Functions

Given a function $f(x)$, how does it transform under a change of
coordinates?
If it's $x=\phi(u)$, it's easy:
$\phi^* f(u) = f(\phi(u))$: composition.
If it's $u=\psi(x)$, what you usually do in practice is try to
``express $x$ in terms of $u$''

Concrete EGs:
- if it's $\cos x^2$, and $u=x^2$,
  then it's obviously $\cos u$.
- if it's $x^2 - 2x$, and $u=x+1$,
  then it's $x^2 -2x +1 - 1 = u^2 - 1$;
  of course this case can be done by substituting $x=u-1$

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A natural question is: ``how do -zeros- change''?
Well, $\phi^* f(u) = 0$ iff $f(\phi(u))=0$ iff $x=\phi(u)$
is a zero for $f$.
So if $Z_f$ is the zeros of $f$,
then $Z_{\phi^* f} = \phi^{-1} Z_f$.
(Similarly, $Z_{\psi_! f} = \psi Z_f$,
 and $Z_{\psi_* f} = \set{x | \psi(f(x)) = 0}$
   is just ``level sets of $\psi^{-1}(0)$'',
   so kinda lame)

A particularly interesting case is for polynomials and M\"obius
transforms, b/c of how it acts on coefficients.

Let $\alpha_i$ be the zeros of $p(x)= \sum_{i=1}^n a_i x^i$
Then zeros of $p(\lambda x)$ are just $\lambda^{-1} \alpha_i$,
and the coefficients are $b_i=a_i \lambda^i$.
  (the ``change of coefficients matrix'' is
   $\diag(1,\lambda,\lambda^2,\dots)$)
  Significantly, this lets you normalize to being monic.
For $x+\lambda$, the  zeros are $\alpha_i - \lambda$,
  and $b_i=\sum \binom{n}{k} a_{i+k}$
  (the ``change of coefficients matrix'' is the
   upper triangular Pascal matrix)
  Significantly, this lets you eliminate the $a_{n-1}$ coefficient,
  which corresponds geometrically to moving to the centroid of the
  zeros.

For $1/x$, the zeros are $1/\alpha_i$ and
  $p(1/x) = x^{-n}(a_0x^n + a_1 x^{n-1} + \dots + a_n)$:
  you read the coefficients in the opposite order,
  and have a pole at zero (corresponding to the pole at infinity of
  the original polynomial);
  if you remove this, you just get the coefficients in the reverse
  order.

The two amusing cases are:
 $-x$: change the sign of coefficients in an alternating way
$1/x$: reverse the order of coefficients!
  (this is what got me started)
  Observe that if there is no constant term, this causes the degree to
  drop (the drop equals the degree of vanishing at zero).
  This can be verified for small degree,
  and this could lead someone to a conjecture -- cool exercise, no?
  $ax+b$ has zero $-b/a$;
  $bx+a$ has zero $-a/b=(-b/a)^{-1}$
  $ax^2+bx+c$ has zeros $(-b\pm \sqrt{b^2-4ac})/2a$,
    and if you switch the order, presumably you get a reciprocal!
    [you can easily do this for simple cases]

Now, this isn't just sophistry or fun:
you can also use this to put restrictions on what the quadratic
formula could be (and likewise on cubic etc.), as it must have certain
symmetries.
