Taylor Series Expansion of Analytic Function at x0 = 0

[lolgarithms]

you know this, right?

$$f(x) = \sum^{\infty}_{k=0} \frac{f^{(k)}(x_0) (x-x_0)^k}{k!}$$

for an analytic function, at x0 = 0, you have to say that 0^0 equals 1 for the constant term. if 0^0 is indeterminate then how can you just say it's 1 in this case?

 

[arildno]

Because we define it to be so, merely out of notational convenience.

If we didn't do so, we would have to write something like:
$$f(x)=f(x_{0})+\sum_{k=1}^{\infty}\frac{f^{(k)}(x_{0})(x-x_{0})^{k}}{k!}$$

That expression is unnice.

 

[lolgarithms]

arildno, but the expression that u gave is most mathematically rigorous, isn't it? is the instance of 0^0 called abuse of notation?

 

[Cantab Morgan]

The question comes down to continuity. First, consider the expression $x^0$ and take the limit as x goes to zero. If we define $0^0=1$, then the function is continuous. But if we defined $0^0=0$, then it would not be continuous.

Now consider the expression $0^y$ and take the limit as y goes to zero. If we define $0^0=1$, then the function is not continuous. But if we define $0^0=1$, then it would be continuous.

It turns out to be much more convenient for $x^0$ to be continuous than $0^y$ to be continuous, so we define $0^0$ to be the limit of $x^0$ as x passes to zero. I suggest that it's not so much an abuse of notation as a shorthand for the limit.

(hey, 200th post )

 

[arildno]

arildno, but the expression that u gave is most mathematically rigorous, isn't it? is the instance of 0^0 called abuse of notation?

Without any EXPLICIT definition of that in this particular case, the expression [itex](x-x_{0})^{0}[/tex] is to be understood as identically equal to 1, yes, then you might call it an "abuse of notation".

However, due to the continuity argument given in the last post, most would regard such explicit definition as needlessly pedantic, and that it is "self-evident" what is meant by the expression.

 

[protonchain]

One question I'd like to ask here then is regarding calculators. My ipod touch's calculator for example (because it was the first thing I reached for, not that I used it for undergraduate astrophysics homework or something) does 0^0 as 1. However I had a graphing program installed on it (for giggles) and it plots x^0 at x = 0 as 0.

Then I plotted it on my regular graphing calculator (TI 83 plus Silver), and X^0 is obviously 1 at all points except 0. Where it says "Y = ". Aka nothing at all.

When you type in 0^0 on that calculator it shows up as error:domain.

So is 0^0 just a fad? Some people say its 1, some people say its indeterminate, no one says its 0?

 

[Hurkyl]

There are three things you can do:

(1) Don't think about it; just accept it as convention and move on
(2) Redefine exponentiation of real numbers to include 0⁰=1
(3) Do a more careful treatment of mathematical grammar

I much prefer (3) -- I don't like (1) and exponentiation of real numbers isn't really what's going on here, so I don't find (2) justified.

 

[arildno]

So is 0^0 just a fad? Some people say its 1, some people say its indeterminate, no one says its 0?

And what, exactly, would you say that $\lim_{x\to{0}^{+}}0^{x}$ ought to be?

 

[protonchain]

And what, exactly, would you say that $\lim_{x\to{0}^{+}}0^{x}$ ought to be?

I have no answer to that (aka it's indeterminate), nor was I attempting to be a smart alec as I feel you are judging me to be. I'm just asking if various people treat the value differently because I noticed various calculators treat it differently.

 

[arildno]

Hint:

The limiting value of that expression is NOT indeterminate. It provably exists. See if you can determine it!

 

[protonchain]

The answer is 0

 

[arildno]

The answer is 0

Quite so.