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This explicit numerical method is a two-stage Runge-Kutta method and a one-step method. For a sufficiently smooth ordinary differential equation, the chain rule gives $y^{\prime \prime }=f_x+f_{y}f$, and Taylor's theorem shows that its local truncation error is $O(h^3)$. Uniform Lipschitz continuity of $f$ in its solution argument gives an increment-function Lipschitz constant at most $L(1+hL/2)$. The Lipschitz one-step error bound then proves second-order numerical convergence on a fixed time interval. This method uses an intermediate prediction from $y_n$; the explicit midpoint linear multistep method instead requires a separate previous value $y_{n-1}$.

For $f^{\prime }=\Phi (x,f)$, the standard explicit midpoint method isThe implicit midpoint rule instead uses $y_{i+1}=y_i+h\Phi (x_i+h/2,(y_i+y_{i+1})/2)$. When the derivative depends only on $x$, both give $y_{i+1}=y_i+h{f}^{\prime }(x_i+h/2)$.

The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\frac{h}{2}[\Phi (x_i,y_i)+\Phi (x_{i+1},y_{i+1})]$. This is the trapezoidal rule, and generally differs from the implicit midpoint rule. A further use of “midpoint” is the two-step explicit midpoint linear multistep method, $y_{i+1}=y_{i-1}+2h\Phi (x_i,y_i)$, which needs a second starting value. The printed question supplies no convention to distinguish these meanings; (e) spells out the resulting alternatives.

Put $h=(b-a)/n$, $g_i=g(x_i)$, and $z_0=0$. Under the endpoint-slope convention suggested by the requested connection, the recurrence is $z_{i+1}=z_i+\frac{h}{2}(g_i+g_{i+1})$. Summing telescopes, soSince $G(b)={\int }_a^{b}g(x)dx$ by the fundamental theorem of calculus, this recurrence gives exactly the composite trapezoidal rule. It replaces the average derivative over each strip by the arithmetic mean of its endpoint derivatives.

For the standard midpoint method, the literal endpoint-only request is not valid for general $g$. Both the explicit and implicit one-step midpoint formulas here give insteadthe composite midpoint quadrature rule. It requires half-grid samples. These samples cannot be recovered exactly from $g_0,\dots ,g_n$ for arbitrary smooth $g$: the zero function and $g(x)={\prod }_{j=0}^n(x-x_j{)}^2$ have identical grid values but different midpoint sums. For a smooth $g$, replacing each half-grid sample by $(g_i+g_{i+1})/2$ is an additional approximation, giving $T_n$ with an $O(h^2)$ total difference, rather than an identity. A precise useful relation, obtained by splitting the finer trapezoidal sum into old nodes and midpoints, isHere $T_{2n}$ uses the finer half-grid, so it does not eliminate the missing data.

If “Mid-Point Method” means the two-step explicit midpoint linear multistep method, the recurrence is $z_{i+1}=z_{i-1}+2h{g}_i$. It gives $z_{2r}=2h{\sum }_{j=1}^r{g}_{2j-1}$ and $z_{2r+1}=z_1+2h{\sum }_{j=1}^r{g}_{2j}$, so an odd final index additionally needs a starting value $z_1$. For even $n=2r$, the result is the midpoint quadrature on $r$ double-width strips; equivalently $z_n=2T_n-T_{n/2}$. The question specifies neither even $n$ nor a starter. Thus the trapezoidal expression above is the natural endpoint-average interpretation, while the other common midpoint meanings require the qualifications just given. The ambiguity is present in the original PDF.