XI. Power Series: Polynomials and Series.( draft version in progress)
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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
|c_k  x^k | = | c_k | |x^k| = | c_k | |x| ^k
so
| c_k \ x^k | < | c_k | b^k < | c_k | s^k   {b^k}/{s^k}.

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

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
|{{x^{2k+2}}/{(2k+2)!}}/{{ x^{2k}}/{(2k)!}}| = { | x^{2k+2} \ (2k)! | }/{ |x^{2k} (2k+2)!  |} = {x^2}/{(2k+2)(2k+1)} .
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
{ | {x^{k+1}}/ (k+2) | }/{|{x^k}/{(k+1)} | } ={ | x^{k+1} \ (k+1) | }/{ | (k+2) \ x^k | } ={ |x| (k+1)}/{k+2} .
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.2  Suppose  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.

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
P'(x) =  \sum _{k=1}^\infty \  2k \frac{x^(2k-1)}{(2k)!}= \sum _{k=1}^\infty \ \frac{x^(2k-1)}{(2k-1)!} = x + {x^3}/{3!}+ ....
Notice that the theorem and corollary also can be applied to P'(x) so we have
P''(x) = sum _{k=1}^\infty \(2k-1) \frac{x^(2k-2)}{(2k-1)!}=sum _{k=1}^\infty \ \frac{x^(2k-2)}{(2k-2)!} 
= \sum _{k=0}^\infty \ \frac{x^(2k)}{(2k)!} = 1 + {x^2}/2 + {x^4}/{4!} + ...=P(x).

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
|{{x^{k+1}}/{(k+1)}}/{{ x^{k}}/{(k)}}| = { | x^{k+1} \ (k) | }/{ |x^{k} (k+1) |} = {|x| k}/{k+1} ->|x| as k->oo,
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!} - ...
and
cos(x)=sum_{k=0}^{oo)(-1)^k{x^{2k)}/{(2k)!}= 1 - {x^2}/2 +{x^4}/{4!} - ....
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.
 sum_{k=0}^{oo){2^k x^k}/{3k+1} sum_{k=0}^{oo)(-1)^k{x^{k+1)}/{3^k (k+1)!} sum_{k=0}^{oo)(-1)^k{2^kx^{k+1)}/{(3k)!} sum_{k=0}^{oo){x^{k)}/{(k+1)^2 2^k} sum_{k=0}^{oo){(x-2)^{k)}/{k+3} sum_{k=0}^{oo)(-1)^k{(x+1)^{2k+1)}/{(2k+1)} sum_{k=0}^{oo){x^{2k+1)}/{(2k+1)!} sum_{k=0}^{oo){x^{2k)}/{(2k)!} sum_{k=0}^{oo)(-1)^k{3^k (x-2)^{2k+1)}/{(2k+1)!} sum_{k=0}^{oo){(x+4)^{2k+1)}/{3^k(2k+1)!}
11-20. For each series in 1-10, write the first four nonzero terms and find the derivative series.
1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. Apply exercise 24 to the geometric series for 1/{1+x^2} to find a series representation for arctan(x) = int 1/{1+x^2}dx`.