Solving the Functional Equation f(f(x)+y)=x/(1+xy)

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[karik]

Set y = -f(x) + x

Then you get 2nd order equation for f. It has two roots, the other is not valid.

[benjaminvatovez]

Thanks for the solution. However, I think there is a limitation we should take into account: if f(x+f(1))=1/(x+1) for all positive real x, then f(x)=1/(x-f(1)+1) for any x GREATER THAN f(1) only. In the end, you showed that f(x)=1/x for all x>1 and we should still show that it applies for ]0, 1[. Am I right?

[Jha-s-kitchen]

Excellent, First solving the question then teaching how to come up with the question.
Absolutely Genius 😎

Hey Syber, wanted to ask something... how can I get started with making my own math videos (topics etc?)

[GandalfTheWise]

Another possible method. Let y = 0, then f(f(x))=x and f is an involution or a self-inverse function. A few possible solutions are f(x) = x or f(x)=1/x or f(x) = -x or f(x)=-1/x. Simple substitution of f(x)=1/x shows that f(f(x)+y) = f(1/x+y) = 1/(1/x+y) =x/(1+xy) works.

[arbenkellici]

Thanks for your excellent videos profesor. I am having difficulties in solving this equation:
g(x)= {y=4x+12 and y=-1/2x +12
Find g[g(0)]
Could you show me the method to solve it and the solution?
Thank you in advance!

[ddmm]

Concerning the function f(x) such as: f(f(x)+y)=x/(1+xy) and for which you find that f(x)=1/x, I think you forgot the solution trivial f(x)=0

Indeed, as nothing prohibits trying x=0, we find that:

f(f(0)+y)=0

We set f(0)=c

f(c+y)=0

If we take y=-c we find that f(0)=c=0

So f(y)=0 i.e. f(x)=0

[dionysus]

f(x)=y
f(y+y)=f(2y)=x/(1+xy)
If we asigned as 2y=x
Then, f(x)=y=2y/(1+xy)
Cancel y both sides
1=2/(1+xy)
1+xy=2
xy=1
y=1/x
f(x)=1/x

[orchestra]

I solved it using differentiation and the chain rule if a nested function. Like: t(x, y) = f(x) + y, and f(t(x, y)) = x/(1+xy). Then, just differentiate over x and y separately, and substitute df/dt, and you get the solution. I don't see any need to restrict the domain of the function f(x) to positive numbers only. I think the solution holds over C. Even at x=0 the initial functional equation holds for f = 1/x, if you use the limit (right or left). Can someone explain please?

[BurgoYT]

I tried 1/x straight away and it worked

[MrLidless]

Second method: set y = -f(x). Not any quicker, but an elegant trick.

[carlyet]

How do you get the domain? ... the only restriction I see is y=/= -1/x

[arbenkellici]

Anyone of the viewers could tell me the solution to that exercise please

[christianthomas]

i think i did it intuitively: first one can see that x/(1+xy) can be written as: 1/((1/x)+y). taking the f^(-1) of each member I get:
f(x)+y = f^(1)((1/x)+y) . Now, the simpliest way to have something(x) separated from y from the last equation is to try f(x)+y = (1/x)+y=(1+xy)/x an then: f(f(x)+y) = f((1/x)+y) = x/(1+xy) so that f(x) = 1/x is a solution.

[angelmendez-rivera]

The task is to find all f : (0, ∞) —> (0, ∞) such that f(f(x) + y) = x·(1 + x·y)^(–1) everywhere. We know f(x) + y > 0, but if y = (–1)·f(x) + x, then f(x) = x·(1 + x·((–1)·f(x) + x))^(-1) everywhere. This means f(x)·(1 + x·((–1)·f(x) + x)) = x everywhere. As such, f(x) + (–1)·x·f(x)·f(x) + x·x·f(x) = (–1)·x·f(x)·f(x) + (x·x + 1)·f(x) = x everywhere. Since x > 0, this is equivalent to f(x)·f(x) – (x + x^(–1))·f(x) = –1 everywhere. Thus = f(x)·f(x) – (x + x^(–1))·f(x) + x·x^(–1) = (f(x) – x)·(f(x) – x^(–1)) = 0 everywhere. This means that for some S a subset of (0, ∞), f(x) – x = 0 for all x in S, and f(x) – x^(–1) = 0 for all x in (0, ∞)\S. Hence, for all (x, y) in (0, ∞)^2 such that f(x) + y is in S, f(f(x) + y) = f(x) + y = x·(1 + x·y)^(–1), so f(x) = (–1)·y + x·(1 + x·y)^(–1), meaning that x = (–1)·y + x·(1 + x·y)^(–1) for all such x with x in S, and x^(–1) = (–1)·y + x·(1 + x·y)^(–1) for such x with x in (0, ∞)\S. Otherwise, (f(x) + y)^(–1) = x·(1 + x·y)^(–1), meaning 1 + x·y = x·(f(x) + y) = x·f(x) + x·y, which is equivalent to f(x) = x^(–1). Noticing that this is true for all y, it means S = {}. Therefore, f(x) = x^(–1) everywhere.

[maryvonnedenis]

Pour y = 0, f(f(x)) = x et en dérivant f'(x).f'(f(x)) = 1(*)
Sinon (f(f(x) + y) - f(x))/y = (x/(1 + xy) - x)/y = (x - x(1+xy))/(1 + xy)y = - x²/(1 + xy)
D'où lim (f(f(x) + y) - f(x))/y quand y -> 0 = - x² = f'(f(x))
Or f'(x).f'(f(x)) = 1 (*) donc f'(x) . (-x²) = 1 et f'(x) = - 1/x² soit f(x) = 1/x + k...Comme f(f(x)) = x, k = 0 CQFD