# Quick series check

1. Aug 21, 2006

### Benny

Hi, I've run into a bit of a problem. I don't know why I didn't resolve this issue when I first learned about series but the following is bugging me.

$$e^x = \sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}}$$

I also know that exp(0) = 1.

But if I plug x = 0 into the above series, the term corresponding to k = 0, involves 0^0 which isn't defined as far as I can recall so I don't know how exp(0) = 1 comes about from plugging x = 0 into the above series. This is something I should know by now but I can't see why at the moment. Any help would be good thanks.

Last edited: Aug 22, 2006
2. Aug 21, 2006

### d_leet

I read in a calculus textbook once, I think it might have been stewarts and it probably was since that what my calc 2 class used, but it said something along the lines of that when we work with series we take 0^0 to be equal to one.

3. Aug 21, 2006

### Benny

I guess that would make sense. However, if anyone has some different input, please feel free to post it in this thread.

4. Aug 21, 2006

### shmoe

d_leet is correct, they are taking 0^0 to be defined as 1, it makes the notation a little simpler (it's not some kind of profound mathematical statement about 0^0).

Some authors will avoid having to define 0^0 by writing the series as

$$e^x = 1+\sum\limits_{k = 1}^\infty {\frac{{x^k }}{{k!}}}$$

and likewise for all series. This is a rare approach in modern first calculus type texts (if it exists at all).

5. Aug 21, 2006

### quasar987

Since 0^0 is not defined (there is no general consensus), I like to regard the 0^0 terms in a series as just a notation for the number 1. The authors were lazy and instead of writing

$$1+\sum\limits_{k = 1}^\infty {\frac{{x^k }}{{k!}}}$$,

they just wrote

$$\sum\limits_{k = 0}^\infty {\frac{{x^k }}{{k!}}}$$

as a notation for the above series.

6. Aug 21, 2006

### StatusX

Another way to look at it is that 0^0 is not defined in general because the limit:

$$\lim_{x \rightarrow 0, y \rightarrow 0} x^y$$

does not exist. This just means that the result you get depends on how x and y approach 0. For example, if you hold x=0 and let y approach 0, you get terms like 0^0.1, 0^0.01, 0^0.001, ..., which are all 0, so the limit is 0. But if you hold y=0 and let x approach 0, you get terms like 0.1^0, 0.01^0, 0.001^0, ..., which are all 1, so the limit is 1.

That is why 0^0 as a symbol is not defined. But in your example, we clearly have the latter case above, where y=0 and x is approaching 0, and so we can safely say the limit is 1.

7. Aug 22, 2006

### Benny

Thanks for the input guys, much appreciated.