crubats4b5p

2023-03-15

Give the Maclaurin series for sin x.

plumingagcqr

Beginner2023-03-16Added 7 answers

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

$f\left(x\right)=f\left(x0\right)+f\text{'}\left(x0\right)(x-x0)+\frac{f\u201d\left(x0\right)}{2!}(x-x0{)}^{2}+\frac{f\u201d\text{'}\left(x0\right)}{3!}(x-x0{)}^{3}+\dots ..$

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

$1.{f}^{\prime}(0)=\mathrm{cos}(0)=1\phantom{\rule{0ex}{0ex}}2.f"(0)=-\mathrm{sin}(0)=-0=0\phantom{\rule{0ex}{0ex}}3.{f}^{\u2034}(0)=\mathrm{sin}(0)=0\phantom{\rule{0ex}{0ex}}4.{f}^{\u2057}(0)=\mathrm{sin}(0)=0$

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 $.

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

$f\left(x\right)=f\left(x0\right)+f\text{'}\left(x0\right)(x-x0)+\frac{f\u201d\left(x0\right)}{2!}(x-x0{)}^{2}+\frac{f\u201d\text{'}\left(x0\right)}{3!}(x-x0{)}^{3}+\dots ..$

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

$1.{f}^{\prime}(0)=\mathrm{cos}(0)=1\phantom{\rule{0ex}{0ex}}2.f"(0)=-\mathrm{sin}(0)=-0=0\phantom{\rule{0ex}{0ex}}3.{f}^{\u2034}(0)=\mathrm{sin}(0)=0\phantom{\rule{0ex}{0ex}}4.{f}^{\u2057}(0)=\mathrm{sin}(0)=0$

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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