How can you find the taylor expansion of #f(x) =sinx# about x=0?

Question

Isabella Ackerman · Accepted Answer

See the explanation.

#f(x-x_0)=sum_(k=0)^n f^((k))(x_0) (x-x_0)^k/(k!) + f^((k+1))(epsilon)(x-x_0)^(k+1)/((k+1)!)# In our case #x_0=0# so: #f(x)=sum_(k=0)^n f^((k))(0) x^k/(k!) + f^((k+1))(epsilon)x^(k+1)/((k+1)!)# #f(x)=sinx# #f(0)=0# #f'(x)=cosx => f'(0)=1# #f''(x)=-sinx => f''(0)=0# #f^((3))(x)=-cosx => f^((3))(0)=-1# #f^((4))(x)=sinx => f^((4))(0)=0# and so on.... #sinx= 0 + 1*x/(1!) + 0 - 1*x^3/(3!) + 0 + 5*x/(5!) +0 - 1*x^7/(7!)...# #sinx = sum_(k=0)^oo (-1)^k x^(2k+1)/((2k+1)!)#

Emily Ammerman · Answer

undefined To find the Taylor expansion of $ f(x) = \sin(x) $ about $ x = 0 $, we can use the formula for the Taylor series expansion:

$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dotsb $$

For the function $ f(x) = \sin(x) $, the derivatives at $ x = 0 $ are as follows:

- $ f(0) = \sin(0) = 0 $
- $ f'(0) = \cos(0) = 1 $
- $ f''(0) = -\sin(0) = 0 $
- $ f'''(0) = -\cos(0) = -1 $

Substitute these values into the Taylor series expansion formula:

$$ \sin(x) = 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} - 1 \cdot \frac{x^3}{3!} + \dotsb $$

Simplify the terms:

$$ \sin(x) = x - \frac{x^3}{3!} + \dotsb $$

Thus, the Taylor expansion of $ f(x) = \sin(x) $ about $ x = 0 $ is:

$$ \sin(x) = x - \frac{x^3}{6} + \dotsb $$