Archimedes' Dilemma: The Misunderstood Concept of Pi = 4

[FlexGunship]

That can't possibly be right, since $$e^{\pi} - \pi = 20$$

Well, unless your calculator doesn't work right. At the recommendation of a hilarious XKCD comic, I told students in a programming class to use that as a way to check pi, e, and their exponent functions for a calculator they were writing. If everything were programmed correctly, it should be 20.

 

[JaredJames]

Well, unless your calculator doesn't work right. At the recommendation of a hilarious XKCD comic, I told students in a programming class to use that as a way to check pi, e, and their exponent functions for a calculator they were writing. If everything were programmed correctly, it should be 20.

Mine comes out at 19.99909998. I take it that ain't a good thing?

 

[BobG]

Mine comes out at 19.99909998. I take it that ain't a good thing?

My Post 1460 Versalog comes out to 20. My Faber Castell 2/83N comes out to 20.

 

[turbo]

I went into engineering in the '60s. I don't care if you had a K&E slip-stick that was X feet long - you had rounding errors. The fictional "precision" afforded by calculators and computers today isn't exemplified in application.

 

[Integral]

My reasoning is that 0.9999... is NOT equal to 1.

Nope, it is equal.

 

[Mathnomalous]

Nope, it is equal.

The numbers meet somewhere at infinity? I think it is the same reasoning used for π = 4: the square and the circle merge somewhere at infinity. Then again, it is like a modified Zeno's Dichotomy Paradox.

 

[BobG]

The numbers meet somewhere at infinity? I think it is the same reasoning used for π = 4: the square and the circle merge somewhere at infinity. Then again, it is like a modified Zeno's Dichotomy Paradox.

The pi=4 result is obtained by choosing a bad method to solve the problem. The proper method is to average the perimeter of the outside square and the inside square.

The perimeter of the outside square is 4. The perimeter of the inside square is 2.8284. Average them and you get an approximate value of 3.4142 for pi.

To get a more accurate approximation, cut the corners of the square to make an octagon and make a similar octagon on the inside and average the perimeters (approximate value of 3.188). And so on.

 

[Mathnomalous]

Thanks for the correction, BobG.

 

[Danger]

The numbers meet somewhere at infinity?

That reminds me of another Johnny Hart stroke of brilliance before he degraded into a Jesus-freak. BC set out on a circumnavigation of the world (which is a bit weird because Hart believed that it was flat), dragging a forked stick in order to prove that parallel lines never converge. By the time he returned, of course, it had worn down to being only a single stick.

 

[disregardthat]

The numbers meet somewhere at infinity? I think it is the same reasoning used for π = 4: the square and the circle merge somewhere at infinity. Then again, it is like a modified Zeno's Dichotomy Paradox.

In both cases this is a question of limits. http://en.wikipedia.org/wiki/Limit_(mathematics )

Numerical limits are never reached, they are values which e.g. a sequence or a function may approach.

0.9999... is a description of the infinite sum $$\sum^{\infty}_{n=1}9 \cdot 10^{-n}=0.9+0.09+...$$

An infinite sum $$\sum^{\infty}_{n=1} 9 \cdot 10^{-n}$$ is not the result of summing an infinite collection of rational numbers (that is simply nonsense!), but rather defined as the limit $$\lim_{N \to \infty} \sum^{N}_{n=1}9\cdot 10^{-n}$$, which happens to be 1. This is because the partial sums $$\sum^{N}_{n=1}9\cdot 10^{-n}$$ approach 1 as N grows without restriction.

Hence it makes sense as you mentioned to divide 0.999... by 3, as 0.9999/3 is the limit of the partial sums $$(\sum^{N}_{n=1}9\cdot 10^{-n})/3 = \sum^{N}_{n=1}3\cdot 10^{-n}$$ which approaches 1/3, and may be described by 0.33333...

 

[disregardthat]

The pi=4 result is obtained by choosing a bad method to solve the problem. The proper method is to average the perimeter of the outside square and the inside square.

The perimeter of the outside square is 4. The perimeter of the inside square is 2.8284. Average them and you get an approximate value of 3.4142 for pi.

Both methods will yield 4, not pi. You can't assume that a graph "inside" another will give a smaller length, in this case that a jagged path inside a square will have a lesser perimeter than the circle. It is simply not true.

The proper answer is that the length of a differentiable curve is defined as the supremum of the lengths measured by connecting points on the curve with straight lines, that is the supremum of the sum of the lengths of these lines. Approximating with regular polygons accomplishes this, but jagged paths does not.

Although the jagged paths converge uniformly towards the circle, the lengths will not converge to the length of the perimeter. In general we cannot expect this to be true as this example shows.

 

[BobG]

The pi=4 result is obtained by choosing a bad method to solve the problem. The proper method is to average the perimeter of the outside square and the inside square.

The perimeter of the outside square is 4. The perimeter of the inside square is 2.8284. Average them and you get an approximate value of 3.4142 for pi.

To get a more accurate approximation, cut the corners of the square to make an octagon and make a similar octagon on the inside and average the perimeters (approximate value of 3.188). And so on.

Both methods will yield 4. You can't assume that a graph "inside" another will give a smaller length, in this case that a jagged path inside a square will have a lesser perimeter than the circle. It is simply not true.

The proper answer is that the length of a differentiable curve is defined as the supremum of the lengths measured by connecting points on the curve with straight lines, that is the supremum of the sum of the lengths of these lines. Approximating with regular polygons accomplishes this, but jagged paths does not.

Although the jagged paths converge uniformly towards the circle, the lengths will not converge to the length of the perimeter.

Sloppy terminology, just assuming people would understand what I meant by outside and inside.

The outside square is the square outside the circle with the sides tangent to the circle. The inside square is inside the circle with the corners touching the circle. Likewise the outside octagon and the inside octagon.

In other words, it is approximating with regular polygons since that's the only method Archimedes had available in his era.

(Technically, you could just work with either the outside polygons or the inside polygons, but I think he used both.)

 

[disregardthat]

The outside square is the square outside the circle with the sides tangent to the circle. The inside square is inside the circle with the corners touching the circle. Likewise the outside octagon and the inside octagon.

Ok, I thought you were suggesting that you would obtain the correct result by approximating with jagged paths (not polygons) "inside" the circle as well and take the mean value of the limit of the lengths.

(Technically, you could just work with either the outside polygons or the inside polygons, but I think he used both.)

You would only get a lower bound by approximating with inside polygons (as you cannot be sure that the lengths actually converge towards the supremum). The outside polygons would yield an upper bound as well. Both are converging to the same value, so you are sure that both converge to the correct value.

 

[rottenseed]

If you draw a straight line of distance 1 inch, is it the same linear distance as a 1 ince line composed of extremely tiny square sine waves like this? _|^—|_|^—|_|^—|_

No, it will not...no matter HOW SMALL those tiny rectangle are. It's the same instance here. Essentially the general fractal outline are not the same as the sum of all the tiny parts!

 

[mccarthyp64]

Problem being that repeating gives you an octagon http://upload.wikimedia.org/wikipedia/commons/6/66/Regular_octagon.svg

Instead of a circle, because the side of a circle isn't straight etc.

P.S. I only figured this out 2 days ago

 

[Cbray]

1/3 = .3333333333...

(3) 1/3 = (3) .3333333333...

3/3 = .9999999999...

1 = .9999999999...



I have earned a place in the Math Hall of Fame! Btw, I, too, would have a problem if my name was Archimedes; what kind of parent names his kid Archimedes!?

Hopefully this is a joke, if not start learning about infinity..

 

[Drakkith]

Hopefully this is a joke, if not start learning about infinity..

Hopefully what is a joke?

 

[nitsuj]

Hopefully this is a joke, if not start learning about infinity..

I think the "..." reflects that.

 

[Char. Limit]

Hopefully this is a joke, if not start learning about infinity..

Why does this thread keep getting necro'd?

 

[Drakkith]

Why does this thread keep getting necro'd?

Because necromancers are sexy?

 

[shockwave121]

Not rigorous, but accessible and points out the major flaw of using different metrics "freely".

 

[Jacob_V]

Hah!

 

[Danger]

I banned Micromass.
There is something beautiful about all infractions of whatever nature. - lisab & CL
Division by zero is a one-to-many operation, and thus is not allowed.
.999... is equal to 1.
A number can have more than one decimal expansion.
The rule a√b√=ab−−√ only holds when a and b are both positive.

Of course, if you were a linguist rather than a mathematician, you would have spelled "one-to-many" properly as "one too many".

Because necromancers are sexy?

Hey, now... just between friends... Who doesn't like to crack open a cold one once in a while?

 

[Drakkith]

Hey, now... just between friends... Who doesn't like to crack open a cold one once in a while?

Buahaha!

 

[jreelawg]

Pi does = 4. Don't you understand relativity? Curvature affects the metric and makes distances seam shorter. Sure a circle appears to be 3.14... because the circle is experiencing curvature. In the frame of reference of the square, pi is 4.

 

[Drakkith]

Pi does = 4. Don't you understand relativity? Curvature affects the metric and makes distances seam shorter. Sure a circle appears to be 3.14... because the circle is experiencing curvature. In the frame of reference of the square, pi is 4.

I hope this is a joke lol.

 

[jreelawg]

It's sort of interesting. The perimeter of a digital circle is 8r, and the surface area is approximately 6pir^2.

 

[meldraft]

Btw, I, too, would have a problem if my name was Archimedes; what kind of parent names his kid Archimedes!?

Continuing the necro spirit

Archimedes in ancient Greek translates into "the one who gives leading advice"