Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Functions

Could someone please give me a small hint for part c? I don't really know what to do.

My answers so far are:

Thanks

Last edited by Daniel123 (2008-10-30 23:56:31)

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Functions

You have a given g(x) function, when it says to find a function g, does it mean to find a new function?

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

Well I think it must, otherwise it wouldn't be a question!

I'm pretty sure it's to do with the fact that g(g(x)) must equal f(x) for all x ∈R.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Functions

Maybe im missing something...

We need to find g such that g(g(x)) = f(x) and f(x) = x + 1

So then let g(x) = x + 1/2 and:

g(g(x)) = (x + 1/2) + 1/2 = x + 1 = f(x)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Functions

(c) and (d) don't seem to make sense.  You're asked to find a g, but g is given?  Even if we relabel the g that is given to a different letter, the question still doesn't seem to make sense.

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"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..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Functions

This question is like the first two, but now g(x) has a restriction on it as well. g(x) = x+1/2 wouldn't work, because that doesn't equal 2x+1/3 for 0<x<1/3.

Let's try some values and see where it gets us.
g(0) = 1/3 by definition. Also, g(1/3) = 1.
So g(g(0)) = 1 = 0+1 = f(0), which is good.

g(1/6) = 2/3, and f(1/6) = 7/6, which means our function g must satisfy g(2/3) = 7/6.
g(1/3) = 1, so g(1) must equal 4/3.

With a bit of experimentation, we find that in general we must have g(x) = x/2 + 5/6, for 1/3<x<1.

This new definition will determine the value of g(x) for 1<x<4/3.
Experimenting with that tells us that g(x) = 2x - 2/3, for 1<x<4/3.
This then determines g(x) for 4/3<x<2.

We are beginning to see a pattern here.
Investigating this further gets us to the general definition:

Where I define [x] to be x rounded down to the nearest integer (eg. [2] = 2, [1.7] = 1, [-3.4] = -4)
and (x) to be x-[x]. ( (2) = 0, (1.7) = 0.7, (-3.4) = 0.6 )

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Why did the vector cross the road?
It wanted to be normal.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

I've yet to go through this in detail, but it looks like what I want. Thanks mathsyperson

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

I couldn't quite see how you'd got to the very last answer, so I've come up with my own. It's pretty similar, but a little more complicated:

where [x] = x rounded down to the nearest integer, {x} = x rounded up to the nearest integer, and (x) = x - [x]. It's worked for every value I've tried.

Last edited by Daniel123 (2008-11-02 04:23:37)

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

For part d, after some investigating I get:

I can't really spot a pattern I can write algebraically. Can anyone help?

Thanks.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Functions

I think there's something wrong with the second definition of your version of g(x).
As a random example, try x=0.1.

Then g(x) = 2*0.1 + 1/3 - 0 = 8/15.
g(g(x)) is therefore 4/15 + 1/3 + 1 = 1.6 ≠ 0.1+1 = f(x).

You've done well for d, but you've included more than you needed to.

g(x) maps:
(0,1] to (1,2]
(1,2] to [-1,0)
[-1,0) to [-2,-1)
[-2,-1) to (0,1]

There's a loop here, and so the domain of g(x) that is determined by the question is [-2,2] \ {0}.
For all other values, you can have whatever you like.

Having said that, it's quite a hard question to solve even then. Trying to use the answer from a or b doesn't work because you get g(x) = ix, which isn't real-valued.
I think you're probably meant to extend the part of the function that you already know, but I can't  see how yet.

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Why did the vector cross the road?
It wanted to be normal.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

I think there's something wrong with the second definition of your version of g(x).
As a random example, try x=0.1.

Then g(x) = 2*0.1 + 1/3 - 0 = 8/15.
g(g(x)) is therefore 4/15 + 1/3 + 1 = 1.6 ≠ 0.1+1 = f(x).

Oops, I typed it incorrectly. It should have said:

I'll look at what you've written for part d as soon as I can. Thanks

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Functions

That's better - now we both have the same thing.

2x + 1/3 - [x]
2([x]+(x)) + 1/3 - [x]
[x] + 2(x) + 1/3

x/2 + 1/3 + {x}/2
([x]+(x))/2 + 1/3 + ([x]+1)/2*
[x] + (x)/2 +1/3 + 1/2
[x] + (x)/2 + 5/6

*I can do this because {x} = [x]+1 whenever x isn't an integer.
This is the same as when (x) ≠ 0, and this is clearly the case whenever (x) >1/3.

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Why did the vector cross the road?
It wanted to be normal.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Functions

For part d), I've drawn a graph. It just continues to spiral outwards:

Which gives:

EDIT: where [x] is x rounded down to the nearest integer. Obviously this could be written differently i.e. without (mod 2), but I think it's nicer this way.

Last edited by Daniel123 (2008-11-12 13:52:57)