Before introducing Taylor series (which are a complicated thing),
introduce -formal- power series.

Why?
(1) It -breaks down- the infomation needed to be learned into bite-sized
  chunks. Once you understand power series, a particular kind (Taylor series)
  are easy.
(2) Kids -love- formal manipulations -- it's what they've been doing forever.
  (honestly, that's how symbolic calculus works)

"""
Power series are a generalization of polynomials
"""

Key EG: geometric series
(easiest non-trivial example, and is interesting to look at)

"""
Power series can be purely formal objects,
-but- it's (very) interesting to see if they represent functions.
"""
(the function/power series relationship is very rich, and leads into
 19th century analysis, especially Weierstrass)
(it's a great example of how different perspectives can tell you
 interesting and complimentary things)

formal power series ------------------------------

We can work with polynomials -formally-, i.e., not as functions.
EG $(elephant)x^3 + 3(apple)x$
makes sense formally, even if it's garbage qua function.

We can add, subtract, etc.

A -formal power series- is just an infinite polynomial.
We can (purely formally) manipulate 'em:
- add/subtract
- multiply
  (this takes a little doing -- illustrate using my "box" model
   of polynomial multiplication, summing along anti-diagonals)
  (i.e., "Cauchy product" (i.e., convolution of functions N -> R))
- differentiate
  termwise!
- integrate
  termwise!
- divide??
  yup: two ways to do this:
  do polynomial long division (but $x$ goes in the weird direction:
    corresponds to $x=\epsilon$ for ordering...)
  invert power series (do-able), and then multiply
    (valuable for algebra: $f/g=fg^{-1}$: think of quotient rule:
      thinking of division as invert then multiply is helpful)
- compose
  tricky: can do $f\circ g$ if $g$ has no constant term;
    (b/c get finite sum for coefficients)
  if $g$ has a constant term, needn't be defined
    (e.g. of $f(x)=1+x+x^2+\dotsb$ (geometric) and $g(x)=1$)
    b/c get divergent series for coefficients
  ...to understand this, need to look at analytic functions on $\bC$

EG:
$1+x$, $x^3$ -- any polynomial is a power series
$1+x+x^2+\dotsb$ -- the simplest infinite power series
  (the geometric series)

notation ----------------------------------
The $\Sigma$ notation in power series is one of the big obstacles to
understanding (in particular, to actually manipulating).
One interesting EG is to differentiate $1+x+x^2+\dotsb$ termwise,
both as a $+\dotsb$ and as a $\Sigma$.
Also, use \emph{reindexing} (i.e., changing $k=0$ to $k=1$ to
$k=0$, either chopping off low terms that die or usefully changing
where you start counting).
-- rote -- :: students need some practice with this

geometric series --------------------------

Note that if $y=1+x+x^2+\dotsb$, then $y'=y^2$, so if we have a function
that satisfies this (and our formal manipulations make sense), it
satisfies the differential equation.
This is separable, and easily solved (careful about the constant of
integration: use the initial condition $x=0 \mapsto y=1$), to get
$y=1/(1-x)$.
Of course, we knew this -- it's the geometric series.

NB:
the power series makes -formal- sense alone
$1/(1-x)$ is a function that makes sense so long as $x\neq 1$
...and the power series -converges- (i.e., makes sense -qua- function)
  so long as $\abs{x}<1$ -- but not for $x=2$, for instance, even
  though $1/(1-x)$ makes sense for $x=2$
-- weird -- !!
It's obvious (looking at $1/(1-x)$) why the series doesn't make sense at
$x=1$, but that doesn't explain why it doesn't work for $x=-1$
  (yes, this is obvious looking at the power series, but not looking at
   the function it represents)

Moral: to understand behaviour of power series, need to look at geometry
  of analytic function it represents
  (over complex plane! look at $\arctan$: it's unclear looking at
   the reals why it blows up).

Now qua power series,
$(1-x)(1+x+x^2+\dotsb)=1$
completely formally.
...but qua functions, this only makes sense for $\abs{x}<1$.

(try plugging in $x=1,2,-1$)
In some sense, you might want
$1-1+1-1+1-\dotsb=\frac{1}{2}$
...if you're Abel or Caesaro.
I mean, that's what it is -on average-, right?
...and that's how you define Caesaro sums!
  (makes you think more about what we mean when we write infinite sums
   ...we defined it to be the lim of partial sums, but that's just
   a choice -- we could also define it to be the lim of averages)

intro to convergence questions -------------------------------
power series converge on discs, if they converge
look at geometric series, and draw picture (over $\bC$):
- converges on unit disc
- diverges outside
- behaviour on boundary is -complicated-

...in this case it's particularly interesting:
for $x=1$, blows up to infinity
for everything else (e.g., $-1, i$), it bounces around in a bounded way.
hence:
- we can look at the Abel sum ($\lim_{r\to 1}$)
- we can look at the Cesaro sum (lim of averages)
