Solving f(x)+f(y)=f[(x+y)/(1-xy)]

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PLAYLISTS 🎵 :

[WahranRai]

Recall : tan(A + B) = (tanA + tanB) / (1- tanA*tanB) with A and B angles ---->
A = arctanX and B = arcctanY read it as angles with tan equals to X, Y -->
tan(arctanX + arctanY) = (tan(arctanX) + tan(arctanY)) / (1- tan(arctanX) * tan(arctanY)) --->
after simplification ---> *arctanX + arctanY = arctan((X+Y)/ (1-XY))* that is our functional equation

[WahranRai]

Variant : Let x=y ----> *2f(x) = f(2x/(1-x^2))*
Recall : tan(2A) = 2tanA / (1- tanA^2) ---->
x = tanA --> tan(2A) = 2x/(1-x^2) ---> by taking arctan of two sides ----> arctan(tan(2A)) = 2A = arctan(2x/(1-x^2)) --->
*2arctan(x) = arctan(2x/(1-x^2))* that is our functional equation

[angelmendez-rivera]

Notice that the function tan : (–π/2, π/2) —> R is bijective. As such, it is also surjective, so for every real u, there exists some real t such that u = tan(t). Therefore, let x = tan(p) and y = tan(q). Hence f[tan(p)] + f[tan(q)] = f([tan(p) + tan(q)]/[1 – tan(p)·tan(q)]) = f[tan(p + q)]. Let g = f°tan. Hence g(p) + g(q) = g(p + q). Since f is differentiable, g is differentiable. Therefore, g(x) = A·x for all x, for some real A. Therefore, A·z = f[tan(z)]. The extension of tan to R has period π, so tan(z) = a implies z = arctan(a) + m·π for m in Z. Therefore, A·z = f[tan(z)] implies A·arctan(a) + A·m·π = f(a). So A·arctan(x) + A·m·π + A·arctan(y) + A·m·π = A·arctan[(x + y)/(1 – x·y)] + A·m·π, which implies m = 0, so f(a) = A·arctan(a).

[antonyqueen]

No need to do partial differentiation over y and f’(x)/f’(y) in the first method. Just set x=0 in df(x, y)/dx
=>f’(0)=f(y)*(1+y^2) =>f(y)=f’(0)arctan(x)+k
K=0 since f(0)=0 (by plugging x=y=0 in original equation)

[manjoker]

please solve this
[ integral ( log(x lnx))/x dx ]

[JR]

f(1)+f(x) = f((1+x)/(1-x))
You get different values when RHS when x approaches 1 left or right but you get same value for LHS.
Informally, 2f(1) should approximate both f(1000) and f(-1000) but that doesn't work for arctan.
So only solution is the zero function.

[petereziagor]

I really need to learn this function thing from scratch and learn some basic terms used in solving them
X is odd, the function is differentiable and stuff

[ianmathwiz]

I guessed the arctangent was going to be in the solution when I first saw the problem, because the original equation looked like the tangent addition formula.

[neuralwarp]

I don't like the inconsistency of the tan‐¹(x) notation. tan²(x) means tan(x)×tan(x) not tan(tan(x)) . I prefer arctan.

[bollyfan]

How can you cancel the first derivative "f'(....)" of that expression, since in one case it is differentiated w.r.t. x and in the other case w.r.t. y?
Maybe I am missing something, but can you explain that part?

[rob]

let x = tan A and y = tan B
then f(tan A) + f(tan B) = f([tan A + tan B]/[1 - tan A tan B]) = f(tan [A + B])
this works when f(x) = k arctan x

[cavendishcavendish]

That's great. We want more mathematics tricks. Thank you for your efforts

[holyshit]

f(x)=arctan(x), but for arctan this equation is not always true

[bigjazbo]

I can see by inspection that the zero function is also a solution.

[giuseppemalaguti]

Assomiglia a f(x) =arctg.... dalla formula di addizione della tangente