Answer to Question #339825 in Calculus for mia

Question

use newton raphson method :

sin ( x + π/2) - ln |x| =0

Answer 1

We remind the Newton-Raphson method. Suppose that $f(x)$ is a differentiable function on the interval $(t_0,t_1)$, $t_0<t_1$, $r\in(t_0,t_1)$ is a root of equation $f(x)=0$ and $x_0=r+h\in(t_0,t_1)$ is a point near the root $r$. Define: $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$, $n\in{\mathbb{N}}$. Then, the sequence $x_n$ converges to the root $r$. The error after $n+1$ iterations is proportional to the square of error after $n$ iterations. The method works in case there are no other roots in the neighborhood of $r$.

Consider the function $f(x)=\sin(x+\frac{\pi}2)-\ln(|x|)=\cos(x)-\ln(|x|)$. For $|x|>e$ there are no roots, since $ln(|x|)>1$ and $|\sin(x+\frac{\pi}{2})|\leq1$. Analysis of the graph of the function $f(x)$ with the help of mathematical packages shows one root at the interval $(-e,0)$ and one root at the interval $(0,e)$. Consider the function for $x>0$. We get: $f(x)=\cos(x)-\ln(x)$. $f'(x)=-\sin(x)-\frac1x$. $f'<0$ for $x\in(0,1)$. Set $x_0=1$ and perform the method:

$x_1=x_0-\frac{f(x_0)}{f'(x_0)}\approx1.29341$, $x_2=x_1-\frac{f(x_1)}{f'(x_1)}\approx1.30296$, $x_3=x_2-\frac{f(x_2)}{f'(x_2)}\approx1.30296$.

The method can be stopped at $x_4$. The root is: $r_1\approx1.30296.$ All values are rounded up to $5$ decimal places. Since the function is even, another root is: $r_2\approx-1.30296.$

Answer: the roots are $r_1\approx1.30296$ and $r_2\approx-1.30296.$ The values are rounded to $5$ decimal places.