Give a fixed point iteration

Question

consider the following equation $$x=\sin{x}-ax+30$$ (a) Give a fixed point iteration scheme that will converge for all values of $a$ (b) How do you know that the scheme will work for all values of $a$?

Please help me as I have tried different $g(x)$ for $x=g(x)$ but none of them worked for all values of $a$.

Answer 1

hint

You must write your equation as

$$x=\phi (x). $$ to be sure the schema works and iterations will converge to the root, you should have

$$|\phi'(x)|\leq k <1$$.

If $a\geq 0$, you could take

$$x=\frac {\sin (x)+30}{1+a} $$

Answer 2

Here is a solution :

$x_{n+1}=f_a(x_n) \ \ $ with $ \ \ f_a(x):=\dfrac{\sin(x)-x\cos(x)+30}{1+a-\cos(x)}.$

Indeed

• the fixed point equation $x=f_a(x)$, i.e., $x=\dfrac{\sin(x)-x\cos(x)+30}{1+a-\cos(x)}$ is equivalent to the given equation.

• As $f_a'(x)=\dfrac{ \sin(x) [a x+x- \sin{x} -30]}{(a-\cos{x} +1)^2}$ is equal to $0$ when $x$ is equal to the fixed point (because then the quantity between brackets is $0$!), we can say that $|f'(x)|<1$ for all $x$ in a certain neighborhood of this fixed point, thus the convergence is guaranteed. But we must have a function that is always defined. If $a=0$, the denominator becomes $1-cos(x)$, thus can be zero, which makes the function tends to $\infty$ (which is not at all desirable...), whereas, if $a>0$, this will never happen.

It is very instructive to see the curves of some functions $y=f_a(x)$ (look below): the fixed point is classicaly obtained as the abscissa of intersection of the curve with the line whose equation is $y=x$ : at this point, the curve has a zero derivative as shown upwards ; thus the curve is very flat in such a point.

Remark: I have built function $f_a$ by considering Newton's convergence scheme $x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$ which, with the initial function, gives:

$$x_{n+1}=\underbrace{x_n-\dfrac{x_n+ax_n-\sin x_n -30}{1+a-\cos(x)}}_{f_a(x_n)}$$