2. 10 pts; 5 per iteration complete two iterations of newtons method ($x_{n + 1}=x_n-\frac{f(x_n)}{f(x_n)}$)…

2. 10 pts; 5 per iteration complete two iterations of newtons method ($x_{n + 1}=x_n-\frac{f(x_n)}{f(x_n)}$) to find an approximate solution of the equation $\\sin(x)-\\frac{3}{5}=0$. start with $x_1 = 1$.

2. 10 pts; 5 per iteration complete two iterations of newtons method ($x_{n + 1}=x_n-\frac{f(x_n)}{f(x_n)}$) to find an approximate solution of the equation $\\sin(x)-\\frac{3}{5}=0$. start with $x_1 = 1$.

Answer

Explanation:

Step1: Find (f(x)) and (f^{\prime}(x))

Given (f(x)=\sin(x)-\frac{3}{5}), then (f^{\prime}(x)=\cos(x))

Step2: First - iteration ((n = 1))

We know (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}), with (x_{1}=1) (f(1)=\sin(1)-\frac{3}{5}\approx0.8415 - 0.6=0.2415) (f^{\prime}(1)=\cos(1)\approx0.5403) (x_{2}=x_{1}-\frac{f(x_{1})}{f^{\prime}(x_{1})}=1-\frac{\sin(1)-\frac{3}{5}}{\cos(1)}) (x_{2}=1-\frac{0.2415}{0.5403}\approx1 - 0.447=0.553)

Step3: Second - iteration ((n = 2))

(f(0.553)=\sin(0.553)-\frac{3}{5}\approx0.526 - 0.6=- 0.074) (f^{\prime}(0.553)=\cos(0.553)\approx0.851) (x_{3}=x_{2}-\frac{f(x_{2})}{f^{\prime}(x_{2})}=0.553-\frac{\sin(0.553)-\frac{3}{5}}{\cos(0.553)}) (x_{3}=0.553-\frac{-0.074}{0.851}\approx0.553 + 0.087=0.64)

Answer:

The first - iteration approximation is (x_{2}\approx0.553) and the second - iteration approximation is (x_{3}\approx0.64)