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
Looks good, assuming your arithmetic is ok.
To find the Taylor series for f(x) centered at a = -1, we need to compute the derivatives of f(x) at x = -1, and then use these values to construct the series.
First, let's compute the derivatives of f(x):
f(x) = x^3
f'(x) = 3x^2
f''(x) = 6x
f'''(x) = 6
Now, let's evaluate these derivatives at x = -1:
f(-1) = (-1)^3 = -1
f'(-1) = 3(-1)^2 = 3
f''(-1) = 6(-1) = -6
f'''(-1) = 6
So, the values of the derivatives at x = -1 are:
f(-1) = -1
f'(-1) = 3
f''(-1) = -6
f'''(-1) = 6
Now we can construct the Taylor series centered at -1 using the derivatives and the terms of the series:
f(x) = f(-1) + f'(-1)(x - (-1))^1/1! + f''(-1)(x - (-1))^2/2! + f'''(-1)(x - (-1))^3/3! + ...
Substituting the values of the derivatives at x = -1, we have:
f(x) = -1 + 3(x + 1) + (-6(x + 1)^2)/2! + 6(x + 1)^3/3! + ...
Simplifying this expression:
f(x) = -1 + 3(x + 1) - 3(x + 1)^2 + 2(x + 1)^3 + ...
So, the Taylor series for f(x) centered at a = -1 is:
f(x) = -1 + 3(x + 1) - 3(x + 1)^2 + 2(x + 1)^3 + ...
Note that in this response, I did not show the error term Rn(x) approaching zero because you mentioned not to include it. However, keep in mind that the Taylor series approximation is only valid within a certain range around the center value.