Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle e^{- x}= \frac{41 x^{3}}{125} - 1$ using $LaTeX: \displaystyle x_0=1$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{41 x_{n}^{3}}{125} + 1 + e^{- x_{n}}}{- \frac{123 x_{n}^{2}}{125} - e^{- x_{n}}}$ Using $LaTeX: \displaystyle x_0 = 1$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{41 (1.0000000000)^{3}}{125} + 1 + e^{- (1.0000000000)}}{- \frac{123 (1.0000000000)^{2}}{125} - e^{- (1.0000000000)}} = 1.7692101895$ $LaTeX: x_{2} = (1.7692101895) - \frac{- \frac{41 (1.7692101895)^{3}}{125} + 1 + e^{- (1.7692101895)}}{- \frac{123 (1.7692101895)^{2}}{125} - e^{- (1.7692101895)}} = 1.5704909155$ $LaTeX: x_{3} = (1.5704909155) - \frac{- \frac{41 (1.5704909155)^{3}}{125} + 1 + e^{- (1.5704909155)}}{- \frac{123 (1.5704909155)^{2}}{125} - e^{- (1.5704909155)}} = 1.5467433796$ $LaTeX: x_{4} = (1.5467433796) - \frac{- \frac{41 (1.5467433796)^{3}}{125} + 1 + e^{- (1.5467433796)}}{- \frac{123 (1.5467433796)^{2}}{125} - e^{- (1.5467433796)}} = 1.5464286221$ $LaTeX: x_{5} = (1.5464286221) - \frac{- \frac{41 (1.5464286221)^{3}}{125} + 1 + e^{- (1.5464286221)}}{- \frac{123 (1.5464286221)^{2}}{125} - e^{- (1.5464286221)}} = 1.5464285675$