Derivation of Local Error Terms for Integration Formulae

5.1. Derivation of Local Error Terms for Integration Formulae#

Local truncation error terms associated with various integration formulae can be derived by using the interpolating formulae or by the method of undetermined coefficients. For this module we focus on the former through the following examples.

Example 5.1

Given a general sequence \(\{f_j=f(x_j)\}_0^{\infty}\) with \(x_j=jh\), use the Gregory-Newton forward-interpolation formula:

\[\begin{split}\begin{aligned} f_{j+s} =~ f_j + s\Delta f_j & + \frac{s(s-1)}{2!}\Delta^2 f_j + \frac{s(s-1)(s-2)}{3!}\Delta^3 f_j \\ ~ & + \frac{s(s-1)(s-2)(s-3)}{4!}\Delta^4 f_j + \cdots \\ \end{aligned}\end{split}\]

to derive Simpson’s rule and its error term \(T\) (local truncation error):

\[\begin{aligned} \int_{x_0}^{x_2} f(x)\,\dx ~=~ \frac{h}{3}(f_2 + 4f_1 + f_0) + T \end{aligned}\]

Solution

\[\begin{split}\begin{aligned} &Using & s ~&=~ \frac{x - x_0}{h}, & \dx ~&=~ h \ds\\ &\therefore & x ~&=~ x_0, & s ~&=~ 0\\ & & x ~&=~ x_2, & s ~&=~ 2\\ \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \therefore \quad \int_{x_0}^{x_2} f(x)\,\dx ~&=~ h \int_{0}^{2} \left( f_0 + s\Delta f_0 + \frac{s(s-1)}{2!}\Delta^2 f_0 \right)\,\ds + T \\[1em] \text{Substituting} \quad \Delta f_0 ~&=~ f_1 - f_0, \quad \text{and} \quad \Delta^2 f_0 ~=~ f_2 - 2f_1 + f_0 \quad \text{gives} \\[1em] \int_{x_0}^{x_2} f(x)\,\dx ~&=~ h \left[ sf_0 + \frac{s^2}{2}(f_1 - f_0) + \frac{1}{2}\left( \frac{s^3}{3} - \frac{s^2}{2} \right)(f_2 - 2f_1 + f_0) \right]_0^2 + T \\ &=~ h \left[ 2f_0 + 2(f_1 - f_0) + \frac{1}{3}(f_2 - 2f_1 + f_0) \right] + T \\ &=~ \frac{h}{3}\bigl[f_0 + 4f_1 + f_2\bigr] + T \end{aligned}\end{split}\]

For the derivation of the error term we integrate the terms involving third and fourth differences (i.e. \(\Delta^3\) and \(\Delta^4\)), thus

\[\begin{split}\begin{aligned} T ~&=~ h \int_{0}^{2} \left\{ \frac{s(s-1)(s-2)}{3!}\Delta^3 f_0 + \frac{s(s-1)(s-2)(s-3)}{4!}\Delta^4 f_0 \right\}\ds \\ &=~ h \int_{0}^{2} \left\{ \frac{s^3 - 3s^2 + 2s}{3!}\Delta^3 f_0 + \frac{s^4 - 6s^3 + 11s^2 - 6s}{4!}\Delta^4 f_0 \right\}\ds \\ &=~ h \left\{ \frac{1}{6} \left[ \frac{s^4}{4} - \frac{3s^3}{3} + \frac{2s^2}{2} \right]_0^2 \Delta^3 f_0 + \frac{1}{24} \left[ \frac{s^5}{5} - \frac{3s^4}{2} + \frac{11s^3}{3} - 3s^2 \right]_0^2 \Delta^4 f_0 \right\} \\ &=~ h \left\{ 0 + \frac{1}{24} \left[ \frac{32}{5} - 24 + \frac{88}{3} - 12 \right] \Delta^4 f_0 \right\} \end{aligned}\end{split}\]

Use the Mean Value Theorem and LET \(\Delta^{4} f_0 = h^4f^{(iv)}(z)\) where \(x_0 \leq z \leq x_2\)

\[\begin{split}\begin{aligned} \therefore \quad T ~&=~ h^5 \left\{ \frac{1}{24} \left(-\frac{4}{15}\right) \right\} f^{(iv)}(z) \\ T ~&=~ -\frac{1}{90}h^5f^{(iv)}(z). \end{aligned}\end{split}\]

Theorem 5.1 (Mean Value Theorem)

Let \(f=f(x)\), \(x \in \R\) be continuous over the closed interval \([a, b]\) and differentiable over the open interval \((a, b)\). Then, there exists at least one point \(z\in (a,b)\) such that

\[ f'(z)=\frac{f(b)-f(a)}{b-a} \]

Note

Let \(a=x\) and \(b=x+h\), we can obtain

\[ f'(z)=\frac{f(x+h)-f(x)}{h} = \frac{\Delta f(x)}{h} \quad \text{for} \quad z \in (x, x+h) , \]

re-arranging the equation gives

\[ \Delta f(x) = h f'(z). \]

For high-order differences, we have

\[\begin{split}\begin{aligned} \Delta^2 f(x) & = h^2 f''(z), \\ \Delta^3 f(x) & = h^3 f'''(z), \\ ~ & \vdots \\ \Delta^n f(x) & = h^n f^{(n)}(z) \quad \text{for} \quad z \in (x, x+nh). \\ \end{aligned}\end{split}\]

Example 5.2

Use the Gregory-Newton forward-interpolation formula:

\[\begin{split}\begin{aligned} f_{j+s} =~ f_j + s\Delta f_j & + \frac{s(s-1)}{2!}\Delta^2 f_j + \frac{s(s-1)(s-2)}{3!}\Delta^3 f_j \\ ~ & + \frac{s(s-1)(s-2)(s-3)}{4!}\Delta^4 f_j + \cdots \\ \end{aligned}\end{split}\]

to show that for Milne’s formula:

\[\begin{aligned} \int_{x_{-1}}^{x_3} f(x)\,\dx ~=~ \frac{4h}{3} (2f_0 - f_1 + 2f_2) + T \end{aligned}\]

its error term (local truncation error)

\[\begin{aligned} T ~=~ \frac{14}{45}h^5 f^{(iv)}(z) \quad for \quad x_{-1} \leq z \leq x_3 \end{aligned}\]

can be obtained by integrating the terms involving third and fourth differences.

Solution

\[\begin{split}\begin{aligned} &Using & x ~&=~ x_0 + sh & \dx ~&=~ h\,\ds\\ &\therefore & x ~&=~ x_{-1}, & s ~&=~ -1\\ & & x ~&=~ x_3, & s ~&=~ 3 \end{aligned}\end{split}\]

For the derivation of the error term we integrate the terms involving third and fourth differences (i.e. \(\Delta^3\) and \(\Delta^4\)), thus

\[\begin{split}\begin{aligned} T ~&=~ h \int_{-1}^{3} \left\{ \frac{s(s-1)(s-2)}{3!}\Delta^3 f_0 + \frac{s(s-1)(s-2)(s-3)}{4!}\Delta^4 f_0 \right\}\ds \\ &=~ h \int_{-1}^{3} \left\{ \frac{s^3 - 3s^3 + 2s}{3!}\Delta^3 f_0 + \frac{s^4 - 6s^3 + 11s^2 - 6s}{4!}\Delta^4 f_0 \right\}\ds \\ &=~ h \left\{ \frac{1}{6} \left[ \frac{s^4}{4} - \frac{3s^3}{3} + \frac{2s^2}{2} \right]_{-1}^3 \Delta^3 f_0 + \frac{1}{24} \left[ \frac{s^5}{5} - \frac{3s^4}{2} + \frac{11s^3}{3} - 3s^2 \right]_{-1}^3 \Delta^4 f_0 \right\} \\ &=~ h \left\{ 0 + \frac{1}{24} \left[ \left( \frac{243}{5} - \frac{243}{2} + 99 - 27 \right) {-} \left( -\frac{1}{5} - \frac{3}{2} - \frac{11}{3} - 3 \right) \right] {\Delta^4 f_0} \right\} \\ &=~ \frac{h}{24} \left[ \frac{244}{5} - \frac{240}{2} + \frac{11}{3} + 75 \right] \Delta^4 f_0 \\ &=~ \frac{14}{45} h {\Delta^4 f_0} \end{aligned}\end{split}\]

Let \(\Delta^4 f_0 ~=~ h^4 f^{(iv)}(z)\) where \(x_{-1} \leq z \leq x_3\)

\[\begin{split}\begin{aligned} \therefore \quad T & =~ \frac{14}{45}\,h^5f^{(iv)}(z) \\ & =~ \frac{14}{45} h^5 y^{(v)}(z) \end{aligned}\end{split}\]