Function f(1/x)

lex said:
@JeffM I should have added that often people use the phrase 'the inverse of a function' to mean 'the inverse function', which I'm sure is perfectly acceptable if the context allows.
However e.g. in allowing my trivial proof that for a function [MATH]f, \,(f^{-1})^{-1} = f[/MATH] it is very convenient to use 'inverse' to mean inverse as a relation, since we do not care and it doesn't matter whether the inverse of [MATH]f[/MATH] is itself a function.
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Yes. I probably was one of those people, but I’ll try not to be in the future. For one thing, it is useful to keep in mind that there are many types of inverse so being exact in which kind you are talking about can only improve communication.

I am still a little perturbed by the implication from jomo that the square root function does not have an inverse function. I keep thinking that this can be cured with some language about domains.
 
@JeffM
You are right to be perturbed.
When in doubt about what something is, test it against the definition.
A function is a set of ordered pairs, and a set, such that [MATH] ( (x,y)\in f \text{ and }(x,y') \in f ) \rightarrow y=y'[/MATH]
Is the inverse of [MATH]f: (f \text{ being defined by) } f(x)=\sqrt{x}, x≥0[/MATH] a function?
The function [MATH]f=\{ (x,\sqrt{x}): x≥0\}[/MATH][MATH]f^{-1}=\{ (\sqrt{x},x): x≥0\}[/MATH]
Is [MATH]f^{-1}[/MATH] a function?
It is a set of ordered pairs. Is it true that for each first number there is a unique second number? (To express it informally).
Yes! (For, otherwise [MATH]\hspace1ex \exists \text{ distinct } [MATH][/MATH] x\text{ and } y ≥0: \sqrt{x}=\sqrt{y} )[/MATH]

Therefore [MATH]f^{-1}[/MATH] is a function.
That decides the issue.

Now that we know, we can look back at:
Jomo said:
Note that if f(g(x)) =x does not mean that g(f(x))= x. In this case f and g are not inverses of each other.
Consider f(x) = x^2 and g(x) = sqrt(x). We have f(g(x)) =x but what does g(f(x))=?
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That simply means that [MATH]f[/MATH] is not the inverse of [MATH]g[/MATH], not that [MATH]g[/MATH] doesn't have an inverse.
[MATH]f(x)=x^2[/MATH] is not a function (It's not a set of ordered pairs...)
[MATH]f=\{(x,f(x)): f(x)=x^2, x\in \mathbb{R}\}[/MATH] is presumably the function
[MATH]g=\{(x,g(x)): g(x)=\sqrt{x}, x≥0\}[/MATH] (presumably)
(We can easily check that f is not the inverse of g, by reversing the elements of g to get [MATH]g^{-1}[/MATH]. Now f has elements where the first value is negative. No such elements exist in [MATH]g^{-1}[/MATH], therefore they are different sets. [MATH]f ≠ g^{-1}[/MATH]. The question is settled).

Now looking at the issue from the perspective of composite functions:
[MATH]f\circ g=\{(x,x): x≥0\}[/MATH][MATH]g\circ f=\{(x,|x|): x\in \mathbb{R}\}[/MATH][MATH]g\circ f[/MATH] has elements of the form (x,-x), when x is negative, therefore f, g fail this test for being inverses
(i.e. that [MATH]f \circ g (x)=x= g \circ f (x)[/MATH] for values of x in their respective domains).

However the function [MATH]h=\{(x,h(x): h(x)=x^2, x≥0\}[/MATH] is the inverse function of [MATH]g=\{(x,g(x)): g(x)=\sqrt{x}, x≥0\}[/MATH](We can very easily check that by writing the inverse - reversing the pairs - and seeing that it is the other function).
However looking at the issue of [MATH]h \circ g \text{ and } g \circ h[/MATH]
[MATH]h\circ g=\{(x,x): x≥0\}\\ g\circ h=\{(x,x): x≥0\}\\ h \circ g (x)=x= g \circ h (x)[/MATH] for their respective domains.
(They also happen to be identical functions).
 
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lex said:
@JeffM
You are right to be perturbed.
When in doubt about what something is, test it against the definition.
A function is a set of ordered pairs, and a set, such that [MATH] ( (x,y)\in f \text{ and }(x,y') \in f ) \rightarrow y=y'[/MATH]
Is the inverse of [MATH]f: (f \text{ being defined by) } f(x)=\sqrt{x}, x≥0[/MATH] a function?
The function [MATH]f=\{ (x,\sqrt{x}): x≥0\}[/MATH][MATH]f^{-1}=\{ (\sqrt{x},x): x≥0\}[/MATH]
Is [MATH]f^{-1}[/MATH] a function?
It is a set of ordered pairs. Is it true that for each first number there is a unique second number? (To express it informally).
Yes! (For, otherwise [MATH]\hspace1ex \exists \text{ distinct } [MATH][/MATH] x\text{ and } y ≥0: \sqrt{x}=\sqrt{y} )[/MATH]

Therefore [MATH]f^{-1}[/MATH] is a function.
That decides the issue.

Now that we know, we can look back at:


That simply means that [MATH]f[/MATH] is not the inverse of [MATH]g[/MATH], not that [MATH]g[/MATH] doesn't have an inverse.
[MATH]f(x)=x^2[/MATH] is not a function (It's not a set of ordered pairs...)
[MATH]f=\{(x,f(x)): f(x)=x^2, x\in \mathbb{R}\}[/MATH] is presumably the function
[MATH]g=\{(x,g(x)): g(x)=\sqrt{x}, x≥0\}[/MATH] (presumably)
(We can easily check that f is not the inverse of g, by reversing the elements of g to get [MATH]g^{-1}[/MATH]. Now f has elements where the first value is negative. No such elements exist in [MATH]g^{-1}[/MATH], therefore they are different sets. [MATH]f ≠ g^{-1}[/MATH]. The question is settled).

Now looking at the issue from the perspective of composite functions:
[MATH]f\circ g=\{(x,x): x≥0\}[/MATH][MATH]g\circ f=\{(x,|x|): x\in \mathbb{R}\}[/MATH][MATH]g\circ f[/MATH] has elements of the form (x,-x), when x is negative, therefore f, g fail this test for being inverses
(i.e. that [MATH]f \circ g (x)=x= g \circ f (x)[/MATH] for values of x in their respective domains).

However the function [MATH]h=\{(x,h(x): h(x)=x^2, x≥0\}[/MATH] is the inverse function of [MATH]g=\{(x,g(x)): g(x)=\sqrt{x}, x≥0\}[/MATH](We can very easily check that by writing the inverse - reversing the pairs - and seeing that it is the other function).
However looking at the issue of [MATH]h \circ g \text{ and } g \circ h[/MATH]
[MATH]h\circ g=\{(x,x): x≥0\}\\ g\circ h=\{(x,x): x≥0\}\\ h \circ g (x)=x= g \circ h (x)[/MATH] for their respective domains.
(They also happen to be identical functions).
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Thank you. My intuition about domains was correct. The definition of a function must include the domain obviously.
 
Yes. It's often a matter of restricting a domain.

To define a function you simply need to specify what ordered pairs it contains.
This can be done by listing them: e.g. f={(1,2), (3,4), (7,-2)}
or by giving a recipe for generating them i.e. giving the set of 'first values' and a way of producing the corresponding 'second values':
e.g. [MATH] f=\{(x,f(x))\}, x\in \text{ D(omain) }, f(x)=x^2 \hspace1ex[/MATH] (the convention being if only the rule is specified, then the domain is taken to be the largest possible for which the rule 'makes sense'),
or by drawing a line on a page (a 'graph') - in theory, if not in practice
f={(p,q): p, q prime}
in words: f is the set of ordered pairs of non-zero real numbers, and their reciprocals, (in that order)
or any other way you like.
 
lex said:
That simply means that [MATH]f[/MATH] is not the inverse of [MATH]g[/MATH], not that [MATH]g[/MATH] doesn't have an inverse.
[MATH]f(x)=x^2[/MATH] is not a function (It's not a set of ordered pairs...)
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Yes, you are correct that f is not the inverse of g, not that g does not have an inverse. Thanks for pointing that out!
Although I am sure that you are correct, I am missing why f(x)=x^2 is not a function. It passes the vertical line test!
 
JeffM said:
@Jomo I am not arguing. I am curious. I restricted my more formal proof to intervals that were monotonically increasing or decreasing. So what I view as a theorem applicable to certain domains, you view as a definition. In other words, your definition denies that it is proper to say that the square root function has an inverse at all.

I suspect that we are both missing something.
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We never argue! I agree with everything you wrote. Nice job.
I am just pointing out, coincidentally after your post, that one must show that both f(g(x))= g(f(x)) = x (or just show that one composition = x and consider both graphs)
 
I am not sure how much of this thread helped the original poster, but the contributions of @Cubist and @Jomo and especially @lex certainly educated ME. Having had only a basic education in mathematics (I went to a school that essentially taught nothing but languages and mathematics through applied calculus and then went to Columbia, which back then demanded that you have some competence in mathematics), there is always a possibility that I shall stumble even when I think I know what I am talking about.
 
Jomo said:
Although I am sure that you are correct, I am missing why [MATH]f(x)=x^2[/MATH] is not a function. It passes the vertical line test!
Click to expand...
Yes, that statement certainly needs clarification!
If we think of a function as a set of ordered pairs, clearly a rule e.g. [MATH]f(x)=x^2[/MATH] does not fulfil that definition, not being a set, for one thing! It doesn't specify a function either, as we still don't know what the ordered pairs are. We know how to calculate the 'second value', should we happen to know a 'first value'. The point I was making is that there are many functions f, with the 'rule' [MATH]f(x)=x^2[/MATH], but with different domains and in fact that is the very issue in this case. A function with rule [MATH]f(x)=x^2[/MATH] and an appropriate domain, will actually be the inverse of [MATH]g: g(x)=\sqrt{x}, x≥0[/MATH].
Of course we all speak of [MATH]f(x)=x^2[/MATH] as a function, and that's acceptable e.g. when the domain doesn't particularly matter. In fact when we don't specify the domain, the convention is that we mean it to be the set of all values for which the rule makes sense, but it's worth remembering the function has not explicitly been fully specified.
Here we might be tempted to say that we need to state a rule and a domain to specify a function, but even that is not true, as indicated in post #25.
I hope that sounds a bit more acceptable.

Just as I mention and re-read post #25 I notice f = {(p,q) with p and q prime}. This of course is not a function, only a relation. Apologies for that.
f={(p,q): p<q, p and q consecutive primes} is a function.
 
@JeffM
There's not much evidence of stumbling.
We all stumble however - as I have just evidenced. Humanum est errare.

I see John Wesley expressed what you are saying - "although every man necessarily believes that every particular opinion which he holds is true (for to believe any opinion is not true, is the same thing as not to hold it); yet can no man be assured that all his own opinions, taken together, are true - every thinking man is assured they are not, seeing humanum est errare et nescire..."

It's worth thrashing these issues out; even quite fundamental ones such as what a function is.
 
lex said:
@JeffM
There's not much evidence of stumbling.
We all stumble however - as I have just evidenced. Humanum est errare.

I see John Wesley expressed what you are saying - "although every man necessarily believes that every particular opinion which he holds is true (for to believe any opinion is not true, is the same thing as not to hold it); yet can no man be assured that all his own opinions, taken together, are true - every thinking man is assured they are not, seeing humanum est errare et nescire..."

It's worth thrashing these issues out; even quite fundamental ones such as what a function is.
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Well, that is doubly interesting to me.

I know little of Wesley, but that quotation is enough to raise him quite a bit in my admittedly ignorant estimation. About a year ago, I was reading an essay on Luther and Calvin. In the essay, quotations from both used a phrase translated as “reasonable charity.” Given their backgrounds, it is extremely doubtful that both would both have struck on a phrase that subtle unless it was used by at least the Roman Catholic scholastics by whom both were educated. It raised my estimation of Luther and Calvin (neither of whom I find particularly sympathetic) and the Scholastics. Even Christian charity is to be ”ruled by the Lord Apollo’s golden mean.” Makes sense to me, but I am utterly pagan.

What is also interesting to me is the grammar of “humanum est errare.” Presumably “humanum” is accusative because using a neuter makes no sense. So it is masculine, which, given the conceptions of the time or the conceptions of wives at all times, does make sense. But then it cannot be a subject because then it would be “humanus.” If I were analyzing such a construction in English, I’d say there was an ellipsis, but I do not remember ellipsis as a thing in classical Latin grammar. “Esse humanum est errare” seems like good classical Latin grammar so, if the saying is from classical Latin, ellipsis must have been a thing. On the other hand, there are aspects of the saying that seem alien to classical Latin, namely the continuation “perseverare autem est diabolicum,” which seems quite Christian.
 
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@JeffM
Likewise I know little of the trio mentioned. Although I haven't heard the phrase "reasonable charity" and I am by no means an expert in things religious, it certainly is a concept I am familiar with. By strange coincidence, this very Sunday, the reading from the writings of St. Paul in Catholic churches covered this very topic:

2 Corinthians 8:7,9,13-15

You always have the most of everything – of faith, of eloquence, of understanding, of keenness for any cause, and the biggest share of our affection – so we expect you to put the most into this work of mercy too. Remember how generous the Lord Jesus was: he was rich, but he became poor for your sake, to make you rich out of his poverty. This does not mean that to give relief to others you ought to make things difficult for yourselves: it is a question of balancing what happens to be your surplus now against their present need, and one day they may have something to spare that will supply your own need. That is how we strike a balance: as scripture says: The man who gathered much had none too much, the man who gathered little did not go short.

As for the form of 'humanum est errare', it is very simple. The infinitive is considered a neuter (verbal) noun, in the sense of 'erring'. Humanum is just the nominative of the neuter adjective.
It seems it is often attributed to Seneca but that doesn't really make sense for the reasons you mention.
(I hope we don't get 'struck off' for being off-topic; accused of chit-chat).
 
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