[evan_au]

the circumscribed n-gon tends to 2 * pi faster than the inscribed n-gon

It depends what you mean by "faster":
- The circumscribed polygon has twice the error of the inscribed polygon. So if it is to converge on the same answer (π), it is quite possible that the circumscribed polygon might take larger steps towards the "right" answer
- Graphing and tabulating the successive estimates, it appears that as the subtended angle halves, the step size reduces to one quarter.
- So, in this sense, both the inscribed & circumscribed polygons converge at the same rate

Polygon_Convergence.png (39.98 kB . 609x675 - viewed 1070 times)

There is a way of calculating the sum of an infinite number of steps of reducing size:
- a: Initial step size
- r: ratio of successive step sizes (0.25, in this case)
- s: sum = a/(1-r) = a(1-0.25) = 4a/3

From this, it is possible to estimate the limit as the number of sides approaches infinity (ie subtended angle approaches 0), based on the results with only 96 sides (something the Greeks found practical to calculate)
- And using the results for 96 sides, we can extrapolate the value of π more accurately than if we calculated the perimeter of a polygon with 1536 sides.

[evan_au]

I thought I'd have one last bash at this question of approximating π with polygons....
- I took the table of perimeters of successive polygons
- I extrapolated some of the figures for polygons > 1536 (in grey)
- I applied a geometric series to the convergence

I came up with the table on the right, which shows the successive estimates to π.
- It's pretty close; but it helps that we know the "exact" answer, which was unavailable to the Greeks

Extrapolated_polygons.png (38.15 kB . 1145x295 - viewed 869 times)

Note that to apply the geometric series, you must have calculated the perimeter for 3 successive numbers of polygons, eg 96, 192 & 384 

[Hayseed]

pi is super conditional.  Any wobble in the rotation and pi disappears.  One can set and adjust the ratio of circumference to diameter, with wobble.  Something(a marble) spinning(helixing) in a curved tube can experience this.  And one gets a pi value of 4.  4 diameters for a circumference. Or even more with multiple wobbles.  Like a ratio of 10 diameters for a circumference.

Pi is just the minimum ratio of diameter to circumference, for a rotation.

And if you like analog computers, check out these mechanical calculus machines.

A mechanical way of hitting a moving target while on a moving platform.

Mechanical relativity.

[Colin2B]

pi is super conditional.  Any wobble in the rotation and pi disappears.  One can set and adjust the ratio of circumference to diameter, with wobble. 

Pi is defined as the ratio of a circle's circumference to its diameter. A circle is defined as a curve (the circumference) traced out by a point that moves in a plane so that its distance from a given point (the centre) is constant. So, the circumference of a circle cannot have wobble. So Pi is a constant.

[trushinalexander49 (OP)]

Could you please check the following thing on the computer. It turns out that in my research I came to an interesting conclusion - strictly speaking, an inscribed quadrilateral with unlimited doubling of the sides does not tend to a circle in absolute accuracy, and that's why. Doubling is a harmoniously correct process, which means that in the limit the number of points is "somewhat limited" in comparison with a circle. Strictly it sounds like this: we draw a circle of radius 1 and in the usual way we will inscribe in it first a quadrilateral, then 8, 16, and so on ad infinitum. I argue that any difference of consecutively doubled sides (like - sqrt (2) -sqrt (2-sqrt (2)), sqrt (2-sqrt (2)) - sqrt (2-sqrt (2 + sqrt (2) )) and so on - with further subradical additions, sqrt (2)) will show each time the functional measure of the lag behind the real aspiration of the side to all points of the circle. Indeed, let us first imagine, for convenience, that when we successively reduce the sides of the inscribed polygon, drawing them parallel to the sqrt (2) side for convenience, we seem to rush in the limit to the side, and not to the point, that is, for a while we stopped at one from the sides. We can apply this prototype to each of such sides, which, in the limit, when decreasing, fall on such points that are included in the area of ​​doubling the sides — and only on them. But there will also be points that do not enter the doubling region - this is evident from the fact that with each doubling (starting from the second), when the side is still in the limit, not a point, there will still be an opening between two points taken relative to the axis of symmetry (in this case, the radius), but this side will no longer belong to the original inscribed polygon, which means that if we “scroll” this side for a very small distance to the left (without breaking the inscribed in the circle), there will certainly be in the proper interval and a point located on one side between two points included in the doubling area of ​​this polygon, and on the other side, due to the appropriate selection of a side not belonging to this polygon, not included in the doubling area. On the other hand, it is obvious that the "number of points" missing to the circle in the limit is nothing but an infinitely small difference selected for each of the "main" sides, obtained in the above way. And only now we have the right to say that the circumference consists of the sum of the limit to which the regular inscribed quadrilateral tends with unlimited doubling of the sides and the limit of the missing, expressed through the above difference. In the limit, this difference is expressed as follows: x [2] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2)))) [n-1 times] - sqrt (2-sqrt (2 + sqrt (2 + sqrt (2)))) [n times]). The usual aspiration to the limit of the inscribed quadrilateral gives x [1] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2 + sqrt (2))))) [ n times]. Then 2 * pi will no longer be what we are used to thinking, but it will be 2 * pi = x [1] + x [2] and this should be very close to the usual result. The problem of infinite subradicality of sqrt (2) is eliminated as follows. Let s [1] = sqrt (2 + sqrt (2)) [n-1 times]. Then s [2] = s [1] ^ 2-2. Now let s [3] = sqrt (2-sqrt (2)) [n-1 times], then s [4] = 2-s [3] ^ 2. It remains to deduce the relationship between s [1] and s [3]. It is derived from the equation s [1] ^ 2-s [3] ^ 2 = sqrt (s [1] ^ 2-2) + sqrt (2-s [3] ^ 2). The relationship between s [2] and s [4] is similar to- s [2] ^ 2-s [4] ^ 2 = sqrt (s [2] ^ 2-2) + sqrt (2-s [4] ^ 2) .. Could you use a computer to check this for the number 2 * pi - will the result be close to the usual value? Thank you!

[Bored chemist]

pi is super conditional.  Any wobble in the rotation and pi disappears.  One can set and adjust the ratio of circumference to diameter, with wobble. 

Pi is defined as the ratio of a circle's circumference to its diameter. A circle is defined as a curve (the circumference) traced out by a point that moves in a plane so that its distance from a given point (the centre) is constant. So, the circumference of a circle cannot have wobble. So Pi is a constant.

I think you just defined a sphere, and I think that was part of Hayseed's point.
If you don't restrict things to a plane, you can get weird stuff.

[Hayseed]

My point was to never assume that a rotation IS a circle.  The circumference of a planetary orbit is a helix.  The circumference of a moon's orbit is a helix.  We first saw this watching moon trails thru volcanic debris fields.  It was painting the orbit's history.  But hundreds of years of observation and measurement(without debris fields) showed us it was an elliptic.  When ever science says that it has evidence.......be wery wery careful.

ALL of our gravity theories and vector equations must produce an elliptic.......which does not exist.

[Bored chemist]

We first saw this watching moon trails thru volcanic debris fields.

No, we didn't.

[Colin2B]

1

Pi is defined as the ratio of a circle's circumference to its diameter. A circle is defined as a curve (the circumference) traced out by a point that moves in a plane so that its distance from a given point (the centre) is constant. So, the circumference of a circle cannot have wobble. So Pi is a constant.

I think you just defined a sphere, and I think that was part of Hayseed's point.
If you don't restrict things to a plane, you can get weird stuff.

I thought I had, see bold

My point was to never assume that a rotation IS a circle. 

But we don’t.
My point was that pi is very specific, any closed line that is not a circle will produce a ratio greater than 3.14159265359 etc, but it is not pi.

[evan_au]

Could you please check the following thing on the computer?

How are you accessing this website without a computer?
- Most smartphones will support multiple apps that allow you to do calculations.
- My phone comes with a spreadsheet program that supports simple calculations on its small screen.



 

[trushinalexander49 (OP)]

I am a mathematician, but not a computer specialist, so when I entered the indicated equations into wolfram alpha and even wolfram mathematica, I got that there are no real solutions - in dcode.fr the same story. And you, as far as I understand correctly, know a lot of computational "tricks" - so if it doesn't make it difficult, please help. And by the way, what do you think of my new theory that I wrote in the previous post?

[chiralSPO]

x [2] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2)))) [n-1 times] - sqrt (2-sqrt (2 + sqrt (2 + sqrt (2)))) [n times]). The usual aspiration to the limit of the inscribed quadrilateral gives x [1] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2 + sqrt (2))))) [ n times].

There is no way that these equations can be expected to converge to any finite number. (there must be an error somewhere if you expect it to be converging). In the first limit, we see the product of 2ⁿ⁺² (which diverges quite spectacularly as n approaches infinity) and a infinitely recurring square root (in the limit), that is equal to 2 (see here: https://www.math.toronto.edu/mathnet/questionCorner/infroot.html ). So, x [2] = 2^∞+2 × 2 = not a machine sized real number

I am a mathematician, but not a computer specialist, so when I entered the indicated equations into wolfram alpha and even wolfram mathematica, I got that there are no real solutions - in dcode.fr the same story. And you, as far as I understand correctly, know a lot of computational "tricks" - so if it doesn't make it difficult, please help. And by the way, what do you think of my new theory that I wrote in the previous post?

I think your new theory is incorrect.

[trushinalexander49 (OP)]

Thank you very much for your opinion, but it seems to me that you misunderstood. Yes, 2 ^ (n + 2) in the limit gives infinity, but (sqrt (2-sqrt (2 + sqrt (2)))) [n-1 times] just tends to 0 in the limit, not 2- note the sign after the first 2 is minus but not plus. Therefore, in my opinion, this is still subject to verification, so please check if you can. Thank you so much!

[chiralSPO]

Apologies for misreading your equation.

Can you make it easier to read by posting it using the built-in LaTeX tools, or by attaching an image of the equations?

[trushinalexander49 (OP)]

Normally this equation: s= lim (n->infinity) 2^(n+2)*(sqrt(2-sqrt(2+sqrt(2))))[n-1 times]-sqrt(2-sqrt(2+sqrt(2+sqrt(2))))[n times])+ lim (n->infinity) 2^(n+2)*(sqrt(2-sqrt(2+sqrt(2+sqrt(2)))))[n times] where s must be very closed to 2*pi and if so the real value of pi number. Recommendation about how to avoid radicals you already seen. Thanks!

[evan_au]

x [2] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2)))) [n-1 times] - sqrt (2-sqrt (2 + sqrt (2 + sqrt (2)))) [n times])

When I tried calculating the sides of an inscribed octagon symbolically, I got an answer containing a sqrt(3), which doesn't appear anywhere in your equation.

I have evaluated the inscribed polygon numerically, and it does converge to pi, and doubling the number of sides reduces the error by a factor of 4.

inscribed_polygon.png (23.3 kB . 584x274 - viewed 643 times)

The geometrical construction is shown below, with the derivation of the side c (side of a 2*n-sided polygon) from 2a (the side of an n-sided polygon). It uses Pythagoras' theorem.

inscribed_polygon_diagram.png (15.78 kB . 367x296 - viewed 645 times)

I'll need to do some more work on a circumscribed polygon - I haven't worked out the derivation of the side length, yet.

[trushinalexander49 (OP)]

For your diagram, imagine that 2 * a = d, or equivalently a = d / 2. Then we have c = sqrt ((d / 2) ^ 2 + (1-b) ^ 2), b = sqrt (1- (d / 2) ^ 2). Hence we have c = sqrt (2 - sqrt (4 - d ^ 2)). For a rectangle, insert sqrt (2) instead of d, we get c = sqrt (2-sqrt (2)). For an octagon, instead of d (counting c as the side of the quadrangle), insert sqrt (2-sqrt (2)), we get c = sqrt (2-sqrt (2 + sqrt (2))), then also by induction we make sure that only + sqrt (2) under the radicals will go further.

[trushinalexander49 (OP)]

Yes, and please forgive me for inattention - it is quite obvious that we see every time from which side to start doubling - but inattention is a harsh thing - I mean that in the expression for the difference x [2], after sqrt (2 -...) follows in the reduced + sqrt (2) n times (but not n-1), and in the subtracted n + 1 times (but not n). Despite all the obviousness of this after my above theory, we must be very careful with such things - every "little thing" matters in mathematics.

[evan_au]

I have evaluated the circumscribed polygon numerically, and it does converge to pi, and doubling the number of sides reduces the error by a factor of 4 (but the error is larger than the inscribed polygon with the same number of sides).

circumscribed_polygon.png (23.54 kB . 584x274 - viewed 571 times)

The geometrical construction is shown below, with the derivation of the side 2z (side of a 2*n-sided polygon) from 2x (the side of an n-sided polygon). It uses Pythagoras' theorem and similar triangles.

Circumscribed_polygon_diagram.png (33.08 kB . 641x510 - viewed 568 times)

[trushinalexander49 (OP)]

Yes, we know perfectly well that the inscribed 4-gon tends to the number 2 * pi, but it should be so. Because the computer perceives as the number 2 * pi not the length of a circle of radius 1, which, according to my new theory, is expressed by the sum x [1] + x [2], but only the expression I have given for x [1]. Read carefully my new theory in one of the previous posts and you will understand what I am talking about, that is, the question of what to count as a number 2 * pi-I count as a real number 2 * pi not the usual meaning, but the circumference (x [1] + x [2]).