An Olympiad Problem from Poland 🇵🇱

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1 июл 2024

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Комментарии : 42

@s1ng23m4n 6 дней назад

Replace y with -y and with given eqn we will get a system: { f(x+y) - f(x-y) = f(x)f(y) { f(x-y) - f(x+y) = f(x)f(-y) Sum these eqns: 0 = f(x)(f(y) + f(-y)) Suppose f(x) = 0 then its done. Suppose f(y) + f(-y) = 0 then f(y) - odd function, stop what? But we already prove that f(x) is even(at 4:11) ))) So f(x) and even and odd function. What function it can be? Right, f(x) = 0.

@krishnanadityan2017 4 дня назад

After all the arguments, we prove that f(x)=0

@SyberMath 3 дня назад

😮😜

@GirishManjunathMusic День назад

f(x + y) - f(x - y) = f(x)f(y) Setting y = 0, x is free: f(x) - f(x) = f(x)f(0) either f(x) = 0 for all x, or f(0) = 0. setting x = 0, y is free: f(y) - f(-y) = f(0)f(y) ∴ f(-y) = f(y) setting y = -y, x is free: f(x - y) - f(x + y) = f(x)f(-y) ∴ f(x + y) = f(x - y) ∴f(x)f(y) = 0 for all f(x),f(y) setting x = y: f(2y) - f(0) = f²(y) but we know that f²(y) = 0 for all y ∴ f(2y) = f(0) for all 2y, and as f(0) = 0; f(y) is a constant function = 0.

@cicik57 4 дня назад

- substitute 0,0 : f(0) - f(0) = f²(0) => f(0) = 0 - substitute 0,n:f(n) -f(-n) = f(0)f(n) =0 so f(n) = -f(n), so f is symmetrical - substitute x = a+b, y = a-b: f(a) * f(b) = f(2x) - f(2y) = f(b+a)(b-a) (because f is symmetrical) = f(2y) - f(2x) so 2f(2x) = 2f(2y) or f is constant, but we figured out above that f = 0 - answer: f(anything) = 0

@scottleung9587 6 дней назад

Nice!

@SyberMath 6 дней назад

Thanks!

@Cow.cool. 2 дня назад

i found that f(x) was zero but kept trying because i thought there was another solution. Its like a math rickroll

@majora4 2 дня назад

I finished up from step 4 in a slightly different way. • -f(2x) = f(x) * f(-x) • -f(2x) = f(-x) * f(-x) from step 2 • -f(2x) = [f(-x)]^2 • -f(2x) = f(-2x) from step 3 Previously we deduced that f(x) is even, and we just now deduced f(x) is odd too. The only function that's both even and odd at the same time is f(x) = 0.

@orchestra2603 6 дней назад

Beautiful👍 I did it in a bit different way though. I figured out that f(0)=0, but missed out on the even/odd function argument However, I noticed that if we put y=∆x and ∆x->0, then in the limit the LHS gives 2∆x*f'(x) and the RHS gives f(x)*f'(0)*∆x, because f(dx)=f(0+dx)=f(0)+f'(0)dx=f'(0)dx. This all of course rests on the assumption that this limit exists, and f is differentiable at x. Then, we have 2 f'(x) = f(x)*f'(0). The solution of this ODE is f(x) = f(0)*exp(f'(0)*x/2). But because f(0)=0, then f(x) = 0 for all x.

@SyberMath 2 дня назад

Great job!

@piotrsz1359 6 дней назад

Polska Górą 🇵🇱🇵🇱🇵🇱🇵🇱🇵🇱🇵🇱

@roman_roman_roman 5 дней назад

Bobr kurwa!!!11

@davidsousaRJ 2 дня назад

If we substitute y = 0 and keep x we get f(x) = 0 as a solution since the beginning. But then we have to keep trying other substitutions until we verify no other solutions are possible.

@michaelfaccone5811 День назад

Maybe I don't get how you're doing this. If I substitute y=0 leaving x free, I get f(x)-f(x)=f(x)*f(0), which simplifies to 0=0, not f(x)=0.

@davidsousaRJ День назад

@@michaelfaccone5811 you get 0 = f(x)*f(0), therefore, either f(x) = 0 or f(0) = 0. I did not know that f(0) = 0 yet, this was the first substitution I have made.

@DutchMathematician 3 дня назад

The way I solved it. In such questions it is often beneficial to insert x=0, y=0 (or both), as well as x=y and/or x=-y. I proceeded as follows. Fill in x=0 and y=0. This gives: f(0)-f(0)=0=f(0)*f(0) Hence f(0)=0. Substituting x=y, we get (for arbitrary x): f(2*x)-f(0)=f(x)*f(x) or: f(2*x)=f(x)*f(x) This means that f(2*x) is the product of two identical real numbers, hence f(2*x) is non-negative. Since x was arbitrary, we can conclude that f(2*x) is non-negative for all x and hence the same holds for f(x). Now substitute y=-x in the general equation. We get: f(0)-f(2*x)=f(x)*f(-x) or (since f(0)=0): -f(2*x)=f(x)*f(-x) Since we've concluded that f(x)≥0 for every x, this only can mean that f(x)=0 for every x.

@phill3986 6 дней назад

😊😊😊👍👍👍

@waldenherz9944 День назад

I feel betrayed😅

@neuralwarp 6 дней назад

What if f(x,y) = dy/dx ?

@SilviuBurcea1 6 дней назад

It's a function in one variable, not two.

@dominikwolski2274 6 дней назад

it's Poland, not Polland

@MateusMuila 6 дней назад

You right , But your correction wasn't that needed.

@SyberMath 6 дней назад

That’s right!

@jadali4150 6 дней назад

L or double l ....not that important

@rob876 6 дней назад

@@jadali4150 You're probably not from Holland.

@jadali4150 6 дней назад

Maybe from hell

@leonidfedyakov366 3 дня назад

This video should be shortened to 2 minutes. Too many words.

@SyberMath 2 дня назад

Why?

@leonidfedyakov366 2 дня назад

@@SyberMath слишком детские объяснения. Кто такие видео смотрит, и так быстро всё уловит. Тем более те, кто функциональные уравнения такие решает. Я хоть и закончил Прикладную математику в МАИ, но отродясь такие уравнения не решал. Большая часть рассуждений кажется излишний, слишком много слов от (f(0))^2=0 до вывода f(0)=0, например, или о пользе сохранения предыдущих записей. 10 с половиной минут - слишком много для такой задачи.

@tontonbeber4555 5 дней назад

f(x+y)-f(x-y) = f(x)f(y) (a) y=0 => f(x)-f(x) = 0 = f(x)f(0) => f(x)=0 or f(y)=0 the second in inside the first, so f(0)=0 (b) x=0 => f(y)-f(-y) = f(0)f(y) = 0 => f(y)=f(-y) (c) y=x => f(2x)-f(0) = f(2x)=f(x)² (d) y=-x => f(0)-f(2x) = -f(2x) = f(x)f(-x) = (b) f(x)² (c,d) => f(2x)=-f(2x) => f(2x)=0 and so f(x)=0 forall x

@alextang4688 6 дней назад

f(x+y)-f(x-y)=f(x)*f(y) Put x=x, y=0 f(x)-f(x)=f(x)*f(0) f(0)*f(x)=0 Therefore f(0)=0 or f(x)=0 In short f(x)=0 answer. 😋😋😋😋😋😋

@AltAaltonnov 5 дней назад

But we can show that f(0) = 0, so at that point f(x) can still be anything... As you say it's or not and.

@andypandy6063 2 дня назад

What a useless function that is always zero.. :D

@SyberMath 2 дня назад

Exactly! 😀

@dominiquelarchey-wendling5829 13 часов назад

f(x+y)-f(x-y) = f(x)f(y) x,y := 0 gives f(0)-f(0) = f(0)f(0), hence f(0)² = 0. Thus f(0) = 0 x := 0 gives f(y)-f(-y) = f(0)f(y) = 0. Hence f(y) = f(-y) for any y. y := -y gives f(x-y)-f(x+y) = f(x)f(-y) = f(x)f(y) for any x,y Thus we have for any x,y we have f(x+y)-f(x-y) = f(x)f(y) f(x-y)-f(x+y) = f(x)f(y) adding both gives 0 = 2f(x)f(y) hence f(x)f(y) = 0 for any x,y Hence with y := x, we get f(x)² = 0, thus f(x) = 0 for any x