A Functional Equation from Samara Math Olympiads

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[@ioannismichalopoulos4173]

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?

[@dominiquelarchey-wendling5829]

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.

[@leif1075]

Wjy would younwrite g(g×) at.all that sems and I think is random.and out of nowhere is it not??

[@thecoconutstar9327]

We got sybermath lore 🗣️🗣️🗣️🔥🔥🔥🔥

[@freddyalvaradamaranon304]

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. 😊❤😊.

[@olympiadeduworks]

thank you for sharing

[@Igorious92]

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

[@РайанКупер-э4о]

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.

[@Blaqjaqshellaq]

The solution can also be written as f(x)=[x^2*(x-3)*(x-1)]/[-3*(x-1)^2]

[@narfharder]

Whoops, don't forget the 2 in front of x^4, the numerator has complex factors. You're right that he should pull out x^2, anyway.

[@mega_mango]

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

[@КириллГарбузов-й8м]

САММАТ😍

[@SyberMath]

😍

[@Chrisoikmath_]

Love Turkey!❤ From Greece!

[@VicenteMReyes-vs9nh]

Nice problem!

[@cicik57]

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

[@solaymanelbellaj2608]

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

[@dsafdaa8874]

I love u're vdo form Cambodia 🇰🇭

[@SyberMath]

Thank you! Love from the US

[@Ahwke]

greatly❤❤

[@climbnexplore1187]

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

[@KeshkKeshk-l4z]

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.

[@SyberMath]

Totally agree! 😍

[@scottleung9587]

Nice!

[@ForGu-dv7wz]

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

[@jesusk1358]

Year 11 Mathematical Methods in Australia.

[@MadScientyst]

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

[@SyberMath]

Glad to hear that!

[@surendrakverma555]

Good 👍

[@SyberMath]

Thanks

[@徐瑞斌-i8o]

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

[@YassJ-jd5sz]

That's what he did using t/t-1. X and t are the same. But in order to find out what to put in the equation you need to do the first steps he did, you can't guess it right away

[@jesusk1358]

@@YassJ-jd5sz You don't guess it. You can actually see it right away.

[@leif1075]

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..

[@shantanav31]

Putting x=2 we get f(x)= -x^2

[@andrew-watts]

Hey it's good one

[@SyberMath]

Hey thanks!

[@heastel]

САММАТ САММАТ САММАТ

[@phill3986]

✌️👍👏😎😎👏👍✌️

[@annanemustaph]

nice proof 🌵

[@SyberMath]

Thank you

[@pjb.1775]

ben de diyom bu adam kesin türk geröekten de öyleymiş

[@SyberMath]

Ahaha. Senden kacmiyor üstat 😁

[@Kaan.-]

cidden türkiyede mi büyüdün türkçe biliyor musun???

[@douglassmith1466]

Prove solution is unique, or not.

[@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)] = f{[x/(x-1)]/[x/(x-1)-1]}-x²/(x-1)² ⟹ 2*f[x/(x-1)] = f{[x/(x-1)]/[(x-x+1)/(x-1)]}-x²/(x-1)² ⟹ (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*x²*(x²-2x+1)-x²]/[3*(x-1)²] = [-2*x^4+4x³-2x²-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]

[@tonifreitas5695]

TA ERRADA X3-X2-2X=0

[@satrajitghosh8162]

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) + z^2........ 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

[@smusic677]

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

[@gopalakrishnapillai5911]

It is not at all an intelligent explanation.

[@SyberMath]

Thanks for the feedback!

[@shaileshs.3177]

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