# fixed point function (nonlinear equation)

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.

• 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 $g(x) = x - 1.08^x + \frac{125}{81}$. – M. Vinay Mar 20 at 7:16
• Thank you, it works on my code ;). By the way, is there a general way to find $g(x)$ ? – user587779 Mar 20 at 7:52
• 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. – M. Vinay Mar 20 at 7:54
• The solution is indeed $\ln(125/81)/\ln 1.08\approx5.6374572930016$. – Yves Daoust Mar 20 at 8:37

## 1 Answer

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 $$|g'(x)| = |1-b a^{-x}|< 1$$