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Are there mathematical areas/situations in which defining $0^{0}$ as $0$ is useful/sensible and convention (in contrast to the "common" definition as $1$) ?

user2103480
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    Cheeky answer: How about in the zero ring? –  Apr 29 '15 at 21:37
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    Even in the zero ring, it's still basically 1, but now the multiplicative identity is the same as the additive identity. – jgon Apr 29 '15 at 21:38
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    I cannot think of any such situation. You may be interested in reading through the reasons why $0^0$ is commonly taken as $1$ here. Depending on context, it is taken to be either $1$, indeterminate or undefined/nonexistent; but never $0$. – Mufasa Apr 29 '15 at 21:40
  • @jgon That's why I said it was the cheeky answer. –  Apr 29 '15 at 21:42
  • @Mufasa Because I read those reasons I asked myself if there was any context as described above @ others: Very interesting! Never heard of it before as I'm still in school – user2103480 Apr 29 '15 at 21:45

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The $p$-norm is a generalization of the standard Euclidean norm for vectors. It is defined as:

$${\parallel\mathbf{x}\parallel}_p = \left(\sum_{i = 1}^{n} \left|x_i\right|^p\right)^{1/p}$$

This is for $p > 0$. What about $p = 0$?

Well, maybe we can call that the $0$-norm. The $0$-norm, is very useful in measuring the "sparsity" of a vector. A vector is $s$-sparse if only $s$ of its components are non-zero. It's useful to have $0^0$ be defined as $0$, while any other $r^0$ (where $r \neq 0$) is $1$. Okay, so the $0$ norm is not quite a $p$-norm (because we drop the $1/p$, but it is analogous. Actually, the zero norm is not even a norm (it's not scalable). Oops.

Anyway, so the $\ell_0$-norm (the common name for this measurement), for measuring the sparsity of a vector, looks something like:

$$ {\parallel\mathbf{x}\parallel}_0 = \left(\sum_{i = 1}^{n} \left|x_i\right|^0\right)$$

Why is sparsity a big deal? Well, compressed sensing is one good reason why.

Basically, compressed sensing tells us that under certain conditions (the signal is sparse, and the sensing matrix is randomly distributed), we can reconstruct a signal with even less information than is required by Nyquist-Shannon theorem.

bzm3r
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