What are the taylor polynomials p4 and p5 centered at π6 for f(x)=cos⁡(x)

The general formula for the Taylor series is as follow:∑n=0Nfn(a)n!(x−a)nWhere fn(a) being the nth derivative of f(x) at x→aLet us go to n=4f0(x)=f(x)=cos⁡xf′(x)=−sin⁡xf″(x)=−cos⁡xf‴(x)=sin⁡xf⁗(x)=cos⁡xNow, writing it out, we get:f(a)0!(x−a)0+f′(a)1!(x−a)1+f″(a)2!(x−a)2+f‴(a)3!(x−a)3+…=cos⁡a0!(x−a)0+−sin⁡a1!(x−a)1+−cos⁡a2!(x−a)2+sin⁡a3!(x−a)3+…At a=π6 the fifth p5 degree Taylor polynomial will be=cos⁡(π6)−sin⁡(π6)(x−π6)−cos⁡(π6)2(x−π6)2+sin⁡(π6}{3!}(x−π6)3+cos⁡(π6)4!(x−π6)4−sin⁡(π6)5!(x−π6)5+…

"What are the taylor polynomials p4 and p5..." was answered by our experts

Ernstfalld

Answered question

2021-09-07

What are the taylor polynomials

p_{4}

and

p_{5}

centered at

\frac{π}{6}

for

f (x) = \cos (x)

Answer & Explanation

Maciej Morrow

Skilled2021-09-08Added 98 answers

The general formula for the Taylor series is as follow:

\sum_{n = 0}^{N} \frac{f^{n} (a)}{n!} {(x - a)}^{n}

Where

f^{n} (a)

being the

n^{t h}

derivative of f(x) at

x \to a

Let us go to n=4

f^{0} (x) = f (x) = \cos x

f^{'} (x) = - \sin x

f^{″} (x) = - \cos x

f^{‴} (x) = \sin x

f^{⁗} (x) = \cos x

Now, writing it out, we get:

\frac{f (a)}{0!} {(x - a)}^{0} + \frac{f^{'} (a)}{1!} {(x - a)}^{1} + \frac{f^{″} (a)}{2!} {(x - a)}^{2} + \frac{f^{‴} (a)}{3!} {(x - a)}^{3} + \dots

= \frac{\cos a}{0!} {(x - a)}^{0} + \frac{- \sin a}{1!} {(x - a)}^{1} + \frac{- \cos a}{2!} {(x - a)}^{2} + \frac{\sin a}{3!} {(x - a)}^{3} + \dots

a = \frac{π}{6}

the fifth

p_{5}

degree Taylor polynomial will be

= \cos (\frac{π}{6}) - \sin (\frac{π}{6}) (x - \frac{π}{6}) - \frac{\cos (\frac{π}{6})}{2} {(x - \frac{π}{6})}^{2} + \frac{\sin (\frac{π}{6}} {3!} {(x - \frac{π}{6})}^{3} + \frac{\cos (\frac{π}{6})}{4!} {(x - \frac{π}{6})}^{4} - \frac{\sin (\frac{π}{6})}{5!} {(x - \frac{π}{6})}^{5} + \dots}{}

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