How do you find maclaurin series for f(x) = x^2 ln(1 + x^3)?
1 Answer
Well, I only went to the fifth derivative (actually, I went to the third derivative, and used Wolfram Alpha on the rest), but this is just a bad problem.
It should be:
#f(x) = x^5 - x^8/2 + x^11/3 - x^14/4 + x^17/5 - . . . #
but it would never be doable on an exam.
You would find it just like a Taylor series, but centered at
A Taylor series is given by
#f(x) = sum_(n=0)^(oo) (f^((n))(a))/(n!) (x-a)^(n)#
and so a Maclaurin series is given by
#f(x) = sum_(n=0)^(oo) (f^((n))(0))/(n!) x^n#
That means you'll need to take some derivatives.
#color(green)(f^((0))(x) = f(x) = x^2ln(1+x^3))#
#f'(x) = x^2 cdot (3x^2)/(1 + x^3) + 2xln(1+x^3)#
#= color(green)((3x^4)/(1 + x^3) + 2xln(1+x^3))#
#color(green)(f''(x)) = 3 cdot [x^4 cdot -1/(1+x^3)^2 cdot 3x^2 + (4x^3)/(1 + x^3)] + 2x cdot (3x^2)/(1 + x^3) + 2ln(1+x^3)#
#= 3 cdot [-(3x^6)/(1+x^3)^2 + (4x^3)/(1 + x^3)] + (6x^3)/(1 + x^3) + 2ln(1+x^3)#
#= -(9x^6)/(1+x^3)^2 + (18x^3)/(1 + x^3) + 2ln(1+x^3)#
#= (-9x^6 + (18x^3)(1 + x^3))/(1+x^3)^2 + 2ln(1+x^3)#
#= color(green)((9x^6 + 18x^3)/(1+x^3)^2 + 2ln(1+x^3))#
And I'll leave it as an exercise to verify the third, fourth, and fifth derivatives.
#color(green)(f'''(x) = (6x^2(x^6 + 2x^3 + 10))/(1 + x^3)^3)#
#color(green)(f^((4))(x) = -(6x(x^9 + 60x^3 - 20))/(1 + x^3)^4)#
#color(green)(f^((5))(x) = (12(x^12 - 5x^9 + 240x^6 - 230x^3 + 10))/(1 + x^3)^5)#
We find:
#f(x) = (f^((0))(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + . . . #
#= cancel((0^2ln(1+0^3))/(1) x^0)^(0) + cancel(((0^4)/(1 + 0^3) + 0*ln(1+0^3))/(1)x^1)^(0) + cancel(((0^6 + 0^3)/(1+0^3)^2 + 2ln(1+0^3))/(2)x^2)^(0) + cancel(((0^6 + 0^3)/(1+0^3)^2 + 2ln(1+0^3))/(6)x^3)^(0) + cancel((-(0(0^9 + 0^3 - 20))/(1 + 0^3)^4)/24 x^4)^(0) + ((12(0^12 - 0^9 + 0^6 - 0^3 + 10))/(1 + 0^3)^5)/120 x^5 + . . . #
#= color(blue)(x^5 - x^8/2 + x^11/3 - x^14/4 + x^17/5 - . . . )#
It sure takes a while to overcome those zero terms... I didn't want to bother with the sixth derivative and on.
