How do you find the Maclaurin Series for #Sin(x^2)#?

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Answer 1 · 2017-11-12T10:35:12

#x^2 - x^6/(3!) + x^10/(5!) - ....#

#sum_(n=0 )^oo x^(4n+2)/((2n+1)!) * (-1)^n#

First we must find the series for #sin(x)#

let# f(x) = sin(x) #
#f(0) = sin(0) = 0#
#f'(0) = cos(0) = 1#
#f''(0) = -sin(0) = 0#
#f'''(0) = -cos(0) = -1#

Now we can apply to the macluarin series;

#f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f'''(0)x^3)/(3!) + ...#

Hence #sin(x) = x - x^3/(3!) + x^5/(5!) - ...#

Hence for #sin(x^2)# we replace each #x# by #x^2# in the series for #sin(x)#

#sin(x^2) = (x^2) - (x^2)^3/(3!) + (x^2)^5/(5!) - ...#

# = x^2 - x^6/(3!) + x^10/(5!) - ....#

What we can write in sigma summation notation as;

#sum_(n=0 )^oo x^(4n+2)/((2n+1)!) * (-1)^n#