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.
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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. $\displaystyle 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, $\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{x^{2n-2}(-1)^{n+1}}{(2n+1)!}$
Originally Posted by ThePerfectHacker $\displaystyle 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, $\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{x^{2n-2}(-1)^{n+1}}{(2n+1)!}$ $\displaystyle 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} $
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