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A friend of mine was talking about how finding MacLaurin series for functions with variables in the denominator might prove difficult without tables. We started making lots of crazy problems, but one that particularly intrigued me was the following: $$ f(x)=\frac{3}{x}\sin\left(\frac{x^3}{3}\right). $$ Due to the $\frac x 3$, we thought there might be some periodic properties to these kinds of functions. But first of all: how would one work out the MacLaurin series for this function without using our favorite aid Wolfram|Alpha?

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If $y>0$ there are only finitely many $x$ so that $f(x)=y$. So $f$ cannot be periodic.

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First start with $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$ Then replacing $x$ with $\frac{x^3}3$ we get $$\sin\left(\frac{x^3}3\right)=\sum_{n=0}^\infty \frac{(-1)^n\left(\frac{x^3}3\right)^{2n+1}}{(2n+1)!}=\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{6n+3}}{3^{2n+1}(2n+1)!}$$ Now multiply by $\frac3x$ to get $$\frac3x\sin\left(\frac{x^3}3\right)=\frac3x\sum_{n=0}^\infty \frac{(-1)^nx^{6n+3}}{3^{2n+1}(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^nx^{6n+2}}{3^{2n}(2n+1)!}=\sum_{n=0}^\infty \frac{(-1)^nx^{6n+2}}{9^{n}(2n+1)!}$$

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