- #1
kudoushinichi88
- 129
- 2
I have a shaky understanding of problems concerning Taylor Series. For example, the question below.
Let [itex]f(x)=\tan^{-1}\left(\frac{1+x}{1-x}\right)[/itex] where [itex]-\frac{1}{2}\leq x \leq \frac{1}{2}[/itex]. Find the value of
[tex]f^{2005}(0)[/tex]
the Taylor Series of [itex]\tan^{-1}[/itex] is
[tex]\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}[/tex]
my friend told me to set [itex]k=2n+1[/itex] and after an attempt to solve this, I got
[tex]f^{2005}(0)=\left(\frac{1+x}{x(1-x)}\right)^{2005}(2004!)[/tex]
I'm not sure if I'm doing this right.
Let [itex]f(x)=\tan^{-1}\left(\frac{1+x}{1-x}\right)[/itex] where [itex]-\frac{1}{2}\leq x \leq \frac{1}{2}[/itex]. Find the value of
[tex]f^{2005}(0)[/tex]
the Taylor Series of [itex]\tan^{-1}[/itex] is
[tex]\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}[/tex]
my friend told me to set [itex]k=2n+1[/itex] and after an attempt to solve this, I got
[tex]f^{2005}(0)=\left(\frac{1+x}{x(1-x)}\right)^{2005}(2004!)[/tex]
I'm not sure if I'm doing this right.