Question

Find the Taylor series for the function ƒ (x) centered at the given value of a.
f(x) = x^4 -4x^3+3x^2+5x-3

I know that in a Maclaurin series, the answer would end up just being f(x). With this question, though, I end up with
2+3(x-1)-3(x-1)^2+(x-1)^4
and then expanded it out to get
x^4-4x^3-x^2+9x-3

Could anyone help me figure out the right method to approach this problem, and if the answer I got is correct or not?

Answer 1

you just need to find
f(1)
f'(1)
f''(1)
f^(3) (1)
and f^(4) (1)

Answer 2

the given value of a is equal to 1 ?

Answer 3

Sorry, yes the value of a is equal to 1.
f(1) = 2 f'(1)= = f''(1) = -6, f'''(1) = 0, and f^4(1) = 24 is what I got that led me to what I posted in the question.

Answer 4

good, so
f(x) = 2 + 3(x-1) - 6/2! (x-1)^2 + 24/4! (x-1)^2

Answer 5

sorry the last term should be to the 4th power

Answer 6

Doesn't that simplify to 2+3(x-1)-3(x-1)^2+(x-1)^4

Answer 7

yes, so your answer was right. When you expanded it, you probably made an algebratic mistakes somewhere.

Answer 8

https://www.wolframalpha.com/input/?i=expand+2%2B3%28x-1%29-3%28x-1%29%5E2%2B%28x-1%29%5E4

Answer 9

Thank you