Hello forum! I was wondering if somebody knows of a fancy way of determining the Taylor series expansion of $\displaystyle \arctan (x)$ about $\displaystyle x = 1$ without computing all the derivatives.

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- Jan 16th 2013, 07:10 AMMathCrusaderTaylor series arctan(x) without computations
Hello forum! I was wondering if somebody knows of a fancy way of determining the Taylor series expansion of $\displaystyle \arctan (x)$ about $\displaystyle x = 1$ without computing all the derivatives.

- Jan 16th 2013, 07:52 AMebainesRe: Taylor series arctan(x) without computations
Determining the series does indeed require computing derivatives, which can appear fairly daunting at first. But after determining the first, second and third derivatives at x=1 you'll see that a simple pattern emerges: arctan(1)= 1/3 + 1/5 - 1/7 +1/9 -....

- Jan 16th 2013, 07:58 AMBariothRe: Taylor series arctan(x) without computations
none that I can think of sorry!

You should never be afraid of derivating altough! Don't be lazy and get to it! - Jan 16th 2013, 02:09 PMMathCrusaderRe: Taylor series arctan(x) without computations
Okay thanks!

- Jan 16th 2013, 04:00 PMTheSaviourRe: Taylor series arctan(x) without computations
It can be derived from geometric series, if that's okay.

$\displaystyle \begin{aligned}\arctan{x} & = \int_{0}^{x}\frac{1}{1+t^2}\ dt = \int_{0}^{x}\sum_{n=0}^{\infty}(-1)^n t^{2n} \ dt \\& = \sum_{n=0}^{\infty}(-1)^n\int_{0}^{x} t^{2n}\ dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}. \end{aligned}$ - Jan 16th 2013, 04:08 PMTheSaviourRe: Taylor series arctan(x) without computations
Just noticed that you wanted the expansion at about x = 1. Shouldn't be hard to tweak the above.