its integral has the power series representation
Let's find the limit of the absolute ratios of this series:
But note that this is exactly the same expression we run into when we compute the limit of the absolute ratios of the Taylor series of f(x). Thus both series have the same radius of convergence!
The Taylor series with center of a polynomial is the polynomial itself! (Check this by taking your favorite polynomial and use the formula for the Taylor series with center !)
What if the center is not 0? Then one really rewrites the polynomial, using instead of as its "building blocks".
Here is an example: Find the Taylor series with center of the polynomial . Let's take its derivatives and plug in :
, so p(-1)=1+3+5=9.
p'(x)=2x-3, so p'(-1)=-2-3=-5.
p''(x)=2, so p''(-1)=2.
All higher derivatives are 0, so the Taylor series terminates!
Using our formula we obtain:
From this it is not hard to see that a function has a terminating Taylor series, if and only if the function is a polynomial.
Let's try to find the Taylor series expansion of with center .
To find the Taylor series for , use that . You obtain the expansion:
As in the example above we can rewrite
Next we multiply both expressions:
Without getting into the technical details: If you write out the first terms of these series, you can combine corresponding powers of (x-1), to obtain the beginning of the Taylor series for f(x):
Click here for the answer, or to continue.