Provided that g : R -> R is such that g(x) = x^3 + 3·x^2 + 3·x everywhere, and that id : R -> R is such that id(x) = x everywhere, find all f : R -> R such that f°g = id. This is a well-defined search, since g is surjective. As is, f is a left-inverse of g. For such a left-inverse to exist, g must be injective. The injectivity of g would normally be tested by looking at monotonicity and continuity, and you would test for monotonicity by looking at the critical points. However, here, notice that x^3 + 3·x^2 + 3·x = (x + 1)^3 - 1, so if a, b : R -> R are such that a(x) = x - 1, b(x) = x^3, then g = a°b°a^(-1), and here, notice that a, b, a^(-1) are all injective and surjective, and thus bijective. Hence g is indeed injective, and in fact, bijective, which implies f exists, and is unique, and it also implies that f°g = g°f, meaning that (f(x) + 1)^3 - 1 = x everywhere. This simplifies the task significantly, and we can immediately conclude f(x) = cbrt(x + 1) - 1 everywhere. Q. E. D.
Note: it actually is important to establish that g is surjective. If g were not surjective, then f *would not* be unique. For instance, let us replace the equation f(x^3 + 3·x^2 + 3·x) = x everywhere with f(arctan(x)) = x everywhere. Here, g : R -> R is such that g(x) = arctan(x) everywhere, and so we want to find all f such that f°g = id. g is injective, but not surjective, with range(g) = (-π/2, π/2). Why is this important? Because we can conclude f(x) = tan(x) for all x in (-π/2, π/2), but you *cannot* conclude f(x) = tan(x) everywhere. As an example, let f0(x) = tan(x) for all x in (-π/2, π/2), but f0(x) = 0 for all other x. Then, even though f is not equal to f0, f0 also satisfies the equation, in that f0°g = id in this case. Indeed, f0(arctan(x)) = x everywhere. Therefore, there are at least two functions satisfying the equation. In fact, there are infinitely many. Consider an arbitrary h : R\(-π/2, π/2) -> R. Let f[h] : R -> R be such that f[h](x) = tan(x) for all x in (-π/2, π/2), f[h](x) = h(x) for all other x. Then, for all such h, f[h]°g = id is true, and f[h] is distinct for each h. This means there are card(R)^card(R\(-π/2, π/2)) = Beth(1)^Beth(1) = 2^Beth(1) = Beth(2) functions that solve the equation. This can be made completely general. In the most general situation, consider some given, known function g : R -> R. The task is to find all f : R -> R such that f°g = id, where id : R -> R such that id(x) = x everywhere, a.k.a find every left inverse of g. Let g^λ : range(g) -> R be such that g^λ(g(x)) = x everywhere. g^λ is well-defined, since there is only one function satisfying this property. Consider an arbitrary function h : R ange(g) -> R. Now, let f[h] : R -> R such that f[h](x) = g^λ(x) for all x in range(g), f[h](x) = h(x) otherwise. Thus, for all such h, f[h]°g = id, and for each distinct h, f[h] is distinct as well. The number of distinct f[h] that exist is equal to card(R)^card(R ange(g)) = Beth(1)^card(R ange(g)). In the special case that g is surjective, range(g) = R, so R ange(g) = {}, meaning that card(R ange(g)) = 0, so Beth(1)^card(R ange(g)) = Beth(1)^0 = 1, and so f is unique. Otherwise, there are infinitely many f: at least Beth(1) of them, and if R ange(g) is uncountable, then exactly Beth(2) of them. Of course, this all assumes g is injective. If g is not injective, then f simply does not exist, and there is nothing else to discuss. So, in summary: if g is not injective, then there are 0 functions f satisfying the equation; if g is injective, and surjective, then there is 1 function f satisfying the equation; if g is injective, and not surjective, and range(g) is cocountable in R, then there are Beth(1) functions satisfying the equation; if g is injective, and not surjective, and range(g) is not cocountable in R, then there are Beth(2) functions satisfying the equation. In this video, though, g is surjective, so f is unique, and in fact, f = g^(-1), keeping it simple.
This is how I solved it: we know f(g(x))=x so g(x)=f^(-1)(x) That is, it is the inverse of the function f y=ax³+bx²+cx+d If b²-3ac=0 If this condition is met, the inverse of the function is equal to : y^(-1)=[(x-d)/a+(b/3a)³]⅓-(b/3a) f(x)=cbrt((x+1)-1
*General strategy* set y = x^3 + 3x^2 + 3x and then calculate x in terms of y. y = (x+1)^3 - 1 => x = (y+1)^(1/3) - 1 So, using x instead of y, f(x) = (x+1)^(1/3) - 1 *Simple* Right ?
I took the derivative to look for any extrema. After noticing that there was exactly one horizontal tangent and that the a value of the cubic was 1, I knew it would be in the form f(x)=cbrt(x-h)+k where (k,h) is the point at which the horizontal tangent intersects the cubic. Finding this was easy due to my first step.
The equation f(g(x)) = x is asking the inverse function of g(x). I would not call that a functional equation. A functional equation requires f to occur at least twice and not of a constant like f(1).
Also, strictly speaking, the equation is only asking for a left-inverse. It is conceivable that not only multiple left-inverses could exist, if no right-inverses exist, but also that none may exist at all. g is bijective, and that is really the key observation to solve this equation uniquely. However, g could have been chosen so that it was injective, but not surjective on R, or that it was not injective to begin with. So, the problem is not as simple as you are dismissing it to be. Syber kept it easy by choosing g to be bijective, but strictly speaking, you still have to at least state that explicitly as a premise in your proof, and explain why that matters. In this case, g being bijective means that f°g = g°f = id, and so we can rewrite f°g = id as g°f, meaning that (f(x) + 1)^3 - 1 = x everywhere, and now the solution is obvious. You could not assert this if g was not injective, though. You also could not assert f is unique without indicating g is surjective (if g is not surjective, then f is definitely not unique).
@@angelmendez-rivera351 Thank you for your kind and thorough stipulations. It is this kind of interaction with people more knowledgeable than me that makes it worthwhile to participate in a group like this. I notice that most people don’t like to be wrong about what they say. I love being wrong sometime, that is when I learn. I studied physics (BSc & MSc) and I notice this my attitude is a scientist’s attitude.
Mr, it's just an inverse function of the said expresion. Namely, take the cubic root of ((...)十1)，and then -1. The correct wordings are therefore f(y)=(cu rt(y十1))-1，when y=..., f(y)=x.
I feel like the question requires more parameters. There are infinite functions that solve this expression. For example, f:Pn(R) -> P1(R) can be defined as f(a + bx + cx² + dx³ + ...) = (1/3)bx and we are done
Another way: The given equation: f(x^3 + 3x^2 + 3x) = x We denote: u = x^3+3x^2+3x and we have to express x as a function of u . Therefore, we have to solve for x the equation: x^3 + 3x^2 + 3x = u Or: x^3 + 3x^2 + 3x + 1 = u + 1 Or: (x + 1)^3 = u + 1 Or: x + 1 = ∛(u + 1) ⟹ x = ∛(u + 1) - 1 Therefore we obtain: f(u) = ∛(u + 1) - 1 Or: f(x) = ∛(x + 1) - 1
This is dumb the answer is negative 1. I looked at that seeing three larger x values = x. so it must be negative. Seen negative one was a candidate to work solved it using negative one as x and the answer comes out as negative one.
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
Provided that g : R -> R is such that g(x) = x^3 + 3·x^2 + 3·x everywhere, and that id : R -> R is such that id(x) = x everywhere, find all f : R -> R such that f°g = id. This is a well-defined search, since g is surjective.
As is, f is a left-inverse of g. For such a left-inverse to exist, g must be injective. The injectivity of g would normally be tested by looking at monotonicity and continuity, and you would test for monotonicity by looking at the critical points. However, here, notice that x^3 + 3·x^2 + 3·x = (x + 1)^3 - 1, so if a, b : R -> R are such that a(x) = x - 1, b(x) = x^3, then g = a°b°a^(-1), and here, notice that a, b, a^(-1) are all injective and surjective, and thus bijective. Hence g is indeed injective, and in fact, bijective, which implies f exists, and is unique, and it also implies that f°g = g°f, meaning that (f(x) + 1)^3 - 1 = x everywhere. This simplifies the task significantly, and we can immediately conclude f(x) = cbrt(x + 1) - 1 everywhere. Q. E. D.
Note: it actually is important to establish that g is surjective. If g were not surjective, then f *would not* be unique. For instance, let us replace the equation f(x^3 + 3·x^2 + 3·x) = x everywhere with f(arctan(x)) = x everywhere. Here, g : R -> R is such that g(x) = arctan(x) everywhere, and so we want to find all f such that f°g = id. g is injective, but not surjective, with range(g) = (-π/2, π/2). Why is this important? Because we can conclude f(x) = tan(x) for all x in (-π/2, π/2), but you *cannot* conclude f(x) = tan(x) everywhere. As an example, let f0(x) = tan(x) for all x in (-π/2, π/2), but f0(x) = 0 for all other x. Then, even though f is not equal to f0, f0 also satisfies the equation, in that f0°g = id in this case. Indeed, f0(arctan(x)) = x everywhere. Therefore, there are at least two functions satisfying the equation. In fact, there are infinitely many. Consider an arbitrary h : R\(-π/2, π/2) -> R. Let f[h] : R -> R be such that f[h](x) = tan(x) for all x in (-π/2, π/2), f[h](x) = h(x) for all other x. Then, for all such h, f[h]°g = id is true, and f[h] is distinct for each h. This means there are card(R)^card(R\(-π/2, π/2)) = Beth(1)^Beth(1) = 2^Beth(1) = Beth(2) functions that solve the equation.
This can be made completely general. In the most general situation, consider some given, known function g : R -> R. The task is to find all f : R -> R such that f°g = id, where id : R -> R such that id(x) = x everywhere, a.k.a find every left inverse of g. Let g^λ : range(g) -> R be such that g^λ(g(x)) = x everywhere. g^λ is well-defined, since there is only one function satisfying this property. Consider an arbitrary function h : R
ange(g) -> R. Now, let f[h] : R -> R such that f[h](x) = g^λ(x) for all x in range(g), f[h](x) = h(x) otherwise. Thus, for all such h, f[h]°g = id, and for each distinct h, f[h] is distinct as well. The number of distinct f[h] that exist is equal to card(R)^card(R
ange(g)) = Beth(1)^card(R
ange(g)). In the special case that g is surjective, range(g) = R, so R
ange(g) = {}, meaning that card(R
ange(g)) = 0, so Beth(1)^card(R
ange(g)) = Beth(1)^0 = 1, and so f is unique. Otherwise, there are infinitely many f: at least Beth(1) of them, and if R
ange(g) is uncountable, then exactly Beth(2) of them.
Of course, this all assumes g is injective. If g is not injective, then f simply does not exist, and there is nothing else to discuss. So, in summary: if g is not injective, then there are 0 functions f satisfying the equation; if g is injective, and surjective, then there is 1 function f satisfying the equation; if g is injective, and not surjective, and range(g) is cocountable in R, then there are Beth(1) functions satisfying the equation; if g is injective, and not surjective, and range(g) is not cocountable in R, then there are Beth(2) functions satisfying the equation. In this video, though, g is surjective, so f is unique, and in fact, f = g^(-1), keeping it simple.
Wow! You are amazing, Angel!
Chapeau.
This is how I solved it:
we know f(g(x))=x so g(x)=f^(-1)(x)
That is, it is the inverse of the function f
y=ax³+bx²+cx+d
If b²-3ac=0
If this condition is met, the inverse of the function is equal to :
y^(-1)=[(x-d)/a+(b/3a)³]⅓-(b/3a)
f(x)=cbrt((x+1)-1
Given:
f(x³ + 3x² + 3x) = x
To find:
f(x)
Setting x³ + 3x² + 3x = y:
Adding 1 to both sides:
x³ + 3x² + 3x + 1 = y + 1
(x + 1)³ = y + 1
Raising both sides to the ⅓ power:
x + 1 = (y + 1)↑⅓
x = (y + 1)↑⅓ - 1
Thus:
f(y) = (y + 1)↑⅓ - 1
Nice
*General strategy* set y = x^3 + 3x^2 + 3x and then calculate x in terms of y.
y = (x+1)^3 - 1 => x = (y+1)^(1/3) - 1
So, using x instead of y, f(x) = (x+1)^(1/3) - 1 *Simple* Right ?
I took the derivative to look for any extrema. After noticing that there was exactly one horizontal tangent and that the a value of the cubic was 1, I knew it would be in the form f(x)=cbrt(x-h)+k where (k,h) is the point at which the horizontal tangent intersects the cubic. Finding this was easy due to my first step.
The equation f(g(x)) = x is asking the inverse function of g(x). I would not call that a functional equation. A functional equation requires f to occur at least twice and not of a constant like f(1).
Hey! Long time, no see! 😁
*A functional equation equation requires f to occur at least twice...*
According to who?
Also, strictly speaking, the equation is only asking for a left-inverse. It is conceivable that not only multiple left-inverses could exist, if no right-inverses exist, but also that none may exist at all. g is bijective, and that is really the key observation to solve this equation uniquely. However, g could have been chosen so that it was injective, but not surjective on R, or that it was not injective to begin with. So, the problem is not as simple as you are dismissing it to be. Syber kept it easy by choosing g to be bijective, but strictly speaking, you still have to at least state that explicitly as a premise in your proof, and explain why that matters. In this case, g being bijective means that f°g = g°f = id, and so we can rewrite f°g = id as g°f, meaning that (f(x) + 1)^3 - 1 = x everywhere, and now the solution is obvious. You could not assert this if g was not injective, though. You also could not assert f is unique without indicating g is surjective (if g is not surjective, then f is definitely not unique).
@@angelmendez-rivera351 Thank you for your kind and thorough stipulations. It is this kind of interaction with people more knowledgeable than me that makes it worthwhile to participate in a group like this. I notice that most people don’t like to be wrong about what they say. I love being wrong sometime, that is when I learn. I studied physics (BSc & MSc) and I notice this my attitude is a scientist’s attitude.
6:21 in Poland we call it "shooting a fly with a cannon"
Wow! That's cool!
Once I saw that the LHS was a difference of cubes the rest was easy. A general solution of f(cubic in x) = X is much more complex.
Mr, it's just an inverse function of the said expresion. Namely, take the cubic root of ((...)十1)，and then -1.
The correct wordings are therefore f(y)=(cu rt(y十1))-1，when y=..., f(y)=x.
Nice problem with original Solution .Thank you .Bravooo 👍👍👍👍👍👍👏
You are welcome
I feel like the question requires more parameters. There are infinite functions that solve this expression. For example, f:Pn(R) -> P1(R) can be defined as f(a + bx + cx² + dx³ + ...) = (1/3)bx and we are done
Let's try something like ...
y = x3 + 3x2 + 3x = (x+1)3 -1
(y+1) = (x+1)3
f(y) = x = cubicroot(y+1) -1
The Master has done it again. Bravo.
Aww, thanks for the kind words!!! 🥰🤗💖
2:29 YOU SHOULDN'T HAVE GIVEN ME THIS POWER!!!
Too late! 😜
Sup! One question, why, in the minute 1:56, you can move “y” to the left side? I didn’t understand
Today I viewed both solutions...
Very, very good...
🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷
Thank you! 💖
Another way:
The given equation:
f(x^3 + 3x^2 + 3x) = x
We denote: u = x^3+3x^2+3x and we have to express x as a function of u .
Therefore, we have to solve for x the equation:
x^3 + 3x^2 + 3x = u
Or:
x^3 + 3x^2 + 3x + 1 = u + 1
Or:
(x + 1)^3 = u + 1
Or:
x + 1 = ∛(u + 1) ⟹ x = ∛(u + 1) - 1
Therefore we obtain:
f(u) = ∛(u + 1) - 1
Or:
f(x) = ∛(x + 1) - 1
"remind myself to slow down"
hhh nice
😜
Hi Syber! I spent some time trying but I fell into the temptation to see how you solve it XD
Hi Juan! Some temptations are not that bad after all! 😁🤩
That's really one of your best problems. Great!!!!
Glad you think so! The fact that it's homemade makes me more proud 😉🥳
@@SyberMath just multiply both sides by 0
Duh
@@science_nepal_lover you are so smart 🤓
f(x) =sqrt3(1+x)-1
f(x)= cube root (x+1)-1
Easy-peasy. :)
Abi ingilizcen gayet gelişmiş. Daha demin 1 yil onceki videonu izliyodum, türk oldugun hayvan kadar anlasiliyordu
Saol! 😁😂
@@SyberMath zekani harbiden apprieciateliyorum abi. content cok kafa aciyor, cok ilgi cekici ayni zamanda.
Bakıdan salamlar.Əla həll etdiniz.
Çox sağ ol! Amerika Birləşmiş Ştatlarından salamlar!
Just send every Polynomial to the Polynomial x.
This is dumb the answer is negative 1. I looked at that seeing three larger x values = x. so it must be negative. Seen negative one was a candidate to work solved it using negative one as x and the answer comes out as negative one.
For what?
Nice video👍🏻
6/10/2022
-1
I found 'f' in front of (
f(x)=0
No, I can't
👍
sirr are you from Ukraina ?
No, I'm not
It's an incredible solution sir thank's very much, i like your channel.
Thanks and welcome! 🥰
Zero
同じことやん
are you türk?
Senden kaçmıyor abi hiçbir şey! 😜
@@SyberMath bazen çıkmış sorular bazen de aksan çok hafif belli belirsiz buradayım diyor. Elinize aklınıza sağlık başarılarınızın devamını dilerim
Boring
F=ma