# URGENT, NEEDED BY 3:30! Power Series and error.

• May 3rd 2008, 10:40 AM
dark_knight_307
URGENT, NEEDED BY 3:30! Power Series and error.
I was having trouble with this problem:

Use a power series to approximate the value of the integral with an error less than 0.0001. Assume that the integrand is defined as 1 when x=0. $\displaystyle \int_{0}^{1/2}{{arctan(x)\over{x}}}$
• May 3rd 2008, 11:01 AM
TheEmptySet
Quote:

Originally Posted by dark_knight_307
I was having trouble with this problem:

Use a power series to approximate the value of the integral with an error less than 0.0001. Assume that the integrand is defined as 1 when x=0. $\displaystyle \int_{0}^{1/2}{{arctan(x)\over{x}}}$

The series for

$\displaystyle \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$

$\displaystyle \frac{\tan^{-1}(x)}{x}=\frac{1}{x} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}$

$\displaystyle \int_{0}^{1/2}\frac{\tan^{-1}(x)}{x}dx=\int_{0}^{1/2} \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}=$

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)^2}|_{0}^{1/2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n+1}(2n+1)^2}$

This is an alternating series so the error is always less than the absolute value of next term

so $\displaystyle |a_n|=\frac{(-1)^n}{2^{2n+1}(2n+1)^2}$

We need $\displaystyle a_n < .0001$

$\displaystyle \left| \frac{(-1)^n}{2^{2n+1}(2n+1)^2} \right|< .0001$

Solving I get n = 3.24 so we need to take the first four terms.

Good luck
• May 3rd 2008, 11:41 AM
dark_knight_307
I'm confused about how you create the MacLauren Series though, because I'm having difficulty coming up with the derivatives of f(x).
• May 3rd 2008, 12:40 PM
o_O
I think using geometric series would be the way to go because finding the derivatives of arctan(x) and generalizing looks like it's going to be a pain.

Recall the geometric series: $\displaystyle \sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}$

Looking at the derivative of arctan(x), we have $\displaystyle \frac{1}{1+x^{2}}$ which resembles the geometric series with $\displaystyle a = 1, r = -x^{2}$

This gives us:
$\displaystyle \frac{1}{1+x^{2}} = \sum_{n=0}^{\infty}(-1)^{n}(x)^{2n} = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + ...$
$\displaystyle \arctan(x) = \int \sum_{n=0}^{\infty}(-1)^{n}(x)^{2n}dx$

Integrating gives you the power series for arctan(x), then the rest follows as TheEmptySet showed you.

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If you need to find the maclaurin series for arctan x by finding all the derivatives, it's done here: Maclaruin Series (Scroll down a bit)

Gets messy ...