Section C.5 The Error Behaviour of the Secant Method
Let \(f(x)\) have two continuous derivatives, and let \(r\) be any solution of \(f(x)=0\text{.}\) We will now get a pretty good handle on the error behaviour of the secant method near \(r\text{.}\)
Denote by \(\tilde\varepsilon_n=x_n-r\) the (signed) error in \(x_n\) and by \(\varepsilon_n=|x_n-r|\) the (absolute) error in \(x_n\text{.}\) Then, \(x_n =r+\tilde\varepsilon_n\text{,}\) and, by (C.4.1),
\begin{align*}
\tilde\varepsilon_{n+1}
\amp = \frac{x_{n-1} f(x_n) - x_n f(x_{n-1}) }{f(x_n)-f(x_{n-1})} -r \\
\amp = \frac{[r+\tilde\varepsilon_{n-1}] f(x_n) - [r+\tilde\varepsilon_n] f(x_{n-1}) }
{f(x_n)-f(x_{n-1})} -r \\
\amp = \frac{\tilde\varepsilon_{n-1} f(x_n) - \tilde\varepsilon_n f(x_{n-1}) }
{f(x_n)-f(x_{n-1})}
\end{align*}
\begin{align*}
f(x_n)\amp = f(r) + f'(r)\tilde\varepsilon_n +\frac{1}{2} f''(c_1) \tilde\varepsilon_n^2 \\
\amp = f'(r)\tilde\varepsilon_n +\frac{1}{2} f''(c_1) \tilde\varepsilon_n^2 \\
f(x_n)-f(x_{n-1})\amp = f'(c_2)[x_n-x_{n-1}] \\
\amp = f'(c_2)[\tilde\varepsilon_n-\tilde\varepsilon_{n-1}]
\end{align*}
for some \(c_1\) between \(r\) and \(x_n\) and some \(c_2\) between \(x_{n-1}\) and \(x_n\text{.}\) So, for \(x_{n-1}\) and \(x_n\) near \(r\text{,}\) \(c_1\) and \(c_2\) also have to be near \(r\) and
\begin{align*}
f(x_n)\amp \approx f'(r)\tilde\varepsilon_n +\frac{1}{2} f''(r) \tilde\varepsilon_n^2 \\
f(x_{n-1})\amp \approx f'(r)\tilde\varepsilon_{n-1} +\frac{1}{2} f''(r) \tilde\varepsilon_{n-1}^2 \\
f(x_n)-f(x_{n-1})\amp \approx f'(r)[\tilde\varepsilon_n-\tilde\varepsilon_{n-1}]
\end{align*}
and
\begin{align*}
\tilde\varepsilon_{n+1} \amp = \frac{\tilde\varepsilon_{n-1} f(x_n) - \tilde\varepsilon_n f(x_{n-1}) }
{f(x_n)-f(x_{n-1})} \\
\amp \approx \frac{ \tilde\varepsilon_{n-1} [f'(r)\tilde\varepsilon_n +\frac{1}{2} f''(r) \tilde\varepsilon_n^2]
- \tilde\varepsilon_n [f'(r)\tilde\varepsilon_{n-1} +\frac{1}{2} f''(r) \tilde\varepsilon_{n-1}^2] }
{ f'(r)[\tilde\varepsilon_n-\tilde\varepsilon_{n-1}] } \\
\amp = \frac{\frac{1}{2}\tilde\varepsilon_{n-1}\tilde\varepsilon_nf''(r)[\tilde\varepsilon_n-\tilde\varepsilon_{n-1}]}
{ f'(r)[\tilde\varepsilon_n-\tilde\varepsilon_{n-1}] } \\
\amp = \frac{f''(r)} {2f'(r)} \tilde\varepsilon_{n-1}\tilde\varepsilon_n
\end{align*}
Taking absolute values, we have
\begin{equation*}
\varepsilon_{n+1}\approx K \varepsilon_{n-1}\varepsilon_n\qquad
\text{with }K = \left|\frac{f''(r)} {2f'(r)} \right|
\tag{E7}
\end{equation*}
We have seen that Newton’s method obeys a similar formula — (E3) says that, when \(x_n\) is near \(r\text{,}\) Newton’s method obeys \(\varepsilon_{n+1}\approx K\varepsilon_n^2\text{,}\) also with \(K = \left|\frac{f''(r)} {2f'(r)} \right|\text{.}\) As we shall now see, the change from \(\varepsilon_n^2\text{,}\) in \(\varepsilon_{n+1}\approx K\varepsilon_n^2\text{,}\) to \(\varepsilon_{n-1}\varepsilon_n\text{,}\) in \(\varepsilon_{n+1}\approx K\varepsilon_{n-1}\varepsilon_n\text{,}\) does have a substantial impact on the behaviour of \(\varepsilon_n\) for large \(n\text{.}\)
To see the large \(n\) behaviour, we now iterate (E7). The formulae will look simpler if we multiply (E7) by \(K\) and write \(\delta_n=K\varepsilon_n\text{.}\) Then (E7) becomes \(\delta_{n+1}\approx\delta_{n-1}\delta_n\) (and we have eliminated \(K\)). The first iterations are
\begin{alignat*}{2}
\delta_2\amp \amp \amp \approx \delta_0\delta_1 \\
\delta_3\amp \approx \delta_1\delta_2
\amp \amp \approx \delta_0\delta_1^2 \\
\delta_4\amp \approx \delta_2\delta_3
\amp \amp \approx \delta_0^2\delta_1^3 \\
\delta_5\amp \approx \delta_3\delta_4
\amp \amp \approx \delta_0^3\delta_1^5 \\
\delta_6\amp \approx \delta_4\delta_5
\amp \amp \approx \delta_0^5\delta_1^8 \\
\delta_7\amp \approx \delta_5\delta_6
\amp \amp \approx \delta_0^8\delta_1^{13}
\end{alignat*}
Notice that every \(\delta_n\) is of the form \(\delta_0^{\alpha_n}\delta_1^{\beta_n}\text{.}\) Substituting \(\delta_n=\delta_0^{\alpha_n}\delta_1^{\beta_n}\) into \(\delta_{n+1}\approx\delta_{n-1}\delta_n\) gives
\begin{equation*}
\delta_0^{\alpha_{n+1}}\delta_1^{\beta_{n+1}}
\approx \delta_0^{\alpha_{n-1}}\delta_1^{\beta_{n-1}}
\delta_0^{\alpha_n}\delta_1^{\beta_n}
\end{equation*}
and we have
\begin{equation*}
\alpha_{n+1}=\alpha_{n-1}+\alpha_{n} \qquad
\beta_{n+1}=\beta_{n-1}+\beta_{n}
\tag{E8}
\end{equation*}
The recursion rule in (E8) is famous. The Fibonacci sequence (which is \(0\text{,}\) \(1\text{,}\) \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) \(5\text{,}\) \(8\text{,}\) \(13\text{,}\) \(\cdots\)), is defined by
1
Plug “Fibonacci sequence in nature” into your search engine of choice.
2
Fibonacci (1170-1250) was an Italian mathematician who was also known as Leonardo of Pisa, Leonardo Bonacci and Leonardo Biglio Pisano.
\begin{align*}
F_0\amp =0 \\
F_1\amp =1 \\
F_n\amp =F_{n-1}+F_{n-2}\qquad\text{for }n>1
\end{align*}
So, for \(n\ge 2\text{,}\) \(\alpha_n = F_{n-1}\) and \(\beta_n=F_n\) and
\begin{equation*}
\delta_n \approx \delta_0^{\alpha_n}\delta_1^{\beta_n}
= \delta_0^{F_{n-1}}\delta_1^{F_n}
\end{equation*}
One of the known properties of the Fibonacci sequence is that, for large \(n\text{,}\)
\begin{equation*}
F_n\approx\frac{\varphi^n}{\sqrt{5}}\qquad\text{where }
\varphi=\frac{1+\sqrt{5}}{2} \approx 1.61803
\end{equation*}
This \(\varphi\) is the golden ratio. So, for large \(n\text{,}\)
3
Also worth a quick trip to your search engine.
\begin{align*}
K\varepsilon_n \amp = \delta_n
\approx \delta_0^{F_{n-1}}\delta_1^{F_n}
\approx \delta_0^{\frac{\varphi^{n-1}}{\sqrt{5}}}
\delta_1^{\frac{\varphi^n}{\sqrt{5}}}
= \delta_0^{\frac{1}{\sqrt{5}\varphi}\times\varphi^n}
\delta_1^{\frac{1}{\sqrt{5}}\times\varphi^n}\\
\amp = d^{\varphi^n}\qquad\text{where}\quad
d=\delta_0^{\frac{1}{\sqrt{5}\,\varphi}}\delta_1^{\frac{1}{\sqrt{5}}}\\
\amp \approx d^{1.6^n}
\end{align*}
Assuming that \(0\lt \delta_0=K\varepsilon_0\lt 1\) and \(0\lt \delta_1=K\varepsilon_1\lt 1\text{,}\) we will have \(0\lt d\lt 1\text{.}\)
By way of contrast, for Newton’s method, for large \(n\text{,}\)
\begin{gather*}
K\varepsilon_n\approx d^{2^n}\qquad\text{where}\quad
d=(K\varepsilon_1)^{1/2}
\end{gather*}
As \(2^n\) grows quite a bit more quickly than \(1.6^n\) (for example, when n=5, \(2^n=32\) and \(1.6^n=10.5\text{,}\) and when \(n=10\text{,}\) \(2^n=1024\) and \(1.6^n=110\)) Newton’s method homes in on the root quite a bit faster than the secant method, assuming that you start reasonably close to the root.
