Question

How do you find the Maclaurin Series for `sin(2x^2)`?

Answer 1

`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`,

`sinx = x-x^3/(3!)+x^5/(5!)-x^7/(7!) + x^9/(9!) -... \ \ x in RR`

Then the series for `sin(2x^2)` is:

`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)!)`