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2.4: Order of Convergence

  • Page ID
    96043
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    Let \(r\) be the root and \(x_{n}\) be the \(n\)th approximation to the root. Define the error as

    \[\epsilon_{n}=r-x_{n} \nonumber \]

    If for large \(n\) we have the approximate relationship

    \[\left|\epsilon_{n+1}\right|=k\left|\epsilon_{n}\right|^{p}, \nonumber \]

    with \(k\) a positive constant, then we say the root-finding numerical method is of order \(p\). Larger values of \(p\) correspond to faster convergence to the root. The order of convergence of bisection is one: the error is reduced by approximately a factor of 2 with each iteration so that

    \[\left|\epsilon_{n+1}\right|=\frac{1}{2}\left|\epsilon_{n}\right| . \nonumber \]

    We now find the order of convergence for Newton’s Method and for the Secant Method.

    2.4.1. Newton’s Method

    We start with Newton’s Method

    \[x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \nonumber \]

    Subtracting both sides from \(r\), we have

    \[r-x_{n+1}=r-x_{n}+\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \nonumber \]

    Or

    \[\epsilon_{n+1}=\epsilon_{n}+\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \nonumber \]

    We use Taylor series to expand the functions \(f\left(x_{n}\right)\) and \(f^{\prime}\left(x_{n}\right)\) about the root \(r\), using \(f(r)=0\). We have

    \[\begin{aligned} f\left(x_{n}\right) &=f(r)+\left(x_{n}-r\right) f^{\prime}(r)+\frac{1}{2}\left(x_{n}-r\right)^{2} f^{\prime \prime}(r)+\ldots, \\ &=-\epsilon_{n} f^{\prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime}(r)+\ldots ; \\ f^{\prime}\left(x_{n}\right) &=f^{\prime}(r)+\left(x_{n}-r\right) f^{\prime \prime}(r)+\frac{1}{2}\left(x_{n}-r\right)^{2} f^{\prime \prime \prime}(r)+\ldots, \\ &=f^{\prime}(r)-\epsilon_{n} f^{\prime \prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime \prime}(r)+\ldots \end{aligned} \nonumber \]

    To make further progress, we will make use of the following standard Taylor series:

    \[\frac{1}{1-\epsilon}=1+\epsilon+\epsilon^{2}+\ldots, \nonumber \]

    which converges for \(|\epsilon|<1 .\) Substituting \((2.2)\) into \((2.1)\), and using \((2.3)\) yields

    \[\begin{aligned} \epsilon_{n+1} &=\epsilon_{n}+\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \\ &=\epsilon_{n}+\frac{-\epsilon_{n} f^{\prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime}(r)+\ldots}{f^{\prime}(r)-\epsilon_{n} f^{\prime \prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime \prime}(r)+\ldots} \\ &=\epsilon_{n}+\frac{-\epsilon_{n}+\frac{1}{2} \epsilon_{n}^{2} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots}{1-\epsilon_{n} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots} \\ &=\epsilon_{n}+\left(-\epsilon_{n}+\frac{1}{2} \epsilon_{n}^{2} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots\right)\left(1+\epsilon_{n} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots\right) \\ &=\epsilon_{n}+\left(-\epsilon_{n}+\epsilon_{n}^{2}\left(\frac{1}{2} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}-\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}\right)+\ldots\right) \\ &=-\frac{1}{2} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)} \epsilon_{n}^{2}+\ldots \end{aligned} \nonumber \]

    Therefore, we have shown that

    \[\left|\epsilon_{n+1}\right|=k\left|\epsilon_{n}\right|^{2} \nonumber \]

    as \(n \rightarrow \infty\), with

    \[k=\frac{1}{2}\left|\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}\right| \nonumber \]

    provided \(f^{\prime}(r) \neq 0 .\) Newton’s method is thus of order 2 at simple roots.

    2.4.2. Secant Method

    Determining the order of the Secant Method proceeds in a similar fashion. We start with

    \[x_{n+1}=x_{n}-\frac{\left(x_{n}-x_{n-1}\right) f\left(x_{n}\right)}{f\left(x_{n}\right)-f\left(x_{n-1}\right)} \nonumber \]

    We subtract both sides from \(r\) and make use of

    \[\begin{aligned} x_{n}-x_{n-1} &=\left(r-x_{n-1}\right)-\left(r-x_{n}\right) \\ &=\epsilon_{n-1}-\epsilon_{n} \end{aligned} \nonumber \]

    and the Taylor series

    \[\begin{aligned} f\left(x_{n}\right) &=-\epsilon_{n} f^{\prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime}(r)+\ldots, \\ f\left(x_{n-1}\right) &=-\epsilon_{n-1} f^{\prime}(r)+\frac{1}{2} \epsilon_{n-1}^{2} f^{\prime \prime}(r)+\ldots, \end{aligned} \nonumber \]

    so that

    \[\begin{aligned} f\left(x_{n}\right)-f\left(x_{n-1}\right) &=\left(\epsilon_{n-1}-\epsilon_{n}\right) f^{\prime}(r)+\frac{1}{2}\left(\epsilon_{n}^{2}-\epsilon_{n-1}^{2}\right) f^{\prime \prime}(r)+\ldots \\ &=\left(\epsilon_{n-1}-\epsilon_{n}\right)\left(f^{\prime}(r)-\frac{1}{2}\left(\epsilon_{n-1}+\epsilon_{n}\right) f^{\prime \prime}(r)+\ldots\right) \end{aligned} \nonumber \]

    We therefore have

    \[\begin{aligned} \epsilon_{n+1} &=\epsilon_{n}+\frac{-\epsilon_{n} f^{\prime}(r)+\frac{1}{2} \epsilon_{n}^{2} f^{\prime \prime}(r)+\ldots}{f^{\prime}(r)-\frac{1}{2}\left(\epsilon_{n-1}+\epsilon_{n}\right) f^{\prime \prime}(r)+\ldots} \\ &=\epsilon_{n}-\epsilon_{n} \frac{1-\frac{1}{2} \epsilon_{n} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots}{1-\frac{1}{2}\left(\epsilon_{n-1}+\epsilon_{n}\right) \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots} \\ &=\epsilon_{n}-\epsilon_{n}\left(1-\frac{1}{2} \epsilon_{n} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots\right)\left(1+\frac{1}{2}\left(\epsilon_{n-1}+\epsilon_{n}\right) \frac{f^{\prime \prime}(r)}{f^{\prime}(r)}+\ldots\right) \\ &=-\frac{1}{2} \frac{f^{\prime \prime}(r)}{f^{\prime}(r)} \epsilon_{n-1} \epsilon_{n}+\ldots, \end{aligned} \nonumber \]

    or to leading order

    \[\left|\epsilon_{n+1}\right|=\frac{1}{2}\left|\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}\right|\left|\epsilon_{n-1}\right|\left|\epsilon_{n}\right| \nonumber \]

    The order of convergence is not yet obvious from this equation, and to determine the scaling law we look for a solution of the form

    \[\left|\epsilon_{n+1}\right|=k\left|\epsilon_{n}\right|^{p} . \nonumber \]

    From this ansatz, we also have

    \[\left|\epsilon_{n}\right|=k\left|\epsilon_{n-1}\right|^{p} \nonumber \]

    and therefore

    \[\left|\epsilon_{n+1}\right|=k^{p+1}\left|\epsilon_{n-1}\right|^{p^{2}} \nonumber \]

    Substitution into \((2.4)\) results in

    \[k^{p+1}\left|\epsilon_{n-1}\right|^{p^{2}}=\frac{k}{2}\left|\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}\right|\left|\epsilon_{n-1}\right|^{p+1} \nonumber \]

    Equating the coefficient and the power of \(\epsilon_{n-1}\) results in

    \[k^{p}=\frac{1}{2}\left|\frac{f^{\prime \prime}(r)}{f^{\prime}(r)}\right| \nonumber \]

    and

    \[p^{2}=p+1 \nonumber \]

    The order of convergence of the Secant Method, given by \(p\), therefore is determined to be the positive root of the quadratic equation \(p^{2}-p-1=0\), or

    \[p=\frac{1+\sqrt{5}}{2} \approx 1.618 \nonumber \]

    which coincidentally is a famous irrational number that is called The Golden Ratio, and goes by the symbol \(\Phi\). We see that the Secant Method has an order of convergence lying between the Bisection Method and Newton’s Method.


    This page titled 2.4: Order of Convergence is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Jeffrey R. Chasnov via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.