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here's the following problem, I'm trying to find a real root by fixed-point iteration method but I can't find a properly $g(x)$ that meets the condition that $|g'(x_0)|<1$. Well, my nonlinear equation is $f(x)=1.08^x-\frac{125}{81}=0$ so I add +x in both sides, $g(x)=1.08^x-\frac{125}{81}+x=x$ but when I do $g'(x) =d/dx(1.08^x - 125/81 + x) = 0.076961*1.08^x + 1$ and I try to solve inequality to know a initial value $x_0$ I got this after set this equation $g'(x)=0.076961*1.08^x + 1 < 1$: $Im(x) = (π (2 n + 1))/(3 log(3) - 2 log(5))$,$n \in Z $ . PS: My real root of $x \approx 5.637457293$.

Thanks in advance.

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  • $\begingroup$ If you have $x + \text{increasing-function-of-$x$}$, obviously you're going to get $1 + \text{positive-value}$ as the derivative, which will therefore not satisfy the condition. So shift everything over to the right side and take $g(x) = x - 1.08^x + \frac{125}{81}$. $\endgroup$ – M. Vinay Mar 20 at 7:16
  • $\begingroup$ Thank you, it works on my code ;). By the way, is there a general way to find $g(x)$ ? $\endgroup$ – user587779 Mar 20 at 7:52
  • $\begingroup$ I know of no such method (this is not really my area of study, so someone better might actually know some systematic or semi-systematic way). That might very well be a deficiency of the method that makes others like Newton-Raphson method superior to it. $\endgroup$ – M. Vinay Mar 20 at 7:54
  • $\begingroup$ The solution is indeed $\ln(125/81)/\ln 1.08\approx5.6374572930016$. $\endgroup$ – Yves Daoust Mar 20 at 8:37
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Calling

$$ f(x) = a^x-b $$

we have

$$ f(x)= f(x_k) + f'(x_k)(x-x_k) + O(|x-x_k|^2) $$

so with $x$ near a root we have

$$ f(x)\approx 0 = f(x_k)+f'(x_k)(x-x_k) $$

thus

$$ x = x_{k+1} = x_k-\frac{f(x_k)}{f'(x_k)} $$

now calling $g(x) = x - \frac{f(x)}{f'(x)}$ we have

$$ x_{k+1} = g(x_k)\\ x_k = g(x_{k-1}) $$

and

$$ x_{k+1}-x_k = g(x_k)-g(x_{k-1}) $$

but

$$ g(x_k) = g(x_{k-1})+g'(x_{k-1})(x_k-x_{k-1})+O(|x_k-x_{k-1}|^2) $$

or

$$ x_{k+1}-x_k = g'(x_{k-1})(x_k-x_{k-1})+O(|x_k-x_{k-1}|^2) $$

now assuming small $O(|x_k-x_{k-1}|^2)$ we can assure convergence if $|g'(x)| < 1$ at the root vicinity.

In our case we have

$$ g(x) = x-\frac{a^x-b}{a^x\ln a} $$

and for convergence (sufficient) we need $|g'(x)| = |1-b a^{-x}|< 1$

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