Taylor Series.
Theorem 8.1.1
Let
be a function analytic at a point
. Denote by
the largest circle centered at
such that
is analytic at all the points interior to
, and let
be its radius. Then there exists a power series
which converges to
in
.
This series is unique; it is called the Taylor series of
at
. The cofficients of this series are determined by the following formula:
The Taylor series of a function about 0 is called the Maclaurin series
of
.
Example 8.1.3
Compute the Maclaurin series of
. Using 1.2,
we have:
We can use Taylor series in order to find limits:
Example 8.1.4
Let
. We wish to compute
.
Using 1.2,
we get:
Therefore
.
The following result is a consequence of Thm 3.5
and Thm 3.7.
Proposition 8.1.5
Let
be a function analytic on a neighborhood of
.
- We get the Taylor series expansion of
by differentiating term-by-term the Taylor series expansion of
.
- If the Taylor series expansion of
is known, we get the expansion of
by integrating term-by-term the Taylor series expansion of
(take care of the additive constant of integration!)
Example 8.1.6
In 1.2,
we saw that
By term-by-term differentiation, we have:
and this fits 1.2.
Example 8.1.7
Let
. The MacLaurin series of
is:
By integration term-by-term, we have:
Important remark: When we studied power series over
the reals we had a surprise: the convergence domain of a power series is not
always obvious.
Take
. This function is defined over
. Its first MacLaurin expansions are given by:
The graphs of
and of these approximations are displayed in Figure
1.
Figure 1: First approximations
of a given function.
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It seems that the visualization shows that the successive approximations tend
to the original function only for
. The condition for a geometric sequence to be convergent supports this
impression. First of all, the function is not defined at 1 (where it has a
singular point) and this point acts as a "barrier". But, does the power series
make sense out of the interval
? Actually not in our frame of study. Maybe in other frames.
Now take
. It is obtained by the substitution of
instead of
. The first MacLaurin expansions are given by:
The graphs of
and of these approximations are displayed in Figure
2.
Figure 2: First approximations
of a given function.
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We have here the same visual impression: the MacLaurin series tends to the
given function
for
. But for the function
, -1 and 1 are not points of discontinuity. So, what happens?
Passing to the complex setting, consider the function of the complex variable
given by
. It is defined over
. The corresponding MacLaurin series is given by
and is convergent for
in the open unit ball centered at the origin, i.e. on the largest ball centered
at 0 at not touching the two points where
fails to be defined
Figure 3: The largest ball for
the function to be defined.
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This example shows the importance of working in a complex setting. Without
exaggeration, we could say that the complex setting is "more natural" than the
real one.
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