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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0what if f(x) is the integral of (1/1+x^2)? – 2012-12-05
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1If that's all the integral is, then its primitive is just $\arctan x + C$ where $C$ is a constant. – 2012-12-05
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0i would like to know what the taylor series is when f(x) is the integral of (1/1+x^2). – 2012-12-05
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0Take a look at my fun alternative approach, which may give you some insight for the problem. $\int\frac{1}{1+x^2} \ dx = \arctan x + C$. The Taylor Series for $\arctan x$ is usually one of the ones you need to know and you should be able to find it in your book (or above in my post). $$\arctan x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$ – 2012-12-05
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0@question: then you can integrate the Taylor series of $\frac 1{1+x^2}$ term by term, getting $\sum_{n=1}^{\infty} (-1)^n \frac {x^{2n+1}}{2n+1}$ – 2012-12-05
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0what if the taylor series is the integral from x to 0 of (1/1+x^2)? – 2012-12-05
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0$$\int_0^x \frac{1}{1+x^2} \ dx = \arctan x + C \implies \int_x^0 \frac{1}{1+x^2} \ dx = -\arctan x + C$$ – 2012-12-05
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0ok.. nevermind thannks for the help though – 2012-12-05
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0If I didn't answer your question, if you could rephrase it perhaps, I may be able to help. – 2012-12-05
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0i figured it out, but i might fail my finals. – 2012-12-05
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0@Joe : You should put the $(-1)^n/2n+1$ inside the sum at some point in your answer. – 2012-12-05
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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