# Taylor series arctan(x) without computations

• Jan 16th 2013, 08:10 AM
Taylor series arctan(x) without computations
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.
• Jan 16th 2013, 08:52 AM
ebaines
Re: 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, 08:58 AM
Barioth
Re: 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, 03:09 PM
\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}