Home > Calculus > Constructing a Taylor Series

What is the Maclaurin series of #f(x) = cos(x)#?

Answer 1

A Maclaurin series is simply the Taylor series centered around #a = 0# specifically.

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

Thus, we need to take derivatives until we see a unique pattern.

#f^((0))(x) = color(green)(f(x) = cosx)# #color(green)(f'(x) = -sinx)# #color(green)(f''(x) = -cosx)# #color(green)(f'''(x) = sinx)# #color(green)(f''''(x) = cosx)#

So stopping at #n = 3# is fine. We can still figure out the rest by the pattern. We get:

#= (cos(0))/(0!)x^0 + cancel((-sin(0))/(1!)x^1)^(0) +(-cos(0))/(2!)x^2 + cancel((sin(0))/(3!)x^3)^(0) + ...#

#= 1/(0!) - x^2/(2!) + x^4/(4!) - x^6/(6!) + ...#

#color(blue)(= 1 - x^2/2 + x^4/(24) - x^6/(720) + ...)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

The Maclaurin series expansion of ( f(x) = \cos(x) ) is given by:

[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} ]

This series represents the cosine function centered at ( x = 0 ), also known as the Maclaurin series, where each term is derived from the function's derivatives evaluated at ( x = 0 ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.