Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{523 x^{3}}{1000} - 4$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{523 x_{n}^{3}}{1000} + 4 + e^{- x_{n}}}{- \frac{1569 x_{n}^{2}}{1000} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{523 (1.0000000000)^{3}}{1000} + 4 + e^{- (1.0000000000)}}{- \frac{1569 (1.0000000000)^{2}}{1000} - e^{- (1.0000000000)}} = 2.9850897064$ $LaTeX: x_{2} = (2.9850897064) - \frac{- \frac{523 (2.9850897064)^{3}}{1000} + 4 + e^{- (2.9850897064)}}{- \frac{1569 (2.9850897064)^{2}}{1000} - e^{- (2.9850897064)}} = 2.2823174698$ $LaTeX: x_{3} = (2.2823174698) - \frac{- \frac{523 (2.2823174698)^{3}}{1000} + 4 + e^{- (2.2823174698)}}{- \frac{1569 (2.2823174698)^{2}}{1000} - e^{- (2.2823174698)}} = 2.0266470510$ $LaTeX: x_{4} = (2.0266470510) - \frac{- \frac{523 (2.0266470510)^{3}}{1000} + 4 + e^{- (2.0266470510)}}{- \frac{1569 (2.0266470510)^{2}}{1000} - e^{- (2.0266470510)}} = 1.9929344523$ $LaTeX: x_{5} = (1.9929344523) - \frac{- \frac{523 (1.9929344523)^{3}}{1000} + 4 + e^{- (1.9929344523)}}{- \frac{1569 (1.9929344523)^{2}}{1000} - e^{- (1.9929344523)}} = 1.9923819728$