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Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{123 x^{3}}{250} - 4\) using \(\displaystyle x_0=3\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 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{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974\end{equation*} \begin{equation*}x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693\end{equation*} \begin{equation*}x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158\end{equation*} \begin{equation*}x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083\end{equation*} \begin{equation*}x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{123 x^{3}}{250} - 4$ using $x_0=3$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 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{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974\end{equation*}
\begin{equation*}x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693\end{equation*}
\begin{equation*}x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158\end{equation*}
\begin{equation*}x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083\end{equation*}
\begin{equation*}x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204\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{123 x^{3}}{250} - 4 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B123%20x%5E%7B3%7D%7D%7B250%7D%20-%204%20" alt="LaTeX: \displaystyle e^{- x}= \frac{123 x^{3}}{250} - 4 " data-equation-content=" \displaystyle e^{- x}= \frac{123 x^{3}}{250} - 4 " /> 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{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 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%7B123%20x_%7Bn%7D%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B369%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{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 x_{n}^{2}}{250} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{123 x_{n}^{3}}{250} + 4 + e^{- x_{n}}}{- \frac{369 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{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%283.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B123%20%283.0000000000%29%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20%283.0000000000%29%7D%7D%7B-%20%5Cfrac%7B369%20%283.0000000000%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%283.0000000000%29%7D%7D%20%3D%202.3074575974%20" alt="LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974 " data-equation-content=" x_{1} = (3.0000000000) - \frac{- \frac{123 (3.0000000000)^{3}}{250} + 4 + e^{- (3.0000000000)}}{- \frac{369 (3.0000000000)^{2}}{250} - e^{- (3.0000000000)}} = 2.3074575974 " />
<img class="equation_image" title=" x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.3074575974%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B123%20%282.3074575974%29%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20%282.3074575974%29%7D%7D%7B-%20%5Cfrac%7B369%20%282.3074575974%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.3074575974%29%7D%7D%20%3D%202.0630491693%20" alt="LaTeX: x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693 " data-equation-content=" x_{2} = (2.3074575974) - \frac{- \frac{123 (2.3074575974)^{3}}{250} + 4 + e^{- (2.3074575974)}}{- \frac{369 (2.3074575974)^{2}}{250} - e^{- (2.3074575974)}} = 2.0630491693 " />
<img class="equation_image" title=" x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.0630491693%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B123%20%282.0630491693%29%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20%282.0630491693%29%7D%7D%7B-%20%5Cfrac%7B369%20%282.0630491693%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.0630491693%29%7D%7D%20%3D%202.0329307158%20" alt="LaTeX: x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158 " data-equation-content=" x_{3} = (2.0630491693) - \frac{- \frac{123 (2.0630491693)^{3}}{250} + 4 + e^{- (2.0630491693)}}{- \frac{369 (2.0630491693)^{2}}{250} - e^{- (2.0630491693)}} = 2.0329307158 " />
<img class="equation_image" title=" x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%282.0329307158%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B123%20%282.0329307158%29%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20%282.0329307158%29%7D%7D%7B-%20%5Cfrac%7B369%20%282.0329307158%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.0329307158%29%7D%7D%20%3D%202.0324989083%20" alt="LaTeX: x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083 " data-equation-content=" x_{4} = (2.0329307158) - \frac{- \frac{123 (2.0329307158)^{3}}{250} + 4 + e^{- (2.0329307158)}}{- \frac{369 (2.0329307158)^{2}}{250} - e^{- (2.0329307158)}} = 2.0324989083 " />
<img class="equation_image" title=" x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%282.0324989083%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B123%20%282.0324989083%29%5E%7B3%7D%7D%7B250%7D%20%2B%204%20%2B%20e%5E%7B-%20%282.0324989083%29%7D%7D%7B-%20%5Cfrac%7B369%20%282.0324989083%29%5E%7B2%7D%7D%7B250%7D%20-%20e%5E%7B-%20%282.0324989083%29%7D%7D%20%3D%202.0324988204%20" alt="LaTeX: x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204 " data-equation-content=" x_{5} = (2.0324989083) - \frac{- \frac{123 (2.0324989083)^{3}}{250} + 4 + e^{- (2.0324989083)}}{- \frac{369 (2.0324989083)^{2}}{250} - e^{- (2.0324989083)}} = 2.0324988204 " />
</p> </p>