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