How do you find the Maclaurin Series for #f(x)= x / (1-x^4)#?
1 Answer
#x + x^5 + x^9 + . . . # .
A Maclaurin series is just the Taylor series expansion around
A Taylor series is written as:
#= sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n# where:
#f^((n))(a)# is the#n# th derivative of#f(x)# evaluated at#x = a# .#(x-a)# is the expansion point of the Taylor series. Note that the#x# in here is still#x# ; you do not have#(x - a) = (a - a) = 0# . You have#a = 0# , so#(x - a) = x# .#n!# is the factorial of#n# , i.e.#1*2*3cdots*(n-1)*n# .#n# increments from#0# to some arbitrarily large number, such that the series converges onto your desired function.
Since you will have to take many derivatives, let's just do them right now. We can try three of them, but even that's a lot, as there will be a lot of quotient rule and chain rule!
#color(green)(f^((0))(x)) = f(x) = color(green)(x/(1-x^4))#
#color(green)(f'(x)) = ((1-x^4)(1) - (x)(-4x^3))/(1-x^4)^2#
#= color(green)((1 + 3x^4)/(1-x^4)^2)#
And I'll skip the work for the second and third derivatives because you shouldn't need to do this on a test. I just did this in Wolfram Alpha in this case.
#color(green)(f''(x)) = ((1-x^4)^2(12x^3) - (1 + 3x^4)(2(1-x^4)*-4x^3))/(1-x^4)^4#
#[ . . . ] = color(green)((-12x^11 - 8x^7 + 20x^3)/(1-x^4)^4)#
#color(green)(f'''(x)) = ((1-x^4)^4(-132x^(10) - 56x^6 + 60x^2) - (-12x^11 - 8x^7 + 20x^3)(4(1-x^4)^3(-4x^3)))/(1-x^4)^8#
#[ . . . ] = color(green)((60x^26 + 24x^22 - 636x^18 + 1104x^14 - 636x^10 + 24x^6 + 60x^2)/(1-x^4)^8)#
Alright, and now the Taylor series centered around
#color(blue)(sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n)#
#= (f^((0))(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + . . . #
#= f(0) + f'(0)x + (f''(0))/(2)x^2 + (f'''(0))/(6)x^3 + . . . #
#= cancel(0/(1-0^4))^(0) + (1 + cancel(3(0)^4)^(0))/(1-cancel((0)^4)^(0))^2 x + cancel((-12(0)^11 - 8(0)^7 + 20(0)^3)/(1-(0)^4)^4)^(0) x^2 + cancel((60(0)^26 + 24(0)^22 - 636(0)^18 + 1104(0)^14 - 636(0)^10 + 24(0)^6 + 60(0)^2)/(1-(0)^4)^8)^(0) x^3 + . . . #
So, naturally, we needed to do more derivatives... Great. Anyways, you would (eventually) get:
#= color(blue)(x + x^5 + x^9 + . . . )#
