January 18

Why are Asians so good at math? --70.129.187.105 (talk) 01:03, 18 January 2010 (UTC)[reply]

Who says they are? --Tango (talk) 01:14, 18 January 2010 (UTC)[reply]

The vast majority of students at national math programs like MathCamp and AwesomeMath are Asian. Also, China almost always wins the IMO. --75.33.218.22 (talk) 02:45, 18 January 2010 (UTC)[reply]

Asians have a record of doing tremendously well in international mathematics Olympiads, but such says little about how an individual will approach "real mathematics". Olympiads are designed to evoke mathematical interest in high school students, and thus typically they test "problem solving", but with simple mathematical content (thus, the problems are readily understood by an average high school student, but not necessarily easily solved). However, it is quite possible to do very well in Olympiads, but struggle while doing higher level mathematics, and it is also quite possible to be an excellent mathematician, but struggle to solve Olympiads problems (in fact, few contestants in the IMO actually pursue a mathematical career). I would also note that much of the Olympiad subsumes "finite mathematics"; for instance, basic combinatorics, number theory and Euclidean geometry.

With regards to your question, note mathematicians such as

Please define "Asians". Thanks, hydnjo (talk) 04:16, 18 January 2010 (UTC)[reply]

If you were asking this to the OP, he/she might as well direct you to the article Asian. If you were asking this to me, I was merely using "Asian" in the context of the OP's question, the purpose of which I am unable to comprehend. --PST 05:57, 18 January 2010 (UTC)[reply]

With regards to why the Chinese and people from other Asian countries seem to be so good at certain types of math, like olympiads, it's probably cultural. Chinese culture and Chinese institutions emphasize certain skills and values that many Western countries don't as much, and vice versa. That influences how people develop and what kinda of goals they pursue. Even people who have emigrated are still highly influenced by the values encouraged by their families. Rckrone (talk) 04:24, 18 January 2010 (UTC)[reply]

You might like to look at

Olympiads are not the same as real math. Olympiads typically involves speed, and real math is "sit there and think". Money is tight (talk) 18:56, 18 January 2010 (UTC)[reply]

In fact, I would say that Olympiads have little to do with real mathematics, even in terms of content. Many problems in the IMO really subsume finite mathematics, if anything, but even that "finite mathematics" is a poor excuse for real "finite mathematics". Sure, Olympiads require a particular kind of intelligence, but it cannot be concluded that that intelligence is mathematical intelligence. Typically, Olympiads test one's ability to apply known techniques to solve a problem, whereas real mathematics focuses on inventing new techniques to develop intuition. --PST 02:30, 19 January 2010 (UTC)[reply]

Indians are clearly better at math than Western people:

http://www-history.mcs.st-and.ac.uk/Projects/Pearce/Chapters/Ch9_3.html

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

This is probably due to genetics. Indians have lived in tolerant civilizations continuously for more than 3000 years. In Europe people were still living in caves 2000 years ago. In the Middle Ages, people in Europe were persecuted, there was little room for free thought. Count Iblis (talk) 23:31, 18 January 2010 (UTC)[reply]

I strongly disagree; there is no concrete criteria using which we may compare the mathematical abilities of two different races. Could you please state what exactly you mean by "better at math"? --PST 02:30, 19 January 2010 (UTC)[reply]

I think (hope) this is trolling. Rckrone (talk) 04:35, 19 January 2010 (UTC)[reply]

Well, there are more people from China than people from any other country so, assuming fairly equal access to olympiads, you should expect more Chinese winners. Other differences might be explained by differences in culture, education, nutrition etc. See Race and intelligence for more. Zain Ebrahim (talk) 10:31, 20 January 2010 (UTC)[reply]

It may be a matter of prestige of math and specific educational system. E.g. 20 years ago USSR score on IMOs was better Russia, i suppose. My school was the best in math in my city (it provided IMO medalists literally every year). A half or even more of children there were of mixed Jewish-Russin descent. It cannt be explained in terms of genetics: a half or even more of children who tried but failed to pass through exams to our school were also of the same Ashkenazi descent. Well, may be the genes made them _interested_ in math education. I bet it were their parents but not the genes.--95.84.241.53 (talk) 19:09, 20 January 2010 (UTC)[reply]

An IMO winner is not necessarily "good at math"; there is a wealth of qualities that one must possess, not only ability to "solve problems", to become a mathematician. As I noted, IMO can only comprise subject matter relating to grade 12 mathematics at the most; it does not subsume many branches of mathematics studied at university, for instance. In terms of intelligence, one needs to be well acquianted with the techniques used to solve IMO problems, and often this requires intensive practice just like any other skill. But knowing techniques and applying them to a given situation is not what "real mathematics" is about. --PST 02:32, 21 January 2010 (UTC)[reply]

OP asked about math-education and competitions and IMOs specifically. There was a joke: the three major competitors at IMOs were Russian team (jews taught by jews), American team (chinese taught by jews) and China team (Chinese taught by chinese). I think, one have to be smart enough (yes, i cannt define 'smart') to win and one should love math to participate. Both qualities help one with math. At least there shoud be correlation. May be (maybe), such competitions aren't good as educational tools, the problems lack beauty and sence IMHO. Probably they require essentially different kind of creativity than 'pure math' does. Back to OP's question - I just see how moscow jewish children tend to concentrate around sciences and math-education whether they are good at math or bad. There exist cultural reasons. Possible underlying genetics is obscured by them as it usually happens. BTW, abilities necessary to get it through exams into some prestigeous university are never inentical to those one must possess to be good at anything--95.84.241.53 (talk) 08:14, 21 January 2010 (UTC)[reply]

I guess there exist some papers relevant to subject. Stats, sociology, etc. If nobody can point to one, OP may repost the question in humanities section. Another my guess: american chinese may seem 'better at math' than chinese chinese. Let's consider a child exellent in problems solving. Whether the child will be interested in advancing and whether (s)he will be informed of appropriate educational possibilities and whether (s)he will decide to participate in the aforementioned national math-programs depends on the parents, the teachers, the classmates etc. A talented child belonging to chinese diaspora may have better chances to be noticed and to notice him/herself and to realize value and importance of his/her interest at math than 'common' chinese or american child. A 'common' child have chances to be ignored or to chose conventional ways of getting education and work or just to never consider it cool to be a nerd. BTW excuse my English:)

for shm equation ${\displaystyle P''+k^{2}P=0}$, when ${\displaystyle k=0}$, ${\displaystyle P=c\phi +d}$ is a solution. However if you are solving some sort of problem in polar coordinates, and have the restriction ${\displaystyle P(\phi )=P(\phi +2m\pi )}$ for an integer m, is the solution ${\displaystyle P=2\pi \times floor({\frac {c\phi +d}{2\pi }})}$ a solution? —Preceding unsigned comment added by 129.67.39.49 (talk) 01:03, 18 January 2010 (UTC)[reply]

They are not solutions. The solution has sine waves. See

Dmcq (talk) 08:06, 18 January 2010 (UTC)[reply

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In your differential equation ${\displaystyle P''+k^{2}P=0}$, what exactly are you differentiating with respect to ? If this is simple harmonic motion then ${\displaystyle P''}$ normally denotes ${\displaystyle {\frac {d^{2}P}{dt^{2}}}}$, but you seem to be using it to denote ${\displaystyle {\frac {d^{2}P}{d\phi ^{2}}}}$. Gandalf61 (talk) 10:47, 18 January 2010 (UTC)[reply]

To respnd to both of you in one post;

Dmcq, I was considering only k=0, or ${\displaystyle P''=0}$, the sine wave is not a solution.

Gandalf, yes I am actually looking at laplaces equation in spherical polars, which when seperated gives ${\displaystyle P''+k^{2}P=0}$ where is defined by ${\displaystyle V(r,\theta ,\phi )=R(r)T(\theta )P(\phi )}$where k is the seperation constant. Since when solving equations involves summing over all k, I was considering the case where k is zero. (It was my understanding that ${\displaystyle P''}$ denoted a non specific differnetial, with context usually used to give the variable, and that it was ${\displaystyle {\ddot {P}}}$ denoted specifically ${\displaystyle P_{tt}}$). Anyway, any answers would be lovely! —Preceding unsigned comment added by 129.67.39.49 (talk) 16:15, 18 January 2010 (UTC)[reply]

(edit conflict) Actually ${\displaystyle P=c\phi +d}$ is the correct solution, not a sine wave: the OP specified that ${\displaystyle k=0}$ so ${\displaystyle P''=0}$. However, that's not what SHM or simple harmonic motion refers to: SHM means ${\displaystyle P''+k^{2}P=0}$ with non-zero ${\displaystyle k}$ and gives sine-waves as Dmcq says.

As for your actual question: differentiating shows us that it solves the differential equation except at multiples of ${\displaystyle \pi }$ where the sudden jump in ${\displaystyle P}$ means that the derivative is undefined: there's no sensible answer to the question 'what's the derivative at this point?'. It also fails your requirement that ${\displaystyle P(\phi )=P(\phi +2k\pi )}$ - for example, if ${\displaystyle c=1}$ and ${\displaystyle d=2\pi }$ then ${\displaystyle P(0)=2\pi }$ but ${\displaystyle P(2\pi )=4\pi }$.

The most general solution is in fact just ${\displaystyle P(\phi )=d}$, since any non-zero ${\displaystyle c}$ will give discontinuity problems as above.

Also, Gandalf: I disagree that ${\displaystyle P'}$ usually indicates the time-derivative - I'd say it would mean differential with respect to whatever variable ${\displaystyle P}$ is a function of, or the non-time variable if it includes two dependencies. Still, it's best to always say what variable you're differentiating with respect to to avoid confusion. Olaf Davis (talk) 16:32, 18 January 2010 (UTC)[reply]

(after edit conflict)Well, the reference to shm was somewhat confusing. Anyway, the only solutions to ${\displaystyle P''=0}$ that also satisfy ${\displaystyle P(\phi )=P(\phi +2m\pi )}$ are the constant functions ${\displaystyle P(\phi )=d}$. The proposed solution ${\displaystyle P=2\pi \times floor({\frac {c\phi +d}{2\pi }})}$ is piecewise constant, but it has discontinuities at ${\displaystyle \phi ={\frac {2m\pi -d}{c}}}$, where it is not differentiable, and it does not in any case satisfy ${\displaystyle P(\phi )=P(\phi +2m\pi )}$ for all values of φ. Gandalf61 (talk) 16:43, 18 January 2010 (UTC)[reply]

Yes I didn't read the k=0 part. That's a strange way to state the problem. And yes the general solution is P=${\displaystyle P=c\phi +d}$. You can consider k=0 as meaning an infinite wavelength, perhaps they were trying to make it periodic? For that you'd have m infinite in ${\displaystyle P(\phi )=P(\phi +2m\pi )}$ which doesn't lead you anywhere fast.

Dmcq (talk) 17:33, 18 January 2010 (UTC)[reply

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Sorry my apologies all, but I have just realised I have made a rather bad typo. I would actually like to consider the solution ${\displaystyle P=c\phi +d-2\pi \times floor({\frac {c\phi +d}{2\pi }})}$. This I believe does fit the boundary ${\displaystyle P(\phi )=P(\phi +2m\pi )}$. And fulfills the equation P′′=0. What is the behavious at the discontinuity, does it negate it as a possible solution? —Preceding unsigned comment added by 129.67.39.49 (talk) 19:08, 18 January 2010 (UTC)[reply]

That doesn't solve the differential equation. It has discontinuities. Why do you keep thinking it is a solution? It doesn't have a deriviative at the discontinuity so it doesn't have a second deriviative and the differential equation doesn't hold. The second deriviative is not equal to 0 at the discontinuities, if it was there wouldn't be a discontinuity.

Dmcq (talk) 19:57, 18 January 2010 (UTC)[reply

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Maybe the OP is confusing having a periodic domain with having a periodic codomain. For instance if this was a function P:S¹ → S¹, then for c an integer P(φ) = cφ + d = 2π⌊(cφ + d)/2π⌋ would be continuous and a solution to P′′=0. But there's not any reason to think the codomain is anything other than R unless that's a detail the OP neglected to mention. Rckrone (talk) 03:46, 19 January 2010 (UTC) changed some apostrophes into primes to fix the formatting in mine and another post[reply]

I don't see that. It would still be discontinuous at φ = 0 on S<sup[>1 unless it is just the constant c.

Dmcq (talk) 11:06, 19 January 2010 (UTC)[reply

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Oh you're right sorry. I was thinking of the fractional part instead of the floor. I think the OP made that mistake too though. What I meant then was P(φ) = cφ + d = cφ + d - 2π⌊(cφ + d)/2π⌋ in S¹.

2π⌊(cφ + d)/2π⌋ is continuous in S¹ also, it's just uniformly zero. Rckrone (talk) 15:56, 19 January 2010 (UTC)[reply]

Alright dmcq, since you seem to have fun only by lampooning my attempts, you do my fucking homework for me. Is there a solution to P''=0, P'≠0 when restrained to P(φ)=P(φ+2kπ)?

The above post is vulgar, in any context whatsoever, and should be removed (the edit was made by the OP although it is unsigned). You are strongly discouraged to post comments of this nature; if you feel that you have been unfairly treated by Dmcq, note it politely. I do not see Dmcq using vulgar language so neither should you. --PST 02:47, 20 January 2010 (UTC)[reply]

Comments generally should not be removed. Eric. 131.215.159.171 (talk) 06:08, 20 January 2010 (UTC)[reply]

The perils of trying to help someone with their homework.

Dmcq (talk) 10:48, 20 January 2010 (UTC)[reply

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This issue need go no further, but I would not accuse Dmcq of being unfair, at best tactless and at worst arrogant. Whether these were justified is hard to tell, as tonal context is lost. However, I would not call my response a personal attack, as Dmcq did when threatening me with a ban, but I would call it rude; as rude as I percieved his, if more vulgar. Obviously the sarcasm was lost on him as well. —Preceding unsigned comment added by 129.67.39.49 (talk) 14:38, 20 January 2010 (UTC)[reply]

I do not see where you got the idea I was making fun of you. I believe I answered your query politely. I pointed out to you on your talk page that attacking people personally can lead to you being banned as per

Dmcq (talk) 15:19, 20 January 2010 (UTC)[reply

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Bien. —Preceding unsigned comment added by 129.67.39.49 (talk) 22:25, 20 January 2010 (UTC)[reply]

Please sign your post by typing four tildas (~~~~) after your comment, so that we can readily identify you. With regards to your post, I have nothing to say more than common sense should forbid you from posting such comments. It is also forbidden per

WP:AN/I, should you file a complaint. --PST 02:24, 21 January 2010 (UTC)[reply

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The only complaint I have regards your inability to get over it. Seriously it was just a throw away comment made to jibe at Dmcqs outstanding unhelpfulness. I really couldn't care for any of this overblown fallout.

The second derivative being zero means it must be linear. The only linear functions which are periodic are constants, which you rule out, so no. I'm sure you could have worked that out yourself if you had tried. If not, you shouldn't be taking whatever course this is from - you need to go back and learn the prerequisites. --Tango (talk) 04:16, 20 January 2010 (UTC)[reply]

I have the equation for working out the uptake of CO2 in concrete, where the maximum depth is D.

D = K * the square root of t.

To work out K, I need to multiply k1 * k2 * k3. k1 is a particular figure with the suffix (year)-1/2 (where the "-1/2" is always in superscript. (k2 & k3 are simply other figures).

So an example equation is given as

K = 6 * 0.7 * 1.1 * (years)-1/2 = 4.6mm (years)-1/2

What does the (years)-1/2 mean? I know the (years) is the time the concrete exists for (or a derivative thereof) but the "-1/2" is a mystery - I assume it's the square root or 1/the square root but am unsure.

Thanks. 157.203.42.175 (talk) —Preceding undated comment added 10:34, 18 January 2010 (UTC).[reply]

Do you mean years^-1/2. In that case it means years to the power of -0.5. So it means divide my the square root of years.

Taemyr (talk) 11:26, 18 January 2010 (UTC)[reply

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Yes, y^-1/2 = 1/(y^1/2) = ${\displaystyle 1/{\sqrt[{2}]{y^{1}}}}$ = ${\displaystyle 1/{\sqrt {y}}}$. In general, y^-n = 1/yⁿ and y^a/b = ${\displaystyle {\sqrt[{b}]{y^{a}}}}$. PrimeHunter (talk) 12:16, 18 January 2010 (UTC)[reply]

The would only make sense if the other constants have units that make the product come out in units of distance per square root time. For example cm/year^1/2. The D value must come out in the end to be in units of distance. See Dimensional analysis.--RDBury (talk) 14:43, 18 January 2010 (UTC)[reply]

Thanks all, that's really useful :) 157.203.42.175 (talk) 16:32, 18 January 2010 (UTC)[reply]