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This question already has an answer here:

We know that $0^0$ is indeterminate.

But if do this:

$$(1+x)^n=(0+(1+x))^n=C(n,0)\cdot ((0)^0)((1+x)^n) + \cdots$$

we get $$(1+x)^n=(0^0)\cdot(1+x)^n$$

So, $0^0$ must be equal to $1$.

What is wrong in this? Or am I mistaking $0^0$ as indeterminate?

Other threads are somewhat similar but not exactly the one I am asking.

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marked as duplicate by Asaf Karagila, Najib Idrissi, Surb, Christopher, user126154 Apr 21 '15 at 15:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ No, you get $0^0\cdot (1+x)^n$. Did you forget the $n$? $\endgroup$ – Thomas Andrews Apr 21 '15 at 12:53
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    $\begingroup$ And yes, $0^0=1$ and $0^0$ is an indeterminate form. Indeterminate doesn't mean undefined. $\endgroup$ – Thomas Andrews Apr 21 '15 at 12:56
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    $\begingroup$ @YvesDaoust I see it, because the first and last term are $a^3b^0$ and $a^0b^3$. ;-) You should explain why a “wrong convention” always gives the right result. I'm not talking about limits, which are a completely different matter: this is algebra and no limit is involved, so defining $0^0=1$ in this context is perfectly sound. $\endgroup$ – egreg Apr 21 '15 at 13:23
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    $\begingroup$ @AsafKaragila: or one of the list of answers linked to that one. $\endgroup$ – robjohn Apr 21 '15 at 13:29
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    $\begingroup$ @YvesDaoust I've yet to find a context in which $0^0=1$ fails. Can you mention one? Not limits, of course, we're not talking about them. $\endgroup$ – egreg Apr 21 '15 at 13:42
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There's a common yet subtle misconception among mathematics students that algebraic laws are prior to the numerical values that they describe. For example, "why" is it that:

$$5(3+2)=5\cdot 3 + 5\cdot 2\tag1$$

The tendency is to say, oh, this is true because of the distributive law: $a(b+c)=ab+ac$. But actually, equation (1) is true because it's true, not because of an abstract law. That is, you can actually calculate the left hand side and the right hand side and check that they're equal. And it's the fact that that always works that justifies the abstract law.

In your case, the binomial theorem is considered true because it always works when you plug in actual numerical values. You know how to calculate $(2+3)^8$ without the binomial theorem, but the result you get is consistent with the binomial theorem. To compute an expression using a general theorem, and then conclude that a specific numerical expression has a specific value is backwards reasoning.

If we're starting from the point of view that $0^0$ is undefined, then when we apply the binomial theorem to $(0+a)^n$ and realize that it's asking us to compute $0^0$, the conclusion is not that the binomial theorem is true all of the time, and that therefore $0^0$ must have a certain value, the conclusion is that therefore the binomial theorem does not apply to $(a+b)^n$ when $a$ or $b$ is zero. If we want it to, we have to come up with a definition for $0^0$, and then re-prove the binomial theorem, making sure that our proof is consistent with our new definition.

And, in fact, if we define $0^0=1$, then it's possible to do that. Your reasoning almost amounts to a proof that given that definition, the binomial theorem works, but it's important to recognize that you're verifying that a given definition leads to a given theorem, you are not concluding that a definition is "true" as a consequence of a theorem.

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    $\begingroup$ This is a great answer. $\endgroup$ – Simon Rose Apr 21 '15 at 13:14
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    $\begingroup$ This is really a better answer than mine, actually. I'd prefer that this be the accepted answer, because my answer has some fairly technical philosophical difficulties, while this really gets to the heart of the matter. $\endgroup$ – Ian Apr 21 '15 at 13:21
  • $\begingroup$ This is another version similar to the above answer: askamathematician.com/2010/12/… $\endgroup$ – Bristol Apr 21 '15 at 14:26
  • $\begingroup$ Perhaps you can add that with a proper rigorous definition of integral powers, it is most natural and very elegant to define the zeroth power to be $1$ because it represents the empty product or more abstractly the identity element in the group of multiplying transformations. I think the main problem is that too many students have never known a formal proof of the binomial theorem, and so they do not realize that even the statement already needs a formal definition of exponentiation, and the value of $0^0$ is thus already decided at that stage. $\endgroup$ – user21820 Apr 22 '15 at 6:40
  • $\begingroup$ Furthermore, the natural choice would be to use 'actual integers' in such a definition, which are exactly the same as integral powers of an invertible function, rather than the isomorphic copy within the reals, because then the binomial theorem would generalize trivially and automatically to other structures like general fields (not just powers of real numbers) and operators (such as $(R-x)^n$ where $R$ is the right-shift operator or $(D_x+D_y)^n$ where $D$ is the differential operator). $\endgroup$ – user21820 Apr 22 '15 at 6:46
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For the binomial theorem, the convention $0^0=1$ is natural, because that's exactly what is necessary for the binomial theorem to extend to $(x+0)^n$, which is basically what you noticed. In fact, in all "discrete" contexts I know of, $0^0=1$ is natural. Ultimately this interpretation boils down to the set theory interpretation, where $x^y$ is the number of functions from a set with $y$ elements to a set with $x$ elements. When $x=y=0$, there is one function from the empty set to itself, namely the empty set.

When dealing with limits, a $0^0$ limit form is equal to $1$ only when the base and exponent go to zero at essentially the same rate. If they don't, then you'll get something else. That is to say, the function $f(x,y)=x^y$, which is uniquely defined for positive $x$ and $y$, can't be continuously extended to $(0,0)$. We can see this by calculating $\lim_{x \to 0^+} x^x$ and $\lim_{x \to \infty} \left ( e^{-x} \right )^{1/x}$. You have probably seen that the former is $1$. For the latter, call the limit $y$, then $\ln(y) = \lim_{x \to \infty} \frac{1}{x} \ln \left ( e^{-x} \right ) = -1$, so $y=e^{-1} \neq 1$. Yet $y$ is clearly a $0^0$ form.

As a consequence of this disparity, all you can really do is either leave it completely indeterminate or carefully adopt a convention, making sure not to use it where it is not applicable.

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    $\begingroup$ The key is that there is a difference between "indeterminate form" and "undefined." That word "form" is important - it means that if you a limit of the form $0^0$, the limit is uncertain - we cannot determine it just knowing this. $\endgroup$ – Thomas Andrews Apr 21 '15 at 12:58
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    $\begingroup$ Your statement about "continuous contexts" does not make sense. $0^0$ by itself has nothing to do with limits, so it is irrelevant whether we can make quantities approach zero such that the exponentiation has some arbitrary limit or none at all. $\endgroup$ – user21820 Apr 21 '15 at 12:58
  • $\begingroup$ @user21820 At least if you are dealing with real numbers and not integers, our choice of definition (or lack thereof) of $0^0$ has everything to do with limits, because we are reluctant to define exponentiation in a discontinuous way. But I edited somewhat to clarify anyway. $\endgroup$ – Ian Apr 21 '15 at 13:01
  • $\begingroup$ @Ian What do you mean, "we," kemosabe? $\endgroup$ – Thomas Andrews Apr 21 '15 at 13:02
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    $\begingroup$ @Ian: I think you are quite mistaken. We don't by default have any continuous thing like $x^y$. It had always begun from discrete powers, followed by extension to rational powers, and finally to real powers. It happens that one can jump straight to the continuous version via the complex exponential, but that itself does not specify anything for $0$ because it is an essential singularity. So there is nothing to gain at all by refusing to define $0^0 = 1$, whereas it solves all the problem with discrete theorems like empty product, binomial theorem, ... $\endgroup$ – user21820 Apr 21 '15 at 13:39
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In the binomial expansion formula for $(a+b)^n$, the generic term is used for convenience so you don't have to write the first and last terms separately. The exponent of $0$ doesn't really appear if you expand the product; it is simply a convenient placeholder with the understanding that no factors with that base are present in the term.

So in this case, you would interpret $0^0$ as $1$. However this is only a matter of convention in this special case. That's the whole reason the form is indeterminate--in other cases, it is more natural to interpret $0^0$ as $0$. Considering the expressions $0^x$ and $x^0$ for very small positive $x$ illustrates the ambiguity, so the form can't be assigned a consistent value.

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    $\begingroup$ Actually, it would be a special case if we decided $0^0$ was undefined... It's not ambiguous just because it is not continuous. $\endgroup$ – Thomas Andrews Apr 21 '15 at 13:00
  • $\begingroup$ @ThomasAndrews: It is generally regarded as undefined precisely because any definition would have to be context-dependent. My point is that, in this context, it makes sense to define it as $1$, but that this definition can't be taken as universal, or as a "proof" that the symbol's value must be $1$, just because it works here. It is indeed ambiguous, because there are two continuity arguments for two different values, both plausible. That's ambiguity, right? $\endgroup$ – MPW Apr 21 '15 at 13:06
  • $\begingroup$ I disagree that it is more natural to interpret $0^0$ as $0$ "in other cases". If you are talking about $0^x$ for real $x$, it is a poor argument, since $0^x$ already cannot be defined for negative $x$, so why is it more natural for it to be $0$ for $x = 0$? If you are talking about $x^x$, complex exponentiation gives $x^x = e^{x \ln(x)}$ with some branch cut, which leaves it undefined at $x = 0$. But nothing goes wrong if we define it to be $1$ when $x = 0$. In fact, we do exactly the same kind of thing with the complex square-root, which would otherwise be undefined at $0$. $\endgroup$ – user21820 Apr 21 '15 at 13:10
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    $\begingroup$ Sorry, but in the decades I've been doing mathematics, I've never found a place where it's “more natural” to define $0^0=0$. Can you point to one, except the pretty useless function $x\mapsto 0^x$? $\endgroup$ – egreg Apr 21 '15 at 13:19
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    $\begingroup$ @columbus8myhw And why should this make $0^0=0$ to appear “natural”? I see no $0^0$ in any of the computations necessary for that limit. “Indeterminate form $0^0$” is just a phrase; if we were used to call it, more properly, “infinitesimal to infinitesimal” there would be no problem (where “infinitesimal” stands for “function with limit $0$”). $\endgroup$ – egreg Apr 21 '15 at 13:37
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The binomial theorem states that: $$(a+b)^n=\sum_k\binom nka^kb^{n-k}$$ (assuming I made no typo). What you noticed is basically that, when $a=0$, this only works if $0^0=1$.

More specifically: When $k=0$, you're supposed to evaluate: $$a^0b^n$$ when $a=0$. Now, while $0^0$ is an indeterminate form, it makes sense to assume that it's $1$ in this case, because $\displaystyle\lim_{a\to0}a^0=1$. In other words, while $0^0$ is indeterminate in general, in this case, it makes the most sense to take it as $1$.

At least, that's my understanding of it.

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  • $\begingroup$ I would be proven wrong if, for example, someone were to find a formula of the form $f(a,\dots)=\sum\limits_k(\dots k^a\dots)$, where the formula only works at the $a=0$ case if you let $k^a=1$ when $k=0$. $\endgroup$ – Akiva Weinberger Apr 21 '15 at 13:20
  • $\begingroup$ When $k=0$, you evaluate $b^n$, that's it. $\endgroup$ – Yves Daoust Apr 21 '15 at 14:03
  • $\begingroup$ @YvesDaoust What about that $a^0$ term, then? $\endgroup$ – Akiva Weinberger Apr 21 '15 at 14:17
  • $\begingroup$ Where would it come from ? Expand $(a+b)^2$ and try to find it. There is no $a^0$ factor. $\endgroup$ – Yves Daoust Apr 21 '15 at 14:30
  • $\begingroup$ @YvesDaoust From the $\sum_k\binom nka^kb^{n-k}$ where $k=0$. I think I see your point now — you're straight up saying that $(a+b)^n$ doesn't equal $\sum_k\binom nka^kb^{n-k}$ when $a=0$? That the formula isn't really true? $\endgroup$ – Akiva Weinberger Apr 21 '15 at 14:35
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When you establish the Binomial theorem, you multiply homogeneous factors $(a+b)$ together and never get any $0$ exponent.

$$(a+b)^1=a+b\\ (a+b)^2=a^2+2ab+b^2\\ (a+b)^3=a^3+3ab^2+3a^2b+b^3\\ \cdots$$

These are introduced artificially to form a compact expression, and this formula cannot serve to "prove" the value of $0^0$.

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  • $\begingroup$ Why this downvote ? What is wrong ? $\endgroup$ – Yves Daoust May 7 '15 at 14:17

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