Find the solution of this equation correct to $5$ decimal places using Newton's method

Question

We have the equation $x^3+2x-1=0$.Find the solution of this equation correct to $5$ decimal places using Newton's method.

Could anyone help me how to use the newton's method when we want to find the solution with correct to five decimal places?

Answer 1

Try an initial guess: $x_0=1$. Now the derivative of the equation is $3x^2+2$, so the iteration we use is $$x\leftarrow x-\frac{x^3+2x-1}{3x^2+2}$$ (x by itself is on the left, subtracted by function over derivative.) Iterating this we find $$x_1=0.6$$ $$x_2=0.464935\dots$$ $$x_3=0.453467\dots$$ $$x_4=0.453397\dots$$ $$x_5=0.453397\dots$$ and so we find the root of the equation to 5 decimal places as 0.45340.

Answer 2

Make an initial guess. My guess would be $x=0$, so that $f(x) = -1$. Note that $f'(x) = 3x^2+2$.

Now, by the Newton Raphson method, let us start our process:

$x_0=0$

$x_1=0 - \frac{f(0)}{f'(0)} = 0.5$.

$x_2=0.5- \frac{f(0.5)}{f'(0.5)} = 0.5-\frac{1}{22} = \frac{5}{11}=0.45454545$.

$x_2=0.45454545-\frac{f(0.45454545)}{f'(0.454545)} = 0.453398$.

$x_2=0.453398-\frac{f(0.453398)}{f'(0.453398)} = 0.453398$.

Hence my answer is coming to $0.453398$, which when rounded to five decimal places comes to $0.453340$.

Answer 3

The equation $x^3+2x=1$ has a unique real solution since the derivative of $f(x)=x^3+2x-1$ is always positive, hence $f(x)$ is an increasing function. Since $f\left(\frac{1}{2}\right)=\frac{1}{8}>0$ and $f$ is convex over $\mathbb{R}^+$, Newton's method with starting point $x_0=\frac{1}{2}$ converges quadratically to the root of $f$. Newton's iteration in this case is $$ x \mapsto \frac{1+2x^3}{2+3x^2} $$ so the sequence of approximations is given by $$ \frac{1}{2},\;\frac{5}{11},\; \frac{1581}{3487},\; \frac{50302634185}{110945952227} $$ and the last two terms yet differ by less than $7\cdot 10^{-7}$. It is not difficult to bound $f'$ on $\left(0,\frac{1}{2}\right)$ and deduce that the last term gives the wanted digits: $$ \frac{50302634185}{110945952227}=\color{red}{0.45339}76515\ldots $$ If we replace the starting point $x_0=\frac{1}{2}$ with the starting point $x_0=\frac{4}{9}$ suggested by the secant method, we achieve the same accuracy in just $\color{red}{\large 2}$ steps.

By the (hyperbolic) trigonometric form of the roots of a cubic polynomial, the exact root is given by $$ \sqrt{\frac{8}{3}}\,\sinh\left(\frac{1}{3}\text{arcsinh}\sqrt{\frac{27}{32}}\right).$$

Answer 4

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Yesterday, I posted this answer in another question of $\texttt{@Mantol Inis}$, who is the current OP. However, that question was related to the Bisection Method and I had to delete it. $\texttt{@Claude Leibovici}$ calls this question to my attention by remarking that my answer was in the wrong post. Anyway, I just moved it to the right place.

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Since $\ds{\verts{x^{3} + 2x - 1}_{\ x\ =\ 0}\,\,\,\,\, =\,\, 1}$ and $\ds{\expo{\ic\theta}}$, with $\ds{\theta \in \mathbb{R}}$, is not a root; there is a root $\ds{z \in \mathbb{C}}$ with $\ds{\verts{z} < 1}$. Anyway, we can check that $$ \left.\vphantom{\Large A}x^{3} + 2x - 1\right\vert_{\ x\ =\ -1} = -4 < 0 \qquad\mbox{and}\qquad \left.\vphantom{\Large A}x^{3} + 2x - 1\right\vert_{\ x\ =\ 1} = 2 > 0 $$ You can start with the guess $\ds{x = 0}$. $\texttt{Newton-Rapson}$ finds a real root with $\ds{5}$ decimal places in $\ds{\large\color{#f00}{3\ !!!}}$ steps.

This is a 'short $\texttt{javascript}$ code':

/* javascript example */
"use strict";
var maxIter = 4; // Iterations
var       n = 0; // Iteration index
var       x = 0; // Guess
var      x2 = null;

while (n < maxIter) {
      document.write(x + "\n");
      x2 = x*x;
      x -= (x*(x2 + 2.0) - 1.0)/(3.0*x2 + 2.0); // Newton-Rapson
      ++n;
}

document.write(x + "\n");

The result is given by: \begin{align} &0 \\ &0.5 \\ &0.454545 \\ \imp\quad &\color{#f00}{0.45339}83366790937769 \\ &\color{#f00}{0.45339}76515166477918 \end{align}

and $\ds{\quad\left.\vphantom{\Large A}x^{3} + 2x - 1\right\vert_{\ x\ =\ 0.45339}\,\,\,\,\,\, =\,\,\,\,\,\, -2.00217 \times 10^{-5}}$.

We can modify the code to check the 'tolerance' ( with some definition ) in each iteration. The present example used ( to be brief ) the old 'by inspection method".