Zero Raised To A Power

Updated! February 5, 2017

The value of goose egg raised to the cypher power, $(0^0)$ , has been discussed since the time of Euler in the 18th century (1700s). There are iii reasonable choices: 1,0, or "indeterminate". Despite consensus amongst mathematicians that the correct answer is one, computing platforms seem to have reached a multifariousness of conclusions: Google, R, Octave, Cherry-red, and Microsoft Calculator choose 1; Hexelon Max and TI-36 calculator cull 0; and Maxima and Excel throw an error (indeterminate). In this article, I'll explain why, for discrete mathematics, the correct answer cannot be anything other than 0^0=i, for reasons that go beyond consistency with the Binomial Theorem (Knuth's argument).

Context of the Debate: Continuous Mathematics

The three choices for the value of $0^0$ appear because $x^y$ , as a function of two continuous variables, is discontinuous at (0,0) and takes three different values depending on the direction of approach to the aperture:

Fixing y=0, we have $x^0=1$ for all $x \neq 0$ . (Proof: $x^0 = x^{(1-1)} = x^1 x^{-1} = x/x = 1$ , each statement holding for all $x \neq 0$ ). Indeed, $x^y \rightarrow 1$ as $x \rightarrow 0$ , approaching from left or right, with y=0. (This was Euler'southward reason.)
Fixing x=0, we take $0^y=0$ for $y >0$ . (When y < 0 we have division by zero which is undefined in the reals and $+\infty$ in the extended reals). Taking limits, $x^y \rightarrow 0$ every bit $y \searrow 0$ , approaching from above only, with ten=0.
Fixing x=0, we have an undefined value when y < 0 due to division by nix.

Notice that the discontinuity is not a simple (point) discontinuity, but rather a pole aperture due to the arroyo from beneath. (Exercise: what happens as the origin is approached from 45 degrees?)

Principles for a Determination in Mathematics: Extension and Consistency

In mathematics, when there is more than one choice, a determination is typically made past extending an existing precedent to maintain consistency with the show that is already accumulated and accepted.

An elementary instance is the way ordinary multiplication is extended from ii positive numbers to a positive and a negative number, and so to two negative numbers, i.due east. $(-1)(-1) = 1$ .

"Minus times minus is plus.
The reason for this we need not discuss!"
— Westward.H. Auden

Empirically, multiplication of two positive numbers has a well-defined, tangible meaning as repeated addition. This significant holds when one of the numbers is negative. But when both are negative, the empirical meaning fails.

For the mathematician, declaring something to be undefined (throwing an mistake) means a loss of efficiency because every instance now has to be checked for the undefined example, and this must be treated separately. If a definition could be found that remains consequent with all other empirically obtained rules, and if that definition means that calculation can keep indifferent to the decision, and so that is a big win.

The consistency in this item example is the distributivity of multiplication over improver, a police force which, for positive numbers, tin can be accepted on entirely empirical grounds. (See the footnote for the total argument.¹.)

Turning to Detached Mathematics – Consistency with the Binomial Theorem

In discrete mathematics, at that place is no notion of "approaching" — one is either at $0^0$ or away from it, in which case $1^0 = 1$ or $0^1 = 0$ .

The case of $0^0$ can be decided on consistency grounds with respect to the binomial theorem, i.e. loss of computational efficiency to have to treat this instance separately. This is the argument of Knuth (of The Art of Calculator Programming, and TeX fame), based on maintaining consistency with the binomial theorem $(a+b)^x$ when ten=0, due to its cardinal place in both detached and continuous mathematics:

"Some textbooks leave the quantity $0^0$ undefined, because the functions $0^x$ and $x^0$ have different limiting values when $x$ decreases to 0. But this is a mistake. Nosotros must define $x^0=1$ for all $x$ , if the binomial theorem is to be valid when $x=0, y=0$ , and/or $x=-y$ . The theorem is likewise important to be arbitrarily restricted! By contrast, the office $0^x$ is quite unimportant."
– from Physical Mathematics, p.162, R. Graham, D. Knuth, O. Patashnik, Addison-Wesley, 1988

Different Conventions Among Mathematical Computing Platforms

Given the universality of the $0^0=1$ convention amidst mathematicians, it is surprising to find that diverse computing platforms take implemented different values:

An Alternative Decision Criteria – tangible computation with verifiable count that requires the answer

While Knuth'south statement of convenient extension works, the finite summation of integer powers provides us with a real, tangible effect (a finite sum), whose value (an empirically determinable fact) depends unavoidably on the chosen value of $0^0$ . So here we accept a consistency statement that does non rely on efficiency.

The crucial step in this argument occurs in the derivation of (*1b) from (*1a) in Finite Summation of Integer Powers, Office two.

Extracting the relevant role of that derivation, we take:

$\sum_{k=1}^{N-1} (N-K)(K+1)^{p-1}\ \ \ \ \ (*1a)$

After expanding the binomial power using the binomial formula and farther manipulation, we get in at:

$= \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=0}^{N-1} \left[NK^j -K^{j+1}\right]$
(Pull the $K=0$ term out of both summations. Notation: $0^0 = 1\ \ \ \ \ \mbox{(***)}$ )
$= N + \sum_{j=0}^{p-1} \binom{p-1}{j} \sum_{k=1}^{N-1} \left[NK^j -K^{j+1}\right]$
(which, afterward additional manipulation, yields)
$= N + N(N-1) - S_p(N-1) + \sum_{j=1}^{p-1} S_j(N-1)\left[N\binom{p-1}{j} - \binom{p-1}{j-1}\right]$
$= N^2 - S_p(N-1) + \sum_{j=1}^{p-1} S_j(N-1)\left[N\binom{p-1}{j} - \binom{p-1}{j-1}\right]\ \ \ \ \ \mbox{(*1b)}$

The central pace happens in (***) to a higher place: we peel off the $K=0$ term of the inner summation to get: $N0^j - 0^{j+1}$ . Peeling this out of the outer summation requires considering the expression for all $j$ . Now, 0 raised to any positive ability is 0, so we tin can dispel the example of $j>0$ . But what is $0^0$ ? A decision must be made: it is either $0$ or $1$ . Indeterminacy is not an pick, since the situation is real and is required to go on the simplification.

The Argument for $0^0 = 1$

What are the consequences of choosing the other definition, i.e. $0^0 = 0$ ? In this case, the final formula for $S_p(N)$ is off past a linear abiding $N$ , while the selection $0^0 = 1$ leads to the exact formula and a computed value that matches a fauna force summation.

For $S_5(10) = 1^5+2^5+3^5+4^5+5^5+6^5+7^5+8^5+9^5+10^5$ , the divergence is betwixt 220,825 (the right, verifiable answer), and 220,815 (verifiably NOT correct). The correct definition is clear: 0^0 = ane is for empirical reasons that have to do with counting and summing. While information technology is the binomial theorem that provides the detail, the argument is ane of verifiable necessity and not one of consistency.

For detached mathematics, the empirical evidence shows that 0^0=1 is required:²

References
The Math Forum

(If yous're a software programmer of a mathematical bundle, I'd be interested in how you arrived at your decision. You can send me an email using the Comments link below.)

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