Let’s discuss if the power tower equation x^x=0 is solvable or not? Of course, we will discuss the real cases and the complex cases. Then we talk about the limit.
The limit of x^x as x goes to 0+:
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Finally 0^0 approaches 0:
https://youtu.be/X65LEl7GFOw?si=QRgxgdaL5If2FWnP
-∞^-∞
X is dual number known as epsilon. Nilpotent, infinitesimal.
How can both 1^0 and 1^1 be both 1. … therefore 1^0 = 1^1. Is 0^0 also equal to 0^1 bydefinition? 😂😂😂 0^0 is 1 by definition. Anything to 0 is 1. For 0^1 -anything to 1 is itself by definition.
Yes, there's a solution
x^x=0 -> x^x+x^x=x^x -> 2x^x=x^x -> ln(2x^x)=ln(x^x) -> ln(x^x)+ln(2)=ln(x^x) -> ln(2)=0 =><= contradiction, unless I'm doing some faux pas
0^x ÷ 0^x =0^(x-x)=0^0. But 0^x=0, and 0/0 is undefined.
Thank you for showing how exciting math is.
Can't you use octonions to solve this?
TLDR: Not solvable (no real value in answer)
I can see why a lot of mathematicians hate Geometry. It makes them use logic instead of algebra.
x = 0.333333 platypus divided by 4 kitchen sinks x i …. proof pending.
Please forgive the hugs…
Wrong
0^0 is 1
If we do limit
X approaching 0
x^x =1
You will see 0^0 is 1
minus infinity was my first guess before watching
2^0=1 is a mathematical convention that trades off one nonsense against another. In reality, outside of maths, it is clearly not 1.
Without lims it can't be solved
For x!=0 x^x = e^(xlnx) = e^a > 0 for whatever value of a in R, so no it has no solution
If you type out 0 to the 0 as 0^0 and then turn it upside down it looks like a little smiley face so 0^0=🙃
-1
-infinity^(-infinity) should approach zero.
it’s 0i because of the equation used for imaginary self powers.
why can't you just write out what x^x actually means, which is x • x • x •… • x, x times, then equal it to 0. there's only 1 number that when multiplied gives 0, which is 0. that concludes the fact that there's not a single solution that would give 0 other than 0^0, but 0^0 is undefined so there's no solution
The complex case needs more than "I just don't like it" 😂
would -inf*i work?
since were talking about it ill argue 0^0=1 again
1^0 = null = *1 = 1
2^0 = null = *1 = 1
the calculation does not change, it completely ignores the base.
whereas ;
0^1 = 0
0^2 = 0*0 = 0
this DOES change which means it cannot overwrite properties of the exponent that completely change the way the calculation works such as not using the base.
hense [anything]^0 = 1
Quite late on this one, but is there a problem just switching to base e? That is, it becomes e^(x ln x) which is non-zero for all x, whether complex or real, hence it has no solutions.
Actually you can just guess the answer. -(infinity) to the power of -(infinity) is zero
This is awesome
If you consider your actual video of Dual numbers, it is solvable
0^0=1
what about the dual numbers?
7:40
cant we just say if x => -inf, |x| -> inf
so 1/|(-x)^(-x)| = |x|^x = |x^x| also goes to zero,
thus as at the limit |x^x| converges to 0, then so should x^x because |z|=0<=>z=0 for z any complex number
I don’t think I quite agree with not being able to say -infinity approaches 0 even with irrational numbers, considering the i factor would also be shrinking quite fast, no?
I don’t think I quite agree with not being able to say -infinity approaches 0 even with irrational numbers, considering the i factor would also be shrinking quite fast, no?
0^0=indererminate. It is every solution, until defined
dude just made a 9:54 video for nothing 😭
x = -∞
Lim x→-∞ of x^x = 0
I may have posted this before here, but the way I think of x^y is not "x multiplied by itself y times", but instead as "1 multiplied by x, y times".
So, for example, take 3^4. That's 1 multiplied by 3, 4 times, i.e. 1 x 3 x 3 x 3 x 3.
3^3 = 1 x 3 x 3 x 3
3^2 = 1 x 3 x 3
3^1 = 1 x 3
3^0 = 1
Now apply this to 0^x
0^3 = 1 x 0 x 0 x 0
0^2 = 1 x 0 x 0
0^1 = 1 x 0
0^0 = 1
See? It …. actually makes sense (kind of).
T={[0,1,2,3] / *: (a*b)%4 , +:(a+b)%4 a,b=0,1,2,3 }
So:
x in T
x^x=0 then x=2 🤔🤔🤔????
0^0 = 0 ^ ( 1 – 1 ) = 0 ^ 1 * 0 ^ (-1) = 0 / 0 thus undetermined.
add another equation sign plug it into python x=2
6:19 oil identity
Ahh the classic problem of 0 * N = 0 vs N^0 = 1.
X =0 basically it's any number multiply by any number has to be zero. So zero multiplied by the power of zero is zero. No calculation is needed.
0^0 is an empty product, you arent multiplying anything, and an empty product is 1.
Yes ± Yes is the real answer
Let x belong to C: X^x = 0 <==> e^(xln(x)) = 0
Let y = xln(x) with y belonging to C: X^x = 0 <==> e^y = 0
And we know that the equation e^y = 0 has no solution in C
So the equation X^x = 0 has no complexe solution.
x^x=0 x=invalid
1:22 I would love a really detailed explanation for why x^0=1. The expansion method I was always taught of a^x = 1 * a * a… and so on always felt very weak to me.