How do you find the taylor series series at a=-3a=3 of f(x) = x - x^3f(x)=xx3?

1 Answer
May 21, 2015

There's three ways it can be done:

1) Use the definition of the Taylor series at x=ax=a:
f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots.

In the present example, f'(x)=1-3x^2, f''(x)=-6x, f'''(x)=-6, and all higher-order derivatives are zero. Since f(-3)=24, f'(-3)=-26, f''(-3)=18, f'''(-3)=-6, and all higher derivatives are zero, the answer is:

24-26(x+3)+9(x+3)^2-(x+3)^3, and this does indeed equal f(x)=x-x^3 for all x.

2) You could do a change-of-variables that has the effect of translating the origin to the left by 3 units: u=x+3 so that x=u-3. The coefficients of f(u-3) as a Taylor series centered at u=0 would be the same coefficients of f(x) as a Taylor series centered at x=-3. Here's the expansion calculation:

f(u-3)=(u-3)-(u-3)^3=u-3-(u^3-9u^2+27u-27)=24-26u+9u^2-u^3

Hence, f(x)=24-26(x+3)+9(x+3)^2-(x+3)^3.

3) You could realize that since f(x)=x-x^3 is cubic, its Taylor series about any point will be cubic as well. In particular, about x=-3 it will be of the form a+b(x+3)+c(x+3)^2+d(x+3)^3. You can then expand this out:

a+bx+3b+cx^2+6cx+9c+dx^3+9dx^2+27dx+27d= (a+3b+9c+27d)+(b+6c+27d)x+(c+9d)x^2+dx^3

Set these coefficients equal to the corresponding coefficients of x-x^3 to get a system of 4 equations and 4 unknowns: d=-1, c+9d=0, b+6c+27d=1, and a+3b+9c+27d=0. This system is already in "triangular form" (if your form the corresponding matrix ) and is easily solved by "back substitution": d=-1, c=-9d=9, b=1-6c-27d=1-54+27=-26, and a=-3b-9c-27d=78-81+27=24 to get the same answer: 24-26(x+3)+9(x+3)^2-(x+3)^3.