A Cool Functional Equation

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[@erikroberts8307]

When you had all three equations listed on the board at one time. I would have added both the top and bottom equations and, at the same time, subtracted the middle equation from the other two. Once that was done, divide both sides by two. Ta-dah!! the final answer.

[@mrl9418]

I disagree that you have the solution. In the first equation, substituting x= 0, you get f(0) + f(1) = 0. So f is defined in 0 and 1 abd has opposite values there. So there are ab infinite number of solutions, defined by yours outside of 0 and 1, snd by any couple of opposite values on 0 and 1. As an interesting side note, none of the solutions is continuous.

[@FisicTrapella]

But then, take x = 1 and you get f(1) + f(1/0) = 1... How is this possible if 1/0 is not defined??

[@paullaurent-levinson130]

​@@FisicTrapella in a real functional equation problem, your equation would be defined for x in a certain set, and said set would not include 1.

[@FisicTrapella]

Yes, but in this case, if it doesn't work for x = 1, it doesn't work for x = 0 either.

[@paullaurent-levinson130]

@@FisicTrapella why would that be true?

[@Mal1234567]

6:44 only person I know who states x(x-1) correctly was my high school algebra teacher. “X times X minus one” literally translates to x(x)-1. The correct way to say it is “x times the quantity x minus one.”

[@jackyzhang588]

At 5:22, all you need to do add equations 1&3 minus equation 2 and then the answer divide by 2 will be the final result.

[@thexavier666]

"What's in the box?" 🔫😡

[@SidneiMV]

f(x) = -x³/[2x(1 - x)] - (1 - x)/[2x(1 - x) f(x) = (1/2)[x²/(x - 1) + 1/x]

[@scottleung9587]

Wow, what a journey!

[@coreyyanofsky]

let g(x) = 1 / (1 - x) inverting, we find g⁻¹(x) = 1 - (1/x) iterating application, we find g(g(x)) = 1 - (1/x) = g⁻¹(x) therefore g(g(g(x))) = x --- f(x) + f(g(x)) = x ① applying x → g(x) to ① yields f(g(x)) + f(g(g(x))) = g(x) ② applying x → g(x) to ② yields f(g(g(x))) + f(x) = g(g(x)) ③ from ½[① - ② + ③] we find f(x) = ½[x - g(x) + g(g(x))]

[@BlaqRaq]

😂😂😂 I didn’t think to make another sub, so I was stuck.

[@phill3986]

👍😎👍😎👍

[@MathsScienceandHinduism]

hi

[@MrGeorge1896]

with t = 1 / (1 - x) we get f(t) + f((t - 1) / t) = (t - 1) / t and now with x = t: f(x) + f((x - 1) / x) = (x - 1) / x we subtract this eq. from the original one: f(1 / (1 - x)) - f((x - 1) / x) = (x² - x - 1) / x repeat the first sub t = 1 / (1 - x): f(t) - f(1 / (1 - t)) = (t² - t + 1) / (t (t - 1)) again x for t: f(x) - f(1 / (1 - x)) = (x² - x + 1) / (x (x - 1)) add this eq. with the original one: 2 f(x) = (x³ - x + 1) / (x (x - 1))

[@46swa]

This is a useless waste of time. Nobody needs this calculation.

[@MrGeorge1896]

may be but it is both fun and relaxing and that is what I am here for.