How do you find the first three terms of the Taylor's series for #sin(x^2) #?

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Answer 1 · 2016-12-29T10:43:31

#sin(x^2) ~= x^2-x^6/6+x^10/120#

We take the MacLaurin series for #sint#:

#sint = sum_(n=0)^oo (-1)^n(t^(2n+1))/((2n+1)!)#

We substitute #t=x^2#

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

and we take the terms for #n=0,1,2#:

#sin(x^2) ~= x^2-x^6/6+x^10/120#

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