The Newton-Raphson Iteration

Calculus Level 3

Suppose you need the cube root of 3.79, but your calculator only does the four basic operations. You can use the Newton-Raphson method to calculate the root to any amount of accuracy needed.

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

Find f(x).

$x_{0} = 1.8$

$x_{1} = 1.589917695$

$x_{2} = 1.559713388$

$x_{3} = 1.559120922$

$x_{4} = 1.559120697$

x^{3} - 3.79

3^{x - 3.79}

\sqrt[3]{x - 3.79}

\sqrt[3]{x}

1 solution

Brandon Stocks
May 13, 2016

Let $x_{\omega}$ represent the exact root.

$x_{\omega} = \sqrt[3]{3.79}, \; \; \; x_{\omega}^{3} = 3.79, \; \; \; x_{\omega}^{3} - 3.79 = 0$

$f(x_{\omega}) = 0 = x_{\omega}^{3} - 3.79$ Thus

$f(x) = x^{3} - 3.79$

You can test a few guesses to get a number that is in the neighborhood of $\sqrt[3]{3.79}$ . In the example calculations listed in the question, I chose my first test value to be $x_{0} = 1.8$ .

The Newton-Raphson Iteration

1 solution

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