Solving a Rational Functional Equation 

That's right

That's what actually confused me. Thanks for the correction david! Author, please double check your video at the end for possible errors 😊

That really really made this confusing to watch :/ until I rewound. Should've looked at the comments I guess

-👌

Came here looking for an explanation regarding this as well. Thanks.

They are elegant and that's a good idea!

@@SyberMath yes!

@@SyberMath have you done it I cannot see

@@SyberMath Have you done it?

Thank you!

I'm glad!

Great!

Thank you!!! 🥰

Thank you too! 😊

Sure!

💖

😊 thanks

Good question. Because of the expression 1/(x-1), x cannot be 1

@@SyberMath Oh my god. How didn't I notice that? It's literally in the thumbnail too.

That is correct!

@Bruno levi Levi &/- it does so only because . . . it’s an odd number of times fofof . . . of = x It wouldn’t work with an even number of times. Yes, that’s how it can be generalized. There’s another video with fof = x i.e. an even number of times . . . BUT that one works because it’s a product of terms with different powers, not the sum of terms, so the cancellation still works.

Awesome, thank you!

Who is Matt Penn?

www.amazon.com/Introduction-Functional-Equations-Problem-solving-Mathematical/dp/0821853147 www.amazon.com/Topics-Functional-Equations-Third-Xyz/dp/099934286X www.math.uci.edu/~mathcircle/materials/M6L2.pdf

if you've practiced millions of maths problems, you will get that instinct; you know how to solve the problems. There are limited kinds of problems and limited ways of solving them, just try watch as many problem solving videos as you can.....

@@karljo8064 yes I've noticed this too many times. Experience is 👑

a bit too late, but i can answer you question: it`s not intuition, it`s logic. in functional equations, you are, basically, trying to get useful info. to achieve this, there are a few, let`s say, tricks. to know what tricks to use, you analyse the function. in this case, i analysed and can tell you that there are really few tricks one can use. in fact, there is only one: notice that f(x) is linked to f(1/1-x). therefore, if you know what f(1/1-x) is, you get your solution (that`s why he plugged f(1/1-x)). after this, you get f(1/1-x) is linked to f(x-1/x). applying the same logic, you want to know what f(x-1/x) is, so you plug it. then, you have f(x-1/x) is linked to f(x). do you see what happened? f(x) is linked to f(1/1-x), which is linked to f(x-1/x), which is linked to f(x). thus, you have f(x) linked to itself (the actual goal of a functional equation). now, you just solve the system of equation and get the value of f(x). btw, this is called the circle method.

@@caiodavi9829 Ahh I see, thanks for the detailed explanation! One thing I still don’t understand is that how you know the circular method works before even beginning to substitute. Like what is the insight behind x -> 1/1-x -> x-1/x -> x? Does it only work for 1/1-x or is there other fractions in the form p(x)/q(x) that works and why?

Saludos

He doesn't generally do proof type problems but nevertheless it's a good question

Thank you! That's a good problem but a bit too hard for the channel, I think

It is

@@SyberMath IT IS WHAT?

i think you will find that regardless of the initial value of x, cos(x) and sin(x) are in bounds(-1,1) , sin is montonic in the range, cos needs also to consider the value 0. so take two montonic ranges. then each successive application of the function narrows the range which i think converges on sin(x)=x and cos(x)=x those values are different and the ranges of answers for the two different functions do not overlap. You will need a calculator.

I saw that too, but he copied equation #2 wrong. If he had copied it correctly, 4 terms drop out and leave 2f(x).

No need to be so harsh. Besides, another commenter has commented and that comment is pinned.

Siz turkler cok soru soruyor 😂

@@SyberMath o kadar yanıt verince ve aksanından ben de türk olduğunu sandım :D

Thanks

Thanks

If f(x) = 1/2, it works!

😂

yes

🙃😊

Absolutely!

@@SyberMath WHY do that replacement why not plug in values for x like zero, one half and 2 and make equationa..you could do it that way..why wiuld anyone ever rhink of doing that replacement..i don't see why...

@@SyberMath Waitna minute you cant add the equations like that because you made a substitution..so what is f(x) in ome equation is not equals to f(x) in another and same with f(x-1/x)..see what I mean? So this can't be correct. When doing subsittution it's generally better to pick a different variable like u not the same one, as you know.

It's like taking the composition of two functions. For example f(2x+1) is the same as f(g(x)) where g(x)=2x+1

@@SyberMath can you respond to my comment above pretty sure you made a mistake when you added the two equations though..becsuse f of x for one equals f pf 1/1-× or whatever for the others..

Thanks!

Reminds me the movie 😁

because of (x-1)/x

@@SyberMath Thanks :D

Thank you! I know. I make mistakes

More like competition/olympiad level

I did not copy from Michael Penn bruh 😂

@@SyberMath oh sorry, he did the same problem a few months ago though

😂