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

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- Aug 23rd 2006, 08:11 PMgary wartonNatural 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 - Aug 24th 2006, 05:11 AMgalactus
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:

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

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

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

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

The natural domain is $\displaystyle [1,+\infty)$ - Aug 24th 2006, 05:13 AMCaptainBlackQuote:

Originally Posted by**galactus**

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.

SEE ALSO: Analytic Continuation, Domain, Natural Boundary"

RonL - Aug 24th 2006, 05:19 AMgalactus
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?. - Aug 24th 2006, 05:26 AMCaptainBlackQuote:

Originally Posted by**galactus**

RonL - Aug 24th 2006, 05:37 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

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 - Aug 24th 2006, 11:44 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- Aug 24th 2006, 11:49 AMThePerfectHacker
Here is how I think of functions.

A*real function*is,

$\displaystyle f:\mathbb{R'}\to\mathbb{R''}$

where,

$\displaystyle \{\}\subset \mathbb{R'}\subseteq \mathbb{R}$

$\displaystyle \{\}\subset \mathbb{R''}\subseteq \mathbb{R}$

The set $\displaystyle \mathbb{R'}$ is called the domain.

The set $\displaystyle \mathbb{R''}$ is called the co-domain.

The set,

$\displaystyle \phi[\mathbb{R'}]$ is called the range.

Therefore,

$\displaystyle f=g$

if and only if,

there domains are equal AND,

$\displaystyle f(x)=g(x)$ for any $\displaystyle x$ in domain. - Aug 24th 2006, 11:59 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle \mathbb{R}'$ under $\displaystyle f$, which I would ususly denote by $\displaystyle f(\mathbb{R}')$ or some such.

RonL - Aug 25th 2006, 04:25 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- Aug 25th 2006, 07:42 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

function or does it explicitly use the notation of $\displaystyle f$ for a function

on the domain, and $\displaystyle \phi$ 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 - Aug 25th 2006, 08:52 AMThePerfectHackerQuote:

Originally Posted by**CaptainBlank**

Quote:

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

Quote:

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

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.