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Use Newton's method to find the first 5 approximations of the solution to the equation \(\displaystyle e^{- x}= \frac{17 x^{3}}{20} - 5\) using \(\displaystyle x_0=1\).
Using the formula for Newton's method gives \begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - 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{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228\end{equation*} \begin{equation*}x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840\end{equation*} \begin{equation*}x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347\end{equation*} \begin{equation*}x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552\end{equation*} \begin{equation*}x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733\end{equation*}
\begin{question}Use Newton's method to find the first 5 approximations of the solution to the equation $e^{- x}= \frac{17 x^{3}}{20} - 5$ using $x_0=1$.
\soln{9cm}{Using the formula for Newton's method gives
\begin{equation*}x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - 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{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228\end{equation*}
\begin{equation*}x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840\end{equation*}
\begin{equation*}x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347\end{equation*}
\begin{equation*}x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552\end{equation*}
\begin{equation*}x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733\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{17 x^{3}}{20} - 5 " src="/equation_images/%20%5Cdisplaystyle%20e%5E%7B-%20x%7D%3D%20%5Cfrac%7B17%20x%5E%7B3%7D%7D%7B20%7D%20-%205%20" alt="LaTeX: \displaystyle e^{- x}= \frac{17 x^{3}}{20} - 5 " data-equation-content=" \displaystyle e^{- x}= \frac{17 x^{3}}{20} - 5 " /> 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{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - e^{- x_{n}}} " src="/equation_images/%20x_%7Bn%2B1%7D%20%3D%20%20x_%7Bn%7D%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20x_%7Bn%7D%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20x_%7Bn%7D%7D%7D%7B-%20%5Cfrac%7B51%20x_%7Bn%7D%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20x_%7Bn%7D%7D%7D%20%20%20" alt="LaTeX: x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - e^{- x_{n}}} " data-equation-content=" x_{n+1} = x_{n} - \frac{- \frac{17 x_{n}^{3}}{20} + 5 + e^{- x_{n}}}{- \frac{51 x_{n}^{2}}{20} - 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{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228 " src="/equation_images/%20x_%7B1%7D%20%3D%20%20%281.0000000000%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20%281.0000000000%29%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20%281.0000000000%29%7D%7D%7B-%20%5Cfrac%7B51%20%281.0000000000%29%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20%281.0000000000%29%7D%7D%20%3D%202.5483434228%20" alt="LaTeX: x_{1} = (1.0000000000) - \frac{- \frac{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228 " data-equation-content=" x_{1} = (1.0000000000) - \frac{- \frac{17 (1.0000000000)^{3}}{20} + 5 + e^{- (1.0000000000)}}{- \frac{51 (1.0000000000)^{2}}{20} - e^{- (1.0000000000)}} = 2.5483434228 " />
<img class="equation_image" title=" x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840 " src="/equation_images/%20x_%7B2%7D%20%3D%20%20%282.5483434228%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20%282.5483434228%29%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.5483434228%29%7D%7D%7B-%20%5Cfrac%7B51%20%282.5483434228%29%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20%282.5483434228%29%7D%7D%20%3D%202.0081053840%20" alt="LaTeX: x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840 " data-equation-content=" x_{2} = (2.5483434228) - \frac{- \frac{17 (2.5483434228)^{3}}{20} + 5 + e^{- (2.5483434228)}}{- \frac{51 (2.5483434228)^{2}}{20} - e^{- (2.5483434228)}} = 2.0081053840 " />
<img class="equation_image" title=" x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347 " src="/equation_images/%20x_%7B3%7D%20%3D%20%20%282.0081053840%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20%282.0081053840%29%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20%282.0081053840%29%7D%7D%7B-%20%5Cfrac%7B51%20%282.0081053840%29%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20%282.0081053840%29%7D%7D%20%3D%201.8402304347%20" alt="LaTeX: x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347 " data-equation-content=" x_{3} = (2.0081053840) - \frac{- \frac{17 (2.0081053840)^{3}}{20} + 5 + e^{- (2.0081053840)}}{- \frac{51 (2.0081053840)^{2}}{20} - e^{- (2.0081053840)}} = 1.8402304347 " />
<img class="equation_image" title=" x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552 " src="/equation_images/%20x_%7B4%7D%20%3D%20%20%281.8402304347%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20%281.8402304347%29%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20%281.8402304347%29%7D%7D%7B-%20%5Cfrac%7B51%20%281.8402304347%29%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20%281.8402304347%29%7D%7D%20%3D%201.8245056552%20" alt="LaTeX: x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552 " data-equation-content=" x_{4} = (1.8402304347) - \frac{- \frac{17 (1.8402304347)^{3}}{20} + 5 + e^{- (1.8402304347)}}{- \frac{51 (1.8402304347)^{2}}{20} - e^{- (1.8402304347)}} = 1.8245056552 " />
<img class="equation_image" title=" x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733 " src="/equation_images/%20x_%7B5%7D%20%3D%20%20%281.8245056552%29%20-%20%5Cfrac%7B-%20%5Cfrac%7B17%20%281.8245056552%29%5E%7B3%7D%7D%7B20%7D%20%2B%205%20%2B%20e%5E%7B-%20%281.8245056552%29%7D%7D%7B-%20%5Cfrac%7B51%20%281.8245056552%29%5E%7B2%7D%7D%7B20%7D%20-%20e%5E%7B-%20%281.8245056552%29%7D%7D%20%3D%201.8243741733%20" alt="LaTeX: x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733 " data-equation-content=" x_{5} = (1.8245056552) - \frac{- \frac{17 (1.8245056552)^{3}}{20} + 5 + e^{- (1.8245056552)}}{- \frac{51 (1.8245056552)^{2}}{20} - e^{- (1.8245056552)}} = 1.8243741733 " />
</p> </p>