# Where do I start? McLauren series

• Feb 27th 2009, 03:51 PM
thehollow89
Where do I start? McLauren series
f(x) = (x - sinx)/x^3 if x doesn't equal 0
1/6 if x = 0

Where do I start? I was thinking 1/x^2 - sinx/x^3 and then manipulate the sinx series, but I'm not sure if that's what I'm supposed to do.
• Feb 27th 2009, 04:36 PM
ThePerfectHacker
Quote:

Originally Posted by thehollow89
f(x) = (x - sinx)/x^3 if x doesn't equal 0
1/6 if x = 0

Where do I start? I was thinking 1/x^2 - sinx/x^3 and then manipulate the sinx series, but I'm not sure if that's what I'm supposed to do.

$x - \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \right) = \frac{x^3}{3!} - \frac{x^5}{5!} + ...$.

Therefore, $f(x) = \sum_{n=1}^{\infty} \frac{x^{2n-2}(-1)^{n+1}}{(2n+1)!}$
• Feb 27th 2009, 05:03 PM
manjohn12
Quote:

Originally Posted by ThePerfectHacker
$x - \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \right) = \frac{x^3}{3!} - \frac{x^5}{5!} + ...$.

Therefore, $f(x) = \sum_{n=1}^{\infty} \frac{x^{2n-2}(-1)^{n+1}}{(2n+1)!}$

$f(x) = \begin{cases} \sum_{n=1}^{\infty} \frac{x^{2n-2}(-1)^{n+1}}{(2n+1)!} \ \ \text{if} \ x \neq 0 \\ 1/6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x = 0 \end{cases}$