# Using Maclaurin series to find 2005-order derivative

1. Apr 14, 2012

### hqjb

1. The problem statement, all variables and given/known data

Let $f(x) = \arctan(\frac{1+x}{1-x})$
Find $f^{2005}$(0)

2. Relevant equations

I'm guessing this has to do with maclaurin's?

3. The attempt at a solution
...
$f(x) = \pi /4 + \sum^∞_{n = 0} \frac{(-1)^n}{2n+1}x^{2n+1}$

$\sum^∞_{n = 0}\frac{f^n(0)x^n}{n!} = \pi /4 + \sum^∞_{n = 0} \frac{(-1)^n}{2n+1}x^{2n+1}$

So anyone knows how I go about from here? The answer is 2004!(factorial)
How do you compare two infinite series can you cancel them out?

2. Apr 14, 2012

### Harrisonized

(d/dx)xn = nxn-1

(dp/dxp) xn = n!/(n-p)! xn-p
(can you see where this formula comes from?)

Also, (d/dx)C=0

They want the value of the 2005th derivative at 0. If you think about it, the term involving x2005 will become a constant after 2005 differentiations. All terms before x2005 will disappear due to (d/dx)C=0, and all terms after are irrelevant since we are calculating at 0.

The term at x2005 is supplied by the formula you found:

Ʃ (-1)n/(2n+1) x2n+1

Of course, the π/4 term also disappears.

Last edited: Apr 14, 2012
3. Apr 14, 2012

### lol_nl

You can compare the two series since they are polynomials in x, and what you are interested in is only one of the terms of the polynomial. f^(2000) occurs exactly in the term of x^(2005), so you can ignore all other terms and compare the coefficients of x^2005 on both sides.

4. Apr 14, 2012

### hqjb

Ah...i get it now so,
$f^{2005}(0)=\frac{2005!}{2005} * (-1)^{1002}$