How do you find the Maclaurin series of #f(x)=sin(x)#
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Question

Evelyn Agard · Accepted Answer

Maclaurin series of #f(x)=sin(x)# is
#sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}#. Maclaurin series for #f(x)# can be found by
#f(x)=sum_{n=0}^{infty}{f^{(n)}(0)}/{n!}x^n# So, we need to find derivatives of #f(x)=sin(x)#.
#f(x)=sin(x) Rightarrow f(0)=0#
#f'(x)=cos(x) Rightarrow f'(0)=1#
#f''(x)=-sin(x) Rightarrow f''(0)=0#
#f'''(x)=-cos(x) Rightarrow f'''(0)=-1#
#f^{(4)}(x)=sin(x) Rightarrow f^{(4)}(0)=0#
#cdots# Since #f^{(4)}(x)=f(x)#, the cycle of {0, 1, 0, -1} repeats itself, which means that every derivative of even degree gives #0# and that every derivative of odd degree alternates between 1 and -1.  So, we have
#f(x)={1}/{1!}x^1+{-1}/{3!}x^3+{1}/{5!}x^5+cdots#
#=sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}#

Emily Ammerman · Answer

undefined To find the Maclaurin series of $ f(x) = \sin(x) $, you can use the Taylor series expansion centered at $ x = 0 $. The Maclaurin series is a special case of the Taylor series expansion where the center is $ x = 0 $.

The Maclaurin series for $ \sin(x) $ is:

$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots $$

In general, the Maclaurin series for $ \sin(x) $ is the summation of odd powers of $ x $ divided by their respective factorial, alternating in sign starting with positive.

How do you find the Maclaurin series of #f(x)=sin(x)# ?