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XI.A
Definitions, Examples and Key Theorems for Power Series.
XI.A.1
The
Interval
and
Radius
of Convergence.
As we saw in the last chapter if we consider the convergence of an
infinite series involving powers of $x$ it is sometimes possible to
determine all those values of $x$ for which the series converges.
For example, ` sum _{k=0}^{oo} x^k` converges for all $x$ in the
interval `(-1,1)` and diverges for all $x$ outside that interval.
Similarly, by a simple substitution we can see that ` sum
_{k=0}^{oo} (x-3)^k` converges for all $x$ in the interval `(2,4)` and
diverges
for all $x$ outside that interval. This situation is typical of the
behavior of infinite series involving powers of `x`. This feature of
power series is summarized as follows: Theorem
XI.A.1.
The
series
` sum _{k=0}^{oo} c_k x^k` converges
for all `x` in an interval I and diverges for all x outside of that
interval.
In fact if the interval I is bounded
it will have endpoints
$-r$ and $r$ for some real number $r$, called the radius of convergence
for the series .
Notice that in this case the interval I has length $2r$. If the
interval I is unbounded then the radius of convergence for the series
is said to be infinite.
Preface to proof: In the proof we will actually show that if the
series
converges when $x = x_0$ with $s = | x_0 | > 0$ then it converges
absolutely for all $x$ in the interval $(-s,s)$ by comparing the series
of absolute values using `x` to a geometric series. So, here we go. Proof:
Since the series `\sum _{k=0}^\infty \ c_k \ x_0^k ` converges, we can
apply the divergence test to conclude that `c_k x_0^k -> 0 `
as `k -> oo`.
Thus there is some number $B$ that is larger
than ` | c_k x_0^k | ` for any $k$.
Now suppose that $-s < x < s$
so that for some $b>0$, $ |x | < b < s $. Then
Now this last
inequality allows us to compare the series of absolute values to a
geometric series:
`|c_k \ x^k | < B {b^k}/{s^k}`.
Since $0 < b<s$, `0< b/s < 1` so the geometric series ` sum
_{k=0}^{oo} B (b/s)^k`
converges and therefore by comparison the series ` sum _{k=0}^{oo}
c_k x^k`
converges absolutely for any $x$ in the interval $(-s,s)$. EOP Comments:
1. By a simple substitution, we can
state a corresponding result for the power series ` sum _{k=0}^{oo}
c_k (x-a)^k` when the series converges for $x_0$. The interval in
that case will be $(a-r,a+r)$ where $r = |x_0 - a |$.
2. In many cases the radius of convergence is infinite,
and the series converges for all `x`. In these cases the interval of
convergence is said to be the interval $(-\infty, \infty)$.
Example
XI.A.1. Determine the interval of convergence for
the power series` sum
_{k=0}^{oo}{x^{2k}}/{(2k)!} = 1 + {x^2}/2 + {x^4}/{4!} + ... ` Solution. Using the ratio test
we consider the ratios
As $k \rightarrow \infty$ we see that these
ratios have a limit of $0$ for any $x$. So the series will converge for
any $x$ and the interval of convergence is $(-\infty, \infty)$. In many cases the radius of
convergence for a power
series is determined using the Ratio Test. The following example
illustrates the fact that in determining the interval of convergence
for a power series, it is important
to consider separately the behavior
of the series at the the endpoints of an interval of convergence when
the radius of convergence is finite.
Example XI.A.2.
Determine the interval of convergence for the power series ` sum
_{k=0}^{oo} {x^k}/{k+1} = 1 + x/2 + {x^2}/3 + ... `. Solution. Using the ratio
test we consider the ratios
As $k
\rightarrow \infty$ we see that these ratios have a limit of $| x |$.
So the series will converge for $-1
< x <1$ and diverge
for $| x | > 1$.
This leaves the endpoints of the interval
$(-1,1)$ still to be considered.
When $x = 1$ the series is ` sum _{k=0}^{oo} {1}/{k+1} = 1 + 1/2
+ 1/3 + ... ` , the harmonic series. So the series
diverges when x = 1.
However when $x=-1$ the series is ` sum _{k=0}^{oo}
{(-1)^k}/{k+1} = 1 - 1/2 + 1/3 + ... `. This is an alternating
harmonic series
that converges ( as we have seen in X.B ).
Therefore the interval of
convergence for the series ` sum _{k=0}^{oo} {x^k}/{k+1}
` is `[-1,1)`.
XI.A.2
Functions
and
power
series.
In the last section we found that any
power series `\sum_{k=0}^\infty c_k
(x-a)^k` will converge for all
values of $x$ inside an interval.
So for each $x$ in the interval of convergence, the
power series gives us a number, namely the value of the series for that
$x$.
That is for each $x$ in the interval
of convergence we can define
a function $P$ at that $x$ by `P(x) =
\sum_{k=0}^\infty c_k (x-a)^k`. Some immediate questions arise
concerning this function $P$:
Is $P$ a differentiable function?
If so, what is the derivative
of $P$?
Is $P$ a `C^{oo}`
function?
If so, what are the Taylor
polynomials and the Taylor series for $P$?
What is
the relation of functions defined by power series to the problem of
solving differential equations?
For now I'll give you some of the
answers to these questions. These should help put the calculus of power
series into the big picture of solving differential equations and
estimation. To make the implications of the theory
more concrete, we will consider an example or two
in this and the next section. You can then turn to the appendix to find
the arguments
to justify these results. Theorem XI.A.2Suppose `P(x) =
\sum_{k=0}^\infty c_k x^k`
converges on an interval I.
Then P is differentiable for all x inside the interval and
`P'(x)=\sum_{k=1}^{\infty
}kc_{k} x^{k-1}` for all x inside the interval I.
In
particular, this means the series `
\sum_{k=0}^{oo} k c_k x^{k-1}` converges to $P'(x)$ for all $x$
inside $I$.
Corollary XI.A.3 Suppose `P(x) =
\sum_{k=0}^{oo}c_k x^k` converges for all x in an interval I.
Then P is `C^{oo}`
for all x inside the interval, $P^{k}(0)=k! c_{k}$ for all $k$, and
thus the MacLaurin polynomial for P of degree
n is `
\sum_{k=0}^n c_k x^k` for all $x$ inside the interval I.
Comments.
1.Well, this looks very good. If we start with a function defined by a
power series, then the Taylor series for the function is exactly the
same as the power series. That is, the Taylor series for the function
gives the same values as the function for each x inside the interval of
convergence.
2. Again the result can be
generalized
easily to polynomials in "`x-a`" so that if
$P(x)=\sum_{k=0}^{\infty }c_{k}(x-a)^{k}$ converges on an interval I,
then P is differentiable for all x inside the interval and
`P'(x)=\sum_{k=1}^{\infty
}kc_{k}(x-a)^{k-1}` for all x inside the interval I.
In
particular, this means the series $\sum_{k=1}^{\infty
}kc_{k}(x-a)^{k-1}$ converges to $P'(x)$ for all $x$ inside $I$.
3. Likewise the corollary generalizes : Suppose
$P(x)=\sum_{k=0}^{\infty
}c_{k}(x-a)^{k}$ converges on an interval I.
Then P is `C^{oo}`
for all x inside the interval, $P^{k}(a)=k!\ c_{k}$ for all $k$, and
thus the Taylor polynomial for P of degree
n is $\sum_{k=0}^{n}c_{k}(x-a)^{k}$ for all $x$ inside the interval I.
4. It would be nice if this situation were true for any `C^{oo}`
function. What do I mean by this?
I mean it would be nice
if any `C^{oo}` function were equal to the function defined by its
Taylor series on the interval of convergence of the Taylor series.
Unfortunately, this result is not true.
There
are functions that are `C^{oo}` for all real numbers but which
agree in value with the series determined by their Maclaurin
polynomials only at `x=0`.
One such function is given by `f(x) = e^{- x^{-2}}` when `x!=0` and
`f(0)= 0`. The graph of this function is very flat near `(0,0)`.
It can be shown that for this function `f^{(n)}(0) = 0` for all `n`, so
its Maclaurin polynomial `P_n(x,f(x))= 0` for all `n`. However, we do have a
simple
result that is almost as nice as what one might hope for:
Theorem XI.A.4 Suppose f
is a `C^{oo}` function on an interval I containing $a$.
Let `P_{n}(x;a,f)=\sum_{k=0}^{k=n}\ f^{k}(a)\frac{(x-a)^{k}}{k!}`,
`R_{n}(x;a,f) = f(x)-P_{n}(x;a,f), and
`P(x;a,f)=\sum_{k=0}^{\infty }\f^{k}(a)\frac{(x-a)^{k}}{k!}`
Then $f(x)=P(x;a,f)$ if and
only if $lim_{n\rightarrow \infty }R_{n}(x;a,f)=0$ for each $x$ in the
interval of convergence of $P(x;a,f)$.
Graph
of
`f(x)
=
e^{-
x^{-2}}`
Example
XI.A.1(Continued). Let
$P(x) = \sum _{k=0}^\infty
\frac{x^{(2k)}}{(2k)!} = 1 + {x^2}/2 + {x^4}/{4!} + ...
$.
We have shown that this series converges for all `x` and the interval
of convergence is `(-oo,oo)`. Thus `P(x)` is defined for all real
numbers `x`. We can now apply Theorem XI.A.2 and its Corollary
XI.A.3 to conclude that for all `x`,
`P` is a `C^{oo}` function and
So, we have found that the function `P` satisfies the second order
differential equation: `y''=y`.
This example illustrates how the
solution
to a differential equation can be characterized as a function defined
by a power series.
Furthermore, if we let `Q(x) = P'(x)`, then we have also that `Q'(x)
=P''(x)=P(x)` and so `Q''(x) = P'(x)=Q(x)`. Quite remarkable- `Q` is
also a solution to the differential equation `y''=y`! What
distinguishes these two solutions from each other is their values at
`0`: `P(0) = 1`, `P'(0)= 0` whereas `Q(0)=0` and `Q'(0)=1`.
Example
XI.A.5. Geometric and Harmonic
Series Revisited.
From our early work with geometric series we know already that the
series `P(x)=sum_{k=0}^{oo}x^k` has `(-1,1)` for its interval of
convergence and that the
function `P(x)` is actually `1/{1-x}` for `x in (-1,1)`.
From Theorem XI.A.2 we have then that `P'(x)=sum_{k=1}^{oo}kx^{k-1}=
{-1}/{(1-x)^2`. for all `x in (-1,1)`.
Let's consider the series `Q(x)=sum_{k=1}^{oo}{x^{k}}/k
=
x
+
{x^2}/2
+ {x^3}/3 + ...`. Using the ratio test we have
that
so the series converges for `x in (-1,1)` and diverges when `|x|>1`.
When `x=1` the series is the harmonic series- so the series diverges.
When `x=-1` the series is the opposite of the alternating harmonic
series and it converges. So the interval of convergence for `Q(x)` is
[-1,1).
Now applying Theorem XI.A.2 to `Q(x)` we have for all `x in (-1,1)`.
`Q'(x) =sum_{k=0}^{oo}x^k = 1 + x + x^2+...= P(x) = 1/{1-x}`.
But any function that has P(x) for its derivative on `(-1,1)` must be
`-ln(1-x) + C` for some constant `C`. Now `Q(0)= 0= -ln(1-0) + C` so
`Q(x) = -ln(1-x)` and `ln(1-x) = -Q(x) = - x - {x^2}/2 - {x^3}/3 - ...
` for all `x in (-1,1)`.
Now what about when `x=-1`? When `x->-1^+`, `-Q(x)-> 1 - 1/2 +
1/3 ... = sum_{k=0}^{oo}{(-1)^{k+1}}/{k+1}` while `ln(1-x) ->ln(2)`.
Putting these two facts together we have that `sum_{k=0}^{oo}{(-1)^{k+1}}/{k+1} = 1 -
1/2 + 1/3 - 1/4 + ... = ln(2)`.Example
XI.A.6
Sine
and
Cosine
revisited with series.
From our previous work with Maclaurin polynomials and the sine and
cosine functions it should make sense that for all `x`,
`|R_n(x, sin(x))| =
|sin^{(n)}(c)|{|x|^{n+1}}/{(n+1)!}<={|x|^{n+1}}/{(n+1)!} -> 0` as
`n->oo`.
Similarly `|R_n(x, cos(x))| =
|cos^{(n)}(c)|{|x|^{n+1}}/{(n+1)!}<={|x|^{n+1}}/{(n+1)!} -> 0` as
`n->oo`.
Thus by Theorem XI.A.4,
`sin(x) =
sum_{k=0}^{oo)(-1)^k{x^{2k+1)}/{(2k+1)!}= x - {x^3}/{3!} + {x^5}/{5!} -
...`
It is a good exercise in finding the derivatives of these series using
Theorem XI.A.2 to verify that `sin'(x) = cos(x)` and `cos'(x)=-sin(x)`. Exercises:
For each of the series in exercises 1- 10, determine the interval of
convergence.
11-20. For each series in 1-10, write the first four nonzero terms and
find the derivative series.
Suppose `P(x)= sum_{k=0}^{oo)(-1)^k {(3x)^{2k+1)}/{(2k+1)! }`.
Show this series converges on the interval `(-oo,oo)`.
Compute the series for `P'(x)` and `P''(x)`. Show that `P''(x) + 9P(x)
= 0` for all `x`.
Suppose `P(x)= sum_{k=0}^{oo){(3x)^k}/{k!}`. Show this series
converges on the interval `(-oo,oo)`.Compute the series for `P'(x)`.
Show that `P'(x)=3P(x)` for all `x`.
Suppose `P(x)= sum_{k=1}^{oo){x^k}/{k!}`. Show this series
converges on the interval `(-oo,oo)`.Compute the series for `P'(x)`.
Show that `P'(x) - P(x)= e^x` for all `x`.
Suppose $Q(x)=\sum_{k=0}^{\infty }c_{k}{x^{k+1}}/{k+1}$ converges
for all `x` in an interval I.
Show
that $P(x)=\sum_{k=0}^{\infty }c_{k}x^k$ also converges for
all `x` in I and `int P(x) dx = Q(x) + C`.
Apply exercise 24 to the geometric series for `1/{1+x}` to find a
series representation for `ln(1+x) = int 1/{1+x}dx`.
Apply exercise 24 to the geometric series for `1/{1+x^2}` to find
a series representation for arc`tan(x) = int 1/{1+x^2}dx`.