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!

Thanks in advance!

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- May 15th 2010, 01:24 PMs3aPower 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!

Thanks in advance! - May 15th 2010, 01:46 PMRandom Variable
$\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} + ... $ - May 15th 2010, 04:14 PMlilaziz1
wait. what does $\displaystyle c_n $ and $\displaystyle x^n $ represent?

- May 15th 2010, 07:13 PMs3a
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, 02:50 AMHallsofIvy
- May 16th 2010, 02:54 AMHallsofIvy
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. - May 16th 2010, 03:03 PMs3a
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, 03:19 PMs3a
I had what seemed like an epiphany and I did what I'm attaching but I'm stuck still :(.

- May 16th 2010, 04:10 PMRandom Variable
$\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 $\displaystyle 2x^{2} \arctan x^5 = 2 \sum^{\infty}_{n=0} (-1)^{n}\frac{x^{10n+7}}{2n+1}$ - May 16th 2010, 04:27 PMs3a
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, 04:37 PMRandom Variable