Answer:
[tex]x_2=2.82658[/tex]
[tex]x_3=2.60457[/tex]
Explanation:
Approximation of Roots of Equations
Newton's method is widely used to compute approximate values of the roots of equations which cannot be solved with algebraic methods.
Suposse we wanted to solve the equation
[tex]f(x)=0[/tex]
To find the approximate value of x, we use the following recursive formula
[tex]\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}[/tex]
Each value will determine the next value which should be closer to the solution
We want to solve
[tex]5sin(x)=x[/tex]
Which can be rewritten as
[tex]f(x)=5sinx-x=0[/tex]
We are also given an initial value for x
[tex]x_1=2[/tex]
Let's compute the derivative of f
[tex]f'(x)=5cosx-1[/tex]
[tex]f(2)=5sin(2)-2=2.5465[/tex]
[tex]f'(2)=5cos(2)-1=-3.08073[/tex]
[tex]\displaystyle x_{2}=x_{1}-{\frac {f(x_{1})}{f'(x_{1})}}[/tex]
[tex]\displaystyle x_{2}=2-{\frac {2.5465}{-3.08073}}[/tex]
[tex]x_2=2.82658[/tex]
We now find the third approximation
[tex]f(2.82658)=5sin(2.82658)-2.82658=-1.277463[/tex]
[tex]f'(2)=5cos(2.82658)-1=-5.753969[/tex]
[tex]x_{2}=2.82658-{\frac {-1.277463}{-5.753969}}[/tex]
[tex]x_3=2.60457[/tex]
This approximation gives a value of
[tex]f(2.60457)=5sin(2.60457)-2.60457=-0.04667[/tex]
It's accurate to the second decimal
