How do you find the taylor series for #f(x)=1/(1+x^3)#?

Answer 1

The general process I would use is to take all the derivatives you need to do, first.

That way you can simplify them first out of context, when things aren't yet complicated (you'll see). Then, plug them into the Taylor series formula, and use a calculator to numerically simplify the result.

The result would be this.

The formula for the Taylor series is:

#sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n#

Some things to remember are:

So first, I would take the #n#th derivative of this. I want to stop at #n = 3#, though (you'll see why). Here goes:
#f^0(x) = color(blue)(f(x) = 1/(1+x^3))#
#color(blue)(f'(x)) = -1(1+x^3)^(-2)(3x^2) = (-3x^2)(1+x^3)^-2#
#color(blue)( = -(3x^2)/(1+x^3)^2)#
#color(blue)(f''(x)) = -[3x^2(-2(1+x^3)^-3(3x^2)) + (1+x^3)^-2(6x)]#
# = 18x^4(1+x^3)^-3 - 6x(1+x^3)^-2#
#color(blue)( = (18x^4)/(1+x^3)^3 - (6x)/(1+x^3)^2)#
#color(blue)(f'''(x)) = [18x^4(-3(1+x^3)^-4(3x^2)) + ((1+x^3)^-3)72x^3] - [6x(-2(1+x^3)^-3(3x^2)) + ((1+x^3)^-2)6]#
# = [-162x^6(1+x^3)^-4 + 72x^3(1+x^3)^-3] - [-36x^3(1+x^3)^-3 + 6(1+x^3)^-2]#
# = -162x^6(1+x^3)^-4 + 72x^3(1+x^3)^-3 + 36x^3(1+x^3)^-3 - 6(1+x^3)^-2#
# = -162x^6(1+x^3)^-4 + 108x^3(1+x^3)^-3 - 6(1+x^3)^-2#
#color(blue)( = -(162x^6)/(1+x^3)^4 + (108x^3)/(1+x^3)^3 - 6/(1+x^3)^2)#
I'm not going to simplify these anymore though :-P Alright, let's use them. Plug in #a# and whip out your calculator!
#sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n = sum_(n=0)^(oo)f^n(a)*(1/(n!))(x-a)^n#
#= (1/(1+a^3))(1/(0!))(x-a)^0 + # #(-(3a^2)/(1+a^3)^2)(1/(1!))(x-a)^1 + # #((18a^4)/(1+a^3)^3 - (6a)/(1+a^3)^2)(1/(2!))(x-a)^2 + # #(-(162a^6)/(1+a^3)^4 + (108a^3)/(1+a^3)^3 - 6/(1+a^3)^2)(1/(3!))(x-a)^3# #+ ...#
At #a = 1# for simplicity (you see why!):
#= 1/2 - 3/4(x-1) + (18/8 - 12/8)/2(x-1)^2 + (-162/16 + 216/16 - 24/16)/6(x-1)^3#
#= color(blue)(1/2 - 3/4(x-1) + 3/8(x-1)^2 + 5/16(x-1)^3 + ...)#

If you look here, Wolfram Alpha gives the same result! YEAH!

A fuller answer is: #= color(blue)(1/2 + (-3/4)(x-1) + 3/8(x-1)^2 + 5/16(x-1)^3 + (-21/32)(x-1)^4 + 21/64(x-1)^5 + ...)#
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Answer 2

To find the Taylor series for ( f(x) = \frac{1}{1 + x^3} ), you would start by determining the derivatives of ( f(x) ) at a point ( a ). Then, you can use those derivatives to construct the Taylor series expansion centered at ( a ), which is given by:

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

where ( f^{(n)}(a) ) denotes the ( n )-th derivative of ( f(x) ) evaluated at ( x = a ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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