Difference between revisions of "OSH/Example/3"
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====Example problem statement==== | ====Example problem statement==== | ||
− | Estimate $\sqrt{15}$ to | + | Estimate $\sqrt{15}$ to 8 decimal places using Newton's method. |
− | + | ||
====Solution==== | ====Solution==== | ||
+ | |||
+ | In order to use Newton's method, I needed a polynomial that has a zero at $\sqrt{15}$. I used the function | ||
+ | :\(f(x)= x^2-15.\) | ||
+ | |||
+ | For Newton's method, I also needed the derivative of this function: | ||
+ | :\(f'(x)= 2x.\) | ||
+ | |||
+ | The last thing I needed before starting with Newton's method was an initial guess. The nearest perfect square to 15 is 16 so $x=4$ was a good candidate for an initial guess. Graphing $f(x)$, it was clear that $x=4$ is positioned well so that Newton's method should converge to $\sqrt{15}$. | ||
+ | |||
+ | I used a spreadsheet to calculate the iterates by entering "4" into cell A1, "=A1-(A1^2-15)/(2*A1)" into cell A2 and copying the formula in A2 to A3 through A10. Examining the decimal places, it was clear that by cell A4 the first 8 decimal places were no longer changing. The Newton's method iterates were | ||
+ | |||
+ | <math> | ||
+ | \begin{align} | ||
+ | x_{1} &= 3.875\\ | ||
+ | x_{2} &= 3.872983871\\ | ||
+ | x_{3} &= 3.872983346\\ | ||
+ | x_{4} &= 3.872983346 | ||
+ | \end{align} | ||
+ | </math> | ||
+ | |||
+ | As the first 9 decimal points are the same, our estimate $x_4$ should be correct to 8 decimal places. | ||
+ | |||
+ | Note: this example does not include a part 2 like OSH 3. |
Latest revision as of 15:26, 25 September 2014
Example problem statement
Estimate $\sqrt{15}$ to 8 decimal places using Newton's method.
Solution
In order to use Newton's method, I needed a polynomial that has a zero at $\sqrt{15}$. I used the function
- \(f(x)= x^2-15.\)
For Newton's method, I also needed the derivative of this function:
- \(f'(x)= 2x.\)
The last thing I needed before starting with Newton's method was an initial guess. The nearest perfect square to 15 is 16 so $x=4$ was a good candidate for an initial guess. Graphing $f(x)$, it was clear that $x=4$ is positioned well so that Newton's method should converge to $\sqrt{15}$.
I used a spreadsheet to calculate the iterates by entering "4" into cell A1, "=A1-(A1^2-15)/(2*A1)" into cell A2 and copying the formula in A2 to A3 through A10. Examining the decimal places, it was clear that by cell A4 the first 8 decimal places were no longer changing. The Newton's method iterates were
\( \begin{align} x_{1} &= 3.875\\ x_{2} &= 3.872983871\\ x_{3} &= 3.872983346\\ x_{4} &= 3.872983346 \end{align} \)
As the first 9 decimal points are the same, our estimate $x_4$ should be correct to 8 decimal places.
Note: this example does not include a part 2 like OSH 3.