Using Newton's method, find the root of $x^3+2x-1$ with an accuracy of $5$ decimal places.
I think the solution is $0.45339$. Am I wrong?
[Math] Using Newton’s method, find the root of $x^3+2x-1$ with an accuracy of $5$ decimal places
newton raphsonnumerical methods
Related Solutions
Since $f$ is convex, you can show that if $x_0>1$ then $x_n \geq 1$ for all $n$. So we can bound the term $\frac{1}{x_n} \leq 1$. The Newton update for $f$ is $x_{n+1} = \frac{1}{2} (x_n + \frac{1}{x_n})$. So we have the bound $x_{n+1} \leq \frac{1}{2} x_n + \frac{1}{2}$. Working through the details gives $x_n \leq \frac{1}{2^n} x_0 + \frac{1}{2^n} + \cdots + \frac{1}{2} = \frac{1}{2^n} (x_0-1) + 1$. To estimate the number of iterations to get an error of less than 1. we want to find the smallest $n$ such that $2>\frac{1}{2^n} (x_0-1) + 1$, or equivalently, $n > \log_2 (x_0-1) \approx 33.2$. Hence, using this estimate, it will take (assuming I have made no mistakes) 34 iterations to get within an error of 1.
This is an elaboration on the bound above: Since $x_{n+1} \leq \frac{1}{2} x_n + \frac{1}{2}$, we have $x_{1} \leq \frac{1}{2} x_0 + \frac{1}{2}$. For the induction step, suppose $x_n \leq \frac{1}{2^n} x_0 + \frac{1}{2^n} + \cdots + \frac{1}{2}$. Then we have $x_{n+1} \leq \frac{1}{2} x_n + \frac{1}{2} \leq \frac{1}{2} (\frac{1}{2^n} x_0 + \frac{1}{2^n} + \cdots + \frac{1}{2}) + \frac{1}{2} = \frac{1}{2^{n+1}} x_0 + \frac{1}{2^{n+1}} + \cdots + \frac{1}{2}$, hence the formula is true for all $n$.
Summing the geometric series, we have $\frac{1}{2^n} + \cdots + \frac{1}{2} = \frac{1}{2}(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}) = 1-\frac{1}{2^n}$, which gives $x_n \leq \frac{1}{2^n} x_0 + \frac{1}{2^n} + \cdots + \frac{1}{2} = \frac{1}{2^n} x_0 + 1-\frac{1}{2^n} = \frac{1}{2^n} (x_0-1) + 1$.
Some hints.
- Newton's method is a root-finding algorithm (i.e. given $f(x)$ it finds $x^{*}$ such that $f(x^{*}) = 0$) so you need to find a function $f(x)$ which has a root at the same point that $e^{-x} = \sin x$.
- Newton's method will find a root close to your initial guess. Where do you think the smallest root may lie?
Best Answer
Does $x^3+2x-1$ have different signs for $x=0.453385$ and $x=0.453395$? If so, that's s root with 5 digits accuracy.
But actually, according to my calculator we have
$$\begin{array}{c|c} x & x^3+2x-1 \\ \hline 0.453385 & \approx -3\times 10^{-5} \\ 0.453395 & \approx -7\times 10^{-6} \\ 0.453400 & \approx +6\times 10^{-6} \end{array} $$ so the root is between $0.453395$ and $0.453400$, and should therefore be given with 5 significant digits as $0.45340$.
Since the exercise specifically asks you to use Newton's method, however, most of what you will be judged on is probably setting up the right expression to iterate, showing your successive approximations, and demonstrating when to stop iterating.