I want to construct the classical trapezoidal rule by using a collocation method. I am in the time interval $[t_n, t_n + \tau]$.
I know that I have the following conditions for the polynomial $p$
$$ p(t_n) = y_n \\ p'(t_n) = F(t_n,p(t_n)) \\ p'(t_n + \tau)= F(t_n + \tau, p (t_n+ \tau) ) $$
Then, $Y_{n+1}$ will be given by $p(t_n + \tau)$
In the handouts I have
$$ p(t_n +x )= Y_n - \frac{1}{\tau} x (x - \tau) f(t_n,p(t_n)) - \frac{1}{2\tau} x^2 [f(t_n,p(t_n)) + f(t_n+ \tau,p(t_n + \tau))]$$ and the plugging $x=\tau$ I obtain the classical trapezoidal rule.
Question: how is this polynomial built? It seems "almost"a Taylor expansion where the second derivative is estimated by taking a forward finite difference, but I'm sure this is not the case since there's a term more in the first derivative.
It can't also be standard Lagrange interpolation since here I have also derivative values.
My feeling is that since I have three independent conditions, a standard polynomial of degree $2$, $$p(X)= aX^2 + bX +c$$ has been built and the unknowns $a,b,c$ are determined by solving the linear system derived from the three conditions in the first lines.
Yes, that's it, essentially. Take the quadratic polynomial and start to solve the equations for the coefficients. Trivially $$ p(t_n+x)=p_n+F_nx+cx^2\tag1 $$ from the first two conditions relating to the values on the left end of the interval. Then one has to solve the remaining third, usually non-linear, equation \begin{align} \underset{\parallel}{p'(t_n+τ)}&=\underset{\parallel}{F_{n+1}=F(t_n+τ,p(t_n+τ))}\tag{2a} \\ F_n+2cτ&=F(t_n+τ, p_n+F_nτ+cτ^2)\tag{2b} \end{align} for $c$. Once this is solved, it can be used to write $c=\frac{F_{n+1}-F_n}{2τ}$, and with this $$ p(t_n+x)=p_n+F_nx+\frac{F_{n+1}-F_n}{2τ}x^2=p_n+F_n\frac xτ(τ-x)+\frac{F_{n+1}+F_n}{2τ}x^2.\tag3 $$