Mean Value Theorem

- MVT: Mean Value Theorem
- FTC: Fundamental Theorem of Calculus
- Taylor's Theorem
...are all very similar/related,
meaning that any one can easily be proved from another
by essentially formal manipulations.

My preference is to favor the Fundamental Theorem of Calculus.
Why?
B/c it's the least calculusy!
That is, the FTC is a -formal- statement:
it says derivative and integration are inverse (both ways).
The other statements are much pickier point-set analysis-y.

More seriously, FTC is more fundamental.
Also, follow Curt (McMullen): use integrals to understand/prove things
in analysis!
(Integrals and derivatives are (for maps R -> R) really just shifting
 coordinate systems, from position space to velocity space etc.
 WTF?
 Integration and differentiation are -linear isomorphisms-
 (up to a dimension or 2), so they really are "change of
 coordinates"!!)

Similarly, you can prove l'Hopital's rule using FTC,
and that's -much- clearer to my taste.

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MVT

This is a key local to global theorem.
If $f$ is (semi)increasing, then $f'$ is semipositive
(increasing functions can have derivative 0).
MVT lets you conclude the converse:
if $f'>0$ (or $f'\geq 0$, only equals $0$ on sets of local measure 0),
  then $f$ is increasing.
*********
That said, a far better way to understand that is:
$f(b)-f(a) = \int_a^b f'$,
so if $f'>0$, then $f$ is increasing.

[There's a subtlety here:
 why is the integral of a positive function -positive-?
 If $f'$ is continuous, then it's easy,
 but general derivatives are complicated.
 OIC: in this case it's still easy: we assume $f'>0$ for all $x$
 in the region, so it must be $>1/\epsilon$ on some set of positive
 measure.
 Hmm...
 So in order to use the FTC to prove things,
 you really need the Lebesgue integral.]

Actually, this is more powerful:
MVT does not deal with the case where $f'=0$ at one (or measure zero
many) points in the interval,
so it's actually the -weaker- result.

Basically MVT is opaque and elementary:
you avoid integral technicalities.
*********


This is -very close- to the IVT for derivatives;
it's slightly more technical.

Anyway, here it is:
If f is continuous on $[a,b]$ and differentiable on $(a,b)$,
then there exists $c \in (a,b)$ s.t.
$f'(c) = \frac{f(b)-f(a)}{b-a}$:
there's a point s.t. tangent = secant.
[The "interior/exterior" is easy to remember once you just
 remember "there is an issue, and don't need to take derivs at ends"]

Here's another way to say it:
$f'$ satisfies a mean value property:
$\frac{f(b)-f(a)}{b-a} = \frac{1}{b-a} \int_a^b f'$ 
is the -average value- of $f'$,
so MVT says that $f'$ assumes its average (mean) value
at some point in the interval.

The Cauchy Mean Value Theorem is just:
"Throw in another function on the bottom":
$\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$
(and is a good EG of a formula that you should -understand-,
 not just rotely -memorize-)

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BTW, you should not confuse this with the MVP for analytic functions
(called by some, like Math World, Gauss's MVT),
which says that the value of $f$ at a point is -equal-
to the average value on a circle.

(...and note that if this is true for a function,
 it's obviously harmonic etc.;
 for a real function, this means exactly that it is linear.)

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BTW, Rudin uses the MVT (and the Cauchy MVT?) extensively,
to give opaque and medium-strength proofs of results.
(Opaque b/c MVT is generally used very technically.)

Basically he can say "if f is everywhere differentiable",
instead of "if if is everywhere differentiable and has integrable
derivative".

These should almost all be proven using integration instead:
if you have a strong integral (s.t. any everywhere differentiable
function is integrable), then you can prove most such results by
integration.
