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.
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.
- 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
I hoped there was a non-trivial solution for some other field - e.g. one of characteristic 2, but the proof is even quicker: For any x, we have 0 = f(0)-f(0) = f(x+x)-f(x-x) = f(x)^2, and hence f(x) = 0.
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.
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 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.
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 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.
@@SyberMath the length is right as it is. Actually, it is impossible to satisfy everybody: there's always somebody complaining about some particular point. These people complaining about length can use fast forward to skip details. But people at not that high level in maths will like your explanations. Congrats for your channel, It is great!
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.
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
After all the arguments, we prove that f(x)=0
😮😜
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.
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.
I was able to follow the steps. It went great! I love watching the functional equation videos. I feel like my brain gets more flexible. Thanks!
My pleasure!
Awesome 👍👍
- 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
I hoped there was a non-trivial solution for some other field - e.g. one of characteristic 2, but the proof is even quicker: For any x, we have 0 = f(0)-f(0) = f(x+x)-f(x-x) = f(x)^2, and hence f(x) = 0.
i found that f(x) was zero but kept trying because i thought there was another solution. Its like a math rickroll
Could you please say, how one would know that the answer satisfies the condition...Thank you😊
You can use substitution to check if I understood correctly
Thank you
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.
Nice!
Thanks!
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.
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.
@@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.
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.
Great job!
Good one
Thanks
Polska Górą 🇵🇱🇵🇱🇵🇱🇵🇱🇵🇱🇵🇱
Bobr kurwa!!!11
Bolzga gurom
it's Poland, not Polland
You right , But your correction wasn't that needed.
That’s right!
L or double l ....not that important
@@jadali4150 You're probably not from Holland.
Maybe from hell
I feel betrayed😅
So close. If it had only been f(x+y) + f(x-y) = f(x)*f(y).
What if f(x,y) = dy/dx ?
It's a function in one variable, not two.
😊😊😊👍👍👍
f(0)=0
(f(x+h)-f(x-h))/(2h) = f(x)/2 * f(h)/h
h -> 0
f’(x) = f(x)/2 * (f(h) - 0)/(h - 0) = f(x)/2 * (f(0 + h) - f(0))/(h - 0) = f(x)/2 * f’(0)
But when x = 0
f’(0) = f(0)/2 * f’(0) = 0 because f(0) = 0
f’(x) = 0
==> f(x) = 0
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
f(x)*f(0)=f(x+0)-f(x-0)=0 => f(0)=0
f(0+y)-f(0-y)=f(y)*f(0)=0 => f(y)=f(-y)
f(x+x)-f(x-x)=f(x)*f(x) => f(2x)+0=f(x)^2
f(x-x)-f(x-(-x))=f(x)*f(-x) => -f(2x)-0=f(x)*(-f(x))= f(x)^2
So
f(2x)=f(x)^2-f(2x) => f(2x)=0
This video should be shortened to 2 minutes. Too many words.
Why?
@@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.
@@SyberMath the length is right as it is. Actually, it is impossible to satisfy everybody: there's always somebody complaining about some particular point. These people complaining about length can use fast forward to skip details. But people at not that high level in maths will like your explanations. Congrats for your channel, It is great!
@@moonwatcher2001 Thank you! 😍
What a useless function that is always zero.. :D
Exactly! 😀
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. 😋😋😋😋😋😋
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.
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.
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
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