Abhishek's Questions About Solving Nonlinear Equations

[Prove It]

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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

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

 

[jvicens]

So our final iteration gives us $\displaystyle \begin{align*} y(0.1) \approx \frac{407}{448} \approx 0.908\,482\,142\,9 \end{align*}$.

Thank you for sharing your examples of using Newton's Method and Euler's Method for solving nonlinear equations and differential equations, respectively. It is always interesting to see these methods being applied in different scenarios.

I would like to point out that while Newton's Method is a powerful tool for finding roots of nonlinear equations, it may not always converge to the desired solution. This can happen if the initial guess is too far from the actual root, or if there are multiple roots. In these cases, it is important to check the convergence of the method and possibly adjust the initial guess.

Similarly, Euler's Method is a simple and intuitive way to approximate solutions to differential equations, but it may not always give accurate results. Other numerical methods, such as Runge-Kutta methods, can provide better approximations with smaller step sizes.

Overall, these methods are valuable tools in the scientific community and it is important to understand their limitations and use them appropriately.

Thank you for bringing up this discussion and I look forward to seeing more examples from other forum members.