Taylor Series
Taylor Series
Let's compute the derivatives at 
:
, so f(0)=1.
, so f'(0)=0.
, so 
.
, so f'''(0)=0.
, so 
.
Only the even order derivatives are non-zero (since 
is an even function!). We can model this for the general term by using the counter 2n instead of n (use 2n+1, if you want to catch odd terms!).
Since the power of the term 2 equals the order of the derivative, the general term has the form
What about the alternating sign? If n=0, we want to create a positive term, so let's use 
. For n=1, this produces the desired negative term (associated with the second derivative). Thus
This yields the Taylor series expansion
Use the ratio test to show that this series converges for all x.
Computing a Taylor series by successively taking derivatives is often not very practical, and should really be considered a last resort. An elegant way to derive Taylor series for functions is to use substitution.
Consider for example the function 
. It is not much fun to compute its derivatives. Instead we will use the formula for the geometric series:
We can rewrite
which suggests the substitution 
. Substituting in the series yields
Since the formula for the geometric series holds for |q|<1, the new series will converge for
Let's find the Taylor series for 
with center 
.
The tricky part is to rewrite this expression to exploit the geometric series
once again. Note that we need the term 
to show up in our substitution to end up with the right center! So, let's start by writing
In order to use the geometric series, we will have to "replace" the 3 in the denominator by 1. Here is how:
This suggests the substitution 
.
The Taylor series will represent the function as long as
Thus the series will work for -1<x<5.
This works for other series than the geometric series as well. If you know that
then you can easily compute the Taylor series for 
.
Set 
, to obtain
for
. Where does the Taylor series coincide with the function?
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