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

1 Answer

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)!)#