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Topic: f(xf(y)) = yf(x) (Read 6707 times) |
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Aryabhatta
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f(xf(y)) = yf(x)
« on: Dec 17th, 2004, 10:56am » |
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This is from some math olympiad (probably an old IMO).
Find all functions f: [bbr]+ [mapsto] [bbr]+ such that
1) f(xf(y)) = yf(x) [forall] x [in] [bbr]+
2) Ltx[to]0 f(x) = 0
[bbr]+ is the set of postive reals.
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Icarus
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Re: f(xf(y)) = yf(x)
« Reply #1 on: Dec 18th, 2004, 11:36am » |
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::Let a=f(1). Then it quickly follows that f(anx) = f(x) for all x, and all integers n. If a [ne] 1, then either anx [to] 0 as n [to] [infty] or as n [to] -[infty]. In either case, the limit condition gives us f(x) = 0 for all x.
If a = 1, then letting x=1 gives f(f(y)) = y. I.e. f is it's own inverse. Let u = f(y) (so y = f(u)) in the original equation, and we have f(xu) = f(x)f(u). Thus, f(xr) = f(x)r for all rational numbers r. if x<1, letting r [to] [infty], we must have f(x)r [to] 0, which requires f(x) < 1. So if x < y, f(x) = f(x/y)f(y) < f(y), so f is strictly increasing.
Let f(e) = b. Since e > 1, b > 1. The values of er with r rational are dense in [bbr]+. The values of br = f(er) are also dense. The only way that f can be strictly increasing is for it to be continuous. And so f(et) = bt for all real t. So f(x) = f(eln x) = bln x = x ln b = xc. But since f(f(x)) = x, we must have c2 = 1. And since f(x) is increasing, c > 0. I.e. c = 1.
So the only functions satisfying the conditions are f(x) = 0, and f(x) = x.::
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| « Last Edit: Dec 18th, 2004, 11:45am by Icarus » |
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Aryabhatta
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Re: f(xf(y)) = yf(x)
« Reply #2 on: Dec 19th, 2004, 11:01am » |
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on Dec 18th, 2004, 11:36am, Icarus wrote:| Let f(e) = b. Since e > 1, b > 1. The values of er with r rational are dense in [bbr]+. The values of br = f(er) are also dense. The only way that f can be strictly increasing is for it to be continuous . |
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Can you give a brief proof of this assertion? It does not look very obvious to me. The name of the theorem will do, if it is a known fact.
Also, f(x) = 0 is not a solution as the range is the set of positive reals.
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Eigenray
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Re: f(xf(y)) = yf(x)
« Reply #3 on: Dec 20th, 2004, 12:48pm » |
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Suppose f,g are increasing, g is continuous, and f(x)=g(x) for x in D, a dense set. Then for any a,
f(a) [le] inf { f(x) : x>a } = inf { f(x) : x>a, x in D } = inf { g(x) : x>a, x in D } = g(a);
similarly f(a) [ge] g(a), and it follows f=g is continuous.
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| « Last Edit: Dec 20th, 2004, 1:51pm by Eigenray » |
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Aryabhatta
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Re: f(xf(y)) = yf(x)
« Reply #4 on: Dec 20th, 2004, 1:56pm » |
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Nicely done Icarus and Eigenray.
Here is the solution I had in mind.
Let us first show that f(1) = 1.
Now f(xf(y)) = yf(x) means f(f(y)) = yf(1)
We also have that
f(f(xf(y)) = f(yf(x)) = xf(y)
This implies that f(1) = 1.
So we have that f(f(x)) = x.
We can easily see that:
i) f(xn) = f(x)n for all natural numbers n.
ii) f(x)f(y) = f(xy)
if 0 < x < 1, the xn -> 0 as n-> oo. Thus we must have that f(x) < 1.
Similarly if x > 1 then f(x) > 1.
Now we have that f(1/x) = 1/f(x)
Now assume x < f(x).
Consider f(x/f(x)) = f(xf(1/x)) = f(x)/x.
This is not possible as x/f(x) < 1 while f(x/f(x)) = f(x)/x > 1.
Similarly we see that f(x) < x is not possible.
Hence f(x) = x.
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Icarus
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Re: f(xf(y)) = yf(x)
« Reply #5 on: Dec 20th, 2004, 3:35pm » |
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on Dec 19th, 2004, 11:01am, Aryabhatta wrote:Can you give a brief proof of this assertion? It does not look very obvious to me. The name of the theorem will do, if it is a known fact.
Also, f(x) = 0 is not a solution as the range is the set of positive reals. |
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Yes, after coming up with the f(x) = 0 solution in the a [ne]1 case, I overlooked that it was artificially excluded.
EigenRay's argument works fine for what I needed, but it is possilbe to show that f is continuous without presupposing the existance of a continuous function that matches f on a dense subset:
If U and V are intervals in [bbr] and f: U [to] V is monotone, and if there exists D [subset] U such that f(D) is dense in V, then f is continuous.
Proof: Assume f is increasing (the argument for f decreasing is similar). For all x [in] U, f(x-) = sup {f(y) : y <x}, and f(x+) = inf {f(y): y>x} (both infimum are supremum must exist, as the sets are bounded above and below respectively by f(x)). So we have f(x-) [le] f(x) [le]f(x+). If f(x-) < f(x), then there exists d [in] D such that f(d) [in] (f(x-), f(x)), but f(d) < f(x) [bigto] d < x [bigto] f(d) [le] f(x-), a contradiction. So f(x-) = f(x). Similarly, f(x+) = f(x), and therefore f is continuous at x. Since x was arbitrary, f is continuous on all of U.
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Icarus
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Re: f(xf(y)) = yf(x)
« Reply #6 on: Dec 20th, 2004, 3:48pm » |
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So, how many solutions are possible if you weaken the restrictions?
In particular, find all functions f: U [to] [bbr] for some maximal interval U, such that
1) f(xf(y)) = yf(x) for all x, y in U
2) f is continuous.
(The maximal condition on U means that U is the largest possible interval to which f can be extended without violating the other conditions.)
Obviously f(x) = x and f(x) = 0 work. Are there any other solutions?
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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