the greatest functional equation of all time.
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- Опубликовано: 28 май 2022
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Another idea: From the case x=y, we see that xf(x) is a fixpoint. In particular, f has *at least* one fixpoint. What if we plug in two fixpoints x and y? Then x² (f(x)+f(y))=(x+y) f(f(x)y) becomes f(xy)=x² and by symmetry also =y², in other words x=y, i.e., f has *exactly* one fixpoint. Then xf(x) must be constant, i.e., f(x)=c/x. Plug this in and solve x²(c/x+c/y)=(x+y)c/(yc/x) for c
dope. as soon as he wrote xf(x) is a fixed point I knew it was 1/x.
Also more efficiently, if f(x)=c/x then c is a fixed point of f then c = 1.
Is there a proble with your symmetry where you get =y^2?
@@vinbo2232 Just exchange the arbitrary symbols x and y to get f(yx) = y^2, and note f(yx) = f(xy) because yx = xy.
@@Xeroxias Right, thank you.
That's really pretty.
Here's what's amusing. If you add 0 to the domain and range, the the function f(x)=0 is the only solution (fairly easily proved, by letting x=1 and y=0 and following your nose). Very strange, that the addition of one point to the domain and range can cause the functional equation to be satisfied by something completely different. [Also: If you only add 0 to the range, then the function f=0 is a second solution; if you only add 0 to the domain, then there is no solution. Very weird.]
If you only add 0 to the range, then f(f(x)*y) becomes f(0) when f(x)=0, which is invalid. So, 0 can't be part of the range if it isn't part of the domain.
7:35 Homework
7:46 To je res dober kraj za ustavitev… Btw, one of my friends spent one year at University of Ljubljana. She liked the country but she had some health issues which impacted her grades.
To je res dober kraj za ustavitev :D
@@lovrodrofenik1455 Edited, thanks !
🥲
Slave!
Watching Janja Garmbret is just so motivating, all in all the bouldering competitions are my favourite sport events to follow. I hope she comes back as well as the YT livestreams..
Greetings from Slovenia :) So glad to hear you include Janja Garnbret!
7:30 When fixing one variable but not the other, it is necessary to say, " . . . but Y remains wild and free."
My approach: from x=y, we find that f(xf(x))=xf(x), i.e. xf(x) is a fixed point. Special case: f(f(1))=f(1). Fixed points are sometimes useful in solving functional equations. So let's suppose we consider an arbitrary fixed point z. Plug x=z(=f(z)) and y=1 into the equation to get z^2(z+f(1))=(z+1)z => z^2+(f(1)-1)z-1=0.
This last equation is true for ALL fixed points of f, so in particular for z=f(1). Plug that in to find f(1)=1 (using the fact that f(1)>0). But then the equation becomes z^2-1=0, hence z=1 (because z>0). Using now the fixed point z=xf(x), we get f(x)=1/x. And indeed this is a solution to the functional equation.
You’re amazing, Michael. Always love watching your videos.
I love functional equation. Especially when professor penn uploads them!
i absolutely love when you do functional equations and integrals 💞💞
2:02 Helpful tip just in case anyone thought that maybe 2=0 😄
the real number _2_ does not equal the real number _0,_ but sometimes we do have _2_ = _0_ , e.g. the integer-modulo-2 _2_ equals the integer-modulo-2 _0_ :p
@@schweinmachtbree1013 what about small values of 0?
Nicely done!
When is the collab with Janna Garmbret coming?
Functional equations are really fun!
Here's another fun way to do it! Suppose there exists an a such that f(a) = 1. Setting x = a gives a^2(1+f(y)) = (y + a)f(y) and we can solve for f(y) = a^2/(y - a^2 + a).
Then setting x = y (like you did) shows xf(x) is a fixed point, which means x = 1 is a fixed point, so f(1) = 1 = a^2/(1 - a^2 + a), so 1 - a^2 + a = a^2, which gives the quadratic you got (but with a the input instead of the output!) 2a^2 - a - 1 = 0, which gives a = 1.
I was surprised to see the result. This ones different yet still simple.
This was actually surprisingly easy.
Excellent!!!
Nice and succinct.
by symmetry/commutativity: x^2*f(f(y)x) = y^2*f(f(x)y)
This suggests f(x) = 1/x because
(x/y)^2 * f(f(y)x) = f(f(x)y)
if we invert x and y to get
(y/x)^2 * f(f(1/y)/x) = f(f(1/x)/y)
Hence if f inverts then it will simply invert inversions(reciprocals) and all that matters is that it all works out which it does. One can almost read off this solution after reducing by symmetry. It isn't very rigorous and requires a bit of insight but it's quick. If you want you can multiple the last two together to get
f(f(y)x) * f(f(1/y)/x) = f(f(1/x)/y) * f(f(x)y)
And this is more obvious that inversion is a symmetry in the functional equation(if f is a homomorphism then it will have to carry inverses to inverses). This doesn't give us all functions but it's another quick and dirty approach.
I' m used to seeing Janja in thumbnails, but it took me some time to realize this wasn't a World Cup video
What about f(x)?
I'm finally starting to get how to approach functional equations!
Makes me wonder if it's even possible to have a sort of unified theory of functional equations. (If that even makes any sense.)
I think it always includes some trial and error, starting with plugging in pairs like (x,x) , (0,y) , (x,0) , and the like. Michael only shows the successful route, but it's totally normal to step into some dead ends before that.
By the time you're good at it, you'll have that theory nailed down in your head, but putting it into a comprehensive written text won't be easy.
@@taufiqutomo Fair but the process of trying to do so seems like it would be very elucidating.
Implicit Function Theorem is gonna be a good start for it
Unified theory of functional equations is otherwise known as Computer Science since the lambda calculus is Turing complete
Didn't expect to hear Janja get mentioned on this channel lol, though I thought i heard she took the rest of the season off to project La Dura Dura.
The thumbnail lol
So (0,oo) means that 0 is not included. Sorry i didn't know the american/english notation for interval ( in french [ to include, ] to exclude left bound).
Thanks for your quick answer.
Some notes from solving before playing the video, so I didn't know f was only on (0, infinity):
Given the presence of f(af(b)), I went straight to looking for a monomial solution. The presence of an x+y term means x and y have the same dimension. Let the dimension of x and y be [x], and [f(x)] = [x]^a. Plugging this into the equation gives a+2 = 1+a(a+1), or a = 1 or -1. We therefore know we're looking at the form f(x) = Ax + B/x. The cases x=y=1 and x=y=i give B+A and B-A as fixed points, and A=0, B=1 follow as the only monomial solution. Not a thorough proof, but it works on C\{0}.
Functional Equation is a very popular math lesson
Why is that the only one?
This is a pretty nice method to start functional equations 🍺🍺🍻.
asnwer= 1 oo isit
there is another solution f(x)=0. if we put x=0, y=y=any real number. then LHS=0, RHS=y*f(f(0)*y)..now since y=any real value so..f(f(0)*y)=0 =>f(0)*y=a root..now if f(0) not=0, then for every 'y', there is a root, that is entire positive real number domain. hence f(x)=0 is the only function which satisfies it.. so f(x)=0.
Yeah, 0 is not in our range
Nice
Is there a way to make a system of functional equations where the answer is f(x)=sin(x)? You can get the periodicity by f(x+2*pi) = f(x+pi) = f(x) but that could just also be a constant function. Adding f(x+pi/2)= -f(x+3*pi/2) = 1 means it's non-constant at least, but idk.
The closest I know is f(x - y) = f(x)g(y) + f(y)g(x). If you require f and g are continuous (at just a single point) then for some number a it must be f(x) = cos(ax) and g(x) = sin(ax). I don't think it's easy to force a = 1 to be the unique solution in a functional because the thing that distinguishes period 2pi from all other periods are "nice" differential properties like y'' + y = 0
@@CraigNull Cool, thanks! I’ve been messing around with this in maple and it looks like there will might always be an infinite number of a this will work for, when I solve it gives about 20 and says solutions may have been lost, that’s usually what that means anyway.
Hello Sir did I spot a paradox in your solution to the cubic equation? You divide by a nonzero variable a, yet you come up with two nonzero solutions meaning the third solution of the cubic was the a = 0 you divided by? I'm intrigued. Amazing functional equation by the way thank you, your explanation was accessible to someone who has had classroom instruction on finite difference methods surely. PS) I have checked we get a 0/0 indeterminate form. Does this always happen with cubics that go through zero? Elsewhere in the comments it appears we don't need to solve it that way.
Still not sure, what is the best approach to solve such functional equation. Obviously one needs a starting point and can iterate through the results. Practically this is just a simplification of doing the same with the taylor series representation, since the search is restricted to continuous functions. Well. But just setting "x=y" while these values are not even symmetric, sounds odd to me.
a reminder that checking is always important, as i got to the x(f(x)=f(x(fx)), and then set xf(x)=a, so f(a)=a... and thought that was the function, but if i would have plugged that in it would give x²=y², soonly is true if x=y... checking is important
Isn't f(x) = x also a solution?
Why can't you use 2:50 directly to infer f(1)=1?
we don't yet know that f is injective
Can we assume there exists x0 s.t. f(x0)=1? Not sure if f:(0,infinity)->(0,infinity) means that every a€(0,infinity) is an image of f.
If not, then rearranging the equation to (f(x)+f(y))/(x+y)=f(f(x)y)/x^2=f(f(y)x)/y^2, by the symmetry in x,y of the LHS. Setting y=1 and f(1)=a gives f(f(x))=x^2*f(ax). In the original equation, setting y=1 and using this result gives f(x)+a=(x+1)*f(a*x). Then setting x=1/a and we have f(1/a)+a=(1/a+1)*a=a+1 or f(1/a)=1. Finally setting x=1/a in the original equation, we can solve for f(y) depending on a. a is fixed by setting f(1/a)=1.
initially i included 0 in the domain, which led me to some hasty conclusions, unfortunate.
sub 1: y=x. get 2 x^2 f(x) = 2x f(x f(x)).
-> divide by 2x to get x f(x) = f(x f(x)), therefore x f(x) is a fixed point of f.
sub 2: y=x/f(x). get x^2 f(x) + x^2 f(x/f(x)) = (x+x/f(x)) f(x).
x^2 f(x) + x^2 f(x/f(x)) = x f(x) + x. divide all by x f(x).
x + f(x/f(x)) x/f(x) = 1 + 1/f(x)
i have no idea where to go from this.
sub 3a: y=1. f(1) = 1 f(1) = f(1 f(1)) = f(f(1)), therefore f(1) is also a fixed point.
sub 3b: y=1. x^2 f(x) + x^2 f(1) = (x+1) f(f(x)) = x f(f(x)) + f(f(x))
divide by x^2(x+1): (f(x)+f(1))/(x+1) = f(f(x))/x^2. nowhere to go.
sub 4: y=1/f(x). x^2 (f(x) + f(1/f(x))) = (x+1/f(x)) f(1). again nowhere, along with y=1/x
sub 5: x=1/y. f(1/y) + f(y) = y^2 (y+1/y) f(y f(1/y)). nothing.
okay this is getting annoying.
sub 6: y=f(1). x^2 (f(x) + f(1)) = (x+f(1)) f(f(x)f(1))
x=f(1) would be redundant, so maybe x=1
2f(1) = (1+f(1)) f(f(1)^2)
terrible.
f(x)=x also satisfies the equation. Substitute y=x in the original equation and we get x^2(2*f(x)) = 2x*(f(f(x)x)). Since f(x)=x; 2*(x^2)*x= 2x*(f(x*x)) ; 2x^3=2x*f(x^2); Since f(x)=x; f(x^2)=x^2; So the final equation is satisfied. 2x^3=2x*x^2 on both the LHS and RHS.
If x≠y and f(x)=x, our equation becomes: x²(x+y)=(x+y)xy which is false.
It's cute that a seemingly complex functional equation has a simple function as its solution!
This is typically the case on nearly all Olympiad FEs
Since f(1) = f(f(1)), is it not true that 1 = f(1), so there is no need to introduce "a" and solve for "a"?
For instance, say that f(1) not equal 1. Then say that f(1) = b, where b not equal 1. Then we have b = f(1) = f(f(1)) = f(b). Since b was arbitrary, this can only be true if f is the constant function. But f cannot be the constant function (say f = c) because then we would have x*x (2 c ) =(x+y) c, which cannot hold for all values of x and y.
So f must not be the constant function, and f(1) = 1. If we set x = 1, the original equation becomes 1*1 (1 + f(y)) = (1 + y) f(y) or 1 + f(y) = f(y) + y f(y) or 1 = y f(y) or f(y) = 1/y
"Determining all the functions" requires to prove that there are no other functions also... f(x)=1/x solves the equation following the video, where f(x)=1/x > 0 for all x in (0, infinity).... but the equation f(x)=0 also solves the equation, which is not considered.
0 is not in the range of the function, which is the open interval (0, infinity).
The work done in the video proves that whatever f(x) is, it has to satisfy the equation f(x) = 1/x. So uniqueness is proven as well as existence.
The greatest functional equation of all time is still Feigenbaum's equation: a g(g(x/a)) = g(x), with g(x)=g(-x) and g(0)=1, which can be solved for g and a simultaneously, giving the behavior of period doubling cascades.
Just set y=0 and you get the result
How we can ensure that this function is the only function which satisfy the functional equation
I guess we can be sure because we haven't done anything to loose generality and so we get the whole family of solutions (for example, for some functional equations you can get answers involving arbitrary constants which is gonna be a family of solutions, and I bet it's gonna be complete unless you've done something to loose generality)
Can anyone explain why the first conclusion is that there is a fixpoint (new concept to me) instead of the function being the identity function? xf(x) = f(xf(x)) seems like the identity function to me.
You can only conclude this if you know the argument (xf(x)) spans the entire range of f (0 to infinity). Otherwise you only know that for some argument you get the argument as the result. In this particular case it turns out that f(x)=1/x so the argument xf(x) is always 1.
One way of thinking about it is that the identity function has fixed points for all arguments but here we have only found a single fixed point.
Why single? X and Y are arbitrary, right? He even replaces it with a, and concluded and continue from f(a)=a.
xf(x)=f(xf(x)) only holds for x=y.
If x≠y and f(x)=x, our equation becomes: x²(x+y)=(x+y)xy , which is false.
So we can't take y and x equal to zero ? Because if it's the case then, with y=0 and x=1 we have f(1)=0 and with x=1 we have (E) : f(y)=(1+y)f(0) cause f(f(1)y)=f(0), so we have f(0)=0 cause f(1)=0 in (E) so (E) : f(y) = 0...
the domain of f doesn't include 0
@@schweinmachtbree1013 thanks
but wait... you defined f on [0,inf[ but your solution is not defined on 0. Should it be accepted?
Looking at specific values (0,1...) you can get:
* x=0 => 0=y.f(y.f(0)) for any y so f(y.f(0))=0 for any y, in particular when y=0 f(0)=0
* y=0 => x2f(x)=0 for any x so f(x)=0 for any x
and there you go, another function (f:x->0) matching the equation.
you can easily verify it as the equation becomes: x2(0+0)=(x+y)*0 which is true no matter x and y
3:46 x=1, a=x, this means a =1, I think all the substitution of a in the equations and solving the quadratic equations for a, to find it is 1, is unnecessary
And to be more accurate, we do not even the "a" at all,
f(1) = f( f(1) ), that means f(1) = 1, then we continue from the part of substitution ( x=1, y=y)
I'm sure I'm being nitpicky here, and I know it's commutative, but wouldn't it make more sense to write y=x when setting all the y's to x's, instead of writing x=y? When I read x=y, I read that to mean take all the x's and set them to y. I know I know, who cares!
Had to do a double take on the title of the video after seeing Janja lmao
This was an easy one
Oh, f😂ck, it sounds like a goat)))
Is it indeed the greatest of all time?
Certainly not. There are way better ones.
In terms of hardest routes sent, there’s several who have done better. But the claim is that she’s the greatest competition climber of all time, and in that regard she’s the undisputed goat. Nobody else comes close to her level of domination, men or women.
Flammy probably proud you said "2 is not equal to 0"
Plugging x = 1 and y = 0, we have f(1) + f(0) = (1 + 0)f(f(1)0) = f(0) ==> f(1) = 0 !!!
I think you forget the case a = 0 in the equation 2a^3 = a(a+1).
Plugging x=1, whe have f(1) + f(y) = (1 + y)f(f(1)y) , as f(1) = 0 we obtain f(y) = (1 + y)f(0) ==> f(0) = f(y)/(1 + y). set y =1 ==> f(0) = 0
Plugging y = 0, whe have x^2f(x) = xf(0) = 0 => f(x) = 0. So according to me, the only solution in domain [0, oo] is f(x) = 0
What is your opinion ?
you cannot set y=0...
Stating something is "the greatest of all time" is sensationalistic.
Still not a fan of functional equations.
3rd view
First view 🤩😍
I disagree with the title, pretty much an average functional equation with straightforward solution.
I probably do too, but you know... gotta get the clicks. At least this one doesn't have f(x)=x as the only solution!
And it's a good reference to arguably the best competition climber of all time.
@@wavyblade6810 I figured that was the reason for the title