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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Here's a challenge someone set me yesterday - doesn't seem like it can be done at first, but I can assure you there is a way!

Find a function f(x) such that f(x) is defined for every value of x, and continuous for exactly one value of x (and, hence, discontinuous everywhere else).

To clarify, f(x) is continuous at a point *c* if:

(or some other equivalent condition)

*Last edited by Dross (2007-01-24 12:11:50)*

Bad speling makes me [sic]

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I know a possible answer, because one of my lecturers is crazy about stuff like this.

I'll wait a bit before I say it though.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Too hard for me, I'll come back to see later.

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm assuming these functions must be on the real numbers?

The answer, well, my answer, is one which once you see it, you think, "Oh come on!". But I'll wait for someone else to post theirs first.

I find functions which are continuous everywhere but differentiable nowhere to be much more interesting.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Dross****Member**- Registered: 2006-08-24
- Posts: 325

Please do post any solutions you have, all - I'd like to see any others that are out there.

The one I got (after a lot of time and a bit of luck!) is

Bad speling makes me [sic]

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Yours is the one that I had as well. There are many other functions like that that would work though. As long as the two functions you pick equate for precisely one value of x, then it will work.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Right mathsyperson. But, the question is, are there other dense properties in the real numbers? The only ones I know of (without thinking too much into it) are rational and irrational. What about algebraic and transcendental? Interesting question if you ask me.

Oh, and mine was exactly the same, Dross.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Dross, if you don't mind, I'm going to post my own little challenge. If you want me to move this to another thread, just say the word and it's gone.

Find a non-trival function such that when it's input doubles, it's value cubes. That is:

f(2x) = f(x)³

Good luck!

Note: non-trival means non-constant.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oh, right. I wondered what you meant by non-trivial, before you edited that note in. So, we're not allowed to have f(x) = -1, 0 or 1 then.

Hmm. That's quite a tricky problem. Nothing jumps out at me, but I'll have a think about it.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yes, those would be way too easy. I'm fairly certain the only way you can solve this problem is to come up with a strategy, a method of solving. Just trying different functions won't help.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I'm going to start posting hints. This first hint is obvious, but it does reveal a way to think about the problem.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ah, of course. Strange that posting something so obvious would help out so much.

Why did the vector cross the road?

It wanted to be normal.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Here is a "find the function" problem that a friend of mine came up with:

Let f(x) = 1 - 2x^2. Find a function g(x) such that g(g(x)) = f(x) for all x, or prove no such function exists.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

i didnt come up with this myself, and its not quite a complete answer, however:

take an example:

g(x) = x+1

gg(x) = x+1+1

g(x) = x^2 + 1

gg(x) = x^4 + 2x^2 + 1 + 1

thus to yield a degree of x^2 in gg(x), g(x) would have to be a radical with degree between 1 and 2, and would be multivalued and hence not a real function

unless you can magicly get a function with logarithms/hyperbolics/trigonometrics.

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

It is possible, however. For example, here is the corresponding g(x) for f(x) = x²:

Then we have

Also, if you didn't come up with that, then where did it come from!?

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

the magical world of professor rythm

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

He teaches music, not math! Enough of your lies.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

no, thats just his nickname, he is at university studying maths, he helps me out when i have trouble with something, and i help him out when he has trouble with something (usually something not to do with maths)

The Beginning Of All Things To End.

The End Of All Things To Come.

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