Fixed point method I am trying to find why when using fixed point method to find a root, I cannot find the convergent value that I need.
I have the function: $\dfrac{\cos(x\pi)}{9}+\sin(x\pi)$
I solved for x, and I tried two approaches:
$$x= \frac {9\cos^-1(-8\sin(\pi x))}{\pi} \text{ or } x=\frac{\sin^{-1}(\frac{1}{8} (\cos\frac{(x\pi)}{9}))}{\pi}$$
My initial guess is $1$ so my $x_0=1$
But know when I am trying to find the x1 for:
1) $x= \dfrac {9\cos^{-1}(-8\sin(\pi x))}{pi}$ , $x_1=4.5$
and
2) for $x=\dfrac{\sin^{-1}\left(\frac{1}{8}(\cos\frac{(x\pi)}{9})\right)}{\pi}$ , $x_1= 0.037$ and the value converges for $x=0.39$ and I know by observing the graph of $f(x)$ that is not true.
Can anyone help me to find what is wrong with my calculations? I tried other methods to find the first root, and all of them converges for a unique value.
 A: Are you trying to solve $\cos(\pi x)/9 + \sin(\pi x) = 0$? Then you might try
$$\eqalign{\cos(\pi x) &= - 9 \sin(\pi x)\cr
   \pi x &= \arccos(-9 \sin(\pi x))\cr
   x &= \dfrac{1}{\pi} \arccos(-9 \sin(\pi x))\cr}$$
(but this will not have a real value if $|\sin(\pi x)| > 1/9$), or
$$ \eqalign{\sin(\pi x) &=  -\cos(\pi x)/9\cr
       \pi x &= \arcsin(- \cos(\pi x)/9)\cr
          x &= \dfrac{1}{\pi} \arcsin(-\cos(\pi x)/9)\cr}$$
Whether or not either of these will converge will depend on the derivative...
A: You should be much more careful when presenting your problem, but maybe you have not given it so much thought. It seems you are given the expression ${1\over9}\cos(\pi x)+\sin(\pi x)=:f(x)$. You tell us you want to find a root. But there is no equation to be seen. Mabe you want to find a root of the equation $f(x)=0$, who knows. You then tell us you want to use the "fixed point method". Therefore you would have to set up another map $g:\>x\mapsto g(x)$ such that a fixed point of this new map would automatically be a "root", i.e., a solution of the given equation, supposedly $f(x)=0$. And on and on.
Do you get me?
