Taylor series arctan(x) without computations

**MathCrusader** · January 16th 2013, 07:10 AM

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

**ebaines** · January 16th 2013, 07:52 AM

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 -....

**Barioth** · January 16th 2013, 07:58 AM

none that I can think of sorry!

You should never be afraid of derivating altough! Don't be lazy and get to it!

**MathCrusader** · January 16th 2013, 02:09 PM

Okay thanks!

**TheSaviour** · January 16th 2013, 04:00 PM

It can be derived from geometric series, if that's okay.

$\begin{aligned}\arctan{x} & = \int_{0}^{x}\frac{1}{1+t^2}\ dt = \int_{0}^{x}\sum_{n=0}^{\infty}(-1)^n t^{2n} \ dt \\& = \sum_{n=0}^{\infty}(-1)^n\int_{0}^{x} t^{2n}\ dt = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}. \end{aligned}$

**TheSaviour** · January 16th 2013, 04:08 PM

Just noticed that you wanted the expansion at about x = 1. Shouldn't be hard to tweak the above.

Thread: Taylor series arctan(x) without computations

LinkBack

Thread Tools

Display

Taylor series arctan(x) without computations

Re: Taylor series arctan(x) without computations

Re: Taylor series arctan(x) without computations

Re: Taylor series arctan(x) without computations

Re: Taylor series arctan(x) without computations

Re: Taylor series arctan(x) without computations

Similar Math Help Forum Discussions

taylor expansion of arctan(x) at infinity

Series; arctan

Maclaurin Series with arctan

Using elementary taylor series for taylor polynomial

Power Series Computations and More

Search Tags