Why is 0^0 undefined?
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- เผยแพร่เมื่อ 29 ม.ค. 2019
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zero to the zero power, undefined vs. indeterminate: • x^sqrt(x) when x is 0 ...
0^0=1 by convention in power series: • 0^0=1 is "seriesly" us...
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blackpenredpen | 曹老師
WHERE IS THE *L I M I T*
This is getting off limits
The limit is another problem
It's on-line
Given x ≠ 0, if you define x^n and x^-n , you could get x^0 as x^(n - n) = (x^n) / (x^n) = 1.
Isn't that a correct way to get away with not defining x^0 ?
Going further with the x^(n - n) thought and allowing x = 0 :
0^0 = 0^(n - n) = (0^n) / (0^n) = ( n*ln(0) ) / ( n*ln(0) ) = ln(0) / ln(0)
ln(0) is undefined and therefore ln(0) / ln(0) is also undefined, which could imply 0^0 is undefined.
@@armwrestling_nerd I see that you explained this beautifully, but the person was joking really
"You guys are gonna debate in the comments. It's TH-cam-I understand."
This guy gets it.
Yup!!
Hello teacher I'm studying at the same time.
Yes, that's the funny thing about freedom of speech... there's no requirement on being correct to express it.
No
@@blackpenredpen THis video
I don t know but i m wondering why i like this " i will do it for u guys" keep doing it broo
LOLLLLLL, I am very very very glad to hear this! And I will keep doing this for you guys!!!
@@blackpenredpen 😂😂😂😂😂😂😂😂😂😂😂 i wish this comment be posted on instagram
@@xaxuser5033 I agree. I smile at every "isn't it?" and "well, well, well" and "let me write this down...right HERE".
let me simplify: it's 1 to the 0th dimension
It always puts smile on my face, when you call set of points, numbers or expressions as "EVERYBODY"
What a nice profile you have ;)
@@quentinl7021 u too m8
What if you average out both results and have 0^0 = 0.5?
that's dumb
That is not dumb. That is how maths is made. Think Cesaro summation, seems dumb as well but works anyway.
It's like how numberphile does 1-1+1-1... and gets 1/2
@@ThinkDifferentlier Cesaro summation is a load of purposeless pure-math garbage. I agree it's fun to screw around with math and do stupid things like that but it's still dumb, because of how misleading and pointless it is.
nope. the second suggestion is not as nice as the first.. honest would be "something between 0 and infinity", not just 0
I have zero packages of cookies which all have zero cookies and I wanna do multiplication. How many friends do I have?
Best comment XD
Friends are undefined if you study math
You have as many friends as you can help with their math homework.
Fred
@@ffggddss So true
Reijo P.
Wait, I’ll ask my big bro:
*NI-SAAAAAAN !!!!*
(In case you thought about it, no, I know that BPRP’s Chi-.... uh.... Ko-... uh.... not Japanese ! The contrast is the joke)
We should not confuse the limit indeterminate form 0^0 with the 0^0 value itself. The fact that 0^0 in limit sense is indeterminate, does not mean that is undefined. This value, depends on the area, is defined as 1 (e.g. set theory) or it is simply left undefined as some math analysis authors do
Yes it’s better to call 0/0 indeterminate and 7/0 for, example as, undefined.
@@JohnSmith-pv1jq No. Did you not read the comment? Read it again, because I think you missed the point.
Consider the following proof:
X^0=x^(1-1)=x^1/x^1=x/x=1 therefore
X^0=1.
However, using this proof we divide by x, meaning we have to say that x≠0, otherwise this becomes undefined because
0^0=0^(1-1)=0^1/0^1=0/0= undefined, you cannot divide by 0.
@@randomname9291 You are right. But precisely, you cannot even conclude that 0^0=0/0 since for the second step you need the denominator to be distinct to 0 (and also 0/0 is undefined so what does that equality mean?)
As explained in the video, "undefined" means that it is simply not defined so you cannot carry out proofs from an undefined term. If you would like to define 0^0=1, which is often a notational convention, then you should be aware that some exponentiation properties fail, just like the one you provided. This is not a contradiction, this just means that regular properties or theorems that you expect to hold aren't true anymore.
@@lucianosalvetti5852 you need the denominator to be distinct to zero precisely because it is undefined unless it isn’t zero. This equality means that 0^0 is undefined (or at the very least not equal to 1), as you cannot get to 0^0=1 without concluding that it is an undefined term.
Please give 0^0 a rest.
klong1972
This deserves to be the best comment ever!
Thanks for reminding, it's 3am already and I'm watching TH-cam
@@@blackpenredpen Of course in cardinal arithmetic 0 raised to the 0 power is actually 1 - the number of functions with an empty domain i.e. the empty function.
@@blackpenredpen Why don't you pin it?
@@AyushGupta-yj8jz lol same its 3:25 AM here.
"I don't want to talk about limits"
*Proceeds to manually form an intuition to the limit as x goes to 0 for y=x^0 and y=0^x*
elementary level math
In combinatorics, 0^0 is often defined to be 1 by convention. It's similar to 0!=1. eg, the number of permutations of an n-element set is n!, and the number of functions from an n-element set to itself is n^n - we'd like both of these to equal 1 when n=0. so 0! = 0^0 = 1.
That only works for doing the power series like e^x.
It's not even really a convention. You can prove that 0^0 = 1 in discrete mathematics just from the discrete meaning of exponentiation.
math
Since 1^0 is 1 and (-1)^0 is (-1) you want define 0^0 cuz it’s a neutral number if so the number must be somewhere between -1 and 1 but not 0 WHICH IS IMPOSSIBLE
@@penguincute3564 but (-1)^0 is 1!
BPRP, may I make a suggestion? In the past when I have explained to students the meaning of the 0th power and negative powers, I say look at this:
For a ≠ 0, a³·a² = a⁵. In other words, notice that multiplying, means adding the exponents. Conversely, we see that dividing means subtracting exponents. For example, a⁵/a³ = a⁵ˉ³ = a², "isn't it?" So, that means a³/a³ should equal a⁰ by subtracting exponents. But we know that a number divided by itself is 1. So a⁰ must be 1. And a³/a⁵ should equal aˉ². But we know that it's 1/a². So, a negative exponent means take the reciprocal. And I would say this isn't "by definition," it's "by extension!" We have extended the meaning of exponents to include 0 and negative numbers in a way that is consistent with the behavior that we're already familiar with (namely, adding and subtracting corresponds to multiplying and dividing). My point is, the definition isn't totally arbitrary, it's an extension of what we already know.
What do you think?
Ken Haley yup I agree. Because we extended exponents to 0 and negative exponents. And when we say a^-2=1/a^2 we are giving a meaning to this extension. And when we are giving a meaning to something, we call it a definition.
And I think as long as we don't call this a proof or theorem, then we are okay.
but you cant prove something is undefined in the first place. the only use of calling it a proof would be to state that it is undefined because it isnt defined
Ken Haley Extensions are definitions, so there is no distinction in declaring a rule by extension or by definition. It is not arbitrary, but it still is a definition.
@@angelmendez-rivera351 I think what he tried to claim is it isnt totally arbitrary.
It's TH-cam, I understand
Mukund M yup
Well it's like saying "if you don"t agree with my me you're a troll." Bprp can do better.
here's something i have to say. whenever i click on one of his videos, there's a jolt of happiness that passes through me in the first two seconds, I feel energetic to do math, he is a LEGEND! BLACKPENREDPEN
Awww thank you!!!! When I get comments like yours, they make my days!! I really appreciate it!
@@blackpenredpen thanks for these videos, man! keep it up :D
For better understanding 0 exponents, could you also say that;
2^3 = 1*2*2*2,
2^2 = 1*2*2,
2^1 = 1*2,
Thererfore,
2^0 = 1
I use this as another model to make sense of zero exponents (as well as the continuous division of 2 as shown in the video). Dunno if this is legit, but helps me anyway ¯\_(ツ)_/¯
and then for negatives it becomes 1/2, 1/2/2, 1/2/2/2 etc
An empty set by definition has a product of 1, regardless of what the set is empty of.
Exactly
Then can you say why it is defined like that.This statement is not a proof,in reality there is no proof ,2^0 is 1 due to definition , we can use this pattern to be useful ,if we define it has 0 ,we cannot use it efficiently.
In the context of multiplication, combining a set of factors with an empty set of factors doesn't change the product. That's equivalent to multiplying by 1, by definition: 1 is the multiplicative identity, which is defined as the factor that doesn't change the product.
By the way, 0^0 must be defined if you want to include tetration in our number system. According to the arithmetic-geometric conversion, any 0's get converted to 1's, as 0 is the arithmetic identity number, and 1 is the geometric identity number. The operations get increased by 1 hierarchical order during an arithmetic-geometric conversion. Therefore, 0^0 is equal to 1 tetrated to 1. However, failing to define 0^0 causes tetration base 1, 0, and negative numbers to become undefined as well, including 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of 1's with 1 entry, which is just 1. Therefore, 1 is undefined. All integers can be multiplied by 1, and then all integers are undefined. All rational numbers are ratios of integers, and each integer can be multiplied by 1, and the rational numbers get lost in the black hole of undefinedness. Irrational numbers will eventually fall, first the square and cubic roots, then pi and e, and finally the complex numbers until the entire number system gets annihilated except 0. And finally, 0 will accept its fate of being undefined as being a product of 1 and 0, and the entire Mathsverse will collapse.
If 0^0=1, then 1 tetrated to anything is equal to 1, including fractional and irrational numbers. 0 tetrated to anything is 1 if the tetrating number is even, and 0 if the tetrating number is odd. Negative numbers tetrated to anything are:
Defined for all integer values if the negative number, written as a fraction, has both an odd numerator and denominator
Defined for integer values to 2 if the negative number, written as a fraction, has an even numerator and an odd denominator
Defined for integers to 1 if the negative number is either irrational or has an even denominator.
-1 tetrated to anything equals -1 if n is not 0 and is equal to 0 if n is -1, and 1 if n is 0.
This doesn't prove that 0^0 is undefined. You've taken the limits, but that doesn't prove anything for discontinuous functions. What you're trying to define is the behavior exactly at 0^0, not what's happening around 0^0. You can just as easily define the function 0^x to be 1 at x=0, given that it's already discontinuous there, and then you'd have no conflict between the results.
EXACTLY. That is the fundamental mistake behind the video.
exactly. 0^0 = 1.
why?
*where x > 0,*
0^x = 1 * 0
this pattern holds with any x^y
x^y = 1 * x^y
x^2 = 1 * x * x
therefore, 0^0 = 1.
it’s the same reason why x^0 where x != 0 is 1.
It doesnt serve any purpose defining 0^x to be 1....or 0 at x=0. Thats why it better left undefined in algebra.
@Hansel McDonald,
Why are you saying that it doesn't serve any purpose defining 0^x to be 1 at x=0 ? How about the binomial expansion
(a+b)^n = ∑ ((n choose k) * a^k * b^(n-k)) , from k=0 to k=n
for any natural number n, and setting (for example) n=2, a=3, b=0 ?
Or how about the Taylor-series
e^x = lim { ∑ (x^k)/k! , from k=0 to k=N }, for N--> +infinity
and evaluating at x=0 ?
How will you evaluate those expressions without using 0^0 = 1 ?
Consider the following proof:
X^0=x^(1-1)=x^1/x^1=x/x=1 therefore
X^0=1.
However, using this proof we divide by x, meaning we have to say that x≠0, otherwise this becomes undefined because
0^0=0^(1-1)=0^1/0^1=0/0= undefined, you cannot divide by 0.
Oh man, this is on my video to do list! Sometimes students will ask about this during a totally unrelated lesson, so depending on the lesson its hard to stop everything to explain this. This is a nice video to refer to - thanks!
Thanks!!!
I watch your videos!!
I like your cross multiplication video!
@@naveensundar4765 hey thanks :)
We were talking about the zero exponent property *yesterday* in Algebra 2.
The Bloxxer haha nice!!
2^(-1) is the multiplicative inverse of 2.
1/2 is an element of the quotient ring.
We find they are the same.
The empty product is 1 as the empty sum is 0.
😁
I am revisiting this video two years later, and I must say, I have a major pet peeve with trying to use limits to define the values of a function at a point. What the argument in this video essentially concludes is that for f(x, y) = x^y, lim f(x, y) (x -> 0, x > 0, y -> 0, y > 0) does not exist, so f(0, 0) is undefined, which is actually a bad argument, if you think about it mathematically. By definition, the limit of a function to a point has to do with the values of the function NEAR that point, not the value of rhe function AT that point. The relationship between the two is given by a statement about CONTINUITY of a function at a point, or lack thereof. If lim f(x, y) (x -> 0, x > 0, y -> 0, y > 0) does not exist, then you can conclude is discontinuous at (0, 0), but it does not imply f(0, 0) is undefined. That is nonsense.
Here is a simpler example. You all know the floor function, right? BPRP has made videos on it before, and Michael Penn does videos on it all the time. Those of you who are familiar with the floor function would know that lim floor(x) (x -> 1) does not exist. Does this mean floor(1) is undefined? No, it does not. Why do we know it is not undefined? Because from the definition of the floor function, that floor(x) = m for integer m iff m =< x < m + 1, it can be proven that floor(1) = 1. So floor(1) = 1. Period. That is all there is to it, and there is no need for there to be any further discussion.
In our case, f has a definition too, and if we want to know what f(0, 0) is equal to, then we need to apply that definition to x = 0, y = 0. Period. That is all there is to it. There is no point in worrying about limits, in worrying about f for x > 0, y = 0, or x = 0, y > 0, because all we simply need is to plug x = 0, y = 0 into that definition. Problem solved. We have a definition that works for any complex number x whenever y is a cardinal number. 0 is a cardinal number. So for y = 0, we apply that definition, and the apply x = 0 to that. Multiple people in the comments have explained this too, so this is by no means a fringe idea: this is just the single idea that actually agrees with mathematical rigor. Doing this gives you the unambiguous result 0^0 = 1, and no, this is not inconsistent with calculus whatsoever, because this would not mean that the indeterminate limits are equal to 1, since the indeterminate limits are just that: limits, not an arithmetic operation that is either defined or not.
If a person wants to ignore that definition of exponentiation and somehow decide that 0^0 should be undefined, even though there is no legitimately valid argument that logically entails this, then okay, that type of arbitrariety does somewhat go against mathematical rigor, but I will let it slide. However, I seriously need teachers to stop teaching students that a function at a point must be undefined because the limit to that point of the function does not exist or is indeterminate, since this is just straight up incorrect, and all it does is create a bad understanding of what limits are and how they work. It does get tiring when hordes of students have come to believe that evaluating the limit is the same as just plugging the point into the function because this is what they were taught by their teachers. And if you are doing this to then try to justify something as capriciously unfounded as "0^0 is undefined," it makes it that much worse. I dislike it just as much as when teachers say "1/0 is undefined because lim 1/x (x -> 0) does not exist," because again, that is just not how these things work. If you want to explain why 1/0 is undefined, then you need to tell them that the equation 0·x = 1 has no real or complex solutions, and if you try to invent a new type of number that solves this equation, then problems emerge that make it impossible. If you think this is too complicated to explain to a child, then I would rather that you just tell them "you will be better equipped to understand the answer in a few years," rather than pulling up a limit argument that only reinforces a bad understanding of what limits are.
Beautifully stated.
How are you defining x^y? I would define it as e^(ln(x)*y). Plugging in 0 for x any y, we get e^(ln(0)*0). This is basically e^(-infinity*0), which is undefined. Infinity*0 and -infinity*0 are undefined.
@@GoogleAccount-if6pu ln(0) is undefined. It is not -♾, because -♾ is not a number. ln is, by definition, a function from (0, +♾) to R. Of course, your definition does not work, becaause this makes 0^y always undefined, since there is always a ln(0) to deal with. This can easily be fixed if you let x^y := lim exp[y·log(s)] (s -> x), or any other proper definition of exponentiation. Here, 0^y = lim exp[y·log(s)] (s -> 0), and if y = 0, then this evaluates to 1, while if y > 0, this evaluates to 0. This can be extended to the complex numbers as well.
@@GoogleAccount-if6pu I would define it as e^(ln(x)*y) forall X NOT EQUAL TO 0
@@angelmendez-rivera351 bro do u have insta like i wanted to talk to a good mathematician hope u answer
the fact that you consider 0 like x when comparing it to 0^x is something I never considered
Excellent proof!
The proof is pretty bad, because the argument that 0^x = 0 for every x > 0, therefore 0^0 = 0, is really invalid.
When I am teaching about the value of x^0, I tend to use the exponent rules such that a^b / a^c = a^(b-c).
e.g. 2^3 / 2^3 = 2^(3-3) = 2^0
2^3 / 2^3 = 8/8 = 1
Therefore, 2^0 = 1
This can be expanded to any base and power.
a^b / a^b = 1 (a number divided by itself is 1.)
a^b / a^b = a^(b-b) = a^0 (a number subtracted by itself is 0.)
Therefore, a^0 = 1.
I generally use a piece-wise function to define 0^x.
= {0, if x > 0
{1, if x = 0
{undefined, if x
The way you talk and teach math really makes me love math... I still suck a it and hate it but you're turning me around :) you're awesome!
You can define 2^n = 1/(2^-n)
Then 2^0 = 1/2^(-0)
(2^0)^2 = 1
2^0 = sqrt(1) = 1
So 2^0=1 follows from this definition
at line 2 you are assuming 2^0 is nonzero
instead you could define
2^n = 2^(n+1)/2 and 2^1 = 2
then it follows that 2^0 = 1
but if youre gonna change the definition of 2^n then you might as well define 2^0 = 1
so if you use the "counting" definition of 2^n, theres no way to get around defining 2^0 (and consequently having to define negative/fractional powers)
i think the takeaway here is to not use the "counting" definition of 2^n that you learn in elementary school in the first place. just define 2^n in a way that allows you to derive all real powers. and for that matter, you might as well define 0^0 while youre at it.
"Its youtube, I understand"
Hahaha,you're great man😂
Keep up the good work👏
“This doesn’t involve calculus”. Proceeds to show you an informal definition of a limit as an example
Limits are from math analysis and from that are a tool/technique used in calculus. Limits are not inherently part of calculus.
@@jmwild1 Limits are calculus. Calculus is a comsequence of analysis, and to some extent, synonymous with analysis. The difference is in the focus of the teaching of the material, not the actual contents of the theory.
@@angelmendez-rivera351 Limits are not calculus, they are a tool used in analysis and calculus. There's really not an argument there.
@@jmwild1 That is like saying "wheels are not auto parts, they are just a tool used to assemble automobiles." You are contradicting yourself. By definition, any tool used in calculus and analysis IS analysis. A theory of study is defined entirely by its contents of study. Nothing else. And you are right: there is really not an argument here, so I have no idea why you insist on being a contrarian about this.
@@angelmendez-rivera351 By "analysis" I meant the category of math analysis. No there's no contradiction here, and I'm not being the contrarian, I'm correcting the *original* contrarian above. He chose to challenge blackpenredpen's claim and I simply refuted that.
When learning discrete math, namely cardinal arithmetic, we concluded that 0^0=1, since there is only one function that you can make from the empty set to the empty set (when powers are basically how many functions there are from the group with the cardinality of the exponent to the group with the cardinality of the base).
Something else I wanted to point out is that in 7:15, I don't think that you can conclude that 0^0=0 from that because its discontinues at 0 since the limit on the left is different from the limit on the right and I'm not sure you can ignore calculus when talking about graphs like that, but i'm not too sure.
great video as always btw! :)
It drives me nuts to hear people say "it's defined in certain areas". 0^0=1. The area you are in is not a parameter to the calculations. There are no fields where it comes up with a different answer. It's only that people confuse infinitesimal values with zero. This is because limits are a waffly concept, rather than using algebraic infinitesimals. People are also doing blackboard/paper mathematics, rather than writing computer code, where they have to face the music that 0^0=1 has to be used for the code to actually work; and the hand-wringing over it being defined can't be avoided.
It's just that in some fields, where people confuse 0^dx with 0^0; where dx*dx=0 (ie: infinitesimal); they refuse to define 0^0. This is kind of like finitists refusing to define any kind of infinity.
as for "0^0=0".
smallFinite*smallFinite > 0. smallFinite>0
0^smallFinite=0
But...
smallInfinitesimal*smallInfinitesimal = 0. smallInfinitesimal>0.
0^smallInfinitesimal = 1
0 isn't the same as smallInfinitesimal, and that's not the same as smallFinite.
0 < smallInfinitesimal < smallFinite
This is the Limit comment everybody is searching in the comments
lim x ->0 x^x=1
LOL
Here we have it!!!
@@blackpenredpen :-)
@@bhend1, yess!
and how about
lim (x↓0) 0^(x) = 0 .
?
lim x->0 (x^(2x))
See the problem?
Edit: yes I know me from a year ago was shitty at math you can stop correcting me now
But 0^0=1, too!
The full behaviour of x^y around (0,0) would be interesting.
I put it in wolfram alpha, it's a 3D plot.
Big thank from a math teacher!
Nice video!
Thank you so much!
Even though you are not using limits, isn’t it more fit to say that 0^0 is indeterminate, meaning it cannot be determined? Undefined, we use for expressions like 1/0 or the slope of a vertical line.
SyberMath
Yeah, it fits a bit better, but honestly, if you need it, you will define it somewhere. For example, in Analysis 1, we defined x^0 as 1 even if x=0 for Taylor series and alike
Well, expressions like 1/0 or the slope of a vertical line is infinity.
Hello, im new to the channel and like a lot your way of dealing with mathematical problems. As a highschool student I want to ask if it’s possible to learn this power ?
In my opinion, 0^0 is best to defined to be 1. If we leave 0^0 undefined, then a general polynomial must be written as a_0 + sum from i=1 to n (a_i x^i) instead of sum from i=0 to n (a_i x^i), (because the second expression would be undefined at x=0), which seems unnecessarily awkward. In addition, 0^0 is definitely equal to 1 in applications to combinatorics. Sure, defining 0^0=1 makes x^y discontinuous at x=y=0, but that's not a problem so long as we are aware of it.
Kronecker Delta function. These functions show up all over the place in nature. f[x : x >= 0] := 0^x is fine... f[0]=1. f[x : x > 0]=0. The weirdest thing of all is to define it per mathematics area, as if the notation means different things in different areas. The problem is confusing infinitesimals with 0, because limits are a waffly concept.
Nothing says that your formulas have to be convenient. It's very convenient to define 1/0 = infinity and that also simplifies many formulae. Doesn't make it right.
As many people have pointed out, for the set theorist 0^0=1, but this is also true for the algebraist, since we use integer powers, and often use definitions like: a^n is the product of a with itself n times, when n >= 0. Then 0^0 is the empty product, which always has value one (assuming our operation is associative and has identity). This video is misleading at best. There are likely a few fields in math in which 0^0 is undefined, but when it is defined, it seems that it is usually defined to be 1.
You are correct.
7:40 Why you want to make it continuous at x=0? It is already undefined for xZ s.t e(x,0) = 1 and e(x, y+1) = e(x, y).x
In the same way, for rationals, it is the function e:Q²->Q s.t e(x,0) = 1 and e(x, y+1) = e(x, y).x and e(x,y.z) = e(e(x,y),z)
And, finally, for reals numbers, it is the extension of this concept by cauchy sequences, where it is continuous (in (x,y) in R² s.t. not (x = 0 and y
Consider the polynomial p(x) = 3x^2+8x+6. A polynomial in general is Sum i = 1 to n a_i*x^i. So in the case of p, a_2=3, a_1=8, and a_0 = 6, the polynomial is 3x^2+8x^1+6x_0. Substitute 0 for x. We get 3*0^2+8*0^1+6*0^0. Since 0^1=0 and 0^2 = 0, we get 3*0+8*0+6*0^0= 6*0^0. But when we substitute it into the original form, we get 3*0^2+8*0+6 = 6. This means that 6*0^0 = 6, which implies that 0^0=1. I believe that in most algebraic applications, 0^0 winds up being 1, usually by defining it as such.
This is circular logic: by saying the polynomial can be written as that sum you are already implying that 0^0 = 1, so basically what you just said is: "since 0^0 = 1, 0^0 has to be 1"
@@luxo1035 It's also explicit in the binomial theorem that x^0 = 1 and y^0 = 1, even if x = 0 or y = 0. In fact, 0^0 must equal 1 for the binomial theorem to be correct when x = 0 or y = 0. Not to mention that 0^0 must equal 1 in Euler's power series formula for e^x.
SOMETHING ABOUT LIMITS
SOMETHING ABOUT CALCULUS
Wtf does this all mean?!
0^0 is 1. Maclaurin series expansion of e^x = x^0/0! + x^1/1! + x^2/2! + x^3/3! + ...
Taking x=0, all terms from x^1/1! cancel out and become 0, while x^0/0! term gives 1. 0! is 1 and even 0^0 should be 1 too. If 0^0 is undefined, then e^0 will be undefined too. As e^0=1, then 0^0 should be 1 for the Maclaurin series expansion to hold true
Also if we draw graph of y = x^x, we will see that y=1 when x=0. Even if we type 0^0 on Google, we get the answer as 1
So what exactly is the contradiction if you define 0^0=1? Wouldn't the 0^x simply be discontinuous?
Exactly. There is no contradiction at all.
@@MuffinsAPlenty Totally agree. A function being discontinuous in one point of their domain is no contradiction because functions are allowed to be discontinous.
If we take it as the formula of summation of e^x....
e^0=(0^0)/(0!)+(0^1)/(1!)+(0^2)/(2!)+...
Which is
e^0=(0^0)+0+0+0+....
Which says
1=0^0
0^0 = 0^(1-1) = (0^1)/(0^1) =0/0
@@JensenPlaysMC Same logic: 0^2=0^(3-1)=(0^3)/(0^1)=0/0. So according to your logic, 0 to any power should be undefined.
@@kartheyansivalingam7927 Hmm, any idea why the standard rules dont apply in this situation
0^x is 0.
e^(-1)? If you do your rule for e on that, can you tell me what (-1)! Is? XD
that was brilliant loved the joke at the end!!!
Thank you!
You are the best mathematician on TH-cam.The best of the best of the best👍👍👍😊😊
Have you also considered the curve of x^x? If you plot the curve, you will find 0^0 is 1. In fact, when we consider x^0 is from x^1/x^1, then x^(1-1) is x^0. If we put x equal 0, then 0^0 is 0^1/0^1=0^(1-1)=0^0. But the above calculation contains a number divided by zero, 0^1, 0^0 is undefined.
I have the exact same thinking. Anything else doesn't make sense when you think about it this way. The only thing I saw in another comment was that maybe 0/0 is actually defined, then some value could possibly make sense.
Sure, by that definition. If you use the set theory definition for natural numbers the answer is just flat out 1, because there is precisely one function from {} to {}.
So, when you embed the natural numbers into the real numbers using a set theory scheme like ZF then pretty clearly 0^0=1 is valid.
Joshua Hillerup Please elaborate. I’m incredibly curious to know why, but my understanding of set theory is limited.
@@TheGeneralThings So, I'll give a sketch at least, although fully explaining it from first principles takes most of a math course.
The ZF set theory method of constructing natural numbers makes use of Peano arithmetic. You build up the natural numbers by taking the cardinality of sets constructed in a specific way, so 0 = |{}|, 1 = |{{}}|, 2 = |{{}, {{}}}|, and so on (the exact method of constructing it isn't important for this explanation).
You do the various arithmetic operations by doing operations on the sets. For exponentiation, for sets A and B you have |A|^|B| is the number of possible functions that can go from A to B. The number of possible functions that go from {} to any non-empty set is 0 (it's impossible to construct such a function), but the function that go from {} to {} is the identity function (and only the identity function), which exists. So 0^0 = |{}|^|{}| = 1.
Joshua Hillerup how do you work non integers and even complex numbers into these definitions?
@@dospaquetes even integers require an extension. But you basically define -a to be 0 - a (where a is a natural number), fractions based on the ratio of two integers with a procedure to reduce them (and anything divided by 0 is undefined), and so on. My second argument that it is valid the other sets of numbers because it is valid for the natural numbers is on a bit more shaky ground, and it might be better to say that 0^0 = 1 in the same way that (-1)^(1/2) = i, where you're taking a principal value, and certainly if you're using a different approach to creating numbers it might not be true.
Joshua Hillerup the very fact that using a different approach can yield different results is why 0^0 is considered to be undefined
Hey, we can also define x^0 as such:
x^0 = x^(y-y), where let's say y > 0
For eg:
2^0 = 2^(3-3).
Leading this example forward,
2^(3-3) = 2^3/2^3 = 1.
That way, 0^0 = 0^3/0^3 = 0/0 which is undefined.
Shit I just realised that it means any power of 0 is undefined.
0^3 = 0^(5-2) = 0^5/0^2 = 0/0 !!
@@VivekYadav-ds8oz thus you're wrong :)
“It’s TH-cam; I understand.” Everyone should say this in every video ever.
You cannot conclude that 0^x should equal to zero when you only see the graph at the right hand side of it. We know that 0^x is 0 for positive x-es, undefined for negative x-es, but we don't know anything else about the point where x = 0. Therefore it shouldn't be applied in a proof. We could simply define the 0^x function to be equal to one at x=0, and there would be no problem.
Btw. when you enter |x|^x into geogebra, you can see a nice curved path that intersects 1 at x = 0.
Also: 0^0=0^(1-1)=0/0 and 0n=0 so it can hold any value
I used to agree with you until I was shown that 0^2=0^(3-1)=(0^3)/0 which by the same argument should be indeterminate as well.
@@jacobrandell4992 oh yeah but I know for sure that 0^0 is equivalent to 0/0 idk how to explain
@@jacobrandell4992 that's new to me thanks
@@helloitsme7553 Define a^b as a^(b+c)/a^c. a^b/a^c=a^(b-c). For "b-c" to equal "0", "b" must be equal to "c" and a is "0". therefor 0^0=0^b/0^b. The problem I find with this is that with this logic 0^m (let m be any real number) is equal to 0^(m+c)/0^c.
In summary 0 is equal to any number, it does not have to be 0^0
Note: I have not finished my education and like to think I know what I am talking about but honestly have no clue, this may or may not be true.
@@helloitsme7553 how is it equivalent?
"We can't deal with it, let's call it undefined"
Wish I could do this for my math problems
Well it *is* all undefined until you define it. It’s just that you can’t do very much without any definitions...
@@Erik20766,
the problem with math problems is i think in this case that in general other persons have already decided to define all kind of things and well very very much ones, so that the only thing which is left *to me* to define is the pretty creative “this thing, we do not yet know its value, now let's call it x”.
But we need to define it. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
@@Erik20766 But we need to define it. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
It is easier to explain by 2^m/2^n = 2^(m-n). If m=n, then 2^n/2^n =1, and if n>m, then 2^m/2^n = 1/2^(n-m).
note that the power series for e^x is usually written as:
sum 0->inf of x^n/n!
the first term of this series would be:
x^0/0!
0! = 1 so thats no problem, but
x^0 = 1, IF x isnt 0
so based on the power series, e^0 should be undefined.
but you might also see the power series of e^x written like this:
1 + x + x^2/2 + x^3/6 + ...
and in this case it works fine. So does that mean we just cant use sigma notation? or do we have to write it as:
1 + (sum 1->inf of x^n/n!)
That is only true by convention saying that 0^0=1 for power series. In general, we say that 0^0 is indeterminate.
@@justabunga1 0! is also indeterminate in some contexts.
@@AlbertTheGamer-gk7sn no 0!=1.
On TH-cam: Everyone tells me that 0⁰ is undefined or problematic.
In University and literature: Have yet to come across a single situation where 0⁰ is not defined or anything else than 1.
I feel like people either don't understand the situation themselves (having heard plausible sounding arguments in the past and never really examined counterarguments) or feel that calculus students won't be able to handle the difference between values of a function and limits of a function, despite the fact that we teach them the difference the first time we introduce limits.
BriTheMathGuy understands!
a to the power of 0 equals a/a. It is undefined for 0. It is 1 for any small number approaching zero.
When I studied mathematics at UQ a decade ago, 0^0 was defined as 1 in some of the courses I did.. It seems a sensible definition. My Pari-GP calculator gives 0^0 =1.
0^0 = 1 is the "correct" value in a very meaningful sense, but some not-so-great arguments about 0^0 are still widespread. I think that these not-so-great arguments persist because they mimic how we often think of expanding definitions of functions.
Let's say we have a discrete way to define a function f(x) on a domain D. And D is a subset of E. Is there a discrete way to extend f(x) to be defined on E? Sometimes, there isn't. But then, sometimes there is an analytic way to extend f(x) to be defined on E. So we use this analytic way on this extended domain E. If there is no discrete way and no analytic way to extend f(x) to E, we may simply abandon the endeavor to extend f(x) to E altogether.
The arguments saying that 0^0 is undefined follow this basic outline, except they skip the step where they check whether f(x) can be extended in a discrete way. They simply jump to the analytic way. There isn't an analytic way to make sense of 0^0, so they declare it undefined. But there is a discrete way to make sense of 0^0, and it gives a value of 1. We shouldn't ignore that!
This is exacerbated by a larger philosophical issue where a lot of people view analysis is the "most legitimate" branch of mathematics, and that discrete math is fake mathematics - or, rather, not _as_ legitimate as analysis. (This is often seen when people explain 0! = 1 from a discrete perspective, and people respond by saying that this isn't the _real_ reason 0! = 1; the _real_ reason is the gamma function.) So even though discrete math tells us 0^0 = 1 will always work, analysis can't make sense of 0^0 on its own, and since analysis is the "most legitimate" branch of mathematics, the "real, true" answer is that 0^0 is undefined.
Luckily this viewpoint is slowly getting corrected, but it will still take a while.
@@MuffinsAPlenty All of this controversy stems from having a waffly definition of limits. 0^0=1 is true. There isn't a branch of mathematics, or some situation, where you come up with a different answer; it's just that when your intuition is formed by limits, rather than algebraic infinitesimals; you refuse to give an answer. It's like finitists refusing to work with infinity. Or like constructivists refusing to accept the law of the excluded middle.
What amazes me is that people see 0^x and don't think about f[0]=1, f[x : x > 0]=0 and say... oh. that function has a name. Kronecker Delta (for non-negative values). And it's really common. And it's discontinous. There are weird things all over the place like non-commutative objects, discontinuous functions, etc.
@@robfielding8566 Well said.
There's something about 0ˣ that doesn't sit well with me in general.
For defining 2⁰, you divide higher powers by 2 until you get there. But you can't do that with 0 to begin with. It feels like 0ˣ already has a limit 0 and won't ever reach there.
Won't it therefore make sense to say that x⁰ is the "right" way to do it, and assign 1 to it, since 0ˣ doesn't have a solution for it?
(Thanks for your videos, they are super interesting to go through!)
Another way to show that n^0=1 where (n is NOT 0):
say a number A belongs to N or Z
we know: A/A=1
and by definition 1/A=A^-1
also A^m * A^n =A^(m+n)
therefore, A/A=A^1*A^-1=A^(1-1)=A^0=1
If you look any modern book about mathematics, youll see they accept that its 1.
So why not just define 0^0=1? That would make x^0 continuous for all Reals, which 0^x already is not.
Wait if f(x)=0^x then Df=(0,+oo) then it's continuous.We check continuity in the space where the function is defined
In some contexts, we do. For example, the Taylor series expansion for exp(0) is just 0^0/0!, which we know is 1, so we must treat 0^0 as 1 here. In other contexts, however, this definition just doesn't make sense (like with the example of 0^x), so we can't say mathematically that 0^0 = 1 is true (as the video demonstrates). If we always assumed this it would sometimes give incorrect/weird results. Hence, it might be better to call 0^0 indeterminate rather than undefined (although in the limit x -> 0^- of 0^x it's indesputably undefined).
Toby Hawkins that’s not really a “definition” of 0^0 so much as it is a notational convenience. Otherwise you would have to write one term of the Taylor series separate from the usual summation.
I say why not define 0^-1 as 0^-1
@@tobyhawkins 0^x isn't continuous in R anyway, so putting a single dot at (0,1) for its graph wouldn't hurt.
thanks
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You said at the start of the video that you weren't talking about calculus or taking limits, but isn't "connecting the dots" from 5:25 onward just a disguised form of taking limits?
Sush, it is disguised, don't blow the cover!
Not really. Think about in the nice good old days where we graphe parabolas. Say x^2. We first get the points (-2,4),(-1,1),(0,0),(1,1),(2,4) then we connected the dots with a curve. This video is meant to be no calculus.
Phil P No, connecting the dots has nothing to do with limits or calculus for that matter. Connecting the dots simply involves defining some map such that its domain is a superset of the set for which the operation in question is defined, and then postulating axiomatically that this map satisfies the exact same recursion as the previous map for every element in its domain, which is the super set. That is how you extend operations from the natural numbers to the integers, rational numbers, real numbers, and complex numbers, etc.
Please explain how connecting the dots have something to do with Calculus
@@blackpenredpen What about extending the line of 0^x to the y-axis when there was no part of the graph on the other side?
Good discussion!
Nice channel!
Love this! This is exactly what I do when I teach students about exponents.
Sometimes I mention to students that 2^3 = 1*2*2*2 a one multiplied by the number 2 a total number of 3 times.
2^0 = 1 multiplied by 2 a total of zero times
2^-1 =1 times 1/2 once
Yes, but this also does result in 0^0 = 1 as a theorem. Also, while it makes sense to multiply 1 by 0 exactly 0 times, it does not make sense to talk about -1 multiplications. 0 is a natural number, -1 is not.
@@angelmendez-rivera351 0 is not a natural number.
@@noahali-origamiandmore2050 Yes, it is. And most mathematicians would agree. For example, the Peano axioms include 0 as a natural number. So do set theory and combinatorics, where 0 is the empty set. Algebraists also treat 0 as a natural number, since it makes the natural numbers a commutative semiring.
@@noahali-origamiandmore2050 Besides, just as a matter of principle and intuition, you literally use 0 to count. You can distinguish between having 0 chickens and 0 cows, 1 chicken and 0 cows, and, 0 chickens and 1 cow, all because you are using 0 to count. How many elements does the empty set have? 0. It is literally a counting number. The very fact that OP is able to talk about doing something (like multiplication) "zero times" proves my point. Because while it is coherent to say something is done 0 times, it is *not* coherent to say that something is done -1 times, for example.
Yes, I know some mathematicians will treat the natural numbers as starting with 1. They do this during situations where they want to work with the nomzero natural numbers, and they get sloppy with the language, so they just omit the "nonzero" part. This is analogous to how number theorists and analysts are working with algebraic structures where you have to always attach the "nonzero" moniker in front: the nonzero integers, the nonzero rational numbers, the nonzero real numbers. Often, the "nonzero" part is implied by context and omitted from the explicit language, so it almost looks like these mathematicians are saying 0 is not a number at all, which is not what they are actually doing. Other mathematicians like holding on to ancient traditions when 0 was not considered a natural number, so they do that as well, but these are in the minority. If your work is based on works that are much older, then as a matter of convenience, treating 0 as non-natural may be more convenient than not, for the purposes of that particular publication. So, yes, notational convention varies from work to work. But, if we move on from notation, and we focus on the question conceptually, most mathematicians will undoubtedly agree that it makes far more sense to *think* of 0 as a natural number, than as not. The notions of mathematical structure just work out a lot more naturally in that case.
@@angelmendez-rivera351 Natural numbers are counting numbers. Your example of cows an chickens doesn't prove anything. It's not about being able to distinguish having 0 cows or 0 chickens. If you have 0 chickens, there was absolutely no counting involved, and that holds true for whenever 0 is used because you don't count 0 of something because that means no counting at all. Distinguishing between 0 chickens and 1 chickens is its own thing. Furthermore, you don't count chickens by going "0 chickens," "1 chicken," "2 chickens;" you count them like "1 chicken," "2 chickens," "3 chickens." You don't use 0 to count because that means no counting at all. What 0 IS is a whole number. This is when having 0 chickens or 0 cows would work as an example but not for natural numbers.
He really reads comments
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I find it way easier to say n^2/n^2=n^(2-2)=5^0 and n^2/n^2=1 therfore n^0=1
but when applied to n=0
0^2/0^2=0^(2-2)=)0^0 and 0^2/0^2=0/0 therefore 0^0 is an indeterminate undefined number because it essentially becomes a case of 0/0 by the laws of exponents
This law of exponents doesn't hold when 0 is the base, though. By the same reasoning, you could get _any_ power of 0 to be 0/0.
0^2 = 0^(3-1) = 0^3/0^1 = 0/0, for instance.
But this is not true. The issue is that the law stating a^(b-c) = a^b/a^c doesn't hold when a = 0. So, as tempting as it is, the argument you gave here doesn't work.
what if you try to see the meaning of 0^0 in the complex plane with euler's formula?
0^0 = upside down owl. (ovo) 🦉
x^0
e^ln(x)*0
e^0ln(x)
x=0
e^0 * ln (0)
ln (0) is undefined
However, if you consider the limit case:
e^-0*inf
1/e^(0*a/0)
1/e^a
completely wrong
Felipe Lorenzzon You did the limit wrong. lim x ln(x) (x -> 0) = lim ln(x)/(1/x) (x -> 0) = lim (1/x)/(-1/x^2) (x -> 0) = - lim x (x -> 0) = 0. Therefore, e^[0 ln(0)], as a limit, is simply 1 anyway. If we define 0^0 = 1, then the function f(z) = z^z on the complex numbers is continuous everywhere.
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Don Solaris
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Is 0^0=1 because it describes "inverting/ turning inside out" the unit circles origin so that it becomes its perimeter?
=1
8:20 reminds me of another video of you: undifound. it was an april fools video
Christopher Stark yup!!! :))))
Interesting way to present this. Thanks. I'll encorporate this the next time I explain it.
This is how I have come to explain this. If you assume log properties are valid, then you get y=x^0 transformed into log y = 0*log x which means log y = 0 ( or y = 1) unless you get x to be less than or equal to zero --- because the log would become undefined. Which essentially is proof by contradiction (right?) assuming y=x^0 is valid produces a contradiction when x=0.
You will never be able to produce a valid contradiction by assuming 0^0 = 1. (0^0 = 0 is also consistent with with exponential rules, but is not consistent with the empty product convention, the set theoretic definition of exponentiation, or any formula involving discrete exponents where 0^0 can arise.)
If you ever think you can produce a contradiction by assuming 0^0 = 1, see how your argument works with 0^1 or 0^2.
Let's go through it with 0^1 here.
If you assume log properties are valid, then you get y=x^1 transformed into log y = 1*log x, which means log y = log x (or y = x) unless you get x to be less than or equal to zero --- because the log would become undefined. Which essentially is proof by contradiction. Assuming y = x^1 is valid produces a contradiction when x = 0.
The issue with this argument is that there are actually two assumptions here:
1) The log rules are valid
2) x^1 has a value when x = 0
The second assumption is indeed valid, so we must conclude that the first assumption is invalid. The log rules do not generally hold on powers of 0 or powers of negative numbers. When we go back to your argument, we see that you made two assumptions:
1) The log rules are valid
2) x^0 has a value when x = 0
But we've already seen that assumption 1 is not valid when discussing powers of 0. So since we know assumption 1 is invalid, we are unable to conclude that assumption 2 must be invalid. This argument fails to say anything about assumption 2 at all.
There is no reason that 0^0 has to be undefined. 0^0 = 1 is a perfectly consistent and useful definition and is the value that "should" be used in any situation where 0^0 appears in a formula. The only issue that can arise for 0^0 is when we talk about continuity. 0^0 is an indeterminate form because f(x,y) = x^y has a pretty bad discontinuity at (0,0). But just because a function is discontinuous at a point does not mean it must be undefined at that point.
In subjects like calculus and analysis, having 0^0 be undefined is more of a matter of convenience than a matter of necessity. In these subjects, continuous functions is a huge point of study, so un-defining 0^0 allows exponentiation to be generally continuous, which is nice if you want to study continuous functions.
But from an algebraic perspective, there really isn't a reason to have 0^0 undefined. Exponentiation (with natural number exponents) is a well-defined operation because of the associative property of multiplication. We can then use the associative property of multiplication to motivate a product of 1 factor and the product of 0 factors. In particular, the product of 0 factors is known as the empty product. If the empty product is to be consistent with the associative property of multiplication, the empty product must result in the multiplicative identity, i.e., 1.
The associative property is key when talking about discrete exponentiation. Exponentiation is well-defined because of the associative property. Our exponential rules are valid because of the associative property. The empty product is 1 because of the associative property. And this is why replacing 0^0 with 1 in any formula involving discrete exponents works - everything about discrete exponentiation is based on associativity.
As I said, though, if you're dealing with continuous exponents, then 0^0 messes with continuity, so you might choose to have 0^0 undefined if you're concerned with continuity. But the real reason to do this is convenience, not necessity.
5:35 "Seems like just connect the dots". Well, if you look at an equation from the left, and from the right, and determine what it might look like at the value it's approaching.. that's a limit. So without using limits, you have used a limit. And approaching a number and actually evaluating that number, i.e. what it seems like at x^0 or 0^x is not the same thing. At 0^0, the function could jump for example. Its clearly not continuous... There is more possibility that 0^0 = 1, given the behaviors of other functions, like the binomial theorem series etc. I don't think this in anyway proves 0^0 it's undefined.
I thought 0^0 was 1.
In pretty much every context where we don't think of exponentiation as a continuous operation, you are correct.
The issue comes from the fact that (0,0) is a non-removable discontinuity of the function f(x,y) = x^y. As such, analysts like to have 0^0 undefined (it's more convenient for them, since they care about continuity). There's also a bit of a historical note here too. Mathematicians used to consider 0^0 to be equal to 1, but then found out that 0^0 was an indeterminate limiting form. Since mathematicians didn't really understand the connection between continuity and limits at that time, this freaked them out, and they un-defined 0^0 because of this. But now we know that 0^0 being an indeterminate form just means that x^y cannot be made continuous at (0,0) and that it's perfectly fine for a function to be defined at a point of discontinuity. So it's rather silly (and one could say wrong) to still use continuity arguments to claim that 0^0 is undefined.
This sort of reasoning (seeing why it's not _really_ problematic to define 0^0 = 1) is becoming more widely accepted in the mathematics community. But it will take time to convince the analysts that this makes sense, and it will take time for this sort of reasoning to make its way into high school mathematics textbooks/classes.
0^0 doesn't always have to equal. It can be any other value. In algebra, 0^0 is indeterminate. He is not talking about calculus in the video. If it does, you will have to compute limits with more work using l'Hopital's rule.
_This Comment is cross-posted!_
1 is a more consistent answer. The *Taylor expansion* for e⁰ will be *0⁰/0! + 0¹/1! + 0²/2! + 0³/3! + ... = e⁰ = 1.* The only term that is not 0 is *0⁰/0!.* There is also the *Taylor expansion* for *cosine.* If *n* objects each have *k* states, then the equation for the number of the set's positions is *n^k.* Think about the number of positions that [a set of 0 objects each with 0 states] has. This is philosophical, but it is one state.
As for *0^x = 0,* that is only true for _positive_ exponents of 0. The Binomial Theorem also relies on the *0⁰ = 1* statement. As for limits, those are only accurate to the true value for continuous functions. Take the piecewise equation *y = x if x ≠ 5, y = 1 if x = 5.* The limit of y as x approaches 5 is 5, but *y = 1* AT *x = 5.* As for the *Product&Quotient Rules* of exponents, under certain circumstances, those are false for 0.
I hope this makes sense.
@@MuffinsAPlenty But we need to define it. "Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true.
OK...
So 0^0 is undefined....
What about
0^0^0^0^...
Infinite exponentiation?
x^x^x^x^x.... is only defined for e^(−e) ≤ x ≤ e^(1/e)
So that would be undefined ya.
0^0=1
@@seroujghazarian6343 I still feel it's one
THe answer is still indeterminate.
@@justabunga1 wrong "0^0"
We're talking about EXACT VALUES, not infinitesimals
There is no discrepancy between 0^x and x^0, though, because 0^x is only defined for x > 0. So the statement 0^x = 0 is not actually remotely close to being a true statement to begin with. Also, think of it this way, which can be made rigorous.
If I have a special printer which can replicate any object, akin to a 3D printer, and the printer has settings so that you can tell it by what factor multiply its input, then consider taking an object and multiplying by 5 in the first print. There will be 5 printed objects, which reflects the fact that 5^1 = 5. Keep the setting the same, and after the second print, you will have 25 objects, given by the fact that 5^2 = 25. This works because this printer prints analogously to how our axioms define multiplication. Now, if you reverse the process, you will notice that the printer shrinks the amount of copies by a factor of 5. This reflects on the definition 5^(-1) = 1/5, involving definition. On your zeroth print, a.k.a when you never press the button on the first place, you have one copy: the original. This uses 5^0 = 1.
Now suppose you change the settings so that it multiplies the original object by a factor of 0 every time it prints new copies. After the first print, which happens after pressing the button 1 time, you get zero copies of the object in total, with the original object gone from existence. This reflects that 0^1 = 0. Press the button again and you still have 0 copies per the settings of the printer. This reflects 0^2 = 0. But what happens if I press the button 0 times? Then it never applies the process by which it nullifies the object to then print literal nothingness in the first place, so my print is simpy the original object unchanged. Hence 0^0 = 1. And this can be made rigorous using functional analysis by defining function compositional iterations.
There are many other reasons why 0^0 = 1 is the only consistent conclusion with the axioms, and it also is extremely unhelpful and limiting if we leave it undefined. Taylor series could not exist if 0^0 was chosen to be left undefined.
taylor series can still exist, you just cant always write them in sigma notation
@@nathanisbored *But it's not the case for set theory*
@@nathanisbored That is what I meant, yes. After all, exponentiation is nothing but a matter of notation. It always has been.
But by the visual logic on both examples can’t we define 0^0 be = 1? because in the second example there is no reason to think that 0 follows the same line, as the negative numbers don’t do so
x^0=1 including x = 0
0^x=
0 when x > 0
1 when x = 0
undefined when x < 0
Limit of 0^0 form is different from 0^0.
This is the real conclusion.
It's 1.
No
_This Comment is cross-posted!_
1 is a more consistent answer. The *Taylor expansion* for e⁰ will be *0⁰/0! + 0¹/1! + 0²/2! + 0³/3! + ... = e⁰ = 1.* The only term that is not 0 is *0⁰/0!.* There is also the *Taylor expansion* for *cosine.* If *n* objects each have *k* states, then the equation for the number of the set's positions is *n^k.* Think about the number of positions that [a set of 0 objects each with 0 states] has. This is philosophical, but it is one state.
As for *0^x = 0,* that is only true for _positive_ exponents of 0. The Binomial Theorem also relies on the *0⁰ = 1* statement. As for limits, those are only accurate to the true value for continuous functions. Take the piecewise equation *y = x if x ≠ 5, y = 1 if x = 5.* The limit of y as x approaches 5 is 5, but *y = 1* AT *x = 5.* As for the *Product&Quotient Rules* of exponents, under certain circumstances, those are false for 0.
I hope this makes sense.
my math teacher showed us another intresting concept of n^0
is the definition of the divition of n^m/n^k=n^(n-k) so if you define as n^0 as n^(m-m) that is n^0 because of m-m=0 and as you know n^m/n^m = 1 then n^0 is equal to 1, idk if that's a prove to n^0=1 but it makes sence, and does not contradicts 0/0 = indef so yeah!
0^0 is defined as the value of the indeterministic value obtained from the power function operated on zeros. While in homework and school math it might not be of much significance, but in the practice of math and computerizations zero to any power is reserved as the default pointer for indeterministic values, similar but different in function to alphabetical designations. Any functions with alphabets can be rewritten in indeterministic form by replacing each letter with zero to the n'th power, n being the number of alphabets in the function. This can be seen as a solution in programming languages for programs to operate completely numerically without alphabetical implications. Simply define 0^0 as a variable set, starting with 0^1 as the first variable.
I feel bad for 0^0
It is undefined . At least, imaginary numbers can be defined...
Rubix Cube DJ I agree
e^x and 1/x say otherwise
"Undefined" is just a code word of saying, "Screw this challenge. I'm turning back". This is very bad as it states that you are fearful and afraid of challenges. This is the exact opposite goal of humanity. Humans are meant to break away from nature using self-awareness, conscience, willpower, and imagination. This is why mankind managed to establish such civilization that sets them apart from all animals. We 21st-century humans must thank our long-gone ancestors by breaking away even more to make them proud. Einstein left in his will saying the first person that uses his theory of relativity to invent time travel must travel back to April 17th, 1955, to make him proud. "Undefined" is basically stating we are not used to those numbers, so let's just don't use them. It all depends on context. If we were living in Minecraft, a world without circles, and all of a sudden, a circle randomly appeared out of the blue, we would call it "undefined", but since in our world, we have polar coordinates, the premium package with the spherical bundle, we are accustomed to seeing circles, and we won't call them "undefined". Also, a long time ago, people worshipped the moon like a god at an "undefined" distance away from us, and they believed the sky's the limit, and everything they see in the night sky are basically pure celestial spheres of light at an "undefined" distance away from us, and the Earth was the point where those "undefined" distances converged to, but we managed to reach the moon and even send space probes outside our solar system, even attempting to reach the end of a universe, making such distances not "undefined" anymore. Finally, infinities are everywhere. Without it, the Big Bang wouldn't have happened, and every time you move, infinities are required to make it happen. Infinities created us, don't disrespect them by calling it "undefined" Divide by 0, spread your wings, learn how to fly, and do the impossible. We need infinities to make our dreams of time travel and superpowers come true. We need to define the number.
Couldn't we relate 0^0 to 0/0 by saying:
0^0 = 0^(1-1) = 0^1*0^(-1) = 0/0 ?
Therefore we conclude that 0^0 is undefined for the same reasons that 0/0 is undefined?
Not a very good arguement. Using the same logic you can say:
0^1 = 0^(2-1) = 0^2*0^(-1) = 0/0
But we already have 0^1 = 0.
I see.
9:08 Compute 2^1000 - 2^999 - ..........- 1 = 1 Because the values keep halving all the time like 2^1000 = 2.2^999 so 2^1000 - 2^999 = 2^999 etc so it will keep halving until it reaches 1.
4:15 for the real proof after explaining definitions
It's wrong!
You can not explain power by division when base is 0.
0^2 is not 0^3/0
0^1 is not 0^2/0
nor is 0^0 0^1/0
I feel like 2^1 = 2 as a definition, and 2^-1 = 1/2 as a definition, but 2^0 is just 2^1*2^-1 = 1, not itself a definition but derived from other definitions.
In fact, the only axiom required is that x^n is x times itself n times. Then you can easily derive x^1=x. You can also derive x^(a+b)=(x^a)*(x^b) because x^(a+b) is x times itself a times multiplied by x times itself b times.
This lets you derive x^0 by starting with x = x^1 = x^(0+1) = (x^0)*(x^1) = x^1. Dividing both of the last two terms by x^1 gives us x^0=1.
You can derive x^(-n) = 1/(x^n) in a similar way starting with 1 = x^0 = x^(1-1) = (x^1)*(x^-1) = 1. Dividing both of last two terms by x^1 gives us x^(-n) = 1/(x^n).
And you can derive x^(1/n) = sqrt(x) by starting with x = x^1 = x^(1/2 + 1/2) = x^(1/2)*x^(1/2) = (x^(1/2))^2 = x. Take the square root of the last two terms to give x^(1/2) = sqrt(x). Derive any x^(1/n) a similar way but with n occurrences of 1/n.
All because x^n is x times itself n times. 😀
0:45 now that’s meta self-reflecting at its best!
Using the binomial theorem, we get that (1+0)^n=1^n * 0^0 + 1^(n-1) * 0^1 * (n chose 1) + ... + 1^0 * 0^n.
Every term that has 0 to a non-zero power will be 0, so we have that the value for (1+0)^n = 1^n * 0^0, but we know what (1+0)^n is, it is 1, so we could draw the conclusion that 0^0 is 1.
But as seen in the video, there are multiple ways to find a value for 0^0, so it is not defined.
Gergő Dénes No, the video did not provide multiple ways to get a value for 0^0. The argument involving 0^x is already invalid since it is not even defined for x < 0, so connecting the dots is not well-defined.
brilliant.org/wiki/what-is-00/
The definition of 0^0 changes depending on the context. In limits it is most useful to classify it as an indeterminate form, but in most other applications, including probability and set theory, 0^0=1 is the most useful definition. For example, suppose I have an opaque bag with x balls, each of which is a different colour. I decide to play a game where I pick a ball at random, remove it from the bag, write down its colour, and then place it back in the bag. I must do this exactly x times (the same as the number of balls), and I win if I never pick the same ball more than once, but I lose (and the game ends immediately) if I ever pick the same ball twice. On a single pick, the probability that I pick an unpicked ball is the number of unpicked balls in the bag divided by the total number of balls in the bag. On the first pick, those two numbers are equal, but on each subsequent pick the number of unpicked balls decreases by 1; therefore, the probability of winning is x!/x^x. For example, if I have three balls, the probability is (3/3)*(2/3)*(1/3)=3!/3^3=6/27=2/9=0.2222.... If I have four balls, the probability is 4!/4^4=24/256=3/32=0.09375. If there are zero balls in the bag, the probability is 0!/0^0, but this is clearly equal to 1 since it's impossible to pick any ball twice. Since 0!=1, that means 0^0 must equal 1 also.
As another argument (for multivariable calculus students), suppose we graph the cross-sections of z=x^y for different values of x. So we could graph z=3^y, z=2^y, z=1^y, z=0.5^y, z=0.1^y, z=0.01^y, z=0.001^y, etc. As the value of x gets smaller and smaller, three patterns emerge. One, the cross-section curves on the left side of the z-axis become increasingly vertical. Two, the cross-section curves on the right side become increasingly horizontal and hug closer to the y-axis (z=0). Three, and most importantly, all of the function curves intersect at exactly one point: (0,1). So, by this logic, it "seems like" the function curve of y=0^x is not simply a ray originating from the origin and extending to the right; instead, there is a jump discontinuity at x=0 and the value of y=0^x there is, in fact, 1. To make further sense of this, if this was another of our cross-sections from before, the intersection point is now still at (0,1), but all the points to the right are now lying on top of the y-axis (z=0) and all the points to the left have moved up to infinity.
@@ballsnoballs4844 First, you got the limits of 0^a and a^0 backwards. Second, yes, my multivariable argument is only a counter argument to his 0^x approach, not a proof.
undefined just means in general it doesnt have a definiton but in some causes it does. 4/0 is technically defined in a reiman sphere but we call it undefined because we havent state if we are working with reiman spheres.
0^0 is undefined but in certain fields and spaces it might have a definition. in caclulus, abstract math and physics, we cannot give it a definition. Undefined doesnt mean cannot be defined. it means it cannot be defined without more context.
How about we consider all the graphs of b^x where b is some non negative real number (you could do negatives but it's not ideal). For b>1, the graph looks like an normal exponential. For b
When you divide 2^2/2^2 =1, same base you substract the power and since they are the same you get zero and therefore 1.
a^b=a*a*a...*a (b times), if b is an integer positive, but it is the same as say: 1*a*a*a...*a (b times), and if b=0 then there is no a and it only stays 1, so a^0 is 1. For the same reason: 0*0...*0 (b times), if b is an integer positive is always 0 becuase 0*something=0, but if there isnt any 0, it doesnt have to me 0, so... 0^0=1 (without multiply by any 0)
Can we think about it as 0^1=0 then we divide by the base, 0 and so 0^0=0/0 which is undefined?
No, because by that same argument, if we think that 0^4 = 0, then 0^3 = 0/0, which is undefined. But 0^3 is obviously 0, not undefined.
Could you explain me a solution for
Integral (from-~ to ~) of 1/(2phi*x) * exp(-1/2((x-m)/p)^2 - 1/2(((y/x)-n)/q)^2) dx
Please help me.
In both cases you’re only showing a pattern and extending it to the 0th case. Instead doing algebra I can show that 0^0 is 1. Using the rules X^(a-b) =X^a / X^b , X/X =1, and X-X=0, in the case where a=b that means X^(a-a) = X^a / X^a. And X^a / X^a =1. That means X^(a-a) = 1. Well a-a = 0 so X^0=1. If X=0, then 0^0=1.
SQRT(0) seems like division by 0 but it is defined as equal to 0.