# How do you find a one-decimal place approximation for root3 30?

Oct 18, 2015

Use one step of Newton's method to find $\sqrt[3]{30} \approx 3.1$

#### Explanation:

To find the cube root of a number $n$, start with an approximation ${a}_{0}$ and apply the following iteration step:

${a}_{i + 1} = {a}_{i} + \frac{n - {a}_{i}^{3}}{3 {a}_{i}^{2}}$

This is based on Newton's method for finding the zero of a function $f \left(x\right)$ using:

${a}_{i + 1} = {a}_{i} - f \frac{x}{f ' \left(x\right)}$

In our case $f \left(x\right) = {x}^{3} - n$ and $f ' \left(x\right) = 3 {x}^{2}$

Since ${3}^{3} = 27$ is not far off, let us use ${a}_{0} = 3$.

Then:

${a}_{1} = {a}_{0} + \frac{n - {a}_{0}^{3}}{3 {a}_{0}^{2}} = 3 + \frac{30 - 27}{3 \cdot {3}^{2}} = 3 + \frac{3}{27} = 3 + \frac{1}{9} = 3. \dot{1}$

So to one decimal place $\sqrt[3]{30} \approx 3.1$