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April 22nd, 2016, 11:07 AM   #1
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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!!
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April 22nd, 2016, 11:13 AM   #2
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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
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April 22nd, 2016, 11:23 AM   #3
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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.
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April 22nd, 2016, 11:57 AM   #4
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Now would I stick a minus one in next to that fraction, or would I place (-x-1) where the y is?
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April 22nd, 2016, 12:34 PM   #5
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You know the answer to that! (It's the first).
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April 22nd, 2016, 12:53 PM   #6
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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!
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April 22nd, 2016, 01:05 PM   #7
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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)?
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April 22nd, 2016, 01:06 PM   #8
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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 $\frac14$).
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April 22nd, 2016, 01:42 PM   #9
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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 $?
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April 22nd, 2016, 01:46 PM   #10
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$\displaystyle \dfrac{1}{4} \sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{4^n}$
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