\(\text{www.the}\beta\text{etafunction.com}\)
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 \(\displaystyle e^{- x}= \frac{119 x^{3}}{250} - 3\) using \(\displaystyle x_0=1\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - 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{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551\end{equation*} \begin{equation*}x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276\end{equation*} \begin{equation*}x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850\end{equation*} \begin{equation*}x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503\end{equation*} \begin{equation*}x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{119 x^{3}}{250} - 3$ using $x_0=1$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - 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{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551\end{equation*}
\begin{equation*}x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276\end{equation*}
\begin{equation*}x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850\end{equation*}
\begin{equation*}x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503\end{equation*}
\begin{equation*}x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115\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{119 x^{3}}{250} - 3 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B119%20x%5E%7B3%7D%7D%7B250%7D%20-%203%20" alt="LaTeX: \displaystyle e^{- x}= \frac{119 x^{3}}{250} - 3 " data-equation-content=" \displaystyle e^{- x}= \frac{119 x^{3}}{250} - 3 " /> 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{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - e^{- x_{n}}} " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20x_%7Bn%7D%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B357%20x_%7Bn%7D%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="LaTeX: x_{n+1} = x_{n} - \frac{- \frac{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{119 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{357 x_{n}^{2}}{250} - 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{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%281.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20%281.0000000000%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%281.0000000000%29%7D%7D%7B-%20%5Cfrac%7B357%20%281.0000000000%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%281.0000000000%29%7D%7D%20%3D%202.6102859551%20" alt="LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551 " data-equation-content=" x_{1} = (1.0000000000) - \frac{- \frac{119 (1.0000000000)^{3}}{250} + 3 + e^{- (1.0000000000)}}{- \frac{357 (1.0000000000)^{2}}{250} - e^{- (1.0000000000)}} = 2.6102859551 " />
<img class="equation_image" title=" x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.6102859551%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20%282.6102859551%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.6102859551%29%7D%7D%7B-%20%5Cfrac%7B357%20%282.6102859551%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.6102859551%29%7D%7D%20%3D%202.0602328276%20" alt="LaTeX: x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276 " data-equation-content=" x_{2} = (2.6102859551) - \frac{- \frac{119 (2.6102859551)^{3}}{250} + 3 + e^{- (2.6102859551)}}{- \frac{357 (2.6102859551)^{2}}{250} - e^{- (2.6102859551)}} = 2.0602328276 " />
<img class="equation_image" title=" x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.0602328276%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20%282.0602328276%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.0602328276%29%7D%7D%7B-%20%5Cfrac%7B357%20%282.0602328276%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.0602328276%29%7D%7D%20%3D%201.8929765850%20" alt="LaTeX: x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850 " data-equation-content=" x_{3} = (2.0602328276) - \frac{- \frac{119 (2.0602328276)^{3}}{250} + 3 + e^{- (2.0602328276)}}{- \frac{357 (2.0602328276)^{2}}{250} - e^{- (2.0602328276)}} = 1.8929765850 " />
<img class="equation_image" title=" x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%281.8929765850%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20%281.8929765850%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%281.8929765850%29%7D%7D%7B-%20%5Cfrac%7B357%20%281.8929765850%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%281.8929765850%29%7D%7D%20%3D%201.8781334503%20" alt="LaTeX: x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503 " data-equation-content=" x_{4} = (1.8929765850) - \frac{- \frac{119 (1.8929765850)^{3}}{250} + 3 + e^{- (1.8929765850)}}{- \frac{357 (1.8929765850)^{2}}{250} - e^{- (1.8929765850)}} = 1.8781334503 " />
<img class="equation_image" title=" x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%281.8781334503%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B119%20%281.8781334503%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%281.8781334503%29%7D%7D%7B-%20%5Cfrac%7B357%20%281.8781334503%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%281.8781334503%29%7D%7D%20%3D%201.8780222115%20" alt="LaTeX: x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115 " data-equation-content=" x_{5} = (1.8781334503) - \frac{- \frac{119 (1.8781334503)^{3}}{250} + 3 + e^{- (1.8781334503)}}{- \frac{357 (1.8781334503)^{2}}{250} - e^{- (1.8781334503)}} = 1.8780222115 " />
</p> </p>