Functional Equation
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From: Harun Al-Rashid on 14 May 2010 16:55 How to show that there is no continuous functions f from (0, +oo) to itself satisfying f(f(x)) = x^x / 2 (x > 0) ?
From: Robert Israel on 14 May 2010 21:14 Harun Al-Rashid <abdul(a)my-deja.com> writes: > How to show that there is no continuous functions f from (0, +oo) to itself > satisfying f(f(x)) = x^x / 2 (x > 0) ? Hint: start by considering fixed points and 2-cycles. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Harun Al-Rashid on 14 May 2010 18:05 > Harun Al-Rashid <abdul(a)my-deja.com> writes: > > > How to show that there is no continuous functions f > from (0, +oo) to itself > > satisfying f(f(x)) = x^x / 2 (x > 0) ? > > Hint: start by considering fixed points and 2-cycles. > > -- > Robert Israel > israel(a)math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada OK. I think there are exactly two fixed points, namely p_1 = 2, and another one, say p_2, with 0 < p_2 < 1. Then ?
From: Robert Israel on 16 May 2010 13:43 > > Harun Al-Rashid <abdul(a)my-deja.com> writes: > > > > > How to show that there is no continuous functions f > > from (0, +oo) to itself > > > satisfying f(f(x)) = x^x / 2 (x > 0) ? > > > > Hint: start by considering fixed points and 2-cycles. > > > > -- > > Robert Israel > > israel(a)math.MyUniversitysInitials.ca > > Department of Mathematics > > http://www.math.ubc.ca/~israel > > University of British Columbia Vancouver, > > BC, Canada > > OK. I think there are exactly two fixed points, namely p_1 = 2, and another > one, say p_2, with 0 < p_2 < 1. Then ? Yes, those are the fixed points of g(x) = x^x / 2. Show that they must be fixed points of f as well (in general a fixed point of f(f(x)) could be either a fixed point of f or a point on a 2-cycle), but f has no other fixed points. Now there is exactly one p_3 <> p_2 with g(p_3) = p_2. What can you say about f(p_3)? -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Harun Al-Rashid on 17 May 2010 00:16
> > > > Harun Al-Rashid <abdul(a)my-deja.com> writes: > > > > > > > How to show that there is no continuous > functions f > > > from (0, +oo) to itself > > > > satisfying f(f(x)) = x^x / 2 (x > 0) ? > > > > > > Hint: start by considering fixed points and > 2-cycles. > > > > > > -- > > > Robert Israel > > > israel(a)math.MyUniversitysInitials.ca > > > Department of Mathematics > > > http://www.math.ubc.ca/~israel > > > University of British Columbia > Vancouver, > > > BC, Canada > > > > OK. I think there are exactly two fixed points, > namely p_1 = 2, and another > > one, say p_2, with 0 < p_2 < 1. Then ? > > Yes, those are the fixed points of g(x) = x^x / 2. > Show that they must be > fixed points of f as well (in general a fixed point > of f(f(x)) could be > either a fixed point of f or a point on a 2-cycle), > but f has no other fixed > points. Now there is exactly one p_3 <> p_2 with > g(p_3) = p_2. What can you > say about f(p_3)? > -- > Robert Israel > israel(a)math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada Showing that f(p_2) = p_2, and that g'(p_2) < 0 should be enough, I think (i.e., without involving p_3). |