Taylor Series of the inverse tangent function

[kudoushinichi88]

I have a shaky understanding of problems concerning Taylor Series. For example, the question below.

Let $f(x)=\tan^{-1}\left(\frac{1+x}{1-x}\right)$ where $-\frac{1}{2}\leq x \leq \frac{1}{2}$. Find the value of

$$f^{2005}(0)$$

the Taylor Series of $\tan^{-1}$ is

$$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$$

my friend told me to set $k=2n+1$ and after an attempt to solve this, I got

$$f^{2005}(0)=\left(\frac{1+x}{x(1-x)}\right)^{2005}(2004!)$$

I'm not sure if I'm doing this right.

 

[HallsofIvy]

What you have doesn't even make sense. The left side, $f^{2005}(0)$, is a specific number, not a function of x.

Given that the Taylor's series for arctangent is
$$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$$
you can then replace that x with $(1+ x)/(x(1- x))$ but then you have to expand that to get a power series and determine n so that x has power 2005.

$f^{2005}(0)$ is the coefficient not entire term.

 

[jackmell]

May I add something too please? I would suggest he first solve the problem for just arctan(u), just that one, say for the fourth or fifth or like me, the seventh derivative. Get that straight, then go for the other one. So:

$$f(u)=\arctan(u)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} u^n=\sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{2n+1}$$

See, that's already a good start. Then that means:

$$\frac{f^{(2n+1)}(0)}{(2n+1)!}=\frac{(-1)^n}{2n+1}$$

or:

$$f^{(2n+1)}(0)=(-1)^n \frac{(2n+1)!}{2n+1}$$

Now, here's the very important part I wish to emphasize: check that in Mathematica, say for the seventh derivative. Then in that case, n=3 right? so:

In[18]:=
D[ArcTan[x], {x, 7}] /. x -> 0
N[((-1)^3/7)*7!]

Out[18]=
-720

Out[19]=
-720.

Now, here's a hint to solve your problem, take:

$$\frac{d}{du} \arctan(u)$$

and:

$$\frac{d}{dx}\arctan(\frac{1+x}{1-x})$$

What do you notice about the two derivatives and then all the other higher order derivatives?

 

[kudoushinichi88]

So... subbing $\frac{1+x}{1-x}$ into,

$$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$$

I have

$$\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{1+x}{1-x}\right)^{2n+1}}{2n+1}$$,

By setting $2n+1=2005$, and comparing with the general representation of a taylor series,

$$\sum_{n=0}^{\infty}\frac{f^n(a)}{n!}(x-a)^n$$

I finally got

$$\frac{1002!}{2005}$$

Is this correct?

 

[vela]

No, that's not correct. I would strongly recommend you consider jackmell's hint about comparing the first derivatives of those two functions.