power series

[rae27]

Please help. I'm completely lost. This question has multiple parts. I have the series summation, n=0 to infinity x^n/n!

1. compute partial sums s2(x) for the values where x is converges (which is for all real numbers):
I have s2(x)=1+x, am I even on the right track with this?

2. esimate the limit as n-->infinity s sub n (1) to five decimal places:
I've taken the limit of 1^n/n! and come up with 0, which doesn't have 5 decimal places. Did I even start this right?

I don't understand these and any help would be greatly appreciated. Thanks.

 

[Deleted member 4993]

Please help. I'm completely lost. This question has multiple parts. I have the series summation, n=0 to infinity x^n/n!

1. compute partial sums s2(x) for the values where x is converges (which is for all real numbers):
I have s2(x)=1+x, am I even on the right track with this?

2. esimate the limit as n-->infinity s sub n (1) to five decimal places:
I've taken the limit of 1^n/n! and come up with 0,

How do you get the sum to be 0, when the first term is 1 and rest of the terms are positive?

What is the Taylor's series expansion of e[sup:3d3360dh]x[/sup:3d3360dh]?



which doesn't have 5 decimal places. Did I even start this right?

I don't understand these and any help would be greatly appreciated. Thanks.

 

[rae27]

I took the limit of s sub n (1) 1^n/n! to get zero and we haven't even studied Taylor series yet so I have no idea what the expanded e^x would be. I'm still completely lost.

 

[galactus]

The series \(\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\) IS the Taylor series for e^x.


For part 2, if x=1, then we have the expansion for \(\displaystyle e\)

\(\displaystyle e=\sum_{n=0}^{\infty}\frac{1}{n!}\)

 

[rae27]

I think I got it. Thanks for the help.