Finding g(x) for Fixed Point Iteration

Question

I am at a complete loss on finding the equation of this function.

$$f(x) = 10e^{-x}\sin(2\pi x)-2.$$

i am looking for a fixed-point iteration $x_{n+1}=g(x_n)$ that finds a root of f that solves $f(x)=0$.

First try was to to change equation with logarithm to $$x=g(x)=−\log(1/(5\sin(2πx))).$$

i would appreciate any help.

Answer 1

Just try it out:

x = 0.5
for k in range(10): print k,x; x = log(5*sin(2*pi*x))

has the output

0 0.5
1 -35.0292711003
/usr/bin/ipython:2: RuntimeWarning: invalid value encountered in log

seems to be exploding.

Try the other way around

x = 0.5
for k in range(10): print k,x; x = arcsin(exp(x)/5)/(2*pi)

with the result

0 0.5
1 0.05348125972
2 0.033833959096
3 0.0331659027955
4 0.0331434275576
5 0.0331426717019
6 0.0331426462824
7 0.0331426454275
8 0.0331426453988
9 0.0331426453978

which looks better.

Now try to incorporate the other branches of the inverse sine leading to solutions around $0.449260830743$, $1.35816544929$ etc. via $$ g(x)=k+\frac{\arcsin(\exp(x)/5)}{2\pi} $$ or $$ g(x)=k+\frac12-\frac{\arcsin(\exp(x)/5)}{2\pi} $$ where $k\in\Bbb Z$.

Answer 2

Let $$g(x):=x-\frac{1}{m}f(x)$$ where $m$ is some constant to be decided later and we want to find an $x$ such that $f(x)=0$ or equivalently $g(x)=x$. This way you turn the problem into a fixed point problem.

Notice that $f(0)=-2$ and $f(1/8)=5\sqrt{2}e^{-1/8}-2>0$ therefore there is at least one zero on the interval $(0,1/8)$. Moreover $f'(x)>0$ for $x\in(0,1/8)$ since $f'(x)=10e^{-x}(2\pi\cos2\pi x-x)>0$ on this interval. Therefore this zero is a unique zero indeed. Also one can check that there are two numbers $m_1,m_2$ such that $0<m_1\leqslant f'(x)\leqslant m_2$ for all $x\in (0,1/8)$. For example $m_1=f'(0), m_2=f'(1/8)$ would do the job since $f''(x)<0$ on this interval. For $m> m_2$ we have $$0<1-\frac{m_2}{m}\leqslant g'(x)\leqslant 1-\frac{m_1}{m}<1$$ Another possibility is when $0<m_2\leqslant 2m_1$. In this case pick $m\in(m_2/2,m_1]$ then $$-1\leqslant1-\frac{m_2}{m}\leqslant g'(x)\leqslant1-\frac{m_1}{m}\leqslant0$$ In either case we have $|g'(x)|<1$ so $g(x)$ is a contraction and then you apply Banach fixed point theorem. Use the iteration $x_{n+1}:=g(x_n)$ with some $x_0\in(0,1/8)$.