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# Talk:Pi/Archive 5

Archive 4 | Archive 5 | Archive 6

## Calculating/measuring pi

The Gregory-Leibniz form is indeed beautiful, but sheds no light on pi without an explanation of why it works, and thus shouldn't appear (with or without explanation) so early in the article. I was going to edit the sections, but the comment sent me here. I would have moved the "measuring pi" section up, merging in the para about simple/complicated formulas and then left discussion of all the series to the following sections and the "computing pi" article. Lunkwill 22:08, 2 February 2007 (UTC)

I've added a brief mention of ${\displaystyle tan(\pi /4)=1}$ to explain the geometrical origins of the series. --Michael C. Price talk 10:40, 3 February 2007 (UTC)
I don't think that really helps, except to readers who know the power series for arctan − which are definitely not the people that section is written for. There has been quite some debate about the desirability of having a section early in the article that explains to "the mathematically non-sophisticated reader" that there are definite mathematical procedures for approximating pi, i.e. that it is \emph{not} just an empiric matter of measuring large circles in physical space. I remain fairly unconvinced that we need such a section, but as long as we do have it, please let it do its job without confusing its target audience with mathematical prerequisites that they don't have. (I'll resist the temptation to revert MichaelCPrice for the moment). Henning Makholm 23:02, 3 February 2007 (UTC)
I think the section is needed because otherwise people will think it is just a mysterious empirical thing. I reason I added the tan and sin reference is that the section says it has a geometrical origin but explains no further. As Lunkwill said above, this is not really satisfying. (I didn't write out the power series for arcsin and arctan, so don't see that I added any mathematical complexity to the section, just a few concepts and pointers.) --Michael C. Price talk 23:30, 3 February 2007 (UTC)
As it is now, the very target audience of the section will be left out puzzled when you expect them to be able to make the connection from tan(pi/4) to an infinite series. Your text does expect them to make that connection, because it clearly has the form "because of A, B must be true". Any reader who cannot see what A and B have to do with each other will conclude that the matter is too complex or advanced for him to understand, and will give up on learning anything from the section. As far as I can see, the text was perfectly allright for its purpose until you added some advanced claim without explaining further. Henning Makholm 01:25, 5 February 2007 (UTC)
Is it any less baffling when there is no indication of where the series comes from? --Michael C. Price talk 12:26, 5 February 2007 (UTC)
Yes, no indication of where the series comes from is vastly less baffling than an indication that will make no sense at all for the intended audience and thus convince the intended reader that he is not the intended reader at all. Henning Makholm 19:15, 5 February 2007 (UTC)
I found the previous lack of explanation baffling. --Michael C. Price talk 13:16, 6 February 2007 (UTC)
The current lack of explanation is far more baffling because now there is something that the unitiated will wrongly believe explains anything. Henning Makholm 19:13, 6 February 2007 (UTC)
I think the recent changes to that section have needlessly complicated what used to be a straightforward example for laypeople. It doesn't matter what the derivation of the Gregory-Leibniz form is. There are many laypeople who want to know if there is a simple way of calculating the value of Pi. I have pointed three or four of my friends/colleagues to the article and to that section in the past, and they all appreciated and understood it as previously stated. But as soon as you throw in the part about the right isoceles triangle and tangents, it's no longer a simple statement about a way to calculate Pi. Putting it in there also contradicts the previous sentence, about how calculating Pi doesn't require understanding of trigonometry. When I was in high school tangents were covered in Trigonometry. The idea that the derivation belongs here or is required in this section is overthinking it.

Another point. I dislike the way the Gregory-Leibniz series has been modified. It used to say:

${\displaystyle \pi =4-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+{\frac {4}{9}}-{\frac {4}{11}}\cdots }$

Now it says:

${\displaystyle \pi =4\left(1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-{\frac {1}{11}}\cdots \right)}$

The first form is the simplest statement to understand. It may not be as mathematically beautiful, but if it's more understandable it should be what's used there. -- Moondigger 18:59, 5 February 2007 (UTC)

I agree, on the form of the series. Readers will either see no difference, or follow the 4's more easily. Septentrionalis PMAnderson 19:19, 5 February 2007 (UTC)
Hmm, I'd change it back to the original one, but the HTML table-based fractions are really ugly. Is there any way to force it to appear in an image? ~ Keiji (iNVERTED) (Talk | Contribs) 19:48, 5 February 2007 (UTC)
Okay, I've tried to please both of you; the gregory-leibniz is there so people can see that there are simple formulae, and I mentioned the algorithm from the computing pi article that uses pythagoras for inside-outside tests so that people get the notion of why a simple formula can yield an irrational number. Lunkwill 20:56, 6 February 2007 (UTC)
I like the section much better now. Consensus above seems to be that the series is probably easier to grasp for unsophisticated readers if we distribute the 4 into each fraction; I have done so. I also tried to switch to slashed fractions; does that fix the rendering problem Inverted alludes to? (On the other hand, I wonder how many readers will now get the precedence wrong and think it's some kind of continued fraction ...) Henning Makholm 23:56, 6 February 2007 (UTC)
I put the brackets back in, but leaving the 4 on each fraction. That makes it appear in an image, and it shouldn't look too odd even without a coefficient in front of the brackets. ~ Keiji (iNVERTED) (Talk | Contribs) 07:12, 7 February 2007 (UTC)
I'd rather suggest the following form. It looks simpler, doesn't it?
${\displaystyle {\frac {\pi }{4}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-{\frac {1}{11}}\cdots =\sum _{k=1}^{\infty }{\frac {{\left(-1\right)}^{k+1}}{2k-1}}}$
Ocolon 09:23, 8 April 2007 (UTC)

I think a mention of how quickly this converges on pi would be interesting. I had a go at calculating pi using this method in a MS Excel spreadsheet and even after 10,000 terms it was only accurate to a few decimal places. Perhaps a mention of formula efficiency would be worth including? --ManInStone 13:22, 25 July 2007 (UTC)

## Remove semi-protection

As per semi-protection policy this is no longer the 'only reasonable option left to deal with vandalism.' It's been semi-protected since 14th January. I'm putting a request in to remove this here. The18thDoctor 13:05, 6 February 2007 (UTC)

## Recommendations

Just a few suggestions I have for this article -- some major, others minor.

1. What is the source for the assertion that "The constant is named "π" because it is the first letter of the Greek words περιφέρεια 'periphery' and περίμετρος 'perimeter', i.e. 'circumference'"?
2. Do we really need to know what Unicode character is used to represent the greek letter pi? That material would be more appropriate for a different article.
3. "The exact value of pi has an infinite decimal expansion" needs to be reworded. The value of pi does not have an infinite decimal expansion -- the number has an infinite decimal expansion. At least that's what I'd say.
4. The second method for approximating pi should include the term "lattice point" somewhere.
5. The paragraph on the Biblical "approximation" of pi needs to be reworded (and sourced). It's almost unintelligible.
6. "For many purposes, 3.14 or 22/7 is close enough" should probably not mention 22/7, since it fits better in the context of the next sentence, which discusses fractional approximations rather than decimal approximations.
7. Most of the formulae involving pi need to be diked out and inserted into either Computing pi or List of formulae involving pi.
• The section Continued fractions exists in Computing Pi, should we remove it here? Deathbob 17:34, 24 February 2007 (UTC)
It is not clear that that section is relevant at computing pi. Are anyone seriously using the systematic continued fractions to approximate pi? I would have pegged them as merely "interesting properties". –Henning Makholm 19:49, 24 February 2007 (UTC)
1. The statement on the Indiana State Senate really, really needs to be correct. As it stands it is unclear whether it is correct.
• See Indiana Pi Bill and the links there named. Septentrionalis PMAnderson 02:45, 14 February 2007 (UTC)
• I meant that the statement didn't actually say what the bill did. It did not attempt to define a new value of pi, as many people have stated incorrectly. It proclaimed that a certain circle-squarer had succeeded, and that the state of Indiana had the exclusive rights to his discovery. I think the article should make this a little more clear. --N Shar 06:15, 15 February 2007 (UTC)
2. Lots of citations are needed

N Shar 02:24, 14 February 2007 (UTC)

## "Guillaume Pelletier" trigonometric limits

Why has the equations : ${\displaystyle \lim _{P\to \infty }{\pi }=P\left[{\frac {\sin \left({\frac {360}{2P}}\right)}{\sin \left({90-{\frac {360}{2P}}}\right)}}\right]}$ and ${\displaystyle \lim _{P\to \infty }{\pi }=P\cos \left({90-{\frac {360}{2P}}}\right)}$ been removed? The reason provided is not clear. The formulae do not require the value of pi to be inserted. The only value required is that of P which is the number of sides on a polygon. The formulae are effective in large values of P, as the perimeter of the polygon approaches that of the circumference of the circle. The approach of the formulae is not that of supreme practicality, but of functionality, simplicity and concise interpretation: replace P by an extremely high numerical value and you have an extremely accurate pi value. Even higher values of P return even more accurate values of pi —Preceding unsigned comment added by Epgui (talkcontribs) 2007-02-25T23:37:32

Your formulae involve sines and cosines of angles that you specify in degrees, such as 90-360/2P. The procedure for computing a definite value for such a sine or cosine has two steps:
1. Convert angle in degrees to radians by multiplying the number by ${\displaystyle \pi /180}$.
2. Apply the definition for the (co)sine of a radian angle, e.g.:
${\displaystyle \sin(x)=x-x^{3}/3!+x^{5}/5!-x^{7}/7!+x^{9}/9!-\cdots }$
That your formulae converge towards ${\displaystyle \pi }$ are entirely due to the step 1, where a preexisting value of pi is supposed to exist. You may not be aware that this is what your calculator does internally when you ask it to compute the sine of a degree value, but it is. The limit has nothing to do with trigonometry; in fact you would get exactly pi by throwing out the trig functions and approximating ${\displaystyle \cos(\pi /2-\epsilon )=\sin(\epsilon )\approx \epsilon }$ and ${\displaystyle \sin(\pi /2-\epsilon )\approx 1}$ for small ${\displaystyle \epsilon }$. –Henning Makholm 23:57, 25 February 2007 (UTC)
By the way, you're not the first to present such formulae; I removed almost the same one from history of numerical approximations of pi a few months ago, and initially thought you were the same editor who added that. Further investigation shows this to be unlikely. Sorry for the snappy edit summary; I should have assumed good faith. –Henning Makholm 00:33, 26 February 2007 (UTC)
Thanks a lot for helping out, I did not realise beforehand that trigonometric functions implied the use of pi. —The preceding unsigned comment was added by Epgui (talkcontribs) 00:44, 26 February 2007 (UTC).
You're welcome. In the interest of telling the complete thruth, I should mention that there are formulae that allow, e.g., going from sin(v) to sin(v/2) without knowing pi (you may be able to figure them out yourself by considering some appropriately chosen congruent right triangles). Then if you start with sin(90°)=1 you could in principle compute your limits for P's of the form 2n without knowing pi already; in essence this is how Archimedes approximated pi, as mentioned briefly in the article. Some of the early trigonometric tables were computed in this way, too. But modern sine computations invariably work in radians. –Henning Makholm 01:25, 26 February 2007 (UTC)
PS - someone pointed out that the limit was not placed correctly in the equation. We did not do limits in school yet, so I would be curious to see the corrected formulae. (Epgui 00:52, 26 February 2007 (UTC))
It may have confused you that the limit of a sequence article alwas has then "lim" sign is on the far left of an equation. What you should know is that "lim" is something that applies to an expression rather to an entire equation. Thus, for example
1. "${\displaystyle P\sin(180^{\circ }/P)}$" is an expression that has some value if P is given.
2. "${\displaystyle \lim _{P\to \infty }P\sin(180^{\circ }/P)}$" is an expression that has some value even if we don't give it a P, because the lim symbol explains what the P in the inner expression is.
3. "${\displaystyle \pi =\lim _{P\to \infty }P\sin(180^{\circ }/P)}$" is an equation that asserts something about the value of the second expression.
Clearer now? –Henning Makholm 01:25, 26 February 2007 (UTC)
Yes, thank you very much :) (Epgui 11:38, 26 February 2007 (UTC))3.14159265358979323

## Transcendental, not just irrational

In the article lead, it says "The mathematical constant π is an irrational real number". Shouldn't that be "... transcendental real number"? It is mentioned 'way below, but I think it should be highlighted up front. (If the page were not semi-protected I would just go ahead and change it, but I cannot be bothered to try to remember my wikipedia password for this.) -- 72.130.182.201 03:23, 15 March 2007 (UTC)

Done--ĶĩřβȳŤįɱéØ 03:27, 15 March 2007 (UTC)

## Gottfried Leibniz or Madhava of Sangamagrama?

For me, it seems that the articles "History of Pi" and "Pi" is a little unclear about the creator of

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}}$

In "pi" it is just stated that "Leibniz' formula (proof): [the expression]" while in the article "History of Pi" Madhava of Sangamagrama is credited. One might intepret (if only reading the "Pi"-article) Leibniz as the creator. Perhaps this needs to be changed somehow - or am i missing something? Sorry for the bad English. --Bilgrau 22:38, 20 March 2007 (UTC)

## What?

I can't edit this page! Why? —Preceding unsigned comment added by Jaunt (talkcontribs)

This page is semi-protected, meaning that only established users can edit it. An established user is one who has an account 4 days old. It seems your account is currently only 1-2 days old, so you will have to wait a few days in order to edit this article. -- Meni Rosenfeld (talk) 12:23, 24 March 2007 (UTC)

I'll wait. Jaunt 11:03, 25 March 2007 (UTC)

## Misperception = 2500 years of Misconception.

I have a different perception of what Pi is and its' "relativity" to Radius. I attribute Archimedes with greater concern for precision than Mathematic's interpretation of Pi r2 gives him. Archimede's Principle is Perfect and Precise when his Principle is correctly perceived. Archimede's Principle gives a Circle exact equivalence with any "Line Shape" The key is in the Transformation of Shapes. Misperception = 2500 years of Misconception.

Archimedes Principle when correctly Perceived results in this. A circumference is 8 times the Radius or 4 times the Diameter. I can prove this through Transformation of a known length. Test it. It must be tested with other than Pi r2 because as I stated Pi r2 is a misperception of his intent, ergo imprecise. I will not answer email. Post queries here. --Layman1 12:40, 24 March 2007 (UTC)

${\displaystyle u_{\square }=4\times d}$${\displaystyle u_{\bigcirc }=\pi \times d}$
You're basically saying that the value of ${\displaystyle \pi ={\frac {circumference}{diameter}}}$ isn't 3.14… but 4, right? Well, I'd really love to see your proof for that, but this isn't the right place to discuss this. This place is for discussing the article and not for discussing your original research. I propose that you either provide a reliable source for your statement (and I cannot imagine you find one) or move this discussion to your talk page. I will watch it. — Ocolon 13:26, 24 March 2007 (UTC)
Sorry, but I can prove that pi < 4. Consider the Leibniz formula. Note that the series is an alternating series of terms of strictly decreasing magnitude to 0, so the result of the series is strictly between any two consecutive partial sums. Thus, we see that 2/3 < pi/4 < 1, whence 8/3 < pi < 4. Eric119 16:18, 24 March 2007 (UTC)
I think Layman1 would argue here that Leibniz' formula describes a ${\displaystyle \pi }$ that is distinct from the ${\displaystyle \pi }$ defined as ratio of a circle's circumference to its diameter. He's basically saying that the black line describing the circle in the image is just as long as the outline of the square it is contained in. Not more and not less. — Ocolon 16:52, 24 March 2007 (UTC)
If no one objects, I'd like to remove this section. I'd hate it to be associated with Wikipedia. Suffice it to say that humans know a fair bit about circles, and this is definitely not worth discussing. Glooper 02:00, 8 July 2007 (UTC)
If you propose to remove this talk page section, I definitely object. We have a commitment to truth and goodness as regards the encyclopedic pages, but there is no point to censoring a talk page simply because the original poster spoke nonsense. It will get archived eventually and does no harm until then. –Henning Makholm 00:43, 9 July 2007 (UTC)
Sure, OK. Sorry, I just figured it was taking up space that could be used for something serious or intelligent. Glooper 04:34, 9 July 2007 (UTC)

## Article name

It looks like the technical problems that made naming this page π have been solved (as evidenced by π (film)). Should this page be moved to π now that it will no longer look like it is titled Π? (I'd say yes.) -- JHunterJ 12:30, 27 March 2007 (UTC)

That article is actually named Π (film) and has a {{lowercase}} template that uses javascript to change the title to lowercase when it is displayed. This works around a symptom of the technical problem but does not solve it; it is still not possible to have separate articles for Π and π. I don't care what the article is titled, frankly, so long as it has something to do with π. CMummert · talk 13:19, 27 March 2007 (UTC)
Just π by itself is problematic; the letter has too many other uses. But I wouldn't necessarily be against a move to π (mathematical constant) (on the model of e (mathematical constant)). --Trovatore 07:51, 30 March 2007 (UTC)
The mathematical constant e is a lot less known than pi. I would prefer to see it left as pi and redirect from π. 199.125.109.27 16:12, 11 July 2007 (UTC)

## Finally!

I can edit Pi! Jaunt 15:21, 27 March 2007 (UTC)

Congratulations! 84.229.139.208 21:02, 29 March 2007 (UTC)

## Square root.

A rather odd topic I knoe, but does anyone know the square root of Pi?--Kevin mills 00:00, 30 March 2007 (UTC)

I believe it's √π. If you want something with digits, a Google search on sqrt(pi) "sqrt(pi) = 1.77245385". -- JHunterJ 00:22, 30 March 2007 (UTC)
Yep. The square root of π is the positive solution to the equation ${\displaystyle x^{2}=\pi }$, which is usually denoted ${\displaystyle {\sqrt {\pi }}}$. Decimal digits are listed as sequence A002161 in the OEIS. — Ocolon 07:11, 30 March 2007 (UTC)
It might also be of interest that ${\displaystyle {\sqrt {\pi }}=\Gamma \left({\frac {1}{2}}\right)=\int _{-\infty }^{\infty }e^{-x^{2}}\mathrm {d} x}$. See Gamma function, Normal distributionOcolon 13:15, 2 April 2007 (UTC)

## T-shirt equation

${\displaystyle \pi ^{4}+\pi ^{5}=e^{6}}$

LS, I encountered the above equation on a T-shirt. It seems to give a correct and interesting relation between the numbers pi and e. I do not know of any proof for it, but am curious... anyone? — Xiutwel (talk) 10:46, 2 April 2007 (UTC)

Left-hand side = 403.42877581...
Right-hand side = 403.42879349...
Fredrik Johansson 11:07, 2 April 2007 (UTC)
It is a mathematical coincidence. --Aleph4 11:09, 2 April 2007 (UTC)

Plus, e^i(pi)=1, where i is euhler's constant of the square root of -1, and e is the natural logarithm.

## World record is wrong!

It says that the current world record for the memorization of pi is 100,000. But Akira Haraguchi just claims that he remembered 100,000 digits, this claim hasn't been verified by the Guinness World Records! (his other claim (80,000 hasn't been verified as well). So the official and verified World Record for remembering pi is still 67,890 digits by Lu Chao from China look here!, or here. So somebody please change that (or give me permission to do so). greetz Admirator —The preceding unsigned comment was added by Admirator (talkcontribs) 06:23, 6 April 2007.

ok i corrected it.. you guys look if it looks right. and someone delete or change that diagram, cause its wrong too!

## The Origin of pi

This might seem obvious but more should written in this article about how pi was first derived by Pythagoras. —The preceding unsigned comment was added by 86.135.248.72 (talkcontribs) 17:12, 9 April 2007 (UTC).

Pythagoras? I'm afraid you are confusing this with something else. See the history section on π for more information. — Ocolon 16:32, 10 April 2007 (UTC)
Were you thinking of the Pythagorean theorem? --LuigiManiac 16:46, 10 April 2007 (UTC)

## Mnemonics

I am quite concerned about the accuracy of the example poem in mnemonics. It appears that you use the number of syllables to remember the next number, so I will break the example phrases into syllables below:
How I need a drink, al-co-ho-lic in na-ture, af-ter the hea-vy lec-tures in-vol-ving quan-tum me-chan-ics.
How I need a drink, al-co-ho-lic of course , af-ter the hea-vy lec-tures in-vol-ving quan-tum me-chan-ics.

Notice that "in nature" and "of course" have a different number of syllables.
Is this a problem, or am I just imagining it? Freedomlinux 00:09, 11 April 2007 (UTC)
What does the number of syllables have to with remembering the next number? It's the number of letters that's used to determine the current number. -- JHunterJ 00:39, 11 April 2007 (UTC)

## 2 pi

manixrock: I am sorry for putting this here, but I'm not experienced with wikipedia's editing processes yet (but still learning). One suggestion I would like to see added is this article expressing the fact that PI (3.14...) is not the correct number that should be given a special name, but it's double (6.28...); the reasons are outlined here: http://www.math.utah.edu/~palais/pi.html and it also suggests an interesting solution by having PI at it's current value and creating another character for 6.28... that looks like a PI with 3 legs. —The preceding unsigned comment was added by 86.124.229.142 (talkcontribs) 01:08, April 12, 2007 (UTC)

Calling п wrong or incorrect is wrong and incorrect by definition. п is defined as what it is defined as. While "п is wrong" serves as an eye-catcher in a newspaper it should not be put here.
I understand the motivation for redefining п as what is now 2п. The interesting link you provided gave some good examples. But there are also many examples where п appears without the factor 2 (see the article). Why confuse everybody by redefining п and then have п/2 in many formulas? We could then have the same discussion with switched roles again.
About giving it an own name… If you introduce a symbol for 2п then certainly someone comes and wants one for п/4 too. And so on. Furthermore it would be like giving the semi-cirlce an "own" name, such as qwerty. Do we really want to learn different symbols or names for things that are connected by a simple correlation such as doubling? I don't want to. Of course you are free to define new symbols in your own and for your own maths papers. That's a privilege of mathematicians.
I wouldn't mind if you put a link to the article into the external links section. But your approach is not accepted widely enough to appear as major information in the article. Please mind that Wikipedia is an encyclopedia. It is neither a place for original research nor for rare opinions. — Ocolon 08:18, 12 April 2007 (UTC)
For previous discussion on this topic, see Talk:Pi/Archive3#2pi? and Talk:Pi/Archive4#π_vs_2π_and_the_unit_circle. -- EJ 10:31, 12 April 2007 (UTC)
manixrock: Hy. Thanks for correcting my message posting mistake. I'm not sure if I'm answering here the right way, so please correct me if I'm wrong, and add a quick explanation on how I should do it next time.
On the subject, I read the links u gave Talk:Pi/Archive3#2pi? and Talk:Pi/Archive4#π_vs_2π_and_the_unit_circle, and it seems to have brought up just about every point I was going to make, including the link I gave (as a resource for my claims). So I rest my case, and I'm off to read/write more 1's and 0's.

## Digits (Link to first 4 million)

Please do not change the link to state it is 4 MB. I own this page, and I can guarantee you that (at maximum) only 2MB of bandwidth are transferred when this page is viewed. Yes, I know that 4 million digts = 4 million bytes = (about) 4 MB. Due to the nature of dumping the huge amount of traffic for this page, the digits are gzipped and the script requests for the browser to use compression. If the browser does not accept compression, only the first 100 digits or so are used. If the browser does support compression, then 1,993.21 KB (2,041,048 bytes) are used. You can verify this yourself (using firefox) by visiting the site and then right clicking the document and selecting "View Page Info." --Ahsile 19:13, 20 April 2007 (UTC)

## Early approximations section

This section has almost no reference. And in the sentence:

"The Indian mathematician and astronomer Madhava of Sangamagrama in the 14th century computed the value of π after transforming the power series expansion of π⁄4 into the form"

What is the power series expansion of π⁄4?

The series below that sentence comes from the power expansion of the arctan evaluated at 1/sqrt(3) giving Pi/6. But maybe he used another method. Can someone check it? Ricardo sandoval 21:46, 22 April 2007 (UTC)

## Is there any document on the web that has a viewable 1 billion digits of Pi?

--Foljiny 21:36, 5 May 2007 (UTC)

That would require 400MB of data transferred to your computer, so I think "downloadable" is the more appropriate term. I am not familiar with any of those, however. -- Meni Rosenfeld (talk) 09:09, 9 May 2007 (UTC)
A program that is readily available on a personal computer that can calculate pi to billions of places seems to be a suitable alternative. 199.125.109.11 19:23, 13 May 2007 (UTC)
Yep, it's PiFast but this calculation would take longer than downloading it This is the site I found for 4.2 Billion digits SuperMarioBrah 01:05, 13 June 2007 (UTC)

## memorizing digits

The world record info in this section is different from Piphilology#History.--mikeu 02:25, 8 May 2007 (UTC)

## "All the succeeding digits were incorrect" ???

Sorry to be a nitpicker (but hey, this is math): The article states, "William Shanks [. . .] spent over 20 years calculating π to 707 decimal places (accomplished in 1873). In 1944, D. F. Ferguson found that Shanks had made a mistake in the 528th decimal place, and that all succeeding digits were incorrect."

If pi's digits are "random" and independent (apparently a good approximation), then the probability that 1 digit chosen at random is wrong is ~ 9/10, and the probability that 707-528 = 179 such digits are wrong is ~ (9/10179 ~ 6.4 × 10-9. (Also, whichever digits were incorrect very likely still are.)Daqu 19:32, 8 May 2007 (UTC)

• Evidently the lexical accuracy of the succeeding digits is irrelevant - their meaning is completely altered by preceding incorrect digits. 223 differs from 123 quite substantially. AlexTiefling 20:03, 8 May 2007 (UTC)
How about just changing it to "... were irrelevant" (meaningless, moot, etc.). -- JHunterJ 20:25, 8 May 2007 (UTC)
Not so fast. If you are quoting D. F. Ferguson it would be ok to quote him. It also would be better to compare the 179 digits to see if they were all wrong, randomly wrong, or even all right. Has anyone taken the time to do that? I would suggest changing it to say "and most of the succeeding digits were incorrect". 199.125.109.27 16:01, 11 July 2007 (UTC)

## Indiana Pi Bill

What happened to this? I readded a See Also; but we really should have a sentence in text, as a preventative to all the erroneous versions of the incident on the web. Septentrionalis PMAnderson 18:50, 13 May 2007 (UTC)

## Numerical approximations

It is important to note that today engineers just hit the pi key on their calculator to use 10 digits of pi. When pocket calculators replaced slide rules for engineering, run of the mill calculations suddenly jumped from 2 or 3 digit accuracy to 10 digit precision (not accuracy), fundamentally changing engineering. While most engineers round off their calculated results to a meaningful accuracy (sometimes plus a few digits for good measure), it is common to see students foolishly include all ten digits. How about adding something like, "prior to the advent of pocket calculators" or "Today's engineers normally use the pi key on their calculator (10 or 12 digit accuracy)." 199.125.109.3 19:47, 13 May 2007 (UTC)

## Calculating \pi

Just changed and greatly reduced the last paragraph of this section, I guess that as one of the first sections on the article we should really keep things simple, so I decided to emphasize an empirical approach to find the area of a circle (and thus find $\pi$) and just give the gist of the approach using squares, that naturally leads itself (comes from?) another empirical approach.

Not sure if the old paragraph can be used in another part of the article, I remember this approach being used by number theory in Hardy's book, any ideas?

This section and the section "Definition" have some points in common, a merger is in order? Ricardo sandoval 20:06, 13 May 2007 (UTC)

I have reverted. The point of the section is to make clear to the reader that there is are definite mathematical procedure which is guaranteed to find approximations to pi with any desired precision; in particular that it is not a matter of opinion or observational inaccuracies which of two offered approximations to pi is the most precise. Speaking about weighing physical circular objects destroys this impression. –Henning Makholm 20:15, 13 May 2007 (UTC)
Ricardo Sandoval posted the following reply on my talk page:
Hi, I didn't understand your comment about the purpose of the section when you reverted my edit, the mathematical explanation on the last paragraph was really unhelptfull for a generall audience unless some picture was added. So I decided to emphasize another empirical approach and the idea of the approach that was there before. If this doesn't belong to this section where does it belong?
Let me try to be clearer: The section is supposed to be about computing pi, not measuring it. Its purpose is to convince the reader that pi can be computed, without resorting to empirical procedures with their inherent fuzziness. Your version talked about weighing physical representations of circles, which is counter to this purpose. It may be that the explanation is confusing (I don't think so, but you're welcome to point out problems), but if so the cure is to make the explanation better, not to replace it with an easier explanation of something entirely different. (Also, if all wants is to approximate pi by measuring real objects, weighing things is really overkill. Then it is much easier simply to measure the diameter and circumference of a circular object using a tape measure.) –Henning Makholm 21:07, 13 May 2007 (UTC)

Thanks for the explanation. What is in now is mathematically precise but it is still an approximation method unless a limiting process is mentioned. I guess some different ways to empirically aproximate pi would be interesting, appart from the straighforward way. Maybe a new section should be made? Ricardo sandoval 21:31, 13 May 2007 (UTC)

A possible strategy might be to expand Pi through experiment to be more than a redirect to Buffon's needle. –Henning Makholm 23:05, 13 May 2007 (UTC)
We can add a reassurance that, done competently, this does converge; but I doubt it's worth even outlining a proof; does this article need to link to Minkowski's theorem? Septentrionalis PMAnderson 01:24, 14 May 2007 (UTC)

## pi=rational?

Rational numbers are of course numbers which can be expressed as a ratio, or fraction. Pi is defined as A/r², which is a fraction. Is pi therefore rational?

71.207.86.54 20:41, 17 May 2007 (UTC)

No. Rational numbers can be expressed as a ratio of two integers. -- JHunterJ 20:47, 17 May 2007 (UTC)
By definition, in the equation π = A/r2, at least one of A and r will be irrational - transcendental, in fact. AlexTiefling 08:40, 18 May 2007 (UTC)
Well, not by definition. It's a theorem.--agr 11:44, 25 May 2007 (UTC)

### 37.68/12

${\displaystyle 37.68/12}$

Many users have confined this equation as pi, why not put it on early approximations?

You need proof? Tell me here. Goldkingtut4 20:51, 20 May 2007 (UTC)

Yes, we need proof. — Arthur Rubin | (talk) 20:54, 20 May 2007 (UTC)
What a horrible approximation. Using 6 digits you end up with 3 significant digits. At least with 22/7 by using 3 you get 3, and with 355/113 you use 6 and get 7. 199.125.109.11 04:08, 25 May 2007 (UTC)
While I agree this approximation method is unsatisfactory, the truth is that it gives 3 significant digits (it just amounts to 3.14). -- Meni Rosenfeld (talk) 11:29, 25 May 2007 (UTC)
That's not nice, quietly fixing the above making everyone (myself included) think I've gone crazy. -- Meni Rosenfeld (talk) 12:55, 20 July 2007 (UTC)

## This means that there is no polynomial with rational coefficients of which π is a root

In the claim "This means that there is no polynomial with rational coefficients of which π is a root" should the word "non-zero" be added before "polynomial"? Jayme 12:11, 25 May 2007 (UTC)

## Pi in the Greek language

Unicode has a code point for "Greek Pi" at 03D6: ϖ, which is an omega (ω) with an overbar. It may be useful to mention this in the section where the representation of pi is discussed but I don't know the proper usage for this symbol in Greek. --B.d.mills 02:52, 1 June 2007 (UTC)

I have known mathematicians and physicists who use this symbol, and call it "pomega". (They're not Greek.) -- Coneslayer 15:05, 1 June 2007 (UTC)
The Unicode name of this character is "Greek pi symbol" ϖ as opposed to "Greek small letter pi" π. The word "symbol" in its name is significant: it means that it is used as a technical symbol, and not as a letter of the alphabet. This is similar to the variant forms of φ: Greek phi symbol ϕ vs. Greek small letter phi φ. In both cases, they are simply variant (cursive) graphic forms which some mathematicians use as distinct symbols. In particular, ϖ really has nothing to do with omega except its superficial appearance; its interpretation as "an omega with an overbar" is just typographical abuse improvisation.... --Macrakis 16:45, 1 June 2007 (UTC)

## Pi...

Click this link for a song about pi. Funny. Gray94wolf 15:50, 10 June 2007 (UTC)

## Expect frequent vandalism.. :(

I guess a lot of people will try to change it back to "exactly equal to 3" because of this post: http://digg.com/general_sciences/Teach_the_controversy_Pi_3_0_textbook_stickers_included ... :( ebi 23:25, 13 June 2007 (UTC)

Good one (from that link): "Be happy they rounded it. Otherwise the bible would never end." I can see it now: "today's reading continues on from last week's: "720589365290934857620349576620003898576099475908401909288455767692229933880" Gzuckier 15:36, 7 August 2007 (UTC)
Should the article mention the disagreement with the bible ? I think so. ._-zro tc 07:42, 7 August 2007 (UTC)
The apparent disagreement is already discussed. The standard explanation (measurements on the outside and inside), which is supported by other calculations, is presented; Another explanation mentioned is that the circumference really is 30 cubits, with the diameter being 9.55, rounded to 10. -- Meni Rosenfeld (talk) 08:06, 7 August 2007 (UTC)
I'm not sure the ancient Hebrews had the concept of the decimal point, so it would have been hard for them to express pi. Even as a fraction; "Three plus about the seventh part of ..."Gzuckier 14:01, 7 August 2007 (UTC)
Duly noted. My bad. ._-zro tc 22:23, 7 August 2007 (UTC)

## Error?

numerical value the first 60 didgets of pi contain an error its actually 3.14159265358974323846264338327905028841979399375105820974544 in the first 50 didgets written on th page you are missing a 0

Your value and the actual value:
3.1415926535897432384626433832790502884197  9399375105820974544
3.141592653589793238462643383279 50288419716939937510
^                ^         ^^

There are several errors in the source you're using, but the "missing" 0 on this article should remain missing; the second number (ignoring the spaces) is correct. -- JHunterJ 11:07, 18 June 2007 (UTC)

## Calculating PI

I believe that the following iterative formula calculates the value of pi very quickly:

${\displaystyle P_{0}\;=\;3,\;P_{n+1}\;=\;P_{n}\;+\;\sin \left(P_{n}\right)}$

I know little about it except that after 2 iterations it is correct to at least 9 places. Perhaps this should be included? It is the fastest method I have seen, except it may be quite slow to find the sin after quite a few iterations. Robert 18:23, 12 July 2007 (UTC)

You have answered your own question. This formula, indeed, triples the number of correct digits with every iteration, but calculating the sine is too expensive to make the formula effective. -- Meni Rosenfeld (talk) 18:30, 12 July 2007 (UTC)

## Physical Formulae: Measure Dependent

The physical formulae mentioned are measure-dependent, and may not contain Pi if another measure is used. Even though this is briefly illustrated at the bottom of the article, in my opinion it should be stated at this point. Alternatively, concerned formulae could be removed. But the present form gives the reader the impression that Pi comes out of the nature of the physics involved, where it is only used for convenience. 62.178.9.201 00:42, 19 July 2007 (UTC)

I'm no physics expert, but I'd say that π does come out of the nature of the physics involved. If you redefine your units\variables to eliminate π from one equation, it will pop up in another, related equation. -- Meni Rosenfeld (talk) 12:54, 19 July 2007 (UTC)
I would say that it really does not have so much to do with the physics as with the approximations and idealizations one is interested in. In many places in physics one is interested in getting formulas for idealized situations such as "imagine an infinitely long, perfectly regular, cylinder of <some stuff>" or "imagine an infinite plane sheet made of <some stuff>", because these idealizations give rise to simple formulas and are often adequate for analysing real-world situations where the stuff does not really extend infintely. In general pi is going to show up in one or the other of such formulas, simply because of the geometry of the idealized situations. For example, if something in the physics depends somehow on the relation between the volume of the stuff and the surface area of the stuff, it is a geometrical fact that there is going to be a factor involving 2pi in difference between the formulas for "infinite plane" and "infinite cylinder". Because physicists want to have formulas for both kinds of idealizations, they are stuck with having pi somewhere. But it's really geometry more than physics that brings it in, once you have decided that you'd like your physical laws to be rotation invariant. –Henning Makholm 21:29, 19 July 2007 (UTC)
That's exactly what I was trying to say. Ie. that it is not due to the physical fact that forces between charges diminish with ${\displaystyle {\frac {1}{r^{2}}}}$ that ${\displaystyle \pi }$ pops up in Coulomb's law in the SI system, but it is chosen so that the unit sphere area will cancel out somewhere else, if you calculate something which is of more interest to you. In other words, ${\displaystyle \pi }$ reflects the geometry of the space the physics take place in, not the physics itself. I consider this a crucial difference. It is good that the article mentions physical formulae containing ${\displaystyle \pi }$ in the SI measure that might be familiar to the reader, but it should be stated that it is not a physical necessity that this particular equation contains ${\displaystyle \pi }$. Georg Wachter 14:46, 24 July 2007 (UTC)

## protection tag

{{editprotected}} Please add a semi-protection tag to this article.--69.118.235.97 23:15, 27 July 2007 (UTC)

Done. — Arthur Rubin | (talk) 23:21, 27 July 2007 (UTC)

## Significant digits in engineering calculations

The article says, practical science and engineering will rarely require more than 100 digits. Can anybody cite anything to support the claim that anywhere near that many digits are ever used in engineering calculations? The moon is about 10^9 feet away from the Earth. If NASA worked the Apollo calculations to 10 digits, they would have hit the moon within one foot of their target, which is clearly far higher precision than was needed. If you told me they used 12 or even 15 digits, I would believe you. But 100? That's absurd. -- RoySmith (talk) 00:55, 25 February 2007 (UTC)

No claim that "anywhere near that many digits are ever used" is made in the article, and therefore there is no need to support such a hypothetical claim. The sentence you quote asserts that the required number of digits for almost everything is less than 101. Last time I checked, 12 and 15 were both indeed less than 101. –Henning Makholm 01:09, 25 February 2007 (UTC)
The word rarely isn't nearly strong enough. Something is rare when 5% of the instances do it that way. I would say, practical science and engineering will rarely require more than 6 digits -- RoySmith (talk) 03:01, 25 February 2007 (UTC)
You might as well say that practical science and engineering only uses the first Million digits of Pi, or that the Earth is bigger than a proton. It may be true, but it's not useful. Paladinwannabe2 18:17, 27 February 2007 (UTC)
I'd agree with Roy that the statement does suggest that there are computations which require 50-100 digits. However, I appreciate it won't be easy to reformulate it in an acceptable (i.e., attributable) way. That won't stop me from giving it a try though. How about
While the value of pi has been computed to billions of digits, practical science and engineering computations typically use only 16 digits. It is hard to envisage a situation where more than a hundred digits will be needed for this purpose.
The 16 digits refers to standard double precision. I think it's fair to say that is the precision typically used, though perhaps we do officially need a reference to back this up. -- Jitse Niesen (talk) 03:29, 25 February 2007 (UTC)
You could say something like, The roughly 16 decimal digits of precision supplied by the ubiquitous IEEE 754 floating point format is more than sufficient for almost all practical scientific and engineering calculations -- RoySmith (talk) 04:20, 25 February 2007 (UTC)

I agree, this is very silly. 5 digits of Pi gets us within 0.0001 inches at my work, and it looks like we got men to the moon on 5 digits as well, but I can't find an authoritative source yet. I'm going to change it, since I can always supply my own reference in the form of mathematics. Paladinwannabe2 17:18, 27 February 2007 (UTC)

Hardly an attributable source, but there's a scene in Apollo 13 where the guys in mission control are working calculations with slide rules, which are typically good to 3 digits. -- RoySmith (talk) 17:27, 27 February 2007 (UTC)
I changed the example to something sensible. Millimeters are something we can see, we know the earth is huge, this will hopefully get the point across better. Plus, I can provide my own numbers for anyone to see and double-check.
(Using 6,378.137km as the Earth's radius (r) at the equator, 2*(r*3.1415926536 - r*pi) ~= 0.130mm.)Paladinwannabe2 18:03, 27 February 2007 (UTC)
I am unhappy that the "circle the size of the galaxy" example has disappeared. It gave a nice direct demonstration that a realtively small number of digits is sufficient for anything anybody in their wildest dreams could imagine computing. The "six digits for an Earth-sized circle" does not impress this as vividly. Non-matematician readers who don't deeply grok how accuracy in a positional system improves exponentially with the number of digits might end up with a fuzzy expectation that to calculate a circle that encompasses a thousand earths, one would need six thousand digits – and the point that computing millions of digits is practically useless would not be driven home as efficiently. –Henning Makholm 21:05, 27 February 2007 (UTC)

This discussion and the current statement in the text that "practical science and engineering will rarely require more than 10 decimal places. As an example, computing the circumference of the Earth's equator from its radius using only 10 decimal places of pi yields an error of less than 0.2 millimeters" smacks of ill informed WP:original research. First of all there are many measurements in physics and astronomy that are accurate to better than one part in 10^10. More importantly, numerical solutions of differential equations and the like are often very sensitive to small errors and must be carried our at much higher precision than the final answer. So the Earth radius to millimeter calculation prove nothing. This sentence should go.--agr 10:18, 14 October 2007 (UTC)

This is exactly what I've been thinking every time these "you only need 5 digits" discussions pop up, but I've never seemed to have any evidence that high precision is commonly used, or even first hand knowledge (in my own calculations I often require thousands of digits, but these are not for what would be called "practical purposes"). I have modified the offending paragraph for now, but if you have any sort of reference it could be useful. -- Meni Rosenfeld (talk) 00:02, 1 November 2007 (UTC)
I agree, a reference to ever needing more digits for "specialized applications" would be nice, or at least, an article on how many digits are acceptable. —Disavian (talk/contribs) 01:40, 1 November 2007 (UTC)
I'd say that the comment could have stayed as I have put it in the meantime, but I understand that you are probably sensitive to those things because of the GA nomination. -- Meni Rosenfeld (talk) 10:11, 1 November 2007 (UTC)
You would be correct :) —Disavian (talk/contribs) 12:50, 1 November 2007 (UTC)

## Is there an error here?

Under the section Analysis, the second example begins with the sentence: Half the circumference of the unit circle: but the formula states it's equal to Pi; which is the whole circumference, not half of it! What am I missing here? -- Daniel B. Sedory 67.150.121.211 21:57, 25 February 2007 (UTC)

The unit circle has radius 1. Therefore its diameter is 2, and its total circumference is ${\displaystyle 2\pi }$. –Henning Makholm 23:17, 25 February 2007 (UTC)

## Neat thing about pi

Pi x 10^9 is equal to the number of seconds in 99.55319 years or 99 and 55319/100000 years. 24.4.131.142 05:09, 7 October 2007 (UTC)

### Another neat thing

Where n is the diameter of a circle in feet, the circumference of said circle in metres will be about n — 5%. knoodelhed (talk) 08:02, 22 November 2007 (UTC)

## Pi also is...

• the circumference of a circle divided by its raduis —Preceding unsigned comment added by Lairtnogaw (talkcontribs) 21:52, 15 October 2007 (UTC)
No, it is not. -- Meni Rosenfeld (talk) 21:58, 15 October 2007 (UTC)

## Pi in Popular Media

Does this article need a section on this? (Or similar) --Earin 16:16, 20 October 2007 (UTC)

No. We used to have one, but it got unwieldy, had questionable encyclopedic value, and needed constant care and trimming because people tended to add listings for every offhand reference to pi in any fictional context whatsoever. I would support adding fictional references to the current article only if some literary use of pi is of high internal importance within the work of fiction and has this importance has been noted by a respectable secondary source. –Henning Makholm 23:12, 21 October 2007 (UTC)

## Quick-failed "Good Article" nomination

Per the the GA reviewing guidelines, if "The article has any cleanup banners, including but not limited to {{cleanup}}, {{wikify}}, {{NPOV}}, {{unreferenced}}, etc..." it must be quick-failed. As there are multiple merge banners now present, and the completion of any merges would very significantly alter its content, I cannot in good faith allow a GA review to take place. Please feel free to remedy this and renominate the article, or alternatively seek a reassessment if you feel this decision was in error. Thank you for your work so far, VanTucky Talk 04:40, 25 October 2007 (UTC)

Okay then. I'll work on that stuff. this looks useful. —Disavian (talk/contribs) 04:57, 25 October 2007 (UTC)
Before I renominate it, I'd like the following accomplished:
1. At least 20 in-line references used in the article. We're not far from that.
2. Less than half of the external links currently on the article
Those two tasks can be combined if you wish, by using ELs to cite a specific portion of the text. Anyone want to chip in with this task? Also, thank you to those of you who have been copyediting the new prose for style/spelling/formatting, etc. :) —Disavian (talk/contribs) 05:49, 29 October 2007 (UTC)

## Sources

I was going to review the article for its GA nom, and have read the article completely, but I think it needs more sources/references. Until that is accomplished, this won't pass GA. Lex94 Talk Contributions Signatures 19:19, 3 November 2007 (UTC)

Is there anything in particular you'd like to see cited, or are you looking for a specific number of sources? —Disavian (talk/contribs) 05:21, 4 November 2007 (UTC)

## GA review

Hello, I am the reviewer of this article. If you have anything to ask, please do so. You can leave a message here or you can leave one on my talk page. Regards, Daimanta 01:46, 6 November 2007 (UTC)

How does the sourcing look so far? —Disavian (talk/contribs) 01:53, 6 November 2007 (UTC)
Up to "Advanced properties" the sources are good, but after that, no sources appear. Can somebody explain that to me? Regar~ds, Daimanta 12:24, 6 November 2007 (UTC)
I've been going through the article referencing things; that's the only section I haven't appreciatively cited since the last reviewer came along. —Disavian (talk/contribs) 13:20, 6 November 2007 (UTC)
Ok, how long do you think this will take? Daimanta 15:20, 6 November 2007 (UTC)
I don't know... I just got sick, and have some projects due on monday/wednesday, but I'll work on it when I can. —Disavian (talk/contribs) 03:39, 8 November 2007 (UTC)
Ok, that's fine with me. I hope the GA committee won't be angry with :( Regards, Daimanta 12:33, 8 November 2007 (UTC)

I see that new sources are inserted. That's great! Another problem I saw was this: "As a practical matter, this approximation is poor and converges very slowly." Can someone elaborate a little bit on that. Is it possible to give a mathematical deviation from pi? Regards, Daimanta 12:21, 9 November 2007 (UTC)

I haven't been able to find one, so I removed that sentence. If someone comes across one for it at a later point, it can be added back. —Disavian (talk/contribs) 13:37, 9 November 2007 (UTC)
I reinserted and expanded it. It is important that we do not support the impression that some popular mathematics articles give, that Buffon's needle is a serious alternative to calculating pi from first principles. Its shortcomings have nothing to do with pi in particular, or with the derivation of the result, but are readily apparent in the form of the formula itself. It's all about not expecting to get more significant digits out of a post-experiment calculation than there are in the raw measurements one put into it. There are vast numbers of science textbooks that teach this general principle, though finding one that uses Buffon's needle in particular as an example of it would probably be harder. However, demanding such an exact citation would be counterproductive; it would be like requiring a particular cite for a claim that the animal Ruwenzorisorex suncoides bears live young instead of accepting a derivation from the well-sourced fact that mammals in general do so. –Henning Makholm 13:36, 10 November 2007 (UTC)

Advanced properties mostly lacks sources. If that is solved, Pi will become a GA. Regards, Daimanta 12:54, 10 November 2007 (UTC)

Well, it looks like it was already promoted, but I added references to that section anyway (diff). —Disavian (talk/contribs) 02:47, 11 November 2007 (UTC)
Ok, that pisses me off. Somebody else has apparently promoted the article to GA but without discussing it and mentioning it. I WILL have a word with this person since I believe he is stepping his bounds. Anyway, the article is now worthy of GA, congratulation and thanks for helping me. Regards, Daimanta 13:04, 11 November 2007 (UTC)

## What is the relationship between the principles of the universe and the nature of the circle?

The following excerpt from the article talks about the nature of the circle, while I contend that the prevalence of π in physics formulas is due solely to trigonometric coordinate systems. Can anyone expound?

The number π appears routinely in equations describing fundamental principles of the Universe, due in no small part to its relationship to the nature of the circle and, correspondingly, spherical coordinate systems.

Michael.Urban 20:08, 8 November 2007 (UTC)

I think "its" refers to π, so the relationship in the title is not implied. -- Meni Rosenfeld (talk) 20:43, 8 November 2007 (UTC)

## Pronunciation

A small point: the article says, " "π" is usually pronounced as pie when referring to the constant, although the letter is properly pronounced pee otherwise. " This wording is misleading. Pi in English is always "properly" pronounced "pie". It is only correct to say that it is "properly pronounced pee" in modern Greek. See the article on the letter pi which gets this right. 24.136.6.69 04:11, 10 November 2007 (UTC)

The problem we have to solve here is to explain how the mathematical constant is pronounced, without getting into a long irrelevant digressions about other ways it might be pronounced. What makes it difficult is that there is apparently a disagreement about how to speak about π when discussing modern Greek spelling and typography in English. Some, like you, say that all pi's are equal in English pronunciation; others seem to want to use the Greek pronunciation when discussing the Greek language. Though this disagreement, in and of itself, is irrelevant to this article, it makes it difficult to craft an explanation of the (undisputed and uncontroversial!) mathematical usage that sounds neutral to both sides in that other dispute. –Henning Makholm 12:47, 10 November 2007 (UTC)
Well, at the very least it should be refined to say that all mathematical usages of the letter in English are pronounced "pie", whether or not they refer to the constant. --Trovatore 22:40, 10 November 2007 (UTC)
OK, my humble attempt is in place. I've replaced "the constant" by "English in a mathematical context"; that addressed my concern above without belaboring the point. Similarly, "pronounced 'pee' in Greek" should be noncontroversial, and the question of how to pronounce the letter in English when talking about Greek is simply ignored as irrelevant to this article. Hopefully this should satisfy everyone who doesn't have a chip on the shoulder, and it keeps the tangential discussion short. --Trovatore 22:57, 10 November 2007 (UTC)

## Standardize on pi or π

Please standardize the articles about pi to use either pi or π. Pi is 3.14159 | Talk 00:50, 13 November 2007 (UTC)

Why? Both are valid names for the constant. —Disavian (talk/contribs) 01:26, 13 November 2007 (UTC)
I tend to agree that standardizing to π would be desirable. I think this is similar to the American\British English issue - while there is no preference to one or the other, consistency within an article is expected. Also, I could be wrong, but I think of "pi" less as a valid name for the constant and more as a standard placeholder when greek letters aren't available. -- Meni Rosenfeld (talk) 11:54, 13 November 2007 (UTC)
I also agree with this suggestion. Ahsile 21:02, 13 November 2007 (UTC)
As I'm not a native speaker of English, the following should possibly be taken with a grain of salt, but: My impression is "pi" is a fully valid name for the constant, whereas "π" it the symbol for it. This distinction is a bit fuzzy, but consider when you would write "the golden section" rather than "φ" (in a context where the latter is somehow unambiguous). In contexts that are concerned with the precise numerical value, or with use of the constant in calculation, it is most proper to use the symbol, but when we speak about about the idea of the constant, the verbal name seems to be most appropriate. For example, if you commemorate this article's subject on the 22th of July, what you are celebrating is "Pi Day" rather than "π Day".
Such a distinction (assuming it is not just my imagination running wild) would seem to point in the direction of continuing to call the main article Pi. –Henning Makholm 23:00, 13 November 2007 (UTC)
I was personally thinking about the appearances of π \ pi inside the article. As for the title, I agree that Pi is good. -- Meni Rosenfeld (talk) 23:03, 13 November 2007 (UTC)
Ah. Inside the articles, I think that my distinction would lead to π in the vast majority of cases, so we're probably not in actual disagreement. –Henning Makholm 23:14, 13 November 2007 (UTC)
By "the articles about pi" I assume you mean Category:Pi? —Disavian (talk/contribs) 22:05, 13 November 2007 (UTC)

## Is Pi Rational

I found a site thatsays they have proof pi is rational. I don't understand all that random math but the pi article and the site disagree. I think this should be looked into by someone who understands math better than I. 96.228.42.233 (talk) 23:26, 20 November 2007 (UTC)

Search for the phrase you can safely assume this page is a joke in the page. --Trovatore (talk) 23:28, 20 November 2007 (UTC)
It actually appears as you can safely rest assured that that this page is a joke (note that the parity of the multiplicity of that is unchanged). The page is not completely wrong though. It states, for example, that if 1=2 then π is rational, which is absolutely true. -- Meni Rosenfeld (talk) 23:35, 20 November 2007 (UTC)
Is it now? Au contraire, I have a beautiful proof that if 1=2 then π is trancendental. Unfortunately it is too large to fit in this talk page. It is also a well-known fact in absolute piology (which investigates those properties of π that are independent of whether 1=2) that π cannot at the same time be rational and trancendental. If Meni's claim were true, we would be able to conclude (by contradiction) that 1≠2. –Henning Makholm 00:20, 21 November 2007 (UTC)
Since 1≠2, then any statement of the form "if 1=2 then X" is true (independent of X). At least in terms of formal logic. See also Logical implication and in particular the following part: "For example, any material conditional statement with a false antecedent is true. So the statement '2 is odd implies 2 is even' is true." But although true, the statement is also meaningless, because 1≠2. -- Why Not A Duck 01:13, 21 November 2007 (UTC)
I think Henning's comment was meant to be a joke (mine, although true, was also intended to have a humorous effect). -- Meni Rosenfeld (talk) 09:25, 21 November 2007 (UTC)

## A CORRECT VALUE FOR Pi - IS OF SIGNIFICANT IMPORTANCE

When a circle's diameter is 1, its circumference is π. This statement is not irrefutably correct. Consider if the correct statement read this way.

When a circle's diameter is 1, its circumference is Gamma. Plain language: In length - Circumference is 3:1 with its' Diameter

When a circle's radius is 1, its circumference is Stigma. Plain language: In length - Circumference is 6:1 with its' Radius

When a circle's radius is 1, Degrees of Arc is Xi. Plain language: In length - Radius equals 60 Degrees of Arc length

  or   Radius length is Ratio 1:1 with Arc length at 60 Degrees.
or   1 degree of Arc length is Ratio 1:60 with the Radius length.
or   1/2 degree of Arc length is 1:120 with the Radius length.

There,I've let the Cat out of the bag. All that is left to prove is that Gamma, which gives the Ratio 1:3 is correct. It is not difficult to prove. Note that 3:1 is 21/7 & Pi is 22/7 (When questioned about how to get to a destination the reply came "You can't git there from here, with that")

The correct value is important for several reasons: Even a child will understand Pi when the value is correct. Determining the "Length of a Circumference" becomes simple math. Determining the "Area of a Circle" becomes simple math. Determiming a length for fractions of "Degree of Arc" simple math. Perceptions relating to geometric shapes will perhaps be altered. There will be a Domino effect within the Fields of Science & Math.

Yes, it is scary. However, anyone that believes that the correct value is of insignificant importance, does not possess the correct value or the perception to appreciate what the correct value means.

Several months ago I entered ratios here that were simply garbage. Undefendable. The headline was "Misperception = 2500 years of Misconception" I have come back to say that I stand by that headline.

The value for Pi was not a finished work for Archimedes and it is not a finished work yet. The correct Value is there if one searches hard enough, long enough, with a particular perception of Geometry. Decide for yourself if gamma should substitute for Pi, to get to the desired destination. I can offer an explanation of how I arrived at Gamma.

My apology to Meni. Please enter your comment again.

You have written this apology in the same edit as removing my comment , which seems to indicate that you have done it deliberately. My assumption of good faith is being tested. -- Meni Rosenfeld (talk) 17:34, 25 November 2007 (UTC)
He said that he'd point out an error in your proof that the ratio between a radius and a circle's circumference is 6 -- as long as it isn't too long -- and I will do the same. What is your proof? Gscshoyru (talk) 17:31, 25 November 2007 (UTC)

Meni: no evil intent, an honest error, I am a newbie with wiki editing Gschoyru: please read my item carefully. I do not claim 6. I claim 3 (gamma) —Preceding unsigned comment added by 58.186.152.192 (talk) 17:57, 25 November 2007 (UTC)

In that case, here are a few tricks which are good to keep in mind. The first doesn't have much to do with wiki editing or Wikipedia per se - if you're making an edit and something goes wrong (some text gets deleted), instead of saving the edit, copy any text you have added (by selecting it and pressing Ctrl+C), and then refresh your browser (usually F5). The original text will be restored, and you can paste your text and make any necessary adjustments.
The second is - by clicking on the "history" tab on the top, you can see the version history of the page, from which you can access older versions of the page. If for some reason something does get deleted, you can look it up there and add it to the page. This is exactly what I will do now:
The correct value of π has been known for centuries. It is π.
One may be interested in some features of this value - for example, some entries of it continued fraction expansion, or some digits of its decimal expansion. Those are readily available to whoever has the will and the way to compute them. -- Meni Rosenfeld (talk) 15:02, 25 November 2007 (UTC)
I will be more than happy to spot the mistake in your alleged proof that the ratio between the circumference of a circle and its diameter, in Euclidean geometry, is 6 - as long as it is not too long. -- Meni Rosenfeld (talk) 15:28, 25 November 2007 (UTC)
Back to the topic at hand, you did claim 6 before, and our offer extends to 3 as well. -- Meni Rosenfeld (talk) 18:16, 25 November 2007 (UTC)
First, thanks for the editing tip. Second, Consider that the value 6 was edited out, and a clear statement was made that 6 was not claimed, but 3 was. The original value of 6 related to Radius and was an error in composition of the entry, from notes. Hence the edit. I stand by my clear statement, I do not claim 6, I claim 3.
If Diameter is 1 then Circumference is Gamma (Not Pi)

To my alleged proof. I can offer two methods. I will offer the one that should, a priori, prove itself, if my claim is correct. All are familiar with the animated wheel on the Pi main page. The distance (length) between the yellow and red arrow on the horizontal line, is nothing more than the wheel's "Circumference length" transformed to a straight line. Accurately mark the line between the two arrows (Circumerence length), in units of your choice, and the Diameter of the wheel, in the same units. The number of Units between the yellow and red arrow will be 3 times, the number of Units of the Diameter, and 6 times, half the number of the Units of the Diameter, if my claim is correct.
Is there any comment that the Area of the Circle is wrong, when using the stated Method to determine it.

--Layman1 (talk) 23:56, 25 November 2007 (UTC) OOPS - an edit for signature --Layman1 (talk) 23:56, 25 November 2007 (UTC)—Preceding unsigned comment added by Layman1 (talkcontribs) 23:48, 25 November 2007 (UTC)

So... I'm confused. Your proof is "measure the distance?" That's not a proof, nor will it turn out to be correct. Gscshoyru (talk) 04:10, 26 November 2007 (UTC)

When a circle's diameter is 1, its circumference is π. That statement,is a statement of Ratio, of 2 lengths.

Diameter length is a straight line, easy to measure. Circumference length is not a straight line and is difficult to measure. If we transform a Circumference length to a straight line it is easy to measure. Dividing the "Diameter length into the Circumference length", will provides the "Ratio of Circumference length to Diameter length". What the animation on the Pi main page is doing, is transforming a Circumference length, into a straight line, with one precise rotation. The yellow & red arrow accurately represent the Circumference length, as an easy to measure, straight line.

Pi 22/7 is a statement of Ratio lengths. Pi 3.14... is a multiplier to be applied to Diameter length to determine Circumference length. Pi is imprecise for the task it has been applied to. --Layman1 (talk) 16:16, 27 November 2007 (UTC)

So... since the length of the line between those two arrows is 3.14159... then that's what pi is. Or do you have a proof for why that length is somehow wrong? Gscshoyru (talk) 20:29, 27 November 2007 (UTC)

No ... the length of the line between those two two arrows is not 3.14... The length of the line between those two arrows is the true circumference of the object(wheel). You cannot mix apples with oranges. If Pi is imprecise, how can a multiple of it, be precisely equal with the real thing which is precise. Impossible. --Layman1 (talk) 21:23, 27 November 2007 (UTC)

Pi is precise. It has a precise value. Just because we can't express it in digits doesn't mean it's not precise.
The square root of two: 1.4142... is just the same. It's the ratio between the side length of the hypotenuse of a right triangle with 45 degree angles to either leg. It has a precise value. But it's irrational, just as pi is.
You're going to have to do more than simply state that when measuring you get 3. That isn't a proof. Hold on a sec and I'll prove that it must be greater than 3... Gscshoyru (talk) 21:31, 27 November 2007 (UTC)
Ok... first of all, you agree that the length of an arc is greater than the length of a chord connecting the two endpoints of the arc, correct? Well, consider the middle image of this picture:
The one with the hexagon inscribed in the circle. Draw spokes from the center of the circle to the corners of the inscribed hexagon. Note that this makes 6 equilateral triangles. If the radius of the circle is 1, then the length of each spoke is 1, then each of the segments of the hexagon must also be of length 1, and total length 6. And since each segment is a chord of 1/6 of the circle, the length of an arc 1/6th the circumference must be greater than 1, so the circumference must be greater than 6, and therefor pi must be greater than 3.
Anything you don't follow, please tell me and I'll explain better. Gscshoyru (talk) 21:40, 27 November 2007 (UTC)
I would simplify our difference in Perception of the inscribed Hexagon with these statements.

You defend Pi as you perceive that the Chord length is 1:1 with what you refer to as a spoke length.

I defend Gamma as I perceive that the ARC length is 1:1 with what you refer to as a spoke length.

Before proceeding further, would you agree that these 2 statements clearly express our difference and is what we are attempting to resolve? Also, can we agree to substitute the term Radial and Radius for spoke. It's GMT +7 here now, and I am mentally refreshed. --Layman1 (talk) 03:59, 28 November 2007 (UTC)

Yes, that is a good way to express the difference. But this has nothing to do with perception - the second statement is wrong, and the first statement is correct and can be proven easily - the triangle composed of a chord and two radii is isoceles (since the radii are both equal to 1) and has an angle of 60°, and is thus equilateral. The chord is therefore also 1. -- Meni Rosenfeld (talk) 08:39, 28 November 2007 (UTC)
Ok - My thanks to both of you for your patience with me. It is evident to me that I need to spend a lot more time at Dr. Math and leave Wiki to the Professionals. I suppose all of my entries just archive so it is not necessary to delete them. My best regards, --Layman1 (talk) 10:21, 28 November 2007 (UTC)

## SQUARING A CIRCLE IS CHILD'S PLAY

Having determined the length of the Circumference do this.

Where C is Circumference length

Where F is 1/4 of the Circumference length

Where A is the Square Area

C/4 = F

F squared = Square Area of the Circle

I'm sure some can write this neater than I have, but the point is made. You can literally square a circumference length. --58.186.152.192 (talk) 18:10, 25 November 2007 (UTC)

How, with compass and ruler, can you make a straight line the length of a circumference of a circle? Gscshoyru (talk) 04:08, 26 November 2007 (UTC)
You cannot. You Transform a Circumference length, to a Linear length, as described in my other reply to you. That should eliminate your confusion.--Layman1 (talk) 16:23, 27 November 2007 (UTC)
I wonder if I misunderstood your question due to the phrasing.

If you are involving Pi, you can but it will be imprecise. Let me give a clear example, presuming Gamma is substituted for Pi and the Statement is 3:1 Simply Divide the Linear line representing Circumference by 6 and set the Compass to that value. 1/6 of the Linear value is the radius value. --Layman1 (talk) 20:22, 27 November 2007 (UTC)

Layman, one conversation at a time. I'd really like to keep these two seperate, so we don't confuse them... unless you were the anon? Gscshoyru (talk) 20:25, 27 November 2007 (UTC)
Ok. No, I wasn't signed anon anywhere. I jumped in when I realized I had probably misunderstood your question, so my 1st reply would be wrong, if I did. I will correct cannot, to can. I gave the example that I did because it is easier to explain. Let me add that I am GMT + 7 here. --Layman1 (talk) 21:09, 27 November 2007 (UTC)

## Daniel Tammet and Pi

I think it's noteworthy to include a sentence or two about Daniel Tammet, under the Memorizing Digits http://en.wikipedia.org/wiki/Pi#Memorizing_digits

Wikipedia article of him is here http://en.wikipedia.org/wiki/Daniel_Tammet#Pi

124.107.125.165 (talk) 13:19, 11 December 2007 (UTC)

Why? His achievement is nowhere near the world record, and is thus irrelevant here. -- Meni Rosenfeld (talk) 14:27, 11 December 2007 (UTC)


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(x)/(b)(x)/(ab)abxax

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this question the speed V of a particle moving along a straight line is given by a + b square + x square ok ok this is the equation for moment of This particle ok whose speed is gonna be ok and where X is the distance from the origin and acceleration of the particle is ok so we need to find the acceleration here ok access what distance from the origin and this is the equation for motion of the particle speed is we so let's get this equation first initiated a + b square it is equal to x square ok so what is the solution we need to find a solution acylation is change in velocity with respect to change the time this is a solution ok so finding for finding a solution we need to differentiate this whole equation with respect to T ok so let's differentiate is considered as it will be zero ok Plus and V square ok so chain rule to ok 20 concert so right bhi first

to be ok and divide it in this will be the differentiation ok year sem schedule 2x days validity ok now look access distance and distance and about time let me Holla disturb distance about time is what velocity ok so let's write this in next step OK I will be to be here we Divya Pondatti ok it will be equal to 2 X and acid this is well City so we hear ok now cancel the terms it to we cancel to and be ok and as a destination is divided t i am getting Dube it so I can I take ok let se tessellation complete away because I am already so let's get a solution ok acceleration this is equal to this is a rare ok acceleration it will be equals to X upon B

suwaiket value of acceleration that is X upon be so much attention when is getting match with answers option was correct answer thank you

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trivial Picard group

let $S=\operatorname{Spec}(A)$ be an affine scheme. For which ring $A$, not field is it known that $H^1(S,\mathcal{O}_S^{*})$ is trivial?

If $X\to S$ is a finite map and $H^1(S,\mathcal{O}_S^{*})$ is trivial, is it true that also $H^1(X,\mathcal{O}_X^{*})$ is trivial?

Thanks

1 Answer

Sample answers to your first question: If $S$ is Spec of a local ring, or of a UFD, then $H^1(S, \mathcal O_S^{\times})$ is trivial.

The answer to your second question is no: $X =$Spec $\mathbb C[x,y]/(y^2 - x^3 +x) \to S =$ Spec $\mathbb C[x]$ gives a counterexample of a geometric nature, and $X =$Spec $\mathbb Z[\sqrt{-5}] \to S =$ Spec $\mathbb Z$ gives a counterexample of an arithmetic nature.

• is this true for coherent cohomology? Namely if $S$ is affine and $\mathcal{F}$ is coherent on $X$, $X\to S$ finite, is it true that $H^i(X,\mathcal{F})=0$ for $i>0$? – ullo Jul 6 '12 at 14:31
• Yes, because $X$ is affine under your hypotheses! – Georges Elencwajg Jul 6 '12 at 15:31
• @Matt E: can you explain how to compute $H^1(X,\mathcal{O}_X^\times)$ in your geometric example? – adrido Jun 5 '14 at 19:19
• @adrido: Dear adrido, Recall that $H^1(X,\mathcal O_X^{\times})$ is the Picard group of $X$. In the geometric example, the curve $X$ is a once-punctured geometric curve, so its Pic is equal to Pic of an elliptic curve modulo the copy of $\mathbb Z$ generated by a single point (i.e. the point we punctured), which is naturally isomorphic to Pic${}^0$ of the elliptic curve. But this is isomorphic to the elliptic curve itself. In other words, in the geometric example, we have $H^1(X,\mathcal O_X^{\times})$ is isomorphic to an elliptic curve. Regards, – Matt E Jun 6 '14 at 5:55
• @MattE: Dear Matt, that's brilliant, thank you! so the projective curve defined by $wy^2-x^3+wx^2$ is elliptic and once you take out a point (w=0) you get the affine curve above. I wonder if you can generalize this trick to compute the Picard group of affine varieties that are cones over a nice projective variety? e.g. V(xy-z^2)? – adrido Jun 6 '14 at 16:26


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## Encyclopedia > Friedrich Engels

Article Content

# Friedrich Engels

Friedrich Engels (born November 28 1820, Barmen[?], Prussia - died August 5 1895, London) was a Socialist philosopher and co-founder (with Karl Marx) of modern Communist theory. Together, they published The Communist Manifesto in 1848 and Engels edited several volumes of Das Kapital (Capital: A Critique of Political Economy) after Marx's death.

### Biography

Engels was the eldest son of a successful German textile industrialist. As a young man his father sent him to England to help manage his cotton-factory in Manchester. Shocked by the widespread poverty in the city he began writing an account which was published in 1844 as Condition of the Working Classes in England.

In the same year, Engels began contributing to a journal called the Franco-German Annals which was edited and published by Karl Marx in Paris. After their first meeting in person, it turned out that both share the same views on capitalism and hence they decided to work more closely together. After Marx was deported from France in January of 1845, they decided to move to Belgium, which permitted a greater freedom of expression than other countries in Europe.

In July 1845 Engels took Karl Marx to England, where he met several of the Chartist leaders including George Harney[?]. They returned to Brussels in January of 1846 where they set up the Communist Correspondence Committee. The plan was to try and unite socialist leaders living in different parts of Europe. Influenced by Marx's ideas, socialists in England held a conference in London where they formed a new organisation called the Communist League. Engels attended as a delegate and had a great impact on the developed strategy of action.

In 1847 Engels and Marx began writing a pamphlet together. It was based on Engels' The Principles of Communism. The 12,000-word pamphlet was finished in six weeks, written in such a manner as to make communist theory understandable to a wide audience. It was named The Communist Manifesto and was published in February 1848. In March both Engels and Marx were expelled from Belgium. They moved to Cologne where they began to publish a radical newspaper, the New Rhenish Gazette.

By 1849 both were forced to leave the country and moved to London. The Prussian authorities applied pressure on the British government to expel the two men but the Prime Minister, Lord John Russell, held liberal views on freedom of expression and refused. With only the money that Engels could raise, the Marx family lived in extreme poverty.

In order to help supply Karl Marx with an income, Engels returned to work for his father in Germany. After Marx's death in 1883 Engels devoted the rest of his life to editing and translating Marx's writings. He died in London on August 5, 1895.

### Other notable works by Engels:

• Anti-Duhring
• Origins of the Family, Private Property, and the State
• The Peasant War in Germany

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# If $p\circ f$ has a pole at $z_0$ then so does $f$, where $p$ is a polynomial

Let $f : U$ \ ${{z_0}} \to C$ be a holomorphic function, where $U$ is an open disk and let $p$ be a non-constant polynomial. Show that the singularity of $f$ at $z_0$ is a pole of $f$ iff it is a pole of $p \circ f$.

I have proved the 'only if' part of the above question but i am having trouble in proving the 'if' part. Suppose $p\circ f$ has a pole at $z_0$ i.e. There is a holomorphic function $g: U\to C$ with $g(z_0) \neq 0$ such that $$p\circ f = \frac{g(z)}{(z-z_0)^n}$$

So, how should I conclude that f has a pole at $z_0$.

• is $U$ an open disk? The result seems untrue otherwise – user254433 Sep 9 '18 at 9:38
• sorry, yes U is a open disk. – user270331 Sep 9 '18 at 9:41

Hint: $f$ has a pole at $z_0$ if and only if $lim_{z\to z_0}|f(z)|=\infty$ and $f$ has an essential season singularity at $z_0$ if and only if the image of $B(z_0,\delta)\setminus \{z_0\}$ under $f$ is dense in $\mathbb{C}$ for all $\delta$ small enough.


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# Non-linear expanders?

Recall that a family of graphs (indexed by an infinite set, such as the primes, say) is called an expander family if there is a $\delta>0$ such that, on every graph in the family, the discrete Laplacian (or the adjacency matrix) has spectral gap $|\lambda_0-\lambda_1|\geq \delta$. (Assume all graphs in the family have the same degree (= valency) $d$.)

If, for every $p$, we specify some maps from $\mathbb{Z}/p\mathbb{Z}$ to itself, then we are defining a graph with vertex set $\mathbb{Z}/p\mathbb{Z}$ for each $p$: a vertex is adjacent to the vertices it is mapped to.

If we consider the maps $x\to x+1$, $x\to 3x$ (say), then we do not get expanders. On the other hand, if we take $x\to x+1$, $x\to x^{-1}$, then, as is widely known, we do get an expander family.

Consider the maps

$x\mapsto x+1$, $x\mapsto x^3$.

Do they (and their inverses, if you wish) give an expander family? (Let $p$ range only over primes $p\equiv 2 \mod 3$, so as to keep the map $x\mapsto x^3$ injective.)

What about the maps $x\mapsto x+1$, $x\mapsto 3 x$, $x\mapsto x^3$? Do they, taken together, give an expander family?

(Is there a way to relate these maps to the action of a linear group? Are there examples of sets of not neessarily linear maps giving rise to expander families?)

UPDATE: What if these were shown not to be expanders? What would be some interesting consequences?

• As for the last question, as you well know, it was proven long ago by Pinsker that any pair of random permutations will form an expander on $\{1,\dots,n\}$ with probability $1-o(1)$. Absent any obvious nilpotent or solvable structure (as with the $x \mapsto x+1, x \mapsto 3x$ case) I would conjecture that $x \mapsto x+1, x \mapsto x^3$ behaves like two random maps for the purposes of expansion, since there do not appear to be any local relations between these maps (in contrast to the former case in which we have $3(x+1) = 3x + 1 + 1 + 1$). – Terry Tao May 28 '14 at 15:55
• ... proving expansion rigorously, though, would probably require substantial new ideas since there does not appear to be any natural group involved other than the free group. But one could imagine that sum-product theorems could at least give some non-trivial bounds on diameter and mixing time, even if they fall short of true expansion. – Terry Tao May 28 '14 at 15:57
• Agreed - yes, these are my reasons for asking! But what about diameter bounds? I actually want expansion, but that might already be somewhat interesting. – H A Helfgott May 28 '14 at 16:59
• It looks like Bourgain-Konyagin type exponential sum estimates may be able to prove a very weak expansion result of the form $|A \cup (A+1) \cup A^{\cdot 3} | \geq |A| + |A|^\delta$ when $p^\varepsilon \leq |A| \leq p^{1-\varepsilon}$, where $A^{\cdot 3} = \{ a^3: a \in A \}$, which would almost give a non-trivial diameter bound, but the improvement is slight and one also would have to treat the case of very small or very large A somehow. [Maybe a Weyl differencing argument would work?] – Terry Tao May 28 '14 at 18:15
• Actually, thinking about it more, Bourgain-Konyagin (which has to do with how arithmetic progressions and geometric progressions interact) is more relevant for the $x \mapsto x+1, x \mapsto 3x$ case. For $x \mapsto x+1, x \mapsto x^3$, it is indeed Weyl differencing which is the key: to show that if A is basically the union of a small number of intervals, then $A^{\cdot 3}$ does not behave like the union of a small number of intervals. – Terry Tao May 28 '14 at 18:21


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# Newtonian force as a covariant or contravariant quantity

Emeritus
PF Gold
P: 5,597
I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke. This is apparently a bit of a cult classic among mathematicians, not to be confused with Div, Grad, Curl, and All That. Burke was killed in a car accident before he could put the book in final, publishable form, but it can be found in various places online, e.g., on scribd.com. At the beginning of section 12, in the context of Newtonian mechanics, he says,

 We now come to the first surprise. Force is not a vector, but a 1-form. The most direct way to see this is to think of the work done by a force. Force is the operator that takes in a displacement, a vector, and tells you how much work was done. This makes forces dual to vectors, i.e., 1-forms.
In the language of Einstein-style index gymnastics, applied in a nonrelativistic context, this amounts to a statement that energy is a scalar, and displacement is a contravariant (upper-index) vector, so force should naturally be considered as a covariant (lower-index) vector.

The first thing I'm unsure about here is whether energy is really a scalar in the context of nonrelativistic mechanics, if "scalar" is taken to have its full Einstein-style interpretation of "invariant under any change of coordinates." Doesn't nonrelativistic energy change when you rescale your coordinates, implying that it's a scalar density rather than a true scalar? (After all, in relativity, rescaling coordinates changes all the components of the energy-momentum tensor, which means it changes the mass-energy.)

The second thing that bugs me is that if you were to reason from Newton's second law, it seems like you would "naturally" conclude the opposite, that force is a contravariant vector.

If we follow the usual but arbitrary convention of saying that upper indices are used for distances measured on a ruler, then this breaks the otherwise perfect symmetry between vectors and their duals. It then seems clear that things like velocity, which can be obtained by differentiation with respect to a scalar, should also take upper indices (be contravariant). But I'm less convinced that this then breaks the duality symmetry in the case of Newtonian force, or a relativistic quantity like the stress-energy tensor...?
 C. Spirit Sci Advisor Thanks P: 5,616 I think here when we talk about $W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i}$ being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with $\mathbb{R}^{n}$, the dual vectors and vectors are naturally identified.
Emeritus
PF Gold
P: 5,597
A similar but slightly different example...

This WP article http://en.wikipedia.org/wiki/Covaria...nce_of_vectors says,

 Examples of vectors with contravariant components include the position of an object relative to an observer, or any derivative of position with respect to time, including velocity, acceleration, and jerk.
But if I follow Burke's philosophy that work is obviously a scalar, then power should be a scalar as well. If there is only one force acting on a particle of mass m, then the power is given by $P=m a\cdot v$. This implies that a and v can't both naturally be contravariant as claimed by WP.

 C. Spirit Sci Advisor Thanks P: 5,616 Newtonian force as a covariant or contravariant quantity Yes if you write it in that form i.e. $P = ma_{i}v^{i}$ then the acceleration shows up as a co - vector but if we are talking about particles moving in $\mathbb{R}^{3}$, the dual space of $\mathbb{R}^{3}$ IS $\mathbb{R}^{3}$ so it isn't wrong to also say it is a vector if we are just talking about euclidean space with the canonical basis.
Emeritus
PF Gold
P: 5,597
 Quote by WannabeNewton I think here when we talk about $W({x}) = \int_{{x_{0}}}^{{x}}F_{i}dx^{i}$ being a scalar, it just means it is a real valued function defined on the image of the curve that the particle is traveling on and not necessarily a coordinate invariant quantity. Note, regarding your second point if I understood the question correctly, that when we are dealing with $\mathbb{R}^{n}$, the dual vectors and vectors are naturally identified.
What you're saying seems sensible, but it would seem to completely contradict the statements by Burke and in the WP article to the effect that this quantity or that quantity "is" covariant or contravariant. That is, they seem to be claiming that the natural identification provided by duality is not so natural.

As a side issue, I'm not sure what you mean by "when we are dealing with $\mathbb{R}^{n}$." As far as I can see, we're always dealing with a tangent space on a manifold, which is always isomorphic to $\mathbb{R}^{n}$. Whenever you have a vector space, you can make duals of vectors.
 Sci Advisor HW Helper PF Gold P: 4,130 I think power is $$F_a v^a$$. I think that to write Newton's Second Law with acceleration, one needs a metric: $$F_a =mg_{ab}a^{b}$$. Using momentum (thought of as a covector), the Second Law is $$F_a= \frac{d}{dt} p_a$$
Emeritus
PF Gold
P: 5,597
 Quote by WannabeNewton the dual space of $\mathbb{R}^{3}$ IS $\mathbb{R}^{3}$
I don't think this is true. They're isomorphic, but not the same. An element of $\mathbb{R}^3$ is a 3-tuple of real numbers. An element of its dual space is a linear function. Physically, elements of the two spaces have different transformation properties.
Emeritus
PF Gold
P: 5,597
 Quote by robphy I think power is $$F_a v^a$$. I think that to write Newton's Second Law with acceleration, one needs a metric: $$F_a =mg_{ab}a^{b}$$.
Sure. If we assert complete symmetry between vectors and their duals, then the whole issue becomes a non-issue. But both Burke and the WP article seem to be asserting that the symmetry is not complete, and that there is some mode of reasoning that tells you which form is preferred, covariant or contravariant, for a given physical quantity.
C. Spirit
Thanks
P: 5,616
 Quote by bcrowell I don't think this is true. They're isomorphic, but not the same. An element of $\mathbb{R}^3$ is a 3-tuple of real numbers. An element of its dual space is a linear function.
Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?
Thanks
P: 4,160
 I recently came across a very cool book called Div, Grad, and Curl are Dead by Burke.
Bill was a remarkable person, and one of my best friends. His biography is here.
Mentor
P: 18,278
 Quote by WannabeNewton Hi ben yes strictly they are not equal but their vector space structures are identical so for all practical purposes they are the same. For vector spaces over the reals, of with which we are concerned with, we can readily go from a vector to a corresponding dual vector via the Riesz representation so why should there be some a priori preference for saying say force must be co - vector as opposed to a vector: is that what is being asked?
If that is the question, then the answer is easy: because you want to define line integrals.

The entire reason for defining 1-forms is to be able to take line integrals. A line integral should only ever be defined on 1-forms (or differential forms). Taking line integrals of vector fields is done in practice, but is not a good standard.

So there is a reasonable process to decide whether something should be a 1-form or a vector. The question you should ask: do you intend to take line integrals of your object? If the answer is yes, then you are dealing with a 1-form.

Since taking line integrals of force fields is something that is very common, this is an explanation of why force should be a 1-form and not a vector.
 Sci Advisor PF Gold P: 5,059 Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.
C. Spirit
Thanks
P: 5,616
 Quote by PAllen Just to add to this, Bergmann, in his 1942 book on SR+GR, treated force as a covariant vector, and velocity and 4-acceleration as contravariant vectors. 4-momentum was treated inconsistently, sometimes contravariant, sometimes covariant.
Just to further add on to this PAllen if you don't mind =D, Soper in his text on classical field theory also defines force as a one form and you can see it in the tensorial nature of the euler lagrange equations which are one - form equations.
Emeritus
PF Gold
P: 5,597
 Quote by Bill_K Bill was a remarkable person, and one of my best friends. His biography is here.
I really like the book, and it's a shame that it seems to have been condemned to this sort of dusty junk-shop existence on the internet. Do you have any idea whether his heirs (Violet John?) would be interested in seeing it spiffed up and distributed for free under a creative commons license? I'd been thinking of making inquiries, but didn't know who to contact initially, was thinking maybe his colleagues at UCSC.
 Sci Advisor HW Helper PF Gold P: 4,130 This point of view about force as a 1-form appears in Burke's Applied Differential Geometry and in Spacetime, Geometry, Cosmology. A related article is a J.Math Phys article on Twisted Differential Forms for Electromagnetism ( http://dx.doi.org/10.1063/1.525603 ) see also http://www.ucolick.org/~burke/home.html Earlier sources include Schouten's Tensor Analysis for Physicists. Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint. ...all of this talk is nudging me to get back to a project of mine to visualize tensors and differential forms.
C. Spirit
Thanks
P: 5,616
 Quote by robphy Momentum as a covector seems natural from a Hamiltonian Mechanics viewpoint.
Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.
HW Helper
PF Gold
P: 4,130
 Quote by WannabeNewton Here I think it is more strictly required than seen as one possible viewpoint in order for the phase space to be the cotangent bundle to the configuration space manifold. Unlike this case that you mentioned, in GR I find it hard to keep track of quantities that are naturally taken to be one - forms or vectors in the context of index based computations because one freely uses the metric tensor to go form one to the other.
In typical presentations of GR or other subjects, there is little attention paid
to what fundamental structures (e.g. metric, volume-form, connection, etc...)
are being used in a construction or calculation.

So, we have this blurring of whether something is naturally a vector, covector (one-form), or maybe even a bivector or two-form. This is especially true when one deals with a space (e.g. Minkowski space or Euclidean space) with lots of symmetries.

The references by Burke and Schouten (and others) try to tease out what could be the fundamental structures... For example, there is work on metric-independent formulations of electromagnetism... where, e.g., the magnetic field is fundamentally a two-form. Then, only with additional structure, can it be thought of as an axial-vector [in three spatial dimensions].

I actually like the [abstract-]index computations because they help me keep track of
what kinds of objects I am dealing with. Ideally, it would be nice to a have a geometrical picture
to go along with the computations to suggest what is physically going on.
 Quote by robphy Using momentum (thought of as a covector), the Second Law is $$F_a= \frac{d}{dt} p_a$$


text:
## LUXE: A new experiment to study non-perturbative QED in $e^{-}$-laser and $γ$-laser collisions

Ruth Jacobs

SciPost Phys. Proc. 8, 088 (2022) · published 13 July 2022

DIS2021

### Abstract

The LUXE experiment (Laser Und XFEL Experiment) is a new experiment in planning at DESY Hamburg using the electron beam of the European XFEL. LUXE is intended to study collisions between a high-intensity optical laser and up to 16.5 GeV electrons from the Eu.XFEL electron beam, or, alternatively, high-energy secondary photons. The physics objective of LUXE are processes of Quantum Electrodynamics (QED) at the strong-field frontier, where QED is non-perturbative. This manifests itself in the creation of physical electron-positron pairs from the QED vacuum. LUXE intends to measure the positron production rate in a new physics regime at an unprecedented laser intensity. Parasitically, the high-intensity Compton photon beam of LUXE can be used to search for physics beyond the Standard Model.

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