Possible Duplicate:
Zero to zero power
According to Wolfram Alpha:
$0^0$ is indeterminate.
According to google: $0^0=1$
According to my calculator: $0^0$ is undefined
Is there consensus regarding $0^0$? And what makes $0^0$ so problematic?
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This question will probably be closed as a duplicate, but here is the way I used to explain it to my students:
Since $x^0=1$ for all non-zero $x$, we would like to define $0^0$ to be 1. but ...
since $0^x = 0$ for all positive $x$, we would like to define $0^0$ to be 0.
The end result is that we can't have all the "rules" of indices playing nicely with each other if we decide to chose one of the above options, it might be better if we decided that $0^0$ should just be left as "undefined".
Old John
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Thanks! I met the problem of trying to explain this to students so many times, that I finally developed an argument that would convince even stubborn students as quickly and simply as possible :) – Old John Dec 15 '12 at 21:24
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16@OldJohn: It completely fails to convince me. Who cares about $0^x$? On the other hand $x^0$ must be $1$ always, or polynomials wouldn't work at $x=0$ ... – hmakholm left over Monica Dec 15 '12 at 21:42
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Doesn't that depend on how you define polynomials? My definition is purely algebraic and does not involve $x^0$. – Old John Dec 15 '12 at 21:48
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@OldJohn: From a purely algebraic point of view, replace "polynomials" by "the systematic derivation of functions $R\to R$ from formal polynomials in $R[X]$". And what about power series? – hmakholm left over Monica Dec 15 '12 at 21:55
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1That is fine, but I stand by my original premise that my explanation was definitely at the right level for the students I was teaching, to whom "systematic derivation" and "formal polynomials in $R[X]$" might just as well have been Klingon ... – Old John Dec 15 '12 at 21:58
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@MichaelHardy I still stand by my premise that my explanation was the most suitable for my students: " ... because it's an empty product ... " would not have been appropriate for students at their level of mathematical maturity – Old John Dec 15 '12 at 23:57
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But mentioning that there are standard identities that fail unless $0^0$ is $1$ is another matter. – Michael Hardy Dec 15 '12 at 23:59
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2But there are also standard identities that fail unless $0^0 = 0$ ... – Old John Dec 16 '12 at 00:01
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3@OldJohn : please show me a standard identity that fails unless $0^0=0$. I just posted a question asking whether there is a good reason not to define $0^0$ to be $1$. – Stefan Smith Dec 14 '13 at 18:16
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Not so much an identity, more like a limit. We have $0^{1/n}=0$ for all $n>1$, so we would want $0^0$ to be zero for the limit process to work nicely. Mostly this is not important, and it generally makes sense to define $0^0$ to be $1$, but however you define it, something is going to break, which is why some people prefer to leave it undefined. – Old John Dec 14 '13 at 18:37
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@OldJohn: Lets look at the logic behind your argument. We would like $P$: $0^0=1$ as well as $Q$: $0^0=0$. But $P$ contradicts $Q$ and so: not ($P$ and $Q$). However, from this you draw the conclusion: (not $P$) and (not $Q$). – Mark Feb 12 '17 at 22:14
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In set theory, where everything is a set, $0$ is represented by the empty set. Exponentiation of sets $\alpha^\beta$, let's call them cardinalities and write $|\alpha|^{|\beta|}$, is defined to be the cardinality (number of elements) of all functions from $\beta \to \alpha$. If both $\alpha$ and $\beta$ are empty, then there is exactly one function $\varnothing \to \varnothing$, hence $0^0 = 1$.
Though this is just a convention, I like how it justifies $0^0 = 1$.
Rudy the Reindeer
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The question is tagged (real-analysis). So set theoretic justifications don't qualify here :) – Ralph Dec 15 '12 at 22:45
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One way of looking at it is that there are two different exponentiation operators that are denoted $a^b$:
Discrete (algebraic): when $b$ is an integer. Then the exponentiation is multiple multiplication or division. This can be used in any group. This operator occurs in Taylor series, the binomial expansion, when calculating the size of the set $A^B$ given sizes of $A$ and $B$ etc. For this operator, $0^0=1$, or in general for any element $a$, $a^0$ is the multiplicative identity.
Continuous (analytic): when $b$ is real, and $a>0$. Then we define $a^b = \exp(b \ln a)$. Note that here $a$ must be positive. Here it is best to leave $0^0$ undefined, as otherwise the function will be discontinuous.
You can have several more variants. In a monoid, you can define exponentiation $a^n$ where $n$ is a nonnegative integer. In a semigroup, you can define exponentiation where $n$ is a positive integer. In complexes (or an algebraically closed field), you can define multi-valued $a^{p/q}$ for a rational exponent. Cardinals and ordinals have their own exponentiations. The "continuous" exponent can be extended to complex numbers: when $a>0$ then you can define $\exp(b \ln a)$. Yet another exponentiation on complex numbers is multi-valued $\exp(b \operatorname{Ln} a)$.
All those operations are different - they have different domains. Mathematicians are unusually sloppy about which exponentiation they are talking about and use context-dependent $a^b$. (Some programming languages have multiple exponentiation operators to deal with this problem.)
sdcvvc
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"Here it is best to leave $0^0$ undefined, as otherwise the function will be discontinuous." I don't understand this. Which function would be discontinuous ? – Kasper Dec 15 '12 at 22:02
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@Kasper: If you extend the function $f:(0,\infty) \times \mathbb R \to \mathbb R$, $f(x,y)=x^y$ to the point $(0,0)$ it will no longer be continuous. – sdcvvc Dec 15 '12 at 22:02
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That argument is based on the "continuity rule" which says that if $f$ is discontinuous at a point $p$ then one should not define $f$ at $p$. However, that rule has been abandoned more than a century ago. – Mark Feb 12 '17 at 22:20
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@Mark Let me put it differently: a continuous function with a simple definition $a^b=\exp(b \ln a)$ would have to be extended with a special case and made discontinuous. This is ugly - I can't say "wrong" because it's hard to call definitions wrong. You can extend $1/x$ to $x=0$ so that (say) $1/0=0$, but it's not useful. All places I know where $0^0=1$ are useful use the discrete operator - show me a noncontrived application where the analytic operator is used and I'll change my mind. – sdcvvc Mar 12 '17 at 14:40
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@sdcvvc People assume $0^0=1$ all the time. Textbooks say that $1/(1-x) = \sum x^n$ for all $|x|<1$. Plug in $x=0$ and you find $1/(1-0)=0^0+0^1+\cdots$. Defining $0^0=1$ is not unusual or contrived, it simply acknowledges what we are already doing. It is the cleanest way to tell our students what an expression like $\sum x^n$ means. We should not tell our students that "sometimes it is this and sometimes it is that" because that's telling our students that math is inconsistent (which it isn't, if we simply let $0^0$ be $1$). – Mark Mar 12 '17 at 15:18
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@Mark $\sum x^n$ uses the discrete operator, not the analytic one (I think I made it clear in the answer because I referred to Taylor series). I asked you about the analytic one. I am not buying the argument about students - it's normal to give simplified accounts or Wittgenstein's ladders as a teaching tool. – sdcvvc Mar 12 '17 at 15:31
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Wherever one encounters $0^0$ (power series, polynomials, combinatorics, set theory, etc.) it is $1$. A good definition must reflect actual usage.
The continuity argument is bogus because, if it were taken seriously, it would also make floor(0) undefined.
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@Mark All the cases you mention use the discrete operator, and as I wrote in the answer, $0^0$ should be $1$ in this case. If there is actual usage of $x^y$ where $y$ varies continuously and $x=0$, then please show it. Otherwise you are making a simple exponentiation $\exp(y \ln x)$ (1) defined piecewise (2) discontinuous while it was continuous before the stitches (3) with no actual usage for the discontinuous part. None of (1)-(3) applies to floor (it's not defined piecewise, it was already discontinuous by its definition, it's actually used on integer arguments). – sdcvvc Mar 12 '17 at 20:59
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I rest my case - I think I've said enough and you can have the last word. – sdcvvc Mar 13 '17 at 23:42
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$0^0=1$, and "$0^0$" is an indeterminate form.
The fact that it's a well defined expression in no way conflicts with the fact that it's an indeterminate form.
$0^0=1$ because it's an empty product. Multiplying by no number is the same as multiplying by $1$; therefore when one multiplies by no number, the product is $1$.
It's indeterminate because one can let the pair $(x,y)$ approach $(0,0)$ along a path that makes the limit of $x^y$ equal to $5$ or to $1$ or to $\infty$, or to any of infinitely many other values.
If one approaches $(0,0)$ along any path that remains between two lines of positive slope, then the limit is $1$.
If $0^0$ were not equal to $1$, then the familiar expansion $$ e^z= \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{z!} + \cdots $$ would fail when $z=0$, since the first term is $\dfrac{0^0}{0!}$.
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The consensus is that if you are going to adopt a convention defining $0^0$, then it should probably be $0^0=1$.
The problem is that there is good reason for basic arithmetic operations on real numbers are continuous, and the real and complex exponentiation operators cannot be continuous at $0^0$.
The solution, IMO, is to honestly recognize that there are multiple exponentiation operators. All of the cases where the convention $0^0 = 1$ is useful are discrete: e.g. in a power series, where we are interested in monomials with integer exponents. The needs we have for discrete exponents are very different from the needs we have for continuous exponents.
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The answer is: The meaning is context-sensitive. This is surprising only if you assume that mathematical terms are context-insensitive. They are not. See x^y by Sam Derbyshire.
Julia Primes
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1It is context-sensitive in the sense that it is $1$ in contexts where no math errors were made. – Mark Feb 19 '17 at 20:20
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$\forall x \not= 0, x^0 = 1$
$\forall x > 0, 0^x = 0$
Those are either definitions or conventions chosen to extend formulas (just like you chose $0!=1$).
We can't have both functions $x\mapsto 0^x$ and $x\mapsto x^0$ continuous at $x=0$ no matter how we define $0^0$
Continuous functions are functions that commute with limit, ie $\lim f( x_n) = f( \lim x_n)$
And in this case it doesn't work for at least one of the cases.
xavierm02
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1Three errors: (1) It is not true that $0^x = 0$ for all $x \neq 0$. (2) Limit arguments are irrelevant for $0^0$ because $x^y$ is not continuous at $(x,y)=(0,0)$. (3) The last line "one of the cases does not work" only matters if you implicitly assume that it implies "both cases do not work". – Mark Feb 16 '17 at 21:27
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@Mark (1) Alright, $x>0$. (2) Precisely. I'm saying that it can't be made continuous because two induced functions have different limits at $(0,0)$. (3) See (2). – xavierm02 Feb 20 '17 at 12:49
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1OK, you fixed one error. However, your conclusion "it doesn't work for at least one of the cases" won't tell us anything about $0^0$ unless you combine it with faulty logic "not ($P$ and $Q) \Longrightarrow($not $P$) and (not $Q$)". The other conclusion $f(x,y)=x^y$ is discontinuous at the origin, this doesn't tell us anything either unless you equate "discontinuity" with "contradiction". – Mark Feb 20 '17 at 15:10
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@Mark I'm saying something stronger that $f$ being discontinuous at $(0,0)$: No matter what value you define $0^0$ to be, at least one of the two induced functions will be discontinuous. Would replacing "it doesn't work for at least one of the cases" by "it doesn't hold for at least one of the two functions" make things clearer? – xavierm02 Feb 20 '17 at 15:21
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1But how is this relevant? Just because you can write down a discontinuous function doesn't tell us anything about any number (If it does, you have to show how by listing the underlying implicit assumptions). – Mark Feb 20 '17 at 15:35
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Prop: Let $f:U\to \mathbb R$ be a continuous real function, $g: y \mapsto f(x_0, y)$ and $h:x\mapsto f(x,y_0)$ (where $x_0$ and $y_0$ are constant and the domains are the projections of $U$). Suppose that $g$ and $h$ can be extended to the closure of their domain while remaining continuous (and call the extended functions $F$ and $G$). Suppose also that $G(y_0)\not= H(x_0)$. Then $f$ can not be extended continuously on the closure of its domain. Proof: If it could, then $\lim_{y\to y_0} G(y) = \lim_{y\to y_0} F(x_0, y)=F(x_0,y_0) = \lim_{x\to x_0}F(x, y_0)=\lim_{x\to x_0}H(x)$. – xavierm02 Feb 20 '17 at 16:11
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You show "$f$ can not be extended continuously". From which one can not conclude: "$f$ can not be extended". – Mark Feb 20 '17 at 16:13
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@Mark A function can always be extended... But when you can, you want to preserve some sort of regularity. What makes you think that I concluded that "$f$ can not be extended?". – xavierm02 Feb 20 '17 at 16:19
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You gave the standard argument against defining $0^0$. The only reason this argument is effective is because many readers intuitively interpret "can not be extended continuously" as "can not be extended". Before the modern definition of a function, this intuitive interpretation was common, which is why many mathematicians decided to undefine $0^0$ in the 1800's. But just because it is historically understandable, that doesn't make it a valid argument. – Mark Feb 20 '17 at 17:34
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Define $f(x)=x^x$ and compute $\lim_{x\to 0} f(x)$.
Amihai Zivan
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@Ralph : Yes. That's a suggestion of a definition ; it is not universal. – Patrick Da Silva Dec 15 '12 at 21:26
:(– Nick T Dec 15 '12 at 23:28SELECT * FROM Posts WHERE Title like '%0^0%';(or similar) works to search for special characters. – Nick T Dec 16 '12 at 00:07