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2023-03-15

Give the Maclaurin series for sin x.

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Step 1: Maclaurin series explanation
A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function.
The Maclaurin series of a function f(x) up to order n may be found using series [f,x,0,n].
It is a special case of the Taylor series when x=0.
The Maclaurin series' general equation is
$f\left(x\right)=f\left(x0\right)+f\text{'}\left(x0\right)\left(x-x0\right)+\frac{f”\left(x0\right)}{2!}\left(x-x0{\right)}^{2}+\frac{f”\text{'}\left(x0\right)}{3!}\left(x-x0{\right)}^{3}+\dots ..$
Step 2: Maclaurin series for sin x.
Given, $f\left(x\right)=\mathrm{sin}x$
Using x=0, the given equation function becomes
$f\left(0\right)=\mathrm{sin}\left(0\right)=0$
Now taking the derivatives of the given function and using x=0, we have
$1.{f}^{\prime }\left(0\right)=\mathrm{cos}\left(0\right)=1\phantom{\rule{0ex}{0ex}}2.f"\left(0\right)=-\mathrm{sin}\left(0\right)=-0=0\phantom{\rule{0ex}{0ex}}3.{f}^{‴}\left(0\right)=\mathrm{sin}\left(0\right)=0\phantom{\rule{0ex}{0ex}}4.{f}^{⁗}\left(0\right)=\mathrm{sin}\left(0\right)=0$
Thus, we get the series as,
$f\left(x\right)=f\left(0\right)+xf\text{'}\left(0\right)+\frac{{x}^{2}}{2!}f\text{'}\text{'}\left(0\right)+\frac{{x}^{3}}{3!}f\text{'}\text{'}\text{'}\left(0\right)+\frac{{x}^{4}}{4!}f\text{'}\text{'}\text{'}\text{'}\left(0\right)+\dots .$
Putting the values in the preceding series yields
$\mathrm{sin}x=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots$
Therefore, the Maclaurin series for sin x is $1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots$.

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