I think I would try x=tan u, and proceed from there.

1-1/x forms a group of order 3 when composed with itself. 1/x forms a group of order 2 when composed with itself. The elements of the group C_2xC_3 correspond each to one of the equations you wrote. I guess this was the way to see it more clearly. I don't see a way to do the simplification in a simpler way.

Great problem . I’ve tried it on my own , but ended up with Pi/4 instead of your 3Pi/8 .

Thank you for your effort.

just wondering is f(x)=arctan(x)/2 +arctan(1-1/x)/2 +arctan(x/x-1)/2 -π/4 ?

if you sub x->1/(1-x) a few times you end up with a linear combination of terms and then you can solve for f(x)

I love this symmetrical substitution

Very nice

I Remembered this question Solved the same question 3 years ago during my JEE Advanced preparation.

Duudeee We can use the result from my college's previous exam that what I call the group of "Mobius transformations of magnitude one" of order 6 forms a group under function compostion, this means that combining enough times we'll eventually loop back to f(x) and we may have up to 6 linear equations, which will aid us to solve for the 6 unknowns, BRUHH THIS IS INSANE!

neat

Couldnt you do ec2 + ec3 - ec4?

Great video! One small question though, if we add 2 * pi/2 and subtract one aren't we left with pi/2 instead of 3*pi/2 at the end? Love your content!

Just to be clear is pathological a common mathematical term, because i dont really know what it means.

Wow interesante

Your videos are amazing keep doing that and could u please say which tablet that u are using mate? thanks

whoa! i feel like there's a certain video about 1-1/x being a rotation in hyperbolic space 😊 yeah, that was a nice connection ❤

Now what does a PUTNAM mean 😂 It is so ridiculous to think that what may be the life of one may be insignificant to another. There might be some people who work their life to clear the PUTNAM exam while other have no idea it even exist The universe is relative And we're here cuz of maths!

how do i integrate into society

Can't you just Laplace transform the equation?

the only thing left is to proove that there exists at least one function solving that equation idk about you, but it's not obvious to me

Let f : R\{0, 1} -> R\{0, 1} such that f + (f ° g) = arctan, where g : R\{0, 1} -> R\{0, 1} such that g(x) = 1 - 1/x everywhere. The challenge is to find the Lebesgue integral of f on ]0, 1[. Notice, (g ° g)(x) = ((x - 1)/x - 1)/((x - 1)/x) = ((x - 1) - x)/(x - 1) = 1/(1 - x) everywhere, and so g ° g = g^(-1). Hence, f + (f ° g) = arctan is equivalent to both (f ° g) + (f ° g^(-1)) = arctan ° g and (f ° g^(-1)) + f = arctan ° g^(-1). As such, f ° g^(-1) = (arctan ° g^(-1)) - f, hence (f ° g) + ((arctan ° g^(-1)) - f) = arctan ° g, which is equivalent to f ° g = ((arctan ° g) - (arctan ° g^(-1))) + f, thus f + (((arctan ° g) - (arctan ° g^(-1))) + f) = arctan, which is equivalent to f = ((arctan - (arctan ° g)) + (arctan ° g^(-1)))/2. To find the Lebesgue integral on ]0, 1[ of arctan is a fairly well-known exercise. However, the integral of (arctan ° g) - (arctan ° g^(-1)) is much more difficult, since you cannot split it into two terms, because the integrals would diverge. Instead, one must work with integration by parts, with the choice to differentiate (arctan ° g) - (arctan ° g^(-1)) into g'/(1 + (g • g)) - (g^(-1))'/(1 + (g^(-1) • g^(-1))) = 1/((id • id) + ((id - 1) • (id - 1))) - 1/(((1 - id) • (1 - id)) + 1 = 1/(2 (id • id) - 2 id + 1) - 1/((id • id) - 2 id + 2), and the choice antidifferentiate the constant function 1 into id + c for some constant c.