Math Help - Natural Domain

1. Natural Domain

Im trying to work out how to find the natural domain of a function.
Does any one know the steps I should take with some examples provided

2. If a function is defined by a formula and there is no domain explicitly stated, then the domain consists of all real numbers for which the formula makes sense. and the function has a real value. This is the natural domain.

Examples:

$f(x)=x^{2}$

This makes sense and has values for all real values of x. Therefore, the domain is ${(-\infty, \infty)}$

$f(x)=2+\sqrt{x-1}$

f is undefined at x=1, because negative numbers do not have square roots.

The natural domain is $[1,+\infty)$

3. Originally Posted by galactus
If a function is defined by a formula and there is no domain explicitly stated, then the domain consists of all real numbers for which the formula makes ense. and the function has a real value. This is the natural domain.

Examples:

$f(x)=x^{2}$

This makes sense and has values for all real values of x. Therefore, the domain is ${(-\infty, \infty)}$

$f(x)=2+\sqrt{x-1}$

f is undefined at x=1, because negative numbers do not have square roots.

The natural domain is $[1,+\infty)$
Natural Domain is a concept from Complex analysis; MathWorld describes it
thus:

"The natural domain of a function is the maximal chain of domains on which it
can be analytically continued to a single-valued function.

RonL

4. Natural domain is also the definition I provided. I just looked it up in a calculus text and the definition was virtually the same as mine.

I suppose it depends on the subject of study.

Gary, what class are you in?.

5. Originally Posted by galactus
Natural domain is also the definition I provided. I just looked it up in a calculus text and the definition was virtually the same as mine.

I suppose it depends on the subject of study.

Gary, what class are you in?.
Maybe a recent invention in calculus - its not known to Morris Kline.

RonL

6. Originally Posted by CaptainBlack
Maybe a recent invention in calculus - its not known to Morris Kline.

RonL
Interesting online definition for real functions (given here ):

Definition 1
Given a function f, the domain of f is the set of numbers x for which
f(x) is defined. For this course, being defined means being a real number
(as opposed to a complex number).

In the cases where the function is provided as an expression of the
independent variable (and no explicit domain is given), then the domain is
the largest subset of the real numbers for which the expression is defined.
In this case, we will often refer to this set as the natural domain of the
function. In cases where the context of the problem limits the domain
to a smaller set of numbers than the natural domain, we will refer to
this smaller set as the contextual domain or problem domain.

If g and f are functions for which the domain of g is a subset of the
domain of f and for which f(x) = g(x) for every x in the domain of g,
then we will say that f is an extension of g or that we have extended g
to obtain f.

The range of a function describes those numbers which can be written
as f(x) for x in the domain of f.

A target of a function describes any set which contains the range of the
function
.

Interesting because it is a redundant definition, as it is constructed to
be a synonym for the way the author wants to define the domain for
a function given by a formula

RonL

7. Originally Posted by CaptainBlack
Natural Domain is a concept from Complex analysis; MathWorld describes it
thus:

"The natural domain of a function is the maximal chain of domains on which it
can be analytically continued to a single-valued function. "

RonL
Nice, nice, nice. Never thought of it that way. How elegant!

8. Here is how I think of functions.

A real function is,
$f:\mathbb{R'}\to\mathbb{R''}$
where,
$\{\}\subset \mathbb{R'}\subseteq \mathbb{R}$
$\{\}\subset \mathbb{R''}\subseteq \mathbb{R}$
The set $\mathbb{R'}$ is called the domain.
The set $\mathbb{R''}$ is called the co-domain.
The set,
$\phi[\mathbb{R'}]$ is called the range.

Therefore,
$f=g$
if and only if,
there domains are equal AND,
$f(x)=g(x)$ for any $x$ in domain.

9. Originally Posted by ThePerfectHacker
Here is how I think of functions.

A real function is,
$f:\mathbb{R'}\to\mathbb{R''}$
where,
$\{\}\subset \mathbb{R'}\subseteq \mathbb{R}$
$\{\}\subset \mathbb{R''}\subseteq \mathbb{R}$
The set $\mathbb{R'}$ is called the domain.
The set $\mathbb{R''}$ is called the co-domain.
The set,
$\phi[\mathbb{R'}]$ is called the range.
$\phi$ appears out of the blue here. What you want. I imagine, is the image of
$\mathbb{R}'$ under $f$, which I would ususly denote by $f(\mathbb{R}')$ or some such.

RonL

10. Originally Posted by CaptainBlack
$\phi$ appears out of the blue here. What you want. I imagine, is the image of
$\mathbb{R}'$ under $f$, which I would ususly denote by $f(\mathbb{R}')$ or some such.
That is exactly it. But my text on abstract math uses $\phi$ instead of $f$ and uses [ ] for an imagine instead of ( ). I hope this is the traditional way, I would be very angry if I find out most of the notations and definitions in my algebra book are uncommon. That would mean that I would have to almost relearn the entire theory. It certainly does not agree with Schaums' outline on group theory, but that is an outline it cannot be trusted.

11. Originally Posted by ThePerfectHacker
That is exactly it. But my text on abstract math uses $\phi$ instead of $f$ and uses [ ] for an imagine instead of ( ). I hope this is the traditional way, I would be very angry if I find out most of the notations and definitions in my algebra book are uncommon. That would mean that I would have to almost relearn the entire theory. It certainly does not agree with Schaums' outline on group theory, but that is an outline it cannot be trusted.
Yes but does it use $\phi$ consistently throughout for the
function or does it explicitly use the notation of $f$ for a function
on the domain, and $\phi$ for the corresponding function on the
power set of the domain?

The matter of which kind of brackets are employed is of no great significance.

(to mis-quote Gauss - its the notions not the notation that are important)

RonL

12. Originally Posted by CaptainBlank
Yes but does it use $\phi$ consistently throughout for the
function?
Yes.

The matter of which kind of brackets are employed is of no great significance.
Of course not. But I am speaking about an accepted notation otherwise it is difficult to understand eachother. For that reason some times (rarely) and am not able to follow an algebra text because it uses old or unusual notations and do not know what it refers to.

(to mis-quote Gauss - its the notions not the notation that are important)
Hope you will appreciate this because it is math histroy.
In the book, Meditationes Algebraicae of 1770, the English mathematician Edward Waring (you know the famous waring problem) stated several theorems. One of which was by his student John Wilson on the Wilson Theorem (about primes and factorials). To this Waring said,
"Theorems of this kind will be hard to prove, because of the absence of a notation of express prime numbers." When Gauss read the passage he said to himself "notationes versus notiones".
(Source: "Elementary Number Theory" by David Burton, rephrased in my own words).

I have difficult believing that Gauss can say something like that because the purpose of such a quote is to say something important and to be poetic simultaneously. But it is poetic in English, can it be poetic in German? Probably not. Nothing in German is poetic.