How would you answer? When I opened Photomath, it said any non-zero expression raised to the power of zero equals one.
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Also it extends $a^b = e^{b \ln a}$ for $b>0$ continuously to the case $b=0$. – nicomezi Apr 24 '21 at 10:59
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@ultralegend5385 In some sense you are right, either we must begin with $a^0=1$ or with $a^{-m}=\frac{1}{a^m}$. Because of the empty product, $a^0=1$ is a natural definition and together with the power rules we can then derive $a^{-m}=\frac{1}{a^m}$. This way we also have established $a^0=1$ for negative $a$. – Peter Apr 24 '21 at 11:00
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@Peter: Yes, I was trying to say the same thing that $b^0=1$ is a definition, and all of its derivations are wrong because the domain of the identities they use doesn't include zero. – ultralegend5385 Apr 24 '21 at 12:42
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Basically it's a definition. It's not a deductive consequence of simpler facts; it's how we choose to interpret $a^0$.
This is not to say that it is just random, though. It's a particularly useful definition because it's the unique value for $a^0$ that makes makes the rule $$ a^{b+1} = a\cdot a^b $$ hold when we set $b=0$. Since that rule definitely holds in the "obvious" case when $a$ and $b$ are positive integers, it is useful to get it to hold as widely as we can make it. This is also the motivation for later defining $a^{-n} = \dfrac1{a^n}$.
This reasoning does not hold with the same force in the case where $a=0$, because then the above rule just requires $0^1 = 0\cdot 0^0$, which will be true automatically no matter what we take $0^0$ to be. That's the reason why the definition in some books deliberately only defines $a^0$ for nonzero $a$. It is actually very common (but not universal) to define the value of $0^0$ to be $1$ too, but that is for more subtle reasons than this.
Troposphere
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Sometimes $0^0$ is defined to be $0$ (in the context of $\ell_0$ "norm", and this is the only one I am aware of) – nicomezi Apr 24 '21 at 11:04
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@nicomezi Usually, $0^0$ is condsidered to be $1$ which has various advantages (for example to establish the binomial theorem without having to rule out zeros). Some computational tools also use it, but there is in fact no convention about $0^0$ – Peter Apr 24 '21 at 11:07
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(+1) Because it says that the rule is a definition. There are lots of resources out there claiming that they can "prove" it, which just becomes false as I said in my comment on the post. – ultralegend5385 Apr 24 '21 at 12:45
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@nicomezi: the only mentions I can find of "$\ell_0$ norm" say it's the number of nonzero coordinates, but I think one would need more than $0^0$ trickery to get that out of $(|x_1|^0+|x_2|^0+\cdots+|x_n|^0)^{1/0}$ ... – Troposphere Apr 24 '21 at 13:17
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I do not understand your remark. It just a convention I saw multiple times to avoid having to write the $\ell_0$ norm as a limit and you can just write $| \mathbf x |_0 = \sum_k |x_k|^0$. It is just for convenience. The definition with the limit I have now deleted does not depend on the convention you chose for $0^0$ (hence my deletion) if it is what you meant. – nicomezi Apr 24 '21 at 13:53
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@nicomezi: I think my point is that it does not make much sense to me to call that function $\ell_0$ in the first place. Usually the $\ell_p$ norm is $\bigl(\sum_k |x_k|^p\bigr)^{1/p}$ -- to get your expression one wouldn't just need to set $p=0$ but also ignore the $p$th root around the sum. – Troposphere Apr 24 '21 at 13:59
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This is not "my" expression, it is commonly defined like that in the sparse signal recovery theory. This book get rid of the $p$th rooth when defining the $\ell_0$ "norm". The definition given in it is as a limit $| \mathbf x |0 = \lim{p \to 0} \sum_k |x_k|^p$ (as you can see, no $p$th root) so there is no need to rely on convention for $0^0$. But sometimes we just define $0^0=0$ so we can get rid of the limit for convenience. – nicomezi Apr 24 '21 at 14:02