How do you find the Maclaurin Series for #f(x)=sin (2x) #?

1 Answer
Oct 12, 2016

# f(x) = 2 x - 4/3 x^3 + 32/125x^5 ... #

Explanation:

A Maclaurin series can be expressed in the following way:

# f(x) = f(0) + f^(1)(0)/(1!) x + f^(2)(0)/(2!) x^2 + f^(3)(0)/(3!) x^3 + f^(4)(0)/(4!) x^4 + ...#
# = sum_(n=0)^(∞) f^(n)(0)/(n!) x^n#

Now, # f(x) = sin(2x) #
# :. f'(x) =2cos(2x) #
# :. f''(x) =-4sin(2x) #
# :. f'''(x) =-8cos(2x) #
# :. f''''(x) =16sin(2x) #
# :. f''''(x) =32cos(2x) #
# :. f'''''(x) =-64sin(2x) #
# ... #

and so when #x=0# we have:

# f(0) = sin0 = 0 #
# f^1(0) = 2cos0 = 2 #
# f^2(0) = -4sin0 = 0 #
# f^3(0) = -8cos0 = -8 #
# f^4(0) = 16sin0 = 0 #
# f^5(0) = 32cos0 = 32 #
# f^6(0) = 64sin0 = 0 #
# ... #

So the Maclaurin series is:

# :. f(x) = 0 + 2/(1!) x + 0 x^2/(2!) -8x^3/(3!) + 0x^4/(4!) +32x^5/(5!) + ... #

# :. f(x) = 0 + 2/1 x + 0 -8/6 x^3 + 0+32/125x^5 + ... #

# :. f(x) = 2 x - 4/3 x^3 + 32/125x^5 ... #