Power series representation of this function.

**s3a** · May 15th 2010, 01:24 PM

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!

**Random Variable** · May 15th 2010, 01:46 PM

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

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

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

**lilaziz1** · May 15th 2010, 04:14 PM

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

**s3a** · May 15th 2010, 07:13 PM

I don't even understand how to get to the first step you wrote

Is it something I am supposed to have memorized?

**HallsofIvy** · May 16th 2010, 02:50 AM

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 $x^n$ .

**HallsofIvy** · May 16th 2010, 02:54 AM

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.

**s3a** · May 16th 2010, 03:03 PM

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?

**s3a** · May 16th 2010, 03:19 PM

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

.

**Random Variable** · May 16th 2010, 04:10 PM

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

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

**s3a** · May 16th 2010, 04:27 PM

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

**Random Variable** · May 16th 2010, 04:37 PM

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.

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

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