Is this summation limit a valid representation of e^x?

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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...
 
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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.)
 
\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x
 
also...

Sorry about the typo,...

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

Do you mean X^(n + 0.5)?
 
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.
 
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 similar to the talyor series for e^x, but not quite the same
 
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