Start value for which an iterative function converges

Question

How do i find a start value for which the function: $x = - \exp(x)$ converges.

I know how to solve it by making a graph and picking a start value but i don't know how to find the start value for which the function converges. Any help will be great.

Answer 1

When you have to solve an equation looking like $A(x)=B(x)$, think that you are searching for the zero of $$f(x)=A(x)-B(x)$$ and analyze

In your case $$f(x)=x+e^x\implies f'(x)=1+e^x\implies f''(x)=e^x$$ The first and second derivatives are always positive meaning that there is only one root to the equation since the function grows from $-\infty$ to $+\infty$. So, basically, you could start from any point and converge to the solution. For sure, the number of iterations will depend on the quality of the starting point.

Let us try with Newton method with two different starting points which are far away from the solution $$\left( \begin{array}{cc} n & x_n \\ 0 & 10.0000 \\ 1 & 8.99959 \\ 2 & 7.99860 \\ 3 & 6.99625 \\ 4 & 5.99077 \\ 5 & 4.97832 \\ 6 & 3.95111 \\ 7 & 2.89542 \\ 8 & 1.79614 \\ 9 & 0.682831 \\ 10 & -0.210718 \\ 11 & -0.541814 \\ 12 & -0.567026 \\ 13 & -0.567143 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & -10.0000 \\ 1 & -0.000499377 \\ 2 & -0.500125 \\ 3 & -0.566314 \\ 4 & -0.567143 \end{array} \right)$$ You see how different is the path to the same solution.

For sure, graphing or inspection would provide a better starting point such as $x=-1$.

There is another point of interest : if you have the choice between two different guesses, select the one which is such that $f(x_0)\times f''(x_0) >0$ in order to avoid any overshoot of the solution (Darboux-Fourier theorem).

Answer 2

The equation can be written as

$$-x=e^{x} $$ or

$$y=e^{-y} $$

define $$y_{n+1}=e^{-y_n}=f (y_n) $$ with $y_0$ near the root.

it will converge since $$|f'(x)|<1$$