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I remember learning about this series in Precalculus the other day but I neglected to get the name of it. It looks something like this:

$ \begin{align*} \frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!} \end{align*} $

All I remember is that it helps in the modeling of $\sin{x}$.

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    since factorials grow very fast, as a practical matter one can actually use a "partial sum" of that series to obtain reasonably good estimates of $\sin(x)$, just 3 terms will give you 3 decimals places, and the first term alone makes it plausible that for $x$ close to $0$, $\sin(x)$ will be very close to $x$ (and why the limit of $\sin(x)/x$ is $1$ as $x$ approaches $0$, as you can see by dividing each term in the series by $x$).2012-05-01

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This $ \frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!} $ Is called a Maclaurin polynomial for the sine function.

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    Thaaaat's the one! I remember that my teacher used a name that wasn't "Taylor" or anything like that. Thank you2012-05-01
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Are you sure you didn't get the summands flipped?

The following series is called the Taylor series expansion of $\sin{x}$:

\begin{align*} \sin{x} &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} \\[8pt] \end{align*}

It's derived here.

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    This is also shown in [this answer](http://math.stackexchange.com/a/125543/).2012-05-01
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I think you're interested in the Taylor expansion of $\sin x$ but the fractions go other way round!

Related Reading on the Site:

  1. Intuition explanation of taylor expansion?
  2. Taylor series for different points... how do they look? (Cool Animations by JM)
  3. On what interval does a Taylor series approximate (or equal?) its function?
  4. Example to help compute Taylor Series of a function: Basic Taylor expansion question
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    @TheChaz It is `blah` And this mark up fails in comments: Test ---Test--- works on chat but not here.2012-05-01