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January 23, 2019, 00:31:41 AM
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# Pi = 4

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brianvds
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 « on: January 20, 2015, 15:31:32 PM »

Something to fight over, for the mathematically inclined:

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BoogieMonster
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 « Reply #1 on: January 20, 2015, 15:36:30 PM »

Since when is the perimeter (circumference) of a circle equal to π?
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Rigil Kent
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 « Reply #2 on: January 20, 2015, 15:40:30 PM »

Since when is the perimeter (circumference) of a circle equal to π?

When the diameter is 1.

I know you can approximate a circle by adding more and more vertices to a polygon, but then all of the vertices are facing outwards. Here the vertices alternately face inwards and outwards, so all we are doing is drawing a zig-zag line. On whatever microscopic scale you choose, we just end up with a circle made up from an infinitely fine zig-zag line, which will always be longer than a smoothly curved line. I've no idea how to put this in robust terms.

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 « Last Edit: January 20, 2015, 15:55:22 PM by Rigil Kent » Logged
BoogieMonster
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 « Reply #3 on: January 20, 2015, 15:48:41 PM »

No, c = 2πr

If I take the graph above at face value (which I shouldn't), r would be 1 and dropped, then c = 2π (for r = 1)

If the "perimeter" of the black border is 4 (and supposedly = c):

4 = 2π   (/2)
2 = π

The person drawing this graph couldn't even bother to correctly do some basic algebra.
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Rigil Kent
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 « Reply #4 on: January 20, 2015, 15:56:53 PM »

No, c = 2πr

If I take the graph above at face value (which I shouldn't), r would be 1 and dropped, then c = 2π (for r = 1)

If the "perimeter" of the black border is 4 (and supposedly = c):

4 = 2π   (/2)
2 = π

The person drawing this graph couldn't even bother to correctly do some basic algebra.

r=D/2
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BoogieMonster
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 « Reply #5 on: January 20, 2015, 15:59:12 PM »

Yeah, I walked away and reasiled i'd substituted r for d.
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Mefiante
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 « Reply #6 on: January 20, 2015, 16:01:06 PM »

I’m sorry to be a stick-in-the-mud spoilsport here, but this type of dull-witted mathematical naïveté does no one any favours, least of all those struggling with calculus.  You have to calculate the (limiting) circumference using the red line portions, not the black ones — that is, use Pythagoras instead of just foolishly adding the x- and y components together.  The diagram’s treatment effectively says that if you go 50 metres east and then 50 metres north, you have gone 100 metres north-east.

There are far more interesting ways to get π, the ratio of a circle’s circumference to its diameter, to be 4 (or any value you like) using non-Euclidean geometry.

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Rigil Kent
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 « Reply #7 on: January 20, 2015, 16:07:14 PM »

You have to calculate the (limiting) circumference using the red line portions, not the black ones
Ah ... so the sum of a gazzilion tiny hypotenuses? Clever!
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BoogieMonster
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 « Reply #8 on: January 20, 2015, 16:10:20 PM »

My idiots explanation: The circle has a "smooth" outline, the straight-line configuration has a rough outline. You can make that texture approach smooth, but it remains rougher than the red line forever, so there are forever white gaps between the black line and the red line. iow: The black line's length will always exceed that of the red one.
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Mefiante
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 « Reply #9 on: January 20, 2015, 16:13:40 PM »

Ah ... so the sum of a gazzilion tiny hypotenuses?
Zigzagtly.

My bugbear is that, having taught calculus, problems like these confuse students by apparently validating intuitional misconceptions.  It’s the sort of problem you give to your star students with the express instruction to identify the flaw.

The circle has a "smooth" outline, the straight-line configuration has a rough outline. You can make that texture approach smooth, but it remains rougher than the red line forever, so there are forever white gaps between the black line and the red line. iow: The black line's length will always exceed that of the red one.
This explanation fails because, as you can easily validate for yourself, the apparent ratio remains 4, regardless how finely you zigzag.

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Rigil Kent
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 « Reply #10 on: January 20, 2015, 16:20:41 PM »

My bugbear is that, having taught calculus, problems like these confuse students by apparently validating intuitional misconceptions.
I thought it would have the exact opposite and pedagogically meaningful effect. The conclusion that Pi=4 is so obviously wrong that the problem actually invites critical thinking. The student will try and find the flaw in such reasoning, which will hopefully serve to deepen understanding.

the apparent ratio remains 4, regardless how finely you zigzag.
Erm ... but that's the point, not so? As I understand it the puzzle presents a clearly incorrect conclusion based on faulty reasoning and invites us to spot the error in the reasoning.

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BoogieMonster
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 « Reply #11 on: January 20, 2015, 16:25:06 PM »

This explanation fails because, as you can easily validate for yourself, the apparent ratio remains 4, regardless how finely you zigzag.

My point is you're zig-zagging, not that the ratio ever changes. The apparent circle is not, and never will be, a circle.
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Mefiante
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 « Reply #12 on: January 20, 2015, 16:29:30 PM »

I thought it would have the exact opposite and pedagogically meaningful effect.
It’s consistently been my experience that on the whole, such problems only serve to confuse all but the most talented students.  I suppose that there would be significant merit if students were presented with such material once they have mastered the basics and are comfortably competent — at which point they wouldn’t need my instruction anymore.

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Rigil Kent
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 « Reply #13 on: January 20, 2015, 16:33:27 PM »

at which point they wouldn’t need my instruction anymore.
Well, there riemann sum doubt about that ...
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Mefiante
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 « Reply #14 on: January 20, 2015, 16:35:25 PM »

The apparent circle is not, and never will be, a circle.
Yes, correct.  The same cannot be said about the “gazillion hypotenuses”.  In the limit, they are the circle.

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