Need help with this problem thanks! **Problems from Section 3.8**6. Use Newton’s method to approximate $ \sqrt[3]{20} $ to 6 decimal places. You will probably want to use $ f(x) = x^3 - 20 $. Make a sensible choice for $ x_1 $. You will probably need a calculator for this problem.

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Description: Need help with this problem thanks! **Problems from Section 3.8**6. Use Newton’s method to approximate $ \sqrt[3]{20} $ to 6 decimal places. You will probably want to use $ f(x) = x^3 - 20 $. Make a sensible choice for $ x_1 $. You will probabl

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Need help with this problem thanks!

**Problems from Section 3.8**

6. Use Newton’s method to approximate $ \sqrt[3]{20} $ to 6 decimal places. You will probably want to use $ f(x) = x^3 - 20 $. Make a sensible choice for $ x_1 $. You will probably need a calculator for this problem.

Expert Solution

Step 1

In Newton’s method, the root of a polynomial is approximated by iteratively applying the following method:

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

We will iterate until the approximate percent relative error falls below 0.000001% (as we need to approximate up to 6 decimal places). The formula to calculate the percent relative error is

$ε_{a} = |\frac{x_{n + 1} - x_{n}}{x_{n + 1}}| \times 100 % or equivalently ε_{a} = |\frac{x^{n e w} - x^{n e w}}{x^{n e w}}| \times 100 %$

Given:

$\begin{array}{l} f (x) = x^{3} - 20 \\ \Rightarrow f^{'} (x) = 3 x^{2} \end{array}$

So, the formula is

$x_{n + 1} = x_{n} - \frac{x_{n}^{3} - 20}{3 x_{n}^{2}}$

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