# Remanian to continous

1. Oct 9, 2006

### quinn

Is it valid to take this summation limit:

sum (from n=0 to n=infinity) = (x^n)/ n(!) --(goes to)--> e^x

?

Or is there a thoerical problem with this? I am fairly certain that this is sound, but I want to get some opinions of possible downfalls of this logic...

2. Oct 9, 2006

### StatusX

Apart from a couple little problems, that's just the taylor series for e^x, which converges everywhere. You might even define e^x by that series. (the problems are that equals sign after the "sum (...)" which shouldn't be there and the "goes to" should just be an equal sign, since there isn't any limit being taken. Alternatively, if you truncate the series at some N, then the partial sums go to e^x as N goes to infinity.)

3. Oct 9, 2006

### CRGreathouse

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$$

4. Oct 11, 2006

### quinn

also....

now:

SUM (n = 0, n goes to infinity) X(n + 0.5)/(n +0.5)! --> goes to ?

What does this small modifcation do to the sum?

At infinity the sum is the same as before, but at small n the behaviour is different...

5. Oct 11, 2006

### CRGreathouse

Do you mean X^(n + 0.5)?

6. Oct 11, 2006

### StatusX

What do you mean by (n+0.5)! ? If you're thinking of the gamma function, write out (n+0.5)! in terms of ordinary factorials.

7. Oct 17, 2006

### quinn

so many typos I make

Again I am sorry for the typos

SUM(n=0 to infinity) [X^(n+1/2)]/(n+1/2)! --> goes to?

(n+1/2)! can be written as GAMMA(n+3/2)

The above sum is very similiar to the talyor series for e^x, but not quite the same