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Consider the following expression $$\sqrt[4]{1296}=6$$

To find the 4th root of $1296$, first we write $1296$ as product of prime factors $$1296=3^4 \times 2^4$$

Now, $$\sqrt[4]{1296}=\sqrt[4]{3^4 \times 2^4}=3\times 2=6$$

But, I find it confusing when we have a decimal, say find $\sqrt[4.5]{1296}$

What does it really mean? How do we calculate it? I assume there is some real meaning for this because all the calculators calculates these.

Note: I know $\sqrt[4.5]{1296}=4.91688$ and $4.91688^{4.5}=1296$. This is not what I am asking. Actually, $4.5$th root or $4.5$th power, all are confusing statements to me and I was trying to understand if this has a real meaning.

Asaf Karagila
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Kiran
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  • It has a real meaning. Approximately $4.91688$. More generally, $\sqrt[9]{1296}$. There are several approximation methods for roots, such as Taylor Series, Newton-Raphson, etc. – barak manos Dec 24 '16 at 21:28
  • my edit and your comment was almost at the same time. :) – Kiran Dec 24 '16 at 21:29
  • In general, $\sqrt[b]{a}$ is a quantity $x$ such that $x^b=a$. The problem is that for non-integer $b$, we have to somehow define $x^b$ in a sensible way. One way is to declare that $x^b=\exp(b\log x)$ where $\exp(t)=\sum_{n=0}^{\infty} \frac{t^n}{n!}$ and $\log $ is the inverse function of $\exp$.

    So then $\sqrt[b]{a}=a^{1/b}=\exp(\frac{1}{b} \log a)$. Although I'm not quite sure, I suspect this series expansion is how calculators compute actual values for such things;

     – user160738 Dec 24 '16 at 21:38

5 Answers5

5

Note that

$$\sqrt[4.5]{1296}\equiv1296^{1/4.5}=1296^{2/9}=x$$

and we want to determine what $x$ is. Thus,

$$x^9=1296^2$$

$$\implies x^9-1296^2=0$$

From here, one usually uses root finding algorithms.

In the scenario that we have something like $x=1296^\pi$, we solve the approximate case $x=1296^{a/b}$ and let $a/b\to\pi$, where $a$ and $b$ are whole numbers.

Simply Beautiful Art
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5

You can think of it as $1296^{1/4.5} = 1296^{2/9}.$ So it's the 9th root squared. But to calculate it, we'd probably use logs. $\ln(1296^{2/9}) = \frac{2}{9}\ln(1296) = 1.592675084.$ Then $1296^{2/9} = e^{1.592675084} = 4.916884430.$

B. Goddard
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5

When the base of the root is rational, say $\frac{p}{q}$, then $\sqrt[p/q]{a}$ is the same as $a^{q/p} = \sqrt[p]{a^q}$. In your case, $\sqrt[4.5]{1296}$ would be $1296^{2/9}$ (because $4.5 = \frac{9}{2}$), or $\sqrt[9]{1296^2}$.

This becomes a bit more complicated for an arbitrary real number $r$. Essentially, you consider the set of all $\sqrt[s]{a}$ where $s$ is a rational number greater than $r$. Then you define $\sqrt[r]{a}$ to be the supremum, or least upper bound, of this set.

Since roots are essentially just a different notation for exponents, this exactly mirrors how we define real exponents: extend them first to the rationals with the convention that $a^{p/q} = \sqrt[q]{a^p}$, and then extend them to the reals using the least-upper-bound property.

Fargle
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4

In general we have that $\sqrt[w]{x}=x^\frac{1}{w}$.

How is exponentiation defined for a positive number $\alpha$?

for rational numbers of the form $\frac{p}{q}$ we define $\alpha^{\frac{p}{q}}$ as $\sqrt[q]{\alpha ^p}$.

And it is defined for the other real numbers via continuity.

Asinomás
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Suppose we have the goal of defining $a^b$ for arbitrary $a,b>0$. If we had such a number and call it $x$, we have $x=a^b$. The, taking logarithm, $$ \log x = \log a^b= b\log a. $$ If we now exponentiate, we get $$ x=e^{\log x} =e^{b\log a}. $$ Because of this, it is customary to define $$ a^b:=e^{b\log a}. $$

Martin Argerami
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