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 November 25th, 2014, 07:37 PM #1 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 UNSW Mathematics Competition: Last Question So there's a competition in south-eastern Australia called the UNSW Mathematics Competition. I decided to enter it this year just to see if I could do it and I was practicing with past papers at home. Admittedly I wasn't getting far. This question caught my eye and I've been trying to work it out on and off for a good few months now. Suppose $f$ is a continuous and strictly positive function defined for all real numbers, and that $$f(2012) = 2012$$ and $$f(x) = f(x + f(x)).$$ Prove that $f(x) = 2012$ for all real $x$. My first thought was that if I could prove $f'(x) = 0$ for all x, then the first condition would imply that $f(x) = 2012$, so I used the chain rule to get$$f'(x) = f'(x + f(x))(1 + f'(x))$$ but I simply don't see any way to go from there. Another possibility given to me by my maths teacher was saying that $f$ is periodic by the second condition, but the period is arbitrary so the function must be a horizontal line. I have no idea how you would even begin a proof of this. Can someone help me out? November 25th, 2014, 08:05 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra We can start with $$x=2012 \implies f(4024) = 2012 \implies f(6036) = 2012 \implies \cdots$$ Perhaps also we can suppose that $g(x)$ is the inverse of $f(x)$ for some part of the domain of $f(x)$. So that $g(f(x)) = g(f(x+f(x))) \implies x = x + f(x) \implies f(x) = 0$. But if $f(x) = 0$ on this part of its domain, it has no inverse, so our assumption that an inverse exists is false. Thus there is no part of the domain of $f(x)$ for which it has an inverse and therefore it is a constant function $f(x) = c$ (or else it is undefined everywhere) and then the first condition gives us $$f(x)=2012 \qquad \forall x$$ Thanks from CaptainBlack and Azzajazz Last edited by skipjack; November 26th, 2014 at 09:48 PM. November 25th, 2014, 08:13 PM   #3
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Quote:
 Originally Posted by v8archie We can start with $$x=2012 \implies f(4024) = 2012 \implies f(6036) = 2012 \implies \cdots$$
I did get this result by just messing around, but I couldn't work out a way to fill in the gaps, so I gave up on that approach.
Quote:
 Originally Posted by v8archie But if $f(x) = 0$ on this part of its domain, it has no inverse, so our assumption that an inverse exists is false. Thus there is no part of the domain of $f(x)$ for which it has an inverse and therefore it is a constant function $f(x) = c$
I don't quite understand why this is true. Is it always true that a function that doesn't have an inverse is constant?

Last edited by skipjack; November 26th, 2014 at 10:07 PM. November 25th, 2014, 08:19 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra It doesn't have an inverse anywhere - on any part of its domain. If a function is piecewise strictly monotonic, it has an inverse on each strictly monotonic piece of its domain. Since this function does not, it cannot be strictly monotonic anywhere. Therefore it is constant or undefined. Last edited by skipjack; November 26th, 2014 at 10:07 PM. November 26th, 2014, 03:38 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra I'm not convinced that this is right. The assumption is that $g(x)$ is defined on the interval $(x,x+f(x))$. November 26th, 2014, 06:12 AM #6 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 The function is periodic with period 2012/k for some positive integer k. So $\displaystyle f'(x) = f'(x+2012) = f'(x + f(x))$. But you found $\displaystyle f'(x) = f'(x + f(x))(1 + f'(x))$. Suppose $\displaystyle f'(x) \ne 0$ Then, as $f'(x + f(x)) = f'(x + f(x))(1 + f'(x))$, $1 + f'(x) = 1$ so $f'(x) = 0$. Suppose $f'(x) = 0$. Then $f'(x) = 0$. From above, we have $f'(x) = 0$. As $f(2012) = 2012$, and $f'(x) = 0$, $f(x) = 0$ everywhere. (This assumes a derivative exists. Allowed?) November 26th, 2014, 07:20 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra I'm not completely convinced that you can claim that the function is periodic. Certainly not with a constant period. Indeed, if it has a constant period then it must directly be constant. So your proof is circular as far as I can see. Moreover, in your first line you replace 2012 with $f(x)$, so you have defined $f(x)$ to be the desired function before proving that it is that function. The problem is a) proving that the function is periodic; and b) proving that the period is constant. Last edited by v8archie; November 26th, 2014 at 07:31 AM. November 26th, 2014, 11:41 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 I see. Initially, I read f(2012) = 2012 as f(x) = x, where x = 2012. Then I substituted x = 2012 just in the RHS just to proceed with Azzajazz' work. Ugh. November 26th, 2014, 06:13 PM   #9
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 Originally Posted by v8archie I'm not convinced that this is right. The assumption is that $g(x)$ is defined on the interval $(x,x+f(x))$.
Do you mean the assumption for the contradiction or a false assumption you have made? November 26th, 2014, 10:42 PM   #10
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It is given (presumably for a reason) that f(x) is defined for all real values of x and that f(x) > 0.

Suppose f(x) has an inverse for all x such that a < x < b.
By definition of an inverse g(f(x)) = x if a < x < b.
It isn't known that a < x + f(x) < b,
so it isn't known that g(f(x + f(x))) = x + f(x).
In fact, a < x + f(x) < b cannot be true.

Quote:
 Originally Posted by v8archie Indeed, if it has a constant period then it must directly be constant.
What did you mean by "directly"? There doesn't seem to be any obvious proof of the claim.

Quote:
 Originally Posted by Hoempa The function is periodic with period 2012/k for some positive integer k.
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