How do you find the Maclaurin Series for #(sinx cosx) / x#?
2 Answers
With much product rule and tedious derivative work, we'd get:
#1 - (2x^2)/3 + (2x^4)/15 - (4x^6)/(315) + . . . #
For this, it is encouraged that you find the derivatives first so that you do not mess them up as easily as you would if you also wrote them in the Maclaurin series at the same time.
I would also not recommend that you study this problem for a test (no, seriously).
ZERO ORDER DERIVATIVE
#color(green)(f^((0))(x) = f(x) = (sinxcosx)/x)#
FIRST ORDER DERIVATIVE
#color(green)(f'(x)) = (sinxcosx)(-1/x^2) + 1/x(-sin^2x + cos^2x)#
#= -(sinxcosx)/x^2 + (x(cos^2x - sin^2x))/x^2#
#= -1/x^2[sinxcosx - xcos^2x + xsin^2x]#
#= color(green)(-1/x^2[sinxcosx + x(sin^2x - cos^2x)])#
SECOND ORDER DERIVATIVE
#color(green)(f''(x)) = (sinxcosx - xcos^2x + xsin^2x)(2/x^3) - 1/x^2(cos^2x - sin^2x - (xcdot-2sinxcosx + cos^2x) + (xcdot2sinxcosx + sin^2x))#
#= (2sinxcosx - 2xcos^2x + 2xsin^2x)/x^3 - 1/x^2(cancel(cos^2x) - cancel(sin^2x) + 4xsinxcosx - cancel(cos^2x) + cancel(sin^2x))#
#= (2sinxcosx - 2xcos^2x + 2xsin^2x)/x^3 - (4xsinxcosx)/x^2#
#= ((2 - 4x^2)sinxcosx + 2x(sin^2x - cos^2x))/x^3#
#= color(green)(2/x^3[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)])#
THIRD ORDER DERIVATIVE
#color(green)(f'''(x)) = 2/x^3[(1 - 2x^2)(cos^2x - sin^2x) - 4xsinxcosx + x(2sinxcosx + 2sinxcosx) + sin^2x - cos^2x] - 6/x^4[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)]#
#= 2/x^3[cancel(cos^2x) - cancel(sin^2x) - 2x^2cos^2x + 2x^2sin^2x - cancel(4xsinxcosx) + cancel(4xsinxcosx) + cancel(sin^2x) - cancel(cos^2x)] - 6/x^4[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)]#
#= 1/x^4[4x^3(sin^2x - cos^2x)] - 6/x^4[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)]#
#= 6/x^4[2/3x^3(sin^2x - cos^2x)] - 6/x^4[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)]#
#= 6/x^4[2/3x^3(sin^2x - cos^2x) - (1 - 2x^2)sinxcosx - x(sin^2x - cos^2x)]#
#= color(green)(6/x^4[(2/3x^3 - x)(sin^2x - cos^2x) - (1 - 2x^2)sinxcosx])#
Since this was soooo long, here's a summary of the results:
#f^((n))(x)# :
#f^((0))(x) = (sinxcosx)/x# #f'(x) = -1/x^2[sinxcosx + x(sin^2x - cos^2x)]# #f''(x) = 2/x^3[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)]# #f'''(x) = 6/x^4[(2/3x^3 - x)(sin^2x - cos^2x) - (1 - 2x^2)sinxcosx]#
And FINALLY, using the formula for the Maclaurin series:
#sum_(n=0)^(N) (f^((n))(0))/(n!)x^n#
#= (f^((0))(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + . . . #
#= lim_(x->0)(sinxcosx)/x#
#+ (lim_(x->0) -1/x^2[sinxcosx + x(sin^2x - cos^2x)])x#
#+ 1/2(lim_(x->0) 2/x^3[(1 - 2x^2)sinxcosx + x(sin^2x - cos^2x)])x^2#
#+ 1/6(lim_(x->0) 6/x^4[(2/3x^3 - x)(sin^2x - cos^2x) - (1 - 2x^2)sinxcosx])x^3 + . . . #
... we're still not done! These limits are not readily evaluated, as they are all of the form
ZERO ORDER TERM
#lim_(x->0) f^((0))(x) = lim_(x->0) cos^2x - sin^2x = 1#
FIRST ORDER TERM
#[lim_(x->0) f'(x)]x#
#= [lim_(x->0) (sinxcosx + x(sin^2x - cos^2x))/(-x^2)]x#
#= [lim_(x->0) (cancel(cos^2x) - cancel(sin^2x) + x(4xsinxcosx) + cancel(sin^2x) - cancel(cos^2x))/(-2x)]x#
#= [lim_(x->0) ((4x^2)(cos^2x - sin^2x) + 8xsinxcosx)/(-2)]x = 0#
SECOND/THIRD ORDER TERM
You can forget about doing the second-order and third-order limits on a test. So I'll just use Wolfram Alpha to obtain:
#1/2[lim_(x->0) f''(x)]x^2 = 1/2 * -4/3 * x^2 = -(2x^2)/3#
#1/6[lim_(x->0) f'''(x)]x^3 = 0#
Thus, the Maclaurin series (that we could bother to figure out, anyway) is:
#= color(blue)(1 - (2x^2)/3 + . . . )#
Wolfram Alpha lists the fuller Maclaurin series as:
#= 1 - (2x^2)/3 + (2x^4)/15 - (4x^6)/(315) + . . . #
See explanation.
Explanation:
The Maclaurin series for 2 sin x cos x = sin 2x is
(sin x cos x)/x=(sin 2x)/(2x)
As
