1. ## Series

Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

$y = e^{x}$

$y = e^{-x}$

$y = e^{-x^3}$

--------

$y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}$

And then the next I am assuming is the same as above but alternating signs?

No idea for the 3rd.

2. Hello,

Originally Posted by alikation
Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

$y = e^{x}$

$y = e^{-x}$

$y = e^{-x^3}$

--------

$y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}$

And then the next I am assuming is the same as above but alternating signs?

No idea for the 3rd.
Ok for the first one and the second.

For the third one, just replace x in the series of e^x by -x^3, just like you should have done for the second one :

$y=e^{-x} = 1+(-x)+(-x)^2/2+..$

$y=e^{-x^3}=1+( ???) + (???)^2/2+..$

3. Originally Posted by alikation
Write the 1st 6 terms of the power series for, along with the general power series using summation notation.

$y = e^{x}$

$y = e^{-x}$

$y = e^{-x^3}$

--------

$y = e^{x} \implies 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + \cdots + \sum_{n=0}^{\infty} \frac{x^n}{n!}$

And then the next I am assuming is the same as above but alternating signs?

No idea for the 3rd.
You're right about the second. For the third, wherever you see $x$, replace it with $-x^3$. So you will get
$1 - x^3 + \frac{(-x^3)^2}{2!} + \frac {(-x^3)^3}{3!} + ...$

which is equal to
$1 - x^3 + \frac{x^6}{2!} - \frac{x^9}{3!} + ...$