A Pi Question: Why do we use the awkward approximation 22/7 ?

In summary, the approximation 22/7 is commonly used for the mathematical constant pi (π) due to its simplicity and relatively close value to the actual figure of approximately 3.14159. While 22/7 is not exact, it provides a convenient way to perform calculations involving pi, especially in educational contexts. The fraction dates back to ancient civilizations and remains popular for its ease of use, although more precise approximations exist.
  • #106
PeterDonis said:
No, it doesn't. I made the difference clear in post #99.


This is also true of the circumscribed polygon, as I showed in post #99.
I have yet to do the math. To my knowledge, ##\frac{22}{7}## is the upper limit of the rational approximation of ##\pi## (the circumscribing polygon) correct to ##2## decimal places. I have no idea what the lower limit is and how correct it is (the inscribed polygon).

I believe for the inscribed polygon, we're looking at the vertices and for the circumscribing polygon we're looking at the tangential points.
 
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  • #107
Agent Smith said:
Insofar as Archimedes' and Zu Chongzhi's ##\pi## approximation is concerned, we're computing the perimeter of the circumscribing and inscribing regular polygons.
So which regular polygon has the level of "circularity" that corresponds to the ##\pi## approximation ##22/7##?
 
  • #108
A.T. said:
So which regular polygon has the level of "circularity" that corresponds to the ##\pi## approximation ##22/7##?
The one on the outside, yes? My hunch is the inside polygon's perimeter is more accurate.
 
  • #109
Agent Smith said:
The one on the outside, yes? My hunch is the inside polygon's perimeter is more accurate.
It's not about inside or outside. How many sides does the regular polygon have, that corresponds to the ##\pi## approximation ##22/7##?
 
  • #110
A.T. said:
It's not about inside or outside. How many sides does the regular polygon have, that corresponds to the ##\pi## approximation ##22/7##?
Well, I don't know. Archimedes used a 96-gon
 
  • #111
Agent Smith said:
Well, what's the explanation for the error then?
I gave the explanation: The limit of the perimeters is not equal to the perimeter of the limit.
Agent Smith said:
##\pi \ne 4, \pi = 3.14159...##. It can only mean that the curve we're assuming is an approximation of the actual curve (the circle, etc.) isn't what we assume/think it is.
We're overmeasuring or overcounting.
No. It means that the limit of the perimeters is not equal to the perimeter of the limit.
Agent Smith said:
We could investigate where the extra ##0.8584073464102067615373566167205...## is coming from. I'm sure that would be easy for you, being a science person. Can you take a look into that.
I already know. The error is in your expectation that the two figures should be equal.
 
  • #112
Agent Smith said:
Well, I don't know. Archimedes used a 96-gon
The point is that not every possible approximation of ##\pi## can be represented by a regular polygon. And when you don't restrict it to regular ones, then you have many differently looking polygons for the same approximation of ##\pi##. Thus the idea that a given approximation of ##\pi## corresponds to a specific amount of "circularity" is flawed.
 
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  • #113
PeterDonis said:
I'm not sure this is actually true. For a polygon that is inscribed in the circle or circumscribed around the circle, the angle between the sides approaches 180 degrees as the number of sides increases without bound. In other words, the polygon approaches being a smooth curve, with no angles at all (each "angle" of 180 degrees is just a tangent line to the circle at the "angle" point).
Before we can decide what is "actually true", we should have a definition in hand. What does it mean for a sequence of shapes to "approach a shape in the limit"?

I have a definition in mind.

We are working in a flat two-dimensional metric space. A "shape" is simply a collection of points in the space. I will not try to impose additional requirements such as connectedness or smoothness for a "shape". Such properties are unimportant to the definition that I am trying to phrase.

Suppose that we have an infinite sequence of shapes, ##S(i)##. We want to take the limit of this sequence.

The limit, if it exists, is the set of points ##p## such that

For every radius epsilon > 0
There is a minimum index ##n## such that
For every ##m > n##
point ##p## is within radius epsilon of some point on shape ##S(m)##

If no points satisfy this criterion then we say that the limit does not exist.

I claim that under this definition, the limit of a sequence of circumscribed stairstep shapes with decreasing step size about a circle of fixed radius is the circle.

I claim that under this definition, the limit of a sequence of circumscribed regular polygons with increasing side count about a circle of fixed radius is the circle.

I claim that under this definition, the limit of a sequence of inscribed regular polygons with increasing side count within a circle of fixed radius is the circle.

There is some speculation to the effect that normalized Pythagorean triples are dense on the unit circle. If so, one could form a sequence of sets, each containing finitely many rational coordinate pairs such that the sequence would converge in the limit to a fully populated unit circle.
 
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  • #114
jbriggs444 said:
We are working in a flat two-dimensional metric space.
I have an improved definition. It is essentially identical to the previous one. But it applies to arbitrary topological spaces, even those for which no metric is provided.

A "shape" is still a set of points in the space.
We are still trying to define the limit of a sequence ##S(i)## of shapes.

The limit is the set of points ##p## such that

For every open ball ##B## containing ##p##
there is a minimum index ##n## such that
for every ##m > n##
##S(m) \cap B## is non-empty

The difference between this definition and the previous one is that the previous one used a ball of radius ##r## centered on ##p##. This one relaxes that and simply uses an open ball containing ##p##.

[Be gentle. I've never taken a formal course in topology]
 
  • #115
jbriggs444 said:
One way of compactly stating the issue is that the limit of the perimeters is not equal to the perimeter of the limit. There is no principle of mathematics by which they should be equal. The two ideas do not "commute".
See also this video:

 
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  • #116
Agent Smith said:
Well, what's the explanation for the error then? ##\pi \ne 4, \pi = 3.14159...##. It can only mean that the curve we're assuming is an approximation of the actual curve (the circle, etc.) isn't what we assume/think it is. We're overmeasuring or overcounting. We could investigate where the extra ##0.8584073464102067615373566167205...## is coming from. I'm sure that would be easy for you, being a science person. Can you take a look into that.
Edit: Disregard this comment per post #133.

If one really wants to find ##\pi## with the "step" method, one must add the hypothenuse length of each step.

From my account, in the following image:


troll-maths-presents-pi-equals-4-imgur-png.png

With the square in the top-right corner of the image, the approximate perimeter of the circle is:
$$4 \times \sqrt{\left(\frac{d}{2}\right)^2 + \left(\frac{d}{2}\right)^2}= 2.828d$$
The approximate perimeter of the circle in the middle-left position would be:
$$8 \times \sqrt{\left(\frac{d}{2}\sin 45°\right)^2 + \left(\frac{d}{2}(1-\cos 45°)\right)^2}= 3.061d$$
The approximate perimeter of the circle in the middle-right position would be:
$$8 \times \left( \sqrt{\left(\frac{d}{2}\sin 22.5°\right)^2 + \left(\frac{d}{2}(1-\cos 22.5°)\right)^2} + \sqrt{\left(\frac{d}{2}(\sin 45° - \sin 22.5°)\right)^2 + \left(\frac{d}{2}((1-\cos 45°) - (1-\cos 22.5°))\right)^2} \right)= 3.1214d$$
By using this method, the error drops really fast, and I'm sure we can show that the limit will be ##\pi d## as the number of steps goes to infinity.
 
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  • #117
What is this thread about? We're over 100 messages in and I have no idea. It's certainly not about the title, which is either incorrect or so vague as to be meaningless.
 
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  • #118
jbriggs444 said:
circumscribed stairstep shapes
As I understand the "stairstep" construction, the shapes are not circumscribed after the initial square; there are points on the "stairstep" that are inside the circle, and only the sides that are the remnants of the original sides of the square are tangent to the circle. But I may be misunderstanding the construction.
 
  • #119
jack action said:
the approximate perimeter of the circle
I don't know what you mean. The circle's perimeter is ##\pi d## always. The perimeter of the square is ##4##, and so is the perimeter of every "stairstep" construction derived from it. What "approximate perimeter of the circle" are you talking about?
 
  • #120
PeterDonis said:
As I understand the "stairstep" construction, the shapes are not circumscribed after the initial square; there are points on the "stairstep" that are inside the circle, and only the sides that are the remnants of the original sides of the square are tangent to the circle. But I may be misunderstanding the construction.
As I understand the construction, we begin with a square within which a circle is circumscribed.

Then we cut out corners so that each pair of two orthogonal sides (e.g. over and down) is replaced by two new pairs (e.g. over and down and then over and down again). The new shape still circumscribes the circle. It does not extend into the interior anywhere. We continue cutting out corners, doubling the number of orthogonal sides at each step, ad infinitum.

The intermediate stairstep shapes all retain the property that every second vertex is positioned on the circle that is circumscribed. At least if we count the 4 points of tangency at the top, bottom and the two sides as vertices. They also retain the property that no part of the stairstep path extends into the interior of the circle.
 
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  • #121
jbriggs444 said:
The intermediate stairstep shapes all retain the property that every second vertex is positioned on the circle that is circumscribed.
Ah, I see, that's what I was missing.

This construction still has the issue I described before, that it does not approach a smooth curve in the limit. Your topological definitions of the limit do not require smoothness, but I'm not sure if those definitions are sufficient for the limit of the perimeter to be well-defined.

I do agree that there is no general rule that the limit of the perimeter must be the same as the perimeter of the limit.
 
  • #122
PeterDonis said:
Ah, I see, that's what I was missing.

This construction still has the issue I described before, that it does not approach a smooth curve in the limit. Your topological definitions of the limit do not require smoothness, but I'm not sure if those definitions are sufficient for the limit to have a well-defined perimeter.
It does approach a curve in the limit. The curve that is approached is a circle. A circle is smooth.

The fact that the "smoothness" of the sequence of curves does not converge upon "smooth" in the limit is irrelevant. The limit of the smoothness is not necessarily equal to the smoothness of the limit. Those two concepts do not commute.
 
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  • #123
jbriggs444 said:
It does approach a curve in the limit. The curve that is approaches is a circle.
Does it? Perhaps that's where my question should have been focused. I know it seems intuitively like it does, but intuitions cannot always be trusted.
 
  • #124
PeterDonis said:
Does it? Perhaps that's where my question should have been focused. I know it seems intuitively like it does, but intuitions cannot always be trusted.
Use the definition.
 
  • #126
PeterDonis said:
What definition?
The one I posted. Either the preliminary one in #113 or the relaxed one in #114.
 
  • #127
PeterDonis said:
I don't know what you mean. The circle's perimeter is ##\pi d## always. The perimeter of the square is ##4##, and so is the perimeter of every "stairstep" construction derived from it. What "approximate perimeter of the circle" are you talking about?
Edit: Disregard this comment per post #133.

The "approximate perimeter of the circle" is the perimeter of the irregular polygon composed by the sum of the hypotenuses of each "step", i.e. ##\sqrt{\text{run}^2 + \text{rise}^2}##, lying outside the circle: a 4-side polygon with the square (top-right corner), an 8-side polygon with the middle-left image, and a 16-side polygon with the middle-right image. The more sides you have, the better the approximation.

f4d7b0d8f6201f482c33e3eeb64ba228.jpg
 
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  • #128
jack action said:
The "approximate perimeter of the circle" is the perimeter of the irregular polygon composed by the sum of the hypotenuses of each "step"
Which has nothing whatever to do with the argument that was referred to that uses the stair step construction. So it is irrelevant to this thread.
 
  • #129
PeterDonis said:
Which has nothing whatever to do with the argument that was referred to that uses the stair step construction. So it is irrelevant to this thread.
The question I was answering was:
Agent Smith said:
Well, what's the explanation for the error then? ##\pi \ne 4, \pi = 3.14159...##. [...] We could investigate where the extra ##0.8584073464102067615373566167205...## is coming from.
And the answer is that it's the difference between the sum of the sides adjacent to the right angles and the sum of the hypotenuses formed by those sides.
 
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  • #130
jack action said:
it's the difference between the sum of the sides adjacent to the right angles and the sum of the hypotenuses formed by those sides
But according to the stairstep construction, both of these things approach the same limit. So how can they give different answers? You do not address this, which is the actual question at issue, at all.
 
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  • #131
jack action said:
The question I was answering was:

And the answer is that it's the difference between the sum of the sides adjacent to the right angles and the sum of the hypotenuses formed by those sides.
The figure in question is ##4-\pi \approx 0.8584073464102067615373566167205##.

That is not the difference in perimeter between any particular circumscribed stairstep shape with ##2^{n+1}## sides and the perimeter of the corresponding inscribed ##(2^n)##-gon. The one perimeter is always 4. The other perimeter is always some under-estimate for ##\pi##.

It is the difference between the limit of the sequence of perimeters of the stairstep shapes and the limit of the sequence of perimeters of the corresponding inscribed polygons. The one limit is 4. The other limit is ##\pi##.
 
  • #132
jbriggs444 said:
As I understand the construction, we begin with a square within which a circle is circumscribed.
A circle inside a square and tangent to the four sides of the square is said to be inscribed. A circle outside the square for which the corners of the square touch the circle is said to be circumscribed.

BTW we've ventured quite a way from the original subject of the thread, of why we use the "awkward" approximation to ##\pi## of 22/7.
 
  • #133
PeterDonis said:
But according to the stairstep construction, both of these things approach the same limit.
Sorry, my bad. I should have done one more step in post #116 to find out it was going over ##\pi##. Disregard my comment. o:)
 
  • #134
Mark44 said:
BTW we've ventured quite a way from the original subject of the thread, of why we use the "awkward" approximation to ##\pi## of 22/7.
In an effort to bring us back somewhat on track, let us explore the question of what regular n-gons have a perimeter to "diameter" ratio closest to 22/7. It turns out that the answer is between 90 and 91.

We begin by noting that ##\frac{22}{7}## is greater than ##\pi##. So we should be looking at circumscribed polygons.

The formula for the perimeter of such a polygon with radius ##r## (measured from centroid to the middle of an edge) is:$$2nr \tan \frac{\pi}{n}$$We can tabulate this for a number of sides going from 3 on up. After writing a bit of code:
Code:
[...]
Number of sides: 4, radius: 0.5, side_length: 1, perimeter: 4 target = 3.14285714285714
[...]
Number of sides: 90, radius: 0.5, side_length: 0.0349207694917477, perimeter: 3.1428692542573 target = 3.14285714285714
Number of sides: 91, radius: 0.5, side_length: 0.0345367179994631, perimeter: 3.14284133795114 target = 3.14285714285714
If we flip tan to sine then virtually the same code can look at inscribed polygons where the radius is measured to the vertices.
Code:
[...]
Number of sides: 6, radius: 0.5, side_length: 0.5, perimeter: 3 target = 3.14
[...]
Number of sides: 56, radius: 0.5, side_length: 0.0560704472371918, perimeter: 3.13994504528274 target = 3.14
Number of sides: 57, radius: 0.5, side_length: 0.0550877603558654, perimeter: 3.14000234028433 target = 3.14
Here is the "circumscribed" version of the code.
Code:
#!perl

use strict;

sub tan { sin($_[0]) / cos($_[0]) };

my $i;
my $radius = 0.5;
my $pi = 3.1415926535897932384626;
my $target = 22/7;
my $perimeter = 999;

for ( $i = 3; $perimeter > $target; $i++ ) {
    my $half_angle = $pi / $i;
    my $side_length = 2 * $radius * tan($half_angle);
    $perimeter = $i * $side_length;
    print STDOUT "Number of sides: $i, radius: $radius, side_length: $side_length, perimeter: $perimeter target = $target\n"
};
 
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  • #135
I don't get it. If the jagged staircase does converge on a smooth curve, ##\pi = 4## shouldn't happen. We would be measuring/computing the same thing. How can ##1## thing have ##2## different lengths.

In the video link provided by @A.T. we see that integration has a jagged staircase element to it (more and more of thinner and thinner rectangles, sum their areas and we get the integral). The only difference here is we're not measuring the length of the curve, but the area. Gracias A.T.

@jack action , merci for the explanation. There's an extra ##4 - \pi = 0.8584073464102067615373566167205## when "computing" ##\pi## using the staircase. This mysterious extra length needs to be explained.

##1## quadrant's (quarter circle) arc length = ##2 \pi r / 4 = 2 \pi \times 0.5 / 4 = 0.78539816339744830961566084581988##.

The staircase has for ##1## quadrant (quarter circle) a length of ##0.5 + 0.5 = 1## and
The difference between the staircase and the quadrant arc of the circle = ##1 - 0.78539816339744830961566084581988 = 0.21460183660255169038433915418012##

We can see that ##4 \times 0.21460183660255169038433915418012 = 0.8584073464102067615373566167205##

Perhaps we might need to go TO INFINITY AND BEYOND to make the staircase argument work! 🤔
 
  • #136
@Vanadium 50 , it was a simple high school curriculum question on ##\pi## and it (d)evolved into the staircase paradox.
 
  • #137
Now it's a "paradox"? When I was learning about it, it was just "making a mistake".
 
  • #138
Agent Smith said:
If the jagged staircase does converge on a smooth curve, ##\pi = 4## shouldn't happen.
You have already been told that this argument is not valid. Limits in general do not have to work that way.
 
  • #139
Agent Smith said:
In the video link provided by @A.T. we see that integration has a jagged staircase element to it (more and more of thinner and thinner rectangles, sum their areas and we get the integral). The only difference here is we're not measuring the length of the curve, but the area. Gracias A.T.
As the video explains: You have to show that the error goes to zero. This is the case for the area, but not for the perimeter.
 
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  • #140
Agent Smith said:
@jack action , merci for the explanation. There's an extra ##4 - \pi = 0.8584073464102067615373566167205## when "computing" ##\pi## using the staircase. This mysterious extra length needs to be explained.
It was a mistake on my part. See post #133.
 

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