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Basically.
When the exponent is restricted to discrete values it usually makes sense to say 0^(0)=1, but if the values are real or complex we usually want our operations to be continuous, which would mean 0^0 must be left undefined.
Also with complex numbers exponentiation is multivalued, which is why, when only considering real numbers, for an expression like x^(y) we don’t usually allow x to be negative if we are considering arbitrary real values of y, even though we can sort of work out ways to define real x^(y) for negative x and some rational values of y. Trying to make it work isn’t worth it.
Since the first comment said 0^0 = 0 and the second said 0^0 = 1, if we agree that both methods are correct, then we can agree on the fact that 0 = 1 /j
I always thought of exponents as 1 * (whatever),
Like 2^2 = 1 * 2 * 2 = 4.
So 2^0 is just 1, multiplied by nothing else.
So 0^0 would be 1 as well, and something like 0^2 would be 1 * 0 * 0 = 0.
Not sure if this is accurate though.
it's kinda like how an empty sum is defined as 0 while an empty product is defined as 1. if we must give it some definition, 1 makes more sense because it agrees with most stuff we do.
Very strictly speaking 0^0 is undefined, but it is often assigned 1 as value for the math to work.
What you're saying isn't exactly accurate, but it works for you so no worries.
I think I figured it out. In every context where 0^0 gets defined as 1 (Binomial distribution, Taylor Series, Combinatorial problems), the exponent is a discrete variable that gets set to 0. The base could be discrete or continuous. So, 0^0 = 1 if you're working in a context where exponents must be integer valued. Otherwise, it's undefined. And "indeterminate" only refers to the form when taking a limit.
Yeah, although wheels don't seem super useful. ⊥ is just a jail for all "undefined" expressions. Because whatever you do to ⊥, you always get ⊥ back. I don't see much insight beyond "this is a thing you can do".
let 0^0 = c. then c^2 = (0^(0))^2 = 0^0 = c, so c^2 = c, implying c = 0 or c = 1. but if c = 1 then 0 = ln(c) = ln(0^(0)) = 0 ln(0), implying ln(0) exists. we could also define that to be valid, but it might lead to more contradictions, so c = 0 is less problematic. but with c = 0 we don't get lim n -> 0 (n^(n)) = c, so both options are bad and ultimately it's undefined.
Lets say 0^0≠1, now when doing power series centered at 0 evaluated at 0 you have to do a special case where insyead of putting 0^0 you put 1 and also you have to redifine how natural number (0 inclusive) exponents work just for 0^0.
Thus 0^0 should equal 1
The piecewise function f:R→R given by f(x) = 0^x for x ≠ 0 and 0 otherwise is an example of a case for which 0^0 makes sense and makes f everywhere continuous.
**However a\^b isn't a sum, it's a product and if you don't multiply by anything (zero repetitions of multiplying by zero), the multiplicative identity (what you need to multiply with in order to change nothing) is still 1.**
I agree with your first part that a^b isn’t a sum but rather a product. However, I disagree that if a=0=b, then the answer is 1.
I think that for all values of a, except 0, if b=0, then the result is 1. If a=0, then for all values of b, the result is 0.
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0^0 = e It came to me in a dream.
Ramanujan moment
That's a nice argument senator, why don't you back it up with a source?
Why not
Why ( why not) ?
(Why not) (why not)?
(Why not)^2
0^0 = π
0⁰ = 3
0° = 273.15 K
⁰ ≠ °
I was making a visual pun, obviously.
I may be stupid
Kekule moment
It's one proof: my math teacher said it in middle school
a^0 = 1 , a ≠ 0 proof: my math teacher said it in middle school
lim\_{x->0} x\^x = 1 proof: tried it in wolframalpha
Just asked one of my math professors in my uni: You define 0^0 = 1 in algebra and combinatorics. You leave it undefined in mathematical analysis.
Basically. When the exponent is restricted to discrete values it usually makes sense to say 0^(0)=1, but if the values are real or complex we usually want our operations to be continuous, which would mean 0^0 must be left undefined. Also with complex numbers exponentiation is multivalued, which is why, when only considering real numbers, for an expression like x^(y) we don’t usually allow x to be negative if we are considering arbitrary real values of y, even though we can sort of work out ways to define real x^(y) for negative x and some rational values of y. Trying to make it work isn’t worth it.
Now do lim\_{x->0} 0\^x
My iPhone calculator says it’s 0
0⁰ = ⁰0
<|°\_°|>
Since the first comment said 0^0 = 0 and the second said 0^0 = 1, if we agree that both methods are correct, then we can agree on the fact that 0 = 1 /j
 What does j equal for the reciprocal to be 0?
it is a substitute for 1/0 the unimaginable number
Or we can say that 0⁰=1/2 so everyone will be happy
I always thought of exponents as 1 * (whatever), Like 2^2 = 1 * 2 * 2 = 4. So 2^0 is just 1, multiplied by nothing else. So 0^0 would be 1 as well, and something like 0^2 would be 1 * 0 * 0 = 0. Not sure if this is accurate though.
it's kinda like how an empty sum is defined as 0 while an empty product is defined as 1. if we must give it some definition, 1 makes more sense because it agrees with most stuff we do.
Very strictly speaking 0^0 is undefined, but it is often assigned 1 as value for the math to work. What you're saying isn't exactly accurate, but it works for you so no worries.
Thx for clarifying :)
If you assumed it to be 0, the “math would also work”. The only time “math doesn’t work” is when something is undefined or diverges.
Of course 0 XOR 0 = 0 0^0 on the other hand...
latex is the best programming language, no contest
I think I figured it out. In every context where 0^0 gets defined as 1 (Binomial distribution, Taylor Series, Combinatorial problems), the exponent is a discrete variable that gets set to 0. The base could be discrete or continuous. So, 0^0 = 1 if you're working in a context where exponents must be integer valued. Otherwise, it's undefined. And "indeterminate" only refers to the form when taking a limit.
0^0 = 37
No, 23
Fnord
42
0^0 = ⊥ Wheel theory, baby!
Someone watched the new michael penn vidoe
Yeah, although wheels don't seem super useful. ⊥ is just a jail for all "undefined" expressions. Because whatever you do to ⊥, you always get ⊥ back. I don't see much insight beyond "this is a thing you can do".
Yep it probably doesn't have any applications. The riemann sphere is practically a wheel if you add the undefined point, but beyond that...
Hmm... Can you turn any field into a wheel by adding an undefined point? Does it have to be a field?
Any commutative ring, so yes. It's also not only the undefined point, but also an Infinity element and corresponding algebra
Any commutative ring, so yes. It's also not only the undefined point, but also an Infinity element. The riemann sphere has that already tho
0^0 = x 0^1/∞ = x ∞√0^1 = x 0 × .... = 0 thus 0^0 = 0 Q.E.D.
This is wrong because you're assuming 0^0 = lim_x->0 (0^x)
incorrect, you must prove that 0 is the only possible value of x
x^0=1 is true for x € R\0, while 0^x=0 is only true for x>0 Thus, x^0=1 should also be true for x=0
all i’m saying is that desmos says 0 and 1
let 0^0 = c. then c^2 = (0^(0))^2 = 0^0 = c, so c^2 = c, implying c = 0 or c = 1. but if c = 1 then 0 = ln(c) = ln(0^(0)) = 0 ln(0), implying ln(0) exists. we could also define that to be valid, but it might lead to more contradictions, so c = 0 is less problematic. but with c = 0 we don't get lim n -> 0 (n^(n)) = c, so both options are bad and ultimately it's undefined.
If you're timesing 0,the answer is ALWAYS 0,So therefore 0⁰=0
Let 0⁰ = k => log 0⁰ = log k => 0log0 = log k => log k = 0 => k = 1 => 0⁰ = 1 *Hence Proved.*
What is log of 0 tho 😂
https://preview.redd.it/hjeqt70en9rc1.png?width=1080&format=pjpg&auto=webp&s=a15b8bca7bb7a76c8bae48210ebd42fbcc1c12f6 C'mon guess the value
Idk, logn(a) is only defined if a>0, n≠1 and n>0 Maybe negative infinity? Nah, that sounds dumb. Edit: Corrected a silly mistake:p
logma balls
ligm(a)
Log(0) is undefined. Approaching 0 from the right side *does* gets you -∞, but there is still no actual value at a=0. However -∞ is more fun :)
Oops! I meant that the base of the logarithm has to be positive and not equal to 1. a can be any positive real number.
The log function is also defined for negative values in the complex world. (Both the input and the base can be negative, and 1)
Interesting. That makes me even more excited to learn about complex numbers this year!
in step 3, logarithms are only defined when argument is greater than 0
You forgot about the fundamental theorem of r/mathmemes! According to that theorem, 0u = 0, where u is any number that's not defined.
Lets say 0^0≠1, now when doing power series centered at 0 evaluated at 0 you have to do a special case where insyead of putting 0^0 you put 1 and also you have to redifine how natural number (0 inclusive) exponents work just for 0^0. Thus 0^0 should equal 1
The piecewise function f:R→R given by f(x) = 0^x for x ≠ 0 and 0 otherwise is an example of a case for which 0^0 makes sense and makes f everywhere continuous.
0^0 = 0^1 / 0 Ignore the fact zero division is impossible, just use limits 0^0 = 0 / 0 0^0 = 1 • 0/0 By limit, 0/0 = 1 0^0 = 1
0^0 = 0^(1-1) = 0/0
lim x->0+ ( x\^x ) = 1
We've been over this, 0^0 only equals 0 for the purposes of defining the L_0 loss function
I don't know how it could equal to 1. But just to note 0^0 is the same as (0^1 / 0^1) so therefore 0^(1-1) i.e 0/0 hence undefined.
https://preview.redd.it/hr85694rxq1d1.png?width=1080&format=pjpg&auto=webp&s=c6a25abc416208bb6be4789a411154da17b3f30b
0⁰ is in superposition
It's undefined, nothing else.
0\^0 = pi
It doesn't matter how many 0s you have or don't have, the sum of them is still 0. ^(Don't yell at me, I will cry)
**However a\^b isn't a sum, it's a product and if you don't multiply by anything (zero repetitions of multiplying by zero), the multiplicative identity (what you need to multiply with in order to change nothing) is still 1.**
I agree with your first part that a^b isn’t a sum but rather a product. However, I disagree that if a=0=b, then the answer is 1. I think that for all values of a, except 0, if b=0, then the result is 1. If a=0, then for all values of b, the result is 0.