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C.1: Richardson Extrapolation

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    91837
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    There are many approximation procedures in which one first picks a step size \(h\) and then generates an approximation \(A(h)\) to some desired quantity \(\cA\text{.}\) For example, \(\cA\) might be the value of some integral \(\int_a^b f(x)\,\, d{x}\text{.}\) For the trapezoidal rule with \(n\) steps, \(\Delta x=\tfrac{b-a}{n}\) plays the role of the step size. Often the order of the error generated by the procedure is known. This means

    \[ \cA=A(h)+Kh^k+K_1h^{k+1}+K_2h^{k+2}+\ \cdots \tag{E1} \nonumber \]

    with \(k\) being some known constant, called the order of the error, and \(K,\ K_1,\ K_2,\ \cdots\) being some other (usually unknown) constants. If \(A(h)\) is the approximation to \(\cA=\int_a^b f(x)\,\, d{x}\) produced by the trapezoidal rule with \(\Delta x=h\text{,}\) then \(k=2\text{.}\) If Simpson's rule is used, \(k=4\text{.}\)

    Let's first suppose that \(h\) is small enough that the terms \(K'h^{k+1}+K''h^{k+2}+\ \cdots\) in (E1) are small enough 1 Typically, we don't have access to, and don't care about, the exact error. We only care about its order of magnitude. So if \(h\) is small enough that \(K_1h^{k+1}+K_2h^{k+2}+\ \cdots\) is a factor of at least, for example, one hundred smaller than \(Kh^k\text{,}\) then dropping \(K_1h^{k+1}+K_2h^{k+2}+\ \cdots\) would not bother us at all. that dropping them has essentially no impact. This would give

    \[ \cA=A(h)+Kh^k \tag{E2} \nonumber \]

    Imagine that we know \(k\text{,}\) but that we do not know \(A\) or \(K\text{,}\) and think of (E2) as an equation that the unknowns \(\cA\) and \(K\) have to solve. It may look like we have one equation in the two unknowns \(K\text{,}\) \(\cA\text{,}\) but that is not the case. The reason is that (E2) is (essentially) true for all (sufficiently small) choices of \(h\text{.}\) If we pick some \(h\text{,}\) say \(h_1\text{,}\) and use the algorithm to determine \(A(h_1)\) then (E2), with \(h\) replaced by \(h_1\text{,}\) gives one equation in the two unknowns \(\cA\) and \(K\text{,}\) and if we then pick some different \(h\text{,}\) say \(h_2\text{,}\) and use the algorithm a second time to determine \(A(h_2)\) then (E2), with \(h\) replaced by \(h_2\text{,}\) gives a second equation in the two unknowns \(\cA\) and \(K\text{.}\) The two equations will then determine both \(\cA\) and \(K\text{.}\)

    To be more concrete, suppose that we have picked some specific value of \(h\text{,}\) and have chosen \(h_1=h\) and \(h_2=\tfrac{h}{2}\text{,}\) and that we have evaluated \(A(h)\) and \(A(h/2)\text{.}\) Then the two equations are

    \begin{align*} \cA&=A(h)+Kh^k \tag{E3a}\\ \cA&=A(h/2)+K\big(\tfrac{h}{2}\big)^k \tag{E3b} \end{align*}

    It is now easy to solve for both \(K\) and \(\cA\text{.}\) To get \(K\text{,}\) just subtract (E3b) from (E3a).

    \begin{align*} (\textrm{E3a})-(\textrm{E3b}):&\quad 0=A(h)-A(h/2) +\big(1-\tfrac{1}{2^k}\big)Kh^k\\ \implies&\quad K=\frac{A(h/2)-A(h)}{[1-2^{-k}]h^k} \tag{E4a} \end{align*}

    To get \(\cA\) multiply (E3b) by \(2^k\) and then subtract (E3a).

    \begin{align*} 2^k(\textrm{E3b})-(\textrm{E3a}):&\quad [2^k-1]\cA=2^kA(h/2)-A(h)\\ \implies&\quad \cA=\frac{2^kA(h/2)-A(h)}{2^k-1} \tag{E4b} \end{align*}

    The generation of a “new improved” approximation for \(\cA\) from two \(A(h)\)'s with different values of \(h\) is called Richardson 2 Richardson extrapolation was introduced by the Englishman Lewis Fry Richardson (1881--1953) in 1911. Extrapolation. Here is a summary

    Equation C.1.1 Richardson extrapolation

    Let \(A(h)\) be a step size \(h\) approximation to \(\cA\text{.}\) If

    \[ \cA=A(h)+K h^k \nonumber \]

    then

    \[ K=\frac{A(h/2)-A(h)}{[1-2^{-k}]h^k}\qquad \cA=\frac{2^kA(h/2)-A(h)}{2^k-1} \nonumber \]

    This works very well since, by computing \(A(h)\) for two different \(h\)'s, we can remove the biggest error term in (E1), and so get a much more precise approximation to \(\cA\) for little additional work.

    Example C.1.2 Richardson extrapolation with the trapezoidal rule

    Applying the trapezoidal rule (1.11.6) to the integral \(\cA=\int_0^1\frac{4}{1+x^2}\, d{x}\) with step sizes \(\frac{1}{8}\) and \(\frac{1}{16}\) (i.e. with \(n=8\) and \(n=16\)) gives, with \(h=\frac{1}{8}\text{,}\)

    \[ A(h)=3.1389884945\qquad A(h/2)=3.1409416120 \nonumber \]

    So (E4b), with \(k=2\text{,}\) gives us the “new improved” approximation

    \[ \frac{2^2\times 3.1409416120 -3.1389884945}{2^2-1}=3.1415926512 \nonumber \]

    We saw in Example 1.11.3 that \(\int_0^1\frac{4}{1+x^2}\, d{x}=\pi\text{,}\) so this new approximation really is “improved”:

    • \(A(1/8)\) agrees with \(\pi\) to two decimal places,
    • \(A(1/16)\) agrees with \(\pi\) to three decimal places and
    • the new approximation agrees with \(\pi\) to eight decimal places.

    Beware that (E3b), namely \(\cA=A(h/2)+K\big(\tfrac{h}{2}\big)^k\text{,}\) is saying that \(K\big(\frac{h}{2}\big)^k\) is (approximately) the error in \(A(h/2)\text{,}\) not the error in \(\cA\text{.}\) You cannot get an “even more improved” approximation by using (E4a) to compute \(K\) and then adding \(K\big(\frac{h}{2}\big)^k\) to the “new improved” \(\cA\) of (E4b) — doing so just gives \(\cA+K\big(\tfrac{h}{2}\big)^k\text{,}\) not a more accurate \(\cA\text{.}\)

    Example C.1.3 Example 1.11.16 revisited

    Suppose again that we wish to use Simpson's rule (1.11.9) to evaluate \(\int_0^1 e^{-x^2}\,\, d{x}\) to within an accuracy of \(10^{-6}\text{,}\) but that we do not need the degree of certainty provided by Example 1.11.16. Observe that we need (approximately) that \(|K|h^4 \lt 10^{-6}\text{,}\) so if we can estimate \(K\) (using our Richardson trick) then we can estimate the required \(h\text{.}\) A commonly used strategy, based on this observation, is to

    • first apply Simpson's rule twice with some relatively small number of steps and
    • then use (E4a), with \(k=4\text{,}\) to estimate \(K\) and
    • then use the condition \(|K| h^k\le 10^{-6}\) to determine, approximately, the number of steps required
    • and finally apply Simpson's rule with the number of steps just determined.

    Let's implement this strategy. First we estimate \(K\) by applying Simpson's rule with step sizes \(\tfrac{1}{4}\) and \(\tfrac{1}{8}\text{.}\) Writing \(\tfrac{1}{4}=h'\text{,}\) we get

    \[ A(h')=0.74685538 % 0.746855379790987 \qquad A(h'/2)=0.74682612 %0.746826120527467 \nonumber \]

    so that (E4a), with \(k=4\) and \(h\) replaced by \(h'\text{,}\) yields

    \[ K=\frac{0.74682612 - 0.74685538}{[1-2^{-4}](1/4)^4} %=-\frac{0.000031211}{(1/4)^4} =-7.990\times 10^{-3} \nonumber \]

    We want to use a step size \(h\) obeying

    \[ |K|h^4\le 10^{-6} \iff 7.990\times 10^{-3} h^4\le 10^{-6} \iff h \le\root{4}\of{\frac{1}{7990}} =\frac{1}{9.45} \nonumber \]

    like, for example, \(h=\tfrac{1}{10}\text{.}\) Applying Simpson's rule with \(h=\tfrac{1}{10}\) gives

    \[ A(1/10) = 0.74682495 \nonumber \]

    The exact answer, to eight decimal places, is \(0.74682413\) so the error in \(A(1/10)\) is indeed just under \(10^{-6}\text{.}\)

    Suppose now that we change our minds. We want an accuracy of \(10^{-12}\text{,}\) rather than \(10^{-6}\text{.}\) We have already estimated \(K\text{.}\) So now we want to use a step size \(h\) obeying

    \begin{align*} |K|h^4\le 10^{-12} &\iff 7.99\times 10^{-3} h^4\le 10^{-12}\\ &\iff h \le\root{4}\of{\frac{1}{7.99\times 10^9}} =\frac{1}{299.0} \end{align*}

    like, for example, \(h=\tfrac{1}{300}\text{.}\) Applying Simpson's rule with \(h=\tfrac{1}{300}\) gives, to fourteen decimal places,

    \[ A(1/300) = 0.74682413281344 \nonumber \]

    The exact answer, to fourteen decimal places, is \(0.74682413281243\) so the error in \(A(1/300)\) is indeed just over \(10^{-12}\text{.}\)


    This page titled C.1: Richardson Extrapolation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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