(.. And other things that
really shouldn’t be in my
likely final {or penultimate,
at best} blog-post ever.)
Blog-post # 636:
(636 = 2 * 3 * 53 * 2.
{^Digital-palindromy!..
on both sides..})
Four (and a half)
art inanimations:
(Crappy, sorry.)
Do I even have to say anymore
that math was indeed involved
in making these images?
——————————-
——————————-
Anagrams, 22:
..
.
It was an egg;
its sphere has bent.
=
Gaps are between
things as this.
—
Megaphones rise.
=
A poem sings here.
—
Hatred’s slick
megaphones lied.
=
Sad men seek
their phallic gods.
—
Smog, peahen,..
=
Megaphones..
=
.. Go shape men..
—
Shining
void/flower/ghost:
=
Wings do so hover
in flight.
—
Kaleidoscopic
thorns:
=
In each slick
pod’s root.
—
Holding our data in,
in-fact, the smog,..
=
Again,
that clouded form
is nothing.
—
Your mortal madness:
=
A mostly random ruse.
—
Betrayal was..
=
.. Yet as a brawl.
—
As Republicans’
treasonous acts:
=
A Russian Beast’s
cancerous plot.
—
Fascists are satanic
fuckers/terrorists.
=
Race-traitors suffer
a sickness as strict.
—
Yet still, Hitler
murders his/our
own people.
=
Unto pride, Whites
or losers yell,
“Heil Trump!”
—
Club-Treason raped us.
“Heil!”
=
Hate’s Republicans
do rule.
—
Asinine activist-judges:
=
Jutting inside via cases.
—
Hags spit.
=
Pigs shat.
—
Societal rapists:
=
Polite as racists.
—
Most things are sad,
churned, stuck, and
lopsided.
=
Those knots did crumple;
each strand did snag us.
—
A mask scans
into our ills.
=
As satanic morons
kill us.
—
Of an eclipse:
=
One lip’s face.
—
Bubbles have
so oscillated.
=
These all do
via cubes/blobs.
—
In stained-glass,
horizons, or shadows:
=
A lens is so hard,
oozing its own shards.
—
Crystalline/stone
auras rise.
=
As in Creation’s
surreal style.
——————————-
——————————-
Palindromes:
‘Sire venom. Mammon ever is.’
‘Slay art. Ebb betrayals.’
——————————-
——————————-
Band-names:
‘Venus-Venom’
‘The Betray-Guns’
——————————-
——————————-
Oh, no thanks. I’ll just have
to turn my nose up at..
getting rhinoplasty..
——————————-
Word on the street is..
“This Lane Right Turn Only”..
and..
“Bike-Lane”..
and..
“Slow: School-Zone”..
——————————-
——————————-
Hey, why is Trump nominating a
far-right Supreme-Court judge,
even though there is no real
need that He do so?
..
Oh,..
in-just case!..
.
(.. In-just case Americans
will ever again be thinking
that they are entitled at all
to any of even the most basic
human/civil-rights/liberties,
that is!
Hey, can’t have us thinking
_that_, now now!
Got to be prudent, see.)
——————————-
(Speaking of Trump just not
ever being able to do being
president injustice..*)
..
.
Yes, indeed, it’s true after
all. There actually has been,
just as President Trump had
claimed, violence on BOTH
sides..
..
So you see, on one side,
violent racists these days
are quite UNABASHed.
But yet, on the other side
though, racists are BASHING
their fists.. and their cars
.. (and their bullets)
into peaceful anti-racism
protesters and minorities!..
(Yep indeed, it’s so.
Violence on BOTH sides..
My, oh my.
Who would’ve even thunk it.)
—
*(Regarding Trump trying
to do Justice:
Oh, so _that’s_ why she’s
blindfolded, then!..)
——————————-
——————————-
(From the stupidest of hatred
.. to the smartest of love..)
..
.
The most learned love-goddess
is..
EruditE!..
——————————-
——————————-
(Now for some.. error-dition,
though..)
Math-puzzles:
(Answers are at the end of
this blog-post.)
Math-puzzle 1:
y = f(x) is some continuous
differentiable real-to-real
function.
And x = g(y) is the inverse
function of f(x).
ie. f(g(x)) = x,
and g(f(x)) = x.
But, for all positive x’s,
f(x) = the derivative of g(x).
(ie. For any particular
positive X, f(X) is the slope
of g(x) at x = X.)
And too, consequently, the
definite integral from 0 to X
(ie. the area between the
curve f(x) and the x-axis,
from x = 0 to x = X) equals
g(X).
Visualized:
(f(x) is the non-straight/
non-circular curve in this
image.
The length of the colorful
horizontal line {measured in
linear units} equals the area
{measured in square-units} of
the colorful region under the
curve.
The circular arc illustrates
that the x-coordinate of the
right edge of the colorful
area equals the y-coordinate
of the {colorful} horizontal
line.)
.
What, then, is a/the function
f(x)?
(I haven’t given this too
much thought, so I don’t
know if the solution is
unique. But there indeed is
at least one solution.)
.
Part-b of math-problem 1:
Now, f(0) = g(0) = 0.
But for which other X does
f(X) = g(X) = X?
.
Note:
Had I instead asked for a
function f(x) that equals
its own derivative, rather
than a function that equals
the derivative of its own
inverse function, then the
solution would have been:
f(x) = C * e^x,
for any constant C.
—-
Math-puzzle 2:
Consider the sequence of
reals, a(0), a(1), a(2),..
where, for any m,
m = 0, 1, 2, 3, 4,..
a(m) =
c^(1 -b^m),
for c = any positive
constant you want, and
b = any constant where
|b| is less than 1.
So, a(0) = 1,
a(1) = c^(1 -b),
a(2) = c^(1 -b^2),
etc.
Now, consider the
sequence of reals
d(0), d(1), d(2),..
defined as follows:
For any particular m = M,
d(M) equals the unique
(positive or negative)
real number that, given
the already-generated
terms
d(0), d(1),.. d(M-1),
makes this equality so:
a(M) =
d(0) +
1/(d(1)
+ 1/(d(2) +
.. + 1/d(M))..)),
Or, in other words,
the continued-fraction
with the (finite number
of) terms
d(0), d(1), d(2),.. d(M)
(where d(0) = 1 is the
initial integer-part)
equals a(M) for any
particular M you choose.
So, d(0) = a(0) = 1.
d(1) = 1/(a(1)-1).
(Because d(0) + 1/d(1)
= 1 + (a(1)-1) = a(1).)
d(2) =
1/(1/(a(2)-1) -1/(a(1)-1)).
(Because,
d(0) + 1/(d(1) + 1/d(2)) =
1 + 1/(1/(a(1)-1)
+ (1/(a(2)-1) -1/(a(1)-1)))
=
1 + (a(2)-1) = a(2).)
Etcetera.
.
The question then is,
given that
a(m) = c^(1 -b^m),
what is the limit,
in terms of b and c, of
d(m)*d(m+1)
as m goes to infinity?
(The answer is a bit
surprising, I think.)
——————————-
——————————-
Math-puzzle solutions:
Solution to math-puzzle 1:
The solution I got is:
f(x) =
phi^(1/phi^2) * x^(1/phi),
where phi is the golden-ratio,
ie. (1+squareroot(5))/2.
.
So, in other words,..
uh,.. other math-symbols,
f(x) = C * x^K,
where the constant C =
((1+squareroot(5))/2)
raised to the
((3-squareroot(5))/2)th power.
And x is raised to the Kth =
((squareroot(5)-1)/2)th power.
Thus,
g(x) = phi^(-1/phi) * x^phi.
And the derivative of g(x)
then is:
phi^(1 -1/phi) * x^(phi-1)
=
phi^(1/phi^2) * x^(1/phi),
which is f(x).
—–
Solution to puzzle 1-b:
(Given my solution for
what is f(x).)
X = phi.
f(phi) =
phi^(1/phi^2) *
phi^(1/phi)
=
phi, the golden-ratio.
And,
g(phi) =
phi^(-1/phi) * phi^phi
=
phi.
————–
Solution to math-puzzle 2:
(This puzzle is somewhat
related to math-puzzle 1.)
I got that the limit, as
m goes to infinity, of
d(m)*d(m+1)
equals:
-b – 2 – 1/b.
I find this interesting
in-part because the limit
is independent of c.
(Though c needs to be
positive {and finite and
not 0}.)
Note 1: Remember, |b| must
not be 1 or greater.
Note 2: If b is positive,
then that means that the
terms of the continued-
fraction (from d(1) onward)
alternate between being
positive and being negative.
Thus, d(m)*d(m+1) is
negative (as too is the
limit).
The solution involves a
specific case of a problem
of mine that was published
years ago in a relatively
layman math-journal.
(I don’t want to say which
journal because I am not
absolutely sure which one
it was.)
Basically, ignoring the many
qualifications on f(x) that
are part of the problem’s
set-up, if a(m+1) = f(a(m))
for every m, m = 0, 1, 2,..,
then d(m)*d(m+1) approaches,
as m goes to infinity,
-f'(X) – 2 – 1/f'(X),
where f'(X) is the derivative
of f(x) at x = X = the limit
of a(m) (and of the continued-
fraction) as m goes to
infinity.
(I will not give the proof
of this^ here.)
So, with a(m) = c^(1 -b^m)
in particular, we have
f(x) =
c^(1-b) * x^b.
(Because, for every
m = 0, 1, 2..,
f(a(m)) =
c^(b -b^(m+1)) + 1-b)
= c*(1 -b^(m+1))
= a(m+1).)
Now, the derivative of
f(x) is:
c^(1-b) * b * x^(b-1).
And the limit of a(m), as
m goes to infinity, equals
c, as long as |b| is less
than 1.
So, we then have:
-f'(X) – 2 – 1/f'(X)
=
-c^(1-b) * b * c^(b-1)
-2
-1/(c^(1-b) * b * c^(b-1))
=
-b – 2 – 1/b.
QED (uh, sort of).
——————————-
——————————-
Leroy Quet (uh, sort of)
Hyperthetically 