# Difference between revisions of "Root (operation)"

The $n$th root of a number $x$, denoted by $\sqrt[n]{x}$, is a common operation on numbers and a partial inverse to exponentiation. (The proper inverse is the logarithm)

## Definition

For any (not necessarily real) numbers $x,y,n$, $y=\sqrt[n]{x}$ if $y^n=x$. Note that we generally take only the positive value of $y$, if we wish to take both the positive and negative roots, we write $\pm\sqrt[n]{x}$.

## How to compute all the roots of $\sqrt[n]{x}$

A known method to compute all the roots of $\sqrt[n]{x}$ is by the DeMoivre's formula. $\sqrt[n]{x}=\sqrt[n]{|x|}\left(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\right)$ , where $k=0,1,2,...,n-1$ and $x\in\mathbb{C}$

See that in $\sqrt[n]{|x|}$ we compute its principal root.

### Example with a real number

Compute all the roots of $z^4-16=0$.

First, we need to rearrange the equation . $z=\sqrt{16}$
See that here the " $x$" would be the number 16.
Then, we compute $|16|=16$.
As 2 is a pure real number, we know that $\theta=0$.
As $n=4$, thus $k=0,1,2,3$
We separately compute the cases $k=0,1,2,3$. $\sqrt{16}=\sqrt{16}\left(\cos\frac{0+2\pi\cdot0}{4}+i\sin\frac{0+2\pi\cdot0}{4}\right)\implies\boxed{2}$ $\sqrt{16}=\sqrt{16}\left(\cos\frac{0+2\pi\cdot1}{4}+i\sin\frac{0+2\pi\cdot1}{4}\right)\implies\boxed{2i}$ $\sqrt{16}=\sqrt{16}\left(\cos\frac{0+2\pi\cdot2}{4}+i\sin\frac{0+2\pi\cdot2}{4}\right)\implies\boxed{-2}$ $\sqrt{16}=\sqrt{16}\left(\cos\frac{0+2\pi\cdot3}{4}+i\sin\frac{0+2\pi\cdot3}{4}\right)\implies\boxed{-2i}$

## How to approximate a root

There's many methods to approximate roots. Here are two:

• $\sqrt{x}\approx\frac{x+x_0}{2\sqrt{x_0}}$, where $x_0$ is the nearest perfect square.
Computing the square root of 5,
We know that the nearest perfect square is 4, so, $\sqrt{5}\approx\frac{5+4}{2\sqrt{4}}$ $\sqrt{5}\approx2.25$
With this method you can get a little good approximation. $\sqrt{5}\simeq2.2360679$

• Also, you can use Newton-Raphson's method: $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ , where $x_n$ is a number close to the root.
The more times you apply this formula (consecutively), the better is the approximation that you can get.
Computing the square root of 2,
See that it would be: $x=\sqrt{2}$ $x^2-2=0$ $\boxed{f(x)=x^2-2}\implies f'(x)=2x$ $x_{n+1}=1.4-\frac{1.4^2-2}{2\cdot1.4}$ $x_{n+1}=1.4142857142857144$
This it's very close to $\sqrt{2}$. Sometimes it can be a "very ugly bashing", but it's a method to get really good approximations. But, if couldn't get a good approximation at first, you can apply it a second time.