## Troll Science: Pi

A forum for good logic/math puzzles.

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### Troll Science: Pi

The guy who came up with this is an expert troll. He stumped dozens, and even convinced a few people, with the seemingly logical process with which he derives pi to be four. However, there IS a solution!

Mathematicians, GO!
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thorgold

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### Re: Troll Science: Pi

Spoiler:
In order for a construction similar to that to be accurate, its limit must correctly approach both position and direction (and maybe further derivatives too, I'm not sure). This one gets position right, but nothing else - the end result is tangent to the circle at exactly 4 points, and no amount of iterating or taking the limit of the process will ever change that.
douglasm

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### Re: Troll Science: Pi

douglasm wrote:
Spoiler:
In order for a construction similar to that to be accurate, its limit must correctly approach both position and direction (and maybe further derivatives too, I'm not sure). This one gets position right, but nothing else - the end result is tangent to the circle at exactly 4 points, and no amount of iterating or taking the limit of the process will ever change that.

Spoiler:
Correct. To word it differently, the troll's derivation of four takes advantage of the difference between countable and un-counted infinite strings. Whereas irrational numbers are uncountable, the infinite degression of the square towards the circle has a countable difference from pi, which results in the answer of four.

It's like a staircase for a ten foot drop. The more steps you have, the less steep each one becomes, but the total height reached is still ten.
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thorgold

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### Re: Troll Science: Pi

Sam Hughes has an excellent deconstruction at Things of Interest.
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The EGE

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### Re: Troll Science: Pi

This thread, also. The upshot: the pointwise limit of those shapes is a circle, but the pointwise limit doesn't preserve many features of functions, including arclength. Other definitions of the limits of functions do preserve arclength, but under those definitions the shapes don't approach a circle in the limit (indeed, they have no limit at all).
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phlip
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### Re: Troll Science: Pi

phlip wrote:The upshot: the pointwise limit of those shapes is a circle, but the pointwise limit doesn't preserve many features of functions, including arclength.
This weirds me out more than the notion of pi being four, to be honest.

Sigh. I've spent the last week wishing I'd done more maths at uni. Now I'm reading that thread and wishing I'd taken up the offer to sit in on topology classes last semester.
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JayDee

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### Re: Troll Science: Pi

Oh, this can be solved visually without any math at all!

Spoiler:
Just imagine zooming infinitely close to the infinite limit. No matter how many times you repeat this process, the edge will never be tangential except at four points, everywhere else it will zigzag, creating a perimeter which is longer than the curve it approximates. The funny thing, though, is that at the limit, the area should be accurately numerically approximated. Cool.
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JoeZ

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### Re: Troll Science: Pi

JayDee: (spoilered for long and rambly... and also because it's a spoiler)
Spoiler:
The whole topology thing is basically: what does it mean for one function to be "close" to another one?

Simple answer is just to take a pointwise limit... lim_{n\to\infty}f_n = f_L iff lim_{n\to\infty}(f_n(x)) = f_L(x) for all x. To explain the difference between this and some of the other ways of doing it, take the sequence of functions f1(x)=atan(x), f2(x)=atan(2x), f3(x) = atan(3x), etc... So, to find the pointwise limit, fL, take for each x:
\begin{align} f_L(x)&=\lim_{n\to\infty}{f_n(x)}\\&=\lim_{n\to\infty}{\arctan(nx)}\\&=\begin{cases}-\frac{\pi}2&x<0\\0&x=0\\\frac{\pi}2&x>0\end{cases}\\&=\frac{\pi}2 \text{sgn}(x)\end{align}
which isn't even continuous...

Now, one thing to notice with this limit is that while each point in the function does eventually tend towards its limit, ones closer to 0 take longer to do so... and for any large n, there are still points of fn(x) that are quite far from fL(x). They will eventually move out and approach it, but even when they do so, there'll be more points even closer to 0 which still haven't. So even though each point individually approaches fL(x), they don't all do it together. In that thread I linked to, Cycle brings up some other ways of taking the limit that do require that... the first is the C0 sup norm, where you need exactly that - under that norm, the limit of those atan functions isn't the sign function, because each one has some point that differs from fL by nearly pi/2, and that norm defines the "distance" between functions as "how far apart are they at their furthest point?". Under that system, our atans have no limit at all. What this extra restriction gives is continuity - the limit of a bunch of continuous functions is itself a continuous function... which wasn't the case in our atan example.

Now, that's just an example to show how different ways of doing the limit can have different results... it doesn't really apply to this case, since those shapes do approach the circle even under the C0 norm. However, to take the length of a curve, you need the derivative - that's how the arc-length formula works. So in order for the length to be consistent, we need to require that the result be the limit both of the curve, and of its derivative. This is what douglasm was talking about in the first post here, and the C1 norm that Cycle mentioned in that post I linked to. For the limit of the initial functions to exist in that norm, not only does every point need to approach its final value, and not only do they need to do it together, but the same has to be true of the derivative.
So now look at the sawtooth graphs from that qntm page again... the derivatives of the sawtooth graphs are always 1 or -1 (or undefined, on the corners). But the final function is just a horizontal line... derivative 0. So under this norm, they don't approach it. And they need to, if we want arclength to be preserved.
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phlip
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### Re: Troll Science: Pi

Spoiler:
Well, the perimeter of the "circle" would be higher than a real circle because it never cuts down, it only slices. A zoom in on that circle would show jagged edges that add to the perimeter of the non-circle.
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math

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### Re: Troll Science: Pi

this one is just as simple as the square with a line:

troll says "take a square with edge length = 1. connect two opposite edges by a staircase of lines. staircase goes all the way to left and all the way up, i.e., distance covered=2. make distance before drop infinitely small by increasing # of stairs. sqrt(2)=2!!!!!"

just like .9999...=1, this infinite series preserves original value and will never reach the straight diagonal proposed
koobz2142

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### Re: Troll Science: Pi

it's simply taxicab geometry. you can't 'infinitely' repeat it to infinity and therefore make the assumption it goes in a curved shape. rather it retains that jagged edge thus retaining the original length of the square, just at an infinitely small level. according to taxicab geometry, it will never be a curved line even if you take the limit to infinity.

Micali

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### Re: Troll Science: Pi

Micali wrote:it's simply taxicab geometry. you can't 'infinitely' repeat it to infinity and therefore make the assumption it goes in a curved shape. rather it retains that jagged edge thus retaining the original length of the square, just at an infinitely small level. according to taxicab geometry, it will never be a curved line even if you take the limit to infinity.

This is not true. For any e>0, one can come up with an iteration N (sufficiently large) after which every point in the constructed shape is within e of the nearest point on the circle. In the limit as N->infinity, e can go to 0. Thus, the circle, is actually the limit of the process.

No, the only logical flaw is in assuming that the limit of the length of a series of curves is the length of the limit curve of that sequence. The above image depicts a counterexample to that premise, and is therefore a proof that it does not hold. See the link to Sam Hughes' blog above.

quintopia

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### Re: Troll Science: Pi

The EGE wrote:Sam Hughes has an excellent deconstruction at Things of Interest.
Do I understand it correctly that 1) "after" infinitely many steps the resulting line is indeed identical to the circle, but 2) the perimeter is only 4 after finitely many steps, but "after" infinitely many steps it's "suddenly" PI?
charonme

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### Re: Troll Science: Pi

charonme wrote:
The EGE wrote:Sam Hughes has an excellent deconstruction at Things of Interest.
Do I understand it correctly that 1) "after" infinitely many steps the resulting line is indeed identical to the circle, but 2) the perimeter is only 4 after finitely many steps, but "after" infinitely many steps it's "suddenly" PI?

Yes. Given that you are taking the limit of the curve to be limit of the points on the curve, then the length of the limit needn't be the limit of the length. You can force those two things to be the same, but then the limit of the rectilinear curve doesn't exist.
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jestingrabbit

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### Re: Troll Science: Pi

I recall seeing the stair-step square-root-of-two version in one of Martin Gardner's anthologies as a child. As far as I'm concerned, the lesson here is that, given an infinite sequence s and a function f that operates on the sorts of things in that sequence, \lim_{i\to\infty}f(s_i) does not necessarily equal f(\lim_{i\to\infty}s_i). And, really, why would it be? It may be intuitive that they'd be equal, but intuition is often false when it comes to limits and infinite series.

(You want proof of my assertion that they're not always equal? You've already got it. It's up there at the top of the page with a trollface on it.)
baf

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### Re: Troll Science: Pi

You could similarly propose a "circle" like a Koch snowflake and use that to prove that the value of Pi is infinite.
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uncivlengr

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### Re: Troll Science: Pi

there are ways of constructing a series of functions where the limit of the lengths will be infinite, but the length of the limiting function will be PI, but the koch snowflake is not the way. The series of finite koch iterations does not converge to anything except the ideal koch snowflake

For example instead going from N sawtooths of height H to 2N sawtooths of height H/2, you can go from (N,H) to (4N, H/2)
charonme

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### Re: Troll Science: Pi

uncivlengr wrote:like a Koch snowflake
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uncivlengr

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### Re: Troll Science: Pi

You don't need some complex fractal-type pattern... you can just do something simple like:
Spoiler:
circle.gif (30.67 KiB) Viewed 13816 times
Points that get arbitrarily smaller and approach a circle, but get arbitrarily pointier, so have total arclength as an arbitrarily high multiple of their net distance travelled around the circle.
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phlip
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### Re: Troll Science: Pi

A circle, by definition, has all of its points equidistant from the center. That is not an actual perfect circle. It is just a n-gon with an infinitely small number of sides. The shape will still retain its vertices. A circle has no vertices, the figure above, has an infinite of them.
zerkrox

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### Re: Troll Science: Pi

zerkrox wrote:That is not an actual perfect circle.

To what does the word "that" refer?

There are (at least) two different things you might be talking about:

--an infinite sequence of polygons (which is not a shape, but an infinite sequence of shapes, each one of which is a polygon)

or

--the pointwise limit of that sequence of polygons. Which is one shape. And which happens to be a circle.
skullturf

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### Re: Troll Science: Pi

Regardless of the sides, no matter how many, a polygon will never become a circle, only a ∞-gon. That will always just be an infinitely close approximation.
The concept is similar to anti-aliasing. From afar it may look like a circle, but if you look closer you will see the sides and vertices of the polygon, of which a circle doesn't have.
zerkrox

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### Re: Troll Science: Pi

You may want to review the principles of calculus before making statements like that.

A regular polygon with an infinite number of sides is a circle.
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uncivlengr

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### Re: Troll Science: Pi

zerkrox wrote:Regardless of the sides, no matter how many, a polygon will never become a circle, only a ∞-gon. That will always just be an infinitely close approximation.

True, a polygon is not a circle.

But, one can have an infinite sequence of polygons, defined in such a way as to have a pointwise limit. If you set up the sequence of polygons in the right way, the limit is a circle.

You might be tempted to reply "But we never get there!" That objection misses the point. For example, consider the sequence { 1, 1/2, 1/3, 1/4, ... }. The limit of that sequence is the number 0. Saying "We never get to 0" misses the point. Sure, if you listed each element, or paused at each element, you would never get to the end. But it's standard in mathematics to consider infinite sets or sequences "as a whole" and to ask about the limit of an infinite sequence.

uncivlengr wrote:A regular polygon with an infinite number of sides is a circle.

I wouldn't say it that way. I would just say it's meaningful to ask about the limit of an infinite sequence of polygons with increasing numbers of sides.
skullturf

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### Re: Troll Science: Pi

http://www.scribd.com/doc/95216567/Solution-to-Circle-Con-Nun-Drum

Not much of a proof, but I didn't want to get buried in the details and wanted it be accessible.
shockwave121

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### Re: Troll Science: Pi

For this infinite sequence of curves: the length of the limit is not equal to the limit of the length.

In this particular case, the area of the limit is equal to the limit of the areas: but even this is not true in all cases. For example, define:
f_n(x) = \begin{cases} n & \text{for $0 < x < \frac{1}{n}$} \\ 0 & \text{elsewhere}
\end{cases}
So f_n(0) = 0 for all n, and for every x > 0 there exists an integer N such that f_n(x) = 0 for all n > N. So:
f(x) = \lim_{n \to \infty}f_n(x) = 0
for all x.
But:
\int_0^1 f_n(x)\, dx = 1
while
\int_0^1 f(x)\, dx = 0

By defining f_n(x) = n^2 for 0 < x < \frac{1}{n} we can make the limit of the areas infinite, while the area of the limit is still zero.
mward

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### Re: Troll Science: Pi

I think that the best way to resolve this is to realize that, even after infinitely iterations, there are still many points (uncountably many, actually) on the circle that never touch the folded in square. This means that 'most' of the circle do not touch the folded in square, so they are entirely different shapes.
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tomtom2357

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### Re: Troll Science: Pi

tomtom2357 wrote:I think that the best way to resolve this is to realize that, even after infinitely iterations, there are still many points (uncountably many, actually) on the circle that never touch the folded in square. This means that 'most' of the circle do not touch the folded in square, so they are entirely different shapes.
After infinity iterations, every point on the folded in square is on the circle, and vice-versa. The limit of the folded in squares is precisely a circle.
mward

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### Re: Troll Science: Pi

I'm reiterating a point that has been made previously, but the key is this:

The limit of a sequence can have different properties from the elements of the sequence.

Here's a simpler example. Consider the sequence {1, 1/2, 1/3, 1/4, ...}. Each element of that sequence is positive. However, the limit of the sequence is the number 0, which is not positive.

Anybody who says things along the lines of "The limit of 0 is only approached, never reached" is missing the point, in my opinion.

The limit of the above sequence is the number 0. The limit isn't "sort of" zero; it's not some kind of nebulous number that's "almost" zero but not quite. The limit is exactly zero.

Similarly, the limit of the above polygons (using reasonable definitions) is a circle. That's not where the problem lies. However, this circle can turn out to have properties that are different from the polygons in the sequence (namely, the length of its perimeter).
skullturf

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### Re: Troll Science: Pi

This is a simple problem derivative from the "Zeno's Aporias". The guy is making equality between continuous set of points and discrete set of points. And this is wrong (as we already know from the quantum physics)

Here is another troll problem which is not the same but can be counted as "similar":

Problem: Which set has more points - the set of all natural numbers or the set of all even numbers?

The answer is simple - you can't really tell!
wattie

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### Re: Troll Science: Pi

This is a simple problem derivative from the "Zeno's Aporias". The guy is making equality between continuous set of points and discrete set of points. And this is wrong (as we already know from the quantum physics).

And again, your objection misses the point; and the quantum physics has no relevance here, as it's a purely mathematical problem. The point-wise limit of the series of curves is exactly the circle; the problem is that the point-wise limit does not necessarily preserve the curve's length.

Problem: Which set has more points - the set of all natural numbers or the set of all even numbers?

The answer is: The set of all natural numbers and the set of all even numbers have the same number of elements! Meaning that there exists a one-to-one correspondence between the two sets; namely, y=2*x. (Set theory 101.)
Mike Rosoft

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### Re: Troll Science: Pi

Mike Rosoft wrote:And again, your objection misses the point; and the quantum physics has no relevance here, as it's a purely mathematical problem. The point-wise limit of the series of curves is exactly the circle; the problem is that the point-wise limit does not necessarily preserve the curve's length.

Yes, it's a math problem and yes - it has relation to quantum physics. The relation is that similar math problem gave birth to to the quantum physics. And the answer to the problem can be given in the Bulgarian high schools already - it is that "you cannot compare discrete (finite) set of points with continuous (infinite) set of points". End of story

Mike Rosoft wrote:The answer is: The set of all natural numbers and the set of all even numbers have the same number of elements! Meaning that there exists a one-to-one correspondence between the two sets; namely, y=2*x. (Set theory 101.)

Again wrong - infinities cannot be compared. You cannot count the number of elements in these sets.
wattie

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### Re: Troll Science: Pi

wattie wrote:
Mike Rosoft wrote:The answer is: The set of all natural numbers and the set of all even numbers have the same number of elements! Meaning that there exists a one-to-one correspondence between the two sets; namely, y=2*x. (Set theory 101.)

Again wrong - infinities cannot be compared. You cannot count the number of elements in these sets.

I'm not sure where you got the idea that there's some kind of "rule" that you can't compare infinities.

There's a well-developed area of mathematics concerned with the sizes of infinite sets. It is, indeed, standard to compare various infinite sets with one another. Some pairs of infinite sets have the same number of elements, and some pairs of infinite sets have different numbers of elements.

http://en.wikipedia.org/wiki/Cardinal_number
skullturf

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### Re: Troll Science: Pi

wattie wrote:Yes, it's a math problem and yes - it has relation to quantum physics. The relation is that similar math problem gave birth to to the quantum physics. And the answer to the problem can be given in the Bulgarian high schools already - it is that "you cannot compare discrete (finite) set of points with continuous (infinite) set of points". End of story

The problem is at best tangentially related to quantum physics. And I don't know what you mean by comparing a finite and an infinite set - you don't seem to understand the concept of a limit.

wattie wrote:Again wrong - infinities cannot be compared. You cannot count the number of elements in these sets.

What I said is the standard way of comparing sets in set theory. Why can I say that the sets {0, 1, 2} and {a, b, c} have the same number of elements? Because there exists a one-to-one correspondence between the two: 0 -> a, 1 -> b, 2 -> c.

The same definition can be used for infinite sets. The set of natural numbers ({0, 1, 2, ...}) can be put into a one-to-one correspondence with its subset, the set of even numbers ({0, 2, 4, ...}); the function is: y=2*x. (There exists a whole class of sets that have the same number of elements, or cardinality, as the set of natural numbers; they are called countably infinite sets.) On the other hand, it can be proven that the set of real numbers cannot be put in a one-to-one correspondence with the set of natural numbers.

Mike Rosoft
Mike Rosoft

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### Re: Troll Science: Pi

Am I the only one who'd love to see a troll childrens math book with lots of these kinds of proofs?
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HungryHobo

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### Re: Troll Science: Pi

HungryHobo wrote:Am I the only one who'd love to see a troll childrens math book with lots of these kinds of proofs?

I'd rather be able to find a website where you can actually find tons of those troll logic/troll sciences.
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Pingouin7

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### Re: Troll Science: Pi

The first time I saw this fallacy, it involved a diagonal line rather than a circle. The lesson I took away from it: you can't just assume in general that the limit of a property of a sequence is equal to the property of the limit of the sequence.

Really, this should be obvious. Consider the sequence 1, 1/2, 1/4, ... 1/2^n ... Every single term in this sequence is positive, but the limit isn't.
Or, heck, consider the sequence 1, 2, 3... Every term in this sequence is finite. The limit isn't.
So, the troll mathematician gives us a sequence of curves. Each term in this sequence has length 4. We're supposed to conclude that the limit of the sequence also has length 4? Why? Should we also conclude that the limit is composed entirely of horizontal and vertical segments?
baf

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### Re: Troll Science: Pi

That's one of the best succinct responses to this that I've seen.
skullturf

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### Re: Troll Science: Pi

baf wrote:Really, this should be obvious. Consider the sequence 1, 1/2, 1/4, ... 1/2^n ... Every single term in this sequence is positive, but the limit isn't.
Or, heck, consider the sequence 1, 2, 3... Every term in this sequence is finite. The limit isn't.
So, the troll mathematician gives us a sequence of curves. Each term in this sequence has length 4. We're supposed to conclude that the limit of the sequence also has length 4? Why? Should we also conclude that the limit is composed entirely of horizontal and vertical segments?

to be fair, at least with 1, 1/2, 1/4, ... 1/2^n each term is at least getting closer to zero.
It makes some intuitive sense.
(Though I've been coding too much that I think in terms of positive and negative zero. )

with the troll pi one it's just 4,4,4,4,4,4,4.... unchanging until suddenly the limit of 3.14...
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HungryHobo

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### Re: Troll Science: Pi

On the other hand, how did ancient mathematicians estimate the value of pi? By constructing inscribed and circumscribed polygons and computing their perimeters.

How come this limit works, but the original one doesn't?
Mike Rosoft

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