Question #339825

use newton raphson method :


sin ( x + π/2) - ln |x| =0


1
Expert's answer
2022-05-12T09:29:02-0400

We remind the Newton-Raphson method. Suppose that f(x)f(x) is a differentiable function on the interval (t0,t1)(t_0,t_1), t0<t1t_0<t_1, r(t0,t1)r\in(t_0,t_1) is a root of equation f(x)=0f(x)=0 and x0=r+h(t0,t1)x_0=r+h\in(t_0,t_1) is a point near the root rr. Define: xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, nNn\in{\mathbb{N}}. Then, the sequence xnx_n converges to the root rr. The error after n+1n+1 iterations is proportional to the square of error after nn iterations. The method works in case there are no other roots in the neighborhood of rr.

Consider the function f(x)=sin(x+π2)ln(x)=cos(x)ln(x)f(x)=\sin(x+\frac{\pi}2)-\ln(|x|)=\cos(x)-\ln(|x|). For x>e|x|>e there are no roots, since ln(x)>1ln(|x|)>1 and sin(x+π2)1|\sin(x+\frac{\pi}{2})|\leq1. Analysis of the graph of the function f(x)f(x) with the help of mathematical packages shows one root at the interval (e,0)(-e,0) and one root at the interval (0,e)(0,e). Consider the function for x>0x>0. We get: f(x)=cos(x)ln(x)f(x)=\cos(x)-\ln(x). f(x)=sin(x)1xf'(x)=-\sin(x)-\frac1x. f<0f'<0 for x(0,1)x\in(0,1). Set x0=1x_0=1 and perform the method:

x1=x0f(x0)f(x0)1.29341x_1=x_0-\frac{f(x_0)}{f'(x_0)}\approx1.29341, x2=x1f(x1)f(x1)1.30296x_2=x_1-\frac{f(x_1)}{f'(x_1)}\approx1.30296, x3=x2f(x2)f(x2)1.30296x_3=x_2-\frac{f(x_2)}{f'(x_2)}\approx1.30296.

The method can be stopped at x4x_4. The root is: r11.30296.r_1\approx1.30296. All values are rounded up to 55 decimal places. Since the function is even, another root is: r21.30296.r_2\approx-1.30296.

Answer: the roots are r11.30296r_1\approx1.30296 and r21.30296.r_2\approx-1.30296. The values are rounded to 55 decimal places.


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