I'd be interested in seeing various proofs that 0^0=1, if anyone has any

[Johnny5]

I am looking for a quick proof that

 

[math] 0^0=1 [/math]

 

some kind of argument, doesnt have to be fancy

 

Thank you

 

e.g.

 

 

let x = 0^0

 

therefore

 

ln x = 0 ln 0

 

It's provable from the field axioms that 0*y=0, for any number y.

 

Hence

 

ln x = 0

 

therefore x=1

 

QED

 

 

I am looking for other proofs.

 

PS: And i know that

 

lim x-->0+ of x = -infinity

 

[math] \lim_{x \to 0^+} ln x = - \infty [/math]

 

That's why I want a different proof, because the above isn't one.

[stevem]

There is no proof. You can define [math]0^0[/math] to be whatever proves convenient. I prefer to say it is indeterminate because [math]\lim_{x \to 0^+}x^0=1[/math] but [math]\lim_{x \to 0^+}0^x=0[/math] but others disagree. See http://mathforum.org/dr.math/faq/faq.0.to.0.power.html for a discussion of this topic.

[Johnny5]

Steve, quick question.

 

Must it be the case that 0! = 1?

 

Regards

[jcarlson]

ln x = 0 ln 0

 

It's provable from the field axioms that 0*y=0' date=' for any number y. [/quote']

 

Its also not a proof because ln 0 is not a number, it is undefined. and therefore the field axiom 0*y=0 does not apply where y = ln 0, because y must equal a complex number.

[matt grime]

0!=1 and 0^0=1 are merely conventions. The first is universal the second isn't.

[Johnny5]

0!=1 and 0^0=1 are merely conventions. The first is universal the second isn't.

 

Matt, doesn't 0! need to be defined to be one, so that certain extremely important combinatorical relations be satisfied?

 

For example the number of 3-permutations of set {a,b,c,d}.

 

P(N,r) = N(N-1)(N-2)... (N-r+1)

 

Or

 

[math] P(N,r) = \sum_{n=1}^{n=N} \frac{N!}{(N-r)!} [/math]

 

So my point is it has to be the case that 0!=1.

 

Do you see why?

 

Regards

 

Example given:

 

Generate all 3-permutations of the set {a,b,c,d}.

 

A three permutation is going to be a three tuple, an ordered triple, with elements chosen from the set {a,b,c,d}. I simply want to list them all.

 

There are four choices for the first component of the tuple, three choices for the second, and two choices for the third, so the total number of 3-permutations is 4*3*2=24. Now I just have to list them out.

 

 

Wolfram on Lexicographic order

 

Here is an online program that generates r-permutations using lexicographic order:

 

Permutation generator

[matt grime]

The reason you cite as "proof" is exaclty why it is convenient and universally accepted that 0! is DEFINED to be 1. If it weren't then we would have to make several exceptional cases for things like permutations.

 

n! is the number of ways of arranging n objects, or n*(n-1)*...*1, etc. None of those descriptions define 0!, and I need not define 0! to give any of the things above. Indeed I can define your P(n,r) without any reference to 0!, however it is a convenience to extend the factorial to include 0.

 

Your argument a priori assumes 0! "exists" and satisfies those relations. How do you know that?

[ydoaPs]

let x = 0^0

 

therefore

 

ln x = 0 ln 0

 

It's provable from the field axioms that 0*y=0' date=' for any number y.

 

Hence

 

ln x = 0

 

therefore x=1

 

[/quote']

doesn't work, iirc, because ln0 is undefined.

 

[math]x^0=1[/math]

let x=0

[math]0^0=1[/math]

[Johnny5]

doesn't work' date=' iirc, because ln0 is undefined.

 

[math']x^0=1[/math]

let x=0

[math]0^0=1[/math]

 

Yep, i know.

[ydoaPs]

does mine work?

[Johnny5]

does mine work?

 

if you define x^0 to equal 1, for any number x, then yes.

 

But that would be a definition, not an axiom, nor a theorem.

 

But mathematics has to be internally consistent, obviously, so one should not choose a definition like that without checking to see first, whether or not it can be a theorem, and additionally what if it contradicts something else?

 

x^3=xxx by definition

x^2=xx by definition

x^1=x (seems to make sense)

 

 

x^-1 = 1/x definition

x^-2 = 1/xx definition

x^-3 = 1/xxx definition

 

but the case x^0, setting it equal to 1... for any x...

 

well

 

it is a theorem of the axioms of the real number system that

 

0*x = 0 for any real number x.

 

So...

 

0^3=000=0

0^2=00=0

0^1=0

 

0^0=? (not so clear)

 

1/0 (will lead to an explicit contradiction)

1/00 (will lead to an explicit contradiction)

1/000 (will lead to an explicit contradiction)

 

I am not sure yet about 0^0, which is why I asked the question.

 

Actually, provided that d/dx(e^x) = e^x, then it must be the case that 0^0=1, but the relation above is what started this in the first place.

[ydoaPs]

isn't x^0 always one? plot a graph of x^0.

[Johnny5]

isn't x^0 always one? plot a graph of x^0.

 

The correct way to answer you, is for me to utilize general order logic. I am still working on it, when I figure out something perfect, you will know.

 

Regards

 

I intend to tie everything in to basic combinatorical formulas.

 

 

If I had to guess now, i would say that it's arbitrary.

[stevem]

Actually, provided that d/dx(e^x) = e^x, then it must be the case that 0^0=1

How do you get that? I repeat what I said - there is no proof that [math]0^0=1[/math]. If you wish to use that definition then feel free to do so; it's just that not everyone likes that definition, whereas 0!=1 is, as Matt says, used by everyone, though, again you are free to leave it undefined.

[matt grime]

Here is a simple reason why 0^0 is not defined ANALYTICALLY.

 

Consider the function x^0 for all x not zero. This is identically 1, hence 0 is a removable singularity so I may define f(x)=x^0=1 for all x as a perfectly good continuous function.

 

Now consdier the function g(x) = 0^x for all x not zero. This is identically 0, so g has a "removable singularity" too so we define g(0)=0 and get a perfectly good continuous function.

 

Hence there is no "best choice" for 0^0 in analytic terms.

 

There are good combinatorical ones for declaring 0^0 to be 1, and we, as a convention, often define empty prodcuts and unions in such fashion, but it is just a convention. Thus I doubt you can actually prove 0^0=1 on its own without more of a context.

[Johnny5]

How do you get that? I repeat what I said - there is no proof that [math]0^0=1[/math']. If you wish to use that definition then feel free to do so; it's just that not everyone likes that definition, whereas 0!=1 is, as Matt says, used by everyone, though, again you are free to leave it undefined.

 

The proof I am thinking of goes like this.

 

First, obtain the definition that:

 

0!=1

 

Based on some combinatorical argument.

 

Next...

 

Analyze the series for e^x.

 

Notice the following:

 

[math] \prod_{k=1}^{k=n} x = x^n [/math]

 

and also

 

[math] \prod_{k=1}^{k=n} \frac{1}{k} = \frac{1}{n!} [/math]

 

Combine them like so:

 

[math] \prod_{k=1}^{k=n} \frac{x}{k} = \frac{x^n}{n!} [/math]

 

Then define e^x as follows:

 

[math] e^x \equiv \sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{x}{k} [/math]

 

Hence it logically follows that:

 

[math] e^x \equiv \sum_{n=0}^{n=\infty} \frac{x^n}{n!} [/math]

 

Now, from memory, there is a thing about the first constant term, but I have to solve a few ordinary differential equations to remember what it was, but it will come back to me. But the point is this:

 

Presuming that the first term is one leads to:

 

[math] e^0 = \frac{0^0}{0!} + 0 +0 +... = \frac{0^0}{0!} [/math]

 

Then using the result based on a combinatorical argument that 0!=1, it follows that:

 

[math] e^0 = \frac{0^0}{0!} = \frac{0^0}{1} = 0^0 [/math]

 

So granted that e^0=1, you would now have a proof that 0^0=1.

 

Now, if e^0 isn't 1, then the first term of the series isn't 1.

 

Regards

[Johnny5]

Suppose you are asked to generate all three permutations of the set {a,b,c,d}.

 

You could attempt to generate the list randomly. But how would you be certain you listed all the elements? If you knew a priori, how many elements there are in it, then you could just keep going until you finally had the list.

 

But there is a systematic method to generate all the r-permutations of a set of n objects, and it is called "lexicographic ordering."

 

In the example problem about to be done, you are given the following set of four elements:

 

{a,b,c,d}

 

The distinction between an n-set, and an n-tuple, is that the order that the elements of a set are listed in from doesn't matter, but the order that the elements of an n-tuple are listed in does matter.

 

The way lexicographic order works is this.

 

Even though we are given a set of random things, we assign increasing numerical value to them, from right to left. Thus, in the example here, a has a lower lexicographic value than b, b has a lower lexicographic value than c, and so on.

 

Now, we generate the set of all r-permutations, by listing out the elements in increasing order, as if they were numbers.

 

So for example suppose you were asked to generate the set of all three permutations of the set {a,b,c,d}, then the lowest valued 3-permutation is:

 

(a,b,c)

 

Now, you want to find the next lowest valued 3-permutation. If you think in terms of numerical quantities, you will eventually realize that it is:

 

(a,b,d)

 

The next lowest is:

 

acb

 

And the rest... in lexicographic order are:

 

acd

adb

adc

bac

bad

bca

bcd

bda

bdc

cab

cad

cba

cbd

cda

cdb

dab

dac

dba

dbc

dca

dcb

 

 

And dcb is the higest valued 3-permutation so we are done.

 

Now, we can just count, to see how many elements there were, the answer is: 24

 

The next question is, can we find a formula that would have told us in advance how many there are, so that if desired, we could generate the list at random, until we had the right list?

 

One way to answer this question, is for us to have access to the multiplication principle. I have a book here and they state the principle as follows:

 

Multiplication Principle: If A is a set of p objects and B is a set of q objects, then the number of ordered pairs of the form (a,b) with a an object of A and b and object of B is equal to the number of elements in the cartesian product A X B... which is p times q.

 

So, using the multiplication principle, we can figure out in advance, how many elements will be generated using lexicographic order, before listing them.

 

In the example just done, we were looking for three tuples.

 

(X,Y,Z)

 

where X,Y,Z are all distinct.

 

There are four ways to choose X.

For each of these ways, there are three ways to choose Y.

And now, for each of these ways, there are two ways to choose Z.

 

So you can draw a tree diagram, which lists them.

 

For example:

 

a

/|\

b c d

/\ /\ /\

cdbdbc

 

So right here we have six different 3-permutations, beginning with a. They are:

(a,b,c),(a,b,d),(a,c,b),(a,c,d),(a,d,b),(a,d,c)

 

Likewise, for 3-permutations beginning with b, we also have six permutations:

 

b

/|\

a c d

/\ /\ /\

bcadac

 

And each of these is different from all of the first six, because they begin with b, not a.

 

And for 3-permutations beginning with c, there are six 3-permutations, and six more beginning with d.

 

So the total number of unique paths in the tree diagrams, is 6+6+6+6=24, which agrees with our earlier result.

 

Now, we just want to summarize this using a formula.

 

The number of ways to choose the first element is n.

Then for each of those ways, the number of ways to choose the next element is (n-1).

 

Then for each of these ways, the number of ways to choose the next element is (n-2).

 

And you keep going until you reach r rows, in your tree diagrams.

 

In the example problem just considered, n=5, and r=3.

 

So that the answer was given by

 

5(5-1)(5-2)

 

So as you can see, the general formula is:

 

n(n-1)... (n-r+1)

 

And r is constrained to be less than or equal to n.

 

We can now introduce product notation for this.

 

[math] \prod_{k=n-r+1}^{k=n} k [/math]

 

Multiplication is commutative, so the same number could be written as:

 

[math] \prod_{k=n}^{k=n-r+1} k [/math]

 

I generally write the smaller number at the bottom, so in this case I will go with:

 

[math] \prod_{k=n-r+1}^{k=n} k [/math]

 

Now, as I said r is constrained to be less than or equal to n.

 

Suppose that r is equal to n. Then we have:

 

[math] \prod_{k=n-n+1}^{k=n} k [/math]

 

Which is equivalent to:

 

[math] \prod_{k=1}^{k=n} k [/math]

 

Which is the customary defintion of n factorial, written n!. That is:

 

[math] n! = \prod_{k=1}^{k=n} k [/math]

 

Now, consider the case where r=1. In this case we have:

 

[math] \prod_{k=n-1+1}^{k=n} k [/math]

 

Which is equivalent to:

 

[math] \prod_{k=n}^{k=n} k [/math]

 

Which is equivalent to n.

 

That is:

 

[math] n = \prod_{k=n}^{k=n} k [/math]

 

Now consider the strange case where r=0.

 

Thus, we are given a set of n objects, and asked to generate all 0-tuples. In other words, tuples with no elements.

 

Using the formula already obtained, we have:

 

[math] \prod_{k=n-r+1}^{k=n} k [/math]

 

[math] \prod_{k=n-0+1}^{k=n} k [/math]

 

[math] \prod_{k=n+1}^{k=n} k [/math]

 

Here is where we run into problems. Because multiplication is commutative, the previous iterated product is equivalent to

 

[math] \prod_{k=n}^{k=n+1} k [/math]

 

and

 

[math] \prod_{k=n}^{k=n+1} k = n(n+1) [/math]

 

But I think there is a problem with this.

 

Suppose you are given a set of 5 objects, and asked how many 0-permutations there are.

 

Then n=5, and r=0, and using the above blindly you have:

 

[math] \prod_{k=5}^{k=6} k = 5(6) =30 [/math]

 

So somewhere we need to stop this error from occurring.

[matt grime]

But the defintion of e^x in that form adopts the convention that 0^0=1, but it is just a convention that is useful there, and there is no universally agreed convention. Jeez you make even the simplest things far too complicated.

 

Note the 0^0 is usually defined to be one since it is an empty product, and it is the cardinality of the set of functions from the empty set to itself, agreeing with m^n being the cardinality of the function set from {1,..,n} to {1,..,m{

[Johnny5]

But the defintion of e^x in that form adopts the convention that 0^0=1' date=' but it is just a convention that is useful there, and there is no universally agreed convention. Jeez you make even the simplest things far too complicated.

 

Note the 0^0 is usually defined to be one since it is an empty product, and it is the cardinality of the set of functions from the empty set to itself, agreeing with m^n being the cardinality of the function set from {1,..,n} to {1,..,m{[/quote']

 

I wasn't planning on dictating that the leading coefficient is 1 though. In fact, I was going to use a Taylor series expansion, where I would be looking for a function which is its own derivative.

 

Regards

[matt grime]

The leading term in ANY taylor series can and usually is defined to be f(0), it is NOT f(0)x^0/0!, this would lead to issues of defining 0^0 directly. It is onotologically unnecessary to define 0^0 in order to define either taylor series or the exponential.

[Johnny5]

The leading term in ANY taylor series can and usually is defined to be f(0), it is NOT f(0)x^0/0!, this would lead to issues of defining 0^0 directly. It is onotologically unnecessary to define 0^0 in order to define either taylor series or the exponential.

 

Actually, I am currently working on exactly this.

 

Maybe you can help me.

 

Suppose we have a set given to us by someone else, and they chose it at random.

 

So you recieve the set from them.

 

Since this can randomly vary, let n denote the number of elements in the set you recieve.

 

Now, suppose you are further asked to generate all r-permutations of the set.

 

Suppose that you are given a set with n=2.

 

Constraint: r cannot be greater than n, but r can be less than or equal to n.

 

e.g. {a,b}

 

Case I: r=2

 

(a,b)

(b,a)

 

Case II: r=1

 

(a)

(b)

 

Case III: r=0

 

How many elements in the generated set, with P(n,r) elements in it?

 

Zero.

 

 

Yes or no?

[matt grime]

Define random, why is a random set finite, from what sample set are they "randomly" choosing these sets?

 

(r-permutation? that also is not something that I would consider common terminology)

 

"since it [the set] can vary randomly, let n denote the number of elements in the set"

 

that is a very strange sentence indeed

[Johnny5]

(r-permutation? that also is not something that I would consider common terminology)

 

Well it's in at least one book of mine. How would you say it?

 

Regards

 

PS: As for the term 'random' I thought it's meaning was clear from the context.

 

Oh, and actually its an r-permuation on an n-set.

[matt grime]

No, "random" isn't clear from context. Random requires probability, which requires a sample space and a measure on it. And, no, i''ve no idea what an r-permutation is - i can think of several things it might be, but seeing as you've not even explained what random means, and apparently can't, then i think you should state all of your assumptions and meanings.

 

So, please, try and explain what it means to "give" a random set to some one; why is that set finite; what is the sample space, and what is the measure on the sample space.

 

However, it isn't even clear if that is important, as with many of the things you include in a post.

 

Are you simply asking how many r perms of a set of cardinality n ( a natural number) there are? what ever an r perm may be. (ie the random thing is completely unnecessary - does the fact that the set is given "randomly" affect the number of r-perms??)

[Johnny5]

(ie the random thing is completely unnecessary

 

I strongly disagree.

 

I thought r-permutation was entirely clear, based upon the context.

 

Let's see... are you familiar with either of the following:

 

P(n,r)

 

C(n,r)

 

 

?

 

 

Combinations, permutations