Taylor Series


To accommodate the center, we rewrite

displaymath89

Next we use the geometric series with tex2html_wrap_inline93 :

displaymath90

The series will converge for -1<x<11.


Taylor series and differentiation

It is easy to take derivatives of Taylor series: Just take the derivative term-by-term. The radius of convergence of the derivative will be the same as that of the original series.

This can be exploited to find Taylor series! Consider the example tex2html_wrap_inline103 . Its Taylor series has the form

displaymath95

We can then find the Taylor series with center tex2html_wrap_inline105 of its derivative:

displaymath96

Since the constant term has derivative 0, the summation starts at n=1. We then just take the derivative of the general term:

displaymath97

N.B. Dividing both sides by -2x yields the maybe more interesting formula

displaymath98


Try it yourself!

Find the Taylor series with center tex2html_wrap_inline105 for the hyperbolic cosine function tex2html_wrap_inline115 by using the fact that tex2html_wrap_inline117 is the derivative of the hyperbolic sine function tex2html_wrap_inline119 , which has as its Taylor series expansion

displaymath111

(If you remember the Taylor expansions for tex2html_wrap_inline121 and tex2html_wrap_inline123 , you get an indication, why their hyperbolic counterparts might deserve the names "sine" and "cosine". )

Click here for the answer, or to continue.


Helmut Knaust
Tue Jul 16 11:25:05 MDT 1996

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