Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{17 x^{3}}{40} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{40} + \cos{\left(x_{n} \right)} + 1}{- \frac{51 x_{n}^{2}}{40} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{17 (1.0000000000)^{3}}{40} + \cos{\left((1.0000000000) \right)} + 1}{- \frac{51 (1.0000000000)^{2}}{40} - \sin{\left((1.0000000000) \right)}} = 1.5269631920$ $LaTeX: x_{2} = (1.5269631920) - \frac{- \frac{17 (1.5269631920)^{3}}{40} + \cos{\left((1.5269631920) \right)} + 1}{- \frac{51 (1.5269631920)^{2}}{40} - \sin{\left((1.5269631920) \right)}} = 1.4088053494$ $LaTeX: x_{3} = (1.4088053494) - \frac{- \frac{17 (1.4088053494)^{3}}{40} + \cos{\left((1.4088053494) \right)} + 1}{- \frac{51 (1.4088053494)^{2}}{40} - \sin{\left((1.4088053494) \right)}} = 1.4011123101$ $LaTeX: x_{4} = (1.4011123101) - \frac{- \frac{17 (1.4011123101)^{3}}{40} + \cos{\left((1.4011123101) \right)} + 1}{- \frac{51 (1.4011123101)^{2}}{40} - \sin{\left((1.4011123101) \right)}} = 1.4010805038$ $LaTeX: x_{5} = (1.4010805038) - \frac{- \frac{17 (1.4010805038)^{3}}{40} + \cos{\left((1.4010805038) \right)} + 1}{- \frac{51 (1.4010805038)^{2}}{40} - \sin{\left((1.4010805038) \right)}} = 1.4010805033$