[evan_au]

Pondering this question a bit further, last night....

Use Radians!
This question of calculating the circumference of the circle gets extremely easy if you specify the angles in radians
- The length of a circular arc with radius r=1, subtending an angle θ radians at the center of the circle is r*θ = θ
- The circumference of a full circle with radius r=1, subtending an angle 2π radians at the center is 2π*r = 2π
- Calculating π this way becomes a simple definition
- Catch 22: You have to know what π is before you can calculate it numerically!

Getting back to the Greek method to approximate π with a polygons around a circle of radius r=1...


Polygon_Trig.png (32.56 kB . 785x355 - viewed 3317 times)
Inscribed Polygon
In triangle OAB, the length of side o = SIN(θ)
- Which is always less than the true arc length a=r*θ

Circumscribed Polygon
In triangle CXY, the length of side c = TAN(θ)
- Which is always more than the true arc length a=r*θ

We can compare these estimates with the Correct answer a=θ as follows:

The graph below compares the True Arc Length (a, in red) with the Circumscribed polygon length (in green) and the inscribed polygon length (in purple).
They diverge considerably from the true arc length when the angle is large (eg for a hexagon)
But as the angle gets small, they all converge on the same values.
So the polygon approximation to a circle converges to the exact circumference of a circle as the number of sides approaches infinity.

Polygon_Errors.png (20.88 kB . 491x399 - viewed 3112 times)

Taylor Series
We know the true arc length is a=θ

For the inscribed polygon, the length of side o is
o = SIN(θ) = θ - θ³/6 + θ⁵/120 - ...
The first term o=θ is the "correct" answer for the length of the arc.
For small positive angles, SIN(θ) is always smaller than the correct answer by an error term θ³/6 + θ⁵/120 - ..

For the circumscribed polygon, the length of side c is
c = TAN(θ) = θ + θ³/3 + 2*θ⁵/15 + ...
The first term c=θ is the "correct" answer for the length of the arc.
For small positive angles, TAN(θ) is always larger than the correct answer by an error term θ³/3 + 2*θ⁵/15 +..

Just taking the first 3 Taylor series for SIN(θ) and TAN(θ), we can tabulate these error terms as follows:


As the number of polygon sides increases, θ gets smaller and these error terms get small much faster than θ (the correct answer).
- So both answers converge on the correct answer as θ approaches zero.

Polygon_Trig.png (32.56 kB . 785x355 - viewed 3317 times)
You will notice that the error term for TAN(θ) is larger than the error term for SIN(θ), so the inscribed polygon gives a more accurate estimate of π than the circumscribed polygon with the same number of sides.

[alancalverd]

It's a bit of a tautology to use trig functions like sin and tan, because they are only tabulated or computed from truncated infinite series. You might just as well use a good series for π, even if it takes longer to converge (like Gregory-Leibnitz) because it is just as rigorous as any practical value of sin.

[trushinalexander49 (OP)]

This is a good study, but it has a curious flaw - 2 * pi will remain constant and in the case when we add to what we used to count as 2 * pi, the above expression for x [2] = lim (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2)))) [n times] -sqrt (2-sqrt (2 + sqrt (2 + sqrt (2)))) [n + 1 time]), and x [2], I repeat, is logically necessary, and here is a logical argument to this: “Not all points of the circle are included in the doubling area, consider this thing: we know that each point of the circle corresponds to a symmetric one when drawing a diameter , that is, the number of points is in a certain sense "even", and now imagine that we have "gouged out" one of the numbers corresponding to some proper point included in the doubling region, then whatever chord we draw from the "punctured" point to another "punctured", there will always be a point in the middle between them, to which they together "reduce" in the limit, which we will call "control", but in turn, the remaining 2 points must be taken as "control" for al ebraic preservation of the integrity of the figure, however, to take them in such a special way, with the help of a functional measure of lag, namely, with the help of the above difference, so that there are no “covering” repetitions of overlapping points ”. The question of the usual meaning of radians, as you understand, directly depends on how this or that calculator computer perceives the number 2 * pi itself, that is, I have no doubts about the correctness of your reasoning in myself, but the question is different - is it correct most people used the method to determine 2 * pi according to the method of the ancients (doubling the number of sides), or it may need the improvement I have given.

[evan_au]

It's a bit of a tautology to use trig functions like sin and tan

It is a tautology to use radians to calculate the circumference of a unit circle (or perhaps, a definition, if you prefer).

I used the first 3 terms of the Taylor expansion of SIN & TAN as:
- The first term is the "exact" answer
- The rest of the terms represent the error in the Greek polygon method
- This is enough to show that both the relative and absolute error declines to zero as the number of sides approaches infinity
- The only reason I showed the spreadsheet-calculated SIN & TAN was to verify that the second and third Taylor terms capture effectively all the Greek errors. (As you say, the spreadsheet Trig functions are generated from a truncated & optimized Taylor series.)
- The Greek error (for small angles and useful polygons) is dominated by just the second term of the Taylor series.

See: https://en.wikipedia.org/wiki/Taylor_series

[evan_au]

Not all points of the circle are included in the doubling area ... now imagine that we have "gouged out" one of the numbers

I don't understand why you think I left out some points on the circle.
- For an n-sided polygon, I calculate 2n triangles
- These triangles "overlap", in the sense that the start and end points of one side are the same as points also counted on the adjacent sides.
- But a point has infinitesimal size compared to the finite side length of any finite polygon, so it doesn't affect the answer

So I can see that you might think I was "double-counting" some points, but I don't see how I could have "gouged out" some point and ignored them?

[alancalverd]

I used the first 3 terms of the Taylor expansion of SIN & TAN as:
- The first term is the "exact" answer

Ummm....

sin x = x - x³/3! + x⁵/5!.....

so sin π/4, say = π/4 -.....

and you need a series for π in order to evaluate π by the polygon method!

[trushinalexander49 (OP)]

First, we prove that lim sin (x) / x (x-> 0) is not equal to 1 contrary to the generally accepted representation. First, if the sine is determined in the broadest way, then for a given angle of a given triangle, the ratio of the opposite leg to the hypotenuse. The attached drawing shows all the inaccuracy of such a definition, since in an understandable sense "parallel" curves are of such a property that if the smaller circle is moved along the bisector of the right angle until both arcs of circles "coincide", then in fact the "true" length of the proper part of the smaller one will not coincide with the "true" length of the proper part of the larger arc, since they are pulled together by different chords, and although when the arc of the smaller circle is swept slightly lower by a larger one, they will coincide, but this only indicates that the length of the circle is proportional to the diameter (note , that 1/4 of the smaller circle is equal by the condition 2 * fi), however, the key point is that while maintaining the original shape, the arcs of the circles will not coincide, and the corresponding chords sqrt (2) and 2 * sqrt (2-sqrt (2)). Then, in the limit, obviously, an equal number of chords corresponding to smaller equal arcs will be selected on both arcs. But due to the fact that each time for each new doubling the corresponding chords are not equal, it follows that in the limit we will also obtain unequal arc lengths if we measure the arcs only by their contracting chords, which, in view of the real equality of the arcs, implies that the method of measuring by chords is incorrect (in the limit, any corresponding chord of the larger arc always remains larger than the corresponding chord of the smaller arc, from which it follows that the difference of arcs in the limit is not zero, which should not be). Accordingly, in such a context, the concept of a sine should also be clarified - that is, it is always constant for a given arc selected for a common measure, the above-mentioned ratio of the opposite leg to the hypotenuse (with the consequent proportional similarity) or something primarily related to any arc of a circle - in fact, the question is open. From this, by the way, it follows that both the decomposition of the sine into an infinite product and the Wallis formula are not absolutely accurate and should be questioned. The same reasoning also serves as a proof that not all points of the circle are included in the doubling region (otherwise the arcs could be measured by chords). But the missing functional measure will be the aforementioned difference x [2], since in the limit it will cover (and even if you look closely at the process without repetitions) the missing points.

[evan_au]

Hi, trushinalexander49...
You are doing very well in English, when I see that your native language uses a Cyrillic script!

I can see that the length of a₁ = √2
- But I am having trouble seeing how Y and a₂ are constructed, beyond the fact that they are parallel to a₁

The reference to the Wallis Product to calculate π was interesting.
- But I did not use the Wallis Product to calculate the SIN function - I used the Taylor series, which is quite well defined for SIN, with infinite range of convergence
- The Taylor series is not quite so well defined for TAN (finite radius of convergence), but still well-behaved in the region 0 < θ < 1 radian, which is the useful range for analyzing the Greek Polygon method
See: https://en.wikipedia.org/wiki/Wallis_product

I am having trouble following the description; some editorial suggestions:
- It will be easier to read if you break down one big paragraph into separate paragraphs, expressing different ideas.
- "parallel" curves": possibly "concentric circles" might be better?
- It is easier to see which lines and triangles are being discussed if you label them with letters. Then you can talk about a line AB, or a triangle ABC, or a circle with center O and radius 1.

[evan_au]

so sin π/4, say = π/4 -.....
and you need a series for π in order to evaluate π by the polygon method!

I did describe the use of radians as making the calculation of π "extremely easy", or even a "tautology" or a "definition".
- In reality, the Greeks could draw a line of length π extremely easily, and they could draw a line of length √2 very easily, using their favorite tools of a compass and straight-edge (in fact, they regarded anything else as "cheating").
- What they couldn't do was to decide if √2 was "male" or "female", because that meant turning it into a rational fraction (which it isn't - it is irrational, the solution of a quadratic equation). The Pythagoreans were apparently quite upset when they discovered √2 was non-binary.
- And the problem is even worse with π, as it is a transcendental number, the solution of a polynomial of infinite degree (∞ is another concept they had trouble with, as do we, frequently).

Anyway, this thread is about trying to use modern mathematical techniques to analyze the ancient Greek method of polygon approximations to π, and to see if it really worked.
- A later writer in 60 AD records that Archimedes went beyond a 96-sided polygon - but then writes down an incorrect value for the answer he got. Whether that was an error by Archimedes or the later writer, we don't know. But I do know that calculating square roots in their number system would have been a nightmare!
- trushinalexander4 asserts that the polygon method gives the wrong answer for π. I am still uncertain about why.
- I have modeled the polygon method numerically and analytically (and estimated its errors) well enough to convince myself that in fact the inscribed and circumscribed polygons do give valid lower and upper bounds for π, and that they converge to the correct answer for π as n → ∞.

If I wanted to calculate π, I myself would not choose to use the polygon method (as you say, I would use an infinite series with a suitable stopping criterion)
- But if I was a Greek mathematician who only trusted a compass and straight-edge, the polygon method would be a very good solution!

See: https://en.wikipedia.org/wiki/Approximations_of_%CF%80#Polygon_approximation_to_a_circle

[evan_au]

we prove that lim sin (x) / x (x-> 0) is not equal to 1 contrary to the generally accepted representation

I don't see how you proved this (it looks like a geometrical argument, but I don't follow it).

If you are interested in the limit as x -> 0, the Taylor series is easily derived by taking the successive derivatives of SIN(x) at x=0 .
- First derivative of SIN(x) = COS(x); at x=0, COS(x) = 1
- Second derivative of SIN(x) = -SIN(x); at x=0, -SIN(x) = 0
- Third derivative of SIN(x) = -COS(x); at x=0, -COS(x) = -1
- Spoiler alert: It keeps cycling 0, +1, 0, -1, 0, +1....

Put it together, and SIN(x) = x - x³/3! + x⁵/5! - ...
...and this is valid for all values around x=0
...and also valid for all real values of x, but we only need small values of x, here

SIN(x)/x = (x - x³/3! + x⁵/5! - ... )/x
= x/x - x³/3!/x + x⁵/5!/x - ...
= 1 - x²/3! + x⁴/5! - ...

LIM(x→0) SIN(x)/x = LIM(x→0) 1 - x²/3! + x⁴/5! - ... = 1 - O(x²) = 1
We don't even have to use limits, as the answer is perfectly well defined at x=0: SIN(0)/0 = 1 - 0 = 1

Please show where this calculation is wrong.

PS: If this wasn't so, your cellphone would not work.

[evan_au]

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

I see this formula quoted here: https://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula#Related_formulas

It does converge to π quite rapidly...
- But you would have problems persuading an ancient Greek geometer that it was a valid way to calculate π!

Vietes_Pi_SQRT.png (10.92 kB . 276x314 - viewed 2937 times)

[Bored chemist]

I love the way that all these calculations glibly throw "sqrt" into the equation, without calculating how much time that function takes to calculate.
How many of you can calculate a square root using a pen and paper?

[trushinalexander49 (OP)]

-I'm talking about arcs AB and CD, which in fact are equal, but with their original concentric shape, they are pulled together by different chords.
-Arc AB is pulled together by the chord a [2] = 2 * sqrt (2-sqrt (2)) - this follows from the similarity of triangles at which the bases contract the arcs fi and 2 * fi I have designated.
-The arc CD is pulled together by the chord a [1] = sqrt (2).
-Now let's go from the opposite point- suppose that we can measure arcs with chords, then the first conditional measurement for arcs AB and CD- if we take into account the axiom of measuring functional continuity- these are just the chords a [2] and a [1] contracting them.
-We will now follow the traditional process of doubling the number of sides (halving each time exactly the arcs AB and CD) - then, as you understand, with each new doubling, the corresponding chords for the arc AB will always be MORE than the corresponding chords of the arc CD.
- Passing to the limit, we note that for an EQUAL number of considered chords, their sum, due to the previous remark, is always MORE for the arc AB than for the arc CD.
-In this case, as you understand, the remark that in the limit when all chords tend to zero, this difference supposedly disappears, it does not work here- imagine 2 different rectilinear segments tending to zero according to different laws, then, obviously, that to obtain a "control" segment, consisting of the sum of various segments tending to zero, it will take a DIFFERENT number of corresponding segments tending to zero. See the advanced drawing below.
-From this it is immediately concluded that measuring arcs ONLY with chords, if we want absolute accuracy, is wrong.
-After these considerations, it should be clear that, among other things, we must literally clarify what a sine and the trigonometric functions associated with it are, because the seeming arc-chord correspondence from the above reasoning did not come true, which means that the definition of the derivative of the sine, and other steps need clarification.
-In my opinion, 2 * pi = limit (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2 +… + sqrt (2 + sqrt (2))…) )) (n times) + sqrt (2-sqrt (2 + sqrt (2 +… + sqrt (2 + sqrt (2))…))) (n times) - sqrt (2-sqrt (2 + sqrt (2 +… + Sqrt (2 + sqrt (2 + sqrt (2)))…))) (n + 1 times)). This is if we assume that 2 * pi is the length of a circle of radius 1.
- Pay attention to one important feature - it is very naive to check the above expression for 2 * pi by the method of a thread wound on a circle, followed by measuring the ruler - this is incorrect due to the fact that the circle is not actually measured by the "exact number" of unit radii. The fact is that if we imagine that we were able to pull a straight line segment onto an arc (equate them), or at least pick up a segment “very close to” the arc, the arc will tend to zero much faster than the segment while doubling.

[alancalverd]

Even quicker: if sinx/x does not converge to 1, what  value does it converge to? I've always admired authors like Kastner & Newman's classic "Mathematics and the imagination" that uses sweeping ideas like "by symmetry" or "obviously" which hide a lot of mathematical rigor!

[trushinalexander49 (OP)]

I mentioned sine only because the traditional definition of sine for a trigonometer is not entirely correct (if you read my previous post carefully, it says why) - and therefore Wallis's formula based on the concept of sine (as well as Taylor's formula - these 2 formulas are really in the known sense, equally basic) is not accurate, but more directly, is incorrect. This is the case in fact, where one should, I repeat, first, EXACTLY define what a sine is - in a SPECIFIC case, this is NOT a nitpicking to terminology, but a completely honest open question - if (as follows from the previous post, I repeat) arcs are NOT measured by chords, either the concept of direct proportionality for the sine of a given angle is inevitably lost and we cannot talk about the similarity of triangles (for the sine, since we consider the sine to be a unique functional characteristic of the arc), or we lose the concept of the exact functional correspondence of the sine-arc.

[alancalverd]

I beg to differ. Sine is defined as opposite /hypotenuse of a right-angled triangle in euclidean space.  Arc is the fraction of circumference subtended by an angle at the centre of a circle. You can propose any number of series approximations to either, but you can't argue with their geometric definitions! 

[trushinalexander49 (OP)]

I absolutely agree with you - however, I am stating exactly the same as you, but with a serious amendment. The fact is that on the logical sense of the very definition of the sine of a given arc, taken as a unit curve segment, that this is the ratio of the opposite leg to the hypotenuse, the geometrically inextricable sine-arc relationship is also based, but since if you nevertheless carefully read my previous post , the chord-arc ratio is not justified due to the incorrectness of measuring arcs only by chords, then the sine-arc ratio is NOT justified, since, as you understand, the chord and sine are taken as a measure of the angle (or what is the same, taken as the measure of the arc) are inextricably linked. This in no way means that the sine is certainly something other than the ratio of the opposite leg to the hypotenuse, but the question is different - we have no right to attribute this definition as a property of any arc, since if we could (imagine) though If we could roughly express the sine in terms of an angle (for example, using an integer polynomial of some degree), we would get a violation of the general functional form for different arcs - since, according to the traditional definition of sine, we are obliged to measure arcs with chords, which, as I have shown, is not true.

[evan_au]

I'm talking about arcs AB and CD

Thanks for breaking the argument down into paragraphs. It leaves room for my mind to breathe.

I lost you on the first sentence. The polygon approximation to π works on a unit circle.
- The arc CD is on a unit circle (radius=1), and a1 calculates the side of an inscribed square = √2
- But the arc AB is on a circle of radius=2, so it does not calculate the side of a circumscribed polygon on a unit circle.
- And a2 calculates the side of an inscribed polygon on a circle with radius=2, not a circumscribed polygon on a circle with radius r=1.
- So I fail to see the significance of the outer circle with radius = 2
- I also lost you on the second sentence I don't see "fi" marked on the diagram (is this meant to be an angle, or the Greek letter Φ?)

(this version of Viete's formula with lots of square roots) does converge to π quite rapidly...

I can see that this formula has some advantages for hand calculation (especially if you are using Newton's method to calculate square roots).

But I later realized that the formula has a fatal flaw when you use it with floating-point arithmetic on a computer:
- You lose a lot of accuracy when you subtract two almost-equal numbers.
- The final steps of the calculation involves subtracting 2 from a number which is almost 2.
- You get serious roundoff errors
- So if you carry on a few more steps, it stops converging to π, jumps around a bit, and then diverges away from π.

Vietes_Pi_SQRT.png (21.5 kB . 348x591 - viewed 2958 times)

the incorrectness of measuring arcs only by chords

Everyone agrees that the perimeter of an inscribed polygon is less than the circumference of a unit circle.
- And the perimeter of a circumscribed polygon is greater than the circumference of a unit circle.
- But the perimeter of an inscribed polygon monotonically increases towards the circumference of a unit circle as you double the number of sides.
- And the perimeter of an circumscribed polygon monotonically decreases towards the circumference of a unit circle as you double the number of sides.

So you have a built-in error detecting mechanism. You know you have made an error if:
- The perimeter of an inscribed polygon with 2n sides is less than the perimeter of an inscribed  polygon with n sides.
- The perimeter of an circumscribed polygon with 2n sides is greater than the perimeter of a circumscribed polygon with n sides.
- The perimeter of an inscribed polygon with 2n sides is greater than the perimeter of a circumscribed polygon with 2n sides.
- You can never calculate π exactly, but you can be pretty sure you are close, and you can measure how close, with a confidence interval
- as BC says, calculating square roots by hand is very error-prone!

You don't have this error-detecting mechanism with a formula like the repeating square roots.
- You don't know when you (or your computer) has made an error, as shown by taking the square roots a few more steps, above.
- Unless you already know the "correct" answer (which the Greeks did not!)

And, contrary to your assertion, on a unit circle, the perimeter of inscribed polygon with n sides does approach the perimeter of a circumscribed polygon with n sides, as n → ∞.
- And since the circumference of a circle is between them, they both approach the circumference of a circle as n → ∞

PS: The first million digits of π are available for free from project Gutenberg. But the first 40 digits are enough for any practical purpose.
See: https://www.gutenberg.org/files/50/old/pimil10.txt

[trushinalexander49 (OP)]

"The polygon approximation to π works on a unit circle.
- The arc CD is on a unit circle (radius = 1), and a1 calculates the side of an inscribed square = √2
- But the arc AB is on a circle of radius = 2, so it does not calculate the side of a circumscribed polygon on a unit circle.
- And a2 calculates the side of an inscribed polygon on a circle with radius = 2, not a circumscribed polygon on a circle with radius r = 1.
- So I fail to see the significance of the outer circle with radius = 2 ".
 -You misunderstood- the fact is that arcs AB and CD are equal by condition, as can be seen from the drawing itself, respectively, following the method of chords, we actually measure both arcs with them, and there is no contradiction with the fact that the length of the first circles 2 * pi, and the second-4 * pi- because the method of chords remains the same - we are interested in the above-mentioned equal arcs.
- "I also lost you on the second sentence I don't see" fi "marked on the diagram (is this meant to be an angle, or the Greek letter Φ?)". In the drawing, fi is marked above the curly braces (and yes, if you like it better, consider it as F - a matter of preference).
- "(this version of Viete's formula with lots of square roots) does converge to π quite rapidly ..." - the point is that I have a different formula, look carefully at each of its symbols - it is not so long-2 * pi = limit (n-> infinity) 2 ^ (n + 2) * (sqrt (2-sqrt (2 + sqrt (2 +… + sqrt (2 + sqrt (2))…))) (n times) + sqrt (2-sqrt (2 + sqrt (2 +… + sqrt (2 + sqrt (2))…))) (n times) - sqrt (2-sqrt (2 + sqrt (2 +… + sqrt (2 + sqrt (2 + sqrt (2)))…))) (n + 1 times)).
- Where did I get this formula, I have already explained, from the difference covering the "missing" points, which for some reason you took for pi (I mean x [2]). But we simply add this difference, as can be seen from my formula, to the "traditional" value 2 * pi.
- Also note that, as I mentioned in the previous post, for the arc there is a reduction of the rectilinear measurement in itself - this is another "subtle" thing - since it is easy to prove that with the conditional equality of an arc = a segment, an arc while reducing by half decreases faster than the segment.

[charles1948]

Do you think there's something disturbingly wrong about Pi not being a whole number?

I mean, why isn't it a simple whole number, like 3.  If it was, that would make sense in a 3-dimensional Universe.

In such a Universe, the 3 dimensions - length, breadth, height  ( or equivalent terms for these properties )
would surely be capable of clear-cut representation by the numbers: 1, 2, 3.  There wouldn't be any "fuzziness"

Doesn't the fact that Pi is not a clear-cut whole number, mean that the Universe is not 3-dimensional?