I've seen enough videos about Lamberts W function that i should have recognized that we can use it here, but i was still dumbfounded with this problem for a bit.
There's another way to write the answer, it's known as the super square root, Ex. 3^3 = 27, √27₄ = 3. The subscript four stands for tetration, as it is the opposite of tetration. so f^-1(x) = √x₄
Well, I’ve been looking for a way to inverse x^x and i’ve found one. I’m just kind of sad that it involves the w lambert function, because it’s not as numeric as other functions like x^2
Wolfram Alpha is only -1/e to 1/e x real solutions. There are no Wolfram Alpha tables extending out to other x numbers like there are tables of other functions in math: ln(x), log(x), 1/x, √x, sinx, cosx, etc. tables. Leaving answers as a Wolfram Alpha answer doesn't give useful data to continue on vs numerical analysis mathematics of interpolated and nested algorithms mathematics used for computer programs.
What are you guys talking about inverse is what gives you x when you ibout y so the answer is f inverse os clearly x^1/× plain and simple because wine you plug in x^× you get x^×*1/×= x so the fingers function equals x^1/×
I've seen enough videos about Lamberts W function that i should have recognized that we can use it here, but i was still dumbfounded with this problem for a bit.
the square superroot got violated💀
There's another way to write the answer, it's known as the super square root,
Ex. 3^3 = 27, √27₄ = 3. The subscript four stands for tetration, as it is the opposite of tetration. so f^-1(x) = √x₄
So in the same manner:
√x₄ = exp(W(ln(x)))
√x₃ = x¹ᐟ²
√x₂ = x/2
√x₁ = x−2
Pretty wise :)
So e^W(lnx) is the super square root, right?
Oh
Well, I’ve been looking for a way to inverse x^x and i’ve found one. I’m just kind of sad that it involves the w lambert function, because it’s not as numeric as other functions like x^2
What do you think about this? x^x^x
What the domain and range for f(x)=x^x ?. Anyone?
D = [0,inf)
R = [e^(-1/e),inf)
@@SyberMath thank you very much. I've searching for this answer about 2 month. You're the saviour bro!
I love It.
Is it e^W(ln(x))?
Rather f⁻¹(x) = exp(Wₙ(ln(x))), I'd say (this way we may also allow for complex solutions).
Wolfram Alpha is only -1/e to 1/e x real solutions. There are no Wolfram Alpha tables extending out to other x numbers like there are tables of other functions in math: ln(x), log(x), 1/x, √x, sinx, cosx, etc. tables. Leaving answers as a Wolfram Alpha answer doesn't give useful data to continue on vs numerical analysis mathematics of interpolated and nested algorithms mathematics used for computer programs.
Bit the right answer is f^-1(×)= x^1/× clearly..that gives you x when you plug in x^× so that is obviously the correct answer??
If you apply logx(x) is not the reverse ? logarithm of base x
Nooooooooo 😂 . Log base x is gone be log base x^x . And btw the fugggk u saying man logx(x)=1
that's just y=1
I was thinking about square root with a x in the top left corner
X root of x
Another nice problem,that i solved in my head😊
Frfr man
ok
f(x) = y = x^x
ln y = x ln x = ln x e^(ln x)
W(ln y) = ln x
x = e^W(ln y)
finv(x) = e^W(ln x)
What are you guys talking about inverse is what gives you x when you ibout y so the answer is f inverse os clearly x^1/× plain and simple because wine you plug in x^× you get x^×*1/×= x so the fingers function equals x^1/×
@@leif1075 Dude! What are you talking about? Plz check your grammar before you hit reply
y = x^x
ln y = x ln x
W(ln y) = ln x
(exp ∘ W ∘ ln)(y) = x
x=e^W(lny)