A functional equation from Kyrgyzstan

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[skerJG]

Letting p(x) = x^2 + x + 3, we have x^2 - 3x + 5 = p(1-x), so the functional equation is f(p(x)) + 2f(p(1-x)) = 6x^2 - 10x + 17. But letting x = 1- x, we get f(p(1-x)) + 2f(p(x)) = 6x^2 - 2x + 13.

These two equations let us solve for f(p(x)) = 2x^2 + 2x + 3 = 2p(x) - 3. Letting y = p(x) gives f(y) = 2y - 3.

So f(2009) = 4015

[lilylikesmarkies]

Hi Michael, are you planning on covering any of the 2022 IMO problems?

[tiripoulain]

There’s a typo at 9:30. You should have q(x)=p(x)-4x+2, meaning that q(x_0)=2011-4x_0 and p(x_1)=4x_1+2007. Then everything checks out.

[kuronno]

I skipped a lot of those steps by starting simply setting x = -x + 1 in the original equation. That kind of made the two polynomials on the left hand side to basically switch places. From there I just solved it as a simple linear system and got:

f(x² - 3x + 5) = 2x² - 6x + 7

Then I found a solution to:
x² - 3x + 5 = 2009

and set x to that in the previous expression to get f(2009) = 4015

[jonpress]

Usually when Michael solves a functional equation, he starts by trying some values of the variable(s). So, start with x = 0, and you get f(3) + 2 f(5) = 17. OK, try another. If x = 1, you get
f(5) + 2 f(3) = 13. That's too symmetrical to ignore. Solving the system gets you f(3) = 3 and f(5) = 7. If you assume that f is linear, you're virtually done, but that's probably too much of a leap. So try more values of x. x = -1 leads to f(3) + 2 f(9) = 30. We already know f(3), so that leads to f(9) = 15. x = 2 also leads to f(9), but x = -2 leads to f(15) = 27.

So we have:

x f(x)
3 3
5 7
9 15
15 27

That's "definitely" linear -- f(x) = 2x - 3 -- though it's not a proof. To prove it, show that it solves the original function equation. That should be good enough for the judges.

[fefeisbored]

doesn't it pretty easily follow from the first step that f(x)=2x-3?

[carstenmeyer]

Try factoring the argument polynomials:

*p(x) := x^2 + x + 3 = (x + 1/2)^2 + 11/4*
*q(x) := x^2 - 3x + 5 = (x - 3/2)^2 + 11/4*

We notice translation- and mirror-symmetry between *p(x), q(x)* :

*x in R: q(x) = p(x - 2) = p(1 - x)*

Using the short-hand *g(x) := f(p(x))* and the mirror-symmetry above the functional equation simplifies to

*x in R: g(x) + 2g(1 - x) = 6x^2 - 10x + 17*

A change of variables " *x -> 1 - x* " leads to a second functional equation:

*x in R: g(1 - x) + 2g(x) = 6x^2 - 2x + 13*

Using both functional equations we can actually solve for *g(x)* :

*x in R: g(x) = 2x^2 + 2x + 3 = 2p(x) - 3 = f(p(x))*

Setting *x := -1/2 + 5/2 * \sqrt{321}* such that *p(x) = 2009* we finally have

*f(2009) = 2 * 2009 - 3 = 4015*

[Abc-npeo]

Nice to see problems from my homecountry!Thx for the video

[khiemngo]

This problem is one of the best. Thanks for the solution! By the way, for the solution of p(x) = 2009 and q(x) = 2009, it's sqrt(321), not sqrt(312).

[izgirft]

Differentiation of both sides by x leads to df/dy=2 when y is composite argument. So, f(y)=2y+C. The substitution f(x)=2x+C to equation makes C=-3.

[zh]

Let r=x^2+x+3, s=x^2-3x+5, t=6x^2-10x+17
Then from above: 2r+4s=t+9
Substitute back to the original expression: f(r)+2f(s)=t=2r+4s-9
This is a simple functional equation problem,
Let (r, s)=(0, 0) then f(0)=-3
Let (r, s)=(free, 0) then f(r)=2r-3

[-Curved]

this was my approach:
let x^2+x+3=X
and x^2-3x+5=Y


suppose 6x^2-10x+17=aX+bY+c

after some algebra involving a, b and c. you get

f(X)+2f(Y)=2X+4Y-9

sub in Y=X to get f(x)=2x-3

[zeggwaghismail]

a, b and c are real numbers st ab+bc+ca=1
Show that :
(a+1/b)^2 +(b+1/c)^2 + (c+1/a)^2 > 16

[julienmallet]

hmmm at 3min10 it's very obvious that f(x) = 2x - 3 satisfies the entire functional equation

so f(2009) = 2*2009 - 3 = 4015

lol

[ianteles]

Hi michael, i have a suggestion : imo 2022 p2, and imo 2022 p5

[maxmartinelli]

In my opinion it is simpler if you perform the change of variable x -> x+1, then the equation writes as f(x^2 + 3x +5) + 2 f(x^2 -x + 3) = 6 x^2 + 2x +13. Then another change of variable x-> -x and now the equation becomes f(x^2 - 3x +5) + 2 f(x^2 + x + 3) = 6 x^2 - 2x +13.
Using the original equation you can find f(x^2 + x +3) = 2 (x^2 + x + 3) - 3. Setting x^2 +x +3 = 2009 you obtain f(2009) = 2 * 2009 - 3 = 4015

[jerrymouse]

I have a completely different approach..but it's a bit calculative!!
at first, put x = ( sqrt(8025) --1)/2 in the given eqn to get
f( 2013-- 2*sqrt 8025) +2f(2009) = a numerical value ....(1)
Then put x =(3--sqrt 8025)/2 in the eqn to get
f(2009) + 2f (2013--2*sqrt 8025)= another numerical value
from the above eqns (1) and (2), we easily get the required value of f(2009)
[this process is a bit calculative]

[Szynkaa]

i think i did it way easier. Put x=y-2 into equation and then put x= -y-1 into equation. This gives us two linear equations with two unknown terms, we solve it to immediately get f(y^2-3y+5] =2y^2-6y+7. Therefore if we want f(2009] we take y such that y^2-3y+5=2009.
So y^2-3y=2004, so 2y^2-6y=4008, so f(2009]=4015

[pNsB]

I substituted x=0, and then x=1, to get two different equations:

f(3) + 2f(5) = 17
2f(3) + f(5) = 13

solving this gives f(3)=3 and f(5)=7. I didn't assume that f(x) was linear, but if I did I could have found that f(x) = 2x - 3 and substituted in 2009 to get 4015.

[figuritout]

I did this problem more numerically, generating values for f(x) by plugging in values for x such as 0 and 1. Then I found the x-value for which the “interior polynomials” were equal, which was at x=0.5, y = 3.75. Using that I was able to get a third vale of f(15/4)=9/2. Using f(3)=3 and f(5)=7, I used a matrix and assumed a quadratic form for f(x). The solution revealed a linear function, f(x)=2x-3 for which I evaluated it at x=2009 to get the solution. Great problem! Love it