My Math Forum Creating a power series expansion

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 April 22nd, 2016, 12:07 PM #1 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Creating a power series expansion I know that when f(x)=$\displaystyle e^x$, the sum of it is written as $\displaystyle x^n/n!$. Knowing this, how do I find the power series expansion for e^(-x)-1? Assume I know nothing about series. Thanks!!
 April 22nd, 2016, 12:13 PM #2 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 P.S. when I plugged this into the taylor series formula I got ( (-1)^n(x^n) ) / (n!), but i'm not positive that's right
 April 22nd, 2016, 12:23 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra As you say, the power series for $e^y = \sum \limits_{n=0}^\infty {y^n \over n!}$. (Here I have just replaced $x$ with $y$ for clarity). You want the power series for $e^{-x}-1$, so into the formula above you put $y=-x$ and then subtract 1. You need to write the summation properly in order to get these correct - especially in this case.
 April 22nd, 2016, 12:57 PM #4 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Now would I stick a minus one in next to that fraction, or would I place (-x-1) where the y is?
 April 22nd, 2016, 01:34 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra You know the answer to that! (It's the first).
 April 22nd, 2016, 01:53 PM #6 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Thank you! I'm actually dealing with a similar problem now, well really a lot of problems, this stuff makes 0 sense to me. But i'm trying not to spam the forum! Anyways, i'm to turn f(x) = $\displaystyle 1/(4+x^2)$ into a taylor series, but again I am lost. I know that $\displaystyle 1/(1-x)$ is put into the form $\displaystyle x^n$. Does that mean that in this case, it would be x^(2n)/4? I'm not too sure how to do it the mathematical way, but if this problem would be a good example in finding its taylor series form I would greatly appreciate that, otherwise i'll just take the same explanation from the previous problem. Thanks again!
 April 22nd, 2016, 02:05 PM #7 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Oops! I meant to say I know that $\displaystyle 1/(1+x)$ is put into the form $\displaystyle (-x)^n$ so would that meant that in this case would it be put into the form (-x/4)^(2n)?
 April 22nd, 2016, 02:06 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra No. You need to turn the function you are given into the form you know. In this case we need to get the numbers the same in $f(x)$ as they are in ${1 \over 1-x}$. Again for clarity we will write ${1 \over 1-y} = \sum \limits_{n=0}^\infty y^n$. Then $$f(x) = {1 \over 4 + x^2} = \frac14 \cdot {1 \over 1 + {x^2 \over 4}} = \frac14 \cdot {1 \over 1 - {-x^2 \over 4}}$$ And now, with $y = -{x^2 \over 4}$ we can create our series (which will be multiplied by $\frac14$).
 April 22nd, 2016, 02:42 PM #9 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Okay thank you, I never knew how to do this. So would the 1/4 go outside the epsilon, or inside it next to the $\displaystyle y^n$?
 April 22nd, 2016, 02:46 PM #10 Math Team   Joined: Jul 2011 From: Texas Posts: 2,781 Thanks: 1432 $\displaystyle \dfrac{1}{4} \sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{4^n}$

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