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We all know that the domain of the constant function is $\mathbb{R}$

$$f(x)=c=c\ x^0$$ Is 0 is in the domain eventhought that $0^0 $ is undefined ?

And If we considered that $f(x)=c$ Is a different from the function $g(x)=x^0c$.

So the domain of $f$ is $\mathbb{R}$ and the domain of $g$ is $\mathbb{R-{\{0\}}}$.

And If we want to express the constant function in terms of the independent variable explicitly What is the exact definition ?

And Is it Logically correct that $f(x)=x^0c\neq c $ ?

Thank you...

Mahmoud Hassan
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  • why you write $x^0$??? There is no $x$ in the definition of a constant function. – Masacroso Mar 19 '17 at 23:01
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    On the contrary, we all know that $$g(x)=0=0^{|x|}$$for all $x\in\mathbb R$, thus $0^0=0$. – Simply Beautiful Art Mar 19 '17 at 23:02
  • Every function has independent and dependent variable So logically x must appear in the function . – Mahmoud Hassan Mar 19 '17 at 23:03
  • @SimplyBeautifulArt That's beautiful :p – Furrane Mar 19 '17 at 23:06
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    @topspin $x$ appears in the definition $f(x)=c$, just not on the right side of the equation. The only reason to write $f(x)=cx^0$ is to illustrate the pattern in standard form for polynomials, which is a very specific context. Otherwise it's just confusion for confusion's sake. – anon Mar 19 '17 at 23:15
  • There was a question very much like this not very long ago that got a huge amount of attention; do the answers there seem relevant to you? And you don't want to talk about the image of $0^0$, unless you're already agreeing that $0^0$ is a real number. – pjs36 Mar 19 '17 at 23:41

1 Answers1

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By convention :

$$\forall x \in \mathbb{R}, x^0=1$$

This is indeed valid for $x=0$.

Furrane
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  • There is a good reason for this convention: $$\lim_{x\to 0}x^0=1$$ – Masacroso Mar 19 '17 at 23:04
  • There is certainly not clear concensus on this "convention", see e.g. this question. Since the limit $$\lim\limits_{(x,y)\to (0,0)} x^y$$ does not exist, in some cases one leaves the expression undefined. Obviously, in this question -- even if it is left undefined -- it is a removable singularity, so it's not a big problem in any case. – Eff Mar 19 '17 at 23:10
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    It's not about concensus, but use cases. Indeed $\lim_{(x,y)\to (0,0)} x^y$ is undefined but apart for very specific cases like this one, we use that convention. It simplifies a lot of work where we would have to constantly treat 0 as a separate case. And the author question certainly falls under the general case. – Furrane Mar 19 '17 at 23:16