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Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{33 x^{3}}{250} - 3\) using \(\displaystyle x_0=3\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{33 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{99 x_{n}^{2}}{250} - 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{33 (3.0000000000)^{3}}{250} + 3 + e^{- (3.0000000000)}}{- \frac{99 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8577080160\end{equation*} \begin{equation*}x_{2} = (2.8577080160) - \frac{- \frac{33 (2.8577080160)^{3}}{250} + 3 + e^{- (2.8577080160)}}{- \frac{99 (2.8577080160)^{2}}{250} - e^{- (2.8577080160)}} = 2.8506761050\end{equation*} \begin{equation*}x_{3} = (2.8506761050) - \frac{- \frac{33 (2.8506761050)^{3}}{250} + 3 + e^{- (2.8506761050)}}{- \frac{99 (2.8506761050)^{2}}{250} - e^{- (2.8506761050)}} = 2.8506594714\end{equation*} \begin{equation*}x_{4} = (2.8506594714) - \frac{- \frac{33 (2.8506594714)^{3}}{250} + 3 + e^{- (2.8506594714)}}{- \frac{99 (2.8506594714)^{2}}{250} - e^{- (2.8506594714)}} = 2.8506594713\end{equation*} \begin{equation*}x_{5} = (2.8506594713) - \frac{- \frac{33 (2.8506594713)^{3}}{250} + 3 + e^{- (2.8506594713)}}{- \frac{99 (2.8506594713)^{2}}{250} - e^{- (2.8506594713)}} = 2.8506594713\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{33 x^{3}}{250} - 3$ using $x_0=3$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{33 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{99 x_{n}^{2}}{250} - 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{33 (3.0000000000)^{3}}{250} + 3 + e^{- (3.0000000000)}}{- \frac{99 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8577080160\end{equation*}
\begin{equation*}x_{2} = (2.8577080160) - \frac{- \frac{33 (2.8577080160)^{3}}{250} + 3 + e^{- (2.8577080160)}}{- \frac{99 (2.8577080160)^{2}}{250} - e^{- (2.8577080160)}} = 2.8506761050\end{equation*}
\begin{equation*}x_{3} = (2.8506761050) - \frac{- \frac{33 (2.8506761050)^{3}}{250} + 3 + e^{- (2.8506761050)}}{- \frac{99 (2.8506761050)^{2}}{250} - e^{- (2.8506761050)}} = 2.8506594714\end{equation*}
\begin{equation*}x_{4} = (2.8506594714) - \frac{- \frac{33 (2.8506594714)^{3}}{250} + 3 + e^{- (2.8506594714)}}{- \frac{99 (2.8506594714)^{2}}{250} - e^{- (2.8506594714)}} = 2.8506594713\end{equation*}
\begin{equation*}x_{5} = (2.8506594713) - \frac{- \frac{33 (2.8506594713)^{3}}{250} + 3 + e^{- (2.8506594713)}}{- \frac{99 (2.8506594713)^{2}}{250} - e^{- (2.8506594713)}} = 2.8506594713\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{33 x^{3}}{250} - 3 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B33%20x%5E%7B3%7D%7D%7B250%7D%20-%203%20" alt="LaTeX: \displaystyle e^{- x}= \frac{33 x^{3}}{250} - 3 " data-equation-content=" \displaystyle e^{- x}= \frac{33 x^{3}}{250} - 3 " /> 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{33 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{99 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%7B33%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%7B99%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{33 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{99 x_{n}^{2}}{250} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{33 x_{n}^{3}}{250} + 3 + e^{- x_{n}}}{- \frac{99 x_{n}^{2}}{250} - 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{33 (3.0000000000)^{3}}{250} + 3 + e^{- (3.0000000000)}}{- \frac{99 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8577080160 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%283.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B33%20%283.0000000000%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%283.0000000000%29%7D%7D%7B-%20%5Cfrac%7B99%20%283.0000000000%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%283.0000000000%29%7D%7D%20%3D%202.8577080160%20" alt="LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{33 (3.0000000000)^{3}}{250} + 3 + e^{- (3.0000000000)}}{- \frac{99 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8577080160 " data-equation-content=" x_{1} = (3.0000000000) - \frac{- \frac{33 (3.0000000000)^{3}}{250} + 3 + e^{- (3.0000000000)}}{- \frac{99 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.8577080160 " />
<img class="equation_image" title=" x_{2} = (2.8577080160) - \frac{- \frac{33 (2.8577080160)^{3}}{250} + 3 + e^{- (2.8577080160)}}{- \frac{99 (2.8577080160)^{2}}{250} - e^{- (2.8577080160)}} = 2.8506761050 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.8577080160%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B33%20%282.8577080160%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.8577080160%29%7D%7D%7B-%20%5Cfrac%7B99%20%282.8577080160%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.8577080160%29%7D%7D%20%3D%202.8506761050%20" alt="LaTeX: x_{2} = (2.8577080160) - \frac{- \frac{33 (2.8577080160)^{3}}{250} + 3 + e^{- (2.8577080160)}}{- \frac{99 (2.8577080160)^{2}}{250} - e^{- (2.8577080160)}} = 2.8506761050 " data-equation-content=" x_{2} = (2.8577080160) - \frac{- \frac{33 (2.8577080160)^{3}}{250} + 3 + e^{- (2.8577080160)}}{- \frac{99 (2.8577080160)^{2}}{250} - e^{- (2.8577080160)}} = 2.8506761050 " />
<img class="equation_image" title=" x_{3} = (2.8506761050) - \frac{- \frac{33 (2.8506761050)^{3}}{250} + 3 + e^{- (2.8506761050)}}{- \frac{99 (2.8506761050)^{2}}{250} - e^{- (2.8506761050)}} = 2.8506594714 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.8506761050%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B33%20%282.8506761050%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.8506761050%29%7D%7D%7B-%20%5Cfrac%7B99%20%282.8506761050%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.8506761050%29%7D%7D%20%3D%202.8506594714%20" alt="LaTeX: x_{3} = (2.8506761050) - \frac{- \frac{33 (2.8506761050)^{3}}{250} + 3 + e^{- (2.8506761050)}}{- \frac{99 (2.8506761050)^{2}}{250} - e^{- (2.8506761050)}} = 2.8506594714 " data-equation-content=" x_{3} = (2.8506761050) - \frac{- \frac{33 (2.8506761050)^{3}}{250} + 3 + e^{- (2.8506761050)}}{- \frac{99 (2.8506761050)^{2}}{250} - e^{- (2.8506761050)}} = 2.8506594714 " />
<img class="equation_image" title=" x_{4} = (2.8506594714) - \frac{- \frac{33 (2.8506594714)^{3}}{250} + 3 + e^{- (2.8506594714)}}{- \frac{99 (2.8506594714)^{2}}{250} - e^{- (2.8506594714)}} = 2.8506594713 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%282.8506594714%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B33%20%282.8506594714%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.8506594714%29%7D%7D%7B-%20%5Cfrac%7B99%20%282.8506594714%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.8506594714%29%7D%7D%20%3D%202.8506594713%20" alt="LaTeX: x_{4} = (2.8506594714) - \frac{- \frac{33 (2.8506594714)^{3}}{250} + 3 + e^{- (2.8506594714)}}{- \frac{99 (2.8506594714)^{2}}{250} - e^{- (2.8506594714)}} = 2.8506594713 " data-equation-content=" x_{4} = (2.8506594714) - \frac{- \frac{33 (2.8506594714)^{3}}{250} + 3 + e^{- (2.8506594714)}}{- \frac{99 (2.8506594714)^{2}}{250} - e^{- (2.8506594714)}} = 2.8506594713 " />
<img class="equation_image" title=" x_{5} = (2.8506594713) - \frac{- \frac{33 (2.8506594713)^{3}}{250} + 3 + e^{- (2.8506594713)}}{- \frac{99 (2.8506594713)^{2}}{250} - e^{- (2.8506594713)}} = 2.8506594713 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%282.8506594713%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B33%20%282.8506594713%29%5E%7B3%7D%7D%7B250%7D%20%2B%203%20%2B%20e%5E%7B-%20%282.8506594713%29%7D%7D%7B-%20%5Cfrac%7B99%20%282.8506594713%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.8506594713%29%7D%7D%20%3D%202.8506594713%20" alt="LaTeX: x_{5} = (2.8506594713) - \frac{- \frac{33 (2.8506594713)^{3}}{250} + 3 + e^{- (2.8506594713)}}{- \frac{99 (2.8506594713)^{2}}{250} - e^{- (2.8506594713)}} = 2.8506594713 " data-equation-content=" x_{5} = (2.8506594713) - \frac{- \frac{33 (2.8506594713)^{3}}{250} + 3 + e^{- (2.8506594713)}}{- \frac{99 (2.8506594713)^{2}}{250} - e^{- (2.8506594713)}} = 2.8506594713 " />
</p> </p>