Zero Exponent Rule
Arithmetic of Zero Power
The number with exponent zero, $x^0$, is highly nonintuitive, and there are at least two ways of looking at it, shown herein. Using Euler's algebra of exponents we know that $a^ma^n$ is $a$ with exponent $m+n$. And, since $0=1-1=(1)+(-1)$, we can construct an alternative and equivalent form of $a^0$. Let $a$ and $n$ be fixed numbers, other than zero, then:
$$ f(a)=a^0=a^{n+(-n)}=a^na^{-n}=\frac{a^n}{a^n}=1 $$
For negative base values, $a\in \mathbb{R}$, with either even or odd intermediate exponent, the negative factors can be separated in numerator and denominator, $(-x)^n = (-1)^n x^n$, and evaluated as independent factors:
$$(-\lvert a\rvert)^n/(-\lvert a\rvert)^n=\lvert a\rvert^0=1$$
So we have for the unity function even symmetry, and it is constant for negative and positive argument. The continuity of $f(x)=1$ around the origin suggests that we can divide by zero, when there is also a zero in the numerator. Any base, $a$, with a zero exponent is equivalent to unity.
Let's look at general patterns of radicals using the arithmetic of the Product Rule in order to appreciate what this arithmetic means in a limit format.
Analysis of Zero Power
We can derive what the zero exponent gives you, by using calculus and analysis to bound and limit the function.
What is a zero power if not the limit of the $n$th-root:
$$\begin{equation}\label{nthroot_intro_eq} b=a^{1/n} \end{equation}$$
Symmetry of Solutions
If $n$ is an even, positive-definite Natural Number, then the radical, $b$, is an even function of the radicand, $a$.
Even indexed radicals of negative radicands are complex, in fact pure-imaginary, which dimensionally orthogonal radicals.
$b$ is the $n$th root of $a$, and to indicate the positive solution of the radical it is annotated with the radical sign, $\sqrt[n]{}$. The positive solution always exists for positive radicand $a$, it's just when the index is even do we have the symmetry of the equation, which requires us to define a principle branch for the inverse to work as a single-valued map, or proper function.
When there are more than one solution for a given ordinate value, $y=f(-x)=f(x)$, and the range of the inverse is restricted to one branch.
| Index $n$ | Solutions |
|---|---|
| even | $ b=a^{\frac{1}{n}} : b = \pm \sqrt[n]{a} $ |
| odd | $ b=a^{\frac{1}{n}} : b = \sqrt[n]{a} $ |
When $n$ is even and the radicand is positive, that is $n=2m$ for m as some positive-definite Natural number, $m\in\mathbb{N}_+$, then the positive $n$th root is $b=\sqrt[n]{a}$ and is a subset of the solutions $b=a^{\frac{1}{n}}\Longleftrightarrow b^n=(b^m)^2=a$.
High Index Radical
$b$ approximates $a^0$ for large $n$, which is argued by the following:
- The roots are magnitudinally ordered according to $n$. Which is to say $b=b(n)$ is the function $b$ of $n$, and that for a given $n=n_1$, we can say $b(n_1)$ is either greater or lesser than another $b(n_2)$ without calculating either the $n_1$'st or the $n_2$'nd root.
- The roots $b(n)$ are monotonic separately considered, each on one side of $1$. Monotonic is to say the curving value $b(n)$ goes in one direction for a direction of change in $b$.
To be demonstrated is that $b$ gets arbitrarily close to $1$ in the large $n$ limit.
The formula for $b$, eq. (\ref{nthroot_intro_eq}), is known as the $n$'th root of $a$, where $b$ has the power of inverse $n$. If the power is a quotient, $\frac{m}{n}$, then using the rules of exponents, the form is equivalent to the root to the power $m$.
To calculate the radical, $a^{1/n}$, the test of verity is that $b$ is the number which when multiplied by $n-1$ factors of itself, equals $a$:
Where the ellipsis substitutes $(b)^{n-2}$ terms in the product, since $2$ are displayed inline (assuming $n\gt 2$).
For example, if we choose $n=5$, and $a=10$, then the (fifth) root of $10$ is given by a calculator to be about (rounded) $b=1.5849$. And, $b^5$ = $10.0002$. If we don't round the number and truncate the fraction to $b=1.5848$ then $b^5$=$9.9970$. In practice, it never comes out perfect ($b^n\approx a$) because the root ($b$) is an irrational number (a.k.a. a "surd", ibid. ) meaning no truncated number will be exact and complete to serve as $b$.
Product Ordering
If the coefficient, $c$ of some number, $x$, is compared to the unit number 1, it can be seen immediately what the order of the product and the lone number, $x$, is. The further from unity the coefficient, the greater the deviation from the identity function, $f(x)=x$
$$ 1 \lt a \to b \lt ab $$
Likewise,
$$ 1 \gt b \to a \gt ab $$
For a given product, $p=c x $, we have:
| coefficient order | product order |
|---|---|
| $c \lt 1$ | $ p \lt x $ |
| $c = 1$ | $ p = x $ |
| $c \gt 1$ | $ p \gt x $ |
Radical Absolute Ordering
Using only the product ordering table above, and the definition of the $n$'th root, eq. \eqref{nthroot_intro_eq}, we shall prove the following Radical Absolute Ordering table.
| $a$ | $n$'th root |
|---|---|
| $a \lt 1$ | $ b \lt 1 $ |
| $a = 1$ | $ b = 1 $ |
| $a \gt 1$ | $ b \gt 1 $ |
Case of Small Radicand
First row of Table 3: If $a$ is less than one and greater than zero, then the $n$'th root is less than one, $b\lt 1$, which is proven by contradiction with the following logic. If it were that $b\ge 1$, then $bb\ge b$, by product ordering, and by repeating the product ordering rule $n-1$ times, we have $b^n\ge 1$ which contradicts $a\lt 1$, therefore $b\lt 1$. Further in this case of $a\lt 1$, $b\gt a$, because $b\lt 1$ means the following:
$$ a = b^n\lt b^{n-1}\lt\dots\lt b $$
Which we will prove by induction on $n$. For the $n=1$ case, we have $b=a\lt 1$, so all we need is to prove that if the ordering is true for the $n=m$th case then it's true for the $(m+1)$th case, which will prove the statement for all $n$. Using Table 2 product ordering and $n\gt 1$, with $c\rightarrow b$ and $x\rightarrow b^{n-1}$, we have $p=cx=b(b^{n-1})\lt b^{n-1}\le b\lt 1$.
Case of Unity
For the second row of Table 3, if $a$ is one then the $n$'th root is one. Proof by contradiction: if the radical were not one, then it is either greater or lesser. If $b\lt 1$ then $b^2\lt b \lt 1$, but $b^n=a=1$, which contradicts the premise that $b\lt 1$. The other candidate for the radical, that it's greater than unity, is similarly contradictory. So the radical is one when the radicand is one.
Case of Large Radicand
The third row of Table 3 is similar to the first row, we have to demonstrate bounding and ordering of the radical. If the radicand is greater than unity, $a\gt 1$, then the radical has to be greater than unity, also. If the radical were one, then any of its powers are simply unity. And if the radical were less than one, then by product ordering its $n$th power is also less than one, which contradicts the premise, $b^n=a\gt 1$.
Further, the radical is less than the radicand in this case. Otherwise, by Product Ordering, the $n$th power of the radical would be strictly larger than the radical, $b^n\gt b\gt a$, which is a contradiction of the premise.
Radical Radicand Ordering
Further, regardless of where $a$ lies in the positive Real Numbers (anywhere between zero and infinity), the square root ($a^{1/2}$) is closer to one than $a$, with the exception of being equivalent at one.
| Radicand | Radical |
|---|---|
| $a \lt 1$ | $ b \gt a $ |
| $a = 1$ | $ b = a $ |
| $a \gt 1$ | $ b \lt a $ |
The fourth root ($(a^{1/2})^{1/2}$) is closer to one than $a^{1/2}$, because it's two more factors of it that are equal to the same radicand, so that the larger $n$ the closer the radical is to one.
$$ 1 \le \left(\left(a^{1/2}\right)^{1/2}\right)^{1/2} \lt \left(a^{1/2}\right)^{1/2} \lt a^{1/2} \lt a $$
Which is obtained by using the ordering rule for the square root ($n=2$), when $a\gt 1$, repeatedly.
To demonstrate this graphically, we'll let $a=x$ and look at the curves of $b(a)=a^{1/n}$ for $n=\{2, 4, 8\}$.
Plot of the identity, square, fourth, and eighth roots, over $[0, 1]$. For $a\lt 1$, the higher root is closer to the flat line above the curves at $b=1$. The trend with higher radical index is flattening the radical to one, and always between unity and the identity, which is the radicand.
In the case $n$ goes to infinity the infinite root is just $1$, for any radicand $a$.
Zero Radicand
You may be skeptical about the zero radicand, infinite root case, $f=0^0$, so let's look at the following equation, with a graph of continuous index, with radicand now inverse-radical index:
$$\begin{equation}\label{zeroToZeroPow} 0^0\stackrel{?}{=}\lim_{x\to 0} x^x \end{equation}$$
This test function is a naturally occuring example of tetration of $x$ of height two, ${}^2x$, as an aside.
We can use the Change of Base Rule to move the variable from base to exponent:
$$\begin{equation}\label{nat_base_eq}x^x=e^{x\ln(x)}\end{equation}$$
The argument to the exponent has one factor going to zero in the limit, while the other factor goes to negative infinity, so let's look at the derivatives and some graphs of the whole domain and then microscopically close to zero. Employing The Chain Rule, and the derivative of the logarithm and exponential, we have:
$$\begin{equation}\label{zeroRadicand_prime} f'=e^{x\ln(x)}(1 + \ln(x)) = f(x)(1 + \ln(x))\end{equation} $$
The second derivative is an application of The Product Rule, and the first derivative, $f'$:
$$\begin{equation}\label{zeroRadicand_dprime} f''=e^{x\ln(x)}\left((1 + \ln(x))^2 + \frac{1}{x}\right)\end{equation} $$
Since we know the $n$th root of a radicand less than one, then by Radical Ordering described above, we know the radical is between one and the number. then we know that the tetration in eq. \eqref{zeroToZeroPow} is bounded and increasing for negative step on the interval, $(0,\frac{1}{e})$. Figure 9, below demonstrates the negative-definite derivative on that open interval, which confirms we can say its moving in the right direction and increasing for argument changes towards zero, and therefore making the function convergent to the boundary, one. The value of the tetration function is increasing, moving in reverse from the low point, because the derivative is only zero at the point $x=\frac{1}{e}$, because $\ln(1/e) = -1$ appears in the derivative, by eq. \eqref{zeroRadicand_prime}, with a positive-definite second derivative, eq. \eqref{zeroRadicand_dprime}.
Under a multiplicative inverse Change of Variable $x=\frac{1}{n}$, in the small $x$ limit we have the inverse limit for the alternate representation $n\to \infty$. In that high $n$ radical index limit, . The test function, eq. \eqref{nat_base_eq},
$$x^0=1:\forall x\in \mathbb{R}$$