Consistency of Adams-Multon Two step method 
Show that  Adams-Multon two-step method is consistent of order 4.
Where the method is defined by:
$$u_0= a,~~ u_1=b,~~
 (*)~~~  u_{i+1}=u_i+\frac{h}{12}(5f(x_{i+1},u_{i+1})+8f(x_i,u_i)-f(x_{i-1},u_{i-1})).$$

Definition:
For all $(x,u)\in G$ define $\eta=\eta(\xi)$ to be the single solution for:
$\eta'=f(\xi,\eta)$ , $\eta(x)=u$ for the initial values $(x,u)$,  then:
$\delta(x,u;h)=1/h[\eta(x+rh)+\sum_{m=0}^{r-1} a_m \eta(x+mh)]-\rho(x,\eta(x),...,\eta(x+(r-1)h);h)$ is called the local discriminatory error.The multi-step method is called consistent if $lim_{h\to 0} \delta(x,u;h)=0$
uniformly for all $(x,u)\in G$ and is consistent of order p if $|\delta(x,u;h)|\leq kh^p$ for all $(x,u)\in G$ and $h>0$ where k is a constant.
That's what I did:
We know that $u_i$ is the estimated/approximated solution of the real solution $u(x_i)$. And $f(x,u)=u'(x)$.
Using Taylor expansion of $u$ about $x_i$ we get
$$
u(x_{i+1}) = u(x_i)+ hu'(x_i)+h^2/2! u''(x_i)+ h^3/3! u'''(x_i)+h^4/4! u''''(x_i)
$$
This gives:
$$
1) u_{i+1}=u_i+hu_i'+ h^2/2 u_i''+h^3/3! u_i''' +h^4/4! u_i''''+...
$$
On the other hand,


*$u_{i+1}=u_i+h/12 [(5u_i '+ 5hu_i ''+5h^2/2u_i'''+ 5h^3/3! u_i ''''+...)+8u_i '- (u_i '-hu_i'' +h^2/2! u_i'''-h^3/3! u_i''''+...)]+O(h^4)$.

Then if we compare between coefficients in equations 1 and 2 we get:
$O(h^4)=u''''(c) (-h^4)/24$.
Tyis is the discriminatory local error and it satisfies that $|O(h^4)| \leq K*h^4$ for a constant K, I am not sure why there exists such a constant!. so by definition, we get that the method described is consistent of order 4.
If we use the definition instead to show this, it shoud lead to the same result.
Is what i did enough accurate?
 A: You computed that
$$
δ_{AM3}(x,u;h)=\frac{u(x+h)-u(x)}{h}-\frac1{12}(5u'(x+h)+8u'(x)-u'(x-h))=-\frac{h^3}{24}u^{(4)}(c)=O(h^3),
$$
which fits the condition for consistency order 3. In contrast, the order 4 method would have the error computed as
$$
δ_{AM4}(x,u;h)=\frac{u(x+h)-u(x)}{h}-\frac1{24}(9u'(x+h)+19u'(x)-5u'(x-h)+u(x-2h))=O(h^4).
$$
there is one more term involved with a different coefficient sequence.
Note that the definition of the consistency error has a factor $\frac1h$ in the leading term for the difference quotient, this lowers the degree in $h$ of the error you computed by one.
For the bound of the higher order derivative note that the derivatives of the exact solution can be computed starting with $u'=f(x,y)$ as
\begin{align}
u''(x)&=f^{[1]}(x,u)=f_x+f_uu'=f_x+f_uf
\\
u'''(x)&=f^{[2]}(x,u)=f^{[1]}_x+f^{[1]}_uu'=f_{xx}+2f_{xu}f+f_uf_x+f_{uu}[f,f]+f_u^2f
\\
u^{(4)}(x)&=f^{[3]}(x,u)=f^{[2]}_x+f^{[2]}_uf=...
\end{align}
This means that you can express $u^{(4)}$ and any other derivative as a polynomial expression in $f$ and its derivatives. Then take a sufficiently large tube around the exact solution to contain all the local exact solutions with IC $u(x_i)=u_i$ used in the construction. You get the standard situation of a continuous function on a bounded/compact set, there is a maximum (of the norm).
You need that $f$ is continuously differentiable to a sufficient order, the 3rd here. This smoothness is assumed in the definition of the error order of the method. If the $f$ in a given problem is not that smooth, then the actual numerical error will have a lower order.
