How do you find the maclaurin series expansion of # f(x) = (1-cosx)/x#?

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Answer 1 · 2016-01-01T02:09:47

Start from the Maclaurin series for #cos(x)# and derive:

#(1 - cos(x))/x = sum_(k=0)^oo (-1)^k/((2k+2)!) x^(2k+1) = x/(2!) - x^3/(4!) + ...#

Note that:

#d/(dx) cos(x) = -sin(x)#

#d/(dx) sin(x) = cos(x)#

#sin(0) = 0#

#cos(0) = 1#

Hence the Maclaurin series for #cos(x)# is:

#cos(x) = sum_(k=0)^oo (-1)^k/((2k)!) x^(2k) = 1 - x^2/(2!) + x^4/(4!) -...#

So:

#1 - cos(x) = sum_(k=0)^oo (-1)^k/((2k+2)!) x^(2k+2) = x^2/(2!) - x^4/(4!) + ...#

Hence:

#(1 - cos(x))/x = sum_(k=0)^oo (-1)^k/((2k+2)!) x^(2k+1) = x/(2!) - x^3/(4!) + ...#