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Here's the work i've done to find the Maclaurin series. However, I'm having a very hard time finding a representation for the series using sum, n for the nth term, and x from g(x).

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  • $\begingroup$ It is absolutely fine. Where do you have a problem? $\endgroup$ – Rohan Dec 12 '16 at 4:17
  • $\begingroup$ @Rohan I need a Maclaurin series in the form of a summation representation: for example, $$\sum_{n=0}^ ∞ sin(x) = \frac{(-1)^n*(x)^{2n+1}}{(2n+1)!}$$ and I wasn't sure if it was sufficient to just alter the x into (x+2) in this notation $\endgroup$ – gticecream8 Dec 12 '16 at 4:27
  • $\begingroup$ @gticecream8 I believe you got that backwards : p. $\endgroup$ – YoTengoUnLCD Dec 12 '16 at 4:52
  • $\begingroup$ Why wouldn't you approximate it at x0 = -2 ? $\endgroup$ – Itay.V Dec 12 '16 at 5:56
  • $\begingroup$ Or you just could change variable such as t = x + 2 and approximate at t=0 which will be maclaurin $\endgroup$ – Itay.V Dec 12 '16 at 5:58
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One way to get the "${+}{+}{-}{-}$" pattern is with $(-1)^{n(n-1)/2}$.

To alternate between $\sin(2)$ and $\cos(2)$: $$\sin(2)\frac{(-1)^n+1}{2}+\cos(2)\frac{(-1)^{n-1}+1}{2}$$

So you could write $$\sum_{n=0}^{\infty}(-1)^{n(n-1)/2}\left(\sin(2)\frac{(-1)^n+1}{2}+\cos(2)\frac{(-1)^{n-1}+1}{2}\right)\frac{x^n}{n!}$$

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