The Taylor series for f about c can be written as follows.
f(x) = f(c) + f'(c)(x-c) +{ f''(c)(x-c)2}/2! + {f'''(c)(x-c)3}/3! + ... + { fn(c)(x-c)n}/n! + ...
When c = 0, the Taylor Series is equivalent to Maclaurin Series.
Example:
Use Taylor Series to expand ex around 1.
Solution:
f(x) = ex and f(1) =e
f'(x) = ex and f'(1)=e
f''(x) = ex and f''(1) =e and so on.
Therefore, f(x) = e + e(x-1) + {e(x-1)2}/2! + {e(x-1)3}/3! + ...
This message was edited by rbraxton on 2003-02-18 12:56