# CLP-1 Differential Calculus

## SectionC.5The 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*}
By the Taylor expansion (3.4.32) and the mean value theorem (Theorem 2.13.5),
\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 1 . The Fibonacci 2  sequence (which is $$0\text{,}$$ $$1\text{,}$$ $$1\text{,}$$ $$2\text{,}$$ $$3\text{,}$$ $$5\text{,}$$ $$8\text{,}$$ $$13\text{,}$$ $$\cdots$$), is defined by
\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 3 . So, for large $$n\text{,}$$
\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 $$n\text{,}$$
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.