• Create Account

## Pi = 4. Discuss.

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

92 replies to this topic

### #41Wavarian  Members

Posted 02 December 2010 - 09:19 PM

This is silly. It's like looking at a parabola and stating that since it has an asymptote of x = 0, it must at some point into infinity have x = 0 actually lying on the parabola's curve.

### #42JoeCooper  Members

Posted 02 December 2010 - 10:37 PM

Can we assume that x^2+y^2=r^2 is a valid test for whether or not a vertex lies on a circle? Is that a given?

If so, in a regular polygon (the method this spoofs that - I think - we're trying to differentiate it from), every specified vertex lies on a circle.

But you draw a line from that vertex to the next and the midpoint of that line must deviate from a circle; x^2+y^2 will be less than r^2.

Clearly, it's not a circle, many points on it will do not satisfy the equation of a circle. But if I do something like this...

This shows the sum total of the mid-point's hypotenuse divergence from r in a regular polygon. As we add vertices, the divergence from a circle approaches zero; the difference between our circle approximation and a hypothetical true circle is reduced.

The excavation method does not have this characteristic.

Both the square excavation method and regular polygon contain vertices which satisfy the points, alternating with ones that do not. But as we iterate the square excavation method, the sum total of the vertice's deviation increases.

I'm not really sure how to make a function for this one - I could try if necessary - but for now I'll but I'll run some numbers here manually.

In the first iteration, showing the square, we have four specified vertices for which their hypotenuse is sqrt(2) (1.41etc) each, so we can say they deviate from r^2 (1) by 0.41etc, or 1.65ish total. (There are also four points that do fit on the circle, at the mid-points.)

On our second iteration, the deviant vertices' average hypotenuse is sqrt(cos(45)^2+1) or 1.22ish, so the deviation is that, minus one; 0.22ish. There are now 8 of these, so the total sum of the midpoint vertices' deviation from r is now increased to 1.76.

This contrasts with the regular polygon which reduces deviation from the definition of a circle by adding vertices.

Thus we can quantify that squares are an inferior approximation of a circle compared to a regular polygon.

Or something.

Quote:
 You've given no good reason why a valid "iterative" approximation of a quantity should necessarily converge (you said "reduce"?) indefinitely to some limiting value

In the above pitch, a correct procedure should converge onto zero when measured this way because a non-zero value shows failure to satisfy the equation of a circle. It should be more like a circle because the goal is to approximate a circle's properties by creating a shape to approximate a circle, then measuring its properties.

Is any of this relevant or am I barking up the wrong tree?

[Edited by - JoeCooper on December 3, 2010 8:37:55 AM]

### #43BitMaster  Members

Posted 03 December 2010 - 01:23 AM

Quote:
 Original post by HodgmanI'm assuming that "1/inf" is the smallest value that is still greater than zero.

Such a value does not exist. Whenever you believe you have found such a value x, then x/2 is smaller, positive and not zero either (this works regardless of whether we are working with real or rational numbers).
In actual textbook math (not school-math, university-level math) there are preciously few points were an "infinity" can be. There are some more specialized branches of mathematics which allow actually using infinity as a more regular symbol, but even there you have to be extremely careful what you are allowed to do with it. There is a Wikipedia article about that.
The rest of the post kinda dies with that.

Maybe I should not have skimmed the thread like that but so far it seems arguments are on the "drawing sketches, waving hands"-level largely.

The only proper definition for the convergence of a sequence S(n) to the limit C I know is
$\forall \epsilon > 0: \exists n_0 \in \mathbb N: \forall n > n_0: |S(n) - C| < \epsilon$
and I don't think I have seen anyone trying to apply this to the problem. That would of course first require us to put the happy sketch into an actual formula you can work with but I feel a bit lazy today.
My personal guess is the result will be that the happy sketch idea simply does not converge because for every finite value the circumference is 4 while the limit would have to be pi. So even for $\epsilon = \frac 1 2$ it's not possible to find an $n_0$ as required.

### #44JoeCooper  Members

Posted 03 December 2010 - 01:39 AM

Quote:
 My personal guess is the result will be that the happy sketch idea simply does not converge because for every finite value the circumference is 4 while the limit would have to be pi.

Isn't that just saying that it doesn't work because the result isn't Pi? Or, it should converge on Pi and it doesn't?

I just got my rear handed to me on a platter for that.

### #45BitMaster  Members

Posted 03 December 2010 - 01:44 AM

Well, as I said that is my personal guess. Actually proofing that would require a bit of work I currently do not have the time (and honestly, also not the motivation) for.

I'd be happy though if the rest of the post steers the topic toward the slightly more mathematical. A lot of the things I read while I glanced over the thread made me wince really, really badly even though my last 'real' math is seven, eight years behind me now. It's a bit frustrating, like watching people who are unwilling to believe that $0.\overline{9} = 1$ but do not have the mathematical background to actually argue their point or even understand the proves people who do know what they are talking about are supplying. :(

### #46nilkn  Members

Posted 03 December 2010 - 04:27 AM

In my first post I asserted that the limit curve is not a circle. I was wrong. It does converge to a circle, and this is clear just from the definitions.

The root issue here is that limits and derivatives do not generally commute, and they fail to do so here for a very clear geometric reason.

Let c_n(t) : [0,1] --> R^2 designate the curve after the nth iteration with c_0 being the square, and let c(t) designate the unit circle.

The length of the limit curve is the integral over [0,1] of |(limn-->infty c_n)'(t)|. The limit of the lengths of the curves is limn-->infty of the integral over [0,1] of |c_n'(t)|. The limits don't commute with the integrals. The reason in this case is that c'(t) is a vector that could point in any direction depending on t whereas c_n'(t) is always either horizontal or vertical. Observe that this does *not* happen when one uses the traditional circumscribed n-gons that we're all familiar with.

[Edited by - nilkn on December 3, 2010 12:27:13 PM]

### #47Antheus  Members

Posted 03 December 2010 - 04:48 AM

I think the trick also comes from fractal nature of the problem - unlike typical shapes, the perimeter and area of a fractal are not necessarily correlated.

Koch snowflake has infinite perimeter, but finite area.

I'd say that the problem here comes from implication that value of Pi follows directly from perimeter, whereas even when using the curve it must be derived from area.

### #48Platinum314  Members

Posted 03 December 2010 - 06:31 AM

The limit curve is 'infinitely' squiggly/pointy (Non-differentiable everywhere).

The area is being monotonically reduced each time with the limit converging to PI*r^2 as any area outside of the circle eventually gets cut out. The perimeter is not getting reduced each time, and has a limit of 4. We created a shape with the same area as a circle, but not a circle.

The limit need not equal the value of the function. For example the piece wise function { 1 if x = 0; 0 otherwise } the limit as x->0 exists and equals 0. The value at 0 is 1.

### #49Luckless  Members

Posted 03 December 2010 - 06:47 AM

Quote:
 Original post by Platinum314We created a shape with the same area as a circle, but not a circle.

But, you didn't do that. You always have a saw tooth edge that comes near the circle, but must always include an area that exists outside the boundary of the circle. It therefore must have an area Larger than the circle itself, not equal to.

### #50nilkn  Members

Posted 03 December 2010 - 06:52 AM

Quote:
 Original post by Platinum314The limit curve is 'infinitely' squiggly/pointy (Non-differentiable everywhere). The area is being monotonically reduced each time with the limit converging to PI*r^2 as any area outside of the circle eventually gets cut out. The perimeter is not getting reduced each time, and has a limit of 4. We created a shape with the same area as a circle, but not a circle.The limit need not equal the value of the function. For example the piece wise function { 1 if x = 0; 0 otherwise } the limit as x->0 exists and equals 0. The value at 0 is 1.

In this case however the curves not only converge to the circle but they do so uniformly. Let c_n(t) be the curve after the nth iteration. For any epsilon > 0 draw an annulus a_epsilon of width epsilon whose central circle is the unit circle. For sufficiently large n the curve c_n(t) is contained entirely within this annulus a_epsilon. This is the condition for uniform convergence to the limit curve (the circle in this case).

The issue in this case is that two functions can be close yet their derivatives can differ by a lot. More precisely, the derivative of a limit of a sequence of functions need not coincide with the limit of the derivatives. I actually think you can say something stronger in this case case: despite the fact that the limit curve is everywhere differentiable the sequence of derivatives c_n'(t_0) for a fixed t_0 in [0,1] does not have a limit.

### #51Platinum314  Members

Posted 03 December 2010 - 06:57 AM

Quote:
Original post by BitMaster
Quote:
 Original post by HodgmanI'm assuming that "1/inf" is the smallest value that is still greater than zero.

Such a value does not exist. Whenever you believe you have found such a value x, then x/2 is smaller, positive and not zero either (this works regardless of whether we are working with real or rational numbers).
In actual textbook math (not school-math, university-level math) there are preciously few points were an "infinity" can be. There are some more specialized branches of mathematics which allow actually using infinity as a more regular symbol, but even there you have to be extremely careful what you are allowed to do with it. There is a Wikipedia article about that.
The rest of the post kinda dies with that.

Maybe I should not have skimmed the thread like that but so far it seems arguments are on the "drawing sketches, waving hands"-level largely.

The only proper definition for the convergence of a sequence S(n) to the limit C I know is
$\forall \epsilon > 0: \exists n_0 \in \mathbb N: \forall n > n_0: |S(n) - C| < \epsilon$
and I don't think I have seen anyone trying to apply this to the problem. That would of course first require us to put the happy sketch into an actual formula you can work with but I feel a bit lazy today.
My personal guess is the result will be that the happy sketch idea simply does not converge because for every finite value the circumference is 4 while the limit would have to be pi. So even for $\epsilon = \frac 1 2$ it's not possible to find an $n_0$ as required.

Exactly, you can't define an infinitesimal as the smallest number that is still greater than zero, because you can always generate a real number between that number and zero, and so on. The set (0,n) is open on the lower bound. It makes no sense to ask what the 'first' element of the set is.

You can however define different types of infinities. You can construct ordinal numbers by defining them in terms of sets. We start with defining zero as the empty set. Where the number x is the set that contains all subsets corresponding to the numbers less then x.Then you can define omega as the set that contains all finite ordinals (It's bigger than any finite ordinal number). However now you are able to construct omega+1 which is bigger than that all the way to 2*omega ... omega^2 ...

Similarly you can define an infinitesimal as 1/omega. Its larger than zero but smaller than any 1/n where n is a finite ordinal. But now you can construct 1/omega+1 ... 1/2*omega ... 1/omega^2 ...

This type of infinity is different than a cardinal infinity, such as the 'amount' of things we have in a set. Aleph Null is defined to be the size of a countable set(Natural numbers, Integers, Rationals) The infinity of the continuum (The real numbers) is uncountable and larger than Alpha Null (The continuum hypothesis says that it is Aleph One) You can always create a larger cardinal by taking the power set of a set (The set that contains all of the subsets of the specified set)

Things then can get really weird from here on out as you can define 'large cardinals' that are inaccessible by a succession of Alephs or even a limit of Alephs. Then you can create hyper-inaccessible cardinals, then hyper-hyper-inaccessible cardinals ...

Sorry I get carried away.

### #52BitMaster  Members

Posted 03 December 2010 - 07:00 AM

Quote:
Original post by Talroth
Quote:
 Original post by Platinum314We created a shape with the same area as a circle, but not a circle.

But, you didn't do that. You always have a saw tooth edge that comes near the circle, but must always include an area that exists outside the boundary of the circle. It therefore must have an area Larger than the circle itself, not equal to.

The problem is, as soon as limits are involved an argumentation like that just does not work anymore. ;) Just take a look at a simple sequence like S(n) = 1/n. Whichever n you put in, it's always bigger than 0. But the sequence is the example of a sequence converging to 0.

### #53Way Walker  Members

Posted 03 December 2010 - 07:57 AM

Quote:
 Original post by JoeCooperIf so, in a regular polygon (the method this spoofs that - I think - we're trying to differentiate it from), every specified vertex lies on a circle.

Actually, that would produce a lower bound. The upper bound is found by placing the midpoints of the edges, not the vertices, of a regular polygon on the circle. I think your analysis wouldn't be changed too much, but I wouldn't be surprised if the equivalent lower bound "spoof" would, like the Koch snowflake, have an infinte perimeter (so infinty < pi < 4 [smile]).

Quote:
 Thus we can quantify that squares are an inferior approximation of a circle compared to a regular polygon.

It's not so simple as showing that it is inferior. What we need is to show that it's inferior in a meaningful way. You can approximate the volume of a cone as a stack of cylinders or as a stack of truncated cones. You can quantify how much better the latter is, but both will produce the same result in the limit as the number of sections in the stack approaches infinity.

One thing that bothers me about this analysis is that it singles out the worst points. If we just look at the best points, the number of points with zero deviation increases linearly for the regular polygons but exponentially for the square excavation. By this metric, the square excavation is quantifiably better.

Perhaps the area enclosed between the two would be a better metric (since it's the sum total deviation for all points normalized by the arc length), but that would seem to be a better measure of how quickly the approximations converge and not how well they converge.

As I said before, I think the best criterion will be related to the convergence of the tangents. See nilkn's post for a good explanation of that issue.

Quote:
 Original post by AntheusI'd say that the problem here comes from implication that value of Pi follows directly from perimeter, whereas even when using the curve it must be derived from area.

It's maybe more robust to derive it from the area, but the most basic definition of pi is a relation between the perimeter and diameter.

### #54Antheus  Members

Posted 03 December 2010 - 08:31 AM

Quote:
 Original post by Way Walkerbut the most basic definition of pi is a relation between the perimeter and diameter.

The geometric definition is either that of circumference or that of area.

Quote:
 It's maybe more robust to derive it from the area,
Except that in this case the curve does not define Pi, if my assumption about difference between perimeter of curve and area holds.

### #55way2lazy2care  Members

Posted 03 December 2010 - 08:41 AM

the problem really boils down to the square/polygon being turned into a concave polygon rather than a convex polygon. Geometrically a lot of things no longer apply that make the op wrong.

### #56Luckless  Members

Posted 03 December 2010 - 10:09 AM

Quote:
Original post by BitMaster
Quote:
Original post by Talroth
Quote:
 Original post by Platinum314We created a shape with the same area as a circle, but not a circle.

But, you didn't do that. You always have a saw tooth edge that comes near the circle, but must always include an area that exists outside the boundary of the circle. It therefore must have an area Larger than the circle itself, not equal to.

The problem is, as soon as limits are involved an argumentation like that just does not work anymore. ;) Just take a look at a simple sequence like S(n) = 1/n. Whichever n you put in, it's always bigger than 0. But the sequence is the example of a sequence converging to 0.

Like you said, 1/n, where n is any positive value, never actually equals zero. The area enclosed by the example shown in szecs's first post is that of one that neither equals the area of the circle, nor is it one who's shape is similar to that of a true circle.

I don't know what kind of kindergarten you went to, but back in my day circles didn't have a series of right angles on them.

### #57JoeCooper  Members

Posted 03 December 2010 - 10:49 AM

Quote:
 Actually, that would produce a lower bound. The upper bound is found by placing the midpoints of the edges, not the vertices, of a regular polygon on the circle.

Err, right. Still, the analysis should yield the same result either way.

Quote:
 One thing that bothers me about this analysis is that it singles out the worst points. If we just look at the best points, the number of points with zero deviation increases linearly for the regular polygons but exponentially for the square excavation.

That bothered me too at first and I actually thought about it for a while before going ahead and posting.

Lemme try to explain why I discounted it, but I'm not very articulate so I apologize in advance.

First, singling out the worst points. I don't necessarily do that, actually. If I add in the valid vertices' deviation, all I do in either analysis is add 0 (and half it, I think. I did this earlier and now I can't remember exactly.)

A focus on the bad points is a natural outgrowth of the fact that they're the only ones that differ.

Secondly, the iteration # means nothing outside of a given procedure.

We're only looking to see if a given procedure converges on 0; where they're at when n=2 isn't important. What's important is what happens when we go from n=2 to n=3 and onward toward infinity.

Therefore, the fact that the rate at which a count of valid points increases with n isn't important.

Third, it's also counterbalanced by the fact that in the square'd circle, there is an exactly equal exponential increase in bad points. In fact, that is how the sum of the deviation increases even though the per vertex deviation increases.

Quote:
 Perhaps the area enclosed between the two would be a better metric

I disagree; the area doesn't predict the shape's perimeter.

Measuring the perimeter is an integral part of the procedure.

We can't divorce the way we plan to use the shape from how we assess its fitness for purpose.

If the procedure were based on using its area and diameter to find Pi, instead of its perimeter and diameter, than the fact that its area does approximate a circle of the given diameter suggests it is a valid approach.

<EDIT>It just occurs to me I might've misunderstood that, but uhh, I'll just let you reply.</EDIT>

Quote:
 It's not so simple as showing that it is inferior. What we need is to show that it's inferior in a meaningful way.

Let me try this differently then.

Nevermind relative descriptions like inferior to.

To assess its fitness for purpose, we must show that it approximates a circle. We must do this because we're trying to find Pi, which is defined as "ratio of the circumference of a circle to its diameter". In other words, we are trying to find a property of a circle. Therefore, we try to approximate a circle, then accept that approximation's circumference as an approximation of a circle's circumference.

Key point; we can assess if the procedure is actually measuring a shape that is an actual approximation of a circle.

This procedure iterates over a square, cutting parts away, so that the vertices are closer and closer to a circle.

My analysis is to examine their hypotenuse, and the deviation from a circle, and see if it actually converges on 0 or not.

It does not, so, it is not an approximation of a circle, and that makes it inappropriate the only meaningful way; if it isn't a circle, than any resemblance between its properties and those of a circle is purely coincidental.

[Edited by - JoeCooper on December 3, 2010 5:49:36 PM]

### #58Way Walker  Members

Posted 03 December 2010 - 02:48 PM

Quote:
 Original post by JoeCooperFirst, singling out the worst points. I don't necessarily do that, actually. If I add in the valid vertices' deviation, all I do in either analysis is add 0 (and half it, I think. I did this earlier and now I can't remember exactly.)

It's easy enough to create a construction that exploits the fact that it singles out the worst points.

For example, let f(n) be this metric for the nth iteration of the square excavation. Now, create a star whose inner vertices are along the limiting circle (radius rinfinity), outer vertices along the circle running through the vertices of the circumscribing regular n-gon (radius rn), and ceil(f(n)/(rn - rinfinty)). This creates a concave figure where the given metric will always be greater than or equal to f(n) but produces the correct result. Modifying this slightly, you could control the metric to behave like any function that doesn't go to infinity.

You could also create a sequence that is the limiting circle at all points except for the worst points of the excavation sequence. This is a discontinuous figure with metric f(n) that also produces the correct result.

For a third example, take the circle circumscribing the regular n-gon. The metric is always infinite but, again, it produces the correct result.

Quote:
 Secondly, the iteration # means nothing outside of a given procedure.

It's as meaningful as comparing an O(n) algorithm to an O(2n) algorithm. The same n isn't really comparable between the two, but the way they behave as n increases is comparable.

Quote:

Quote:
 Perhaps the area enclosed between the two would be a better metric

I disagree; the area doesn't predict the shape's perimeter.

As I said, in this case, the enclosed area is just the weighted deviation of all points normalized by the arc length. It's a generalization of the metric you proposed that takes care of the fact that it singles out the worst points (by taking the weighted contribution of all points) and the fact that arcs of the worst deviation would make it infinite (by normalizing by arc length).

Quote:

Quote:
 It's not so simple as showing that it is inferior. What we need is to show that it's inferior in a meaningful way.

This procedure iterates over a square, cutting parts away, so that the vertices are closer and closer to a circle.

My analysis is to examine their hypotenuse, and the deviation from a circle, and see if it actually converges on 0 or not.

It does not, so, it is not an approximation of a circle, and that makes it inappropriate the only meaningful way; if it isn't a circle, than any resemblance between its properties and those of a circle is purely coincidental.

But why does this metric mean anything? Another metric is how close the limiting figure is to a circle, which it passes. You could also consider whether the enclosed area goes to zero, which it also passes. I gave three examples of false negatives by the metric you proposed, so it's not a necessary condition. It may be sufficient, but it cannot reject an approximation.

### #59Way Walker  Members

Posted 03 December 2010 - 03:03 PM

Quote:
Original post by Antheus
Quote:
 Original post by Way Walkerbut the most basic definition of pi is a relation between the perimeter and diameter.

The geometric definition is either that of circumference or that of area.

Quote:
 It's maybe more robust to derive it from the area,
Except that in this case the curve does not define Pi, if my assumption about difference between perimeter of curve and area holds.

I think I maybe misunderstood what you were saying. There's nothing wrong with deriving pi from the perimeter of a circle, the problem comes in assuming that, if the limit of a sequence of curves is a circle, then the limit of the perimeter of that sequence will be the same as the perimeter of the circle. However, you can still (my confusion is perhaps from the word "must"?) derive pi from the limit of the area of that sequence (since the area enclosed between the circle and the curve must go to 0 so the areas will be identical).

### #60BitMaster  Members

Posted 03 December 2010 - 08:06 PM

Quote:
Original post by Talroth
Quote:
Original post by BitMaster
Quote:
Original post by Talroth
Quote:
 Original post by Platinum314We created a shape with the same area as a circle, but not a circle.

But, you didn't do that. You always have a saw tooth edge that comes near the circle, but must always include an area that exists outside the boundary of the circle. It therefore must have an area Larger than the circle itself, not equal to.

The problem is, as soon as limits are involved an argumentation like that just does not work anymore. ;) Just take a look at a simple sequence like S(n) = 1/n. Whichever n you put in, it's always bigger than 0. But the sequence is the example of a sequence converging to 0.

Like you said, 1/n, where n is any positive value, never actually equals zero. The area enclosed by the example shown in szecs's first post is that of one that neither equals the area of the circle, nor is it one who's shape is similar to that of a true circle.

I don't know what kind of kindergarten you went to, but back in my day circles didn't have a series of right angles on them.

That was exactly my point. As soon as a limit is involved you do not have a guarantee there are still any right angles in there. For example one 'correct' method for calculating pi is fitting an n-gon to the circle (either fitting just inside or just outside) and calculating the area of that n-gon, then letting n go towards infinity. Every single finite n-gon is angular but the limit is a circle.
Similarly, if we pick any single point on the circumference of the happy squares method we can keep exact track of it through the iterations (no problem since the circumference remains constant and 0° would be a fixed start point). Now, watching this single point P(n) through the iterations, it is seems obvious that for a sufficiently high n_0, P(n) with n > n_0 will be as close to the real circle as desired (without ever reaching it in the general case, just like 1/n never reaches 0).
I currently see no flaw in saying that for every single point P(n) on the surface, the limit of that point is exactly on the circle. The error creeps in because we have an infinite number of points on the circumference and I expect the cardinality of infinities (not sure this is the right expression, it's been a long time) on the squares and the circles don't fit. That's why pi using the circumference of the happy squares does not work but should work using the area (again, my guess; proof of that bit to whoever is interested enough).

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.