Originally Posted by
samtrix
any help will be appreciated. thanks!!
Hello,
to #3:
You are supposed to know the Raphson-Newton formula:
If you are looking for the solutions of T(x) = 0 where T(x) is a term in x and you have found an initial approximate value for a solution then the next approximate value is:
$\displaystyle x_{n+1}=x_n-\frac{T(x_n)}{T'(x_n)}$ With your problem:
$\displaystyle x_{n+1}=x_n-\frac{\cos(x_n) - 1.4 \cdot \sqrt{x_n^3}}{-\sin(x_n) - 2.1 \cdot \sqrt{x_n}}$
Start with $\displaystyle x_0 = 1$
Code:
x_0 1
x_1 0.70773205
x_2 0.677232264
x_3 0.676832564
x_4 0.676832494
All following values equal $\displaystyle x_4$ (but that depends on the calculator you use)