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May 24, 2009, 4:01:38 AM5/24/09

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Are there any solutions, R into R, for

f(x) + 2x = f^2(x)

f(x) + 2x = f^2(x)

other than f(x) = -x and f(x) = 2x?

May 24, 2009, 7:01:29 AM5/24/09

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Bonjour,

Three notices:

1°) sum(b(i)f^i(x)) = 0 ,b(i)constant, i integer

Be r(j) the roots of sum(b(i)*a^i) = 0 ,

possible solutions on C f(x)= r(j)x .

2°) One fixed point x=0 ,

3°)a formal relation:

from 2x = f^2(x) -f(x)

x = {(f-I)/2} o f(x) ,I identity

Thence (f-I)/2} = f^ -1

Alain

May 24, 2009, 8:30:34 AM5/24/09

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William Elliot a �crit :

> Are there any solutions, R into R, for

> f(x) + 2x = f^2(x)

>

> other than f(x) = -x and f(x) = 2x?

> Are there any solutions, R into R, for

> f(x) + 2x = f^2(x)

>

> other than f(x) = -x and f(x) = 2x?

-x and 2x are the only two continuous solutions of f(x)+2x=f(f(x)).

But infinitely many non continuous solutions exist.

May 24, 2009, 3:02:32 PM5/24/09

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Patrick Coilland <pcoi...@pcc.fr> writes:

E.g. take any subset A of R such that (t -> -t) maps A to itself and

(t -> 2t) maps R \ A into itself, and let f(x) = -x for x in A and

2x for x in R \ A.

--

Robert Israel isr...@math.MyUniversitysInitials.ca

Department of Mathematics http://www.math.ubc.ca/~israel

University of British Columbia Vancouver, BC, Canada

May 24, 2009, 3:08:43 PM5/24/09

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Hi Robert,

I would rather have my name associated with this triviality:P=NP as

was proven:

http://groups.google.com/group/sci.math/browse_thread/thread/c910051ec44e17c8/9fe84fe28bd2f8f4?lnk=raot#9fe84fe28bd2f8f4

MMM

May 24, 2009, 4:20:53 PM5/24/09

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And I would rather have you fuck off and go trash

another website than trashing this one.

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