prove that the order of convergence of the secant method is approximately 1.618 and asymptotic error constant is $$K= C^{\frac{1}{\alpha}}=(\frac{F"(P_n)}{2F'(P)})^{\alpha -1}$$.

The proof is as follows:- *The secant method is the root finding scheme based on the recurrence relation

$P_{n+1} = P_n - F(P_n) \frac{P_n - P_(n-1)}{F(P_n)- F(P_(n-1))}$

Now subtracting P from both sides

$P_{n+1} - P = P_n - P- F(P_n) \frac{P_n - P_(n-1)}{F(P_n)- F(P_(n-1))}$

and performing remaining steps

that are identical to derive the error evolution equation for false position the end result is

$P_{n+1} - P \approx (P_n -P)(P_{n-1} -P) \frac{F"(P)}{2F'(P)+F"(P)(P_n +P_{n-1}-2P)}$

I need those "steps identical to derive error evolution equation for the method of false position".

Textbook: Brain Bradie - A friendly introduction to numerical analysis.

Thanks in advance.



Make the situation more simple by setting $P=0$. Then $$ F(P_n)=0+F'(0)P_n+\frac12F''(0)P_n^2+O(P_n^3). $$ Insert that into the equation to get, disregarding third order terms in $P_n$ and $P_{n−1}$, \begin{align} P_{n+1} &= P_n-\frac{(F'(0)P_n+\frac12F''(0)P_n^2)·(P_n−P_{n−1})}{(F'(0)P_n+\frac12F''(0)P_n^2)-(F'(0)P_{n−1}+\frac12F''(0)P_{n−1}^2)}\\ &=P_n-P_n\frac{(F'(0)+\frac12F''(0)P_n)}{F'(0) +\frac12F''(0)(P_n+P_{n−1})} \\ &=P_n\frac{F''(0)P_{n−1}}{2F'(0) +F''(0)(P_n+P_{n−1})} \\ &=\frac{F''(0)}{2F'(0) +F''(0)(P_n+P_{n−1})}P_nP_{n−1} \end{align} As $P_{n+1}$ is thus quadratic in $P_n,P_{n−1}$, the non-constant part of the denominator constitutes another $O((|P_n|+|P_{n−1}|)^3)$ term. Leaving it out gives $$ |P_{n+1}|\le C\,\|P_n|\,|P_{n-1}|,\quad C=\frac{|F''(0)|}{2|F'(0)|} $$ Then once $C\,|P_{n}|<1$ for all $n>N$ $$ \ln(C\,|P_{n+1}|)\le \ln(C\,|P_n|)+\ln(C\,|P_{n-1}|) $$ where thus these logarithmic terms have has a (negative) multiple of the Fibonacci sequence as an upper bound.

If one wants a derivation by exact formulas, then define the function $G$ via $F(x)=xG(x)$, $G(x)=\int_0^1 F'(tx)\,dt$. Then the formula for the root of the secant reads as $$ \frac{aF(b)-bf(a)}{F(b)-F(a)}=ab\frac{G(b)-G(a)}{F(b)-F(a)}=ab\frac{G'(c)}{F'(c)} $$ by the extended mean value theorem for some $c$ between $a$ and $b$. Now $$G'(c)=\int_0^1tF''(tc)\,dt=F''(θc)\int_0^1t\,dt,$$ $θ\in(0,1)$ by the weighted integral mean value theorem. As with $a,b\to 0$ also $c\to 0$, the derivative ratio converges again to $\frac{F''(0)}{2F'(0)}$.


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