Talk:Area of a disk

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Mathematics rating:	A Class	Low Priority	Field: Geometry
Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Promoted to A class (discussion) — Carl (CBM · talk) 02:18, 5 June 2007 (UTC)

I know that im stupid but the formatting isn't very good, so if anyone could help with that, I would appreciate it. --Chuck 11:06, 12 April 2006 (UTC)

Someone must erase this article because circles have not area --kiddo 02:40, 29 October 2006 (UTC)

• sorry I didn't mean to erase exactly

--kiddo 05:23, 2 November 2006 (UTC)

You ever think that wikipedia might go too far in explaining concepts? —Preceding unsigned comment added by 67.166.28.81 (talk) 04:52, 15 November 2007 (UTC)

Contents 1 Humor 2 Archimedes 3 Some issues with the current presentation 4 Numeric computation 4.1 Derivation 5 Tarski's rearrangement theorem 6 Ideal sources sought 7 Archimedes triangle 8 comments 9 send it thru FAC then.. 10 Why then do we do it this way 11 Monte Carlo

[edit] Humor

πr²---Pi R Squared
πr²---Pi R Not Squared
πr^o---Pi R Round
cr²---Cornbread R Squared

[edit] Archimedes

Maybe I'll get around to working on the article itself. Meanwhile, here's a description of the way Archimedes proved that the area of a circle must be exactly the same as the area of a right triangle whose base is the circumference and whose height is the radius. This can be found in T. L. Heath's translation of J. L.Heiberg's Greek version of "Measurement of a circle" in The Works of Archimedes, Dover, 2002 (originally Cambridge University Press, 1897), ISBN 978-0-486-42084-4.

---

The proof begins with the claim that if the area of the circle is not equal to that of the triangle, then it must be either greater or less. It then eliminates each of these by contradiction (using regular polygons), leaving equality as the only possibility. The proof uses what today is called the Axiom of Archimedes.

Suppose the circle area, C, is greater than the triangle area, T = ¹⁄₂cr, by an amount Δ. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those segments, S₄ is greater than Δ, split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total area, S₈. Continue splitting until the total segment area, S_n, is less than Δ. Now the area of the inscribed polygon, P_n = C−S_n, must be greater than that of the triangle.

But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides, ns, is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals ¹⁄₂nhs. But since h < r and ns < c, the polygon area must be less than the triangle area, ¹⁄₂cr, a contradiction. Therefore our supposition that C is greater than T must be wrong.

Suppose the circle area is less than the triangle area by an amount δ. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area trapped between the square and the circle, A₄ is greater than δ, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the trapped area is less than δ. The area of the polygon, Q_n, must be less than T.

This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length r. And since the total side length is greater than the circumference, the polygon consists of n identical triangles with total area greater than T. Again we have a contradiction, so our supposition that C might be less than T must be wrong as well.

Therefore it must be the case that the area of the circle is precisely the same as the area of the triangle.

---

The circumference of a circle with radius r is, of course, 2πr (a separate discussion). Therefore the area of the circle is

Archimedes then goes on to compute lower and upper bounds for π, again using inscribed and circumscribed polygons.

One appeal of this proof is that it uses no calculus, no integrals, no limits. It merely depends on the fact that polygon splitting will eventually force an area difference less than Δ or δ. --KSmrq^T 23:07, 13 December 2006 (UTC)

this proof is of historical significance and should definitely be included in the article. seems to me one can make a pretty good case this is the very beginning of analysis. the idea of a limit is already present in the proof. approximating from inscribed and circumscribed polygons is essentially the same as the Riemann lower and upper sums. and this is how many centuries before Newton, et al.? pretty remarkable. Mct mht 01:25, 14 December 2006 (UTC)

[edit] Some issues with the current presentation

There are several different things to prove about the area of a disk:

1. The area of a disk is some constant times the square of its radius. By convention, we can call that constant π.
2. The constant π defined as the area of a unit circle is equal to some other non-circle-related definition of π.
3. The area of a disk is equal to the half the circumference times the radius. Since we know (by some other proof) that the circumference is 2πr, the formula follows.

I think it would be helpful to state more precisely which is being proved in each proof.

The proof by calculus appears to be attempting to prove the first statement, that the area scales as the square of the radius. But this statement isn't really about circles: the area of a figure scaled by a factor of r is r² times the area of the figure, regardless of whether the figure is a circle or not. A proof of this needs to depend more carefully on how we define area, but is a straightforward consequence of the definition e.g. for Lebesgue measure. For that matter the generalization to ellipses doesn't really depend on ellipses, it's a more general fact about affine transformation of a circle. So I don't really see what the point of all the calculus is here.

I like the general approach of the proof using limits, which I view as being about the relationship between area, perimeter, and radius, but I think it might be more clear with more words and less trig formulas:

Consider a sequence of regular 2^k-gons, inscribed in the circle. Subdivide each polygon into triangles by connecting the vertices to the center of the circle. As k increases, each step reduces the area not covered by triangles to less than half its previous value, so the total area of the triangles converges to the area of the disk. At each step, the total area of triangles is (by the triangle area formula) half the perimeter of the polygon times the triangle height. The perimeter converges to that of the circle, while the height converges to the radius of the circle, so half their product converges to the circle's perimeter times half its radius.

What more is needed than that? —David Eppstein 00:51, 14 December 2006 (UTC)

I don't believe anybody has mentioned the nicest reason I've seen for the formula. If you chop up the disc into pie wedges and reassemble it into a parallelogram (it gets closer to a rectangle the thinner the wedges), you notice that area of the parallelogram is approximately the radius multiplied by half the circumference. This clearly is a better approximation the thinner the wedges get. The nice thing about this method is that you can draw a very nice, understandable picture. --C S (Talk) 01:54, 14 December 2006 (UTC)

Let's take a step back and consider what we want to say and to whom we may be saying it. I see a few obvious audiences:

1. Young students
2. Their teachers
3. Early calculus students
4. Curious adults
5. Fans of π

What do we want to say to them, being mindful of the overlap with our pi article? For the young students and their teachers, we prefer visual methods with few formulas and no trigonometry or calculus. For the calculus students we need an explicit integration. Perhaps nothing additional is needed for adults. For the fans, we could mention a mind-bender (and our article):

• Laczkovich, Miklós (1990). "Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem". Journal für die reine und angewandte Mathematik (Crelle’s Journal) 404: 77–117. ISSN 0075-4102. 

As the Archimedes proof makes clear, we need no trigonometry and no limits to connect the area to the circumference. A modern approach to the connection is found in Serge Lang's little book, Math! Encounters With High School Students, Springer, 1985, ISBN 978-0-387-96129-3. He does use limits, but in a simple way that requires no explicit calculus.

A while back, in connection with pi, I created an animation using SVG that depicts a rearrangement method attributed to Leonardo da Vinci. Although it is really much the same mathematics as the inscribed polygon limit, visually it is different. Perhaps I'll polish its rough edges and make it available. --KSmrq^T 16:43, 14 December 2006 (UTC)

Re "For the calculus students we need an explicit integration" — I agree it's helpful for calculus students to be shown the relation between area and integration, but what does the explicit integration actually prove? And how can we tell that it isn't just circular reasoning, unless we state more clearly what we're proving and how those π's got into the integration formulas we're using? I mean, going through the actual integration proof again, I can see that the π comes into it as part of a change of variable from Cartesian to polar, so the π in the formula is the π of the 2πr circumference formula rather than just some constant of convenience that we're calling π, but I think the formula-heavy presentation of the proof obscures that fact. I do think your visual approach could help with the limit based proof. —David Eppstein 18:44, 14 December 2006 (UTC)

The other potential problem with the integration proof is that it uses the antiderivative formula for cosine. Working with trig functions just invites circular definitions, so a lot of care must be taken. CMummert 03:49, 15 December 2006 (UTC)

Nice pun. Nothing about the definition of sine or cosine or the antiderivation formula under discussion requires the formula for the area of a disk. --C S (Talk) 04:38, 15 December 2006 (UTC)

(ARGH! A web site crashed my browser, as I was almost finished composing a large post. I shall try to recreate what I can.)

I agree with David that the appearance of π should be explained. Ultimately, it comes from the connection shown by Archimedes. Previously it was known that the ratio of circumference to diameter was a constant independent of the size of the circle. Archimedes proved that the area involved the same number, and then used area to approximate the number π.

For calculus students, the easiest integration builds in the area–circumference–radius relationship, using the "onion" approach.

Here 2πt is the circle at thickness t. In this approach we have no limits, no trigonometry, no change of variables. This approach also naturally draws us into higher dimensions. (Again Archimedes was there first; he showed that the area of a sphere equals the area of a circumscribed cylinder, (2πr)(2r), which integrates to ⁴₃πr³. He also found ways to show the volume of a sphere without calculus.)

(I like connections for at least four reasons. One is personal; I think that way. A second is pedagogical; research shows we remember connected facts better than isolated ones. A third is promotional; connections draw people into learning more mathematics. And a fourth is philosophical; all of mathematics is connected.)

Two alternative integrals are obvious possibilities, but each brings complications.

The first uses two semicircles.

Usually π makes its appearance during a trigonometric substitution,

via the limits of integration,

The integral becomes

Now we invoke a double-angle trigonometric identity,

to produce

then split and change variables to get a sum of known integrals.

The first integral is a full period of a sinusoid, thus vanishes; the second is trivially π. This gives us the desired πr², but the reader is forced to wade through a mess of calculus, algebra, and trigonometry to get there. (And we silently slipped in the trick of extracting r² to work with a unit circle.) We get little insight, and little inspiration.

A second alternative sums radial wedges.

Although formally correct, this appears completely unfamiliar to the beginning calculus student. We must show

and use a half-angle substitution to get a familiar form,

which we can then trivially integrate. The "we must show" part is an exercise in limits.

But if we're going to do that, perhaps we should instead directly compute the limit of the area of a circumscribed regular polygon. This has the side benefit of bringing us back to Archimedes. --KSmrq^T 20:31, 15 December 2006 (UTC)

[edit] Numeric computation

Here's another bit for the article. This uses ideas of Willebrord Snell (Cyclometricus, Lugduni Batavorum: Elzevir, 1621) followed up by Christiaan Huygens (De Circuli Magnitudine Inventa, 1654), described in

• Gerretsen, J.; Verdenduin, P. (1983). "Chapter 8: Polygons and Polyhedra", in H. Behnke, F. Bachmann, K. Fladt, H. Kunle (eds.): Fundamentals of Mathematics, Volume II: Geometry, S. H. Gould (trans.), MIT Press, pp. 243–250. ISBN 0-262-52094-X (ISBN 978-0-262-52094-2). 
  (Originally Grundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1971.)

Given a circle, let u_n be the perimeter length of an inscribed regular n-gon, and let U_n be the perimeter length of a circumscribed regular n-gon. Then we have the following doubling formulae.

Archimedes doubled a hexagon four times to get a 96-gon. For a unit circle, an inscribed hexagon has u₆ = 6, and a circumscribed hexagon has U₆ = 4√3. We have have luxury of decimal notation and our two equations, so we can quickly double seven times:

n	k	uk	Uk	(uk+Uk)/4	k=6×2n
0	6	6.0000000	6.9282032	3.2320508
1	12	6.2116571	6.4307806	3.1606094
2	24	6.2652572	6.3193199	3.1461443
3	48	6.2787004	6.2921724	3.1427182
4	96	6.2820639	6.2854292	3.1418733
5	192	6.2829049	6.2837461	3.1416628
6	384	6.2831152	6.2833255	3.1416102
7	768	6.2831678	6.2832204	3.1415970

A best rational approximation to the last average is ³⁵⁵⁄₁₁₃, which is an excellent value for π. But Snell proposes (and Huygens proves) a tighter bound than Archimedes.

Thus we could get the same approximation, with decimal value 3.14159292…, from a 48-gon. --KSmrq^T 22:32, 16 December 2006 (UTC)

[edit] Derivation

Let one side of an inscribed regular n-gon have length s_n and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem, this is a right triangle with right angle at B. Let the length of A′B be c_n, which we call the complement of s_n; thus c_n²+s_n² = (2r)². Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is s_2n, the length of C′A is c_2n, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is similar to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us

In the first equation C′P is C′O+OP, length r+¹⁄₂c_n, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,

If we now circumscribe a regular n-gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side S_{n; then this is Sn : sn = 1 : ¹⁄2cn. (We have again used that OP is half the length of A′B.) Thus we obtain}

Call the inscribed perimeter u_n = ns_n, and the circumscribed perimenter U_n = nS_n. Then combining equations, we have

so that

This gives a geometric mean equation. We can also deduce

or

This gives a harmonic mean equation.

Illustration to come. --KSmrq^T 07:32, 20 December 2006 (UTC)

I gladly see how this page had evoluzioned (and revoluzionated some thinkings) to see it for a perfect nomination to an excelent page of the moment :) kid--148.202.11.52 16:32, 12 March 2007 (UTC)

[edit] Tarski's rearrangement theorem

If the proof uses the axiom of choice to construct the pieces, then I would say it does tell "how" to construct them, although the construction won't be completely explicit because of the choice required. The current wording suggests a proof by contradiction in which the nonexistence of a partition is shown to be impossible but no actual partition is constructed. CMummert · talk 11:59, 29 May 2007 (UTC)

I added explicit language about the axiom of choice to clarify. Perhaps it would help if you skimmed the two relevant references (one of which is the paper itself). Your wording conveys the wrong meaning to me, and mine to you, eh?

The original estimate was that 10⁵⁰ pieces would be required, so that alone would make a figure difficult. :-D

But setting that frivolous obstacle aside, wording poses a dilemma. At the level this article is otherwise written, the axiom of choice certainly does not say how. Nor is this a non-constructive proof in the sense that a constructive proof might be possible, as your wording would suggest. The fact that we must use the axiom of choice means we throw up our hands and say we can so choose, and we can never say how. --KSmrq^T 12:23, 29 May 2007 (UTC)

I scanned through the paper this morning. The proof requires the existence of two vectors independent over the rationals with some special property, and proceeds by showing that almost every pair of vectors will do. This isn't the axiom of choice per se. The only possible place I see that AC might be used is in the measure-theoretic facts used throughout the paper. So I rephrased the last sentence to remove the claim that AC is crucial to the proof. CMummert · talk 14:16, 1 June 2007 (UTC)

Without AC it is legitimate to assume all sets are measurable, so the fact that the construction involves non-measurable sets implies that some form of choice is crucial, doesn't it? Geometry guy 22:00, 6 June 2007 (UTC)

[edit] Ideal sources sought

If anyone would like to help, I thought it might be nice to find two more sources to cite. What I'm looking for is:

1. A calculus text or MAA article or some such taking the "onion" approach. Lang is great in some ways, but not ideal for this. The typical calculus text integrates the area under (1−x²)^1/2, which is not what we want. One likely context would be a discussion of multidimensional sphere areas and volumes, for by itself it is out of pedagogical sequence for a text and too trivial for an article.
2. A Monte Carlo article or text that throws darts and that simply explains the variance issue. Buffon's needle is a popular example, but it's about computing π rather than the area of a circle.

In both cases, a scholarly source that is freely available on-line would be extra nice. --KSmrq^T 12:54, 29 May 2007 (UTC)

uh... I am about to go to bed, so I probably won't respond for a while, but check out my list of references (link on user page) and E-mail me if you think any of them would be helpful. Cheers --Cronholm144 13:14, 29 May 2007 (UTC)

[edit] Archimedes triangle

I have removed the image of circle and triangle because it is badly composed. This I know because I've long had a better image (in fact an elaborate SVG animation), but thought it not a benefit to the article, only a distraction. --KSmrq^T 00:59, 2 June 2007 (UTC)

[edit] comments

• Wow, a lot of hard work has been poured into this article. It certainly has no problems with breadth of coverage...
• Does it need a history section? Many minds have apparently struggled with the question...
• "Other approaches are also of interest." Of interest to whom?
• (Archimedes 2002)? Egads. This is not an individual work that has been translated; it is a compilation and translation and editing etc. AFAIK, in cases such as this you're licensed to list Heath as the author e.g. (Heath 2002). Another method might be "... writing in the third century B.C., Archimedes showed that blah blah..." with a footnote which leads to a Notes section. In the Notes you can say "For a modern English translation see Heath (2002)". Finally, in the References section you would have Heath, T. (2002). The Works of Archimedes. New York: Dover Publications.
• "...great mathematician Archimedes". I know he was great. You know he was great. Everyone beyond say 5th grade or so knows he was great. But the language is unencyclopedic. Some would say this is POV. Maybe it is; that's arguable. But I think a stronger case could be made that it is unencyclopedic. Drag out your Encylopedia Brittanica and find an article which mentions Archimedes (not the actual Archimedes article). See whether it has such adjectives...
• What does it mean to use regular polygons "in an essential way"?
• I can come back and look again later; busy today... I'm supposed to be studying for prelims... But I made some edits to the article that show the gist of what I might do... spell everything out clearly ("the first argument that leads to a contradiction...") .. eliminate the word "we" (who is we?) etc. I'll look again another time... Ling.Nut 22:15, 5 June 2007 (UTC)

Point by point:

---

• A brief comment brought me here awhile back, I started editing, and the rest is history. It was fun as well as work. But thanks.
• No, it does not need a history section, because it in many ways it already is a history, and to go further would essentially duplicate the history of numerical approximations of π.
• The wording "are of interest" is not ideal, but another editor wanted to point out that the article went beyond Archimedes.
• Yes, I know 2002 is an abomination. The problem is that the auto-linking of the Harvard citation mechanism has limitations. Ideally I would refer to Heath 2002, but Heath is a translator, not an editor. So the formal citation has Archimedes as the author and 2002 as the date of publication, and that is what the reference must use. We do not have a single definitive source by Archimedes himself; even the recent work with the palimpsest — which itself is a transcription — is limited. And, please, no #%#&*^!! footnotes, which are a worse abomination that should be banned from the Web.
• The "great" serves a purpose, because Archimedes' proof did something radically new, using methods that would not be improved for centuries. There is zero controversy about his greatness, so I have little sympathy for knee-jerk claims of POV should they arise. In fact, scholarly works and histories regularly state that Archimedes is one of the greatest mathematicians of all time. The Encyclopædia Britannica] article begins "… the most famous mathematician and inventor of ancient Greece." Many lay readers will be more familiar with Napoleon than with Archimedes, so I don't share your view that almost everyone knows he was great.
• If you do not understand how the use of regular polygons is essential to the proof, then you do not understand the proof. Sorry.
• I removed every edit you made to the article, because each one (!) made the article worse.
  1. Although a link to pi is a good idea, we have had repeated discussions elsewhere about linking mathematical symbols, and always conclude that a wikilink on the symbol itself should not be used.
  2. You changed active voice to passive voice, changing two simple declarative sentences to a tortured combination.
  3. The section headings make clear what you felt the need to state. If the reader is unable to grasp that, the whole proof is wasted on them. We don't need the clutter.

---

Unfortunately, I've seen many such examples of well-meaning editors who make things worse. Honestly, I appreciate the thought, just not its embodiment. It troubles me that so many writers — especially those with academic and technical backgrounds — have gaping blind spots in their sensitivity. For example, numerous readability studies show that active voice is easier to read, and our mathematics manual of style promotes it.

I'll find a way to add a link for pi, and look for a less awkward notice about other approaches. As for the rest, I think they're best left as they are. --KSmrq^T 15:15, 6 June 2007 (UTC)

We all have blind spots KSmrq! ;) Anyway, I agree with you that the active voice is much more readable, and that a history section is unnecessary. Still, I thought Ling.Nut's comments were more helpful than your response suggests. In particular, "Archimedes 2002" would surely be better replaced by a reference to "Heath 1897", or possibly "Heath 1897, 2002", if that is compatible with auto-linking: I agree with Ling.Nut that this work is more than just a translation.

As you point out yourself, the sentence on regular polygons being "essential" adds nothing, either for the reader who understands the proof, or the reader who doesn't. Concerning the overall style, I know it can be quite a challenge to write engaging prose without using the first person plural, but I think it is worth the effort. Maybe I will give it a go, but I have rather a lot of blind spots of my own and this is already a very polished piece of work... Geometry guy 16:53, 6 June 2007 (UTC)

Much as I appreciate the {{citation}} template, it is still a work in progress, and the reference is as it must be for now. The true dates are a little awkward anyway, because we don't have Archimedes originals, only later transcriptions (as in the palimpsest); then we have especially Heiberg's work; and we have Heath; but the print version (cited) is a Dover Publications reprint of the original Cambridge University Press edition (linked). (Whew!) The way I would like to say it is "(Heath 2002)". Ah well, we do the best we can with what we have.

I certainly did not assert that the "essential" comment adds nothing, since that is the opposite of my view! It orients the reader to the nature of the method, which I strongly feel is helpful.

The use of "we" makes some editors uncomfortable. Apparently WP:MSM says to avoid it, which is dead wrong, and is contradicted by advice elsewhere. Specifically, the main Manual of Style says:

Nevertheless, it is sometimes appropriate to use we or one when referring to an experience that anyone, any reader, would be expected to have, such as general perceptual experiences. For example, although it might be best to write, “When most people open their eyes, they see something”, it is still legitimate to write, “When we open our eyes, we see something”, and it is certainly better than using the passive voice: “When the eyes are opened, something is seen.”

It is also acceptable to use we in mathematical derivations; for example: “To normalize the wavefunction, we need to find the value of the arbitrary constant A.”

Almost every effort to avoid "we" that I have seen is a disaster. It can be misused, but I try to be careful how I use "we". It is not meant to be personal, as the use of "you" or "I" must be. Instead, it refers to "we mathematicians" or "we humans" or the like. I get really annoyed at overuse of the epithet "unencyclopedic". When I was a youngster I spent long hours curled up with an encyclopedia, so I trust my judgment. Why do so many editors think precise writing must also be stilted? I've spent a lifetime fighting this battle.

Ah well, at least we still have pretty pictures to look at. Remember when geometry was about things we could draw? This article brings back some of the nostalgia. :-) --KSmrq^T 21:05, 6 June 2007 (UTC)

[edit] send it thru FAC then..

Eh, sorry, but "between you and me there is a great gulf affixed." I see the article as seriously in need of help. To me it reads exactly like a Mathematics textbook... and one of the older/moldier/dustier/dead-er ones, at that. :-) KSmrq sees it as needing only trivial tweaks; G-guy said it is "highly polished." ... No. Not even close. No no no.

The chief problem is that the article makes no attempt whatsoever to consider its audience. Witness:

• "If the reader is unable to grasp that, the whole proof is wasted on them. We don't need the clutter."
• "If you do not understand how the use of regular polygons is essential to the proof, then you do not understand the proof."

That attitude is clearly reflected in the text of the article itself! I dunno if you can see that, but I can... The article is written for the author(s), not the reader(s).

Finally, I dunno if you noticed, but one of my degrees is in *teaching English* ...the perception regarding the evils of the passive voice is also demonstrably incorrect. Umm.. I could try to track down the literature... genre analysis of academic literature (particularly *hard sciences* literature!) has shown that the passive voice has a wide range of important functions and uses in academic text.. I'm not saying my edits were perfect; they weren't. It was a quick run-through. I *am* saying that they were a step in the correct direction... that is.. a reader-centered rather than writer-centered one. Sorry if you don't see that. :-)--Ling.Nut 00:08, 7 June 2007 (UTC)

Indeed, our views seem to differ. Perhaps teaching mathematics is more demanding than teaching English; it won't work with a passive audience. As for passive voice, certainly it has its uses; and I have been amused and enlightened by the rogue perspective of Richard Lanham's Style: An Anti-Textbook (ISBN 978-0-300-01720-5). But I dare say I've read (and reviewed) far more technical articles than you, and had my fill of excess passivity. I believe Strunk & White said "Use the active voice" for good reason.

I do appreciate your generous impulse to improve the article. Perhaps we can meet again and collaborate on something less mathematical, where our impulses are more closely aligned. --KSmrq^T 16:02, 7 June 2007 (UTC)

[edit] Why then do we do it this way

Why do they teach us in school that the area of a circle is Pi times the squared radius, when this can't give an exact answer but Area=.5 radius times diameter gives us an exact answer? 79.180.238.41 14:42, 17 October 2007 (UTC)

Questions like this belong on the Reference Desk; this page is about editing the article. Besides, the question is based on two false premises. (1) The area is exactly π times the squared radius. (2) The article clearly states that the area is also half the radius times the circumference (not the diameter). --KSmrq^T 05:21, 15 November 2007 (UTC)

[edit] Monte Carlo

The illustration to accompany the Monte Carlo estimation (darts) technique is confusing, as nowhere is it stated that the square has a sidelength of two! Here I was thinking it was a sidelength of unity... —DIV (128.250.80.15 (talk) 08:04, 14 January 2008 (UTC))

I have stated the length. –Pomte 08:13, 14 January 2008 (UTC)