How do you find the Maclaurin Series for #(sinx cosx) / x#?

Answer 1

See explanation.

For 2 sin x cos x = sin 2x, the Maclaurin series is

#(2x)-(2x)^3/(3!)+(2x)^5/(5!)-...+(-1)^(n-1)(2x)^(2n-1)/(((2n-1)!)+..., #
#x in (-oo, oo)#.
#=2x (1-(2x)^3/(3!)+(2x)^4/(5!)-...+(-1)^(n-1)(2x)^(2n-2)/(((2n-1)!)+..., #
#x in (-oo, oo)#. So, sans x = 0,

(sin 2x)/(2x) = (sin x cos x)/x

#=1-(2x)^3/(3!)+(2x)^4/(5!)-...+(-1)^(n-1)((2x)^(2n-2)/(((2n-1)!)))+..., #
#x in (-oo, 0) U (0, oo)#.
As #x to 0, (sin x cos x)/x= (sin x/x) (cos x) to (1)(1)=1.#
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Answer 2

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)#:

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 #0/0#, so we have to evaluate using L'Hopital's rule...

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) + . . . #
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Answer 3

To find the Maclaurin series for (sin(x) cos(x)) / x, you can start by expressing sin(x) and cos(x) as their Maclaurin series expansions. Then, multiply them together, and finally divide by x.

The Maclaurin series expansions for sin(x) and cos(x) are:

sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ... cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

To find the Maclaurin series for (sin(x) cos(x)) / x, multiply these two series together:

(sin(x) cos(x)) / x = (x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...) * (1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...) / x

Then, divide each term by x:

= (1 - (x^2)/3! + (x^4)/5! - (x^6)/7! + ...) * (1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...)

Now, expand the product of these two series, keeping only the terms up to the desired degree.The Maclaurin series for ((\sin(x) \cdot \cos(x))/x) is:

[1 - \frac{x^2}{6} - \frac{x^4}{30} - \frac{x^6}{140} - \frac{x^8}{630} - \frac{x^{10}}{2772} - \cdots]

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Answer 4

To find the Maclaurin series for (\frac{\sin(x)\cos(x)}{x}), you can use the Maclaurin series expansions for (\sin(x)) and (\cos(x)), and then perform the necessary algebraic manipulations. The Maclaurin series expansions for (\sin(x)) and (\cos(x)) are:

[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ] [ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots ]

Then, multiply the series expansions for (\sin(x)) and (\cos(x)), and divide by (x). After performing the multiplication and division, simplify the resulting series to obtain the Maclaurin series for (\frac{\sin(x)\cos(x)}{x}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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