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

1 Answer
Oct 25, 2016

#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)!))#

Explanation:

#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))!)#

Finally 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 first coefficients are:

#c_0=1/6#
#c_1=1/120#
#c_3=1/5040#
#c_4=1/362880#