Taylor Series



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!

Taylor series and polynomials

What is the Taylor series of a polynomial?

The Taylor series with center tex2html_wrap_inline141 of a polynomial is the polynomial itself! (Check this by taking your favorite polynomial and use the formula for the Taylor series with center tex2html_wrap_inline141 !)

What if the center is not 0? Then one really rewrites the polynomial, using tex2html_wrap_inline143 instead of tex2html_wrap_inline145 as its "building blocks".

Here is an example: Find the Taylor series with center tex2html_wrap_inline147 of the polynomial tex2html_wrap_inline149 . Let's take its derivatives and plug in tex2html_wrap_inline147 :

tex2html_wrap_inline149 , 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.

Multiplying Taylor series by polynomials

Let's try to find the Taylor series expansion of tex2html_wrap_inline171 with center tex2html_wrap_inline173 .

To find the Taylor series for tex2html_wrap_inline175 , use that tex2html_wrap_inline177 . 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):


Try it yourself!

Not nearly as complicated: Find the Taylor series representation with center tex2html_wrap_inline141 for tex2html_wrap_inline185 .

Click here for the answer, or to continue.

Helmut Knaust
Tue Jul 16 12:07:49 MDT 1996

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