Question

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

Answer 1

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

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

Explanation:

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`