# OSH/Example/3

#### 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.