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Nov
4
comment What is the value of $\lim_{x\to 0} x^i$?
@amWhy it is insane because it is neither an algebraic expression, nor indeterminacy.
Nov
4
comment What is the value of $\lim_{x\to 0} x^i$?
@did I do not know, but many of the questions I ask indeed difficult.
Nov
4
comment What is the value of $\lim_{x\to 0} x^i$?
@DonAntonio I usually ask difficult questions that little people can answer. I prefer asnwering easy questions myself.
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
@Micah this is interesting information, it would be great if you could summarize this as an answer.
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
@Micah that is I was asking about. Whether the limits exist and whether they agree in all directions.
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
well at least any smooth function of one argument can be extended to dual numbers as well following the rule: $f(a+b\varepsilon)=f(a)+f'(a)b\varepsilon$. So $p^{a+b\varepsilon}=p^a+p^a \ln p \varepsilon$ and $(a+b\varepsilon)^p=a^p+p a^{p-1} b \varepsilon$ so the both notions of exponentiation are well defined for the most of the dual numbers.
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
After all if such things would not be defined there, the set possibly would not be called a number set.
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
I see your point but I believe that the dual numbers are defined enough well so that exponentiation $x^y$ (in non-boundary cases) is defined. Am I wrong? Does square function of constant e differ in dual numbers from exp(2)?
Nov
1
comment In dual numbers, what is the value of expressions $0^\varepsilon$ and $\varepsilon^{\varepsilon}$?
@Steven Stadnicki why not? Exponentiation is defined there
Oct
30
comment A definable hyperreal system
@Ian Mateus yes this is exactly what I saw before making this question.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
You are right but this also gives it as an axiom: faculty.uml.edu/klevasseur/courses/92.421/PS1_Fall2011.pdf
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Well the definition of recursive hyper operator just postulates it to be 1 at n=3, b=0 as a special case. It is just another formulation of the same axiom.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Well according to en.wikipedia.org/wiki/Exponential_field, the E(0)=1 axiom is required from a set to be considered exponential field. This hints at that it may be an independent axiom over real numbers.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
By the way, in abstract algebra $x^0=1$ is usually cansidered an axiom en.wikipedia.org/wiki/Exponentiation#In_abstract_algebra
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"It can be derived from the definitions/axioms for the natural number case." Can you please give a derivation from the standard axioms?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Re limits: usually, but not necessary. Where do you see it is necessary to define it so?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
Re the link: as I see the definitions on that page are in fact axioms or equivalent to axioms.
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"is not a well formed statement because the limit..." How on earth existence or the value of the limit makes the statement about value of a function in a point well formed or not?
Oct
29
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
"true statement because it is an empty product" that empty product is equal 1 an axiom? If not, how to prove it?
Oct
28
comment Is $0^0=1$ postulate independent of all other axioms of complex numbers?
@Graphth Gnome calculator on Linux. Interestingly, the KDE calculater for a while returned 0 while Gnome calculator returns 1. The KDE calculator has been fixed since following a bugreport.