Thread: Maclaurin series of a function

1. Maclaurin series of a function

Obtain Maclaurin series for the given function

2. Originally Posted by racewithferrari
Obtain Maclaurin series for the given function

Easy, if you know the MacLaurin series of $\displaystyle \tan^{-1}(x^3)$:

$\displaystyle x^2\tan^{-1}(x^3)=x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf ty\frac{(-1)^k}{2k+1}x^{6k+5}=:\sum_{n=0}^\infty a_n x^n$
where

$\displaystyle a_n :=\begin{cases}\frac{(-1)^{(n-5)/6}}{2\cdot\frac{n-5}{6}+1}, &\text{if } \frac{n-5}{6}\in\mathbb{Z}\\0, &\text{otherwise}\end{cases}$

3. What have you tried?

Can you do this one? atan(x)? If you can, you are almost done.

This might help.

$\displaystyle \frac{d}{dx}atan(x)\;=\;\frac{1}{1+x^{2}}\;=\;\fra c{1}{1-(-x^{2})}\;=\;1 - x^{2} + x^{4} - x^{6} + ...$

Note: Is there a Uniform Convergence consideration in here, somewhere?

4. Originally Posted by Failure
Easy, if you know the MacLaurin series of $\displaystyle \tan^{-1}(x^3)$:

$\displaystyle x^2\tan^{-1}(x^3)=x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf ty\frac{(-1)^k}{2k+1}x^{6k+5}=:\sum_{n=0}^\infty a_n x^n$
where

$\displaystyle a_n :=\begin{cases}\frac{(-1)^{(n-5)/6}}{2\cdot\frac{n-5}{6}+1}, &\text{if } \frac{n-5}{6}\in\mathbb{Z}\\0, &\text{otherwise}\end{cases}$
where is there 5 when you multiply 3 with 2k+1.

5. It comes from the product of $\displaystyle x^2$ and the series.

$\displaystyle x^2\cdot\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(x^3\right)^{2k+1}=\sum_{k=0}^\inf ty\frac{(-1)^k}{2k+1}x^{6k+3}\cdot x^2=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{6k+5}$