11/17 (late blog)

Sorry I am late, I just hate computers and avoid them at all cost :).

Taylor Polynomials and Approximations:

The form of a convergent power series:

"In this section you will study a general procedure for deriving the power series for a function that has derivatives of all orders. The following theorem gives the form that every convergent power series must take."

If f is represent by a power series f(x) = E an(x-c)^n for all x in an open interval l containing c, then an = f^(n)(c)/n! and
f(x) = f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +... .

Definition of Taylor and Maclaurin Series:

If a function f has derivatives of all orders at x = c, then the series
E(from n=0 to infinity) f^n(c)/n! * (x-c)^n = f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +...
is called the Taylor series for f(x) at c. Moreover, if c = 0, then the series is the Maclaurin series for f.

"If you know the pattern for the coefficients of the Taylor polynomials for a function, you can extend the pattern easily to form the corresponding Taylor series."

The convergence of a Tyalor series will always equal f^n(c)/n! * (x-c)^n if lim(as n -> infinity) Rn = 0.

Guidelines for Finding A Taylor Series:

1.) Differentiate f(x) several times and evaluate each derivative at c.
f(c), f'(c), f''(c), f'''(c), ... , f^n(c), ...

2.) Use the sequence developed in the first step to form the Taylor coefficients an = f^n(c)/n!, and determine the interval of convergence for the resulting power series
f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +... .

3.) Within the interval of convergence, determine whether the series converges to f(x).



Everyone should look at and put to memory the chart of page 684 about power series for elementary functions.