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Suppose $y=0$ in

$$(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$$

Then we get $\binom{n}{0}x^{n-0}0^0$ as the first term of the sum. We treat this as a 1, normally, though $0^0$ isn't well-defined. What am I missing? Or should $x$ and $y$ be required to be nonzero?

Electro
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    $0^0$ is well-defined, it's $1$. – Christopher Aug 14 '14 at 16:20
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    @user73985 No it's not. You'd have to define it beforehand. $0^n=0$ for all $n$ but $n^0=1$ for all $n$ so you'd have a contradiction. – Jam Aug 14 '14 at 16:26
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    http://www.wolframalpha.com/input/?i=0%5E0 – hanugm Aug 14 '14 at 16:27
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    To avoid this the binomial theorem makes sure that $x,y\ne 0$. Otherwise it is just a trivial power. – Ali Caglayan Aug 14 '14 at 16:27
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    $0^0$ should be understood as $1$. This makes the function $0^x$ discontinuous at $x=0$, but discontinuities are not a disaster, especially at an endpoint of the function's domain. The only reason some people say $0^0$ is undefined is to prevent calculus students from mistakenly thinking that, if $f(x)$ and $g(x)$ both approach $0$ when $x\to a$, then $f(x)^{g(x)}\to 0^0=1$. As long as we're not calculus students who assume everything is continuous, we should use $0^0=1$. – Andreas Blass Aug 14 '14 at 16:34
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    I don't know why people are worried about discontinuities in the context of discrete mathematics, which in general is not concerned with continuity issues. – Asaf Karagila Aug 14 '14 at 16:58

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It is often good to define $0^0=1$, but there are some caveats. Namely, $$ \lim_{x\to0}\lim_{y\to0}x^y = 1 \neq 0 = \lim_{y\to0}\lim_{x\to0}x^y. $$ In a combinatorial setting this issue does not arise.

If you want a reason for why $0^0=1$ "combinatorically", consider the following. For $n,m$ positive integers there are $n^m$ different functions from a set of $m$ elements to a set of $n$ elements. What if $n=0$ or $m=0$? If $m=0$ and $n\geq0$, there is only one function (the empty function). If $m>0$ and $n=0$, there is no function, since there is no possible value to assign to your $m$ points.

Joonas Ilmavirta
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The reason $0^0$ is not defined is that there is no consistent way of defining it - the limit of $y^0$ as $y \to 0$ is clearly $1$, while the limit of $0^y$ as $y \to 0^+$ is clearly zero. But it can be formally defined for any particular application.

Here if you let $y$ approach $0$ with $n$ fixed you discover that the limit you need to make the expression continuous in $y$ is $0^0=1$ - so you can formally state that this is the value you are taking for this application.

Mark Bennet
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  • What about $(x^n)'=nx^{n-1}$ for $n=1$ evaluated at $x=0$? – Electro Aug 14 '14 at 23:57
  • @Electro That is related. Also note that the formula is a generalisation, and where it does not give an answer you can always go back to the definition of the derivative and compute directly what the answer should be. – Mark Bennet Aug 15 '14 at 08:59
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In the sum $$(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k={n\choose0}x^ny^0+\sum_{k=1}^n \binom{n}{k}x^{n-k}y^k$$ the first term (which has been extracted above) ALWAYS has $k=0$; $k$ is not a variable (notice there is no $k$ on the LHS). The sum on the RHS has no restrictions on $x,y$, and the first term has the restriction $y\neq0$. But we take $$\lim_{y\to0}{n\choose0}x^n y^0={n\choose0}x^n$$ From this we can define $0^0=1$ so we do not have to take a limit everytime, and receive the same result. By the way it is approached in this context, it is the most natural way to define $0^0$. This is why, for example, we do not define $0^0=\lim_{y\to0}0^y=0$ here. In other contexts however, the just-mentioned definition may be more convenient or natural.

Pauly B
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