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Is there a difference between $(x)^{\frac{1}{n}}$ and $ \sqrt[n]{x}$ ?

I’m confused with this topic. Any ideas or examples ?

If $(x)^{\frac{1}{n}} = \sqrt[n]{x}$

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Consider $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . Is it the same if I write $ x=\frac{-b \pm(b^2-4ac)^{\frac{1}{2}}}{2a}$ ?

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Yes , there is a slight difference as per one of the conventions.

Consider, $x^{1/2}$ and $\sqrt x$, former is written in a general sense(more appropriate to use) but the latter one is used when $x\ge 0$.So according to this $i^{1/2}$ and not $\sqrt i$.

Hope this helps(not fully but to some extent atleast).

In nutshell, $x^{1/n}$ is a multi-valued function but $\sqrt[n]{x}$ is a single valued function.

As a quick example to this consider $\sqrt[3] 1$,and $1^{1/3}$.The former gives $1$(a single value) but the latter gives “cube roots of unity” (three values).

PS: I posted it as an answer so that others can see it easily while going through this page.

EDIT:THIS DIFFERENCE IS NOT UNIVERSALLY ACCEPTED.Actually I thought it’s explicit from my above written sentence -” $x^{1/n}$ is a multi-valued function ” but it doesn’t seem to be .In multi-valued functions,you can take any branch (for example other than principal branch $(-\pi,\pi]$that you want.Now which one is chosen by you $\implies$ above stated difference is not universal.

There is no difference. In the same way as I write $e^x$ when $x$ is simple enough, and $\exp(x)$ otherwise, I also write $\sqrt{x}$ when $x$ is simple enough, and $(x)^{1/2}$ otherwise.

For example, I write

$$

\left(\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}\right)^{1/2}

\qquad\text{and not the ugly}\qquad

\sqrt{\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}}

$$

Just as I write

$$

\exp\left(\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}\right)

\qquad\text{and not the illegible}\qquad

e^{\frac{1+\frac{9}{x^2}}{\sin \frac{\pi}{9}+2}}

$$

There is no universal convention about the meaning of these notations in general. When $x$ is nonnegative, it is fairly common to say that both $x^{1/n}$ and $\sqrt[n]{x}$ refer to the unique nonnegative $n$th root of $x$. Some authors may use other conventions (such as the one mentioned in kilimanjaro’s answer, in which $\sqrt[n]{x}$ refers to the nonnegative root but $x^{1/n}$ could refer to any root), but these are not universally understood and should not be used without explanation. In particular, when $x$ is not a nonnegative real number, you should never use either notation without specifying which $n$th root you are referring to (or saying that you don’t care which $n$th root is picked). In general, in the absence of a specifically stated convention, I would assume that $\sqrt[n]{x}$ and $x^{1/n}$ are synonymous.

They are the same, it’s the exponent representation of “root operator”.

Example $4^{\frac{1}{2}}=\sqrt[2]{4}=2$

@Angelo Mark

In your comment to @Elekko you say

$$x^2=9 \Rightarrow x=9^{\frac{1}{2}}= \sqrt{9}=3$$

however

$x^2=9$ doesn’t imply neigher $x=9^{\frac{1}{2}}$ or $x= \sqrt{9}$, although they usually mean exactly the same thing.

The main question is not whether $x^{1/n}=\sqrt[n]x$ or not but what is an inverse function of $x^n$. The minimal requirement for a function $f(x)$ to be invertible in some given domain is to be injective (or one-to-one, i.e. $f(x)=f(y)$ if and only if $x=y$ ) in that domain.

A function $f(x) =x^n$ is injective for all real numbers, i.e. $\forall x\in\mathbb{R}$ when $n$ is odd, i.e. $n=2m-1,m\in \mathbb{N}$ and therefore $f(x)=x^{2m-1}$ has an inverse function $f^{-1}(x) =x^{1\over{2m-1}},\forall x\in\mathbb{R} =\sqrt[2m-1]{x}$.

However, for an even $n$, i.e. $n=2m,m\in \mathbb{N}$, the function $f(x) =x^n$ isn’t injective in the domain of all real numbers, since $f(x)=f(-x)$ and therefore there is no inverse function.

Fortunately, if we take only a half of the domain, either all positive or all negative real numbers the function $f(x) =x^{2m}$ will be injective in such a half domain and therefore has an inverse. In the negative half domain, the inverse will be $f^{-1}(x)=-x^{1\over 2m}=-\sqrt[2m]x$ and in the positive half domain $f^{-1}(x)=x^{1\over 2m}=\sqrt[2m]x$.

Getting back at your example, when you write

$x^2=9 \Rightarrow $ either $x=9^{\frac{1}{2}}$ or $x= \sqrt{9}$ you automatically assume that $x^2$ has an inverse function in the domain of all real numbers, which is wrong. The problem is easily seen if you write it as following $x^2=(\pm x)^2=9\Rightarrow \pm x=\sqrt 9=9^{1/2}$ or $x=\pm\sqrt 9=\pm 9^{1/2}$

@kilimanjaro

Both $\sqrt{x}$ and $x^{1/2}$ are functions.

A function associates one, and only one, output to any particular input.

A multi-valued functions are not functions in a regular sense and only add confusions for this sort of discussion. Also, neither of the notation of roots is commonly considered multi-valued. You can define one so and one so for a specific very abstract discussion where you need to distinct between them, but outside of that abstract discussion (i.e. when you speak to “normal people”) it is not so.

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