How do you find the Maclaurin Series for #f(x) = (x-sin x)/ x^3 #?

Question

Emily Ammerman · Accepted Answer

#f(x) = (x-sin x)/ x^3=sum_(k=0)^oo(-1)^kc_k x^(2k)#
with
#c_k=1/2 ((2 (k + 1) (2 k + 1))/((2 k + 3)!) - 1/((2 k + 1)!) + 2/((2 k + 2)!))# #sinx=sum_(k=0)^oo(-1)^kx^(2k+1)/((2k+1)!)# and #cosx=sum_(k=0)^oo(-1)^kx^(2k)/((2k)!)# Also #d^2/dx^2((x-sinx)/x)= (2 (x - sinx))/x^3 + sinx/x-(2 (1 - cosx))/x^2# so # (x-sin x)/ x^3 =1/2(d^2/dx^2((x-sinx)/x)- sinx/x+(2 (1 - cosx))/x^2)# Now, #(x-sinx)/x=1-sum_(k=0)^oo (-1)^kx^(2k)/((2k+1)!)# then #d^2/dx^2((x-sinx)/x)=-sum_(k=2)^oo(-1)^k(2k(2k-1))/((2k+1)!)x^(2k-2)= -sum_(k=0)^oo(-1)^k((2k+2)(2k+1))/((2k+3)!)x^(2k)# #sinx/x = sum_(k=0)^oo (-1)^kx^(2k)/((2k+1)!)# #(1-cosx)/x^2=-sum_(k=1)^oo(-1)^kx^(2(k-1))/((2k)!)=-sum_(k=0)^oo(-1)^kx^(2k)/((2(k+1))!)# Ultimately defining #c_k=1/2 ((2 (k + 1) (2 k + 1))/((2 k + 3)!) - 1/((2 k + 1)!) + 2/((2 k + 2)!))# we have #f(x) = (x-sin x)/ x^3=sum_(k=0)^oo(-1)^kc_k x^(2k)# The initial coefficients are: #c_0=1/6# #c_1=1/120# #c_3=1/5040# #c_4=1/362880#

Emilia Aguilar · Answer

undefined To find the Maclaurin series for $ f(x) = \frac{x - \sin(x)}{x^3} $, we can use the Taylor series expansion. The Maclaurin series is a special case of the Taylor series where the expansion is centered at $ x = 0 $.

1. First, let's find the derivatives of $ f(x) $ up to the desired order.

$$ f(x) = \frac{x - \sin(x)}{x^3} $$

2. Compute the first few derivatives of $ f(x) $:

$$ f'(x) = \frac{3x^2 - x\cos(x) - \sin(x)}{x^4} $$

$$ f''(x) = \frac{6x^3 - 6x\cos(x) - 3x^2\sin(x) - 2\cos(x)}{x^5} $$

$$ f'''(x) = \frac{24x^4 - 18x^2\sin(x) - 12x\cos(x) + 6x^3\cos(x) - 6x^2\sin(x)}{x^6} $$

3. Evaluate each derivative at $ x = 0 $ to find the coefficients:

$$ f(0) = 0 $$

$$ f'(0) = \frac{1}{3} $$

$$ f''(0) = -\frac{1}{3} $$

$$ f'''(0) = \frac{1}{30} $$

4. Write the Maclaurin series using the coefficients found:

$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots $$

Substitute the coefficients:

$$ f(x) = 0 + \frac{1}{3}x - \frac{1}{3 \times 2!}x^2 + \frac{1}{30 \times 3!}x^3 + \cdots $$

This gives us the Maclaurin series for $ f(x) $.