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

1. ## 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. $\int_{0}^{1/2}{{arctan(x)\over{x}}}$

2. 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. $\int_{0}^{1/2}{{arctan(x)\over{x}}}$

The series for

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

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

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

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

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

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

We need $a_n < .0001$

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

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

Good luck

3. I'm confused about how you create the MacLauren Series though, because I'm having difficulty coming up with the derivatives of f(x).

4. 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: $\sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}$

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

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

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 ...