Calculus - Taylor #2

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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) = x3, a = -1


and what i've done so far:

f (x) = x^3
f ' (x) = 3x^2
f '' (x) = 6x^1
f ''' (x) = 6x

f (-1) = -1
f ' (-1) = 3
f '' (-1) = -6
f ''' (-1) = -6

using taylor series equation.. my final answer that was wrong:

((-1(x+1)^0)/(0!))+((3(x+1)^1)/(1!))+((-6(x+1)^2)/(2!))+((-6(x+1)^3)/(3!))

.. is this what the question was asking for? if not, what is it then? thank you very much for your assistance.

f ''' (x) = 6

f'''(-1) = 6

  • Calculus - Taylor #2 -

    Looks good, assuming your arithmetic is ok.

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