Wikipedia:Reference desk/Archives/Mathematics/2010 August 15

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August 15

Math Child Prodigies of 21st Century

Can we systematically list and rank the math child prodigies of 2010 or of the 21st century?For some odd reason, we don't anymore hear of math prodigies,the lsat being Terence Tao about 30-40 years ago.Why has the waterfall of child prodigies stopped?Has math become less popular.If it hasn't,and if there's concrete evidence,I'd like a list please of all math child prodigies.Thanks guys.

The future of math is in the hands of child prodigies and we need to start appreciating that IMO.So who's the greatest child prodigy right now?BTW,Typsetting Bug!And can we document any unsuccessful child prodigies?I'd like to hear examples. —Preceding unsigned comment added by 110.20.19.24 (talk) 01:43, 15 August 2010 (UTC)

Mathematics is not in the hands of child prodigies. Sure Terry Tao was a prodigy and one of the best mathematicians alive. But you only hear the successful child prodigies, because many of the so called "prodigies" just work very hard and appear to be very knowledge while still in their teens, and are unable to produce results later on. Money is tight (talk) 04:34, 15 August 2010 (UTC)

Even Tao wasn't doing original research from an early age, from what I can tell. He got his PhD aged 20, which is significantly earlier than most, but only by 4 or 5 years, and his Wikipedia article doesn't mention any papers he published before his PhD. Professional mathematics (in academia) is all about original research (and teaching). You get child prodigies that are brilliant and learning and doing existing mathematics, but hold no particular talent for original research. Maths is in the hands of people that do lots of great original research and starting 4 or 5 years earlier isn't going to make much difference to how much you can do in a career. So, it's the quality of the mathematician, not their speed of development, that matters. --Tango (talk) 20:34, 15 August 2010 (UTC)

110.20.19.24 (aka 110.20.18.61), why are you again changing the IP address in SineBot's sig of your initial post, replacing the first octet with 119? Do you now wish you were in China? (I'd be happy to be in NSW.) -- 115.67.79.109 (talk) 06:26, 15 August 2010 (UTC)

No.There's a annoying typsetting bug on my computer that's making all these changes.As you can see,my punctuation is typed like I'm a 5 year old.It's annoying but I can't do anything about it.But thanks for correcting it for me.

My I suggest registering a username if you wish to hide your ip? Doing things like you did draw more attention than doing nothing at all. There is a List of child prodigies but I wouldn't think it is very complete and personally I can't see much point in the list. Dmcq (talk) 09:22, 15 August 2010 (UTC)

I don't wish to hide my fucking IP!Sheesh,at least fucking ask me before jumping to damn conclusions as the IP above politely did.

The changes you made were explicitly to change the ip. You gave misleading edit comments each time you did it. It had nothing to do with a typesetting bug as the only other thing you did was stick in a new line between two paragraphs. Please do not start saying 'fucking' or 'damn' about reasonable advice or I will remove the question on WP:CIVIL grounds. Dmcq (talk) 09:55, 15 August 2010 (UTC)

Can you explain exactly what the typesetting bug does? The only problem I see with your punctuation is that there are no spaces after punctuation marks. -- Meni Rosenfeld (talk) 12:21, 15 August 2010 (UTC)

Yes I was wondering about that. I guess they must be using a mobile phone with some textbox input method which suppresses the space for some reason or they go onto a new line and it removes the space at the end of a line. In which case perhaps not going onto a new line or sticking a space at the beginning of the new line would help. I'd be interested in who make the browser. Dmcq (talk) 13:06, 15 August 2010 (UTC)

I have heard of Lenhard Ng (I saw him called Lenny Ng). I heard he was the youngest to ever score an 800 on the math portion of the SAT and also that he went on to be a 3 time Putnam fellow (and he only had 3 years of college (4 time Putnam fellows are weird)). And, the article mentions many other feats including 4 time perfect on AHSME and 2 time gold medalist/1 time silver medalist in IMO. Also, I found a web page with a list of his publications and he has quite a few, though not an insane number or anything. And, number of publications is not necessarily that important. Quality is also important. I don't know quality/difficulty of his work so I can't say anything. I don't know if it's groundbreaking or anything. StatisticsMan (talk) 02:29, 17 August 2010 (UTC)

According to our article on him, he got his PhD aged 25, which is a normal age. So, he doesn't seem to have started research early either. --Tango (talk) 07:00, 18 August 2010 (UTC)

25 is really pretty early. Generally you get your Bachelor's at age 22, and five years is pretty fast for a math or pure-science PhD in the States (engineers sometimes get it done a little faster). Certainly 25 is not exceptionally early, but it's definitely fast. --Trovatore (talk) 09:10, 19 August 2010 (UTC)

Partial fractions

Hi,

I'm having difficulty in solving (to me) a fiendish partial fraction:-

(8x^3 - 22x^2 + 6x - 4) / (x - 2)^2 (x^2 + x +2)

whatever I try, I end up with multiple unknowns - I've found 'B', but after that I end up with 'A', 'C' or 'D' in the same equation. Is it a case of solving the later parts simultaneously?

Thanks for any helpVs60t (talk) 10:08, 15 August 2010 (UTC)

Yes, solving a system of linear equations is the normal way to find partial fractions. It's certainly easier than going out of your way to try avoid it. -- Meni Rosenfeld (talk) 11:28, 15 August 2010 (UTC)

I try to avoid using that method. I call that method the "stupid high school method" :) . My preferred method works by finding the poles of the function in the complex plane. Then you compute the truncated Laurent expansion fp(x) around each pole p that contains precisely all the singular terms and none of the regular terms. The sum of all the fp(x) is then the partial fraction expansion, if the degree of the numerator is smaller than the dgree of the denominator. If that's not the case, you can simply perform a long division in the usual way. Note that doing that amounts to adding the singular terms from the expansion around infinity to the sum of the fp, so this fits in nicely with the picture that a partial fraction expansion is simply the sum of all the singular parts of all expansions around all the singular points.

Proof. If f(x) is a rational function we want to expand in patial fractions, then g(x) = f(x) - sum of fp(x) is also a rational function, but without any singularities. Hence it is a polynomial. If the degree of the numerator is less than the degree of the denominator g(x) will tend to zero for x to infinity, so g(x) must be identical to zero. If the degree of the numerator is larger than or equal to that of the nhumerator, we can compute the polynomial g(x) by expanding f(x) around infinity (i.e. in powers of w = 1/x). Only the coefficients of the negative powers of the expansion parameter w will be nonzero (otherwise g(x) would not be a polynomial). Count Iblis (talk) 14:56, 15 August 2010 (UTC)

Would you like to show us how to compute the principal part of the Laurent series at x = ½(–1 ± i√7), without using Matlab, Mathematica, Maple, or any other computer algebra package? I just tried and I ran out of paper very quickly. — Fly by Night (talk) 18:54, 15 August 2010 (UTC)

We can write the function as:

$f(x)={\frac {8x^{3}-22x^{2}+6x-4}{(x-2)^{2}(x-a)(x-b)}}\,$

with

${\begin{aligned}a&={\frac {1}{2}}\left(-1+i{\sqrt {7}}\right)\\b&={\frac {1}{2}}\left(-1-i{\sqrt {7}}\right)\end{aligned}}\,$

The principal part of the Laurent expansion at x = a is:

$f(x)=\left({\frac {8a^{3}-22a^{2}+6a-4}{(a-2)^{2}(a-b)}}\right){\frac {1}{x-a}}\,$

The question is then how to simplify the coefficient of 1/(x-a) efficiently. A simple method is to exploit the fact that since:

$a^{2}+a+2=0\,$

you can reduce all algebraic expressions Modulo(a^2 + a + 2).

So, we can write:

$8a^{3}-22a^{2}+6a-4=-30a^{2}-10a-4=20a+56$

And

$(a-2)^{2}=a^{2}-4a+4=-5a+2$

With these simplifications it is now trivial to insert the value for a in the expression and to simply it to get the square roots out of the numerator.

Then the expansion around x = b is simply the complex conjugate of the above expansion (taking x real), so we don't have to do that expansion. Count Iblis (talk) 21:21, 15 August 2010 (UTC)

OP - thanks for the help and adviceVs60t (talk) 11:44, 16 August 2010 (UTC)

Rearranging a formula - getting confused!

Hi folks,

How can you get from:

${n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}]}$

to here:

${\sqrt {[n(\sum _{i=1}^{n}{X_{i}}^{2})-(\sum _{i=1}^{n}{X_{i}})^{2}]\times [n(\sum _{i=1}^{n}{Y_{i}}^{2})-(\sum _{i=1}^{n}{Y_{i}})^{2}]}}$

I'm stuck! One thing I know is that:

${\bar {X}}={{\sum _{i=1}^{n}{X_{i}}} \over n}$

and similarly:

${\bar {Y}}={{\sum _{i=1}^{n}{Y_{i}}} \over n}$

Anyone have any suggestions? - 114.76.235.170 (talk) 12:59, 15 August 2010 (UTC)

Should have noted that I've tried the following:

${n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}]}$

$={n[{\sqrt {\sum _{i=1}^{n}(X_{i}-{{\sum _{i=1}^{n}{X_{i}}} \over n})^{2}}}]\times n[{\sqrt {\sum _{i=1}^{n}(Y_{i}-{{\sum _{i=1}^{n}{Y_{i}}} \over n})^{2}}}]}$

But I totally got stuck - I tried expanding the formula to see if I can see a pattern, and it's just not happening for me :( 114.76.235.170 (talk) 13:03, 15 August 2010 (UTC)

Did you get as far as

$\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}=\sum _{i=1}^{n}{X_{i}^{2}}-\sum _{i=1}^{n}{{\bar {X}}^{2}}$

Dmcq (talk) 13:28, 15 August 2010 (UTC)

By the way I think the n should be squared going under the square root, are you sure you haven't copied something wrong about n? Dmcq (talk) 13:39, 15 August 2010 (UTC)

That will only be half a fix, note there is another summand. -- Meni Rosenfeld (talk) 13:47, 15 August 2010 (UTC)

There's a small error in your problem statement; it should have been

${{\sqrt {n\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {n\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}.$

This equality actually comes from the equality

$n\sum _{i}(X_{i}-{\bar {X}})^{2}=n\sum _{i}X_{i}^{2}-\left(\sum _{i}X_{i}\right)^{2}$

applied once for X and once for Y. It is equivalent to a useful way to compute variance, $\mathbb {V} [X]=\mathbb {E} [X^{2}]-\mathbb {E} [X]^{2}$. To prove it, start with

$n\sum _{i}(X_{i}-{\bar {X}})^{2}=n\sum _{i}(X_{i}^{2}-2X_{i}{\bar {X}}+{\bar {X}}^{2}),$

And remember what we taught you about terms that don't depend on i. -- Meni Rosenfeld (talk) 13:45, 15 August 2010 (UTC)

By the way, I see that in Talk:Pearson product-moment correlation coefficient you managed to transform the numerator. You have a few mistakes there and you wrote it in a way more complicated than necessary, but you got the main ideas right, and what you have here is very similar.

Another hint: Writing ${\bar {X}}$ is easier than ${{\sum _{i=1}^{n}{X_{i}}} \over n}$, so you'll want to turn the latter into the former, rather than the other way around, as long as possible. -- Meni Rosenfeld (talk) 13:57, 15 August 2010 (UTC)

Whoa, you are both correct! I made an error in the way I copied in the TeX... sorry, I'm really new to this! You are both so excellent, Meni, let me read and digest your answer(s) :-) And yeah, I'm trying to understand how to get from one forumla to another from the Pearson product-moment correlation coefficient :-)

One day I hope to be able to contribute to Wikipedia articles, or even to this reference desk :-) For now, it's a journey of discovery...

Incidentally, I'm not doing a Uni/College course - I'm trying to (slowly) teach myself statistics. It's heavy going... - 114.76.235.170 (talk) 14:01, 15 August 2010 (UTC)

Another attempt... still a bit stuck

OK, I'm trying again... this time I've gotten quite a bit further, but still can't quite get to the next stage...

The goal

Prove that:

$r={n(\sum {{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}}) \over {\sqrt {[n(\sum _{i=1}^{n}{X_{i}}^{2})-(\sum _{i=1}^{n}{X_{i}})^{2}]\times [n(\sum _{i=1}^{n}{Y_{i}}^{2})-(\sum _{i=1}^{n}{Y_{i}})^{2}]}}}$

is the same as:

$r={\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})}{{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}}$

Proof

OK, so the formula here is:

$r={\frac {\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})}{{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}}}$

Now if you multiple the numerator and the denominator by n then this gives you:

$r={\frac {n({\sum _{i=1}^{n}(X_{i}-{\bar {X}})(Y_{i}-{\bar {Y}})})}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Then you expand the numerator:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-{\bar {X}}{Y_{i}}-{\bar {Y}}{X_{i}}-{\bar {X}}{\bar {Y}}}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Which is the same as:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}X_{i} \over n})\times {Y_{i}}-({\sum _{i=1}^{n}Y_{i} \over n})\times {X_{i}}-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Which is:

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}{X_{i}}{Y_{i}} \over n})-({\sum _{i=1}^{n}{Y_{i}}{X_{i}} \over n})-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

This cancels a term...

$r={\frac {n({\sum _{i=1}^{n}({X_{i}}{Y_{i}}-({\sum _{i=1}^{n}X_{i} \over n})({\sum _{i=1}^{n}Y_{i} \over n})}))}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Leading to the simplified numerator...

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}(X_{i}-{\bar {X}})^{2}}}]\times {\sqrt {\sum _{i=1}^{n}(Y_{i}-{\bar {Y}})^{2}}}\right]}}$

Now Meni suggests that I don't expand the X-bar to sum notation... so I'll try it without. It leads me to:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}{X_{i}}^{2}-2.{X_{i}}.{\bar {X}}+{\bar {X}}^{2}}}\times {\sqrt {\sum _{i=1}^{n}{Y_{i}}^{2}-{Y_{i}}.{\bar {Y}}+{\bar {Y}}^{2}}}\right]}}$

Well... I think I do need to expand the X-bar and Y-bar now.

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left({X_{i}}^{2}-2.{X_{i}}.{{\sum _{i=1}^{n}{X_{i}}} \over n}+{{\sum _{i=1}^{n}{X_{i}}} \over n}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left({Y_{i}}^{2}-2.{Y_{i}}.{{\sum _{i=1}^{n}{Y_{i}}} \over n}+{{\sum _{i=1}^{n}{Y_{i}}} \over n}^{2}\right)}}\right]}}$

Which is:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left({X_{i}}^{2}-{{\sum _{i=1}^{n}2.{X_{i}}^{2}} \over n}+{{\sum _{i=1}^{n}{X_{i}}} \over n}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left({Y_{i}}^{2}-{{\sum _{i=1}^{n}2.{Y_{i}}^{2}} \over n}+{{\sum _{i=1}^{n}{Y_{i}}} \over n}^{2}\right)}}\right]}}$

This leads to:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left(n.{X_{i}}^{2}-{\sum _{i=1}^{n}2.{X_{i}}^{2}}+{\sum _{i=1}^{n}{X_{i}}}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left(n.{Y_{i}}^{2}-{\sum _{i=1}^{n}2.{Y_{i}}^{2}}+{\sum _{i=1}^{n}{Y_{i}}}^{2}\right)}}\right]}}$

or also:

$r={\frac {n(\sum _{i=1}^{n}{{X_{i}}{Y_{i}}})-(\sum _{i=1}^{n}{X_{i}})(\sum _{i=1}^{n}{Y_{i}})}{n\left[{\sqrt {\sum _{i=1}^{n}\left(n.{X_{i}}^{2}-{\sum _{i=1}^{n}{X_{i}}}^{2}\right)}}\times {\sqrt {\sum _{i=1}^{n}\left(n.{Y_{i}}^{2}-{\sum _{i=1}^{n}{Y_{i}}}^{2}\right)}}\right]}}$

Am I on the right track? Seems to be that way... any help getting to the next step would be appreciated. :-) 114.76.235.170 (talk) 15:32, 15 August 2010 (UTC)

No, you are on wrong track. After expanding the numerator (correcting the sign error) you should notice that

$\sum _{i=1}^{n}({X_{i}}{Y_{i}}-{\bar {X}}{Y_{i}}-{\bar {Y}}{X_{i}}+{\bar {X}}{\bar {Y}})$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-(\sum _{i=1}^{n}{\bar {X}}{Y_{i}})-(\sum _{i=1}^{n}{\bar {Y}}{X_{i}})+(\sum _{i=1}^{n}{\bar {X}}{\bar {Y}})$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-{\bar {X}}(\sum _{i=1}^{n}{Y_{i}})-{\bar {Y}}(\sum _{i=1}^{n}{X_{i}})+{\bar {X}}{\bar {Y}}(\sum _{i=1}^{n}1)$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-{\bar {X}}(n{\bar {Y}})-{\bar {Y}}(n{\bar {X}})+{\bar {X}}{\bar {Y}}(n)$

$=(\sum _{i=1}^{n}{X_{i}}{Y_{i}})-n{\bar {X}}{\bar {Y}}$

Bo Jacoby (talk) 06:30, 16 August 2010 (UTC).

When you add a new sum, try giving it a new index (i.e. j, k, l) so that you are sure to to confuse the i in the outer sum with the i in the inner sum. You can not mix them and this mistake appears several times in your algebra. 0¹⁸ (talk) 17:35, 16 August 2010 (UTC)

Pivoting

Is the pivoting operation in the simplex algorithm the same as the pivoting operation in Gauss–Jordan elimination? Yaris678 (talk) 21:33, 15 August 2010 (UTC)

Nearly. In terms of the actual row operations they are the same. In Gauss-Jordan you're trying to write all the variables in terms of a subset of them but in simplex you already have that and are making a new choice of which subset is to be used. This might make a difference in how you store the matrices when you implement the two algorithms.--RDBury (talk) 12:10, 16 August 2010 (UTC)

Brilliant explanation. Thanks. Yaris678 (talk) 05:28, 17 August 2010 (UTC)