\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{41 x^{3}}{125} - 1\) using \(\displaystyle x_0=1\).
Using the formula for Newton's method gives \begin{equation*}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}}} \end{equation*} Using \(\displaystyle x_0 = 1\) and \(\displaystyle n = 0,1,2,3,\) and \(\displaystyle 4\) gives: \begin{equation*}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\end{equation*} \begin{equation*}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\end{equation*} \begin{equation*}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\end{equation*} \begin{equation*}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\end{equation*} \begin{equation*}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\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{41 x^{3}}{125} - 1$ using $x_0=1$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}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}}} \end{equation*}
Using $x_0 = 1$ and $n = 0,1,2,3,$ and $4$ gives:
\begin{equation*}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\end{equation*}
\begin{equation*}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\end{equation*}
\begin{equation*}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\end{equation*}
\begin{equation*}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\end{equation*}
\begin{equation*}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\end{equation*}
}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use Newton's method to find the first 5 approximations of the solution to the equation <img class="equation_image" title=" \displaystyle e^{- x}= \frac{41 x^{3}}{125} - 1 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B41%20x%5E%7B3%7D%7D%7B125%7D%20-%201%20" alt="LaTeX: \displaystyle e^{- x}= \frac{41 x^{3}}{125} - 1 " data-equation-content=" \displaystyle e^{- x}= \frac{41 x^{3}}{125} - 1 " /> using <img class="equation_image" title=" \displaystyle x_0=1 " src="/equation_images/%20%5Cdisplaystyle%20x_0%3D1%20" alt="LaTeX: \displaystyle x_0=1 " data-equation-content=" \displaystyle x_0=1 " /> . </p> </p><p> <p>Using the formula for Newton's method gives
<img class="equation_image" title=" 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}}} " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20x_%7Bn%7D%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B123%20x_%7Bn%7D%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="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}}} " data-equation-content=" 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 <img class="equation_image" title=" \displaystyle x_0 = 1 " src="/equation_images/%20%5Cdisplaystyle%20x_0%20%3D%201%20" alt="LaTeX: \displaystyle x_0 = 1 " data-equation-content=" \displaystyle x_0 = 1 " /> and <img class="equation_image" title=" \displaystyle n = 0,1,2,3, " src="/equation_images/%20%5Cdisplaystyle%20n%20%3D%200%2C1%2C2%2C3%2C%20" alt="LaTeX: \displaystyle n = 0,1,2,3, " data-equation-content=" \displaystyle n = 0,1,2,3, " /> and <img class="equation_image" title=" \displaystyle 4 " src="/equation_images/%20%5Cdisplaystyle%204%20" alt="LaTeX: \displaystyle 4 " data-equation-content=" \displaystyle 4 " /> gives:
<img class="equation_image" title=" 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 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%281.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20%281.0000000000%29%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20%281.0000000000%29%7D%7D%7B-%20%5Cfrac%7B123%20%281.0000000000%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%281.0000000000%29%7D%7D%20%3D%201.7692101895%20" alt="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 " data-equation-content=" 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 " />
<img class="equation_image" title=" 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 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%281.7692101895%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20%281.7692101895%29%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20%281.7692101895%29%7D%7D%7B-%20%5Cfrac%7B123%20%281.7692101895%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%281.7692101895%29%7D%7D%20%3D%201.5704909155%20" alt="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 " data-equation-content=" 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 " />
<img class="equation_image" title=" 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 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%281.5704909155%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20%281.5704909155%29%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20%281.5704909155%29%7D%7D%7B-%20%5Cfrac%7B123%20%281.5704909155%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%281.5704909155%29%7D%7D%20%3D%201.5467433796%20" alt="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 " data-equation-content=" 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 " />
<img class="equation_image" title=" 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 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%281.5467433796%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20%281.5467433796%29%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20%281.5467433796%29%7D%7D%7B-%20%5Cfrac%7B123%20%281.5467433796%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%281.5467433796%29%7D%7D%20%3D%201.5464286221%20" alt="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 " data-equation-content=" 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 " />
<img class="equation_image" title=" 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 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%281.5464286221%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B41%20%281.5464286221%29%5E%7B3%7D%7D%7B125%7D%20%2B%201%20%2B%20e%5E%7B-%20%281.5464286221%29%7D%7D%7B-%20%5Cfrac%7B123%20%281.5464286221%29%5E%7B2%7D%7D%7B125%7D%20-%20e%5E%7B-%20%281.5464286221%29%7D%7D%20%3D%201.5464285675%20" alt="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 " data-equation-content=" 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 " />
</p> </p>