What is the Taylor Series generated by #f(x) = x - x^3#, centered around a = -2?

Question

Charles Arocha · Accepted Answer

The answer is #6-11(x+2)+6(x+2)^2-(x+2)^3#

There are two methods to get the answer.

1) Use the expression #f(-2)+f'(-2)(x+2)+(f''(-2))/(2!)(x+2)^2+(f'''(-2))/(3!)(x+2)^3+\cdots#

Since #f(x)=x-x^3#, we get #f(-2)=-2+8=6#, #f'(x)=1-3x^2# so that #f'(-2)=1-12=-11#, #f''(x)=-6x# so that #f''(-2)=12#, and #f'''(x)=-6# so that #f'''(-2)=-6#. All higher-order derivatives are zero since #f# is a cubic.

The answer is therefore #6-11(x+2)+6(x+2)^2-(x+2)^3#.

2) Let #u=x+2# so that #x=u-2# and expand #f(x)=f(u-2)# before replacing #u# with #x+2# at the end.

Here are the details, which follow from the binomial theorem (Pascal's triangle ).

#f(u-2)=(u-2)-(u-2)^3#

#=u-2-(u^3-6u^2+12u-8)=6-11u+6u^2-u^3#.

The answer is therefore, as we saw before,

#6-11(x+2)+6(x+2)^2-(x+2)^3#.

Charles Anctil · Answer

undefined The Taylor series generated by $ f(x) = x - x^3 $ centered around $ a = -2 $ is:

$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots $$

First, find the derivatives of $ f(x) $ up to the third derivative:

$ f(x) = x - x^3 $

$ f'(x) = 1 - 3x^2 $

$ f''(x) = -6x $

$ f'''(x) = -6 $

Now, evaluate these derivatives at $ a = -2 $:

$ f(-2) = -2 - (-2)^3 = -2 + 8 = 6 $

$ f'(-2) = 1 - 3(-2)^2 = 1 - 12 = -11 $

$ f''(-2) = -6(-2) = 12 $

$ f'''(-2) = -6 $

Now, plug these values into the Taylor series formula:

$$ f(x) = 6 - 11(x+2) + \frac{12}{2!}(x+2)^2 - \frac{6}{3!}(x+2)^3 + \cdots $$

This is the Taylor series representation of $ f(x) = x - x^3 $ centered around $ a = -2 $.