Question

How do you find the maclaurin series expansion of `f(x)=cos3x`?

Answer 1

Refer to explanation

Explanation:

A Maclaurin series is a Taylor series expansion of a function about 0,
hence

`f(x)=f(0)+f'(0)x+(f''(0))/(2!)*x^2+(f'''(0))/(3!)+...+(f^(n)(0))/(n!)*x^n`

where `f^(n)(0)` is the n-th order derivative of `f(x)`.

Hence we have to calculate some derivatives around zero so

`f(x)=cos3x=>f(0)=1`
`f'(x)=-3sin3x=>f'(0)=0`
`f''(x)=-3^2*cos3x=>f''(0)=-3^2`
`f'''(x)=3^3*sin3x=>f'''(0)=0`
`f''''(4)=3^4*cos3x=>f''''(0)=3^4`

So the maclaurin series can be written as

`cos3x=Σ_0^oo (-1)^n*(3)^(2n)*(x^(2n))/((2n)!)`