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Regarding the finding of domains of functions

by d1ngell
Tags: function domain
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d1ngell
#1
Sep9-13, 01:33 PM
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First of all, I would like to appologize if I'm posting this in the wrong section, although I believe it to fit this area.

I'm usually not convinced when books claim to have found the natural domain of certain functions. For instance, this book I've been reading has defined the natural domain of a function as the largest set of elements whose range is in the real numbers set.

Ever since I've been having trouble accepting that the domain of functions such as f(x) = x˛ is ℝ.

Should I give [itex]\sqrt{-1}[/itex] as input value, the function would return -1, which is a real number. Therefore, I would be taken into accepting that the domain would be at least ℝ[itex]\cup[/itex]{[itex]\sqrt{-1}[/itex]}, for its range is still ℝ. What should be the bound when looking for a function's domain?

Thanks in advance,
d1ngell
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lurflurf
#2
Sep9-13, 01:53 PM
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It is silly that books say "Find the domain of this function." The definition of a function includes the domain. Your approach is not much better. By defining the function for all objects for which it makes sense we expand the function in a way that is not necessarily helpful. For example if we know x is real it is not helpful to define x^2 for non-real x. We might go further and define x^2 for matrices and all sorts of other object which is still not helpful. The point your book is making (poorly) is that given a set that contains the domain we can find the domain by throwing out the elements for which the function is undefined. Often we want to define functions on subsets of the reals, so we define it for all reals that are reasonable.
d1ngell
#3
Sep9-13, 10:18 PM
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What you're saying is that the domain of a function may vary depending upon application? I can see that, for example, when calculating the area of a circle given a radius. A(r) = πr˛, and the domain would be [itex]\Re[/itex][itex]^{+}_{*}[/itex]

I'm sorry if I'm bothering by asking, but I have had trouble understanding the following: "The point your book is making (poorly) is that given a set that contains the domain we can find the domain by throwing out the elements for which the function is undefined."

What I understand from that is that, disconsidering the unhelpfulness of the expansions of the functions, the [itex]\Re[/itex], commonly called domain of functions such as f(x) = x˛, would not be in fact the largest set of numbers to whom the function returns real values, hence not being the natural domain of the function.

(I would like to apologize once again if I'm sounding stupid. I'm brazilian so I eventually have a little bit of difficulty understanding some things. Rephrasing usually solves it.)

chiro
#4
Sep9-13, 11:10 PM
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Regarding the finding of domains of functions

The domain refers to inputs and the range/co-domain refers to outputs.

That's pretty much all there is to worry about.
lurflurf
#5
Sep10-13, 12:18 AM
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Do you know Portuguese? Do you live near a beach? To define the natural domain we need to start with a set of possible numbers for the domain and range. I take it you book is using the real numbers for the domain and range. The complex numbers and integers are other possible choices. In the real case the natural domain is the set of all real numbers x for which f(x) is real.

for x2
If the starting set is real numbers the natural domain is all real numbers
If the starting set is complex numbers the natural domain is all complex numbers
If the starting set is complex numbers, but we require the function to be real the natural domain is a all real and pure imaginary numbers.

The trouble when we ask about all elements such that x2 is real is where do you stop? Besides real and imaginary numbers we could have dual numbers, split complex numbers, quaternions, and any number of other types of numbers. Usually we have some number system in mind like real or complex numbers and we do not care about the values of the function for other types of numbers. So to answer your original question is that $$\sqrt{-1}$$ should be in the domain if you are working with complex numbers.
HallsofIvy
#6
Sep10-13, 09:29 AM
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I strongly suspect that your text book had already, implictely or explicitely, said that it was talking about real valued functions defined on the real numbers.
d1ngell
#7
Sep10-13, 02:19 PM
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I think chiro didn't really understand my question, which has been properly solved by lurflurf with HallsofIvy's contribution. Now I understand what you're saying. As HallsofIvy mentioned, my book does indeed talk about real valued functions, although I never really thought about that as being an inferrence that I should always consider the real numbers as the largest set when inspecting all the possible input values for the functions. Thank you for your answers.

@lurflurf
Yes, Portuguese is my main language. I currently live in the brazilian capital, Brasília, which isn't a costal city. I have lived for about 8 months in front of the beach, though.
Mark44
#8
Sep10-13, 02:41 PM
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Quote Quote by lurflurf View Post
It is silly that books say "Find the domain of this function." The definition of a function includes the domain.
But the domain might not be given explicitly. At the precalculus level it's reasonable to ask for the domain of the function f(x) = √(3 - x2).
Quote Quote by lurflurf View Post
Your approach is not much better. By defining the function for all objects for which it makes sense we expand the function in a way that is not necessarily helpful. For example if we know x is real it is not helpful to define x^2 for non-real x. We might go further and define x^2 for matrices and all sorts of other object which is still not helpful. The point your book is making (poorly) is that given a set that contains the domain we can find the domain by throwing out the elements for which the function is undefined. Often we want to define functions on subsets of the reals, so we define it for all reals that are reasonable.


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