Solving a Nice Functional Equation

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#ChallengingMathProblems #FunctionalEquations

PLAYLISTS 🎵 :

[pwmiles]

It looks easier if you substitute x = u^2

f((u +1)/(u -1)) = u^2 - 1 = (u -1)(u +1)

Put y = u +1

u - 1

This is easy to invert, pretend the coefficients are a 2x2 matrix and forget about the determinant (trust me it works)

u = - y - 1

- y + 1

u = y + 1

y - 1

u -1 = 2/(y - 1)
u+1 = 2y/(y - 1)
f(y) = 4y/(y - 1)^2

f(x) = 4x/(x - 1)^2 :-))

[damiennortier]

Here's an alternative :
Since (sqrtx + 1)(sqrtx - 1) = x - 1 we can use that :
Multiply the top and the bottom by sqrtx - 1, you'll got f([x-1]/[sqrtx - 1]^2) = x - 1
Now, just take the first expression again and multiply by sqrtx + 1, you'll got f([sqrtx + 1]^2/[x - 1]) = x - 1
Set x - 1 = y and solve for sqrtx in terms of y :
Sqrt x = sqrt(y+1)
Just replace into the 2 equation :
f(y/[sqrt(y+1) - 1]^2) = y
f([sqrt(y+1) + 1]^2/y) = y
...

[andylee]

I enjoyed watching this solution and also the brief digression at the beginning about solving inverse-f problems.

I don't remember learning to solve functional equations in school. Either I forgot or I never learned it. Either way, it feels new now. I've enjoyed seeing solutions to this class of problems. Hopefully eventually some of these lessons will actually stick in my head. 🙂

[stevenlitvintchouk]

That was relatively easy. No mathematical theorems or creative flashes of insight required.

[pageboysam]

One small miss on the domain of t: t ≠ 0

Given (sqrt(x) + 1 )/ (sqrt(x) - 1), then

sqrt(x) - 1 ≠ 0
sqrt(x) ≠ 1
x ≠ 1

Given 4t/(t-1)^2 = x - 1 and x ≠ 1, then

4t/(t-1)^2 + 1 = x ≠ 1
4t/(t-1)^2 ≠ 0
t ≠ 0

The reason t ≠ 0 is because the original parameter was defined with a denominator of sqrt(x) - 1, and no other reason.

[advaykumar]

Replacing LHS with f(y) and using componendo divendo we get value of x in terms of y
Then we substitute value of x in terms of y in RHS
Then we will get function as 4x/(x-1)^2

[Jha-s-kitchen]

When you said "2t or not to 2"...
really enjoyed that 😂

[kanankazimzada]

I had done like this. This was easy👍🏻👍🏻👍🏻I appreciated it.

[narfharder]

5:21 Missed "2t fruity" opportunity.

[angelmendez-rivera]

[sqrt(x) + 1]/[sqrt(x) – 1] = 1 + 2/[sqrt(x) – 1]. Notice that x must lie in [0, 1) or (1, ♾). If x lies in [0, 1), then sqrt(x) –1 lies in [–1, 0), and 2/[sqrt(x) – 1] lies in (–♾, –2], hence 1 + 2/[sqrt(x) – 1] lies in (–♾, –1]. Meanwhile, if x lies in (1, ♾), then sqrt(x) – 1 lies in (0, ♾), and 1 + 2/[sqrt(x) – 1] lies in (1, ♾).

As such, consider the function g : [0, 1) + (1, ♾) —> (–♾, –1] + (1, ♾) such that g(x) = 1 + 2/[sqrt(x) – 1]. This is an injective and surjective function, and so it is invertible, which is to say, that g^(–1) : (–♾, –1] + (1, ♾) —> [0, 1) + (1, ♾) exists and is well-defined, with g^(–1)(x) = [1 + 2/(x – 1)]^2. So consider h : [0, 1) + (1, ♾) —> R such that h(x) = x – 1. We want to solve for f : (–♾, –1] + (1, ♾) —> R such that f°g = h. Since g^(–1) exists, we can say that f = h°g^(–1). The explicit expression gives us that f(x) = [1 + 2/(x – 1)]^2 – 1 = 1 + 4/(x – 1) + 4/(x – 1)^2 – 1 = 4/(x – 1) + 4/(x – 1)^2.

[moshamomomd]

How can you write a function without telling its domain of definition. Which in this case is the main issue

[maxamedaxmedn]

This was actually a nice and straight forward problem I loved it

[neuralwarp]

Shouldn't you restrict the domain to x>0 ?

[hadibendj]

Two t or not two t that's the question