Calculating Newton's Method: Step-by-Step Guide

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SUMMARY

The discussion focuses on applying Newton's Method to calculate the second iterate for the function defined piecewise as $$f(x) = \begin{cases} \dfrac{x^3-1}{\sqrt{x}-1}, & x > 1\\ \cos(x-1)-x^2, & x \leq 1\end{cases}$$. The initial guess is set at $$x_0 = 1$$, and the formula for Newton's Method is correctly stated as $$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$. The first iteration was calculated as $$x_{1}=x_{0}-\frac{\cos(0)-1}{-\sin(0)-2}$$, and the next step involves evaluating $$\cos(0)$$ and $$\sin(0)$$ to proceed with the calculations.

PREREQUISITES
  • Understanding of Newton's Method for root-finding
  • Basic calculus concepts including derivatives
  • Familiarity with piecewise functions
  • Knowledge of trigonometric functions and their values
NEXT STEPS
  • Calculate the derivative $$f'(x)$$ for both cases of the piecewise function
  • Perform the evaluation of $$\cos(0)$$ and $$\sin(0)$$ to finalize the first iteration
  • Compute the second iterate $$x_2$$ using the result from $$x_1$$
  • Explore convergence criteria for Newton's Method
USEFUL FOR

Students studying calculus, mathematicians interested in numerical methods, and anyone looking to understand iterative methods for finding roots of functions.

jisbon
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Homework Statement
Let $$f(x) =
\begin{cases} \dfrac{x^3-1}{\sqrt{x}-1}, & x > 1\\
cos(x-1)-x^2, & x \leq 1\end{cases}$$

Use Newton's method with ##x_{0} =1##, compute the second iterate to approximate value ##c## where ##c## is a stationary value that lies in the x-axis for some ##0<c<1##
Relevant Equations
-
Since the Newton's method is as follows:

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$

$$x_{1}=x_{0}-\frac{cos(0)-1}{-sin(0)-2}$$

Is this correct? What should I proceed on from here?
 
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jisbon said:
Homework Statement: Let $$f(x) =
\begin{cases} \dfrac{x^3-1}{\sqrt{x}-1}, & x > 1\\
cos(x-1)-x^2, & x \leq 1\end{cases}$$

Use Newton's method with ##x_{0} =1##, compute the second iterate to approximate value ##c## where ##c## is a stationary value that lies in the x-axis for some ##0<c<1##
Homework Equations: -

Since the Newton's method is as follows:

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$

$$x_{1}=x_{0}-\frac{cos(0)-1}{-sin(0)-2}$$

Is this correct?
Yes, as far as you went
jisbon said:
What should I proceed on from here?
Evaluate cos(0) and sin(0) and substitute in the value for ##x_0##.
 

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