A Functional Equation from Samara Math Olympiads

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❤️ f(x/(x-1)) = 2f(x) + x^2

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PLAYLISTS 🎵 :

[ioannismichalopoulos]

3:24 no need to replace x in f(x/(x-1)) with t/(t-1), as you stated already t=x/(x-1). Remember, komsu?

[Igorious]

It's a full family of functional equations:
f(ax + b ÷ cx – a) = λf(x) + g(x)
Its solution is:
f(x) = [λg(x) + g(ax + b ÷ cx – a)] / (1 – λ²)
In our case:
a = 1, b = 0, c = 1, λ = 2, g(x) = x².
So,
f(x) = –[2x² + x²/(x – 1)² ]/3

[dominiquelarchey-wendling]

With g(x) := x/(x-1) we get g(g(x)) = x. Hence f(g(x)) = 2f(x)+x² gives f(x) = f(g(g(x)) = 2f(g(x))+g²(x) = 2(2f(x)+x²)+g²(x) = 4f(x)+2x²+g²(x). So f(x) = (-1/3)*(2x²+g²(x)) = -x²(2+1/(x-1)²)/3. We can then check that such f satisfies the original equation.

[mega_mango]

b = 2a + x²
a = 2b + x/(x-1)²
(2x² + x/(x-1)²)/3 = f(x)

[РайанКупер-эо]

Someone in the internet knows Samara😮
The equation is pretty neat, but I don't remember solving functional equations when I was there solving math Olympiads over four years ago.

[freddyalvaradamaranon]

Muy interesante ejercicio de funciones, muchas gracias por compartir tan buen video 😊❤😊. Con cambio de variable y procedimientos de simplificación y reducción, muy bien explicados. Saludos y bendiciones Profesor. 😊❤😊.

[cicik]

common way is to have x/(x-1) = t ; x = tx-t; x(t-1)=t; x = t/(t-1); with that
f(t) = 2f(t/t-1) + (t/(t-1))^2 you can replace this on one variable and have a system
2f1 = 4f2 + 2x
f2 = 2f1 + (x/(x-1))^2
f2 = 4f2 + 2x + (x/(x-1))^2
3f(x) =-2x-(x/(x-1))^2

[solaymanelbellaj]

it is very essay we can replace with X=x/x-1 we find f(x)=2f(x/x-1) + sq( x/x-1 ) and we have f(x/x-1)=2f(x)+ sq(x) it's a very easy system

[KeshkKeshk-lz]

so what I learnt from my journey in mathematics, never let urself be afraid from taking a step in solving even if u think it is idiot, stupid . No, No, every mistake will be a new lesson.

[climbnexplore]

Please add domain and target I assume Reals? For people who want to solve before watching! Thx!!

[ForGu-dvwz]

This is a very common question for a normal grade 10 student in China.

[徐瑞斌-io]

In fact, we don't need t. Just plug in (x/x-1) into each x in the original equation.

[MadScientyst]

Great problem & solution!
Thx for pointing out the Turkish heritage too.
Some of my favorite YT Food Channels are from

[leif]

Why woud thst ne misleading at 2:09..How? It would just give 2 particular value but I don't see how it's misleading..

[pjb.]

ben de diyom bu adam kesin türk geröekten de öyleymiş

[Kaan.-]

cidden türkiyede mi büyüdün türkçe biliyor musun???

[gelbkehlchen]

Solution:
(1) f[x/(x-1)] = 2*f(x)+x² |-x² ⟹
2*f(x) = f[x/(x-1)]-x² |I replaced x/(x-1) for x ⟹
2*f[x/(x-1)] = ⟹
2*f[x/(x-1)] = ⟹
(2) 2*f[x/(x-1)] = f(x)-x²/(x-1)² ⟹
2*(1)-(2) = (3) 0 = 4*f(x)+2x²-f(x)+x²/(x-1)² ⟹
(3a) 0 = 3*f(x)+2x²+x²/(x-1)² |-3*f(x) ⟹
(3b) -3*f(x) = 2x²+x²/(x-1)² |/(-3) ⟹
(3c) f(x) = -2/3*x²-x²/[3*(x-1)²] = [-2*x²*(x-1)²-x²]/[3*(x-1)²]
= =
= [-2*x^4+4x³-3x²]/[3*(x-1)²] ⟹
(3d) f(x) = x²*(-2*x²+4x-3)/[3*(x-1)²]

[satrajitghosh]

writing z = x /( x - 1) one gets
z = 1 + 1/( x - 1)
or (z - 1)( x - 1) = 1
Hereby f (z) = 2 f (x) + x ^2 ...Eq_1
Again
f ( z /( 1 - z) ) = 2 f ( z) + z ^2
or f ( x) = 2 f ( z) + Eq_2
Eq (1).& Eq (2) gives
f ( x) = 2 ( 2 f ( x) + x^2) + z ^2

3f (x) + 2 x^2 + ( x / ( x - 1)) ^2 = 0

f (x) = - (x/( x -1))^2 [1 + (x -1)^2] /3
= -(x/( x -1))^2 [x^2 - 2 x + 2] /3

[smusic]

Buruwanta kisima deyak terenne na.a nisa wedakuth na.ap dete.harakge.chik.uba kawda passen yana watayak taram wath ganan ganne.

[shaileshs.]

So you are Turkish . Currently India & Turkiye are not friendly. I think I should unsubscribe your channel. 😂😂