Archimedes' Dilemma: The Misunderstood Concept of Pi = 4

[JaredJames]

http://www.lolblog.co.uk/wp-content/uploads/2010/11/1290616506315.jpg

Just thought I'd share this. Don't know if anyone has seen it, but I found it rather amusing.

^{Disclaimer: I know why it doesn't work and am not trying to push this as some "new" maths.}

 

[OmCheeto]

http://www.lolblog.co.uk/wp-content/uploads/2010/11/1290616506315.jpg

Just thought I'd share this. Don't know if anyone has seen it, but I found it rather amusing.

^{Disclaimer: I know why it doesn't work and am not trying to push this as some "new" maths.}

Oh dear god. Don't get me started again...

https://www.physicsforums.com/showthread.php?t=450364

And could someone please tell me whether I was wrong or right?!

I'm very insecure.

------------------------------------
NOT!

I just can't remember what that was called. Something accelerating towards infinity, infinitely faster than infinity. (hic!)

Sorry...

:)

 

[JaredJames]

Ah, nevermind then.

Pity, didn't think it would be in the maths section, given what it is.

 

[G037H3]

I always respond to it with "Archimedes here. Yes, I do have a problem with it, *explanation about circumscribing and inscribing* you should view my work."

 

[Mathnomalous]

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!?

 

[JaredJames]

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

I know, poor bugger named after a pump.

 

[The legend]

1/3 = .3333333333...

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

3/3 = .9999999999...

1 = .9999999999...

The same thing can be done by greatest integer function too...




WHAT!
THIS ISN'T THE MATH SECTION??

 

[Mathnomalous]

I know, poor bugger named after a pump.

At least they did not name him Pi.

 

[Jimmy Snyder]

I suspect that by using appropriate geometrical shapes and a similar limiting argument, you could prove the circumference to be any real number greater than or equal to pi and even prove that the circumference is infinite.

 

[Chi Meson]

http://www.lolblog.co.uk/wp-content/uploads/2010/11/1290616506315.jpg

This reminds me of the following logic:

"So,
If she weighs the same as a duck,
then she's made of wood,
and therefore,
A WITCH!"

(Also reminds me of most political logic)

 

[Danger]

At least they did not name him Pi.

Hey, now! That's the number that I put on my baseball jersey.

 

[jobyts]

What I don't understand is how does it prove Pi = 4!. I didn't get the factorial part.

 

[leroyjenkens]

I always round Pi down to 3.

 

[G037H3]

What I don't understand is how does it prove Pi = 4!. I didn't get the factorial part.

it's part of the trolling

it's saying that pi = 4, not pi = 4!

 

[Mathnomalous]

Hey, now! That's the number that I put on my baseball jersey.

That is one loooooooooooooooooooooooooooooooooooooooooooong baseball jersey...

 

[Danger]

That is one loooooooooooooooooooooooooooooooooooooooooooong baseball jersey...

What can I say? I used to be fat.

 

[Mathnomalous]

What can I say? I used to be fat.

Infinitesimally fat?

 

[G037H3]

Infinitesimally fat?

His blubber was a fractal. :X

 

[Danger]

I'll have you know, you young whippersnappers, that I ballooned up to 132 lbs. when I was playing ball. That extra 5 lbs. above my off-season weight was pure muscle.

 

[Hepth]

it's part of the trolling

it's saying that pi = 4, not pi = 4!

man I hope you just got trolled...

 

[Stan Marsh]

Archimedes will be really be pissed off when he see his name appear with that face in the last picture.

 

[Mathnomalous]

Archimedes will be really be pissed off when he see his name appear with that face in the last picture.

I see a resemblance.

 

[JaredJames]

It's the spitting image of him.

 

[chaoseverlasting]

At least they did not name him Pi.

Have you read 'The life of Pi' ? Brilliant book!

 

[Integral]

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!?

Sorry I don't get it? What is the problem? Why are you surprised at this result? Am I missing some sarcasm?

 

[lisab]

Sorry I don't get it? What is the problem? Why are you surprised at this result? Am I missing some sarcasm?

I'm in the dark too...

 

[FlexGunship]

Is there such a number as 0.99999...98?

Where the "9" portion repeats forever but the last digit is an "8"??

 

[JaredJames]

Where the "9" portion repeats forever but the last digit is an "8"??

Doesn't that contradict itself?

 

[FlexGunship]

Doesn't that contradict itself?

...no...

 

[disregardthat]

Where the "9" portion repeats forever but the last digit is an "8"??

Surely there are many such numbers!

Take for example 0.8 = 0.79999...

...

 

[Drakkith]

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

Dividing 1 by 3 is an operation that never completes. You have to infinitely continue the operation. Seems to me that it is similar to dividing by 0. You can't actually do it. The operation isn't allowed or isn't possible. We just use "shorthand" if you will, and say that dividing by 0 is just 0 because it avoids the whole question of why you can't do it.

However, the fraction 1/3 represents 1 part out of 3 parts. Each part is a "Whole Part". 3 pieces put together equal 3/3, or 1 "Whole Thing".

If you slice a pie into 3 pieces, you don't get 0.333...x3, you get 3/3. 3/3 divided by 3 equals 1/3.

Thats just my take on it though. I'm sure many others would disagree. =)

 

[disregardthat]

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

You'll only get this thread locked. Consult the other of 0.9999 != 1 threads on this forum.

 

[Drakkith]

You'll only get this thread locked. Consult the other of 0.9999 != 1 threads on this forum.

If my one post gets this thread locked, then so be it. I'm not that concerned about being incorrect, especially here in the General forum.

Edit: After reading a few posts, I'm still not convinced. 0.333... x 3 is also an operation that never ends. How can 0.3333...x3 = 0.999... if you never get a result? Am i missing some kind of rule or something in math that says otherwise?

Edit 2: I don't want to get this post sidetracked, so don't worry about responding to me. =)

 

[Danger]

I feel crappy about this, but I feel obliged to repost something:

http://yfrog.com/2hpf2jbj

 

[BobG]

http://www.lolblog.co.uk/wp-content/uploads/2010/11/1290616506315.jpg

Just thought I'd share this. Don't know if anyone has seen it, but I found it rather amusing.

^{Disclaimer: I know why it doesn't work and am not trying to push this as some "new" maths.}

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

 

[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