Post by: trushinalexander49 on 05/01/2021 11:52:49
2 * x [3] = (1-x [3]) * sqrt (2), whence x [3] = sqrt (2) / (2 + sqrt (2)), also x [1] = 1-sqrt (1- (2-sqrt (2)) / 4) and x [2] = sqrt (1 + 1 / (3 + 2 * sqrt (2))) - 1. Since x [1]> x [2] and each side of the inscribed n-gon has a side equal to the sum of the two sides of the triangle containing x [1] and x [2] for the described n-gon, we conclude that the limit of the described n -gon, although like the inscribed n-gon has a circle, the circumscribed n-gon tends to 2 * pi faster than the inscribed n-gon. As far as I know, no one paid attention to this fact earlier, trying to approve the concept of calculating the circumference of a circle in the wrong way by finding the supposedly zero limit of the difference between the variable perimeters of the described and inscribed polygons, assuming this while increasing the number of their sides. in this regard, I am sure that the limit of the difference between the circumscribed and inscribed polygons is not equal to 0. Am I right? Thanks a lot!
Post by: alancalverd on 05/01/2021 12:08:10
Take the simplest example of a 45° right-angled triangle ABC where B = 90° and AB = BC = 1. We know the length of the hypotenuse CA is √2.
Now approximate the shape with n smaller triangles of side 1/n: the path along and up the steps is 2 however large we make n.
This has important implications in education: you can't teach analytical geometry with a computer!
Post by: trushinalexander49 on 05/01/2021 12:35:28
Post by: alancalverd on 05/01/2021 12:44:14
Post by: vhfpmr on 05/01/2021 13:40:13
This has important implications in education: you can't teach analytical geometry with a computer!I once said that 4/(0.4recurring) = 9, and got into an argument with someone who said that he had tested it with his calculator, and insisted that it was only an approximation.
Post by: chiralSPO on 05/01/2021 16:05:06
There is a commonly-taught numerical approximation for the area under the curve which can be found by adding the areas of increasingly thinner rectangular slices of heights that touch the curve above them. The narrower the rectangle, the closer the approximate area is to the real one. In the limit, this is simply the definite integral. However, this rectangles approach will never converge to the correct path length of the curve (unless the curve is a straight line with zero slope, in which case the approximation will be perfect). The sum of those edges will always just be the length of the domain over which the value is being calculated.
Switching to the trapezoid method (allowing for different edge lengths and heights within a slice by setting the height of each side to the value of the function at that position--essentially making the slice contain more information) makes only a slight difference for area approximation, allowing the approximated area to converge to the real area slightly faster than the simpler method (and the limit is, in both cases the actual integral). But by allowing for slopes of the line segments approximating the curve to change, this also will allow the path length to be approximated.
Post by: alancalverd on 05/01/2021 16:31:26
Post by: charles1948 on 05/01/2021 19:04:55
But you can teach a lot of physics and maths with an analog computer!When you mention "analog" computers, I've often wondered whether our modern fixation with "digital" computers comes simply from this:
Our present technical inability to make big computers operate reliably with anything more than crude binary "on" and "off" switches. And what disturbs me even more - the switches used are transistors!
The transistor is capable of "analog" functions, just as its predecessor, the "thermionic valve" or "vacuum tube" was.
To see the potential of this beautiful device wasted, by using it as a simple switch, is disappointing.
Post by: alancalverd on 06/01/2021 00:20:54
Post by: evan_au on 06/01/2021 09:27:16
Quote from: charles1948
The transistor is capable of "analog" functionsI was introduced to "hands on" analog and digital computers in the same summer holiday program when I was at high school.
- These analog computers did not use transistors as a building block, but the basic building block was an "operational amplifier", a circuit consisting of many transistors. It also needed precision resistors and capacitors, and wired up with a lot of flying leads. You could hope for an accuracy of perhaps 0.1% (providing the integrators did not drift away to infinity, as Alan observed). Looking back, I think that pickup of Radio Frequency Interference in the wiring would be a real problem for differentiators....
- They used BASIC for the introduction to digital computers, using a digital approximation to an integrator. We quickly learned that low-order integrators are not very useful! But overall, the digital simulation could provide more decimal places (if your algorithm could generate more decimal places!)
- The subject of the exercise was simulating the behavior of a nuclear reactor during transients (and included a visit to 2 research reactors)
- As someone who had played with electronics for many years, the analog computer came more naturally to me. Others found the digital computer easier.
- A time-shared BASIC system could handle far more students at once than an analog computer.
- We know where this ended - digital computers have taken over, and the analog computers are now museum pieces
https://en.wikipedia.org/wiki/Analog_computer#Electronic_analog_computers
However, we know that digital computers cannot match the energy efficiency of the human brain for many real-world problems (eg self-driving cars). Part of this is due to the analog aspects of neurons
- Researchers are looking at low-accuracy analog computing using transistors in a neural network with pre-computed weights
- Unfortunately, our current techniques for computing the weights requires high-power, high-precision digital calculations (unlike the human brain, which dynamically refines the weights without using any more power)
Post by: evan_au on 06/01/2021 09:43:34
Quote from: OP
I am sure that the limit of the difference between the circumscribed and inscribed polygons is not equal to 0. Am I right?I ran the arithmetic through a spreadsheet, assuming a unit circle (radius r=1 unit).
- Perimeter of a unit circle = 2πr = 2π
- Area of a unit circle = πr2 = π
- I did cheat, using trigonometry functions to determine the lengths of sides, instead of exclusively Pythagoras' theorem (no ancient Greek geometer would accept using an infinite series!)
Polygon_pi.png (23.02 kB . 759x232 - viewed 4568 times)It seems to me that the perimeter of the inscribed & circumscribed polygons do converge to a good estimate of π
- And the difference between the inscribed and circumscribed polygon perimeter does converge to 0
- The average of inscribed and circumscribed polygons converges even faster
- The same trend is visible when you look at the polygon areas
See: https://en.wikipedia.org/wiki/Approximations_of_%CF%80#Polygon_approximation_to_a_circle
Quote from: ChiralSPO
Switching to the trapezoid methodEffectively, the polygons do converge to the curve of the circle, so the errors of rectangular integration don't apply here.
- But I did calculate π using both the area and the perimeter. Both work ok.
Post by: trushinalexander49 on 07/01/2021 14:44:46
Post by: chiralSPO on 07/01/2021 15:27:11
So you're right that for any finite value of n, there will be a disagreement between inscribed and circumscribed polygon models of a circle. But as n increases, the difference between these two models decreases.
And it's not just that the difference between them gets smaller (we can define δ = perimeter of circumscribed polygon – perimeter of inscribed polygon). There is no number n for which the inscribed polygon will have a greater perimeter than the circumscribed polygon (ie δ cannot be less than zero). And there is no limit greater than zero for this difference. Find me a number greater than zero, and I will tell you which value of n allows δ to be less than that number, whether it's 0.001, or 0.000000000000000001 or whatever.
Post by: trushinalexander49 on 07/01/2021 16:36:28
Post by: evan_au on 07/01/2021 21:43:38
Quote from: trushinalexander49
1. for fixed initial polygons, we simultaneously (!) Double their sidesThis method of estimating π by using inscribed and circumscribed polygons dates back to ancient Greek mathematician Archimedes.
- The Ancient Greeks only approved of doing geometry with compass and straight-edge.
- They knew of no way to trisect a general angle (and now we know that there is no such method)
- But they did know how to draw a hexagon in a circle, or a hexagon outside a circle
- They knew how to bisect a general angle or a general line
- So this method repeatedly bisects the angles of a hexagon, giving polygons with 12, 24, 48 etc sides for both inner & outer polygons
- The Greeks did not have the option of using different numbers of polygons inside & outside, since they did not know how to construct them
- ...and it wouldn't have made much difference anyway, since a straightedge and compass cannot give accurate numerical values - the width of your pencil line introduces inaccuracy.
- Numerical calculations were extremely tedious with their number system - the abacus would be more effective.
- They could construct square roots geometrically, but there is no finite numerical method to calculate general square roots (as needed for Pythagoras' Theorem)
Quote
the different speed of tending to the circleThe Ancient Greeks didn't know that.
- We have the advantage that we know the "right" answer, which my spreadsheet shows as 3.14159265359...
- We can compare the calculated error for inscribed & circumscribed polygons, and see that they do converge at different rates.
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For the ancient Greeks, it was a huge advance to be able to estimate π as "between 310/71 and 31/7", (3.1408... and 3.1428) using a 96-sided polygon.
- It was good that they could get an upper and lower bound
- This would have been enough to exclude the "Indiana Pi Bill", which implied a value for pi around 3.2
See: https://en.wikipedia.org/wiki/Indiana_Pi_Bill
Mathematicians today go to extreme lengths to derive upper or lower limits on mathematical conditions, if they can't prove an exact answer
- So the Greek method of calculating π is actually very successful, yielding a very narrow bounds for a practical calculation (96 sides)
- And it would have yielded the exact answer as the number of sides approached infinity.
- We only know the "right" answer because our computers uses a method that only gives the right answer as the number of calculations approaches infinity (which can never be reached). So we have no reason to criticize the ancient Greeks!
Editorial comment: Alexander, try breaking your post into paragraphs and sentences - it is a bit breathless, as written.
Post by: syhprum on 07/01/2021 23:19:20
The chief problem was not zero drift but the jittery signals produced the 50 or so consumer grade potentiometers
When I first got to service one I was much relieved to note that the potentiometers were of an open construction so that they could be squirted full of WD40 that got everything working smoothly for a month or so.
I came to HELL from Muirhead that used simple analogue computers but always had a speaker on line so that you could always detect a jittery potentiometer which I tried to convince HELL to incorporate but the Germans would not listen to a technician from a primative third world country
Post by: evan_au on 08/01/2021 20:50:09
Quote from: alancalverd
the simplest example of a 45° right-angled triangle ABCThis is also a case that the Ancient Greeks could have constructed, since they knew how to construct a square inside & outside a given circle.
Here are the results for successively doubling the sides of a square:
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It does converge to the right value. But the recorded result from Archimedes started with a hexagon, perhaps because a square clearly does not fill a circle, but a hexagon starts fairly close
PS: Inscribing a square inside a given circle is very different from producing a square with the same area as a given circle, which can't be done with straightedge & compass (in a finite number of steps).
Post by: charles1948 on 08/01/2021 21:09:13
It just seems to be a number. Why should a particular number have any significance?
I mean, the number of fundamental forces in the Universe is currently supposed to be 4.
Suppose we found the number should be increased to 5. Why would that be important?
Post by: chiralSPO on 09/01/2021 04:19:13
I wonder whether "pi" is all that important anyway.
It just seems to be a number. Why should a particular number have any significance?
There are some numbers that appear to be of particular significance purely mathematically. Pi shows up basically anywhere curves are involved. Another number, e, appears to be even more ubiquitous, appearing across a range of seemingly unrelated contexts. Here's a nifty video that goes through several examples:
Post by: evan_au on 09/01/2021 09:05:24
I wonder whether "pi" is all that important anyway.In it's traditional role as "the ratio of the circumference to the diameter of a circle", π comes out of Euclid's geometry.
However, that is not the only kind of geometry.
- If you imagine the shape of the Earth as an ideal ellipsoid, and draw a circle on it, the ratio of the circumference to the diameter of that circle is < π
- Last I heard, cosmologists were unsure: If you drew a circle on the scale of the observable universe, whether the ratio of the circumference to the diameter of that circle would be less or more than π
Quote
Why should a particular number have any significance?Some numbers are important for life as we know it.
- For example, if the attraction between protons & neutrons was just a little stronger, all hydrogen would have fused in the Big Bang, and we would not have hydrogen to make our organic molecules
- If it were a little weaker, repulsion between protons would have dominated, and we would not have the carbon we need for organic molecules.
There are debates about how much variation is possible, but the universe would look very different if some of these numbers varied even a little.
See: https://en.wikipedia.org/wiki/Fine-tuned_universe#Examples
Quote
I mean, the number of fundamental forces in the Universe is currently supposed to be 4.The discovery here would be discovery of the existence a 5th force, not creation of a 5th force.
Suppose we found the number should be increased to 5
- If such a force exists now, we are living quite happily with it, and its discovery would win a Nobel prize.
- But the discovery won't suddenly destroy all life on Earth (unless we start to use it against each other)
Post by: evan_au on 09/01/2021 21:52:47
Quote from: OP
the circumscribed n-gon tends to 2 * pi faster than the inscribed n-gonIt 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
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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.
Post by: evan_au on 11/01/2021 08:59:03
- 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 3447 times)Note that to apply the geometric series, you must have calculated the perimeter for 3 successive numbers of polygons, eg 96, 192 & 384
Post by: Hayseed on 11/01/2021 17:19:35
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.
Post by: Colin2B on 11/01/2021 23:43:16
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.
Post by: trushinalexander49 on 12/01/2021 17:59:08
Post by: Bored chemist on 12/01/2021 19:18:53
I think you just defined a sphere, and I think that was part of Hayseed's point.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.
If you don't restrict things to a plane, you can get weird stuff.
Post by: Hayseed on 12/01/2021 21:01:04
ALL of our gravity theories and vector equations must produce an elliptic.......which does not exist.
Post by: Bored chemist on 12/01/2021 21:07:11
We first saw this watching moon trails thru volcanic debris fields.No, we didn't.
Post by: Colin2B on 12/01/2021 23:01:33
I thought I had, see boldPi 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.
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.
Post by: evan_au on 12/01/2021 23:45:35
Quote from: trushinalexander49
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.
Post by: trushinalexander49 on 13/01/2021 01:44:44
Post by: chiralSPO on 13/01/2021 03:09:30
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.
Post by: trushinalexander49 on 13/01/2021 03:52:23
Post by: chiralSPO on 13/01/2021 04:11:20
Can you make it easier to read by posting it using the built-in LaTeX tools, or by attaching an image of the equations?
Post by: trushinalexander49 on 13/01/2021 04:36:39
Post by: evan_au on 13/01/2021 08:36:14
Quote from:
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.
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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.
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I'll need to do some more work on a circumscribed polygon - I haven't worked out the derivation of the side length, yet.
Post by: trushinalexander49 on 13/01/2021 16:56:24
Post by: trushinalexander49 on 14/01/2021 02:08:27
Post by: evan_au on 14/01/2021 09:55:24
[ Invalid Attachment ]
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.
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Post by: trushinalexander49 on 14/01/2021 12:53:52
Post by: evan_au on 12/02/2021 07:18:45
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.
Post by: alancalverd on 12/02/2021 11:34:31
Post by: trushinalexander49 on 12/02/2021 17:44:49
Post by: evan_au on 12/02/2021 21:22:09
Quote from: alancalverd
It's a bit of a tautology to use trig functions like sin and tanIt 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
Post by: evan_au on 12/02/2021 21:29:19
Quote from:
Not all points of the circle are included in the doubling area ... now imagine that we have "gouged out" one of the numbersI 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?
Post by: alancalverd on 13/02/2021 11:52:09
I used the first 3 terms of the Taylor expansion of SIN & TAN as:Ummm....
- The first term is the "exact" answer
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!
Post by: trushinalexander49 on 14/02/2021 13:36:39
Post by: evan_au on 14/02/2021 21:20:25
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.
Post by: evan_au on 14/02/2021 21:47:10
Quote from: alancalverd
so sin π/4, say = π/4 -.....I did describe the use of radians as making the calculation of π "extremely easy", or even a "tautology" or a "definition".
and you need a series for π in order to evaluate π by the polygon method!
- 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
Post by: evan_au on 15/02/2021 10:06:04
Quote from: trushinalexander49
we prove that lim sin (x) / x (x-> 0) is not equal to 1 contrary to the generally accepted representationI 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.
Post by: evan_au on 15/02/2021 10:21:14
Quote from:
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!
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Post by: Bored chemist on 15/02/2021 11:09:33
How many of you can calculate a square root using a pen and paper?
Post by: trushinalexander49 on 15/02/2021 11:28:12
-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.
Post by: alancalverd on 15/02/2021 13:03:42
Post by: trushinalexander49 on 15/02/2021 14:25:53
Post by: alancalverd on 15/02/2021 17:56:11
Post by: trushinalexander49 on 15/02/2021 19:44:52
Post by: evan_au on 15/02/2021 21:31:06
Quote
I'm talking about arcs AB and CDThanks 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?)
Quote from: evan_us
(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)Quote
the incorrectness of measuring arcs only by chordsEveryone 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
Post by: trushinalexander49 on 16/02/2021 10:00:53
- 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.
Post by: charles1948 on 21/02/2021 22:53:43
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?
Post by: Colin2B on 21/02/2021 23:19:17
Doesn't the fact that Pi is not a clear-cut whole number, mean that the Universe is not 3-dimensional?Not at all.
3 dimensional means defined by 3 dimensions (each having its own number). It has nothing to do with a single number that just happens to be close to 3.
You are confusing things which are not the same.
Post by: Bored chemist on 22/02/2021 08:53:48
Do you think there's something disturbingly wrong about Pi not being a whole number?If you calculate the ratio of the edge of a square to the diagonal you will find it is also irrational. It's the square root of two, about 1.414......
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?
Post by: evan_au on 12/03/2021 10:19:24
Quote from: Bored Chemist
If you calculate ... the square root of twoThe polygon method also assumes that you are able to calculate square roots accurately
- Unlike addition, multiplication and division, calculating square roots is not something they teach in school - you just press the square root button on the calculator.
- Or, in the bad old days, you looked it up in a book of mathematical tables, or worked it out with a slide rule.
I did not realise that one simple way of calculating square roots was known to the ancient Babylonians.
- And it has a built in error-correcting mechanism with upper & lower bounds
- Method to find SQRT(C) where C is a constant:
1) Have an initial guess at the answer to SQRT(C); call it X
2) Calculate the quotient Q=C/X; if guess X is a bit high (upper bound), Q will be a bit low (lower bound) - and vice-versa
3) Average X (too high) & Q (too low) to give a better estimate X.
4) Go back to step 2, if you think X and Q are still too far apart
I was surprised to see that this converges as quickly as Newton's method
- in fact it gives identical answers to Newton's method applied to find the zeroes of y=x2-C
- For example, to calculate SQRT(16), with an initial guess of 5:
Babylonians_vs_Newton.png (15.09 kB . 679x255 - viewed 2539 times)Those Babylonians really knew their arithmetic (in base 60)!
- It requires fewer arithmetic operations than Newton's method
- It automatically give you upper and lower bounds for SQRT.
So why would anyone use Newton's method?
- I guess it is a very general method, that works for any continuous function for which you can calculate the derivative.
PS: I chose SQRT(16) as an example, because everyone knows SQRT(16) = exactly 4
See: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method
Post by: alancalverd on 12/03/2021 10:59:39
Or, in the bad old days, you looked it up in a book of mathematical tables, or worked it out with a slide rule.Hand caclulation was taught and examined in my schooldays. Slide rules were "only for engineers".
Post by: syhprum on 12/03/2021 14:03:37
Post by: charles1948 on 12/03/2021 19:12:00
My father had very little education but the one thing he was proud of was he could calculate square roots with pencil and paper (it is not very hard !)
Yes, it can be done by iterations. Long ago, back in the primitive 1970's, I bought a Sinclair "Executive" pocket calculator. It wasn't much of a calculator, by modern standards. Only had basic arithmetic functions: add, subtract, multiply, and divide.
But despite these deficiencies, it had a big selling point - it was very slim. Only about a quarter of an inch thick, as I recall. This made it look quite "futuristic". When compared to other calculators of the time, which were much thicker.
Anyway, the Sinclair "Executive" came with an instruction booklet, which explained how it could be used to perform more complex arithmetical operations, such as deriving "Square Roots".
The method set out in the booklet, as far as I can remember, started like this:
1. Suppose you're asked to derive the square-root of 27.
2. Make a "reasonable guess" at what it might be. Such as 5
3. Divide 5 into 27, then see what the result is, as displayed on the calculator's screen.
I can't remember what you were supposed to do after that. Except that it might possibly have involved "doubling" the result, then dividing again by something, in repeated iterations, until the answer came up on the screen.
That was in the bad old days. Thankfully, nowadays calculators are much more sophisticated. They have "square root" keys. You just enter "27", then press the "square root" key, and the answer comes up on the screen instantly.
I suppose the calculator uses its internal micro-processor, to do the same kinds of things that your Dad did with his pencil and paper. Only much faster?
Post by: alancalverd on 12/03/2021 19:49:24
Post by: charles1948 on 12/03/2021 20:54:00
www.freecodecamp.org/news/find-square-root-of-number-calculate-by-hand (http://www.freecodecamp.org/news/find-square-root-of-number-calculate-by-hand) takes me back to the 1960s.
Thanks alan, for the link you very kindly provided. I've studied it, but the method which it describes, such as dividing numbers into "pairs" and so on, is far more abstruse and complicated than what I remember from the Sinclair booklet.
The Sinclair method, as I attempted to describe earlier, was a simple iterative process, which boiled down to:
1. Take a reasonable guess at what the the square root might be;
2. Divide your guess into the number, and look at the result.
3. Do something with the result (what you were supposed to do, I can't remember)
4. Keep repeating steps 2 and 3, until a square root of sufficient accuracy for your purposes, emerges.
It was just simple iteration. It definitely didn't involve the stuff your link sets forth, which by the way slightly reminds me of a reputed old Russian method of multiplying two numbers, by writing down long columns of numbers, and crossing out various entries in the columns, following some arcane rules.
Thanks again for your reply, best wishes!
Post by: evan_au on 12/03/2021 21:48:30
Quote from: alancalverd
...takes me back to the 1960sThat method is not much harder than long-division....
Quote from: charles1948
dividing numbers into "pairs"What they appear to be doing here is making use of the fact that 100 is a perfect square = 10 x 10
- When they group numbers into pairs, they are effectively turning base 10 numbers into base 100 numbers
- So 20 25 = 20x100 + 25
- as distinct from our Base 10 way of saying 2025 = 2x1000 + 2x10 + 5;
- only 1000 & 10 are not perfect squares, and make the answer very messy!
- They proceed to find the √20, without mentioning that they are also multiplying it by √100 =10
- Anything "left over" from √20 must then be included as the leading digits when finding √25
Post by: charles1948 on 15/03/2021 20:10:40
Anyway, it leads me to this question:
Suppose we weren't using decimal, base-10 arithmetic. But were operating in base-2 binary.
On this assumption, and going back to the original OP about PI, when Pi is calculated using decimal base-10, the calculations seemingly go on for billions of decimal figures, without showing any obvious patterns in the figures.
Would this be the same, if we did the calculations in Binary. Might not the masses of Binary figures when printed out, in "1"'s and "0"'s , eventually reveal some kind of geometrical pattern. Like a Circle?
Post by: Colin2B on 15/03/2021 23:18:58
Alan when you mention "long division", this a difficult operation using decimal arithmetic. So difficult that I don't think it's even taught in schools these days. There's some alternative procedure, I think, which is more on the lines of the Russian method I mentioned in a previous post.Long division is not difficult and is taught to 9-11yr olds in UK. In younger years they can be taught chunking which is a method using repeated subtraction which helps them understand the principle of division.
Post by: charles1948 on 16/03/2021 00:21:42
Long division is not difficult and is taught to 9-11yr olds in UK. In younger years they can be taught chunking which is a method using repeated subtraction which helps them understand the principle of division.
Thanks Colin, I knew there was a word for the new procedure, which you've kindly supplied: "chunking".
If you don't mind me saying so, I doubt that this "chunking" business actually helps any of the kids understand long division in any respect whatsoever.
It reminds me of my years in Primary School. Once, the headmaster made us kids get together in a room, and draw
a chalk-marked diagram on the floor.
Then we had to pick up bundles of straw-stalks, and keep putting them down, in varying orders, on the chalk marks. This was meant to physically demonstrate the operations involved in long-division.
But it didn't achieve that at all. It just baffled and bemused everyone except the well-meaning headmaster
Post by: evan_au on 16/03/2021 07:10:49
Quote from: Charles1948
Might not the masses of Binary figures when printed out, in "1"'s and "0"'s , eventually reveal some kind of geometrical pattern.People have searched for patterns in many bases, and not seen a pattern
- This is not surprising if you look at the infinite series for pi
- The first few terms produce a clear pattern
- But later terms come in and modify the pattern in later decimal places
Quote
some kind of geometrical pattern...Like a Circle?This geometrical pattern is certainly present. Pi is the ratio of the Diameter to the Circumference of a circle.
Post by: Colin2B on 16/03/2021 07:58:42
If you don't mind me saying so, I doubt that this "chunking" business actually helps any of the kids understand long division in any respect whatsoever.Don’t mind at all :)
I think the more ways you have of visualising numbers and the various operations the more chance you have of understanding what is really happening. There are some useful ideas out there which help but, as you’ve experienced, quite a few that don’t - or maybe work for some.
I must confess that I really never got on with arithmetic, maths never made any sense to me until we got on to algebra and geometry.
Post by: Bored chemist on 16/03/2021 08:49:57
Would this be the same, if we did the calculations in Binary.Yes.