The Vindication of Computer Simulations

Abstract

The relatively recent increase in prominence of computer simulations in scientific inquiry gives us more reasons than ever before for asserting that mathematics is a wonderful tool. In fact, a practical knowledge (a ‘know-how’) of scientific computation has become essential for scientists working in all disciplines involving mathematics. Despite their incontestable success, it must be emphasized that the numerical methods subtending simulations provide at best approximate solutions and that they can also return very misleading results. Accordingly, epistemological sobriety demands that we clarify the circumstances under which simulations can be relied upon. With this in mind, this paper articulates a general perspective to better understand and compare the strengths and weaknesses of various error-analysis methods.

Appendix: An Example of a Comparative Asymptotic Analysis of the Error of Computational Methods

The following appendix exemplifies how numerical analysts quantify the error resulting from computational methods by means of asymptotic analyses. Let us return to the two methods of numerical integration discussed by Tai. To find the error in the midpoint approximation M, we take the Taylor expansion about the midpoint of the interval [−h∕2, h∕2], which is 0 (note that for convenience we make the standard change of variable a = −h∕2 and b = h∕2):

$$\displaystyle\begin{array}{rcl} I =\int _{ -h/2}^{h/2}f(x)dx& =& \int _{ -h/2}^{h/2}\sum _{ n=0}^{\infty }\frac{f^{(n)}(0)} {n!} x^{n}dx =\sum _{ n=0}^{\infty }\frac{f^{(n)}(0)} {n!} \int _{-h/2}^{h/2}x^{n}dx {}\\ & =& hf(0) +\sum _{ n=1}^{\infty }\frac{f^{(n)}(0)} {n!} \int _{-h/2}^{h/2}x^{n}dx {}\\ \end{array}$$

Note that, when n is odd, ∫ _−h∕2 ^h∕2 x ⁿ dx vanishes to 0 (by symmetry about the point 0). Thus,

$$\displaystyle\begin{array}{rcl} I& =& hf(0) + \frac{f^{(2)}(0)h^{3}} {24} + \frac{f^{(4)}(0)h^{5}} {1920} + O(h^{7}) {}\\ \end{array}$$

Now, for this interval, the midpoint approximant M is hf(0). Thus, the error E _h ^(M) resulting from the midpoint rule on an interval of width h is

$$\displaystyle{E_{h}^{(M)} = I - M =\sum _{ \begin{array}{c}n=2\\ \exists k\in \mathbb{Z}.2k=n\end{array}}^{\infty }\frac{f^{(n)}(0)h^{n+1}} {(n + 1)!2^{n}} }$$

Taking the first two terms of E _h ^(M), we have

$$\displaystyle{E_{h}^{(M)} = \frac{f^{(2)}(0)h^{3}} {24} + \frac{f^{(4)}(0)h^{5}} {1920} + O(h^{7}).}$$

Now, we will find the trapezoidal error, E _h ^(T) = I − T, by expressing T as a combination of Taylor series. Let us first find the Taylor series about x = 0 (the midpoint) for f(−h∕2) and f(h∕2):

$$\displaystyle\begin{array}{rcl} f(-\frac{h} {2})& =& \sum _{n=0}^{\infty }\frac{f^{(n)}(0)} {n!} (-\frac{h} {2})^{n} =\sum _{ n=0}^{\infty }\frac{f^{(n)}(0)(-1)^{n}h^{n}} {n!2^{n}} {}\\ & =& f(0) -\frac{1} {2}f^{(1)}(0)h + \frac{1} {8}f^{(2)}(0)h^{2} - \frac{1} {48}f^{(3)}(0)h^{3} + \cdots {}\\ f(\frac{h} {2})& =& \sum _{n=0}^{\infty }\frac{f^{(n)}(0)} {n!} (\frac{h} {2})^{n} =\sum _{ n=0}^{\infty }\frac{f^{(n)}(0)h^{n}} {n!2^{n}} {}\\ & =& f(0) + \frac{1} {2}f^{(1)}(0)h + \frac{1} {8}f^{(2)}(0)h^{2} + \frac{1} {48}f^{(3)}(0)h^{3} + \cdots {}\\ \end{array}$$

Because of the alternating sign in $f(-\frac{h} {2} )$, we find that the odd powers of n vanish when we calculate T:

$$\displaystyle\begin{array}{rcl} T& =& \frac{h} {2}\bigg(f(-\frac{h} {2}) + f(\frac{h} {2})\bigg) = h\bigg(\sum _{\begin{array}{c}n=0 \\ \exists k\in \mathbb{Z}.2k=n\end{array}}^{\infty }\frac{f^{(n)}(0)h^{n}} {n!2^{n}} \bigg) {}\\ & =& hf^{(0)}(0) + \frac{1} {8}f^{(2)}(0)h^{3} + \frac{1} {384}f^{(4)}(0)h^{5} + \cdots {}\\ \end{array}$$

As a result,

$$\displaystyle\begin{array}{rcl} E_{h}^{(T)} = I - T = -\frac{f^{(2)}(0)h^{3}} {12} -\frac{f^{(4)}(0)h^{5}} {480} - O(h^{7}).& & {}\\ \end{array}$$

So, both rules are accurate to order O(h ³), but the coefficient of the dominant term for the error of the midpoint rule is smaller than that of the trapezoidal rule by a factor of 2.