What is 0^0?

Question

Is 0^0 accepted as 1?
I saw a question, which asks for the interval of convergence of the series Sigma x^n, n= 0 to n = oo
It diverges for all x different from zero, but then I got confused about x = 0. Can the question ask for sigma x^n staarting at n = 0?

sorry, sigma n!x^n, n = 0, n = oo

can you please try answering the question in terms of the series which I have put down? thankyou

the series is sigma n!x^n.
I forgot the n! in my initial writing of the question.

my real question is not what is 0^0 as much as it is " does it make sense for the sigma n!x^n to start at n=0?"
Blackwolf is the closest in answering my question, and I get what he said, that the ratio test leads to a form that no longer contains 0^0, but I still do not know if it is valid to start indexing at 0 in the first place.

Answer 1

0^0 is generally taken as indeterminate but depending on the context it can be set as 1
see
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
for the arguments

as for your question
∑n!x^n
ratio test
lim | [(n+1)!x^(n+1)] /n!x^n |
= lim |(n+1)x| = ∞ for all x except 0
since if x = 0
lim |(n+1)0| = lim 0 = 0 < 1
so it converges if x = 0 and diverges otherwise

your problem is you are trying to treat 0 in the same way you would treat a boundary value, which it's not. In this case x=0 is covered by the ratio test, since it's not on the boundary, so there is no need to consider 0 separately by plugging it into the sum, hence no need to worry about 0^0

the end
.

Answer 2

Based on the algorithm leading to x^0 = 1 for all nonzero values of x, 0^0 is undefined.

The reasoning behind a number raised to the zero power equals one is based on properties of exponents.

2^1 / 2^1 = 2^0 using subtraction of exponents in division of like bases. But, since 2 / 2 = 1, then 2^0 = 1.

Likewise for all real numbers except 0......

0^1 / 0^1 = 0^0, and since 0 / 0 is undefined, then 0^0 would also be undefined.

Answer 3

Hey there!

0^0 is NOT 1. 0^0 is one of the indeterminate forms.

0^0 can equal to 1, in the following limit.

lim x^x --> Write the problem.
x-> 0+
y=x^x --> Set the limit equal to y.
ln(y)=xln(x) --> Take the natural logarithm on both sides of the equation. Use the formula ln(m^n)=nln(m) on the right side of the equation.
lim xln(x) --> Write the above problem in the limit form.
x->0+
lim ln(x)/1/x --> Rewrite xln(x) as ln(x)/1/x.
x->0+
lim (1/x)/(1/x^2) --> Apply l'hopital's Rule.
x->0+
lim (x) --> Simplify (1/x)/(1/x^2).
x->0+
0 --> Use the definition of a limit. Substitute 0 for x.
ln(y)=0 --> Set the above problem equal to the ln(y).
y=e^0 --> Take the "anti-log" on both sides of the equation.
y=1 --> Use the formula a^0=1.
lim x^x=1 Write the answer.
x->0+

That was the proof on why the limit is equal to 1. L'Hopital's Rule needs to be applied, in order to avoid indeterminate forms.

By the definition of a summation, if the index, i.e. n=0, is set on that number, then, it has to start on that number, regardless, if the series converges or diverges.

Hope it helps!

Answer 4

I can see the reason for your confusion, which comes about because of an apparent conflict between two of the power rules:

Rule 1: Any (nonzero) real number - or complex number, for that matter - raised to the zeroth power is 1. By extension, it would seem that zero raised to the zeroth power should also be 1 to be rid of this single exception. However, this comes into apparent conflict with:

Rule 2: Zero to any positive real power is zero. How much trouble would it be to extend this definition to make zero to any nonnegative real power to be zero? Sometimes none, but sometimes that brings us into direct conflict with Rule 1. Zero is also the boundary between the positive numbers and negative numbers. Zero to any negative number is undefined. To define it would bring this expression in direct conflict with the definition of a negative power, i.e., x^(-n) = 1 / x^n, where n is a positive real number and x is any nonzero real (or complex) number. If the restriction on x of not allowing it to be zero is maintained, then the definition for a negative power also applies for any complex value for n, positive or not. If we tried to define zero to a negative power, we would also first need to define division by zero. No one has found a logically consistent way of dividing by exactly zero, though you can come as close as you like, and often do come arbitrarily close in calculus.

According to rule 2, 0^0 falls on the border between zero and an undefined number. This makes 0^0 a dangerous factor that can blow up unexpectedly in your mathematical expressions. For logical safety's sake, 0^0 is usually considered to be undefined. In some contexts, this expression may be taken to be 1 provided that the definition isn't extended to an area in which logical danger lurks; Nevertheless, 0^0 remains a special case and must be handled with caution.

Answer 5

0^0 is undefinided. Not 1

Answer 6

0^0 is usually defined as 1. This is a convenient definition, especially when we deal with Tayllor series and series like the one you mentioned.

By the way, Sigma x^n, n= 0 to n = oo is a geometric series. What you wrote is wrong. It CONVERGES for all x with |x| <1 and DIVERGES for all x with |x| >=1. The convergence interval is, therefore, (-1, 1). Of course, for this series, with n starting at 0, to make sense, we have to define 0^0 If we define 0^0 =1, then, for x = 0, the series converges to 1.

Answer 7

It's indeterminate since 0 raised to any power is 0 and any number raised to the 0 power is 1. Because of this ambiguity, 0^0 is indeterminate, not 1.

Answer 8

I think u r talking like that...
sigma X^n,where X is 0 and n from 0 to oo

If its like that,

Sigma X^n,

u can break it like (1-1/1)^n

by binomial theorem,

1-1+1-1....

if n is even thn answer 0,if odd thn answer 1

Answer 9

0^0 is undefined but, when you are working with series, some authors define it as 1 to avoid this problem

The same happens when you work with polynomials

Lets P(x) = a_nx^n + ... + a_1 x + a_0

You can consider that P(x) = sum [0 to n] a_i x_i

Now, if the independant term is a_0 x^0, how would you be able to find P(0)?

Ana

Answer 10

0^0 is undefined (meaningless) , it never be accepted as 1.

series Sigma x^n , n=0 to n=00(infinity)

is equalto 1/(1-x) fopr each value of n this serese is absolutely converge however at x =1 it becomes undefined.

I hope it will help, if you have any question please let me know.
Thanks,