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
Moreover, we can compute the error of approximating by the polynomial = 1 + x + ... + , using the following formula
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 .
We can use
------- (1)
to obtain power series representations for some other functions.
Algebraic manipulations ---- put the functions in the form of
Example 1 Express as the sum of a power series and find the intervals of convergence.
Replacing by in Equation (1), we have
= = =
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
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
= =
= =
This series converges if and only if , that is, . So the interval of convergence is ( ).
The animation below shows the convergence :
Example 3 Find a power series representation for .
Since , all we have to do is to multiply the series by .
=
=
=
= + .....
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
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)
(iv)
Example 4 Express as the sum of a power series and find the interval of convergence.
Note that . Differentiating both sides of the equation , we get
=
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 ( ) :
Example 5 Find a power series representation for and find the interval of convergence.
Note that , integrating both sides of , we get
= =
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 ( ) :
The convergence for :
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
by integrating both sides of the equation above, we have
put , we get
=
Since and for all in ,
=
Therefore, approaches as approaches . This implies
=
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
=
=
= + .......
for all .
The convergence for x in ( ) :
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
=
that is,
= = + ......
Moreover, by the Alternating Series Estimation Theorem, we get
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
Not bad, right ?
Example 7 Approximate correct to within .
We first express the integrand, , as the sum of a power series.
= = =
This series converges for , that is, for .
Since , now we can integrate term by term :
= =
By the Alternating Series Estimation Theorem, we get
In order that the approximation is correct to within , we need to take so large that
:
:
It suffices to take , that is, is approximately equal to which is about
Power series center at
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
= = =
Therefore,
= = for
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 ( ]
=
Thus, for all such that < , that is for all in ( ],
= =
=
=
=
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
= =
=
=
=
Notice that the above equations hold for , that is, .
Therefore, for
= =
that is,
The radius of convergence of the required series is .
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 .