Solving a Functional Equation

This video is about a Functional Equation
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#ChallengingMathProblems #FunctionalEquations

EXPLORE:

[nicolasmarin]

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?

[theimmux]

I had no idea (1-x)/(1+x) was special in any way. That's an amazing property right there.

[ruggbi]

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

[mackenziekelly]

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.

[satyapalsingh]

Wow, Heart filled with joy . So interesting . God, bless you .

[joaquingutierrez]

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 !!!

[nitayweksler]

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 !!

[Justin-gkhu]

Please do video on how to solve functional equations

[mathsmadeeasywithj.j.d]

Oh this is pretty superb Prof...
One of the most resourceful channel ever❤️❤️❤️❤️

[aristotlesdritsas]

When you rise up both sides at the 2 I think that you lose the solutions of X<0 if I am wrong please correct me, also we know that f(f^-1(X))=X that means that when you had (f(1-x/1+X))^2 => (f(f^-1(X)))^2 =X^2 what is wrong about what I just said lol!

[jaykumarjindal]

Just put (1-x)/(1+x)=y
You will get shine in your eyes after doing this

[Aditya_]

XD same remembered that the substitution results in itself

[mr_angry_kiddo]

Sir please do more problems on function's, differentiatial and Integral calculus😍😍😍😍😍😍😍😍.

[vuyyurisatyasrinivasarao]

Excellent ur real mathematician salute sir

[thomaslangbein]

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.

[conrad]

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.

[carlosharmes]

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))?

[cicik]

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

[atishsawant]

I eventually solved it ooh that satisfaction.. 😁

[InDstructR]

Damn I'm so happy. That's the first functional equation problem I solved without any help!