Difference between revisions of "OSH/Example/3"
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Estimate $\sqrt{15}$ to 4 decimal places using Newton's method. | Estimate $\sqrt{15}$ to 4 decimal places using Newton's method. | ||
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====Solution==== | ====Solution==== | ||
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| + | In order to use Newton's method, we need to find an appropriate associated polynomial that has a zero at $\sqrt{15}$. We'll use the function | ||
| + | :\(f(x)= x^2-15.\) | ||
| + | |||
| + | For Newton's method, we also need the derivative of this function: | ||
| + | :\(f'(x)= 2x.\) | ||
| + | |||
| + | The last thing we need before starting with Newton's method is an initial guess. The nearest perfect square to 15 is 16 so $x=4$ is a good candidate for an initial guess. Graphing $f(x)$, it is clear that $x=4$ is positioned well so that Newton's method should converge to $\sqrt{15}$. | ||
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| + | The Newton's method iterates, calculated by ________ (hand / calculator / spreadsheet) are | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | x_{1} &= 3.875\\ | ||
| + | x_{2} &= 3.872983871\\ | ||
| + | x_{3} &= 3.872983346 | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | As the first 5 decimal points are the same, our estimate $x_3$ should be correct to 4 decimal places. | ||
Revision as of 12:26, 19 September 2014
Example problem statement
Estimate $\sqrt{15}$ to 4 decimal places using Newton's method.
Solution
In order to use Newton's method, we need to find an appropriate associated polynomial that has a zero at $\sqrt{15}$. We'll use the function
- \(f(x)= x^2-15.\)
For Newton's method, we also need the derivative of this function:
- \(f'(x)= 2x.\)
The last thing we need before starting with Newton's method is an initial guess. The nearest perfect square to 15 is 16 so $x=4$ is a good candidate for an initial guess. Graphing $f(x)$, it is clear that $x=4$ is positioned well so that Newton's method should converge to $\sqrt{15}$.
The Newton's method iterates, calculated by ________ (hand / calculator / spreadsheet) are
\( \begin{align} x_{1} &= 3.875\\ x_{2} &= 3.872983871\\ x_{3} &= 3.872983346 \end{align} \)
As the first 5 decimal points are the same, our estimate $x_3$ should be correct to 4 decimal places.
