(The discontinuum-hypothesis,^..
{maybe} indeed..)
Blog-post # 588:
(588 = 2 * 7 * 3 * 7 * 2.)
Two new (sort-of-new)
art-inanimations:
Seen In These
Kaleidoscopes We Seek
Sought Via Those
Kaleidoscope We See
–
And here are some math-related
non-art inanimations
(inart nonanimations?).
But I don’t now feel like
describing the math, though.
If you want math, however, then
keep reading this post.. and
reading it.. and reading..
Zag-Zig
Helio-Spikes
Of That Integerness’ Arcs
Go-primality
–
(I will say, though, that the
bottom image shows, for m and n
both running from 1 to 97, which
coordinates pairs (m,n) are such
that m and n are coprime to each
other. Most of these coprime
pairs are plotted in white.
However, the purple squares
represent those coprime pairs
that would not be accessible
{by starting from the lower-left
at (1,1)} if this plot was a
maze and the white areas were
the paths.
{Albeit, it’s a crappy maze with
multiple possible ways to reach
many of the white squares.}
..
The name “Go-primality” comes
from, had this been a Go-board,
the fact that the purple squares
would have been where white
pieces were before being captured
by black pieces surrounding them.
{But I can’t recall the rules of
Go regarding if your opponent
needs or not to have also placed
his pieces in all the diagonally-
adjacent locations too around
your piece in order to capture
that piece.
But whatever are the real rules
of Go, a white piece is captured
here even if the pieces diagonal
to it are white.})
——————————-
——————————-
Anagrams, 26:
..
.
Hyper-dimensional Tao:
=
Only inside a metaphor.
—
Solar-eclipse:
=
Rise/collapse.
—
The solar-eclipses:
=
Spheres oscillate.
—
Of solar-eclipses:
=
Fires so collapse.
—
A sun’s Earth;
a kaleidoscope’s suns:
=
As such polka-dots are
as seen in us.
—
Helices do poke at..
=
The kaleidoscope.
—
Their kaleidoscopes:
=
Helio-spikes do react.
—
Those kaleidoscopes:
=
To seek loops I chased.
—
The design is distant,
is else a wicked loop.
=
Angles twisted inside
this kaleidoscope.
—
Insight’s clear
quintessence:
=
This science’s
learning quest.
—
This entirety’s vastest
hemispheres arise
amongst all.
=
Its metaphors yet
as strange have their
limitlessness.
—
A night’s starry stains
are celestial.
=
This clarity is as
eternal as strange.
—
The Klein-bottle hourglass
is its own sand.
=
A thing’s tilted lobes
turn as shown, like so.
—
Hot fire’s heat:
=
Of this aether.
=
(.. The Earth is of.)
—
Cosines add ten-by-ten
decagons,..
=
..
Ascending as descent..
to beyond.
—
The circle is of
alternations.
=
Fractions oscillate
therein.
—
All roundest thin
edges of this:
=
These fold unto
radii’s lengths.
—
They are thin singles
in love,..
=
.. Yet having their
loneliness.
—
Subhuman genetics:
=
Such beings un-mate.
=
(Such been mating us..)
=
(She cut abusing men.)
=
Change must be in us..
—
Rigorously a nil:
=
Originally ours.
—
These strange
fractions:
=
Of that integerness’
arcs.
—
Products’ divisions:
=
In void, cusps do stir.
—
As spun those round
inflections,..
=
No truth is one of
cusps and lines.
—
Eyes are too blinded
to see the illusions.
=
Nil does lie, yet nil
does soothe but erase.
—
Infinity has twisted so,..
=
As of its destiny within.
—
Of many tiniest limits:
=
Infinity almost is met.
——————————-
——————————-
Palindromes:
‘Wall-lie: Ill law.’
‘Did nil blind id?’
——————————-
——————————-
Another one of those word-play
word-triplet thingies I talked
about on Feb 26, 2016 in
blog-post # 507:
. . Salt
Table . Water
So going clockwise, we have:
Saltwater
Water-table
Table-salt
..
(These triplets are the
word-play equivalent of
rock-paper-sissors, I
guess.)
——————————-
——————————-
“And what do you do there,
exactly, at Mathematical
Oxymora, Incorporated?”
“Oh, I head up the
products-division.”..
——————————-
“Hey. Did you, you lazy bum,
yet place the 1 before the 2
and the 3 before the 4!?”
“Uhh,.. sort-of, boss.”..
“So, yes, then?”
“Uhh,.. SORT-of.”…
——————————-
A military-coup in the US would
sure be.. un-president-ed!..
——————————-
——————————-
Ignorance of the law
is no excuse…
But THE ignorance of the law
is every excuse!..
..
(^Possession of ignorance
is 9/10ths of the law!..)
——————————-
Republicans’ new slogan:
The Party Of
“They Really Mean ‘Yes'”!..
..
{Hopefully this^ does not count
as joking-about-rape, though.
Two possible of the more obvious
non-jokey meanings to the slogan:
1) The Republicans are the ones
saying/believing this about women
(who are the ‘They’ in this case)
and maybe too about how women are
actually only “lying”, see, when
they claim to want equal rights.
2) ‘They’ refers to Republicans
themselves, since they aren’t
exactly saying “NO!” to Trump
or to His agenda (which even
includes as part of it a
scheme to kill millions of
Americans by taking our
health-care;..
and too, of course, it includes
its grand-scheme to take away
women’s rights).
Republicans now days are acting
more than even ever before like
they are frat-boys who like to
rape girls at parties —
ie. they are acting like their
own president has literally
acted.
..
And NO!!!..
The truth in-fact is — as
Republicans should know, but
apparently just can’t ever get —
that neither do most women..
nor does most anyone else..
actually appreciate it even at
all when Republicans try to force
their agenda into Americans’ lives
.. or into Americans’ bodies.}
——————————-
——————————-
(Shifting-gears quite a bit..)
Two math puzzles (along with
some, if not all, of..
The Answers):
Puzzle 1:
Consider all distinct positive
rationals r along with each r’s
reduced-form denominator d(r).
What is the average, then, of
all (the infinite-number of):
r/d(r) ?
The answer has a “closed”-form,
which here means, there is a way
to write out the answer in-terms
of already-defined mathematical
constants, even if not each of
those constants themselves is
necessarily further expressible
(given what mathematicians now
know, that is) completely
in-terms of still other famous
already-defined mathematical
constants.
(Answer a little below.)
———
———
Puzzle 2:
And consider (unlike above, where
we had considered all positive
rationals {which are each greater
than 0}), all those (infinite
number of) distinct rationals r
that are each strictly greater
than 1.
And, let n(r) equal the numerator
of r when r is represented in its
reduced-form.
But instead of taking the average
of all of the following (which I
guess would be zero), simply take
the sum, summing together every..
r/n(r)^3.
The exact sum may surprise you.
What is it?
(Answer a little below.
And it is definitely expressible
in-terms-of already well-known
constants.)
————–
————–
Answer to (and more regarding)
the first math-puzzle above:
I get that the average of
all the r/d(r)’s is…
.
.
.
.
.
Numerically about..
1.125, which is almost
1 and an 1/8th;
and it is “exactly”:
pi^4/(72 zeta(3)).
(But the closed-form, completely
in-terms of other well-known
constants, for zeta(3) =
1 + 1/2^3 + 1/3^3 + 1/4^3…,
is, or last I read years ago,
not known to mathematicians.)
—
Fun-fact (fact?) regarding^:
If we let x equal a constant
that numerically is about:
1.3964,
and we raise each d(r) to this x,
then the average, over all distinct
positive rationals r, of:
r/d(r)^x
is…
1.
This is because, or I figure,
the average, generally for any
x > 1, of:
r/d(r)^x
is
(pi^/12)*zeta(x+1)/zeta(x+2).
And too because..
(pi^/12)*zeta(x+1)/zeta(x+2)
= 1 at x = about 1.3964.
The exact value of x may have
a closed-form, but I don’t
know what it is.
—
Back to the average of the
r/d(r)’s:
But I am a little confused
about this whole average
thingy, I think, because —
seemingly paradoxically — if
we expand the set of numbers
we are taking the average over
from being just all positive
rationals to all positive
reals, then things don’t go
as I would expect (unless I
erred, which is definitely
possible).
The original result I had
found by considering only all
positive rationals with, in
reduced-form, their numerators
and denominators both being
not greater than some large-
but-finite fixed-but-arbitrary
positive integer m.
The sum of all such r/d(r)’s
divided by m^2 approaches a
nonzero-finite constant as
m increases towards infinity.
And too, the _number_ of such
r/d(r)’s divided by m^2
approaches a nonzero-finite
constant as m increases.
So when considering the average
(the sum divided by the number),
the m^2’s basically cancel-out
as m goes to infinity.
And the average, for any m,
should simply approach, as m
increases, the ratio of these
two non-zero finite constants.
And this average should be the
same even when considering
rationals with numerators and
denominators that are
arbitrarily MASSIVE but finite.
Even if the numerators and
denominators of the r’s are
allowed (as they indeed are) to
have many many more than even
quadrillions and quadrillions
of digits, say, then still, the
average (taken over all positive
r’s) of:
r/d(r)
should still be the same
constant, about 1.125.
(In-fact, the average should even
be _closer_ to this constant than
if the numerators and denominators
had instead been restricted to be
relatively smaller.)
But, though!..
In the case where we include
all positive irrational numbers
as well, so r is taken over all
positive real numbers, then..
We basically have the case
where we include r’s with
infinitely large numerators and
denominators.
But since m can be as large as
we want, as long as it is
finite, and we still get the
same average (and especially
since, as well, it’s even more
so this average, if anything,
the LARGER the finite numerators
and denominators get),..
then I _would_ have thought
then that the average over all
positive real r’s would have
been the same as the average
over all positive rationals.
Maybe so this is indeed the
case, since people’s, and maybe
mine too, intuitive understanding
of averages is a bit off.
(Aside: It is also an apparent
paradox to most people that my
average over all rationals is not
instead 0 or 1, say.
I think this misconception
arises from the fact that the
rationals with different
denominators contribute unequally
to the value of the average.)
..
Though, if r is irrational, then
r/d(r) is zero, since we have a
finite real divided by its
infinitely large denominator.
And — furthermore given too that
the number of all reals is a much
huger infinity than the number
of all rationals (Ask Cantor!) —
we have the average over the
positive reals being equal to..
{0 * (massive number of
positive irrationals)
plus
(sum, over relatively puny
number of positive rationals r,
of the r/d(r)’s)}
all divided by
(huge number of all positive
reals together).
Now, 0 times the number of
irrationals is indeterminate,
and could maybe here be 0, a
finite nonzero number, or
infinity.
But since the size of the
infinity of the number of
irrationals is the same as
the size of the set of all
reals, then I think these
infinities — those on the
left side of the numerator
over the denominator– must
then cancel-out.. to make our
average equal to:
0 plus..
the ratio..
(sum, over relatively puny
number of positive rationals r,
of the r/d(r)’s)
divided by
(huge number of all positive
reals).
Now, this is where things get
tricky..
The sum of the ratio’s numerator
is infinity, and too the
denominator is infinity, but I
think that the denominator’s
infinity is so much larger than
the numerator’s infinity that
this ratio must approache ZERO,
not 1.125.
(But if the actual average of
r/d(r)’s over all real r’s
indeed is 1.125 and not 0,
though, _maybe_ that is due
to..
0 * (massive number of
positive irrationals)/
(huge number of all positive
reals)
equalling 1.125 instead of 0??
But I doubt it, and
I don’t know..*)
*(Indeterminates, irrationals,
yadda-yadda, oh alas..)
————–
————–
Answer to the 2nd math-puzzle
above:
The sum, over all distinct
rationals r that are greater
than 1, of
r/n(r)^3,
where n(r) = r’s reduced-form
numerator, is..
.
.
.
.
.
1.
Yep, 1 exactly.
(Unless _I_ am the 1 that is
not exact..)
.
An interesting related more-
visually-fun puzzle is to
determine if the following
is possible. Is it?
Take, for any rational r where r
is greater than 1, the cuboid*
with side-lengths of:
(1/d(r)) by (1/n(r)) by (1/n(r)).
*(A ‘cuboid’ is a way-too-obscure
fancy way of saying a ‘box’, or a
‘right-parallelepiped’ if you want
to be even fancier.)
d(r) and n(r), as above, are
the denominator and numerator,
respectively, of r in reduced-
form.
Now, each of these cuboids has
a volume, given its particular r,
of..
r/n(r)^3
=
1/(d(r)*n(r)^2).
And the sum of all these infinite
number of cuboids’ volumes is
exactly 1, which is the volume of
a 1-by-1-by-1 cube.
The question is, though, can
these cuboids indeed all be packed
into a 1-1-1 cube, while of course
leaving no air-space and no parts
of any cuboids jutting outside the
cube?
It seems likely that the answer
is yes.
And, true, there are always
cuboids, after a finite number of
them have already been packed into
the cube, that are smaller than
any gaps/spaces left yet to fill,
no matter how small and awkwardly
shaped those gaps/spaces.
However, there is still the
possibility that after filling
the cube to some point as best
as possible, that, maybe after
having necessarily skipped past
some larger cuboids so to pack
them in later as you fill the
cube tightly, there are maybe no
individual gaps/spaces left that
will allow the largest yet-
unpacked cuboid into it.
But there must be a maybe-simple
procedure/algorithm that could
for-sure automatically/successfully
pack the 1-by-1-by-1 cube.
(Maybe even simply by requiring at
each step that any large yet-
unpacked cuboids do indeed still
have enough empty space somewhere
within the cube for them to fit into
when the time comes to do so.)
And you are allowed to orient the
cuboids within the cube in any
manner you wish.
I assume without proof that any
successful packings would require,
though, that the cuboids be oriented
orthogonally, ie, with their sides
parallel/perpendicular to the cube’s
sides and to each others’ sides.
(But maybe the cuboids can indeed be
tilted weirdly within the cube in
some manner, yet I highly doubt it.
But maybe so, though, given that
there are an infinite number of
cuboids.)
By the way, if we consider only the
cuboids that have their longest side
equal to 1 (ie. those representing
r = integer n, and that are
of sides 1-by-1/n-by-1/n),..
then these cuboids’ collective
volume is
pi^2/6 -1,
or about .64493..,
which means that all the other
cuboids, corresponding to all
non-integer fractions greater
than 1, must be packed together
into a total (not necessarily
contiguous) volume of just
.355…,
which is a bit over a third of
the cube’s volume.
——————————-
——————————-
Solution to my previous post’s
penny-puzzle:
The only number of pennies for
what Max originally had that
satisfied the conditions is…
But, hey, let’s first look at
some matrices..
Here is a chart showing the number
of pennies in his pocket after the
first purchase, given 0 through 4
pennies (the columns) he had
to-start, and given 0 through 4
cents and too (respectively) 5
through 9 cents (the rows) that
ends the price (after tax) for one
six-pack:
— {Pocket-pennies}
—– 0 1 2 3 4
————–
0/5 – 0 1 2 3 4
1/6 – 4 0 1 2 3
2/7 – 3 4 0 1 2
3/8 – 2 3 4 0 1
4/9 – 1 2 3 4 0
^{Price-end}
So, if say, he had just 1 penny in
his pocket to-start, and the price
ended in 3 cents, then he wouldn’t
have used that penny since he has
fewer than 3 pennies. But he would
have gotten 2 pennies back as part
of the change. And adding 2 to 1,
he ends up — as we can see by
looking at the entry at column 1
and row 3/8 — with 3 pennies in
his pocket at the end of his first
purchase.
Okay, things get a little more
interesting after the second
purchase, though.
This time, the chart’s column-
numbers are not the number of
pennies he had at the beginning of
the second purchase, which would
simply produce the same chart, but
instead are how many pennies he
had had when first arriving at the
store.
So, to generate the second chart,
we plug the above matrix into
itself, and, given the 1st chart’s
numbers of pocket-pennies, we first
determine from the 1st chart how
many pennies he has just before the
second purchase.
Then we read off the number at the
referred-to pocket-penny column of
the 1st chart, but the number which
is in the same row (since the beer
costs the same for both purchases).
So then, we get the number of
pennies in his pocket after the
second purchase:..
— {Original pocket-pennies}
—– 0 1 2 3 4
————–
0/5 – 0 1 2 3 4
1/6 – 3 4 0 1 2
2/7 – 1 2 3 4 0
3/8 – 4 0 1 2 3
4/9 – 2 3 4 0 1
^{Price-end}
So, for example, as per the first
example, let’s say he first had 1
penny, and the cost ended in 3
cents. So he then had 3 pennies
after the first purchase.
But looking up in the first chart
how many pennies he ends up having
if he had instead started with 3
pennies, and again the price ends
in 3 cents, he should be left with
0 pennies, having used them all up
on the second purchase.
And so, looking up the number at
column 1 and row 3/8 of the _second_
chart, we indeed see 0 pennies is
what he finally leaves the store
with after the second purchase.
Okay,..
(Finally stuff gets interesting..)
Now separate, perhaps with little
horizontal line-segments drawn
between the appropriate numbers,
those parts of each column of the
1st chart into those numbers greater
than Max’s starting number of
pennies (as per the particular
column) and those numbers less than
his starting number.
Those line-segments of the first
chart would then just happen to be
placed between the 0 and 4 of each
column (making a descending
staircase of line-segments).
And we can just ignore the top row,
since it is given that the number
of pennies in Max’s pocket had not
remained the same.
And then do the same for the second
chart, dividing the regions of this
chart where the final-penny numbers
are greater than the number of
pennies he first arrived at the
store with from where the final-
penny numbers are less than the
number he started with. But the
boundary between the greater-than
and less-than regions is more
complicated than for the 1st chart,
with — if we again ignore the top
row — there being either 0, 2, or 3
number of dividing horizontal line-
segments per column.
But the funny thing is,..
If we were to superimpose one chart’s
boundary-line-segments onto the other
chart’s boundary-line-segments (so
that they align as per the proper
column numbers and the proper row
numbers),..
we’d then see there is ONLY ONE
line-segment, which spans just one
column, that they both have in
common.
(Again, ignoring the top row.)
This is the single location on the
graph where had the price instead
been one penny more for the beer
than it actually was, then the
numbers of pennies in Max’s pocket
after BOTH purchases would have
changed each time to being either
more than what he started out with
or to fewer.. but necessarily
changing in the direction that is
the reverse of whatever is the
direction given the actual price.
And so,..
as Min figured out..
(and thus Max owes her some big-
bucks,.. uh,.. some big-pennies..),..
Given that this single common line-
segment lies between rows 2/7 and
3/8 and is in column 2,..
we can conclude that, therefore,..
Max at first had 2 pennies in his
pocket,..
and the beer’s total price (after
tax) ended in either 2 or 7 cents.
So, after the first purchase, he
had 0 pennies in his pocket, a loss
in pennies.
Then after the second purchase, he
therefore had 3 cents in his pocket,
which is greater than the 2 he
started out with.
But though, had the beer’s cost
(after tax) ended instead in 3 or 8
cents, then, during the first
purchase, he would have held onto
his 2 pennies and gotten 2
additional pennies as part of the
change.
So he would then have had 4 pennies,
which is more than the 2 pennies he
began with.
Then on the second purchase he would
have spent 3 of those pennies,
leaving him with 1 penny, fewer than
he began with.
So, on the first trip, Max actually
has FEWER pennies, but alternative-
expensive-beer-universe Max has MORE
pennies.
And,..
on the second trip, Max actually has
MORE pennies, but alternative-
expensive-beer-universe Max has FEWER
pennies.
And there are no possible price-ending/
number-pocket-pennies situations that
lead to any of the other 3 (out of 4)
possibilities for more-vs-fewer I
talked about in my previous post.
(Given, anyway, the 0 to 4 pocket-
pennies to start, and given the
standard pay-by-cash protocol I talked
about in that post.)
So, we have:
2 pennies to start in Max’s pocket.
Then 0 pennies.
Then 3 pennies (which he then “lost”.)
So, pay up, Max!..
(Note: Try not to be confused — yet
I may have somewhere above, despite
trying not to be — by the fact that
the number of pennies changes from
fewer to more than he originally had
after, respectively, the first and
second purchases.
What matters is only that the fewer-vs
-greater situations differ as compared
to had the beer’s price instead been
one cent more.
Then too, what else only matters is
the fact that this difference in
either direction must occur after
both purchases.
..
It could have instead been — I note
for the purpose of understanding the
puzzle-statement, yet this entire
situation never could have actually
ended up occurring — that Max had
more pennies than he began with
after the first purchase; and then
he again had more than he began with
after the second purchase. Or maybe
instead he had fewer pennies both
times.
But had that been the case, then
according to the puzzle set-up, his
expensive-beer-universe counterpart
would have needed to end up with
fewer pennies after both purchases,
if he actually had more both times.
Or he would have had needed to end up
with more pennies both times in the
other universe if he actually had
fewer pennies after both purchases.)
——————————-
——————————-
Leroy
Hyperthetically 