What does 0⁰ equal?
1, 0, or is this not defined?

There could be just a convention for 0⁰ , however I don't remember it, so if anyone could help, thanks!

0⁰ is not defined (It is like asking what is 0/0 the answer could be any real number as zero times any any real numer is zero). However it can be shown that x^x where x can be infinitely close to zero (but not zero) is infinitely close to 1. in Other words: lim x^x when x tends to 0 is 1.

About 0⁰ , note that as x goes to 0, 0^x goes to 0, but x⁰ tends to 1. So I think that there is a convention that 0⁰ is defined as 1 because x⁰ occurs frequently in power series whereas 0^x isn't really a very useful function at all.

Isn't x⁰ defined by x being multiplied by itself 0 times and then put into an equation? I think exponents were meant to abbreviate how many times a term appears in a multiplication, so any x⁰ would be 1. Complex proof: y=x⁰ differentiates to 0, so y always equals 1 so long as the binomial theorem is correct.

I agree with Brad here. It is far more useful to have 0⁰ = 1 rather than 0.

The thing is that with x⁰ , the x actually doesn't feature at all. The whole point of the exponent being 0 is that the power is independent of x. If x = 0, then once you make it dependent on x (by setting the exponent to anything greater than 0) the power does get influenced by the 0 and so the answer is 0. On the other hand if you set it to something less than 0, it becomes unbounded.

The problem is that in this case, division is not a good way to reverse multiplication. But that doesn't mean 0⁰ shouldn't be defined. It is defined by virtue of the fact that the zero exponent means that the zero (being raised to the power) is never multiplied in so can't affect the outcome.

Perhaps a good definition of f(x,y) = x^y for real x,y would be as follows. (I've got no idea if it's done this way).

1) f(x,0) = 1
2) f(x,1) = x
3) f(x,a+b) = f(x,a)f(x,b) (a ¹ -b if x = 0)
4) f(x,ab) = f(f(x,a),b)
5) f(x,y) is continuous on y

That should do the trick I think. To extend to complex numbers you need to look at its derivative, and generalise that too.

Yours,

Michael

The usual way to define x^y is as e^{y ln(x)} , where the exponential function is well defined by its Taylor expansion. You can easily prove from this that it has the five properties above; proving that a function exists and is unique from the above looks more difficult but probably quite easy (haven't tried).

-Dave

Well I think you can do it as follows.

By rule 2): f(x,1) = x.

By rule 3): f(x,a) = x f(x,a-1)

So f(x,2) = x² , f(x,3) = x³ ...

For all integral n:

f(x,n) = xⁿ

Then by rule 4): (letting a = 1/2 and b = 2)

f(x,1/2)² = f(x,1) = x

Similarly by letting a = 1/n and b = n:

f(x,1/n)ⁿ = x

which shows that for all unit fractions it is consistent. Then to get to all the other rationals apply rule 4) again. And then for the irrationals use rule 5) - every irrational can be given a sequence of rationals tending to the irrational limit. And the sequence of f(x,r_n ) will converge uniquely because f(x,r) is continuous on the rationals. Therefore f(x,y) = x^y for all real y.

Maybe rule 1) is redundant...

Michael

Thank you for all your answers.

I wouldn't have thought so many answers would gather.

BRAD, could you tell me what the binomial theorem is?

It is an infinite series used to describe (x+b)ⁿ . It's main use that I have seen is for differentiation. It provides the mathematics to show that for every y=xⁿ , dy/dx=nx^n-1 . This important as it shows that for

y=x⁰

dy/dx=0

And thus y always stays at 1.

Brad

The binomial theroem has many uses in all areas of maths, especially probability, and series expansions (giving approximations to series). If you go through the Special Relativity formulae you can use the binomial theroem to show that for v < < c, E~(1/2)mv² , ie: Newton's mechanic formula.

Neil M

Brad, I'm not sure that quite works. Say y=x⁰ , then dy/dx=0 x x^-1 =0/x, not 0. At x=0 this is indeterminate: 0/0, so y can be defined to behave however you want at x=0.

Help,
My year 12 double extension maths son has asked me and I know that BY DEFINITION 0⁰ = 1 , but don't know how to prove it to him.

His teachers don't know either !!

Any help will be appreciated greatly

$({\log }_e(n))/n\to 0$ in this limit.

which is easy. (consider the expansion of $e^y$)

Kerwin

(PS. This was written before I'd read the message Kerwin has just posted...)

Actually I'd be inclined to say that you can't prove that 0⁰ = 1, it is entirely a matter of what you define 0⁰ to be, which is conventionally 1. You can, however, demonstrate that this is a sensible definition, because x⁰ = 1 for all x ¹ 0, so it makes sense for 0⁰ = 1 as well.

Just to confuse things, there is a certain argument for saying that 0⁰ should be 0, since 0^x = 0 for all positive x, but I think that despite this it makes more sense for it to be 1.

I hope that clarifies things a little,

James.

Well, 0⁰ =0¹ /0¹ =0/0 just to complicate matters..:) Though I would feel more happy about defining 0/0 to be 1 than 0⁰

Multiplication is only defined on the set of real numbers bar 0 so 0/0 is meaningless.

Er, multiplication is perfectly well defined on the real numbers including zero, but 0 has no multiplicative inverse, so (as you say) 0/0 is meaningless, similarly x^y . You can fiddle around with limits in various ways, but (if I remember) you can get pretty much anything you like this way. For instance, there was a recent thread on this board I think where someone showed 1⁰ =-1, his argument was that (-1)²ⁿ =1, therefore 1^1/2n =-1, let n tend to infinity and get 1⁰ =-1. I hope this demonstrates the futility of this sort of argument? The argument for 0⁰ =1 is usually made on the grounds of convenience, for instance in binomial expansions, etc.

The fact that we can call 0⁰ either 'undefined' or let it be something sensible like 1 is, I think, analogous to the idea of Cauchy principal values. We can let the power and the number we are raising to it tend to 0 either at different rates (and from different directions) in which case our answer is not well defined, or we can let both parts tend to zero from above at the same rate, in which case we get 1.

I agree that the value 0⁰ can't be deduced from the value of a^b everywhere else, and also that the most sensible definition is 1.

Here is a demonstration of why 0⁰ = 1 is a good idea:

By the binomial expansion:

(a + b)ⁿ = aⁿ b⁰ + nC1 a^n-1 b¹ + ... + nCn a⁰ bⁿ

Now set a = 1 and b = 0:

1ⁿ = 1ⁿ 0⁰ (all but first term goes to zero).

So 0⁰ = 1.