Problem : Compute the Taylor polynomial at 0 of degree 3 for f (x) = sin(ex).

The first few derivatives are


f'(x)=cos(ex)ex  
f''(x)=cos(ex)ex - sin(ex)e2x  
f'''(x)=f''(x) - [sin(ex)2e2x + cos(ex)e3x]  

Substituting 0 for x yields f (0) = sin(1), f'(0) = cos(1), f''(0) = cos(1) - sin(1), f'''(0) = - 3 sin(1), so the desired Taylor polynomial is:

sin(1) + cos(1)x + + x3    

Problem : Approximate log(0.8) using the Taylor polynomial of degree 3 at x = 1 for log(x).

The desired Taylor polynomial is

(x - 1) - +    

Substituting 0.8 for x, we have log(0.8) -0.222.

Problem : Find the Taylor series for sin(x) + cos(x).

Adding term-by-term, we have

1 + x - - + + ...