newton raphsonnumerical methods
Given $f(x) = x – sin(x)$, I'm asked to solve this using the Newton-Raphson method and give the order of convergence. Now for me this is strange because I need to have a starting value $x_0$ in order to go through the Newton-Raphson method.
However, I really don't know how to calculate the order of convergence. Any help for this formula, and how to do this in general?
Since $$ \sqrt{a}x_{n+1}-1=\sqrt{a}x_n-1+0.5\sqrt{a}x_n(1+\sqrt{a}x_n)(1-\sqrt{a}x_n) \\ =(\sqrt{a}x_n-1)(1-0.5\sqrt{a}x_n(1+\sqrt{a}x_n)) \\ =-(1+0.5\sqrt{a}x_n)(\sqrt{a}x_n-1)^2 $$ you will get quadratic convergence if $$ \frac12<\sqrt{a}x_0<\frac32 \text{ or } \frac14<ax_0^2<\frac94, $$ so that $|1+0.5\sqrt{a}x_0|<\frac74<2$ and $\frac74|\sqrt{a}x_0-1|<\frac78<1$. Which then implies $$ |\sqrt{a}x_n-1|<\frac47\left(\frac74|\sqrt{a}x_0-1|\right)^{2^n}<\frac12\left(\frac78\right)^{2^n-1} $$ Your parameter and initial value fall far away from that condition.
For the graph of $f(x)=x^3-5x$, we have $f'(x)=3x^2-5$ $$f(1)=-4$$ $$f'(-1)=4$$ $$f'(1)=-2=f'(-1)$$
The tangent line at $x=1$ is $y+4=-2(x-1)$ which is $y=-2x-2$.
The tangent line at $x=-1$ is $y-4=-2(x+1)$ which is $y=-2x+2$
Geometrically, what has happened is you are trapped in the following cycle.
Starting from $(1,-4)$, by traveling along the tangent, we intercept the $x$-axis at $x=-1$. From $(-1,4)$, by trveling along the tangent, we intercept the $x$-axis at $x=1$.
Best Answer
The only root is $x=0$ and it has multiplicity $3$. Now apply what you know about the convergence of Newtons method to multiple roots.