Domain of composite function question

[Steven G]

Ok thats really helpful.
One last queston, is it possible that when looking at the simplified fg(x), additional (different) restrictions came into play ( for the maximum possible domain) that have not already been covered by looking at steps 1 and 2 above?

Actually it should be for the minimum possible domain.
I am sure that you know this answer by my last post. Again, if f(x) = g(x) =1/x then f(g(x))=x. Just looking at f(g(x)) it seems that there is no restriction on the domain But in the end x \(\displaystyle \neq\) 0 !

 

[JeffM]

That post wasn’t in response to anything you wrote. It was the answer to my original question! He doesn’t read the forum.

I did not say that he read the forum. What I said is that he disparaged the forum, or at least my responses, without having had the courtesy or intelligence to bother reading it. I suggest you have him answer your questions in the future without bothering us Neanderthals. I at least am setting you to ignore just like harpazoa. Oxonbridge has plenty of real mathematicians. Use them.

 

[pka]

I did not say that he read the forum. What I said is that he disparaged the forum, or at least my responses, without having had the courtesy or intelligence to bother reading it. I suggest you have him answer your questions in the future without bothering us Neanderthals. I at least am setting you to ignore just like harpazoa. Oxonbridge has plenty of real mathematicians. Use them.

I find it absolutely astonishing that this thread is now at 21 replies. Replies #2 & #5 were given by true PhD mathematicians. Now many of you may think that is no matter. But in this case it is everything. This is actually a very subtle and difficult question. #2 & #5 give correct answers. Why would we need more?

 

[Dr.Peterson]

Here it is. He is someone who does a lot of research in Lie Algebra(??)
Hi

Technically you should exclude both.

What does it mean to calculate ‘the domain of g’. Not really anything, as the specification of a function g:X->Y is simply a couple of sets X and Y and a list of pairs (x,y_x) for every x in X and some y_x in Y.

It’s just in our case it’s implicit that we want X and Y to look like the real numbers and we want y_x=2/x whenever that is possible.

Then it makes sense to make Y=\R and X as big as possible, so you set X= \R - {0}.

Now you then want to specify f: Y\to Z, but you can’t make that work without excluding at least -2 from Y. But this entails removing -1 from X.

Another way to do this is to extend the range of Y by adding in infinity. If you do that, then you don’t need to remove 0, since we define 2/0 = \infty.

Then decide whether you want infinity in Z or not. If you’re happy to have infinity in Z, then you don’t need to lose any points. If you do, then you need to remove just the -2 from Y, hence the -1 from X.

So it’s basically about whether you’re happy to have infinity in Y or not.

I don't think anyone has actually responded to this; and my impression is that you are saying you asked your friend independent of our discussion, and perhaps even before asking us, so there is no disparagement at all. Right?

The only issue here is that he ends up going far beyond the context of your question (and potentially just showing off his greater knowledge, without really helping you, though I am not going to presume anything about motivation).

Now, I am assuming you asked exactly the question you submitted here, so that the answer, "Technically you should exclude both," means that the domain is all real numbers except 0 and -1. I'm not sure of the reason for the word "technically". He is calling the domain of g X and the range Y (well, really that would be the codomain). What we call the domain of a function, when none has been specified and the context assumes real numbers, is the largest subset of the reals on which the function can be defined.

Then he is calling the domain and range of f Y and Z (but actually restricting Y, and then X, beyond what is required by f in order to accommodate the composition; I would have been clearer about that). What we have is a chain of functions X--(g)-->Y--(f)-->Z; f forces Y to exclude -2, and then g forces us to exclude from X the value x for which g(x) = -2, namely -1, in addition to 0, which g itself requires.

The stuff about infinity is silly; he's talking about extending the real numbers, which you can ignore since your context is only real numbers. So ignore the last three paragraphs.

 

[Steven G]

So long as we are restricting discussion to the real numbers, there is no point discussing x/0: there is no such real number.

You are making the same mistake as before.

[MATH]f(x) = \dfrac{1}{x} = g(x), \text { and } h(x) = f(g(x)) \text { if } x \ne 0.[/MATH]
It is then true that

[MATH]x \ne 0 \implies h(x) = x[/MATH], but

h(x) is and always undefined at x = 0.

It is true of course that we can define i(x) = x for all real x, but h and i are different functions with different domains. They are not the same. They happen to be equivalent within h's domain.

Ok thats really helpful.
One last queston, is it possible that when looking at the simplified fg(x), additional (different) restrictions came into play ( for the maximum possible domain) that have not already been covered by looking at steps 1 and 2 above?

I misread your question before. No, the final result will not add any new restriction

 

[apple2357]

I don't think anyone has actually responded to this; and my impression is that you are saying you asked your friend independent of our discussion, and perhaps even before asking us, so there is no disparagement at all. Right?

Yes, i asked a week ago. He didn't respond at the time. So i came on the forum

The only issue here is that he ends up going far beyond the context of your question (and potentially just showing off his greater knowledge, without really helping you, though I am not going to presume anything about motivation).

I am not sure his motivation either!

Now, I am assuming you asked exactly the question you submitted here, so that the answer, "Technically you should exclude both," means that the domain is all real numbers except 0 and -1. I'm not sure of the reason for the word "technically". He is calling the domain of g X and the range Y (well, really that would be the codomain). What we call the domain of a function, when none has been specified and the context assumes real numbers, is the largest subset of the reals on which the function can be defined.

yes exactly the same question

Then he is calling the domain and range of f Y and Z (but actually restricting Y, and then X, beyond what is required by f in order to accommodate the composition; I would have been clearer about that). What we have is a chain of functions X--(g)-->Y--(f)-->Z; f forces Y to exclude -2, and then g forces us to exclude from X the value x for which g(x) = -2, namely -1, in addition to 0, which g itself requires.

The stuff about infinity is silly; he's talking about extending the real numbers, which you can ignore since your context is only real numbers. So ignore the last three paragraphs.

Thanks , Thats the bits i didnt get!

 

[apple2357]

I did not say that he read the forum. What I said is that he disparaged the forum, or at least my responses, without having had the courtesy or intelligence to bother reading it. I suggest you have him answer your questions in the future without bothering us Neanderthals. I at least am setting you to ignore just like harpazoa. Oxonbridge has plenty of real mathematicians. Use them.

Apologies, I didn't mean to offend! I sent him the original question, he never responded until last night and in the meantime i posted it here.

I understood the responses on this forum it was much clearer and comprehensible than my friends reply!

I should not have even posted my friends response but i didn't understand all that stuff about infinity!

Apologies again. Lets close this thread before i annoy anyone else!

 

[JeffM]

Actually, I should (and do) apologize. I had a bad day, and the response from Cambridge was irrelevant to your question. So my annoyance should not have been directed at you, but at the response you received.

Mathematicians develop new number systems for various reasons. So, for example, the projectively extended real numbers do permit division by zero. See

Projectively extended real line - Wikipedia

en.m.wikipedia.org

When you are trying to understand functions on the real numbers, discussion of other number systems is not going to be helpful. So the response was largely irrelevant to your question, which had to do with understanding the domain of a composition of functions defined with respect to the real numbers. So, as I said before, my annoyance should have been directed at what I felt was a pretentious and unhelpful response. And as Dr. Peterson said, the response may possibly have been based on a misunderstanding of your question. So put down my response to my having had a bad day.

 

[apple2357]

Ok that projective number line stuff is fascinating!

And yes, we all have days like that and i hope everyone can see i wasn't trying to wind anyone up!
pka is right , the question got answered pretty well very early on. And i am satisfied. Thanks all.