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

1 Answer

Aug 8, 2016

#(sinx-x)/x^3 = sum_{k=1}^oo(-1)^k (x^{2k-2})/((2k+1)!)#

Explanation:

The series expansion for #sin x# is given for

#sin x = sum_{k=0}^oo(-1)^k (x^{2k+1})/((2k+1)!) =#
#= x -1/(3!)x^3+1/(5!) x^5 -1/(7!) x^7 + cdots#

then

#(sinx-x)/x^3 = sum_{k=1}^oo(-1)^k (x^{2k-2})/((2k+1)!)#

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