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 (x-x_{0})^{0}[/tex] is to be understood as identically equal to 1, yes, then you might call it an &quot;abuse of notation&quot;.<br /> <br /> However, due to the continuity argument given in the last post, most would regard such explicit definition as needlessly pedantic, and that it is &quot;self-evident&quot; 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.