Taylor series

[VBDX]

Hi!

I'm trying to expand this function using Taylor expansion, that is, find a way to represent algebraically the coefficients of the expansion of the following function

[MATH]\frac{2x^4-x^3-x^2-x+1}{1-2x+x^5}[/MATH]
I'm trying to do this on the point [MATH]x=0[/MATH], but I really need to know the algebraic form of these coefficients.

Thaks!

 

[topsquark]

Hi!

I'm trying to expand this function using Taylor expansion, that is, find a way to represent algebraically the coefficients of the expansion of the following function

[MATH]\frac{2x^4-x^3-x^2-x+1}{1-2x+x^5}[/MATH]
I'm trying to do this on the point [MATH]x=0[/MATH], but I really need to know the algebraic form of these coefficients.

Thaks!

Are you using a Taylor expansion or a MacLaurin expansion? In a Taylor expansion we expand in terms of x - a where a is small and for a MacLaurin expansion we expand in terms of x where x is small.

Frankly unless you are told to do this by a Taylor (or MacLaurin) expansion I'd avoid it. The derivatives are going to be pretty messy. (And note that x = 1 gives an indeterminate value.) I'd do the long division, then do the expansion.

-Dan

 

[Deleted member 4993]

Hi!

I'm trying to expand this function using Taylor expansion, that is, find a way to represent algebraically the coefficients of the expansion of the following function

[MATH]\frac{2x^4-x^3-x^2-x+1}{1-2x+x^5}[/MATH]
I'm trying to do this on the point [MATH]x=0[/MATH], but I really need to know the algebraic form of these coefficients.

Thaks!

Did you calculate f(0), f'(0), f"(0),.... etc.?

 

[Paul Belino]

The simplest way to do this is to insert values for x and then fit a polynomial to the data. The coefficients are then automatically given. You will need to choose the degree and the number of points carefully for a good fit. You could do this in EXCEL.
Beware it is more accurate to nest polynomials and avoid powers so that, for example the expression becomes
[1 -x(1+x(1+x(1-2x)))] / [1-x(2-x.x.x.x)]
I did this and found a good fit for a 3rd degree polynomial, but chose your X range carefully.

 

[topsquark]

The simplest way to do this is to insert values for x and then fit a polynomial to the data. The coefficients are then automatically given. You will need to choose the degree and the number of points carefully for a good fit. You could do this in EXCEL.
Beware it is more accurate to nest polynomials and avoid powers so that, for example the expression becomes
[1 -x(1+x(1+x(1-2x)))] / [1-x(2-x.x.x.x)]
I did this and found a good fit for a 3rd degree polynomial, but chose your X range carefully.

This is an interesting idea but is not a Taylor expansion.

-Dan