integrate(arctan(x^2)) from 0--->1 with finding a numerical serie,what is the value which makes the error smaller than 0,0001??

thanks for,ı need this as soon as possible.

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- Nov 21st 2008, 01:03 PMsah_mata question about power series
integrate(arctan(x^2)) from 0--->1 with finding a numerical serie,what is the value which makes the error smaller than 0,0001??

thanks for,ı need this as soon as possible. - Nov 21st 2008, 01:18 PMMathstud28
We know that

$\displaystyle \forall{x}\in[-1,1]~\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$

So

$\displaystyle \forall{x}\in[-1,1]~\arctan\left(x^2\right)=\sum_{n=0}^{\infty}\frac{ (-1)^nx^{4n+2}}{2n+1}$

So since $\displaystyle (0,1)\subset[-1,1]$ and a power series is uniformly covnergent and integrable on the interior of its IOC we have that

$\displaystyle \begin{aligned}\int_0^1\arctan\left(x^2\right)dx&= \int_0^1\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n+2}}{2n+1}dx\\

&=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^nx^{4n+2}}{2n+1}dx\\

&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(4n+3)}

\end{aligned}$

Now to find the appropriate error note that if $\displaystyle S=\sum_{n=0}^{\infty}(-1)^a_n$ and $\displaystyle S_n=\sum_{n=0}^{N}(-1)^na_n$ that

$\displaystyle R_N=\left|S-S_N\right|\leqslant{a_{N+1}}$

So the error of your series is less than or equal to the first neglected term, so calculate $\displaystyle a_{N+1}\leqslant{.00001}$