Maths Questions

heffo500

Can someone please help me with this?

1. (a) Give an example of a function with domain R − {1, 2}.
(b) Give an example of a function with domain R − (1, 2).
(c) Give an example of a function with domain R − [1, 2].

LeixlipRed

Get your ass to the Maths Support Centre! Unluckily for you you chose a forum with someone who works in the NUIM maths department as moderator Anyway, if you want help you need to a) tell us what you've done and b)tell us what it is that you don't understand about the question.

heffo500

Thanks will go there tomorrow!

Fremen

Hm, that's a weird question. What's to stop you proposing

f(x) = x on the domain D

where D is any of the above?

MathsManiac

Fremen wrote: »

Hm, that's a weird question. What's to stop you proposing

f(x) = x on the domain D

where D is any of the above?

Exactly what I was thinking!

LeixlipRed

There's nothing to stop the student doing that. We're changing curriculum out here, open ended questions on assignments on the non-rigurous courses. Any answer is acceptable. Anyway, there are more elegant solutions than that, you can use a quadratic to do it all.

MathsManiac

Hmm, so you think a quadratic function is more elegant than the identity function. Intriguing.

Here's a much more elegant solution, also defined on all of the stated domains:
f:x-> (Euclid's proof of that there is no rational number whose square is 2)

LeixlipRed

Yeh, I do think it's more elegant than simply setting f(x)=x or any other function to be defined on those domains. This is the sort of argument or disagreement the lecturer of that particular course hoped would occur.

MathsManiac

LeixlipRed wrote: »

Yeh, I do think it's more elegant than simply setting f(x)=x or any other function to be undefined on those domains. This is the sort of argument or disagreement the lecturer of that particular course hoped would occur.

What do you mean by: "... function to be undefined on those domains"? The functions are defined on the given domains.

LeixlipRed

That's what I meant. Undefined on the set, open interval and closed interval. Relax!

MathsManiac

I'm quite relaxed, but I still don't follow.

A function is defined on its stated domain (assuming it's properly defined). It's obviously not defined anywhere else, since if it was, it would not be the same function, (not being the same set of couples).

LeixlipRed

I simply put in an undefined instead of defined. When I wrote it I was thinking about the question in the sense that these functions were all defined on R except on a) a finite set b) an open interval and c) a closed interval. That's were the undefined slipped into my mind and hence the mistake!

MathsManiac

Ok. So what the original question really meant was something like this:

Given this domain D, which is a proper subset of R, find a function on D which is defined by some relatively straightforward rule, and which has the property that the same rule cannot be used to define a function on any subset of R that is a superset of D.

This rules out straightforward things like constant functions and identity functions.

Is that what you're getting at?

LeixlipRed

No! Not at all. I just made a mistake, I said undefined instead of defined in post no.8 in this thread. The answer can be any function that satisfies the condition. There are infinitely many answers. I originally said that I thought it was more elegant that there was one quadratic that could form part of all 3 answers. You disagreed. I then put in an "un" by accident and now we're circling back around to closing thread time.

MathsManiac

Yes, but let's leave aside the extraneous "un"! I'm sorry if I caused or exacerbated that distraction. What I'm trying to tease out is why the solution you have that is based on a quadratic might be considered to be more elegant. This is the interesting thing, in my view.

I consider Fremen's solution to be very elegant, in the following senses:
1. It uses the identity function, and the concept of an identity element or operator is a powerful and elegant one in mathematics
2. It is supremely generalisable, since it works no matter what set you're given as the domain. In mathematics, generalisable solutions are frequently regarded as more elegant than ones that are not.

I think I can guess what your solution is. It is undoubtedly more complex than Fremen's simple and highly generalisable solution, (which was, effectively: given any set A, the set {(x,x) | x in A} is a function on A.) So, in what sense is this more complicated and less generalisable solution more elegant?

Pondering this is what led me to my last post. I believe that you have in mind, in each case, a function-defining rule for which the following can be said: since we often like to define functions on R, we will start with R and throw out only those numbers for which this rule can't work; doing this, what we are left with is the domain given in the question. The form of the rule has effectively created the domain by preventing some elements of R from being in the domain, other than by setting up a case-by-case rule.

The solution is then deemed elegant because the domains given in the question are, in a certain sense, maximal domains for the rule offered.

And you can make the same quadratic do the job in each case by just using it in three slightly different ways.

Sorry if you find this tiresome, but I am prone to probing what makes certain things seem elegant. And it is interesting to think that a complicated once-off solution could potentially be considered more elegant than a simple general solution, don't you think?

LeixlipRed

No, it's not tiresome at all. Sorry if I came across as seeming irritated. Maybe I should clarify a few things, these students are not mathematicians. The vast majority of them never will be. When I used the word "elegant" I didn't really mean that in the sense a mathematician would mean. I'm not back-tracking on what I said, when I first seen the problem my solution was similar to Fremen's. And it's obviously more general and hence more powerful. But these students need to grasp the basic ideas of range and domain, essentially to try and understand what a function is. And the quadratic solution, to them, is probably more interesting and relevant. That's what I meant by "elegant". I didn't mean it to come across as my opinion of what's elegant or not.

Also, I don't want to get into an argument about whether it's right or not that we're giving them a limited view of functions or whatever way you want to look at it. It's a progression from a question of the form: f(x)=x^2 -5, what is the domain and range of this function? Which is the same old rubbish they've been asking these students to do for years now. Baby steps in the right direction as such.

Final point, I did discuss this with some students this week and mentioned there was other ways of approaching this question and to look into their notes and see if anything jumped out at them and a few (a limited few) actually came up with something similar to Fremen. Anyway, your feedback is interesting and the lecturer who set this question is my supervisor so I'll chat to them this week about it and maybe get back to you if the debate raises any other interesting points.

sponsoredwalk

This seems like a very open-ended question, what's to stop me from defining
a function to be equal to 1 when x is rational and contained within [1,2] and
f(x) = the dirac delta function shifted to that particular x value in between 1
& 2 when x is irrational?
Obviously they want to test your knowledge of the brackets () but still

LeixlipRed

There is nothing to stop you except the fact they wouldn't have a clue how to do that But basically, it's completely open-ended!

Michael Collins

sponsoredwalk wrote: »

This seems like a very open-ended question, what's to stop me from defining
a function to be equal to 1 when x is rational and contained within [1,2] and
f(x) = the dirac delta function shifted to that particular x value in between 1
& 2 when x is irrational?
Obviously they want to test your knowledge of the brackets () but still

But is the Dirac Delta Function a function? (This statement makes more sense than it seems!)

sponsoredwalk

From what I've read it's not strictly a function and I mainly used that
"function" in my question to find out would it work considering the grey
area You know, pushing the limits

https://www.youtube.com/watch?v=4qfdCwys2ew

The extent of my knowledge of this function.

My thinking is that you could define it as Dirac's function on the domain
between 1 & 2 consisting only of irrational values, that way it doesn't
interfere with the constant function 1 I've also defined on the domain of
all rational numbers between 1 & 2. There was no qualification that the
function had to be continuous or anything, we could have defined the
function as being equal to an elephant for all values on 1 to 2 :pac:
I suppose it's a modified dirac delta function I mentioned, or rather just
a function defined to be ∞ when irrational & 1 when rational :rolleyes: