# Power series representation of this function.

• May 15th 2010, 02:24 PM
s3a
Power series representation of this function.
I'm having trouble getting the power series representation of this function.

(My work and the question are attached)

Any help would be greatly appreciated!
• May 15th 2010, 02:46 PM
Random Variable
$\arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... \ \ |x|<1$

$\arctan x^{5} = x^{5} - \frac{x^{15}}{3} + \frac{x^{25}}{5} - \frac{x^{35}}{7} + ...$

$2x^{2} \arctan x^{5} = 2x^{7} - \frac{2x^{17}}{3} + \frac{2x^{27}}{5} - \frac{2x^{37}}{7} + ...$
• May 15th 2010, 05:14 PM
lilaziz1
wait. what does $c_n$ and $x^n$ represent?
• May 15th 2010, 08:13 PM
s3a
I don't even understand how to get to the first step you wrote :( Is it something I am supposed to have memorized?
• May 16th 2010, 03:50 AM
HallsofIvy
Quote:

Originally Posted by lilaziz1
wait. what does $c_n$ and $x^n$ represent?

$x^n$, oddly enough, means the variable x to the nth power. $c_n$ is the coefficient of $x^n$.
• May 16th 2010, 03:54 AM
HallsofIvy
Quote:

Originally Posted by s3a
I don't even understand how to get to the first step you wrote :( Is it something I am supposed to have memorized?

Not necessarily memorized. The important point is that to find the Taylor's series for $2x^2 arctan(x^5)$ you don't have to work directly the entire thing. Start by finding the Taylor's series for arctan(x):

arctan(0)= 0.

$(arctan(x))'= \frac{1}{x^2+ 1}$ (That's a very important formula- you should have it memorized!) so (arctan(0))'= 1.

$(arctan(x))"= \left(\frac{1}{x^2+1}\right)'= \frac{-2x}{(x^2+1)^2}$ which is 0 at x= 0

etc.
• May 16th 2010, 04:03 PM
s3a
Sorry, I am very confused with this stuff. Is Taylor series for arctan(x) the f^(n) (x-a)^n /n! stuff or is that something else?
• May 16th 2010, 04:19 PM
s3a
I had what seemed like an epiphany and I did what I'm attaching but I'm stuck still :(.
• May 16th 2010, 05:10 PM
Random Variable
$\frac{d}{dx} \arctan x = \frac{1}{1+x^{2}} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}, \ |x|<1$

so $\arctan x = \int \sum^{\infty}_{n=0} (-1)^{n} x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n} \int x^{2n} \ dx = \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}$

and $\arctan x^5 = \sum^{\infty}_{n=0} (-1)^{n}\frac{(x^5)^{2n+1}}{2n+1}= \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+5}}{2n+1}$

and finally $2x^{2} \arctan x^5 = 2 \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+7}}{2n+1}$
• May 16th 2010, 05:27 PM
s3a
Are you sure the exponent of your final form on x isn't 10n + 8 since you add 7 to (10n + 1) ?
• May 16th 2010, 05:37 PM
Random Variable
Quote:

Originally Posted by s3a
Are you sure the exponent of your final form on x isn't 10n + 8 since you add 7 to (10n + 1) ?

I'm positive.

$x^{2} (x^{5})^{2n+1} = x^{2} x^{10n} x^{5} = x^{2 + 10n +5} = x^{10n+7}$