Taylor differentition polynomials?

[dagg3r]

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

 

[OlderDan]

You are on the right track with the first one. The series goes on forever. You stop adding terms when the approximation is good enough for any particular application. Take that fourth derivative and apply the general formula for each term in the series.

Be careful with your factorials. This

(-pi/6)sin(pi/6) * (x-(-1)/1!)

should say

(-pi/6)sin(pi/6) * (x-(-1))/1!

 

[Jameson]

Well it's a standard Taylor polynomial that

\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}...

So how would the new center and cosine being evaluated at \frac{\pi{x}}{6} change the polynomial?

 

[arildno]

In my view, the simplest way to do this is as follows:
f(x)=\cos(\frac{\pi{x}}{6})=\cos(\frac{\pi}{6}(x+1)-\frac{\pi}{6})=\cos(\frac{\pi}{6})\cos(\frac{\pi}{6}(x+1))+\sin(\frac{\pi}{6})\sin(\frac{\pi}{6}(x+1))
In this manner, you can easily use the familiar sin/cos expansion around zero.

 

[dextercioby]

I dunno,Arildno,it still looks simpler the way he did it,just by plugging in Taylor's formula...

Daniel.

 

[arildno]

Well, each to his own, I guess.
However I do think it is easier to avoid making mistakes in the series expansion through my way, by reducing it to the most common series expansions of sine and cosine (i.e, about 0).

Besides, since OP found the plugging into Taylor's formula very tough, it might be of advantage for himto see an alternate way of doing it.

 

[dextercioby]

Yeah,whatever.It's good we have divergent opinions converging to a common answer...

"i know the the formula says f^(nth derivative)(a) (x-a)^n / n! "

Daniel.

 

[Hurkyl]

More importantly, it's good to know both techniques.

For example, one trick for finding the n-th derivative of a function is to figure out its Taylor series by alternate means.

For example, for f(x) = e^(x^2), what is f⁽¹⁰⁰⁾(0)?

 

[saltydog]

whats going to be the forth one? confused!

You do have that straight by now right?

Just use the formula:

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

The plot is what you get with four.