Thread: 0^0

**britmath** · October 5th 2012, 07:50 AM

I've recently been involved in a discussion regarding the value of 0⁰ in which I confidently asserted that 0⁰ is undefined, largely because x⁰ is always 1, while 0^x is always 0, leading to a contradiction at 0⁰.

However, I've since seen discussions of the subject that say a consensus has "recently" been reached around 0⁰= 1, seemingly on utility grounds, with x⁰ being a much more important result than 0^x.

I'd be interested to hear some expert views on what the current thinking is. Thanks a lot!

**Plato** · October 5th 2012, 08:44 AM

Originally Posted by britmath

I've recently been involved in a discussion regarding the value of 0⁰ in which I confidently asserted that 0⁰ is undefined, largely because x⁰ is always 1, while 0^x is always 0, leading to a contradiction at 0⁰.

However, I've since seen discussions of the subject that say a consensus has "recently" been reached around 0⁰= 1, seemingly on utility grounds, with x⁰ being a much more important result than 0^x.

I'd be interested to hear some expert views on what the current thinking is. Thanks a lot!

Here is a webpage dedicated to that. Exponentiation - Wikipedia, the free encyclopedia

**ebaines** · October 5th 2012, 11:07 AM

I would point out that $\lim_{x \to 0} x^0 = 1$ , which is a good enough "proof" for me.

**Prove It** · October 5th 2012, 06:14 PM

Originally Posted by ebaines

I would point out that $\lim_{x \to 0} x^0 = 1$ , which is a good enough "proof" for me.

It's not a good enough proof for me. The rule $\displaystyle \begin{align*} a^0 = 1 \end{align*}$ comes from the index law $\displaystyle \begin{align*} \frac{a^m}{a^n} = a^{m - n} \end{align*}$ , with $\displaystyle \begin{align*} m = n \end{align*}$ . However, $\displaystyle \begin{align*} \frac{0^1}{0^1} = 0^{1 - 1} = 0^0 \end{align*}$ . Wouldn't $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ mean "How many 0's go into 0"? Surely there are an infinite number of them that can go into 0, since 0 + 0 + 0 + ... + 0 (infinity times) = 0. This gives THREE possible answers for what $\displaystyle \begin{align*} 0^0 \end{align*}$ is!

So regardless of what the limit of $\displaystyle \begin{align*} x^0 \end{align*}$ is, it will NEVER be able to be defined at that point, and so it is definitely not enough to say that $\displaystyle \begin{align*} 0^0 = 1 \end{align*}$ .

**Deveno** · October 5th 2012, 09:30 PM

Originally Posted by ebaines

I would point out that $\lim_{x \to 0} x^0 = 1$ , which is a good enough "proof" for me.

in general, if $\lim_{t \to 0} f(t) = 0$ and $\lim_{t \to 0} g(t) = 0$ it may still be the case that:

$\lim_{t \to 0} h(t) = f(t)^{g(t)}$ takes on different values depending on what functions we choose for f and g.

for example, if f(t) = g(t) = t, then $\lim_{t \to 0^+} h(t) = 1$ (the right hand limit doesn't exist, as h(t) is undefined, if h is to be real-valued).

however, if $f(t) = e^{-\frac{1}{t}}$ , and g(t) = at, then $\lim_{t \to 0} h(t) = \frac{1}{e^a}$ .

a lot depends on what one MEANS by a^b: is it a function of a, a function of b, or a function of both? what domain for a and b are we talking about, and how do we interpret what the expression a^b is?

for example, if one takes a^b to be $e^{b\log(a)}$ one is faced with the quandry of what $0^*(-\infty)$ should be.

this is related to "what is infinity"? not all infinite quantities are "the same", we can't really treat ∞ as a real number without sacrificing some of the arithmetic structure of the real numbers.

that said, there are some reasons for treating 0⁰ as being 1.

one of the reasons i find most compelling is that:

for finite (non-empty sets) A and B, |B|^|A| represents the number of functions from A to B. if we count the empty set as a finite set, then 0⁰ should represent the number of functions from the empty set TO the empty set, and there is only one such function: the empty function (a "blank sheet of paper as the graph").

another compelling reason is that the power rule:

$\frac{d}{dx}(x^n) = nx^{n-1}$ is not valid for n = 1 at x = 0, unless 0⁰ = 1.

**********

in many situations, "0 is a special case", and deciding how to deal with that special case is context-sensitive. it makes little sense to say "0⁰ = 1", without supplying more information about what kinds of things 0 and 1 are. things they might be:

natural numbers (as in combinatorics)
rational numbers
real numbers
complex numbers
polynomial functions
real-valued constant functions
values of other real-valued functions
bottom and top elements of a boolean lattice

people new to the subtleties of advanced mathematics often think mathematical ideas have some some of "eternal epistemological state", perfect and unchanging. "1" IS a certain something, and nothing else, forever and ever. the truth is, mathematical concepts are a bit more flexible than that: the same string of symbols may mean entirely two different things in different arenas. this is why mathematical definitions can be so intricate: we're trying to define the "scope of discussion" (and what rules apply).

**JJacquelin** · October 6th 2012, 12:02 AM

Originally Posted by britmath

I've recently been involved in a discussion regarding the value of 0⁰ in which I confidently asserted that 0⁰ is undefined, largely because x⁰ is always 1, while 0^x is always 0, leading to a contradiction at 0⁰.

However, I've since seen discussions of the subject that say a consensus has "recently" been reached around 0⁰= 1, seemingly on utility grounds, with x⁰ being a much more important result than 0^x.
I'd be interested to hear some expert views on what the current thinking is. Thanks a lot!

Hi !

Have a look at the paper : "Zero to the Zero-th power", with the link :Scribd

**FernandoRevilla** · October 6th 2012, 02:20 AM

Question: What is $0^0$ ?

Answer: We need to know what is the hidden question behind the question.

**SworD** · October 6th 2012, 10:36 AM

Its a matter of definition and context... I don't think there's a real situation in math where the result of a calculation is gonna be 0^0, leading you to say "Well gee, thats undefined". There's gonna be a limit or something in which case there would be an answer.

BTW. Intuitively, I would still say if we HAD to choose a value for 0^0, it'd be 1... even the case of 0^x supports it. 0 raised to a positive power is 0, but remember 0 raised to a negative power is infinity.. so 0^0 is the bridge in between.

Still, saying 0^0 is equal to 1 would be very misleading in certain situations because the function 0^x would be defined but discontinuous at that point.

**Deveno** · October 6th 2012, 12:52 PM

Originally Posted by SworD

Its a matter of definition and context... I don't think there's a real situation in math where the result of a calculation is gonna be 0^0, leading you to say "Well gee, thats undefined". There's gonna be a limit or something in which case there would be an answer.

BTW. Intuitively, I would still say if we HAD to choose a value for 0^0, it'd be 1... even the case of 0^x supports it. 0 raised to a positive power is 0, but remember 0 raised to a negative power is infinity.. so 0^0 is the bridge in between.

Still, saying 0^0 is equal to 1 would be very misleading in certain situations because the function 0^x would be defined but discontinuous at that point.

to amplify:

there is NO choice of value for f(x,y) = x^y, that makes this function continuous at (0,0).

(interesting historical side-note: Augustin Cauchy (yes...THAT Cauchy) was of the mind, and had even published that 0⁰ = 1, until an anonymous submission (merely signed "S" i believe) exhibited a counterexample (two functions as in my earlier post), and he reversed his position. so this "consensus" the OP speaks of is neither "recent" nor actual. such musings border on the philosophical...there really isn't a "right" answer, but there are several wrong ones).

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