Prove that: for an infinitely differentiable function f(x), the Composite Trapezoid Rule can be written the following formula .

Question

Composite Trapezoid Rule \begin{align*} \int_{_{a}}^{^{b}}f(x)dx=\frac{h}{2}\left(y_{0}+y_{m}+2\sum\limits_{_{i=1}}^{m-1}y_{_{i}}\right)-\frac{(b-a)h^{2}}{12}f''(c) \end{align*} where $h=(b-a)/m$ and $c$ is between $a$ and $b$. If $f$ is an infinitely differentiable function, prove that \begin{align} \int_{_{a}}^{^{b}}f(x)dx=\frac{h}{2}\left(y_{0}+y_{m}+2\sum\limits_{_{i=1}}^{m-1}y_{_{i}}\right)+c_{2}h^{2}+c_{4}h^{4}+c_{6}h^{6}+\cdot\cdot\cdot , \end{align} where the $c_{i}$ depend only on higher derivatives of $f$ at $a$ and $b$, not on $h$. For example, $c_{2}=(f'(a)-f'(b))/12.$

Answer 1

See Error formula for Composite Trapezoidal Rule, Trapezoidal method, deriving error boundary. for the derivation of the first error term.

The general formula is due to the time symmetry of the quadrature method. First assume that the error has the form $c_1h+c_2h^2+c_3h^3+c_4h^4+...$, which you can do due to the infinite differentiability of $f$. Then consider the error for the quadrature of $$ \int_a^bf(a+b-x)dx $$ using the same method. The value and trapezium formula are the same, but the error is $-c_1h+c_2^2-c_3h^3+c_4h^4\mp...$. It follows that the odd-index coefficients all have to be zero.