I was trying to calculate the fourth derivative and then I just gave up.
Is there an easy way of finding the taylor series for $1/(1+x^2)$?
2 Answers
Recall that a geometric series can be represented as a sum by
$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^n \quad \quad|x| <1$
Then we can simply manipulate our equation into that familiar format to get
$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty}(-x^2)^{n} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$
Fun Alternative:
Note that $\frac{d}{dx} \arctan x = \frac{1}{1+x^2}$ and that the Taylor Series for $\arctan x$ is
$\arctan x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$
$\implies \frac{d}{dx} \arctan x = \frac{d}{dx}\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$
Interchange the sum and $\frac{d}{dx}$ (differentiate term-by-term) on the RHS to get
\begin{eqnarray*} \frac{1}{1+x^2} &=& \sum_{n=0}^{\infty}\frac{d}{dx}(-1)^n\frac{x^{2n+1}}{2n+1}\\ &=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} \frac{d}{dx}x^{2n+1}\\ &=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} (2n+1)x^{2n}\\ &=& \sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)x^{2n}}{2n+1}\\ \end{eqnarray*}
Cancel the $(2n+1)$ on the RHS to arrive at $\frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$
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1@Patrick Okay, will do. Was trying to make it easy to show what we were taking the derivative of. I'll add it in now. – 2012-12-05
Yes. $1-x^2+x^4-x^6+x^8-x^{10} \cdots$
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1I think the question was *how* to arrive to that series, not what the series is. – 2012-12-05