Wouldn't you have for x=0, f(0)^2f(1)=0 => f(0) and/or f(1) =0. But we also have if x=1 f(1)^2f(0)=1 which is not possible because it should be equal to zero so no such function from C to C?
So a gold rush🏅⛏️ to be soon expected with lots of prospectors turning up with their pan's for nuggets...... Let's hope none make the accusation of 'Alchemy!'! 😜
Wow, you are the first one that actually cares to explain what you are doing, most math channels assume a lot, and this channel goes step by step, perfect for beginners like me who want to reach this level. Btw any resources you recommend to learn how to do challenging problems? :p
Wow, thanks and welcome to the channel! Absolutely! There are quite a few books out there, depending on what you wanna focus, algebra, geometry, number theory or counting. I will just list a few here for general purposes: EASIER * Challenging Problems in Algebra: www.amazon.com/Challenging-Problems-Algebra-Dover-Mathematics/dp/0486691489 * Challenging Problems in Geometry: www.amazon.com/Challenging-Problems-Geometry-Dover-Mathematics/dp/0486691543 * Competition Math for Middle School: www.amazon.com/Competition-Math-Middle-School-Batterson/dp/1441488871 * More Mathematical Challenges: www.amazon.com/exec/obidos/ASIN/0521585686/ HARDER * Problem-Solving Strategies: www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191 * ARML-NYSML Contests, 1989-1994: ARML-NYSML Contests, 1989-1994: www.amazon.com/exec/obidos/ASIN/0962640166 * Polynomials: www.amazon.com/Polynomials-Problem-Books-Mathematics-Barbeau/dp/0387406271
@@joaquingutierrez3072 download advanced problems in mathematics by Vikas Gupta from the internet. (You’ll find its pdf for free) It’s a book oriented for an entrance exam in india but it’s questions are pure gold.
I did not get that inverse trick :(. But when you got the system of equations I solved it by myself. I wrote the first equation as (f)^2 * fg = L (1) Where L is just the identity function, g(x) = (1 - x)/(1 + x), fg means composition and f * g product And we have (fL)^2 * fgL = L Then the substitution x = g(x) is equivalent to "substituing" L ----) g, then we get (fg)^2 * fgg = g And using the fact gg = L, we get (fg)^2 * f = g f * (fg)^2 = g (2) And instead of dividing I think is better to square (1) as you did and then substituing (2) in the new expression, that is: ( (f)^2 * fg )^2 = L ^2 (f)^4 * (fg)^2 = L^2 (f)^3 * f * (fg)^2 = L^2 (f)^3 * g = L^2 then f = cube-root( L^2/g ) Doing it this way It is only necessary to assume that x =/= -1 (because of g) And it makes clearer that this is true for any g such that gg = L I really liked this equation. I would love to see more functional equations here :). Thanks for the video !!!
if i divide x by (1-x)/(1+x) = x(1+x)/(1-x) should this be the squared function, i miss a multiply by x and the 1/3 power of it in your answer, i thought it was SQRT(x(1+x)/(1-x))?
If (1-y)/((1+y)=f^(-1)(y) -> f((1-y)/(1+y))=y. You could replace y with x. But x is something different. So, your conclusion at 2:51 is not correct. y in this sense has nothing to do with f(x). You let you lead yourself into the woods by an inept choice of nomenclature. I think this is homemade. Let x=0 -> (f(0))^2 * f(1) = 0. Let x=1 -> (f(1))^2 * f(0) = 1. Contradiction. -> There is no solution.
Didn't expect this to happen, but I just assumed f(x)=mx+b. Then after substituting and multiplying out, it's obvious that b=0 because there are no constant terms on the right side. Then solving for m you arrive at m=[(x+1)/[x(1-x))]]^(1/3). Substituting your values for m and b back into f(x) gives you [x^2(x+1)/(1-x)]^(1/3) which was the answer in the video.
@@SyberMath My guess is that this only works since the answer is a single additive term, so your guess would have to have at least as many terms as the actual answer.
Well, you didn't really assume it was a linear equation, you assumed f(x) = xg(x) for some other g(x). You plugged it in and solved for g(x). This is a great method! It appears in differential equations as "variation of parameters" or "reduction of order".
Wow, this is a really confusing task. Just try the initial equation for x=1 and x=0 ... do you see the issue too? At least in the end the pole at x=1 explains everything.
When you reached (f(x))³=x²/1-x/1+x why didn't you multiply 1+x/1-x on both sides you only multiplied it to the right side?. Law of equality?. I would appreciate an answer
The problem's solution implies that f(x) does not exist at: x=1 and accordingly, f((1-x)/(1+x)) does not exist at: x=0. Moreover, (1-x)/(1+x) does not exist at: x=-1. Therefore, the given equation cannot be satisfied at: x=0 or x=±1.
x+(1/x)+4=y [x+(1/x)]^3=(y-4)^3 x^3+(1/x^3)+3(x+(1/x))=(y-4)^3 x^3+(1/x^3)=(y-4)^3-3(y-4) x^3+(1/x^3)+16=(y-4)^3-3(y-4)+16 f(y)=(y-4)^3-3(y-4)+16 ---> replace y with x f(x)=(x-4)^3-3(x-4)+16 ---> replace x with 17 to find f(17)
Bonjour de France! Je ne comprends pas d’où vient « x=(1-y)/(1+y)=f-1(y). Mais en admettant que cette ligne est justifiée, l’egalite x=f-1(y) donne f(x)=y, donc f(x)=(1-x)/(1+x), et la fonction f est déterminée....mais fausse!!! Où est l’erreur?
Je pense qu'il appelle (1-x)/(1+x) comme f(x) momentanément pour illustrer que cette expression est son propre Inverse, mais il ne parle pas de la fonction du problème, mauvais choix du nom :/
J'ai bugger au meme moment, mais comme @Sergio a dis, il a fait un tres mauvais choix de nom de fonction qui mene a confusion.. Limite, si tu ignores le moment ou il essaye d'expliquer que la fonction est involutive, tout le reste s'explique intuitivement.
Yesterday i tried to learn the logic behind this kind of problems by looking at like 3 of your vids on it and this was the first one i tried and i got it right :). Never did something similar before, thanks !!
If you plug in x=0 to the equation you get f(0)^2 * f(1) = 0, meaning at least one of f(0) and f(1) is 0. If you plug in x=1 you get f(1)^2 * f(0) = 1. From plugging in x=0 we know that f(1)^2 * f(0) = 0. Thus we have shown 0 = 1. Doesn't this mean the function f doesn't exist, at least for x=0 or x=1?
You are right. The problem's solution implies that f(x) does not exist at: x=1 and accordingly, f((1-x)/(1+x)) does not exist at: x=0. Moreover, (1-x)/(1+x) does not exist at: x=-1. Therefore, the given equation cannot be satisfied at: x=0 or x=±1.
@@SyberMath Thank you very much for replying! I really enjoy solving the problem which your channel gives! I’ll forever be your fan!!! I’ll always love and support you!
@@SyberMath Do you think most 0people would think of that inverse thing honestly? Because I don't think they would..isnt there another wway of solving..I firdt solved for f(x) to get that f(x) equals x^1/2 times 1/f(1-×/1+×)^1/2 ..now proceed from there..because you know f(x) is some multiple of square root of x..can't you proceed from there..Hope you can respond..
@@leif1075 You can tell me if that is going to lead anywhere. Yes, if you dealt with some functional equation problems, these would be typical substitutions. Of course, functional equations is a huge topic and there are so many types of questions and very many different methods. Sometimes, it's just trial and error and there are no shortcuts!
@@SyberMath Thanks for responding..but if you've never dealt with functional equations before, do you think it would occur to them..I don't see why..hope you can respond again..
right, you may notice at if (1-x)/(1+x) = t then x= (1-t)/(1+t) so substitute it into main equation you will get: f²( (1-x)/(1+x) ) f(x)= (1-x)/(1+x) then you can divide main equasion on this and you will get f(x) / f( (1-x)/(1+x) ) = x (1+x)/(1-x) from where you get f( (1-x)/(1+x) ) = f(x) 1/x (1-x) / (1+x) subsisute in into main equasion you will get f²(x) f(x) 1/x (1-x) / (1+x) = x or f³(x)= x²(x+1)/(1-x) the same result
you had to make a restriction for the problem( x is not equal 1) because at the beginning the student would face a contradiction because f(0)^2.f(1)=0 for x=0 but f(1)^2.f(0)=1 for x=1
Not just x = 1 is not allowed, but x = 0 is not allowed either for the same reason. Also, x = -1 is obviously not allowed, since (1 - x)/(1 + x) does not exist for x = -1. However, the solution has removable singularities at x = -1 and x = 0, but not at x = 0, which is a branch point and a pole simultaneously.
What that contradiction implies is that either x=/= 0 or x=/= 1, when you solve for f(x) you get necessarily x =/= 1, so that get rid of the problem, I think
@@joaquingutierrez3072 After plugging in x=0 into the original equation, you get f(0)=0 or f(1)=0. Now let's plug in x=1. We get [f(1)]^2 * f(0) = 1. Since f(1)=0 or f(0)=0, [f(1)]^2 * f(0) = 0 ≠1. That's a contradiction.
@@Geo25rey The given equation may be written as: F(x) = x where the function F(x) is defined by: F(x) = (f(x))^2 g(x) where the function g(x) is defined by: g(x) = f(u(x)) where the function u(x) is defined by: u(x) = (1-x)/(1+x) Therefore: (a) f(1) does not exist, according to the problem's solution. (b) u(-1) does not exist. (c) u(0)=1 ⟹ g(0)=f(u(0))=f(1)=1 and therefore g(0) does not exist. Therefore, the function F(x) is undefined for x=0, x=1, or x=-1 and therefore, there is no contradiction in this problem.
Let * denote the function composition operator, and g : R\{-1} -> R\{-1} be x |-> g(x) = (1 - x)/(1 + x). Notice that g is a bijection, and that g*g : R\{-1} -> R\{-1} has g*g = x |-> x because g(g(x)) = x for all x elements of dom(g), which you can find just by experimenting. This experiment may be motivated by the general knowledge that rational function with integer coefficients are, in general, cyclic. As -1 is not an element of dom(g) nor of image(g), it is not an element of dom(f). Additionally, note that 0 is an element of dom(f) iff 1 is an element of dom(f). This is because f(x)^2·f(g(x)) = x for all x elements of dom(f) implies f(1)^2·f(g(1)) = f(1)^2·f(0) = 1, so that f(0) = 1/f(1)^2. Therefore, f(0) exists iff f(1) exists. Also, f(x)^2·f(g(x)) = x for all x elements of dom(f) implies f(0)^2·f(g(0)) = f(0)^2·f(1) = 0, which implies f(0) = 0 or f(1) = 0, which implies f(0) does not exist or f(1) does not exist, because of the relationship between f(0) and f(1). As such, 0 and 1 are not elements of dom(f). With this information, you can assume that solutions f : R\{-1, 0, 1} -> R can be found, and those are the ones I will look for, although a similar analysis can be done for f : C\{-1, 0, 1} -> C instead, if done more carefully. The functional equation can be rewritten as f·f·(f*g) = Id, where Id : R\{-1, 0, 1} -> R with Id = x |-> x, giving that g*g = Id. f^2·(f*g) = Id implies (f^2·(f*g))*g = Id*g. Id*g = g, and (f^2·(f*g))*g = (f^2*g)·((f*g)*g) = (f*g)^2·(f*(g*g)) = (f*g)^2·(f*Id) = (f*g)^2·f, thus (f^2·(f*g))*g = Id*g is equivalent to (f*g)^2·f = g. f^2·(f*g) = Id and (f*g)^2·f = g imply (f^2·(f*g))·((f*g)^2·f) = g·Id. (f^2·(f*g))·((f*g)^2·f) = (f^2·f)·((f*g)^2·(f*g)) = f^3·(f*g)^3 = (f·(f*g))^3, so (f·(f*g))^3 = g·Id, or simply that f·(f*g) = (g·Id)^(1/3) (f*g)^2·f = g. f^2·(f*g) = Id and (f*g)^2·f = g also imply f·(f*g)^2 + f^2·(f*g) = Id + g. f·(f*g)^2 + f^2·(f*g) = (f + f*g)·(f·(f*g)), so (f + f*g)·(f·(f*g)) = (f + f*g)·(Id·g)^(1/3) = Id + g implies f + f*g = (Id + g)/(Id·g)^(1/3). f + f*g = (Id + g)/(Id·g)^(1/3) and f·(f*g) = (g·Id)^(1/3) imply f + (Id·g)^(1/3)/f = (Id + g)/(Id·g)^(1/3). This is equivalent to f^2 + (Id·g)^(1/3) = ((Id + g)/(Id·g)^(1/3))·f, which is equivalent to f^2 - ((Id + g)/(Id·g)^(1/3))·f + (Id·g)^(1/3) = f^2 - ((Id + g)/(Id·g)^(1/3))·f + ((Id + g)/(Id·g)^(1/3))^2/4 + (Id·g)^(1/3) - ((Id + g)/(Id·g)^(1/3))^2/4 = (f - ((Id + g)/(Id·g)^(1/3))/2)^2 + (Id·g)^(1/3) - (Id^2 + 2·Id·g + g^2)/(4·(Id·g)^(2/3)) = (f - ((Id + g)/(Id·g)^(1/3))/2)^2 + 4·(Id·g)/(4·(Id·g)^(2/3)) - (Id^2 + 2·Id·g + g^2)/(4·(Id·g)^(2/3)) = (f - ((Id + g)/(Id·g)^(1/3))/2)^2 - ((Id - g)/(2·(Id·g)^(1/3)))^2 = 0. This is equivalent to (f - ((Id + g)/(Id·g)^(1/3))/2)^2 = ((Id - g)/(2·(Id·g)^(1/3)))^2. Thus f = (Id + g)/(2·(Id·g)^(1/3)) - (Id - g)/(2·(Id·g)^(1/3)) = g/(Id·g)^(1/3) or f = (Id + g)/(2·(Id·g)^(1/3)) + (Id - g)/(2·(Id·g)^(1/3)) = Id/(Id·g)^(1/3). f = Id/(Id·g)^(1/3) & f·(f*g) = (g·Id)^(1/3) imply f^2·(f*g) = Id, hence only f = Id/(Id·g)^(1/3) solves the equation. If desired, this can be written in algebraic notation, in terms of x, noting that Id(x) = x & g(x) = (1 - x)/(1 + x), so f(x) = x/(x·(1 - x)/(1 + x))^(1/3). Interestingly, this function has removable singularities at -1 & 0, and this results in the continuation function f~ : R\{1} -> R : f~ = x |-> f~(x) = x^(2/3)/((1 - x)/(1 + x))^(1/3) = x^(2/3)·(1 + x)^(1/3)/(1 - x)^(1/3), so that f~(-1) = f~(0) = 0, and 1/f~(1) = 0, which was actually suggested from the previously derived relation f(0) = 1/f(1)^2, where, had f(1) been defined, f(0) = 0. Solving this when the domain of f forms a wheel instead of a field could have been interesting, and may have actually resulted in a simpler solution, but this was fun either way.
I think I speak for a lot of people here when I say that this was Xtreme! In all seriousness, this was another amazing video, and thank you so much for making it and sharing!
