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I'm having a little trouble doing the part the needs to be proven. The first part is the definition of approximate error and I just need some help showing that it is equal to the second part of this problem.

If the Secant Method converges to $r$, $f'(r)\neq0$, and $f''(r)\neq0$ then the approximate error relationship $$e_{i+1}\approx\left|\frac{f''(r)}{2f'(r)}\right| e_i e_{i-1}$$ can be shown to hold. Prove that, if in addition, $$\lim_{i\rightarrow\infty}⁡\frac{e_{i+1}}{e_i^\alpha}$$ exists and is nonzero for some $\alpha>0$, then $\alpha=(1+\sqrt5)/2$ and $$e_{i+1}\approx\left|\frac{f''(r)}{2f'(r)}\right|^{\alpha-1} e_i^\alpha.$$

I'm not really sure how I am supposed to start this off. I get if the limit exists and is nonzero then the limit must be some number $n$, but other than that I'm lost.

J_Smith88
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1 Answers1

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So you found $$e_{i+1}\approx Ce_ie_{i-1},$$ with $C=\left|\frac{f''(r)}{2f'(r)}\right|$. Multiply with $C$, take the logarithm, use $d_i=\ln(Ce_i)$ to get the classical Fibonacci recursion $$ d_{i+1}\approx d_i+d_{i-1}. $$ Using the Fibonacci sequence with $F_0=0$ and $F_1=1$, $F_{i+1}=F_i+F_{i-1}$, the solution for the error sequence can be written as $$ d_i\approx F_{i-1}d_0+F_id_1\implies e_i\approx\frac1C(Ce_0)^{F_{i-1}}(Ce_1)^{F_i}\le C^{F_{i+1}-1}\max(e_0,e_1)^{F_{i+1}}. $$ or on the step level $$ \frac{d_{i+1}}{d_i}\approx \frac{F_{i}d_0+F_{i+1}d_1}{F_{i-1}d_0+F_id_1} \xrightarrow{i\to\infty}α \implies d_{i+1}\approxαd_i,~~ e_{i+1}\approx C^{α-1}e_i^α. $$

Lutz Lehmann
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