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#Post#: 11929--------------------------------------------------
Blow Your Mind!!!
By: bfitzpatrick Date: September 2, 2014, 2:23 pm
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Post things here that blow your mind.
Here's one:
If you add together all positive numbers infinitely, such as
1+2+3+4+5+6+7..., it adds up to... -1/12.
HTML http://stream1.gifsoup.com/view2/1776032/kevin-butler-mind-blow-o.gif
#Post#: 11934--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: September 2, 2014, 4:59 pm
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Here's one I really like (split into four parts):
It is possible to compare infinite sets
[spoiler]
Suppose you have two sets, and you want to see if they contain
the same number of elements. You can find this out by looking
for a "mapping" between the sets: a set of one-to-one
connections matching them up. For example, {1,2,3} has the same
number of elements as {2,4,6}, because you can connect 1 to 2, 2
to 4 and 3 to 6, and there are no terms left over.
In fact, you can do this with infinite sets as well. For
example, you would think that there are twice as many positive
integers as even positive integers, right? Nope--the two sets
are the same size, because you can map 1 to 2, 2 to 4, 3 to 6,
and so on until infinity. Need more proof? If you pick any
positive integer, you can easily calculate the unique even
number it maps to, and vice versa (for example, 3641 maps to
7282.) Thus, there are just as many positive integers as there
are positive even integers, even though one set contains the
other! This is the first mindblow moment in this topic.
[/spoiler]
There are different types of infinity
[spoiler]
By the proof above, it may seem like all infinite sets contain
the same number of elements. In fact, this is not the
case--there are more real numbers (like 1.34, -2.99999, π,
e, and so on) than there are positive integers.
There are many ways to prove this, but one is particularly
neat--I'll skip over some technicalities, though.
Let's assume that all the real numbers from 0 to 1 (not
including 1) are "countable," meaning that they can be mapped to
the positive integers. Then we can come up with some sort of
ordering, like this for example:
1 → 0.2189571290572169...
2 → 0.5478478304752568...
3 → 0.2735127846139523...
4 → 0.9258193571385763...
and so on.
Now, let's draw a diagonal line through the digits of these real
numbers, and construct a new real number from all the digits on
that line (0.2498...) Take this number and add 1 to every single
digit (obtaining 0.3509...). If we do this, the resulting number
is always less than 1, but it differs from every other
less-than-1 real number there is--for the Nth real number in our
list, our new number has a different Nth digit. Thus, there
exists a real number beyond any countable set we create, so the
real numbers are uncountable.
So we have an infinite number of positive integers, and an
infinite number of real numbers, but there are more real numbers
than positive integers. This means that there must be more than
one "infinity!" The 'countable' infinity is often referred to as
א[sub]0[/sub], or "aleph-nought." Sets of this size
include: the positive integers, the integers, the even integers,
and the "lattice points" (points on the plane with integer
coordinates).
[/spoiler]
There are just as many rationals as there are integers
[spoiler]
It makes sense that there are more real numbers than
integers--you can fit an infinite number of real numbers between
any pair of integers. The same is true of "rational numbers", or
quotients of integers (like 1/3, 9/5, or -7923/35555)--between
two integers, there are infinitely many rationals. But here's
the tricky part: there are just as many rationals as there are
integers!
So the rationals must be countable, right? It seems
impossible--for example, if you start with all the integers and
then work your way to the fractions, you'll have to count
infinitely before you reach 1/2. You can't say "I'll count that
one 10th," or "I'll count that one 1000th," because you'd have
to say "I'll count that one after an infinite amount of time"
(in other words, you'll never do it). So is there a way to map
every rational number to a finite integer?
Yes, there is. Here's how: 1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1,
3/2, 2/3, 1/4, and so on. Essentially, I'm starting with the
fraction whose numerator and denominator sum to 2, then moving
to the ones with a sum of 3, then 4, then 5, and so on. The
series {2,3,4,5,...} is countably infinite, and each step of the
counting process is finite. Imagine putting all the rationals on
a grid, with the x-axis representing numerator and the y-axis
representing denominator--instead of dividing them into infinite
horizontal lines, you can divide them into finite diagonals.
Of course, we would never actually count all the rational
numbers, but now we know that we can. This means that the
rationals are countable, and thus, there are just as many
rationals as integers!
(Note: I only used positive rationals, but it's relatively easy
to extend this proof to all rationals. For example, instead of
diagonals, you could used diamonds centered at 0/0.)
[/spoiler]
An infinite space, and infinite plane and an infinite line are
all the same size
[spoiler]
Take an infinite plane. Every point on that plane is a
combination of two real numbers, like (2.3,3.77) or (4,-π).
We know that the points on this plane are uncountable, since
they contain the real numbers infinity times over...but are
there as many points as there are real numbers? If so, every
point on the plane can be uniquely expressed as a single real
number--if not, the points on this plane transcend to a third
level of infinity, beyond integers and reals.
Our diagonal method (which we used for rationals) will not work
here. If we were to divide the infinite plane into diagonals,
the distances between them would be infinitely small. Let's try
another method: could we weave an infinitely long line such that
it makes an infinitely dense surface? If so, we're done--the
line is our real number line, and the surface is our infinite
plane.
As it turns out, there is a whole class of mathematical
structures dedicated to doing this. They are called
"space-filling curves," curves which follow a repeating pattern
which can be extended infinitely large and shrunk infinitely
small, thus having the ability to produce a perfect surface. And
yes, these do exist--shown below is an iteration of the famous
Hilbert Curve.
HTML http://people.csail.mit.edu/jaffer/Geometry/hilbert.png
The Hilbert Curve is expanded by linking together four rotated
versions of itself with short lines as shown; if its 1st
iteration is a single point, then the figure above shows its 5th
iteration. After a countably infinite number of iterations, this
curve has the fullness of an infinite surface. Thus, the very
existence of this pattern proves that an infinite plane is no
bigger than an infinite line! Taking this even further, we can
use this same curve to bend an infinite plane until it becomes
an infinite space--in other words, everything the universe could
possibly contain can be stored in a single line.
Google "space-filling curve" for more examples--some are quite
simple, others surprisingly beautiful.
[/spoiler]
This is a really neat topic, if hard to understand at first (It
took me a long time to figure it out). But once you discover it,
it's just a fun thing to think about--the more you contemplate
it, the more amazing things you can work out. Let me know if any
of this is overly confusing or wordy, though; I'm still getting
the hang of explaining it.
#Post#: 11939--------------------------------------------------
Re: Blow Your Mind!!!
By: Stickly Date: September 2, 2014, 9:08 pm
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Get some surgery! Replace your brain with dynamite! The results
will BLOW YOUR MIND!
#Post#: 11953--------------------------------------------------
Re: Blow Your Mind!!!
By: Visitor Date: September 3, 2014, 2:57 pm
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This post blows my mind!
#Post#: 12593--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: October 30, 2014, 11:10 am
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Consider the following problem:
Let c(t) be the number of coyotes in a population at time t. Let
r(t) be the number of roadrunners in a population at time t. You
are given the following formulas:
c(t+1) = 0.9c(t) + 0.2r(t)
r(t+1) = 1.2r(t) - 0.1c(t)
Let c = c(0) and r = r(0). Prove the following:
c(t) = 2c - r + (r-c)(1.1)[sup]t[/sup]
r(t) = 2c - r + 2(r-c)(1.1)[sup]t[/sup]
This is a really interesting problem because it lets you take
two functions which are linked together, and separate them into
more manageable formulas. This can be done with any number of
functions; for example, there's a physics problem a lot like
this one which uses FOUR equations.
The easiest way to work out this problem involves eigenvalues,
but if you don't know what those are, try starting the problem
this way:
[list]
[li]What happens if there are the same number of coyotes as
roadrunners?[/li]
[li]What happens if there are twice as many roadrunners as
coyotes?[/li]
[/list]
(Disclaimer: I may not have done the math exactly right...)
#Post#: 13877--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: November 25, 2014, 11:27 am
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I finally took a class where I learned about bfitzpatrick's fact
(1+2+3+4+... = -1/12). It's a really interesting concept, though
unfortunately it turns out to be wrong.
Basically, this fact relies on the Riemann Hypothesis, a widely
accepted but currently unproven theorem. It states that
ξ(s) = ξ(1-s) for all s, where ξ is a certain
weird, complicated function obtained by multiplying other weird,
complicated functions. It turns out that, if you plug in s=2,
you can derive from ξ(2)=ξ(-1) that
Σℤ[sup]+[/sup]=-1/12, which is equivalent to what
bfitzpatrick said.
So, here's the problem with this argument: -1 is outside the
domain of the ξ function, which means that ξ(-1) makes
no sense. This is because ξ(s) relies on ζ(s), where
ζ is a function such that
ζ(s)=Σ[sub]n[/sub](1/n[sup]s[/sup]). ζ(s) only
converges (i.e. is finite) when s>1, so taking things like
ζ(-1) is not allowed, and thus, taking ξ(-1) is not
allowed either.
So 1+2+3+4+... does not exactly equal -1/12, though it's a neat
thought experiment if you know the reasoning behind it. For more
on these infinite sums, see my first post in this thread about
comparing infinities.
#Post#: 14995--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: December 6, 2014, 2:21 am
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Okay, new cool thing:
How to make a sphere vanish without changing its radius
First, some definitions:
An "n-cube," for some value of n, is an n-dimensional cube. For
example, a 1-cube is a line, a 2-cube is a square, a 3-cube is a
cube, a 4-cube is a hypercube, and so on.
An "n-sphere" is defined similarly (line, circle, sphere,
hypersphere, etc.)
Now, consider the following problem: "You inscribe an n-sphere
inside an n-cube. Next, you connect two opposite corners of the
n-cube, forming a diagonal. What fraction of the diagonal is
inside the n-sphere?"
Let's look at n=2 (a circle in a square). Assume the side length
of the square is S. The length of the diagonal is S√2, and
the portion which is inside the circle has length S. Thus, the
fraction is 1/√2, so about 70% of the diagonal is
contained in the circle.
Now let's look at n=3 (a sphere in a cube). The length of the
diagonal is S√3, and the portion which is inside the
circle has length S. Thus, the fraction is 1/√3, so about
57% is...wait, 57%? That's much less than the 70% we got before,
yet all we changed was the number of dimensions...
As it turns out, the general answer is 1/√n (you can use
coordinate geometry to prove this). So if we use a 10-cube and a
10-sphere, only 32% of the diagonal is contained in the sphere.
Once you reach 10000 dimensions, the answer is 1% exactly. But
the really neat thing? As the number of dimensions approaches
infinity, the answer approaches 0%.
So what does this mean? The sphere doesn't get any smaller...the
cube doesn't get any larger...so why do large dimensions seem to
separate the sphere from the diagonal? In fact, as an n-sphere
gains new dimensions, it actually gets bigger (as long as
S>1)--How can the n-sphere appear to vanish, even when it's
growing?
#Post#: 17492--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: December 27, 2014, 1:25 am
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A tricky game...which you'll always win
There are plenty of simple games like this, where one player has
a tricky dominant strategy which always wins. For this post, I
chose a game with literally infinite possibilities, to make it
interesting.
The Rules:
Get a circular table of any size (as long as it's larger than a
penny). Players take turns setting pennies on the table. You
cannot place a penny so that it overlaps with another penny or
hangs off the table. If it's your turn and you cannot place a
penny anywhere, you lose.
The interesting thing? The player who goes first can always win
with the right strategy.
How does it work?
[spoiler]On your first move, place a penny in the exact center
of the table. On each subsequent move, place a penny so that it
opposes your opponent's last move--in other words, reflect the
position of the most recent penny over the center of the table.
If you think about it, this strategy guarantees that you will
always be able to counter your opponent, and that your opponent
will always lose before you do.[/spoiler]
#Post#: 20841--------------------------------------------------
Re: Blow Your Mind!!!
By: Omni_Builder Date: February 21, 2015, 4:01 pm
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Interesting:
1+(1/(1+(1/(1+(1/(1+...)))))) = (1+√5)/2.
Very Interesting:
1+√(1+√(1+√(1+√(1+...)))) =
(1+√5)/2.
Extremely Interesting:
x+(y/(x+(y/(x+(y/(x+...)))))) = (x+√(x[sup]2[/sup]+4y))/2.
Inconceivably Interesting:
Let F[sub]n[/sub] be the nth Fibonacci number. As n increases,
F[sub]n+1[/sub]/F[sub]n[/sub] approaches...
...(1+√5)/2.
Scalebreakingly Interesting:
If ϕ=(1+√5)/2 and Ψ=(1-√5)/2, then
(ϕ[sup]n[/sup]-Ψ[sup]n[/sup])/√5 is always an
integer...
...and is equal to F[sub]n[/sub].
#Post#: 20863--------------------------------------------------
Re: Blow Your Mind!!!
By: BerserkFalcon Date: February 22, 2015, 3:50 pm
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I memorized this word in fifth grade by looking at it for just a
couple seconds: pneumonoultramicroscopicsilicovolcanoconiosis
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