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Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{117 x^{3}}{200} - 2\) using \(\displaystyle x_0=1\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - 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{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749\end{equation*} \begin{equation*}x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443\end{equation*} \begin{equation*}x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630\end{equation*} \begin{equation*}x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195\end{equation*} \begin{equation*}x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{117 x^{3}}{200} - 2$ using $x_0=1$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - 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{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749\end{equation*}
\begin{equation*}x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443\end{equation*}
\begin{equation*}x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630\end{equation*}
\begin{equation*}x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195\end{equation*}
\begin{equation*}x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797\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{117 x^{3}}{200} - 2 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B117%20x%5E%7B3%7D%7D%7B200%7D%20-%202%20" alt="LaTeX: \displaystyle e^{- x}= \frac{117 x^{3}}{200} - 2 " data-equation-content=" \displaystyle e^{- x}= \frac{117 x^{3}}{200} - 2 " /> 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{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - e^{- x_{n}}} " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20x_%7Bn%7D%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B351%20x_%7Bn%7D%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="LaTeX: x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{117 x_{n}^{3}}{200} + 2 + e^{- x_{n}}}{- \frac{351 x_{n}^{2}}{200} - 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{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%281.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20%281.0000000000%29%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20%281.0000000000%29%7D%7D%7B-%20%5Cfrac%7B351%20%281.0000000000%29%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20%281.0000000000%29%7D%7D%20%3D%201.8398401749%20" alt="LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749 " data-equation-content=" x_{1} = (1.0000000000) - \frac{- \frac{117 (1.0000000000)^{3}}{200} + 2 + e^{- (1.0000000000)}}{- \frac{351 (1.0000000000)^{2}}{200} - e^{- (1.0000000000)}} = 1.8398401749 " />
<img class="equation_image" title=" x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%281.8398401749%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20%281.8398401749%29%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20%281.8398401749%29%7D%7D%7B-%20%5Cfrac%7B351%20%281.8398401749%29%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20%281.8398401749%29%7D%7D%20%3D%201.5964664443%20" alt="LaTeX: x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443 " data-equation-content=" x_{2} = (1.8398401749) - \frac{- \frac{117 (1.8398401749)^{3}}{200} + 2 + e^{- (1.8398401749)}}{- \frac{351 (1.8398401749)^{2}}{200} - e^{- (1.8398401749)}} = 1.5964664443 " />
<img class="equation_image" title=" x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%281.5964664443%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20%281.5964664443%29%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20%281.5964664443%29%7D%7D%7B-%20%5Cfrac%7B351%20%281.5964664443%29%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20%281.5964664443%29%7D%7D%20%3D%201.5584587630%20" alt="LaTeX: x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630 " data-equation-content=" x_{3} = (1.5964664443) - \frac{- \frac{117 (1.5964664443)^{3}}{200} + 2 + e^{- (1.5964664443)}}{- \frac{351 (1.5964664443)^{2}}{200} - e^{- (1.5964664443)}} = 1.5584587630 " />
<img class="equation_image" title=" x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%281.5584587630%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20%281.5584587630%29%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20%281.5584587630%29%7D%7D%7B-%20%5Cfrac%7B351%20%281.5584587630%29%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20%281.5584587630%29%7D%7D%20%3D%201.5575942195%20" alt="LaTeX: x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195 " data-equation-content=" x_{4} = (1.5584587630) - \frac{- \frac{117 (1.5584587630)^{3}}{200} + 2 + e^{- (1.5584587630)}}{- \frac{351 (1.5584587630)^{2}}{200} - e^{- (1.5584587630)}} = 1.5575942195 " />
<img class="equation_image" title=" x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%281.5575942195%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B117%20%281.5575942195%29%5E%7B3%7D%7D%7B200%7D%20%2B%202%20%2B%20e%5E%7B-%20%281.5575942195%29%7D%7D%7B-%20%5Cfrac%7B351%20%281.5575942195%29%5E%7B2%7D%7D%7B200%7D%20-%20e%5E%7B-%20%281.5575942195%29%7D%7D%20%3D%201.5575937797%20" alt="LaTeX: x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797 " data-equation-content=" x_{5} = (1.5575942195) - \frac{- \frac{117 (1.5575942195)^{3}}{200} + 2 + e^{- (1.5575942195)}}{- \frac{351 (1.5575942195)^{2}}{200} - e^{- (1.5575942195)}} = 1.5575937797 " />
</p> </p>