# Thread: Power series representation of this function.

1. ## 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!

2. $\displaystyle \arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + ... \ \ |x|<1$

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

$\displaystyle 2x^{2} \arctan x^{5} = 2x^{7} - \frac{2x^{17}}{3} + \frac{2x^{27}}{5} - \frac{2x^{37}}{7} + ...$

3. wait. what does $\displaystyle c_n$ and $\displaystyle x^n$ represent?

4. I don't even understand how to get to the first step you wrote Is it something I am supposed to have memorized?

5. Originally Posted by lilaziz1
wait. what does $\displaystyle c_n$ and $\displaystyle x^n$ represent?
$\displaystyle x^n$, oddly enough, means the variable x to the nth power. $\displaystyle c_n$ is the coefficient of $\displaystyle x^n$.

6. 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 $\displaystyle 2x^2 arctan(x^5)$ you don't have to work directly the entire thing. Start by finding the Taylor's series for arctan(x):

arctan(0)= 0.

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

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

etc.

7. 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?

8. I had what seemed like an epiphany and I did what I'm attaching but I'm stuck still .

9. $\displaystyle \frac{d}{dx} \arctan x = \frac{1}{1+x^{2}} = \sum_{n = 0}^{\infty} (-1)^n x^{2n}, \ |x|<1$

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

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

and finally $\displaystyle 2x^{2} \arctan x^5 = 2 \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+7}}{2n+1}$

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

11. 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.

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