncation error satisfies a bound $$||\tau_n|| \leq \tau$$ for all $n$ (which depends on the exact solution) then 
$$||e_{n+1}|| \leq ||e_n|| + hL||e_n|| + h\tau = (1 + hL)||e_n||+ h \tau$$
$$\leq (1 + hL)^{2}||e_{n-1}|| + (1 + (1 + hl))h\tau $$ 
$$\leq \ldots$$
$$\leq (1 + hL)^{n+1}||e_0|| + \frac{(1 + hL)^{n+1} - 1}{(1 + hL) - 1}h\tau$$
since $(1 + hL) \leq e^{hL} = 1 + hL + \frac{1}{2}(hL)^2 + \ldots$, this gives $$||e_n|| \leq e^{nhL}||e_0|| + \frac{1^{nhL} - 1 }{hL}\tau \leq e^{LT}||e_0|| + \frac{e^{LT} - 1}{L}\tau$$ for $0\leq t_n = nh\leq T$. This shows stability. Now the local truncation error satisfies $$h||\tau_n|| = ||\frac{1}{2}h^2y'(t_n + \theta_n h)|| \leq \frac{1}{2}Mh^2$$ assuming $y$ is $C^2$ and $||y''|| \leq u$. Consistency gives a local bound on $\tau$ and stability allows us to conclude convergence $$|u_n - y(t_n)|| \leq e^{LT}||u_0 - y_0|| + \frac{e^{LT} - 1}{LT}\frac{T}{2}Mh \leq O(|u_0 - y_0|) + O(h)$$ Convergence implies that $u_n\rightarrow y_n$ on the whole interval $[0,T]$ as $h\rightarrow 0$ and $u_0\rightarrow y_0$. A similar proof can be done for any one-step method. A little extra work is required for implicit methods such as implicit Euler $$u_{n+1} = u_n + hf(t_{n+1},u_{n+1})$$ because the right hand side depends on the new unknown $u_{n+1}$. Assume you can solve the implicit equation for $u_{n+1}$ at each step so we can estimate the errors as for explicit Euler $$||e_{n+1}|| \leq e_n + hL||e_{n+1}|| + h||\tau_n||$$ collect each terms on the left, and for $hL \leq \frac{1}{2}$ we can then divide by $1 - hL$ to get $$||e_{n+1}|| \leq \frac{1}{1 - hL}(||e_n|| + h||\tau_n||)$$ using the additional inequality $$\frac{1}{1 - hL}\leq 1 + 2hL \ \ \text{for} \ \ hL \leq 1/2$$ gives stability as for Euler's method, and the consistency computation follows. 
I am not sure how to show 3) any suggestions are greatly appreciated.
Relevant information that may be useful:
The previous convergence theory shows that in the limit $h\rightarrow 0$, method which are accurate of high order $p$ produce highly accurate solutions at less computational expense. Typically all methods cost $O(1)$ work per step to solve systems of ODE's of fixed size $d$(where $y(t)\in\mathbb{R}^d$ for each $t$). In fact the work per step if often conventionally measured in terms of the number of $f$ evaluations, as for numerical integration. Thus the total cost for solving an IVP with a $p^{th}$-order method gives error $\epsilon$ equal to $C_p h^p + O(h^{p+1})$ for some constant $C_p$ depending on the derivations of the solution $y$, the right hand side $f$, and $[0,T]$. Thus the total cost to solve with accuracy $\epsilon$ on $[0,T]$ if of order $O(\epsilon^{-1/p})$. 
 A: For 2) you can combine both estimates to get an error formula for any $θ\in [0,1]$ that gives an error order 1.
For 3) you just have to show that an order of 3 or higher is impossible. Thus is suffices to show via Taylor formula that the local truncation error is always of order 3 or lower. Use the solution with $y^{n}(t_n)=y_n$ to get
\begin{multline}
y^n(t_n+h)=y_n+hf(y_n)+\frac{h^2}2f'(y_n)f(y_n)
\\
+\frac{h^3}6(f''(y_n)[f(y_n),f(y_n)]+f'(y_n)f'(y_n)f(y_n))+O(h^4)
\end{multline}
and compare to $y_{n+1}=y_n+hk$ where
\begin{align}
k&=θf(y_h+hk)+(1−θ)f(y_n)
\\
&=f(y_n)+hθf'(y_h)k+\frac{h^2}2θf''(y_h)[k,k]+O(h^3)
\\
&=f(y_n)+hθf'(y_h)(f(y_n)+hθf'(y_h)f(y_n))+\frac{h^2}2θf''(y_h)[f(y_n),f(y_n)]+O(h^3)
\end{align}
so that in consequence
\begin{multline}
y_{n+1}=y_n+hf(y_n)+h^2θf'(y_n)f(y_n)
\\
+h^3\left(\frac{θ}2f''(y_n)[f(y_n),f(y_n)]+θ^2f'(y_n)f'(y_n)f(y_n)\right)+O(h^4)
\end{multline}
To get identical second order terms you need $θ=\frac12$ which however gives different third order terms. In general, there is no way to satisfy $\frac{θ}2=\frac16=θ^2$.

The general strategy for error estimates of one-step methods is to introduce the update function as 
$$
y_{k+1}=y_k+h·\Phi(x_k,y_k,h)
$$
which covers both explicit and implicit methods. The ultimate trick or tool is to consider the family of exact solutions $y^k(x)$ that have initial values $y^k(x_k)=y_k$. Then by the Gronwall lemma and Lipschitz condition
\begin{align}
\|y^{k+1}(x_n)-y^{k}(x_n)\|
&\le e^{L(x_n-x_{k+1})}·\|y^{k+1}(x_{k+1})-y^k(x_{k+1})\|
\\
&=e^{Lh(n-k-1)}·\|y_{k}+h·\Phi(x_k,y_k,h)-y^k(x_{k}+h)\|
\\
&=e^{Lh(n-k-1)}·h·\left\|\frac{y^k(x_{k}+h)-y_{k}}h-\Phi(x_k,y_k,h)\right\|
\end{align}
where the last factor is the local error for the exact solution $y^k$ at $x_k$. If this error has an estimate $\le M·h^p$, then by the triangle inequality the error to the exact solution $y^0$ of the original IVP is
\begin{align}
\|y_n-y^0(x_n)\|&\le\sum_{k=0}^{n-1}\|y^{k+1}(x_n)-y^k(x_n)\|
\\
&\le  M·h^{p+1}·\sum_{k=0}^{n-1}e^{Lh(n-k-1)}=M·h^{p+1}·\frac{e^{nhL}-1}{e^{hL}-1}
\\
&\le\frac{M}{L}·h^p
\end{align}
