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