# How do you find a one-decimal place approximation for root5 26?

Oct 21, 2015

Use one step of Newton's method to find $\sqrt{26} \approx 1.9$

#### Explanation:

To find an approximation for the $5$th root of a number $n$, choose a reasonable first approximation ${a}_{0}$ then iterate using the following formula:

${a}_{i + 1} = {a}_{i} + \frac{n - {a}_{i}^{5}}{5 {a}_{i}^{4}}$

In our case let $n = 26$ and ${a}_{0} = 2$ (since ${2}^{5} = 32$ is close)

Then:

${a}_{1} = {a}_{0} + \frac{n - {a}_{0}^{5}}{5 {a}_{0}^{4}} = 2 + \frac{26 - 32}{5 \cdot 16} = 2 - \frac{6}{80} = 2 - 0.075 = 1.925$

So $\sqrt{26} \approx 1.9$

If we want more accuracy, just apply the formula again to get ${a}_{2}$ from ${a}_{1}$, etc.

This method is a particular case of Newton's method for finding a zero of a continuous differentiable function $f \left(x\right)$.

In our case $f \left(x\right) = {x}^{5} - n$, $f ' \left(x\right) = 5 {x}^{4}$ and Newton's method says to iterate using the formula:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)} = {a}_{i} - \frac{{a}_{i}^{5} - n}{5 {a}_{i}^{4}} = {a}_{i} + \frac{n - {a}_{i}^{5}}{5 {a}_{i}^{4}}$