# Abhishek's Questions About Solving Nonlinear Equations

#### Prove It

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MHB Math Helper

8. Newton's Method for solving a nonlinear equation \displaystyle \begin{align*} f(x) = 0 \end{align*} is \displaystyle \begin{align*} x_{n+1} = x_n - \frac{f\left( x_n \right)}{f'\left( x_n \right) } \end{align*}.

Here \displaystyle \begin{align*} f(x) = 0.7\,x^4 + x - 4.75 = 0 \end{align*}, so \displaystyle \begin{align*} f'(x) = 2.8\,x^3 + 1 \end{align*}. That means

\displaystyle \begin{align*} x_3 &= x_2 - \frac{f(x_2)}{f'(x_2)} \\ &= 1.471\,147\,220\,2 - \left[ \frac{0.7 \,\left( 1.471\,147\,220\,2 \right) ^4 + 1.471\,147\,220\,2 - 4.75}{2.8\,\left( 1.471\,147\,220\,2 \right) ^3 + 1 } \right] \\ &= 1.471\,146\,714\,1 \end{align*}

The error in any estimate \displaystyle \begin{align*} x_n \end{align*} is never any greater than \displaystyle \begin{align*} \left| x_{n+1} - x_n \right| \end{align*}, so that means that the error in \displaystyle \begin{align*} x_2\end{align*} can be approximated as

\displaystyle \begin{align*} e_2 &\leq \left| 1.471\,146\,714\,1 - 1.471\,147\,220\,2 \right| \\ &= 0.000\,000\,506\,1 \\ &= 5.061 \cdot 10^{-7} \end{align*}

9. \displaystyle \begin{align*} f(x) = 50\,\mathrm{e}^{-\frac{x}{5}} - x^4 = 0 \end{align*}, so \displaystyle \begin{align*} f'(x) = -10\,\mathrm{e}^{-\frac{x}{5}} - 4\,x^3 \end{align*}. Since our starting guess is \displaystyle \begin{align*} x_0 = 0 \end{align*} that means

\displaystyle \begin{align*} x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)} \\ &= 0 - \left[ \frac{50\,\mathrm{e}^{-\frac{0}{5}} - 0^4}{-10\,\mathrm{e}^{-\frac{x}{5}} - 4\,\left( 0 \right) ^3} \right] \\ &= - \left( \frac{50 - 0}{-10 - 0} \right) \\ &= 5 \\ \\ x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)} \\ &= 5 - \left[ \frac{50\,\mathrm{e}^{-\frac{5}{5}} - 5^4}{-10\,\mathrm{e}^{-\frac{5}{5}} - 4\,\left( 5 \right) ^3} \right] \\ &= 5 - \left( \frac{50\,\mathrm{e}^{-1} - 625}{-10\,\mathrm{e}^{-1} - 500} \right) \\ &= 3.795\,649\,063\,1 \\ \\ x_3 &= x_2 - \frac{f(x_2)}{f'(x_2)} \\ &= 3.795\,649\,063\,1 - \left[ \frac{50\,\mathrm{e}^{-\frac{3.795\,649\,063\,1}{5}} - 3.795\,649\,063\,1^4}{-10\,\mathrm{e}^{-\frac{3.795\,649\,063\,1}{5}} - 4\,\left( 3.795\,649\,063\,1 \right) ^3 } \right] \\ &= 2.971\,371\,232\,0 \end{align*}

10. \displaystyle \begin{align*} y' = \frac{4\,y^2 + 6}{x+1} , \, y\left( 0 \right) = \frac{1}{4} \end{align*}. Since we are told \displaystyle \begin{align*} h = \frac{1}{20} \end{align*} and we are trying to move from \displaystyle \begin{align*} x = 0 \end{align*} to \displaystyle \begin{align*} x = 0.1 \end{align*}, we need two iterations. Euler's method is \displaystyle \begin{align*} y_{n + 1} = y_n + h\,y'_n \end{align*}.

Since our initial condition is \displaystyle \begin{align*} y(0) = \frac{1}{4} \end{align*}, that means \displaystyle \begin{align*} x_0 = 0, \, y_0 = \frac{1}{4} \end{align*} and

\displaystyle \begin{align*} y'_0 &= \frac{4\,y_0^2 + 6}{x_0 + 1} \\ &= \frac{4\left( \frac{1}{4} \right) ^2 + 6}{0 + 1} \\ &= \frac{4 \left( \frac{1}{16} \right) + 6}{1} \\ &= \frac{25}{4} \end{align*}

thus

\displaystyle \begin{align*} y_1 &= y_0 + h\,y'_0 \\ &= \frac{1}{4} + \frac{1}{20}\,\left( \frac{25}{4} \right) \\ &= \frac{1}{4} + \frac{5}{16} \\ &= \frac{9}{16} \end{align*}

so moving on to the next iteration with \displaystyle \begin{align*} x_1 = 0 + \frac{1}{20} = \frac{1}{20}, \, y_1 = \frac{9}{16} \end{align*} and

\displaystyle \begin{align*} y'_1 &= \frac{4\,y_1^2 + 6}{x_1 + 1} \\ &= \frac{4\,\left( \frac{9}{16} \right) ^2 + 6}{\frac{1}{20} + 1} \\ &= \frac{4\,\left( \frac{81}{256} \right) + 6}{\frac{21}{20}} \\ &= \frac{20}{21} \,\left( \frac{81}{64} + \frac{384}{64} \right) \\ &= \frac{20}{21} \,\left( \frac{465}{64} \right) \\ &= \frac{5}{7}\,\left( \frac{155}{16} \right) \\ &= \frac{775}{112} \end{align*}

we have

\displaystyle \begin{align*} y_2 &= y_1 + h\,y'_1 \\ &= \frac{9}{16} + \frac{1}{20}\,\left( \frac{775}{112} \right) \\ &= \frac{9}{16} + \frac{1}{4}\,\left( \frac{155}{112} \right) \\ &= \frac{252}{448} + \frac{155}{448} \\ &= \frac{407}{448} \end{align*}