One of my teachers argued today that 0^0 = 1. However, WolframAlpha, intuition(?) and various other sources say otherwise... 0^0 doesn't really "mean" anything..
can anyone clear this up with some rigorous explanation?
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4The problem is that it's ambiguous: there are several things one might actually mean by writing $0^0$, and they don't agree. The most common meanings of the notation $0^0$ would make it undefined, indeterminate, or $1$. – Mar 05 '14 at 00:07
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1I like $0^0=1$. Makes sense for algebraic manipulations. – user2345215 Mar 05 '14 at 00:07
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2http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Scott Lawrence Mar 05 '14 at 00:07
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0^0 = (0^2)/(0^2) = 0/0 = undefined, no? – Dodgie Mar 05 '14 at 00:09
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1As this explains, it really depends on the context. – recursive recursion Mar 05 '14 at 00:11
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@Hurkyl what is the difference between "undefined" and "indeterminate" ? – terrible at math Mar 05 '14 at 00:11
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I'm kind of curious; can someone explain why this question was downvoted? – recursive recursion Mar 05 '14 at 00:13
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1@recursive recursion it's a duplicate – user2345215 Mar 05 '14 at 00:16
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"Undefined" is like gibberish: the symbols simply aren't allowed to be put together that way. "Indeterminate" is an adjective applied to limit forms. e.g. as $x \to 0$, the expression $x/x$ has the indeterminate form $0/0$, which means we can't determine the value of the limit of $x/x$ as $x \to 0$ simply by plugging in the limit of all the parts of the expression. – Mar 05 '14 at 00:59
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Short answer: It depends on your convention and how you define exponents.
Long answer: There are a number of ways of defining exponents. Usually these definitions coincide, but this is not so for $0^0$: some definitions yield $0^0=1$ and some don't apply when both numbers are zero (leaving $0^0$ undefined).
For example, given nonnegative whole numbers $m$ and $n$, we can define $m^n$ to be the number of functions $A \to B$, where $A$ is a set of size $n$ and $B$ is a set of size $m$. This definition gives $0^0=1$ because the only set of size $0$ is the empty set $\varnothing$, and the only function $\varnothing \to \varnothing$ is the empty function.
However, an analyst might not want $0^0$ to be defined. Why? Becuase look at the limits of the following functions: $$\lim_{x \to 0^+} 0^x = 0, \qquad \lim_{x \to 0} x^0 = 1, \qquad \lim_{x \to 0^+} (e^{-1/t^2})^{-t} = \infty$$ All three limits look like $0^0$. So when this is desired, you might want to leave $0^0$ undefined, so that it's a lack of definition rather than a discontinuity.
Typically this is resolved by:
- If you're in a discrete setting, e.g. considering sets, graphs, integers, and so on, then you should take $0^0=1$.
- If you're in a continuous setting, e.g. considering functions on the real line or complex plane, then you should take $0^0$ to be undefined.
Sometimes these situations overlap. For example, usually when you define functions by infinite series $$f(x) = \sum_{n=0}^{\infty} a_nx^n$$ problems occur when you want to know the value of $f(0)$. It is normal in these cases to take $0^0=1$, so that $f(0)=a_0$; the reason being that we're considering what happens as $x \to 0$, and this corresponds with $\lim_{x \to 0} x^0 = 1$.
Clive Newstead
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thanks a lot, this clears up a lot of the confusion and gives me insight to the nature of indeterminate forms. – terrible at math Mar 05 '14 at 00:17
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Consider the following sequences:
$0,0^\frac{1}{2},0^\frac{1}{3},0^\frac{1}{4},\cdots$
$(\frac{1}{2})^0,(\frac{1}{3})^0,(\frac{1}{4})^0,\cdots$
The first evaluates to a sequence of 0s that would imply a limit of $0^0\implies0$ as each term is zero.
The second evaluates to a sequence of 1s that would imply a limit of $0^0\implies1$ as each term is one.
However, as each value is different this would imply that there isn't a value for $0^0$ as generally one would think $0^a=0$ for any positive a, while $x^0=1$ for any non-zero x as a couple of identities that conflict here.
JB King
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1Interesting. I need this for a proof that $\exp(0) = 1$. I was writing out the definition in terms of the infinite series, but the first term is obviously $0^0$ . . . – terrible at math Mar 05 '14 at 00:14
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@JBKing I want to expand on this idea past just $0^0$ - We can take it to be 1, but I haven't seen such possibilities for other indeterminate forms: $\frac{\infty}{\infty} , \frac{0}{0},$ etc are there any other indeterminate forms in which we can reasonably assign a value when we need to? – terrible at math Mar 05 '14 at 00:23
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Ever look into L'Hopital's Rule? http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule may be useful for other indeterminate forms. $f(x)=\frac{k*sin(x)}{x}$ for any constant k evaluated in the limit as x tends to 0 will tend to the constant k. – JB King Mar 05 '14 at 00:29
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@terribleatmath, no problem. Just take $\mathrm{e}^0 = \lim_{x \to 0} \mathrm{e}^x$, then the $x^0 = 0$ all the way. – vonbrand Mar 05 '14 at 00:30
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One of my teachers argued today that $0^0 = 1$.
He probably meant that $\displaystyle\lim_{n\to0}n^n=1$, which is indeed true.
$0^0$ doesn't really "mean" anything...
Of course it does... It's just that its meaning depends on the given context, yielding different results for different cases, as has already been explained above... As opposed to, say, $1+1$, which is always $2$, being independent on context.
Lucian
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The teacher probably meant that $0^0=1$, which doesn't contradict anything, without mentioning limits, which are a different topic. – egreg Mar 05 '14 at 00:47