my recent example of bad freshman calculus
was, heaven knows, something egregious.
but *this*! this is something else again.

on another day, i might have just bitten down
and let it go with full credit… the very
kind of thing i have to do dozens of times
a night when i sit down to earn my right
to go on calling myself a math teacher.

ordinary unambitious undergrads
will *never* be made in classes like ours
to write carefully. and they’ll take any
attempt to do it as a violation of their
basic civil rights *as* undergrads and
threaten to vote with their feet.
so you’ve got to play it pretty cool.
even though it might break your heart.

just mark it wrong, again.
and tell ‘em why, again.
“CODE IS NOT ENGLISH”
usually doesn’t get points-off
*regardless* of how badly the little
dears have mangled plain sense into
worse-than-meaningless mumbo-jumbo
by using “syncopated style”.

but this example, on this day?
i just couldn’t do it. its author
lost a half a point (out of two).

“f(x) = log_a x = (ln x)/(ln a)
f’(x) = 1/(ln a) * 1/x” …
thus far well and good …
“since the derivative of ln x = 1/x”.

and my first instinct, and indeed my
first move, is to circle the “=”,
write “IS” near it, put a big “X” near
the whole mess so far, and write out
“CODE IS NOT ENGLISH”.
and let it go at that.
like usual.

but, finally, dammit.

if that thing deserves a perfect score
then i’m carl friedrich gauss.

if i *ever* encounter “ln x = 1/x”
*without* letting it bother me,
may my right hand lose its cunning.
take away my mathbooks and send me
to seminary or something; i’ll’ve
quit being a *math* teacher altogether.

the student in question will almost certainly
take my comments for raving lunacy. and so,
i suppose, will some portion of readers of this
post. none of *my* business either way, though,
i suppose. sometimes you’ve just gotta get up
on the net and *vent*. thanks for reading.

log_a(x) = ({{ln x}\over{ln a}}) = {1\over{x ln a}} by quotiant rule
EXHIBIT A

but this is *senior level* stuff…
“analysis” *not* freshman-calc.
and even *there*, one seldom encounters
such an *explicit* version of the
natural-log-equals-reciprocal fallacy
ln(x) = {1\over x}.
(the “quotiEnt rule”…
or rather, *the* quotient rule…
our subject has fallen into the
“leaving out articles makes it
harder to understand and so is
in my best interest” trap…
has *no bearing* on this [mis]-
calculation [we aren't diff-
erentiating a fraction (or any-
thing else for that matter)].)

here’s the whole sad story.
math is hard and everybody knows it.
what they *don’t* know is that it’s
nonetheless easier than anything else.
*particularly* when one is trying
to do things like “pass math tests”.

there’s always a widespread (and *very*
persistent) belief abroad in math
classes that trying to understand
what the technical terms mean (for
example) is “confusing” and should
be dodged at every opportunity.

we teachers go on (as we must)
pretending that when we say things like
“an equation” or “the *product* law”
we believe that our auditors
are thinking of things like
“a string of symbols
representing the assertion
that a thing-on-the-left
has the same meaning as
a thing-on-the-right”
or “the rule (in its context)
about *multiplication*”.

but if we ever look at the documents
produced by these auditors in attempting
to carry out the calculations we only
wish we could still believe we have been
*explaining* for all these weeks?
we soon learn that they have been thinking
nothing of the kind.

anyhow. the example at hand.
calculus class is encountered by *most* of its students
as “practicing a bunch of calculating tricks”.

the “big ideas”… algebra-and-geometry, sets,
functions, sequences, limits, and so on…
are imagined as *never to be understood*.
math teachers will perversely insist on
*talking* in this language when demonstrating
the tricks.

well, the “big ideas” that *characterize*
freshman calc are “differentiation” and
“integration”. for example “differentiation”
transforms the expression “x^n” into the
expression “n*x^(n-1)” (we are suppressing
certain details more or less of course).
this “power law” is the one thing you can
count on a former calculus student to have
remembered (they won’t be able to supply
the context, though… those pesky “details”),
in my experience.

anyhow… long story longer… somewhere
along the line, usually pretty early on…
one encounters the weirdly mystifying
*natural logarithm* function. everything
up to this point could be understood as
glorified sets, algebra, and geometry…
and maybe i’ve been able to fake it pretty well…
but *this* thing depends on the “limit” concept
in a crucial way. so to heck with it.

and a *lot* of doing-okay-til-now students
just decide to learn *one thing* about
ell-en-of-ex
(the function [x |----> ln(x)],
to give it its right name
[anyhow, *one* of its right
"coded" names; "the log"
and "the natural-log function"
serve me best, i suppose,
most of the time]):
“the derivative of ln(x) is 1/x”.

anything else will have to wait.
but here, just as i said i would,
i have given the student too much
credit for careful-use-of-vocabulary:
again and again and again and again,
one will see clear evidence that
whoever filled in some quiz-or-exam
instead “learned” that
“ln(x) is 1/x”.

because, hey look.
how am *i* supposed to know
that “differentiation” means
“take the derivative”?
those words have *no meaning*!
all i know… and all i *want* to know!…
is that somewhere along the line in
every problem, i’ll do one of the tricks
we’ve been practicing. and the only trick
i know…. or ever *want* to know!… about
“ln” is ell-en-is-one-over-ex. so there.

it gets pretty frustrating in calc I
as you can imagine. to see it in analysis II
would drive a less battle-hardened veteran
to despair; in me, to my shame, there’s a
tendency… after the screaming-in-agony
moments… to malicious glee. (o, cursed spite.)

because, hey, look. if we all we *meant* by
“ln(x)” was “1/x” what the devil would we have
made all this other *fuss* about? for, in
your case, grasshopper, several god-damn *years*?
did you think this was never going to *matter*?
in glorified-advanced-*calculus*? flunking fools
like this could get to be a pleasure.

but how would i know. i’m just the grader;
it’s just a two-point homework problem.

and anyway, that’s not really the *exact* thing
i sat down to rant about.

next ish: *more* bad freshman calc from analysis II.

3.2.4
Let (n, p) denote an integer and a prime #;
is “x^2 == n (mod p)” solvable for the given
pairs (n, p)?
a. (5, 227) NO
b. (5, 229) YES
c.(-5, 227) YES
d.(-5, 229) YES
e. (7,1009) YES
f.(-7,1009) YES
(The calculations are routine.)

3.2.6
With a TI “grapher”, I’ve just found that
139^2 == 150 (mod 1009).
One simply puts
Y_1 = 1009(fPart(X^2/1009))…
this computes equivalence-classes-mod-1009
of the squares of each of the inputs successively…
and “scrolls down” in the TABLE window
(set to display natural number X-values).

Hence x^2 == 150 (mod 1009)
*does* have a solution.
(In fact, two of them; the other is of course
870 (= 1009 – 139)).

In detail,
139^2 = 19321 = 19*1009 + 150
and
870^2 = 756900 = 750*1009 + 150.
(The point here is that one may easily
*verify* such results by “paper-&-pencil”
methods.)

To find the roots *entirely* by p-&-p methods,
the best line of attack would probably be via
“primitive roots”.

3.2.7
We seek primes p such that x^2 == 13 (mod p)
has a solution.

The case of p = 2 (“the oddest case of all”)
reduces to x^2 == 1 (mod 2); of course this
*does* have a solution. (So we add “2″ to our
list of primes.)

In the case of p = 13, our equivalence is x^2 == 0;
again, this clearly *does* have a solution (and “13″
belongs in our list).

Finally, let p be an odd prime with p \not= 13.
The Gaussian Reciprocity Law now gives us that
(13|p) = (p|13)
(where (_|_) denotes the Legendre symbol).
[Remark:
This happens "because 13 == 1 (mod 4)"...
recall that (p|q) and (q|p) have opposite
signs only when *both* primes are in "4k+3"
form.]

So we need only check the equivalence classes (mod 13)
for “perfect squares”. We know from earlier work that
there are *six* perfect squares (mod 13).
1, 4, and 9 are the “obvious” ones…
4^2 = 16 == 3…
by suchlike computations, one easily arrives at
{1, 4, 9, 3, 12, 10} =
{1, 3, 4, 9, 10, 12};
any odd prime p different from 13 must be
congruent-modulo-13 to one of these “perfect
squares mod 13″.

Summarizing: x^2 == 13 (mod p)
has solutions for p = 2, for p = 13,
and for
p \in { 13k + r |
k \in N ,
r \in {1, 3, 4, 9, 10, 12}
}.
(*)
**************************(end of exercise)******

[
Remark:
Our "list" of primes now begins
L = {2, 3, 13, 17, 23, ...}
.
(Some students had {2, 3, 13})
.
It can be shown
(using "Dirichlet's Theorem on Primes
in Arithmetic Progressions" [DT], section 8.4)
that there are *infinitely many* prime numbers
“for each r” in (*) (i.e., in each of the
“perfect square” equivalence classes mod-13)
.
One is of course not expected to know
about DT already (or to learn it now).
But questions like

*is* there a prime that’s congruent to 4 (mod 13)?

arise in the present context in a very natural way:
it so happens that 17 == 4 (mod 13), and so
in “testing primes” (in their natural order
2, 3, 5, … , 13, 17, …) one soon learns
the answer: there *are* such primes.
(But then… how *many*? And so on.)
]

4.1.2
Clearly with
2^k || 100! and 5^r || 100!,
one has r < k, and so
to "count zeros at the right
end" of Z = 100! we need only
compute "r"
(the largest power of *10* dividing Z…
the number of zeros in question…
is clearly the same as the largest
power of *5* dividing Z).

But for this we need only consult
de Polignac's formula (Theorem 4.2):
r = [100/5] + [100/25] + [100/125] + …
r = 24.

4.1.9
Let BC(x,y) denote the Binomial Coefficient
BC(x,y) = x!/[y!(x-y)!]
(one usually pronounces this object
"x choose y"; see Section 1.4).

One then has the familiar "Pascal's Triangle"
property: a given "entry" can be computed
by "adding the two entries above it".
The object of study in this exercise is then
the "middle entry" of an "even numbered row";
simple algebra gives us

(2n)!/(n!)^2 =
BC(2n, n) =
BC(2n-1, n-1) + BC(2n-1, n) =
2BC(2n-1, n-1).

This is clearly an even integer.

6.
With a TI “grapher”, I’ve just found that
244^2 == 5 (mod 1009). One simply puts
Y_1 = 1009(fPart(X^2/1009))…
this computes the equivalence-class-mod-1009
of the square of each input successively…
and “scrolls down” in the TABLE window
(set to display natural number X’s).

Hence x^2 == 5 (mod 1009) *does* have a solution.
(In fact, two of them; the other is of course
765 (= 1009 – 244)).

In detail,
244^2 = 58536 = 59*1009 + 5
and
765^2 = 585225 =580*1009 + 5.
(The point here is that one may easily
*verify* such results by “paper-&-pencil”
methods.)

To find the roots *entirely* by p-&-p methods,
the best line of attack would probably be via
“primitive roots”.

trouble is, is, that that “5″
was supposed to’ve been “150″.
once more, dear friends!
(or close the wall up with our
english dead!)

Photo on 2014-04-07 at 02.45

i’m going to SPACE next week.

it’s columbus ohio, so the list
narrows down pretty quick:
the “arnold classic”, SPACE,
and, um, let’s see… there
*must* have been somththing
else…

“ameriflora”, maybe, in your
dreams. “miracle mile”, back
from the “miracle” days of
the long-gone late-great
thriving american working class
that shopped (until it dropped)
there. no. never mind.

the chief attraction of columbus
for *me* is that this is where
i *am* (moving around… or
even moving *stuff* around…
is *much* harder than they’d
have you believe…; & of course
the *next* best thing about
columbus is that *madeline*
lives here (and our happy home
*is* our happy home, much to
my surprise). and *staying*
here keeps me this way (happy).
it’s (1) cold (2) cruel
world (3) out there.

the (very existence of)
the billy ireland museum is,
enough to put columbus *somewhere* on
the comics “map”… and there’s already
a better list of local-and-quasi-nearby
talent at the “space” site… so let
me just give a shout-out to ray (!!) t
and “max ink” (still working as far as
i know); one more for glen brewer (even
though i think glen has quit the scene;
his _askari_hodari_ was, for me, very
much a local highlight).

everybody knows about _bone_;
it won’t escape my notice here
that the astonishing paul hornschemeier
lived here, too, when he was getting
started and i met him (and he drew
a cover for my zine gratis… eat
your hearts out). in fact, ghod
*bless* columbus. good night.

i’ve typeface-ized the “formula” stuff
but the point here is the english.
tonight i encountered the passage

Since there are (p-1)/2 quadratic residues & 1^2, 2^2, …, [(p-1)/2)]^2 are all the residues, we need to show that the quadratic residues modulo p are all distinct…

and, after much wailing and gnashing
of teeth, decided that the best spin
i could put on it would be
to *omit* the first “the*
and to replace the second “the”
with “these”:
Since there are (p-1)/2 quadratic residues & 1^2, 2^2, …, [(p-1)/2)]^2 are all residues, we need to show that these quadratic residues modulo p are all distinct…

(which “works” in its context
as the original passage certainly
does *not*).

they should give medals for this kind
of copyediting. this is *hard work*.
not that it does anyone any *good*,
mind you…

handwritingNum'Th'y1

one could simply cite the lemma
and get it over with. but that’s
not how i actually did the exercise…

let p be an odd prime &
let g & g’ be primitive
roots (mod p).

any Primitive Root, h, satisfies
(h^{{p-1}\over2})^2 = 1 (mod p);
since h^{{p-1}\over2}\not= 1 (mod p)
[this uses "h is primitive"],
we can conclude that
h^{{p-1}\over2} ~ -1 (mod p).

etcetera. forget it.
the handwritten stuff
is beautiful though.
\TeX is too hard.
it’s not so bad in the real editor,
of course.

oh, ps. (gg’)^[(p-1)/2]
is now seen to be congruent
to 1… and so is not a
primitive root (which, on
the day, was to’ve been shown).

Photo on 2014-03-14 at 18.56

ladies and gentlemen, PSL(2,7).

Photo on 2014-03-14 at 16.06

here in the middle are the seven colors
in “mister big \pi-oh” (from ohio) order:
MRBGPYO
(mud, red, blue, green, purple,
yellow, orange). i’ve drawn the “line”
(which appears as a triangle) formed by
“marking” the purple vertex and performing
the “two steps forward and one step back”
procedure: one easily verifies that
{G, O, P} is a line as described in
the previous post (“the secondaries”).

all to do with “duality in P^2({\Bbb F}_2)“.
had we but world enough. and time. especially time.

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