Taylor differentition polynomials?

In summary: The next term in the series (the fifth one) would involve the fifth derivative, and so on.In summary, the conversation revolves around finding the Taylor polynomial of degree 4 for the function f(x) = cos (pi*x / 6) about x=-1 and applying the formula f^(nth derivative)(a) (x-a)^n / n! to do so. The conversation also touches on rearranging equations and using the general formula for each term in the series. Additionally, the conversation discusses the benefits of using alternate methods to find the n-th derivative of a function and the confusion around finding the fourth term in the series.
  • #1
dagg3r
67
0
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
 
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  • #2
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!
 
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  • #3
Well it's a standard Taylor polynomial that

[tex]\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}...[/tex]

So how would the new center and cosine being evaluated at [tex]\frac{\pi{x}}{6}[/tex] change the polynomial?
 
  • #4
In my view, the simplest way to do this is as follows:
[tex]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))[/tex]
In this manner, you can easily use the familiar sin/cos expansion around zero.
 
  • #5
I dunno,Arildno,it still looks simpler the way he did it,just by plugging in Taylor's formula...

Daniel.
 
  • #6
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.
 
Last edited:
  • #7
Yeah,whatever.It's good we have divergent opinions converging to a common answer...

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

Daniel.
 
  • #8
More importantly, it's good to know both techniques. :smile:

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)?
 
  • #9
dagg3r said:
whats going to be the forth one? confused!

You do have that straight by now right?

Just use the formula:

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

The plot is what you get with four.
 

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1. What are Taylor differentiation polynomials?

Taylor differentiation polynomials are a type of polynomial function used in calculus to approximate a given function by using its derivatives at a specific point. They are named after the mathematician Brook Taylor, who first introduced them in the 18th century.

2. How are Taylor differentiation polynomials different from regular polynomials?

Taylor differentiation polynomials are different from regular polynomials because they use the derivatives of a function at a specific point to approximate the function, rather than just using the coefficients of the polynomial. This allows for a more accurate approximation of the function, particularly near the specific point.

3. What is the purpose of using Taylor differentiation polynomials?

The purpose of using Taylor differentiation polynomials is to approximate a function and its values, particularly near a specific point, without having to use complicated methods such as limits or integrals. This makes it a useful tool in calculus for solving problems and analyzing functions.

4. How are Taylor differentiation polynomials calculated?

To calculate a Taylor differentiation polynomial, you need to find the derivatives of the function at the specific point and use them to construct the polynomial. The polynomial is then used to approximate the function and its values near the specific point.

5. What are some applications of Taylor differentiation polynomials?

Taylor differentiation polynomials have many applications in mathematics, physics, and engineering. They are used to approximate functions in various fields, such as in optimization problems and in modeling the behavior of physical systems. They are also used in numerical analysis to solve differential equations and in statistics to analyze data.

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