Finding Taylor series for x^3 at a = -1

[mcdowellmg]

Homework Statement

Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1

Homework Equations

f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution

I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0
I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?

 

[LCKurtz]

Homework Statement

Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1

Homework Equations

f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution

I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0
I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?

You have listed f(x), f'(x), f''(x), f'''(x) but I don't see your correct values for

f(-1), f'(-1),f''(-1),f'''(-1)

which you should be using for coefficients.

 

[SammyS]

You can check your work by graphing the result & comparing the result with a graph of x³.

 

[HallsofIvy]

Homework Statement

Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)--> 0.)
f(x) = x^3, a = -1

Homework Equations

f(x) = f(a)+f'(a)(x-a)+(f''(a)/2!)(x-a)^2+(f'''(a)/3!)(x-a)^3+...+(f(nth derivative)(a)/n!)(x-a)^n

The Attempt at a Solution

I found the following derivatives of x^3:
first: 3x^2
second: 6x
third: 6
fourth: 0

Then, I substituted: x^3 = (-1)^3+3^3(x+1)+(-6^3/2!)(x+1)^2+(6/3!)(x+1)^3+0

So f'(-1)= 3, f''(-1)= 6(-1)= -6, f'''(-1)= 6. But did you cube f' and f''?

I reduced to -1-3(x+1)-6(x+1)^2+(x+1)^3, but WebAssign says that is wrong...
am I missing something here?

The simplest way to get this is to let y= x+ 1. Then x= y- 1 and $x^3= (y- 1)^3$. Multiply that out and then replace y with x+ 1.

 

[mcdowellmg]

Thanks! I did cube f' and f'' for some reason. I ended up with -1+3(x+1)-3(x+1)^2+(x+1)^3, which was correct.