\(\text{www.the}\beta\text{etafunction.com}\)
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 \(\displaystyle e^{- x}= \frac{461 x^{3}}{1000} - 5\) using \(\displaystyle x_0=3\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}} \end{equation*} Using \(\displaystyle x_0 = 3\) and \(\displaystyle n = 0,1,2,3,\) and \(\displaystyle 4\) gives: \begin{equation*}x_{1} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192\end{equation*} \begin{equation*}x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567\end{equation*} \begin{equation*}x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742\end{equation*} \begin{equation*}x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647\end{equation*} \begin{equation*}x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{461 x^{3}}{1000} - 5$ using $x_0=3$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}} \end{equation*}
Using $x_0 = 3$ and $n = 0,1,2,3,$ and $4$ gives:
\begin{equation*}x_{1} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192\end{equation*}
\begin{equation*}x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567\end{equation*}
\begin{equation*}x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742\end{equation*}
\begin{equation*}x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647\end{equation*}
\begin{equation*}x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626\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{461 x^{3}}{1000} - 5 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B461%20x%5E%7B3%7D%7D%7B1000%7D%20-%205%20" alt="LaTeX: \displaystyle e^{- x}= \frac{461 x^{3}}{1000} - 5 " data-equation-content=" \displaystyle e^{- x}= \frac{461 x^{3}}{1000} - 5 " /> using <img class="equation_image" title=" \displaystyle x_0=3 " src="/equation_images/%20%5Cdisplaystyle%20x_0%3D3%20" alt="LaTeX: \displaystyle x_0=3 " data-equation-content=" \displaystyle x_0=3 " /> . </p> </p><p> <p>Using the formula for Newton's method gives
<img class="equation_image" title=" x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}} " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20x_%7Bn%7D%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B1383%20x_%7Bn%7D%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="LaTeX: x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{461 x_{n}^{3}}{1000} + 5 + e^{- x_{n}}}{- \frac{1383 x_{n}^{2}}{1000} - e^{- x_{n}}} " />
Using <img class="equation_image" title=" \displaystyle x_0 = 3 " src="/equation_images/%20%5Cdisplaystyle%20x_0%20%3D%203%20" alt="LaTeX: \displaystyle x_0 = 3 " data-equation-content=" \displaystyle x_0 = 3 " /> 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} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%283.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20%283.0000000000%29%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20%283.0000000000%29%7D%7D%7B-%20%5Cfrac%7B1383%20%283.0000000000%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%283.0000000000%29%7D%7D%20%3D%202.4080708192%20" alt="LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192 " data-equation-content=" x_{1} = (3.0000000000) - \frac{- \frac{461 (3.0000000000)^{3}}{1000} + 5 + e^{- (3.0000000000)}}{- \frac{1383 (3.0000000000)^{2}}{1000} - e^{- (3.0000000000)}} = 2.4080708192 " />
<img class="equation_image" title=" x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.4080708192%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20%282.4080708192%29%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.4080708192%29%7D%7D%7B-%20%5Cfrac%7B1383%20%282.4080708192%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%282.4080708192%29%7D%7D%20%3D%202.2419267567%20" alt="LaTeX: x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567 " data-equation-content=" x_{2} = (2.4080708192) - \frac{- \frac{461 (2.4080708192)^{3}}{1000} + 5 + e^{- (2.4080708192)}}{- \frac{1383 (2.4080708192)^{2}}{1000} - e^{- (2.4080708192)}} = 2.2419267567 " />
<img class="equation_image" title=" x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.2419267567%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20%282.2419267567%29%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.2419267567%29%7D%7D%7B-%20%5Cfrac%7B1383%20%282.2419267567%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%282.2419267567%29%7D%7D%20%3D%202.2293865742%20" alt="LaTeX: x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742 " data-equation-content=" x_{3} = (2.2419267567) - \frac{- \frac{461 (2.2419267567)^{3}}{1000} + 5 + e^{- (2.2419267567)}}{- \frac{1383 (2.2419267567)^{2}}{1000} - e^{- (2.2419267567)}} = 2.2293865742 " />
<img class="equation_image" title=" x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%282.2293865742%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20%282.2293865742%29%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.2293865742%29%7D%7D%7B-%20%5Cfrac%7B1383%20%282.2293865742%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%282.2293865742%29%7D%7D%20%3D%202.2293180647%20" alt="LaTeX: x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647 " data-equation-content=" x_{4} = (2.2293865742) - \frac{- \frac{461 (2.2293865742)^{3}}{1000} + 5 + e^{- (2.2293865742)}}{- \frac{1383 (2.2293865742)^{2}}{1000} - e^{- (2.2293865742)}} = 2.2293180647 " />
<img class="equation_image" title=" x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%282.2293180647%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B461%20%282.2293180647%29%5E%7B3%7D%7D%7B1000%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.2293180647%29%7D%7D%7B-%20%5Cfrac%7B1383%20%282.2293180647%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%282.2293180647%29%7D%7D%20%3D%202.2293180626%20" alt="LaTeX: x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626 " data-equation-content=" x_{5} = (2.2293180647) - \frac{- \frac{461 (2.2293180647)^{3}}{1000} + 5 + e^{- (2.2293180647)}}{- \frac{1383 (2.2293180647)^{2}}{1000} - e^{- (2.2293180647)}} = 2.2293180626 " />
</p> </p>