Power series and the value of the expression $0^0$ [duplicate]

I have a doubt regarding the value of the expression $0^0$. I know this value is taken as indeterminate as far as limits are concerned. All was fine upto now. But when I encountered power series, I found out when $x=a$ in the expression summation $[b (x-a)^n]$ where $n=0$ to infinity, of the power series, then the series always converges which is understood. But what bothers me is its value converges to $b$ and not $0$. That is the first term of the power series is written as $b \cdot 0^0$ and $0^0$ is taken as $1$ and not as indeterminate.

Can anyone tell me why this is so? How is it possible at one time we define $0^0$ as indeterminate and at other time its value is taken as $1$? Could anyone help me on this one? Thanks.

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marked as duplicate by Nate Eldredge, Amzoti, Henry T. Horton, amWhy, Shuhao CaoJun 8 '13 at 20:38

"ow is it possible at one time we define $0^0$ as indeterminate and at other time its value is taken as $1$?" It simply reflects discontinuity of $(x, y) \mapsto x^y$ at $(0, 0)$. Nothing special or mind-blowing, you just have to remember what justifies the usual tricks in limit computation. –  Alexei Averchenko May 10 '13 at 2:20
This is no duplicate, but answers of Alex Becker and imranfat are largely off-topical and pertain to math.stackexchange.com/questions/11150/… . The question admits a true answer: we have powers 0, 1, 2, … and, hence, polynomials in any unital ring! Considerations related specifically to real numbers are irrelevant to the question of polynomials and series. –  Incnis Mrsi Nov 9 '14 at 13:40

Generally $0^0$ is taken to be $1$, so that the term of degree $0$ in a polynomial or power series can be written as $c_0x^0$, rather than having some special exception for $x=0$. It makes the notation cleaner.

As an aside, when one talks about an expression being an indeterminant form, say $E(a,b)$, one usually means that if we have functions $f(x),g(x)$ such that $\lim\limits_{x\to x_0}f(x)=a$ and $\lim\limits_{x\to x_0} g(x)=b$, we cannot conclude that $\lim\limits_{x\to x_0}E(f(x),g(x))=E(a,b)$. In the case of $0^0$, this means that even if $\lim\limits_{x\to x_0}f(x)=0$ and $\lim\limits_{x\to x_0} g(x)=0$ we cannot conclude that $\lim\limits_{x\to x_0}f(x)^{g(x)}=0^0$, no matter how $0^0$ is defined. Note that this is not the same as saying that $0^0$ is undefined. It just says the expression does not "play well" with taking limits.

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Alex, if we take $0^0=1$ then we have $\ln(0^0)=\ln(1)$ which implies that $0 \ln(0)=0$. However the right hand side of this expression is undefined well the left is zero. What gives here? –  mtiano May 10 '13 at 1:01
@mtiano Careful, $\log a^x=x\log a$ is true when $a> 0$. –  Pedro Tamaroff May 10 '13 at 1:04
@mtiano No choice of definition for $0^0$ will make every expression work out nicely. No matter what, some definitions involving exponents will involve an exception in this case (i.e. $\ln(x^y)=y\ln x$ as long as $x>0$; in fact, any $y$ will cause problems if $x=0$, regardless of how $0^0$ is defined). The definition $0^0=1$ seems to require the fewest exceptions, making it the most convenient. –  Alex Becker May 10 '13 at 1:05
got it now... thanks for your feedback... –  under-root May 10 '13 at 1:45

As a limit, 0^0 is and remains an indeterminate form. There is no answer to 0^0 in terms of a number. However, in order to make certain formulas "work" the convention is to "let" 0^0 to be equal to 1. While those formulas were never designed to find answers to 0^0, they happen to work only if we accept 0^0 equals 1. In series we find this situation, with the Binomials distribution, this situation also pops around the corner. Look, sometimes we just need to be "flexible" and do certain "modifications" to make applications come out the right way. It is not something that should bother you. As a real number, 0^0 does not exist, in limit form, it's called an indeterminate form, and once in a while we bend (not "break") the rules :)

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