How do you find the Maclaurin Series for #sin(2x^2)#?
1 Answer
Jul 21, 2017
# sin(2x^2) = 2x^2-4/3 x^6+4/15 x^10-8/315 x^14+ ...#
Or in sigma notation:
# sin(2x^2) = sum_(k=0)^oo (-1)^k \ (2^(2k+1) \ x^(4k+2))/((2k+1)!) #
Explanation:
Starting with the Maclaurin Series for
# sinx = x-x^3/(3!)+x^5/(5!)-x^7/(7!) + x^9/(9!) -... \ \ x in RR #
Then the series for
# sin(2x^2) = (2x^2)-(2x^2)^3/(3!)+(2x^2)^5/(5!)-(2x^2)^7/(7!) + ...#
# " " = 2x^2-(2^3 x^6)/(3!)+(2^5 x^10)/(5!)-(2^7 x^14)/(7!) + ...#
# " " = 2x^2-8/6 x^6+32/120 x^10-128/5040 x^14+ ...#
# " " = 2x^2-4/3 x^6+4/15 x^10-8/315 x^14+ ...#
Or in sigma notation:
# sin(2x^2) = sum_(k=0)^oo (-1)^k \ (2^(2k+1) \ x^(4k+2))/((2k+1)!) #
