A Quick and Easy Functional Equation

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[КорнейКвадратных]

For an equation of the kind f(g(x))=h(x) the solution is f(x)=h(g^-1(x))

[davidgillies]

Note that a function of the form f(x) = (a x + b)/(c x + d) with a, b, c, d complex is a Möbius transformation, which is one of those things that crops up _absolutely everywhere_ . The technique of finding the roots (which doesn't help to solve this problem) can still be used to find the fixed points of the transformation, which can be incredibly useful (it's an automorphism).

[sohomsen]

Take z=((2x-1)/(x-3)) and hence we get f(z) in terms of z. Then substitute z=x and we finally get f(x).

[xuepingsong]

I think the reason this works is because the function is defined by a formula to compute any value in the range, it's not an equation with unknowns. For example a function defined by f(a)=a^2 can be also defined using f(x)=x^2, the variable chosen doesn't matter.

[MathElite]

Functional equations are life....
Hehe we both did a functional equation video today

[MushookieMan]

I guess it's valid because you start with an *identity*, and are composing both sides of the equation with the function g(x)=x. It would work for any function where it's defined.

[carlyet]

(3x-1)^2/(x-2)^2 == f(x) the range of the function f(x) = [ 3 + 5/(x-2)]^2. So x<>2

[cunningba]

y->x: good place to introduce the concept of bound variables and the constructs that use them. In this case the implicit lambda() in the function definition. Other familiar constructs that use them: summation, products, quantifiers, integrals.

[namantenguriya]

Nice explanation on the replacement of variables...

[roberttelarket]

Let t = (2x - 1)/(x - 3).
tx -3t = 2x - 1 or x = (3t - 1)/(t - 2).
f(t) = [(3t -1)^2](t - 2)^2.
That's exactly f.

[lollel]

I think the reason why you can replace f(y) with x is because x is just a parameter which passes in values. You just have to make sure that you replace all the y’s in the function with x.

Also there’s one thing I don’t quite get. You said that if you wanted to find the numerical value you do 2x-1/x-3 = x. But I don’t get why you would do this. Why are you allowed to equate that to x?

[AbhishekSingh-qnbz]

Please make a video on ' Strategies for solving Functional Equations ', as you did for Diophantine Eqn...

[azizyosri]

I literally watch your videos as im eating dinner.. got so much better at math thanks to u

[mathisnotforthefaintofheart]

5:30. Very important...What did we actually do? Glad you went into that

[BlaqRaq]

I really like how you did this video. First one I watched and I subscribed.
This is just stoking my addiction to Mathematics being almost 50 and did my Math degree about 30 years ago.
Anyway, great presentation on functions.

[bevobe]

2x-1/x-3=2+5/(x-3)=t so x-3=5/t-2. then x=3t-1/t-2. f(t)=answer.

[mwodka]

Apart from the domain. You cannot simply switch y back with x because they are not equal. y ≠ x, y = (2x-1)/(x-3)

[platformofscience]

Here in iran it's 1 a.m but I'm still watching.

[a]

since (2x-1)/(x-3) = 2 + 5/(x-3)
simply by this method
( 5/(y-2) +3 )^2

[verisap]

I’m new to this field and I have a question please.

If you replace x with (2x-1)/(x-3) and substitute in you get

F(x) =[ (2x-1)/(x-3) ] ^2 = (4x^2 + 4x+1) / (x^2- 6x +9)

So I have f(x) in terms of x. But it is not the solution in the video.

Clearly I’ve done something wrong . Can someone explain it to me please?

Thanks