
π does not equal 4 - iamelgringo
http://www.axiomaticdoubt.com/?p=504
======
cscheid
Geometric convergence is tricky. Specifically, the issue here is that uniform
convergence does not imply convergence in length or area or any other such
measures. (see figure 6 here:
[http://www.sci.utah.edu/~etiene/publications/verifiable-
vis....](http://www.sci.utah.edu/~etiene/publications/verifiable-vis.pdf) .
disclaimer - I'm a co-author)

Imagine a circle and its diameter. Now imagine two circles with half of the
diameter, lined up so that the diameter lines align. Now split those two in
four, etc. The circle becomes a snaking line whose total length doesn't
change, and the snaking line converges uniformly to the line. Clearly,
however, pi is not 1.

What you need is convergence in position _and_ angle. A curve that converges
in position and angle _does_ converge in length: the reference I know which
shows this is reference [9] on the above-mentioned paper. (edit: in case you
don't want to download the gigantic file --- yay for publishing in graphicsy
places --- the reference is: K. Hildebrandt, K. Polthier, and M. Wardetzky. On
the convergence of metric and geometric properties of polyhedral surfaces.
Geometriae Dediacata, (123):89–112, 2006.)

~~~
pixcavator
>>...uniform convergence does not imply convergence in length or area...

Under uniform convergence, the limit of the integrals is equal to the integral
of the limit. So, this works fine for areas, or are you talking about
something else?

~~~
cscheid
Sorry, I should have been clearer.

I was trying to say that convergence in area for two-dimensional surfaces in
R^3 requires convergence in normals in the same way that convergence in length
for one-dimensional curves in R^2 requires convergence in normals.

For area in R^2, volume in R^3, and so on, you're definitely right.

------
extension
In a similar way, you can "prove" that sqrt(2) = 2 by approximating the
diagonal of a unit square (length = sqrt(2)) by iteratively dividing two
adjacent edges of the square (length = 2).

One interesting and useless law that this reveals is that the perimiter of any
convex blob of pixels is equal to the perimiter of its bounding rectangle.

~~~
evandijk70
In fact, just a few days ago, I discovered this law while working on my
master's thesis. I'm modelling cells in a Monte Carlo model on a square grid,
and I needed the contact area with other cells. The contact area seemed to be
to large on diagonal interfaces. After thinking about it a bit, I found the
problem, and I started using a different measure for contact area.

------
pixcavator
Just because the conclusion is wrong it does not mean that there is no point
here. What’s described is exactly what one has to deal with in digital image
analysis. Here is a write-up and a link to a paper:
<http://inperc.com/wiki/index.php?title=Lengths_of_curves>. The paper proves
essentially that there is no good way to approximate lengths of curves with
any grid, even if the size of the mesh goes to 0.

------
RiderOfGiraffes
Nice and fun, and brings into focus the definition of "distance" and how it
interacts with the concept of a metric. Using the L<sub>1</sub> metric gives a
different value of pi than using the L<sub>2</sub> metric. Obviously.

It's easy enough with arguments like this to show the "length" to be anything
you choose. Such demonstrations are instructive.

EDIT: Changed lower-case ell to upper-case for clarity.

~~~
jacobolus
Your comment should be at the top.

Wikipedia has a nice short article about the _ℓ_ ₁ norm (also called “taxicab
distance” or “Manhattan distance”):
<http://en.wikipedia.org/wiki/Taxicab_geometry>

It explains,

> _“The use of Manhattan distance leads to a strange concept: when the
> resolution of the Taxicab geometry is made larger, approaching infinity (the
> size of division of the axis approaches 0), it seems intuitive that the
> Manhattan distance would approach the Euclidean metric [...] but it does
> not. This is essentially a consequence of being forced to adhere to single-
> axis movement: when following the Manhattan metric, one cannot move
> diagonally (in more than one axis simultaneously).”_

------
a-priori
You can "disprove" the Pythagorean equation the same way, by taking the limit
of the Manhattan distance between two points as the grid size approaches zero.
It approaches d_x + d_y, not sqrt(d_x^2 + d_y^2).

------
jcampbell1
Why would anyone assume that because a shape converges, the arc length must
also converge?

A troll can also prove that 2 == 1 by continuously folding the peaks of an
equilateral triangle down to the baseline.

Another fun one:

$1 = 100¢

$.1 = 10¢

$.1^2 = 10¢ ^ 2

$.01 = 100¢

thus

$1 = 1¢

Trollface

~~~
senki
Honoured Sir,

Understanding you to be a distinguished algebraist (that is, distinguished
from other algebraists by different face, different height, etc.), I beg to
submit to you a difficulty which distresses me much.

If x and y are each equal to 1, it is plain that

2 * (x^2 - y^2) = 0, and also that 5 * (x - y) = 0.

Hence 2 * (x^2 - y^2) = 5 * (x - y).

Now divide each side of this equation by (x - y).

Then 2 * (x + y) = 5.

But (x + y) = (1 + 1), i.e. = 2. So that 2 * 2 = 5.

Ever since this painful fact has been forced upon me, I have not slept more
than 8 hours a night, and have not been able to eat more than 3 meals a day.

I trust you will pity me and will kindly explain the difficulty to Your
obliged,

Lewis Carroll.

~~~
xenophanes
> Now divide each side of this equation by (x - y)

You can't divide by 0 you just get nonsense.

~~~
kdeberk
that's not really the point because we are dealing with algebra and not the
numberical values. The error is the assumption that (x-y)^2 equals (x^2-y^2),
which is not the case.

~~~
robinhouston
Hmm, no. Look again. It's using the fact that (x^2 - y^2) = (x - y)(x + y),
which is the case, and a common enough identity (“the difference of two
squares“) that it's used without remark here.

The problem really is that you can't divide by zero, even in an algebraic
expression.

A simpler example of this phenomenon (which blew my mind when I first
encountered it) occurs with the equation x = x^2. If you divide by x, you get
x = 1, which is a solution to the equation, but where did the other solution x
= 0 go??

Whenever you divide an equation by an algebraic expression, you need to
consider the possibility of that expression being zero and treat it as a
special case. So in the case of x = x^2, you can reason as follows: maybe x =
0, in which case … what … ah yes, that's a solution! Or maybe x ≠ 0, in which
case we can divide by it and get x = 1. That doesn't contradict the assumption
x ≠ 0, so it's okay, and x = 1 is the other solution.

------
jbapple
When I first saw a "proof" like this, the explanation of its incorrectness was
something like "limits don't preserve curve length". I wasn't satisfied with
that answer until I took a first course in real analysis, which explained
(some of) the reasons behind calculus.

Real analysis was very satisfying, in somewhat the same way that building low-
level software or libraries is satisfying -- I got to understand the guts. It
was also fun to learn about erroneous historical assumptions made due to
insufficient rigor. IIRC, until at least the 1870s, it was believed that any
continuous function must be differentiable almost everywhere, and that in fact
this should be obvious. It turns out that one can construct continuous
functions that are nowhere differentiable!

This was one of my favorite topics in the class:

[http://en.wikipedia.org/wiki/Construction_of_the_real_number...](http://en.wikipedia.org/wiki/Construction_of_the_real_numbers)

~~~
xyzzyz
What is even more tricky is that the set of functions differentiable in at
least one point is first category, and thus nowhere differentiable continuous
functions are dense in continuous functions space.

~~~
drbaskin
Not only are they dense, but they are "most" continuous functions! (For an
appropriate interpretation of "most", of course)

------
vilhelm_s
Over at Everything2.com we discussed this 10 years ago:
<http://everything2.com/title/2%255E.5+%253D+2>

~~~
user24
mblase's explanation was a very clear way of explaining the problem, with the
two parallel lines never quite touching because you can't get a line to zero
length by cutting it in half repeatedly, no matter how many times you do it.
Thanks for sharing.

~~~
qntm
I actually found mblase's explanation to be just as inaccurate as the one in
the original story. "Infinity is not a number and you can never get there" is
never a good defence in well-formed arguments about limits. It just sidesteps
the question and says that it is meaningless, incidentally doing the same to
all of mathematical analysis.

"Just because a series of curves tends to a limit curve doesn't mean that the
series of lengths of those curves tends to the length of the limit curve" is
the whole of the explanation.

------
martincmartin
For any given n, all the points of the "right angled fractal beast" lie within
some distance epsilon(n) of the circle. As n increases, epsilon(n) approaches
zero, so in the limit, you have a circle, and at every step along the way, the
length is 4. It's just that the length of the resulting shape isn't the limit
of the lengths of the sequence of shapes.

~~~
kwantam
In the limit you do _not_ have a circle.

If we write down area and perimeter error e_A and e_P as a function of n, in
the limit of infinite n e_A is proportional to 1/n. However, e_P is
independent of n, since the perimeter does not change at any step. Thus, if
the perimeter was unequal to begin (clear by inspection), it does not get
closer through this approximation.

The mistake is in accepting that anything that looks like a circle must be a
circle. The fractal beast has area arbitrarily close to that of a circle, but
it's clear once we look at e_A and e_P that it is not an approximation of its
perimeter.

~~~
RiderOfGiraffes
I believe you are mistaken.

Pointwise, the limit is a circle. Every point on the enclosing shape gets
mapped to a sequence of points. Each of these sequences has a limit, that
limit is on the circle. The resulting implied mapping of the original square
to the circle is a continuous bijection. By every sense that we usually talk
of the convergence of lines, this enclosing shape does approach the circle in
the limit.

The point is that the mapping involved is not a length-preserving mapping.

~~~
defen
Wouldn't you only get a countably infinite number of points on the circle?
N.B. I'm not a mathematician

~~~
RiderOfGiraffes
Only countably many points "arrive" at their destination in a finite number of
steps. There are uncountably many points that never "arrive," but whose limit
is on the circle.

The usual picture given to non-mathematicians about sequences and limits can
end up being strongly misleading in cases like this. Just because a sequence
never "gets there," the limit is still the limit. This is the same kind of
murky area that talks about 0.999... recurring never "getting to" 1. It
doesn't have to "get to" 1 because it's never travelling.

It's also the kind of problems that arise when talking about proof by
induction. Talking about dominoes falling down is, in the longer term, very
misleading. We prove P(1), and we prove that P(n) => P(n+1), then they are all
true. They don't become true one by one, they are simply all true - it is what
it is.

I hope that helps.

I'm thinking of starting a blog to talk about things like this - it falls
between the levels of the non-mathematician and true researcher.

~~~
defen
I think I see what you're getting at, but I guess I still don't fully
understand. Any thoughts on proving that the perimeter is pi without using the
knowledge that the limiting shape is a circle? Does that question make sense?
To me it seems strange that for any integer number N (number of times this
chunking operation is applied), the perimeter is 4, but somehow in the limit
the perimeter is not 4.

Edit: Nevermind, reading cousin_it's posting I think I've got a handle on it.
My confusion was exactly the difference between a sequence of approximations,
and the limit itself.

------
huhtenberg
A simpler and a bit more elegant version -
<http://i52.tinypic.com/2daesf4.png>

------
nazgulnarsil
troll math/physics are the only 4chan releated things that consistently make
me crack up.

------
cousin_it
What... ouch. This is not math, this is... I don't even.

Read the post closely. The author introduces a set S of squarey curves
approximating a circle. This set has an obvious correspondence with the
natural numbers: there is curve #1, curve #2, etc. Then the author defines a
function f: S->S that takes curve #n to curve #n+1. Wait, that doesn't sound
right! The function f has no interesting structure whatsoever, it's exactly
equivalent to defining f(n)=n+1 on the naturals. Of course, taking the "limit"
of a function f: S->S makes no sense at all.

What would make sense is taking the limit of a certain function N->C, where N
is the naturals and C is the set of all curves on the plane. That is, the
limit of a sequence of curves (not of a function from curves to curves as the
OP tried to say). To talk about such limits, you need to define what it means
for a sequence of curves to converge - a "topology" on C. There are many ways
to do that, some more outlandish than others. One way is _pointwise_
convergence: assume a parameterization t->C_i(t) on each curves in the
sequence, and require that C_i(t) converges co C_lim(t) for each t separately.

Now, pretty much any reasonable notion of "convergence" on the space of all
curves has to _imply_ pointwise convergence. That is, pointwise convergence is
a very "weak" notion of convergence: if we have a sequence of curves that has
a limit in some reasonable sense, then it had better coincide with the
pointwise limit, dontcha think?

And here we come to the second facepalm moment in the post. Under pointwise
convergence, the sequence of squarey curves under discussion does _not_
converge to some "right angled fractal beast". It converges to the circle. As
n grows, every point on S_n comes closer and closer to some point on the
circle. Ain't nothing more to it.

Now the correct explanation for the original puzzle. Pointwise convergence of
curves doesn't imply that their lengths converge to the limit's length. Hell,
we don't even need 2-dimensional space to show that! A simple _straight line_
will do. Imagine a human traveling a straight road of 1km length in this
fashion: he takes two steps forward, then one step back, then repeats. In the
end he will have traveled about 3km instead of 1km. As we make the human and
his steps tinier and tinier, his movement looks smoother and smoother to an
external observer, but he still travels 3km in total instead of 1. Or maybe
(going back to the 2D space) the human could take a step left, then forward,
then right; this would make his path look like a fine comb that approximates
the straight line more and more closely, but it's always 3x longer. Something
like this is happening in the original puzzle.

Finishing touch: there _are_ notions of convergence where it's true that the
length of the curves in the sequence always converges to the length of the
limit. One such notion says that the _direction of travel_ (velocity vector)
must also pointwise converge to the velocity vector of the limit curve. Under
this definition of convergence, the original sequence of curves does not
converge, because it makes too many sharp turns.

~~~
neilk
On a less trollish note, I had always been puzzled by the related question you
discuss: why Manhattan Distance does not converge to the Euclidean distance.

I'm pretty uneducated about math, so please let me know what trap I'm falling
into.

Your answer isn't yet helping me because you just note that Manhattan Distance
remains the same no matter how many twists and turns there are. Yes, this is
the premise of the question!

But turn up the twistiness to infinity. Now the Manhattan Distance line is
identical to the hypotenuse. There is no point on the Manhattan distance line
isn't also on the hypotenuse and vice versa. They both have an infinite number
of points, of the same aleph-number, I think.

You mention a hypothetical path that doubled back on itself -- it is easy to
see why that would result in a different answer. But the Manhattan Distance
line is not doubling back on itself. Every point is a step towards the goal,
and it doesn't cross itself. All of its deviations from the simple hypotenuse
are infinitely small.

So why are the distances _still_ different?

~~~
simcop2387
The reason it won't converge is that a very large number of the points will
ALWAYS lie off of the line. While you might be able to make something that
looks like a line on a computer screen when you make it smaller and smaller
segments of manhattan distance. it will in fact only ever be able to intersect
the real line (they are infinitely thin remember) at N+1 points at most (where
N is the number of turns you make when following it along the grid in
manhattan distance). because of this you will always have a very large portion
of your approximation lying off of the line.

~~~
neilk
> The reason it won't converge is that a very large number of the points will
> ALWAYS lie off of the line.

No they won't. It moves zero distance horizontally and zero distance
vertically. All the points of the infinitely zigzagging line are on the
hypotenuse.

Of course, now we're back to Zeno; if it's not moving off the line, and it
only makes progress when it's off the line, how can it get _anywhere_?

~~~
RiderOfGiraffes
For every fractally-type enclosing shape, yes they will. And when you talk
about the limit, then you have a circle.

You can't talk about moving "zero distance horizontally and zero distance
vertically" becuase then, as you rightly say, we're back to Zeno. You don't
have an infinitely zigzagging line. That way of trying to think about things
is a dead end, and unhelpful.

Better answers are given elsewhere in this thread.

------
patio11
You can also demonstrate this is incorrect by calculating the area of the
parts of the square cut off, which is the sum of an infinite geometric series.
You will find it doesn't equal the area of the unit square minus the area of
the inscribed circle.

Here, let's try: the square is 1x1. Consider one quarter of the circle: radius
of the circle is 1/2, half a diagonal of the square is sqrt(2) / 2, diagonal
of the removed square is (rt(2) - 1) / 2. Area of the square removed is
((rt(2) - 1) / 2) ^ 2. (This is trivial via the pythagorean theorem, saving
some math.)

Alright, that's the first square we accumulate. Now the magic happens: every
step, we cut the square's side in half, but make two of them. Agree with me so
far? Good. If we cut the side of a square in half, we cut its area to a
quarter, but since we have two squares now the total area is 1/2 of the last
square. Agree with me so far? Good. We can trivially sum infinite geometric
series: t1 / (1 - r), where t1 is the first term and r is the fraction each
term gets multiplied by. In this case, it turns out that in any one quadrant
the sum of the series of squares removed is 2 * ((rt(2) - 1) / 2) ^ 2, or just
(rt(2) - 1)^2 / 2.

Multiply by 4 to get the picture over all four quadrants, and we get 2 *
(rt(2) - 1) ^ 2. A little simplification and we get 2 * (2 - 2 rt(2) + 1) = 6
- 4 rt(2)

So, we've got a unit square, so the area of the square is 1. If we subtract
the area of the infinite series of squares, we get 4 rt(2) - 5 =~ .657. We
expect the area of the circle to be pi / 4 =~ 0.7853975. Thus, the square
minus and infinite series of squares doesn't approximate the known area of a
circle at all.

~~~
spicyj
This isn't quite right; when you divide up your problem, your "quarter circle"
doesn't look the same after each iteration, and so the sizes of the squares
cut off don't follow a geometric sequence. In fact, I'm sure that the area of
the figure does approximate the area of a circle (because no point inside the
circle is ever removed, yet every point outside of it will be removed
eventually).

The problem lies not with the area, but the perimeter; the figure's area turns
out to be unrelated for this problem.

~~~
patio11
Hmh. Its 1 AM here. Give me some time to sleep and think over that. You might
be right: I can't remember why I thought the squares would go geometric. I
thought I had a better reason than "It looks that way on the picture,
kinda...", but it eludes me.

~~~
xyzzyz
If A_n is a sequence of sets bounded by these square like curves, then m(A_n)
-> pi, where m is a Lebesgue measure. Each set is clearly measurable, since it
is closed in Euclidean topology, and thus Borel. You can prove this sequence
converges by seeing that each A_n is enclosed in some circle of diameter 1+e
for some e, and, as n -> \infty, e -> 0.

Thus, argumentation from area will not work here.

------
augustl
This is similar to Zeno's paradox of Achilles and the tortoise:
[http://en.wikipedia.org/wiki/Zenos_paradoxes#Achilles_and_th...](http://en.wikipedia.org/wiki/Zenos_paradoxes#Achilles_and_the_tortoise)

The paradox lies in "infinity" and "never". Achilles will overtake the
tortoise when he's one atom away from the tortoise, and similarly at one point
your corner removal will reach the atom level, where you can no longer reduce
it and maintain a square shape.

At atom level, when your squares consist of three atoms in a L pattern, you
can't reduce it further without distorting the squares

Assuming, of course, that atoms are the smallest particles.

~~~
kd0amg
There is no smallest nonzero distance in Euclidean geometry.

~~~
augustl
My comment was regarding physics in general, not a subset of math/geometry.
Are you saying my comparison to the Zeno paradox is wrong?

~~~
JeremyBanks
I think the analogy is flawed. Using limits we can resolve Zeno's arrow
paradox. For the situation described here limits cannot solve the problem
because it turns out that it is not an accurate model of a circle.

------
csomar
Because in Infinity you can't predict things with sight. Actually, it seems
like the rectangles have become a lined curve, but in reality they aren't.
They are just too small to be noticeable.

4 - π will be that small, tiny difference.

~~~
samstokes
Er, it's about 0.85 - not all that tiny.

~~~
csomar
Oh, sorry

I meant that difference that you no longer see (when you go to infinite), not
0,85.

------
ADRIANFR
The easiest way to informally prove that the demonstration is false is to
imagine starting with a circle in a triangle instead of a circle in a square.
Or with any other weird shape around the circle and follow the same "cutting"
algorithm to infinity. This way you can prove that pi is equal to anything
greater than pi.

------
gyom
Same issue as with fractals. What's the length of Britain's coast ? Infinity ?

------
cies
try to find the perimeter of australia.. depending on the zoom level you find
a different perimeter. :) if the 'troll' from the article adds smoothing to
his 'ad infinitum' squarish circle then he finds pi.

------
finin
Obvously, the science is not settled™.

------
sleight42
There... are... FOUR... lights!

~~~
alanh
Is this a Star Trek TNG reference? Did you just leave that here because, hey,
there’s the number four?

