x^2 * f(x) + f(1-x) = 2x-4

orl

Provided the functional equation f: R->R which satisfies x^2 * f(x) + f(1-x) = 2x-4. Determine f(x).

German IMO Selection Test 1982, problem 3.

**Posted:** Sun Jan 11, 2004 3:25 am

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Moubinool

Change x in 1-x we have a system of two equations with
unknown f(x) and f(1-x)

x^2 * f(x) + f(1-x) = 2x-4
(1-x)^2*f(1-x) + f(x) = -2-2x

**Posted:** Sun Jan 11, 2004 4:48 am

belenos · Riemann Hypothesis Offline Joined: 12 Sep 2003 Posts: 463

Just to finish it :

f(x) = 1 - x�

**Posted:** Wed Jan 21, 2004 12:34 pm

orl

That's wrong. Please plug in your function in the condition. So you will see that your answer is wrong. Have another try ! Smile

**Posted:** Fri Jan 23, 2004 8:16 am

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Math is like love. A simple idea but it can get complicated.

belenos · Riemann Hypothesis Offline Joined: 12 Sep 2003 Posts: 463

I read it :

x^2*f(x) + f(1 - x) = 2x - x^4

like in http://65.36.154.75/Forum/viewtopic.php?t=2015

but here it's x^2*f(x) + f(1 - x) = 2x - 4.

I find f(x) = 2(x^3-4x�+6x-1)/(x^4-2x^3+x�-1) but I don't have the courage to check it Sad

I prefer x^2*f(x) + f(1 - x) = 2x - x^4 Mr. Green

**Posted:** Fri Jan 23, 2004 12:38 pm

amfulger

**Posted:** Mon Jan 26, 2004 7:59 am

belenos · Riemann Hypothesis Offline Joined: 12 Sep 2003 Posts: 463

I don't know (who would like to check that !) if 2(x-4x�+6x-1)/(x^4-2x^3+x�-1) is a solution to the equation .... and I don't want to know Mr. Green

**Posted:** Mon Jan 26, 2004 10:00 am

orl

I think your solution is almost right: f(x) = 2 - (x^3-4x^2+6x-1)/(x^4-2x^3+x^2-1)

**Posted:** Mon Feb 02, 2004 7:15 am

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Math is like love. A simple idea but it can get complicated.