Series--please help

[chopchop]

Problem: for all x-vals for which it converges, the fn f is defined by the series

∞
f(x)= ∑ (x^n) / (n!)
n=0

a. what is f (0)...i know that anything to the power of 0 is 1, but how do u deal with n! .... since it is the product of integers from 1 to n... is the solution 1 because we have x^0 = 1 and then how does n! =1 (it does right?)

b. what is the domain of f? ...this i get lim as n --> infinity is the abolute value of X divided by (n+1) ...the limit is 0 so all reals

but again here i don't really understand the operation with n!

x^(n+1) times n!
________ ___

(n+1) ! x^n is (x) / (n+1)

how do u deal with n! what does it mean in relation to its disapearance?

c. Assuming that f ' can be calculated by diffetiating the series term-by-ter, find the series for f ' (x) ....how do u differentiate this, particularly with the n!
I know that it should be the same series, but I don't really get why... is it because the fn is e^x, probably, but I don't know how to show the work for this question

THANK U ALL

 

[galactus]

Differentiate the series the same as you would anything else.

\(\displaystyle \L\\\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}\)

Now, go from there using what skh showed you on the other site.


0! is defined as 1 so that certain formulas involving factorials are true for all nonnegative numbers

 

[chopchop]

i see that the answer is

nX^(n-1)/n! but how come in the back of my book the answer is listed as X^n/n! ? how are the two the same.. I really don't get this stuff...

 

[skeeter]

f(x) = Sum{n = 0 to inf} (x^n)/n!

list out the series a few terms ...

f(x) = 1 + x + (x^2)/2! + (x^3)/3! + ...

a. f(0) = 1

c. f'(x) = 0 + 1 + x + (x^2)/2! + (x^3)/3! + ...

in other words, f'(x) = f(x) ... what function has this property?

 

[chopchop]

thanku sooo mcuh I love this site; if u hve time, can u answer my other question--jsut to clear up what I have