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Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{x^{3}}{1000} - 2\) using \(\displaystyle x_0=11\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{1000} - e^{- x_{n}}} \end{equation*} Using \(\displaystyle x_0 = 11\) and \(\displaystyle n = 0,1,2,3,\) and \(\displaystyle 4\) gives: \begin{equation*}x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229\end{equation*} \begin{equation*}x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547\end{equation*} \begin{equation*}x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606\end{equation*} \begin{equation*}x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853\end{equation*} \begin{equation*}x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{x^{3}}{1000} - 2$ using $x_0=11$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{1000} - e^{- x_{n}}} \end{equation*}
Using $x_0 = 11$ and $n = 0,1,2,3,$ and $4$ gives:
\begin{equation*}x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229\end{equation*}
\begin{equation*}x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547\end{equation*}
\begin{equation*}x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606\end{equation*}
\begin{equation*}x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853\end{equation*}
\begin{equation*}x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853\end{equation*}
}
\end{question}
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\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{x^{3}}{1000} - 2 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7Bx%5E%7B3%7D%7D%7B1000%7D%20-%202%20" alt="LaTeX: \displaystyle e^{- x}= \frac{x^{3}}{1000} - 2 " data-equation-content=" \displaystyle e^{- x}= \frac{x^{3}}{1000} - 2 " /> using <img class="equation_image" title=" \displaystyle x_0=11 " src="/equation_images/%20%5Cdisplaystyle%20x_0%3D11%20" alt="LaTeX: \displaystyle x_0=11 " data-equation-content=" \displaystyle x_0=11 " /> . </p> </p><p> <p>Using the formula for Newton's method gives
<img class="equation_image" title=" x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 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%7Bx_%7Bn%7D%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B3%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{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{1000} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{x_{n}^{3}}{1000} + 2 + e^{- x_{n}}}{- \frac{3 x_{n}^{2}}{1000} - e^{- x_{n}}} " />
Using <img class="equation_image" title=" \displaystyle x_0 = 11 " src="/equation_images/%20%5Cdisplaystyle%20x_0%20%3D%2011%20" alt="LaTeX: \displaystyle x_0 = 11 " data-equation-content=" \displaystyle x_0 = 11 " /> 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} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%2811.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B%2811.0000000000%29%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20%2811.0000000000%29%7D%7D%7B-%20%5Cfrac%7B3%20%2811.0000000000%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%2811.0000000000%29%7D%7D%20%3D%2012.8429364229%20" alt="LaTeX: x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229 " data-equation-content=" x_{1} = (11.0000000000) - \frac{- \frac{(11.0000000000)^{3}}{1000} + 2 + e^{- (11.0000000000)}}{- \frac{3 (11.0000000000)^{2}}{1000} - e^{- (11.0000000000)}} = 12.8429364229 " />
<img class="equation_image" title=" x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%2812.8429364229%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B%2812.8429364229%29%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20%2812.8429364229%29%7D%7D%7B-%20%5Cfrac%7B3%20%2812.8429364229%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%2812.8429364229%29%7D%7D%20%3D%2012.6038131547%20" alt="LaTeX: x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547 " data-equation-content=" x_{2} = (12.8429364229) - \frac{- \frac{(12.8429364229)^{3}}{1000} + 2 + e^{- (12.8429364229)}}{- \frac{3 (12.8429364229)^{2}}{1000} - e^{- (12.8429364229)}} = 12.6038131547 " />
<img class="equation_image" title=" x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%2812.6038131547%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B%2812.6038131547%29%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20%2812.6038131547%29%7D%7D%7B-%20%5Cfrac%7B3%20%2812.6038131547%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%2812.6038131547%29%7D%7D%20%3D%2012.5992192606%20" alt="LaTeX: x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606 " data-equation-content=" x_{3} = (12.6038131547) - \frac{- \frac{(12.6038131547)^{3}}{1000} + 2 + e^{- (12.6038131547)}}{- \frac{3 (12.6038131547)^{2}}{1000} - e^{- (12.6038131547)}} = 12.5992192606 " />
<img class="equation_image" title=" x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%2812.5992192606%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B%2812.5992192606%29%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20%2812.5992192606%29%7D%7D%7B-%20%5Cfrac%7B3%20%2812.5992192606%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%2812.5992192606%29%7D%7D%20%3D%2012.5992175853%20" alt="LaTeX: x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853 " data-equation-content=" x_{4} = (12.5992192606) - \frac{- \frac{(12.5992192606)^{3}}{1000} + 2 + e^{- (12.5992192606)}}{- \frac{3 (12.5992192606)^{2}}{1000} - e^{- (12.5992192606)}} = 12.5992175853 " />
<img class="equation_image" title=" x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%2812.5992175853%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B%2812.5992175853%29%5E%7B3%7D%7D%7B1000%7D%20%2B%202%20%2B%20e%5E%7B-%20%2812.5992175853%29%7D%7D%7B-%20%5Cfrac%7B3%20%2812.5992175853%29%5E%7B2%7D%7D%7B1000%7D%20-%20e%5E%7B-%20%2812.5992175853%29%7D%7D%20%3D%2012.5992175853%20" alt="LaTeX: x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853 " data-equation-content=" x_{5} = (12.5992175853) - \frac{- \frac{(12.5992175853)^{3}}{1000} + 2 + e^{- (12.5992175853)}}{- \frac{3 (12.5992175853)^{2}}{1000} - e^{- (12.5992175853)}} = 12.5992175853 " />
</p> </p>