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

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The discussion centers on the use of the approximation 22/7 for π, which is often taught in schools despite the existence of alternatives like 314/100 or 157/50 that may be easier to compute with. Participants note that 22/7 is slightly more accurate and easier to memorize, making it suitable for educational purposes. Historical context suggests that 22/7 has been a standard approximation for centuries, particularly before the widespread use of calculators. Some argue that the choice of 22/7 is a relic of past practices, while others emphasize its practical utility in teaching basic arithmetic. Ultimately, the conversation highlights the balance between historical significance and computational ease in mathematical education.
  • #91
This is my favorite* geometric calculation of pi.
Troll Maths presents Pi equals 4! - Imgur.png
*favorite does not mean correct
 
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  • #92
Frabjous said:
This is my favorite* geometric calculation of pi.
View attachment 350043
*favorite does not mean correct
I did a quick google search and it seems this "proof" is erroneous, but I still haven't figured out where the analogy between the staircase and the curve breaks down. The best way to understand the problem with this "proof" is that some of the points on the staircase are not on the curve This issue is not there when we approximate the curve with a polygon (all the points are on the curve), the way Archimedes and Zu Chongzhi did.
 
  • #93
Agent Smith said:
I did a quick google search and it seems this "proof" is erroneous, but I still haven't figured out where the analogy between the staircase and the curve breaks down. The best way to understand the problem with this "proof" is that some of the points on the staircase are not on the curve This issue is not there when we approximate the curve with a polygon (all the points are on the curve), the way Archimedes and Zu Chongzhi did.
Notice that this curve breaks your perimeter/diameter for circle quality construction.
 
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  • #94
Agent Smith said:
I did a quick google search and it seems this "proof" is erroneous, but I still haven't figured out where the analogy between the staircase and the curve breaks down. The best way to understand the problem with this "proof" is that some of the points on the staircase are not on the curve.
Of course the proof is erroneous. We know that going in.

The issue is not that the some of the points on the staircase are not on the curve. Almost none of the points in the circumscribed/inscribed polygons are on the curve either.

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".
 
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  • #95
Frabjous said:
Notice that this curve breaks your perimeter/diameter for circle quality construction.
Oh, that's right, visually that is, but I did stress that the critical ratio has to approximate ##\pi##.

@jbriggs444 , the "perimeter of the limit"?

jbriggs444 said:
Almost none of the points in the circumscribed/inscribed polygons are on the curve either.
The points that matter to the approximation are on the curve (the points get closer and closer as the number of sides increase).

In the case of the staircase, some of the points always stand outside of the curve/line being approximated.
 
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  • #96
Agent Smith said:
@jbriggs444 , the "perimeter of the limit"?
Think about a sequence of stair-step approximations to a circle with more and more sides that are each tinier and tinier. That sequence of stairstep shapes approaches a limiting shape. The limiting shape is the circle, of course.

The "perimeter of the limit" would be the perimeter of the limiting shape, aka the perimeter of the circle, ##\pi d##.

Each of the stairstap shapes will have more and more of the tinier and tinier sides. Each shape will have a perimiter. Each will have the same perimeter as it turns out. ##4d## for all of them.

The "limit of the perimeters" would be the limit of the infinite sequence of perimeters. That limit is obviously ##4d##
Agent Smith said:
The points that matter to the approximation are on the curve (the points get closer and closer as the number of sides increase).
Please stop using fuzzy language. Phrases like "matter to the approximation" are meaningless. The points on a stairstep also get closer and closer as the number of sides increase.
Agent Smith said:
In the case of the staircase, some of the points always stand outside of the curve/line being approximated.
In the case of a polygon circumscribing a circle, some of the points always stand outside of the circle about which they are circumscribed.
 
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  • #97
Agent Smith said:
In the case of the staircase, some of the points always stand outside of the curve/line being approximated.
And some of them always stand inside the curve being approximated.
 
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  • #98
jbriggs444 said:
Please stop using fuzzy language. Phrases like "matter to the approximation" are meaningless. The points on a stairstep also get closer and closer as the number of sides increase.
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.
 
  • #99
jbriggs444 said:
That sequence of stairstep shapes approaches a limiting shape. The limiting shape is the circle, of course.
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).

In the case of the stairstep shapes, however, the angle between the sides is constant at 90 degrees; it does not approach being a smooth curve as the number of stairsteps increases without bound. So I do not think the limit of the stairsteps can be a smooth curve. I think the limit might not be well-defined at all.
 
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  • #100
Agent Smith said:
Well, what's the explanation for the error then?
See my post #99 just now. If the limit of the stairsteps is not well-defined, then the limit of the perimeter is not well defined either.
 
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  • #101
PeterDonis said:
And some of them always stand inside the curve being approximated.
Yes, but as the NumberOfSides(Polygon) increases, all the points are on the curve/circle. This is not true for the staircase. There are 2 sets of points with a staircase, giving us 2 lines, one connects the top edges of the stairs and the other connects the bottom corners of the stairs. Both sets can't be on the same line/curve, producing the error ##\pi = 4##.
 
  • #102
Agent Smith said:
as the NumberOfSides(Polygon) increases, all the points are on the curve/circle.
Only in the limit. At each finite step, the corners of the polygon are not on the circle.

Agent Smith said:
There are 2 sets of points with a staircase, giving us 2 lines, one connects the top edges of the stairs and the other connects the bottom corners of the stairs. Both sets can't be on the same line/curve
For each finite step, this is true. But that doesn't mean it has to also be true in the limit; there is no general rule about limits that entails it.

The correct procedure is to first ask if the limit of the stairsteps is well-defined at all. If it isn't, which is what I argued for in post #99, then it is meaningless to assert any properties of the limit, since there is no well-defined limit for any such properties to apply to.
 
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  • #103
PeterDonis said:
See my post #99 just now. If the limit of the stairsteps is not well-defined, then the limit of the perimeter is not well defined either.
Got it!
 
  • #104
PeterDonis said:
Only in the limit. At each finite step, the corners of the polygon are not on the circle.


For each finite step, this is true. But that doesn't mean it has to also be true in the limit; there is no general rule about limits that entails it.

The correct procedure is to first ask if the limit of the stairsteps is well-defined at all. If it isn't, which is what I argued for in post #99, then it is meaningless to assert any properties of the limit, since there is no well-defined limit for any such properties to apply to.
I should've been clearer. The inscribed polygon's vertices are on the curve being approximated. Archimedes used 2 polygons (circumscribing one and the inscribed one). I have to do the math but my hunch is the inscribed polygon's perimeter more closely approximates the circle's circumference, the outer polygon having the same issue as the staircase approximation. As the number of sides of the polygon increases, the inscribed polygon basically becomes the circle.
 
  • #105
Agent Smith said:
the outer polygon having the same issue as the staircase approximation
No, it doesn't. I made the difference clear in post #99.

Agent Smith said:
As the number of sides of the polygon increases, the inscribed polygon basically becomes the circle.
This is also true of the circumscribed polygon, as I showed in post #99.
 
  • #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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