Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{637 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{637 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 4}{- \frac{1911 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{637 (1.0000000000)^{3}}{1000} + \cos{\left((1.0000000000) \right)} + 4}{- \frac{1911 (1.0000000000)^{2}}{1000} - \sin{\left((1.0000000000) \right)}} = 2.4181084296$ $LaTeX: x_{2} = (2.4181084296) - \frac{- \frac{637 (2.4181084296)^{3}}{1000} + \cos{\left((2.4181084296) \right)} + 4}{- \frac{1911 (2.4181084296)^{2}}{1000} - \sin{\left((2.4181084296) \right)}} = 1.9317802080$ $LaTeX: x_{3} = (1.9317802080) - \frac{- \frac{637 (1.9317802080)^{3}}{1000} + \cos{\left((1.9317802080) \right)} + 4}{- \frac{1911 (1.9317802080)^{2}}{1000} - \sin{\left((1.9317802080) \right)}} = 1.8145977044$ $LaTeX: x_{4} = (1.8145977044) - \frac{- \frac{637 (1.8145977044)^{3}}{1000} + \cos{\left((1.8145977044) \right)} + 4}{- \frac{1911 (1.8145977044)^{2}}{1000} - \sin{\left((1.8145977044) \right)}} = 1.8080581643$ $LaTeX: x_{5} = (1.8080581643) - \frac{- \frac{637 (1.8080581643)^{3}}{1000} + \cos{\left((1.8080581643) \right)} + 4}{- \frac{1911 (1.8080581643)^{2}}{1000} - \sin{\left((1.8080581643) \right)}} = 1.8080383555$