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Numerical Methods with Programming

Leonid Kovalev

Section 12.1 Definition of derivative and the order of error

The definition of derivative says:
\begin{equation*} f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \end{equation*}
This suggests a way to approximate \(f'(x)\text{:}\) take two consecutive values of \(f\text{,}\) subtract, divide by difference of x-values:
\begin{equation} f'(x) \approx \frac{f(x+h)-f(x)}{h}\tag{12.1.1} \end{equation}
Here the value of \(h\) is some concrete number, so the two sides are only approximately equal. One can express this more precisely as
\begin{equation*} \frac{f(x+h)-f(x)}{h} = f'(x) + e(h) \end{equation*}
where the error term \(e\) is a function of \(h\text{.}\) The accuracy of approximation is measured by the order of the error term, which is a number \(d\) such that \(|e(h)|\) is bounded by some multiple of \(|h|^d\text{.}\)
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There is a method to determine the order of error of an approximation to some derivative of \(f\text{:}\)
  1. Let \(n= 0\)
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  2. Plug \(f(x) = x^n\) into the formula and find the error term.
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  3. If the error term is some nonzero multiple of a power of \(h\text{,}\) the exponent of \(h\) is the order of the error term.
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  4. If the error term is zero, increase \(n\) by \(1\) and return to step 2.
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Example 12.1.1. Find the order of error of approximation to \(f'\).

Find the order of error of the formula (12.1.1).
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Solution.
  • With \(f(x) = x^0\) we get \(f'(x) = 0\) and \(\frac{f(x+h)-f(x)}{h}=0\text{,}\) so the error term is zero for this function.
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  • With \(f(x) = x^1\) we get \(f'(x) = 1\) and \(\frac{f(x+h)-f(x)}{h}=\frac{h}{h}=1\text{,}\) so the error term is zero for this function.
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  • With \(f(x) = x^2\) we get \(f'(x) = 2x\) and \(\frac{f(x+h)-f(x)}{h}=\frac{x^2+2xh+h^2-x^2}{h}=2x + h\text{,}\) so the error term for this function is \(h^1\text{.}\)
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In conclusion, the formula has order of error 1.
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The importance of the order of error can be illustrated by an example. Suppose we want to find \(f'(x)\) with absolute error at most \(10^{-12}\text{.}\)
  • If the error term of our formula is \(h^1\text{,}\) we need to use extremely small \(h\text{;}\) that is \(|h|\le 10^{-12}\text{.}\)
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  • If the error term is \(h^2\text{,}\) we need \(|h|\le 10^{-6}\) which is still small but not as extreme.
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  • If the error term is \(h^3\text{,}\) we need \(|h|\le 10^{-4}\text{.}\)
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  • If the error term is \(h^4\text{,}\) we need \(|h|\le 10^{-3}\text{.}\)
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Thus, a higher-order formula should allow us to obtain an accurate result while avoidung extremely small values of \(h\text{.}\) But why do we want to avoid extremely small \(h\text{?}\) This is explained in next section.
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