Is 0^0 Equal to 1? An Explanation of Mathematical Concepts and Terminology

[Ashwin_Kumar]

You can use taylor expansion to prove that 0⁰ is one.

$$(x+1)^{0}=1+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}...$$

x-a= 1
f'(a)= 0
f''(a)=0
.
.
.
So we take x=-1, and we get
0⁰= 1+0+0...
= 1
As simple as that.

 

[Ashwin_Kumar]

The evidence micromass pointed out is sufficient. The idea that $$0^{0}=0$$ doesn't make sense.

And,
$$\frac{1}{0^{2}}$$ is equal to infinity, without the need for limits.

 

[micromass]

You can use taylor expansion to prove that 0⁰ is one.

You can never prove that 0⁰=1. You can only define it as being 1, but you can never prove it.

Your Taylor series argument is merely an indication why it is sometimes useful to set 0⁰=1, but it is no proof of it.

 

[micromass]

The evidence micromass pointed out is sufficient. The idea that $$0^{0}=0$$ doesn't make sense.

It does make sense. But it just isn't useful. (Except in studying limits of functions such as $0^x$).

And,
$$\frac{1}{0^{2}}$$ is equal to infinity, without the need for limits.

Uuh, I was actually pointing out that this was not the case. You can not divide by zero.

 

[Ashwin_Kumar]

Oh- wait- no. Taylor expansion proves that 0⁰ is infinity.

$$f'(a)=0^{-1}=\infty$$

So we get $$0^{0}=\infty$$

 

[micromass]

Uuh, are you serious or just trolling??

Oh- wait- no. Taylor expansion proves that 0⁰ is infinity.

$$f'(a)=0^{-1}=\infty$$

You can't divide by 0, and I have no clue what your f is...

 

[Hurkyl]

I'm to the point of leaving a forum/moderator that uses "power" to get his point across. Intellectual discourse is frowned upon, so adios, auswiedersehn.

And this is why I was closing the discussion. It isn't intellectual discourse to continually insist to others that mathematics should be practiced your way while simultaneously closing your mind to the how and why of mathematics is actually practiced. I put my moderator hat on not to get a point across, but because it's time to clean up.

 

[hillzagold]

Let me start by saying I don't speak French. Pointing to an untranslated French paper has zero meaning to me.

Now, if the idea idea of 0^0 is mystifying to the average person, at what point can one expect to learn how to explain it? Don't say to use google, because it will immediately jump to a confusing answer, and even independent study needs a structure that provides necessary prerequisite knowledge. Just tell me to learn X, then Y, then finally Z

 

[Ashwin_Kumar]

Using, different methods, you get different results for 0⁰
But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

$$\frac{1}{0}=0^{-1} =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty$$

 

[micromass]

Using, different methods, you get different results for 0⁰

You can never obtain different results for 0⁰, you can only define these results. There is no way you can write 0⁰=1 or 0⁰=infinity, with defining things that way. All we are doing here is setting a convention. You cannot prove anything about this.

But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

$$\frac{1}{0}=0^{-1} =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty$$

Sigh... Dividing by 0 is not allowed in the real numbers. Not now, not ever. And infinity is not a real number, so we can't work with that either.

 

[3.1415926535]

Using, different methods, you get different results for 0⁰
But micromass said you cannot divide a number by zero to get infinity(or am i mistaken). And way, you obviously can:

$$\frac{1}{0}=0^{-1} =(\frac{x}{\infty})^{-1}=\frac{\infty}{x}=\infty$$

No no no, you can't divide by 0. You can use limits like this
$$\lim_{x\rightarrow 0}{x^0}= \lim_{x\rightarrow 0}{x^{n-n}}=\lim_{x\rightarrow 0}{\frac{x^n}{x^n}}=\lim_{x\rightarrow 0}1=1$$
or like this
$$\lim_{x\rightarrow0}\left (e^{-\frac{1}{t}} \right )^t=\lim_{x\rightarrow0}e^{-\frac{t}{t}}=\lim_{x\rightarrow0}e^{-1}=\frac{1}{e}\approx 0.367879$$
As you can see 0^0 varies depending the limit, therefore it is an inditerminate form
Without limits however, mathematicians avoid writing 0^0 because it rests undefined

 

[timthereaper]

Okay, I've read all the posts and have tried to figure out the controversy of 0^0=1. From what I understand (correct me if I'm wrong), but it seems that 0^0 is actually indeterminate, but the general consensus is that defining 0^0=1 makes things easier in some ways, but it violates some rules somehow.

I'm no mathematician (I'm an engineer), so why is it convenient to define it this way? I guess I'm trying to figure out what problems arise if we didn't say that 0^0=1.

P.S. I'm not trying to beat a dead horse or rile things up, but I'm trying to learn more about this interesting topic.

 

[micromass]

Okay, I've read all the posts and have tried to figure out the controversy of 0^0=1. From what I understand (correct me if I'm wrong), but it seems that 0^0 is actually indeterminate, but the general consensus is that defining 0^0=1 makes things easier in some ways, but it violates some rules somehow.

That's a very good summary!

I'm no mathematician (I'm an engineer), so why is it convenient to define it this way? I guess I'm trying to figure out what problems arise if we didn't say that 0^0=1.

Well, for example in Taylor series. The Taylor series of e^x is

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$$

No problems so far, but if we want to write it more compactly, we get

$$e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}$$

However, when we evaluate this in 0 (thus if we want to evaluate e⁰), then we get 0⁰ in the first term. That is:

$$e^0=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+...$$

Now the conventions 0!=1 and 0⁰=1 are handy because they allow us to calculate e⁰=1 (which we already knew to be true).
If we didn't set 0⁰=1, then we would had to write

$$e^x=1+\sum_{k=1}^{+\infty}{\frac{x^k}{k!}}$$

which is less elegant. So you see, we only define 0⁰=1, because it is sometimes more elegant to do so. We won't create new mathematics with it, we won't run into problems with it, it just makes things nicer.

In a way, it's the same thing as setting 0!=1. This is just a handy convention that makes a lot of things easier. But setting 0!=2 would have been as good, but it would make the formula's uglier...

 

[3.1415926535]

That's a very good summary!



Well, for example in Taylor series. The Taylor series of e^x is

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$$

No problems so far, but if we want to write it more compactly, we get

$$e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}$$

However, when we evaluate this in 0 (thus if we want to evaluate e⁰), then we get 0⁰ in the first term. That is:

$$e^0=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+...$$

Now the conventions 0!=1 and 0⁰=1 are handy because they allow us to calculate e⁰=1 (which we already knew to be true).
If we didn't set 0⁰=1, then we would had to write

$$e^x=1+\sum_{k=1}^{+\infty}{\frac{x^k}{k!}}$$

which is less elegant. So you see, we only define 0⁰=1, because it is sometimes more elegant to do so. We won't create new mathematics with it, we won't run into problems with it, it just makes things nicer.

In a way, it's the same thing as setting 0!=1. This is just a handy convention that makes a lot of things easier. But setting 0!=2 would have been as good, but it would make the formula's uglier...

Same goes with the taylor expansion of $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$

Generally, If we define $$0^0\neq1$$ most of the Taylor-Mc Laurin expansions and some other power series would not be well defined since
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$

Lastly the function $$f(x)=x^0$$ would be "deprived" of its continuity and differentiability

 

[micromass]

Lastly the function $$f(x)=x^0$$ would be "deprived" of its continuity and differentiability

That's not really an argument, since now the function

$$f(x)=0^x$$

get's deprived of its continuity and differentiability.

 

[timthereaper]

Okay, I think I understand. It's an elegance thing. However, I thought that 0!=1 is true and not just defined that way. Maybe I'm wrong, but I thought the rule was n!=n*(n-1)! and if n=1, then 1!=1*0! and thus 0!=1.

 

[3.1415926535]

That's not really an argument, since now the function

$$f(x)=0^x$$

get's deprived of its continuity and differentiability.

Haha

If we define $$0^0=1$$ then the function $$f(x)=x^0$$ is continuous and differentiable everywhere unlike the function $$g(x)=0^x$$

If we define $$0^0=0$$ then the function $$g(x)=0^x$$ is continuous and differentiable everywhere unlike the function $$f(x)=x^0$$

If we define $$0^0\neq1$$ and $$0^0\neq0$$ then none of the functions are continuous and differentiable everywhere!

 

[3.1415926535]

Okay, I think I understand. It's an elegance thing. However, I thought that 0!=1 is true and not just defined that way. Maybe I'm wrong, but I thought the rule was n!=n*(n-1)! and if n=1, then 1!=1*0! and thus 0!=1.

0! is defined as 1 in order to satisfy the rule you have written and because the Gamma Function $$\Gamma (x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt=(x-1)!$$ has a value 0!=Γ(1)=1

 

[timthereaper]

Okay, so the n!=n*(n-1)! rule is something that we constructed *because* we defined 0!=1. However, I'm wondering if the Gamma function defines the factorial.

 

[3.1415926535]

Okay, so the n!=n*(n-1)! rule is something that we constructed *because* we defined 0!=1.

No, we constructed the rule because it is true for every natural number apart from 0 and so is the gamma function. Defining 0!=1 not only allows us to make more elegant formulas but also satisfies the two equalities mentionned above. Apart from that, I believe that the official definition of the factorial is the Gamma Function since it is not limited to natural numbers.
0! is officially defined as 1 because there are no problems with this unlike 0^0

 

[timthereaper]

Okay...I think I got it. Thanks for all your help.

 

[micromass]

Hmm, I don't think we defined 0!=1 because of the Gamma function. The convention 0!=1 was made before the Gamma function was introduced. But of course, seeing that the Gamma function agrees with our convention 0!=1 is reassuring.

Now, 0!=1 was mainly introduced to make formula's more elegant. For example:

$$\binom{n}{0}=\frac{n!}{0!(n-0)!}=\frac{1}{0!}$$

Now, we want $\binom{n}{0}$ to denote the number of ways we can pick 0 numbers from a collection of n numbers. And we can do that in one way: picking nothing at all.

Another reason is because of the binomial theorem:

$$(a+b)^n=\sum_{k=0}^n{\binom{n}{k}a^kb^{n-k}}$$

if k=0, then we see $\binom{n}{k}$ appearing, and to make the formula work, we want to have $\binom{n}{0}=1$.

If we didn't define 0!=1, then we would have made the definition

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}~\text{k nonzero and}~\binom{n}{0}=1$$

which is certainly less elegant.

The same thing appears in why we defined 2⁰=1, this is also a mere convention that is handy to make (and furthermore, it's a very cool convention because all the rules of exponentiation stay valied, unlike with 0⁰=1).

 

[JJacquelin]

From hillzagold :
Let me start by saying I don't speak French. Pointing to an untranslated French paper has zero meaning to me.

I am so sorry ! I beg in two ways your pardon, since this is another very interesting paper :
http://forums.futura-sciences.com/mathematiques-superieur/166012-finir-0-0-a.html
I suppose that there are similar papers written in English. These papers are only compendiums of what is known for years. 0^0 is an hackneyed subject.

 

[spamiam]

and for those of us who don't read French?

Let me start by saying I don't speak French. Pointing to an untranslated French paper has zero meaning to me.

All right all you belly-achers, here is my humble attempt at a translation of JJacquelin's paper.

Now quit your griping and learn some French!

JJacquelin, I hope you're OK with my translation. If not, PM me and I'll remove my post.

 

[JJacquelin]

I hope you're OK with my translation.

I am very pleased with your translation which appears quite perfect.
I contact you by private message for acknowlegments.

 

[3.1415926535]

I am very pleased with your translation which appears quite perfect.
I contact you by private message for acknowlegments.

Excellent work JJacquelin!

 

[JJacquelin]

Hi !
The paper, now entitled "Zéro Puissance Zero" & "Zero To Zero^th Power" has been updated with the spamiam's English translation. Pubished on Scribd : http://www.scribd.com/JJacquelin/documents
Thanks again,
JJ.

 

[dimension10]

Did you prove my equation for all $0 < j < k$ already? So that you can now complain about notational issue with $0=j$ and $j=k$ cases?

But seriously... I just said that I define $0^0$ to be $1$. Therefore $0^0=1$, by definition. OK? What do you think about dialogue like this:

Person A: "I have defined $f$ so that it is $f(x)=x^2$"

Person B: "I see. I think I'm going to define $f$ so that $f(x)=x^3$... What a minute! Did you just say that $f(x)=x^2$? That's wrong! By (my) definition $f(x)=x^3$, and $x^2\neq x^3$".

Person A: ""

Well, but actually you 0^0 is not 1 or 0. I mean, I thought it was interdeterminate?

 

[Hurkyl]

Well, but actually you 0^0 is not 1 or 0. I mean, I thought it was interdeterminate?

Viewing the expression $0^0$ as a limit form, it is indeed indeterminate.

But the math problem that spawned this thread was not about limit forms.

 

[Orion1]

$$\lim_{x \rightarrow 0} 0^x = 0$$
$$\lim_{x \rightarrow 0} x^0 = 1$$
$$0^0 = \text{indeterminate}$$

 

[micromass]

$$\lim_{x \rightarrow 0} 0^x = 0$$
$$\lim_{x \rightarrow 0} x^0 = 1$$
$$0^0 = \text{indeterminate}$$

Indeed, when used as a limit form, it is indeterminate!
But when working with integer exponents only, then 0⁰ is a convention that is often made!

 

[SteveL27]

$$\lim_{x \rightarrow 0} 0^x = 0$$
$$\lim_{x \rightarrow 0} x^0 = 1$$
$$0^0 = \text{indeterminate}$$

True.

But:

$$\lim_{x \rightarrow 0} x^x = 1$$

and

$$| \emptyset^ \emptyset| =1$$ in set theory.

So in some contexts it makes perfect sense to take $$0^0 = 1$$ as a convention.

I believe these points have already been made several times in this thread.

 

[dimension10]

Let a be a number such that a^a=a/a

a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.

Thus, 0^0=0/0. Since 0/0 is intedeterminate, 0^0 is intedeterminate. However, now we have more information. So, we can say:

0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1

Thus,

0^0=1.

However, with limits, it is both 0 and 1. But when we have more information, can't we claim it to be 1?

 

[disregardthat]

Since this discussion will never end, we should settle on some compromise. So let's declare 0^0 = 1/2.

 

[Hurkyl]

0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1

This argument only makes sense if exponentiation is repeated multiplication. This is why the convention $0^0=1$ works very well when plugging 0 into a polynomial or Taylor series, because the exponentiation there is repeated multiplication. (and for the same reason it is often used in algebra and combinatorics)

However, the exponentiation operator on real numbers is not repeated multiplication, and continuity is rather important for calculus and real analysis, which is why real exponentiation leaves $0^0$ undefined.

Let a be a number such that a^a=a/a

a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.

Ah, the classic fallacies of forgetting your hypotheses, and reversing the flow of logic.

The very premise of your problem requires a to be a positive number. (0 is not a positive number)

Also, what you have proved is

If a is a (positive) number satisfying $a^a=a/a$, then a = 0​

which is very different from

If a=0, then $a^a = a/a$.​