Representations of Functions as Power series
Why do we want to express a known function as the sum of a power series ?
We will see later that this is a good strategy for integrating functions that don't have elementary antiderivatives (
or
for examples), for solving differential equations, and for approximating functions by polynomials. Scientists do this to simplify the expressions they deal with; computer scientists do this to represent functions on calculators and computers.
We now know that for
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Moreover, we can compute the error of approximating
by the polynomial
= 1 + x + ... +
, using the following formula
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Clearly we can see from this formula and the animation below that
(1). For fixed
, the closer
is to
, the better
approximates
.
(2). For fixed
, the larger
is, the better
approximates
.
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We can use
------- (1)
to obtain power series representations for some other functions.
Algebraic manipulations ---- put the functions in the form of
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Example 1 Express
as the sum of a power series and find the intervals of convergence.
Replacing
by
in Equation (1), we have
=
=
=
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for
, that is,
. Therefore, the interval of convergence is (
). Of course, we can also determined the radius of convergence by ratio test, but that is unnecessary here. Because,
if and only if
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Play with the animation below and observe carefully.
Note that the function
is defined for all
, but the power series does not converge for
.
Example 2 Find a power series representation for
.
Put this function in the form of
as follows
=
=
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=
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=
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This series converges if and only if
, that is,
. So the interval of convergence is (
).
The animation below shows the convergence :
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Example 3 Find a power series representation for
.
Since
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, all we have to do is to multiply the series by
.
=
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![]()
=
![]()
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=
![]()
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=
+ .....
The interval of convergence is also (
).
The animation below shows how the series converges :
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>
Differentiation and Integration
For
in the interval of convergence, we can differentiate or integrate a power series just as we would do for the polynomials ---- differentiate or integrate term-by-term .
Theorem If the power series
has radius of convergence
, then the function
defined by
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is differentiable (and therefore continuous) on the interval (
) and
(i) f ' ( x ) =
=
+ ...
(ii)
=
+ ...
The radii of convergence of the power series in Equations (i) and (ii) are both
.
We called this differentiate or integrate term-by-term because that Equations (i) and (ii) can be rewritten in the form :
(iii)
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(iv)
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Example 4 Express
as the sum of a power series and find the interval of convergence.
Note that
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. Differentiating both sides of the equation
, we get
=
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According to the theorem, the radius of convergence of the differentiated series is the same as the original series, namely,
. Since
diverges at
and
(take a good look at the animations below), we get that the interval of convergence of the series is (
).
The convergence for x in (
) :
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Example 5 Find a power series representation for
and find the interval of convergence.
Note that
, integrating both sides of
, we get
=
=
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According to the theorem, the radius of convergence of the series is also equal to
. However, the series
converges at
and diverges at
. Therefore, the interval of convergence of the series is [
).
The convergence for x in (
) :
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The convergence for
:
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We have shown that
for all
in (
). From the animation above, we would conjecture that
for
, that is,
=
+.... .
In fact, it is true indeed. Since
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by integrating both sides of the equation above, we have
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put
, we get
=
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Since
and
for all
in
,
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=
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Therefore,
approaches
as
approaches
. This implies
=
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That is,
=
+ ......
Example 6 Find a power series representation for
and find the interval of convergence.
In Example 1, We have shown
=
for all
. We get
=
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=
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=
+ .......
for all
.
The convergence for x in (
) :
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Notice that the interval of convergence of the series
is [
].
We can show, as in Example 5, that
for all values of
in [
].
In particular, for
we get a formula to express
as an infinite series, namely
=
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that is,
=
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=
+ ......
Moreover, by the Alternating Series Estimation Theorem, we get
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This gives us a good way to approximate
. If we want to find an approximation of
to 4 decimal places, we need to take
to be greater than
; we may take
, and the corresponding partial sum is about
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Not bad, right ?
Example 7 Approximate
correct to within
.
We first express the integrand,
, as the sum of a power series.
=
=
=
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This series converges for
, that is, for
.
Since
, now we can integrate term by term :
=
=
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By the Alternating Series Estimation Theorem, we get
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In order that the approximation is correct to within
, we need to take
so large that
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:
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:
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It suffices to take
, that is,
is approximately equal to
which is about
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Power series center at
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Example 8 Find a power series representation center at 1 for
and find the radius of convergence.
Since
, we can find the require series by integrating the power series center at 1 for
.
Notice that for
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=
=
=
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Therefore,
=
=
for
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The radius of convergence of the required series is
.
The convergence for
:
Note that the interval of convergence of the series
is (
].
The convergence for
:
We can show, as in Example 5, that
for all
in (
].
Another way of doing this is to view
as a translation of
by
. In Example 5, we have shown that for all
in (
]
=
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Thus, for all
such that
<
, that is for all
in (
],
=
=
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=
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=
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=
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Example 9 Find a power series representation center at 2 for
and find the radius of convergence.
Since
, we can find the require series by integrating the power series center at 2 for
.
As in Example 8, we have
=
=
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=
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=
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=
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Notice that the above equations hold for
, that is,
.
Therefore, for
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=
=
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that is,
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The radius of convergence of the required series is
.
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If
is a power series representation center at
for
, what do you think the radius of convergence of
should be ?
From the examples above, we have the following observations :
If function has a power series representation
, then
(1). For fixed
, the closer
is to
, the better
approximates
.
(2). For fixed
, the larger
is, the better
approximates
.