The reason as to why a lot of people define 0^0 as 1 is because there's certain functions that require it to be 1 in order for some proper equalities to hold (ex. the binomial expansion and the power series of e^x). However, some people still classify it as undefined due to 0^0 being considered an indefinite form in limit calculus, where approximating the same "value" yields different results. For example, x^0 as x approaches 0 is equal to one, yet 0^x as x approaches 0 (from the positives, at least) equals 0. Despite both values being "equal to" 0^0, they yield different results. However, these "values" aren't actually the input 0^0, just approximations made using infentesimally adjacent values of x. It's a bit hard to describe, but imagine if I had a function that rounded numbers down to the nearest integer (this is called the floor function). Suppose I want the limit of the floor function on 5 as I approach it from the left. Well, since everything between 4 and 5 is not an integer, it should be rounded down to the nearest integer, so the limit should be 4. However, when I input 5 into the floor function, I get 5 because 5 is already an integer, so I don't need to round it down. So, it's possible to have cases where the limit value does not equal the actual value of the function. In that similar vein, just because approaching 0^0 gives me 0 or 1, that shouldn't necessarily mean that it's has to be both values, or either values even. Still, there do exist contexts where 0^0 works as 1, so it's not just an arbitrary definition or choice either. Hope this helps give some insight.
What one needs to first consider is what exponentiation even is. In discrete contexts, it's repeated multiplication, so that's why in things like the binomial formula, you can get away with 0^0=1, because it simply represents the empty product. However, when considering ^ as a binary operation over real numbers, 0^0 is literally undefined. It's partially due to the limit argument One way to define exponentiation is using limits of rational exponentiation (which is well defined) Another way is using composition of the exponential function and logarithm There are many similar ways to define exponentiation over real numbers, and none of them yield a meaningful result for 0^0 This is why we say it is undefined, because it literally isn't captured by any definition of exponentiation, as it would render the definition inconsistent
I always viewed x^y as "1 is multiplied by x, y times (or divided by x, |y| times if it's negative), so x^2=1*x*x, x^1=1*x, and therefore x^0=1, since there's no x to multiply (or divide) with 1, which would work for 0^0=1
one way to do it is that 0^0 is undefined such that it does nothing this is a strange one, but it's there, and I'm not sure if it's mathematically useful (n*(0^0) = n, n+(0^0) = n)
Well Not really. Imagine 0^0 actually is one. Then to say it is any other number just multiply by c. The 0 just covers it up, but it still happened. Same logic for 0/0, but easier to see. So the infinite forms of 0^0 are just 1* the form. Also with this logic you have to say 0*2 is not the same as 0*1 or 0*c, even though they are all equal. This is why math needs to change to involve these numbers.
@@OOLOIBESOOSSOnope, undefined & indeterminate are both similar terms, but refer to different things. undefined refers to something with 0 solutions, whereas indeterminate refers to something with infinitely many
@@OOLOIBESOOSSO Not at all. Undefined typically is seen in asymptotes, while indeterminate values give you different answers from different approaches.
No, indeterminate. And what if the 0^0=0 is really not the same as 0^0=1 because they are not equal and what if the true 0^0 is one and the 0^0=0 is just 0*c(the true 0^0)=0*c*1=0. This would explain everything about it.
the problem is, any base raised to 0 gives 1 no matter what, even imaginary numbers, but taking 0 to even negative numbers is undefined, that is why the second rule is way less weaker, so i say 0^0 must be 1
That's this stupid kind of math i was tortured with on my college. Anyway, to use these rules you gotta apply what set of number you even use. Whats the lesson here? Don't watch 3 minute video on maths, it will confuse you more than help.
@epry0 if your argument depends on you dividing by zero the math theorists probably aren't the ones who made an error. (weird replying to @epry0 and his name doesn't get tagged)
probably thinking about "approaches infinity when denominator approaches zero." but hard to have it be equal infinity since the top is zero. bank won't let my bill be paid no matter how many pieces of nothing i give them.
I was always taught that 0^0 is undefined. Also it just seems logical because it could be written as 0^x * 0^-x and 0^-x will be a number divided by zero
If you cut 0 into 0 pieces, how large is each piece? The non existent piece can be literally any size because it doesn’t exist in the first place. Therefore 0/0 is equal to all real numbers. Since everything works, it is considered undefined, or more accurately “indeterminate”
@@David280GG i thought it was 0 because if you have nothing to share around then nobody gets anything, and if you share something with nobody then still nobody gets anything
The reason as to why a lot of people define 0^0 as 1 is because there's certain functions that require it to be 1 in order for some proper equalities to hold (ex. the binomial expansion and the power series of e^x). However, some people still classify it as undefined due to 0^0 being considered an indefinite form in limit calculus, where approximating the same "value" yields different results. For example, x^0 as x approaches 0 is equal to one, yet 0^x as x approaches 0 (from the positives, at least) equals 0. Despite both values being "equal to" 0^0, they yield different results. However, these "values" aren't actually the input 0^0, just approximations made using infentesimally adjacent values of x. It's a bit hard to describe, but imagine if I had a function that rounded numbers down to the nearest integer (this is called the floor function). Suppose I want the limit of the floor function on 5 as I approach it from the left. Well, since everything between 4 and 5 is not an integer, it should be rounded down to the nearest integer, so the limit should be 4. However, when I input 5 into the floor function, I get 5 because 5 is already an integer, so I don't need to round it down. So, it's possible to have cases where the limit value does not equal the actual value of the function. In that similar vein, just because approaching 0^0 gives me 0 or 1, that shouldn't necessarily mean that it's has to be both values, or either values even. Still, there do exist contexts where 0^0 works as 1, so it's not just an arbitrary definition or choice either.
Hope this helps give some insight.
What one needs to first consider is what exponentiation even is. In discrete contexts, it's repeated multiplication, so that's why in things like the binomial formula, you can get away with 0^0=1, because it simply represents the empty product.
However, when considering ^ as a binary operation over real numbers, 0^0 is literally undefined. It's partially due to the limit argument
One way to define exponentiation is using limits of rational exponentiation (which is well defined)
Another way is using composition of the exponential function and logarithm
There are many similar ways to define exponentiation over real numbers, and none of them yield a meaningful result for 0^0
This is why we say it is undefined, because it literally isn't captured by any definition of exponentiation, as it would render the definition inconsistent
Another formula that says 0^0 is one is the growth formula A=P(forgot)^nt
I got a major headache trying to understand these zero equations. I think I should go to the l'hopital.
0:34 +c crying in the corner
I always viewed x^y as "1 is multiplied by x, y times (or divided by x, |y| times if it's negative), so x^2=1*x*x, x^1=1*x, and therefore x^0=1, since there's no x to multiply (or divide) with 1, which would work for 0^0=1
Sure, this works if you consider ^ as an operation over integers
But in the context of real analysis, this doesn't work. 0^0 is undefined in analysis
2:44 I guess that makes sense
0/0 is literally the same thing as 0^0
one way to do it is that 0^0 is undefined such that it does nothing
this is a strange one, but it's there, and I'm not sure if it's mathematically useful
(n*(0^0) = n, n+(0^0) = n)
Probably not because it doesn’t follow these properties at all times
0^0 is undefined
*undetermined
and 0⁰ is also undetermined
Well Not really. Imagine 0^0 actually is one. Then to say it is any other number just multiply by c. The 0 just covers it up, but it still happened. Same logic for 0/0, but easier to see. So the infinite forms of 0^0 are just 1* the form. Also with this logic you have to say 0*2 is not the same as 0*1 or 0*c, even though they are all equal. This is why math needs to change to involve these numbers.
0/0 is more accurately “indeterminate” because it can equal any real number.
thats what undefined means 🫠
@@OOLOIBESOOSSOnope, undefined & indeterminate are both similar terms, but refer to different things. undefined refers to something with 0 solutions, whereas indeterminate refers to something with infinitely many
@@OOLOIBESOOSSO Not at all. Undefined typically is seen in asymptotes, while indeterminate values give you different answers from different approaches.
1:21 is that a blackpenredpen reference???
Actually 0^0 is 0 and 1 at the same time and as such is undefined
No, indeterminate. And what if the 0^0=0 is really not the same as 0^0=1 because they are not equal and what if the true 0^0 is one and the 0^0=0 is just 0*c(the true 0^0)=0*c*1=0. This would explain everything about it.
@IsaacDickinson-tf8sf no, someone on RUclips even explained the math on it, id like the video though you cant in youtube comments. 0^0 is both 1 and 0
0^0 is also undefined lmao
If any power base 0 is 0 and any power raised to 0 is 1 then 0 raised to 0 is by definition a mathematical paradox
the problem is, any base raised to 0 gives 1 no matter what, even imaginary numbers, but taking 0 to even negative numbers is undefined, that is why the second rule is way less weaker, so i say 0^0 must be 1
@ 0 to anything is 0 that just makes sense. If 0 is absorbent in multiplication it only makes sense it would be so in exponentiation
Its not about being "weaker", its about satisfying the properties, and 0^0 = any number breaks most of them
shouldn't 0x0 also be undefined? It can for example be interpreted as being "not zero"?
0 x 0 is 0 wdym 💀
0 x 0 is universally zero
@@leffa1 it's literally 0 boxes of zero items... ya can't get more 0 than that
No.
That's this stupid kind of math i was tortured with on my college.
Anyway, to use these rules you gotta apply what set of number you even use.
Whats the lesson here?
Don't watch 3 minute video on maths, it will confuse you more than help.
0^0 is not 1 it's undefined
How can you say that when you don’t even know it it is quite literally one. That’s just a math fact clearly you haven’t been to. 9th grade yet
0^0 is not 1. 0^0 can be rewritten as 0^x / 0^x or 0^1 / 0^1 which is equal to 0/0. therefore it is undefined
@ you’re gonna be looking back and laughing at how dumb you were when you reach ninth or 10th grade depending on where you live
@epry0 if your argument depends on you dividing by zero the math theorists probably aren't the ones who made an error. (weird replying to @epry0 and his name doesn't get tagged)
no one in this comment chain even watched the video to the end or knows what they're talking about
I always thought it was infinity
probably thinking about "approaches infinity when denominator approaches zero." but hard to have it be equal infinity since the top is zero. bank won't let my bill be paid no matter how many pieces of nothing i give them.
Well, in computation, it is, but in the actual concept of math, it’s not
The limit is not the number of
Huh?
I was always taught that 0^0 is undefined. Also it just seems logical because it could be written as 0^x * 0^-x and 0^-x will be a number divided by zero
0^0 is undefined, but your reasoning is flawed
first
also, 0/0 is 0, not undefined
just like 0*0 is 0, if i have nothing and cut it up into 0 pieces i still have 0
You're factually wrong.
If you cut 0 into 0 pieces, how large is each piece?
The non existent piece can be literally any size because it doesn’t exist in the first place. Therefore 0/0 is equal to all real numbers. Since everything works, it is considered undefined, or more accurately “indeterminate”
0/0, what number multiplied by zero equals zero? any, thats what undefined means
@@David280GG i thought it was 0 because if you have nothing to share around then nobody gets anything, and if you share something with nobody then still nobody gets anything
Technically it's one because you are dividing something by the same number. Do this with ANY other number go on I'll wait.
I DIED FROM BOREDOM WHILE WATCHING THIS VIDEO
Of course, because Russians don't want to learn
OF COURSE IT DOESN'T WORK USE COMMON SENSE