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I'm reading about the order of convergence of the method of false position and there is one tricky point in the proof I don't understand. The method itself for finding the minimum $x^*$ of a function $f(x)$ is:

$$x_{k+1} = x_k-g(x_k)\left[ \frac{x_k-x_{k-1}}{g(x_k)-g(x_{k-1})} \right],$$

where $g(x) := f'(x)$. The proposition is the following:

Proposition:

Let $g$ have a continuous second derivative and suppose $x^*$ is such that $g(x^*)=0, g'(x^*)\neq 0$. Then for $x_0$ sufficiently close to $x^*$, the sequence $\{x_k\}_{k=0}^{\infty}$ generated by the method of false position above converges to $x^*$ with order $\tau_1 \simeq 1.618$, which is the golden mean.

In the following I have written the proof from the part I start having troubles:

Defining $\epsilon = x_k-x^*$ we have, in the limit ($k\rightarrow \infty$),

$$\epsilon_{k+1} = M\epsilon_{k}\epsilon_{k-1},\;\;\;\;\;\;(1)$$

where $\displaystyle M=\frac{g''(x^*)}{2g'(x^*)}$. Taking logarithm of both sides of $(1)$, we have with $y_k = \log M\epsilon_{k}$,

$$y_{k+1} = y_k + y_{k-1},\;\;\;\;\;\;(2)$$

which is the Fibonacci difference equation. A solution to this equation will satisfy

$$y_{k+1}-\tau_1y_k \rightarrow 0.\;\;\;\;\;\;(3)$$

Thus

$$\log M\epsilon_{k+1}-\tau_1\log M\epsilon_{k}\rightarrow 0$$

or

$$\log \frac{M\epsilon_{k+1}}{(M\epsilon_k)^{\tau_1}} \rightarrow 0,$$

and hence

$$\frac{\epsilon_{k+1}}{\epsilon_{k}^{\tau_1}} \rightarrow M^{\tau_1 -1 }. \blacksquare$$

This is taken from: Linear and Nonlinear programming, 3rd edition by David Luenberger, page 224.

My questions:

  1. How did $(2)$ follow from $(1)$?

  2. Explanation for $(3)$?

My try:

If I take the logarithm of $(1)$ myself I get:

$$\log \epsilon_{k+1}=\log M\epsilon_k + \log\epsilon_{k-1},$$

so what am I doing wrong here?...

Thank you for any help!

user1868607
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jjepsuomi
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    This is for part A. Multiply the equation with $M$: $$\epsilon_{k+1} = M\epsilon_{k}\epsilon_{k-1}\ \Longrightarrow M \epsilon_{k+1} = (M\epsilon_{k})(M\epsilon_{k-1})\ \Longrightarrow \log(M \epsilon_{k+1}) = \log(M\epsilon_{k})+\log(M\epsilon_{k-1})\ \Longrightarrow y_{k+1} = y_{k}+y_{k-1} $$ – gammatester Jul 17 '14 at 12:50
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    For the connection between the Fibonacci relation (3) and the golden ratio $\tau_1$ or $\varphi$, see here. Assuming $y_{k+1} = y_k + y_{k-1}$ has a geometric solution $y_k = a R^k$ leads to the quadratic $R^2 = R+1$ and hence to the golden ratio (and its reciprocal). – hardmath Jul 17 '14 at 13:52
  • Hi, @gammatester can you post your comment as an answer so that I can accept it? =) – jjepsuomi Jul 21 '14 at 08:48
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    Hi, just did it. – gammatester Jul 21 '14 at 09:25

1 Answers1

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This is for part A. Multiply the equation with M:

$$\epsilon_{k+1} = M\epsilon_{k}\epsilon_{k-1}$$ $$\Longrightarrow M\epsilon_{k+1} = (M\epsilon_{k})(M\epsilon_{k-1})$$ $$\Longrightarrow \log\left(M\epsilon_{k+1}\right) = \log\left((M\epsilon_{k})(M\epsilon_{k-1})\right)$$ Apply the functional equation for the $\log$ of a product $$\Longrightarrow \log\left(M\epsilon_{k+1}\right) = \log(M\epsilon_{k})+\log(M\epsilon_{k-1})$$ and substitute the definition of $y_k$ $$\Longrightarrow y_{k+1} = y_{k}+y_{k-1}$$

gammatester
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