- #1

dagg3r

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**taylor differentition polynomials?**

hi got a question here that involves this extremely difficult question anyone that can point me in the right direction on what to do will be most appreciated :)

Find Exactly the tayor polynomial of degree 4

f(x) = cos ( pi*x / 6 ) about x=-1

i know the the formula says f^(nth derivative)(a) (x-a)^n / n! when trying to apply this equation what i did was

f(x)=cos(pi*x/6)

f'= (-pi/6)sin(pix/6)

f``= (-pi^2 / 36)cos(pix/6)

f```= (pi^3/216)sin(pix/6)

do i do the same for f````?

once i get that

do i apply the formula and sub x=-1 and get example

cos(pi/6) + (-pi/6)sin(pi/6) * (x-(-1)/1!) - this is an example of the first one?

whats going to be the forth one? confused!

and now for the ralph Newton

it says consider the equation

cos x = 1.3x^1.3

what i thought i did was cos x - 1.3x^1.3 = 0

rearrange the equation then i differentitaed the equation therefore got

f`= -sin(x) - 1.69x^0.3

and try to apply the formula

xn+1 = xn - f(xn) / f`(xn)

how do i apply this? thanks