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 $\sin{x}$.

  • 5
    $\begingroup$ I believe you're misremembering the series. If you switch the numerator and denominator and let $n \to \infty$ you get the Taylor series expansion (en.wikipedia.org/wiki/Taylor_series) of $\sin x$. $\endgroup$ Apr 29 '12 at 16:18
  • $\begingroup$ @qiaochu-yuan Yeah, I flipped them on accident. Thanks for the tip $\endgroup$ Apr 29 '12 at 16:36
  • $\begingroup$ 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 $x$). $\endgroup$ May 1 '12 at 17:16

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.

  • $\begingroup$ This is also shown in this answer. $\endgroup$
    – robjohn
    May 1 '12 at 17:17

I think you're interested in the Taylor expansion of $\sin x$ but the fractions go other way round!

Related Reading on the Site:

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  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
  • 1
    $\begingroup$ @TheChaz It is <strike>blah</strike> And this mark up fails in comments: <strike>Test</strike> ---Test--- works on chat but not here. $\endgroup$
    – user21436
    May 1 '12 at 17:32

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.

  • $\begingroup$ Thaaaat's the one! I remember that my teacher used a name that wasn't "Taylor" or anything like that. Thank you $\endgroup$ May 1 '12 at 21:02

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