What is 0^0 (Zero to the power Zero)?

Question

The limit of x^0 as x tends to 0 is 1

In Maple if I put 0^0 it gives an answer of 1.
If I put 0^0 into my scientific calculator it gives an error
If I put 0^0 into google the google calculator gives an answer of 1.

Is there an answer to this or not?

I've got a degree in Mathematics and I get the whole thing about x^0 = 1 for non-zero x.

If it's indeterminate why does Maple give an answer of 1.

Answer 1

So here's the convention that mathematicians use:

0^x=0 for all x besides 0, and 0^0=1. This means the function 0^x is not continuous, but the function x^0=1 for all x is continuous. One of them has to give. One reason this is a good choice (as well as 0!=1), is so that simple formulas for power series expansions hold for all values of x. For example,

e^x=sum{i=0 to infinity} x^i/i!.
e^0=0^0/0! should be 1. (All other terms are 0.)

Answer 2

1

Answer 3

1

Answer 4

Well actually the correct answer is the one that gives your calculator.
I know there is this rule that any number raised to 0 gives 1, but this is not applied for 0. The correct rule is:
x^0 = 1, for any x except 0.
This is because 0^0 can be reduced in the following way:
0^0 = 0^(1-1)= 0/0
As you may remeber, 0/0 is the classical and archetypical indetermined form. Its calue depends on how fast the numerator tends to zero with respect to the denominator.
Maple, and Google are applying the x^0 rule but not in a complete way. They see the exponent = 0 and automatically deliver an answer, but the value of the base is also important to consider

Answer 5

A power tells you how many times a number is multiplied by itself. For example, 2x2x2 = 8 = 2^3 ("two to the power of three").

If you start a sequence going, you get:
16 = 2^4 = 2x2x2x2
8 = 2^3 = 2x2x2
4 = 2^2 = 2x2
2 = 2^1 = 2
? = 2^0 =

As you can see, the last line is a little vague. This is why a calculator might give up and cause an error.
However, if you assume 1xN=N, then we can use
16 = 2^4 = 1x2x2x2x2
8 = 2^3 = 1x2x2x2
4 = 2^2 = 1x2x2
2 = 2^1 = 1x2
1 = 2^0 = 1

which gives the answer 1. This is the mathematically accepted value of anything non-zero to the power of zero.

However, this is a limit of the function. The actual answer of 0^0 is yet undetermined, as it may be a case of divison by zero.


(Please note, the use of 2s in this answer is merely an example of low numbers, but this is the same for any set of numbers you'd like)

Answer 6

saurabh_h_n, Rose, pingu2me, and Jim are correct. 0^0 is an indeterminate form.

An indeterminate form takes place when logical mathematical laws lead to a conflict of answers. If you try the "zero raised to any power" rule, you get 0 for an answer. If you try the "any number raised to the zero power" rule you get 1. Because there is a conflict, 0^0 is indeterminate.

In your details, you note that the limit as x approaches 0 of x^0 is 1, but you could have just as easily gone the other way... the limit of 0^x as x approaches 0 is 0.

There is a subtle difference between mathematical expressions that are undefined versus those that are indeterminate. 5/0, for example, is undefined. There is no number that when multiplied by 0 gives 5 for a product. 0/0, however, is indeterminate because there are an infinite number of possibilities of n, such that n × 0 = 0. Mulitple answers: indeterminate... no answer: undefined.

Answer 7

0^0 is called an _indeterminate form_ because the result is not a specific number--it is not defined.

You know that 8^(1/3)=2 because 2^3=8. The result of calculating 8^(1/3) is a specific number, which in this case is 2.

If we assume that 0^0=x, then 0=x^(1/0), which is meaningless in the mathematical sense.

_Defined_ means that the result is a specific number. 0^0 is not defined. Other indeterminate forms include infinity-infinity, 0/0, infinity/infinity, and so on.

When there is a specific answer to a calculation, the result is _defined_. For those described above, they are not defined and said to be indeterminate. We use calculus and limit laws to evaluate how a function behaves near those values that cause the indeterminate form to occur.

Answer 8

First of all, a^x means a*a*a.... (x times).

Now although it can be shown quite easily that
e^0 = 1, this is not true for a^0.

The most convincing reason for why a^0 = 1 is
not a power series because any such series will
include a^0 so that a^x is defined in terms of
itself. The reason is by virtue of the laws of
exponents:

p = a^n (1)
q = a^n (2)

=> p/q = 1 (1) divided by (2)

In division, we subtract exponents so that

p/q = a^0 (3)

But wew know that p/q = 1, therefore by the transitivity
of equality, we have that a^0 = 1.

Given p = 0^0 => log (base 0) p = 0

But by definition we cannot have a logarithm that
is zero or less, so 0^0 is undefined. The limit
of x^0 as x->0 in this case is irrelevant.

Answer 9

OK, the limit of x^0 as x tends to 0 is 1. But the limit of 0^y as y tends to 0 is 0. And if you consider x^y, where x AND y tend to 0, you've got bubkus (I can make this go pretty much where I want by choosing the spped of the convergence). The limit argument isn't much help here.

This is only a matter of definition then (and I've generally seen this quantity defined as 1, although it's rare you actually SEE this quantity anywhere!).

As an aside, if your calculator is telling you that 0^0=1, it was probably programmed to do so; if not, it is probably trying to compute the answer via some algorithm, but as you see above, this can be troublesome (as it may not be well-defined). This should answer your query about numerical computation of 0^0.

Answer 10

If you look at the graphs of the family of exponential functions y=k^x where k is a positive (real) constant, you will see that the graph passes through (0,1) for every nonzero k, meaning that 1=k^0. If you take the right limit as k approaches 0, you will find that these graphs still pass through (0,1). Therefore, it would be tempting to say that 1=0^0 in the limit that k=0.

0^0 is one of the better-behaved indeterminant forms since it tends to just one value (unlike 0/0 where the limit can approach positive infinity, negative infinity, some other number, or it may not even exist depending on your numerator and denominator).

Answer 11

0^0 is undefined. Every other number raised to the 0th power is 1.

2^-1 = .5 (when raising to the -1 power think of 1/x)
2^0 = 1
2^1 = 2
2^2 = 4

0^-1 = NaN (because 1/0 is indefined)
0^0 = NaN
0^x (if x > 0) = 0

Zero can only be raised to a number that is positive in order to get a real number as an answer.