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April 2

Taylor expansion?

So I'm trying to show that ${\frac {\sin(\pi x)}{(x^{2}-4)^{2}}}$ can be given as the expansion $\sum a_{n}(x-2)^{n}$ The obvious thing would appear to be a Taylor expansion around 2 giving the (x-2)^n form, but this doesn't work as f(2) is a singular point. So I thought maybe a Laurent series of $\sin(\pi x)$ divided by the $(x-2)^{2}(x+2)^{2}$ bottom but then I'm left with the (x+2)^2 terms which isn't the form asked for. Any ideas? —Preceding unsigned comment added by 82.132.136.204 (talk) 02:26, 2 April 2010 (UTC)[reply]

Also note that

\sin(\pi x)=\sin(\pi (x-2))

so the Laurent series expansion of

\sin(\pi x)/(x-2)^{2}

at x=2 comes immediately from the power series expansion of the sine at 0 (make the substitution y=x-2 if you like). As to the other factor

(x+2)^{-2}

, just expand it at

x=2

; to do so quickly write

(x+2)^{-2}={\frac {1}{16}}\left(1+{\frac {x-2}{4}}\right)^{-2}

and use conveniently the series of

(1-z)^{-2}

. --pm a 08:46, 2 April 2010 (UTC)[reply]

de Bruijn's function

Smooth_number#Distribution declares:

“

Let

\scriptstyle \Psi (x,y)

denote the de Bruijn function, the number of y-smooth integers less than or equal to x.

”

while Dickman-de Bruijn function#Definition says:

“

The Dickman-de Bruijn function

\rho (u)

is a continuous function that satisfies the delay differential equation

u\rho '(u)+\rho (u-1)=0\,

with initial conditions $\rho (u)=1$ for 0 ≤ $u$ ≤ 1. Dickman showed heuristically that

\Psi (x,x^{1/a})\sim x\rho (a)\,

where $\Psi (x,y)$ is the number of y-smooth integers below x.

”

So, which one is a de Bruijn's function: $\scriptstyle \Psi (x,y)$ or $\rho (u)$ ? Please clarify the articles. --CiaPan (talk) 06:26, 2 April 2010 (UTC)[reply]

Psi is the de Bruijn function, rho is the Dickman-de Bruijn function. It's horribly common for related (and even unrelated) concepts in mathematics to have similar names based on the people who discovered or investigated them. Compare, for example, Euler-Mascheroni constant and e (mathematical constant), both occasionally called Euler's constant/number. Or, for that matter, List of topics named after Leonhard Euler or List of topics named after Augustin-Louis Cauchy. Confusing Manifestation(Say hi!) 23:41, 6 April 2010 (UTC)[reply]

Thank you. --CiaPan (talk) 06:10, 6 April 2010 (UTC)[reply]

Four-Variable Algebra Problem

Resolved

Just to be clear here, this is honestly not a homework question. This is a question from someone who once knew how to do algebra, but can't for the life of him remember what the next step here is. Given unknown quantities a, b, c and n:

$3na+2nb+nc=1$

$3a+2b+c={\frac {1}{n}}$

How do I solve for n in terms of the other three variables? Is this possible? And if this is something so embarrassingly simple that no one wants to bother with it, could someone please point me in the direction of something that might jog my memory on this stuff? Thanks, everyone. --Brasswatchman (talk) 14:58, 2 April 2010 (UTC)[reply]

You can't. You need four equations to solve for four variables; the solutions now form a plane. 76.230.226.56 (talk) 15:01, 2 April 2010 (UTC)[reply]

You didn't read carefuly: the original poster asked how to solve for one variable, not for all four. Michael Hardy (talk) 17:33, 2 April 2010 (UTC)[reply]

The second equation is equivalent the the first one. As for solution, just write it as n(3a + 2b + c) = 1, and divide both sides by the thing in the bracket:

n={\frac {1}{3a+2b+c}}.

—Emil J. 15:08, 2 April 2010 (UTC)[reply]

That'll do, Emil. Thank you! --Brasswatchman (talk) 15:09, 2 April 2010 (UTC)[reply]

...and notice that if you've written

{\text{something}}={\frac {1}{n}},

then you can write

{\frac {1}{\text{something}}}=n

and then you've solved for n. Michael Hardy (talk) 17:34, 2 April 2010 (UTC)[reply]

Complex Root of Unity

Could anyone please explain this beautiful picture? •• Fly by Night (talk) 18:29, 2 April 2010 (UTC)[reply]

The deobfuscated link to the picture, File:Complex x hoch 3.jpg, takes you to a description page which tells you that it is a graph of the polynomial z³ − 1 in the complex plane, where the colours denote complex values of the function according to the key given there.—Emil J. 18:36, 2 April 2010 (UTC)[reply]

What does "deobfuscated" mean? •• Fly by Night (talk) 18:38, 2 April 2010 (UTC)[reply]

Made clear. Theresa Knott | token threats 18:41, 2 April 2010 (UTC)[reply]

Following standard dictionary practice, we should define each term using other terms at least as obscure as the one we are defining. As such, how about "deobfuscate = to undergo anticomplexification". :-) StuRat (talk) 18:59, 2 April 2010 (UTC) [reply]

Eschew obfuscation! -- Coneslayer (talk) 19:18, 2 April 2010 (UTC)[reply]

:-) Theresa Knott | token threats 22:20, 2 April 2010 (UTC)[reply]

Just from looking at the picture, the idea would appear to be that two points with the same color are complex numbers that have the same cube. And for each color (except the one at 0), there are three points that have that color. Michael Hardy (talk) 00:57, 3 April 2010 (UTC)[reply]

....and so the three black points are the three cube roots of 1:

{\begin{aligned}1^{3}&=1\\[6pt]\left({\frac {-1+i{\sqrt {3}}}{2}}\right)^{3}&=1\\[6pt]\left({\frac {-1-i{\sqrt {3}}}{2}}\right)^{3}&=1\end{aligned}}

Michael Hardy (talk) 01:00, 3 April 2010 (UTC)[reply]

Judging from your comment, my idea of pointing people to the description page apparently didn't get across, hence here's it once again, more explicitly. The picture is a graph of the function f(z) = z³ − 1 in the subset

\{x+iy\mid x,y\in [-3,3]\}

of the complex plane. The markings at the border of the picture tell you which point corresponds to which value of z (the horizontal axis is real, the vertical imaginary, as usual). The colour of that point then represents the value of f(z) at that particular z, and the correspondence of the colours to the complex numbers they represent is given by the picture under the "Summary" subheading, which I repeat here:

. Again, the horizontal axis is real, the vertical axis is imaginary, and the scale is marked on the border (it's nonlinear).—Emil J. 15:19, 5 April 2010 (UTC)[reply]

In case the system being used is unclear from just looking at the picture, the hue of the color represents the argument and the brightness represents the magnitude, with darker colors being smaller and lighter colors being larger. This is a somewhat common way to visualize complex functions since it solves the problem of showing 4 dimensions worth of information. What's nice about choosing the colors this way is that it's easy to pick out most of the features you'd be interested in finding like the zeros (black spots) and poles (white spots), the degrees of zeros or poles (by the number of times the color cycles around the point).Rckrone (talk) 23:45, 6 April 2010 (UTC)[reply]

Properties of Triangles given Coordinates of vertices?

I have a triangle ABC with points A(-28,24) B(-44,14) and C (84,-38)

I need the area, side lengths, circumcentre, [[orthocentre] and centroid of this triangle.

Is there a online program that can give me some/all of these values and save me some work? It's just mundane calculation to prove some property of medians, and I'd rather get the values than spend the next half hour getting them.

If you want to give me some too, that'd be great :)

Thanks in advance,

PerfectProposal 18:42, 2 April 2010 (UTC)[reply]

I can give you a simple formula for the side lengths:

L = sqrt((x₁-x₂)² + (y₁-y₂)²))

If you do the math and post the results, I'll check your work. StuRat (talk) 18:50, 2 April 2010 (UTC)[reply]

Factored lengths are:

2(root)89 4(root)1193 2(root)4097

The other stuff are the ones that take time. Hope I'm right on the lengths though :)

PerfectProposal 19:01, 2 April 2010 (UTC)[reply]

Yea, your lengths look right. StuRat (talk) 19:13, 2 April 2010 (UTC)[reply]

Wolfram alpha is your online maths friend see http://www.wolframalpha.com/input/?i=triangle+%7C+vertex+coordinates%28{-28%2C24}%2C+{-44%2C14}%2C+{84%2C-38}%29 Theresa Knott | token threats 19:08, 2 April 2010 (UTC)[reply]

I didn't know Wolfram Alpha could do that! Thanks: is it possible for it (or any other resource) to calculate the centres for me? PerfectProposal 19:23, 2 April 2010 (UTC)[reply]

Awesome. I'm trying to prove the area=1056 using heron's formula. so far I have lengths

2(root)89 4(root)1193 2(root)4097

and therefore, semiperimeter is (root)89+ 2(root)1193+ (root)4097

so Heron's would look something like

√ √89+ 2√1193+ √4097(√89+ 2√1193+ √4097-2√89)(√89+ 2√1193+ √4097-4√4097)(√89+ 2√1193+ √4097-2√4097)

correct? I can't seem to get an answer for that.

PerfectProposal 19:53, 2 April 2010 (UTC)[reply]

You have made a small error, your second bracketed term should read (√89+ 2√1193+ √4097-4√1193) Also you need to be very careful about brackets.(computers do what you ask them to, not what you want them to, bloody stupid machines). if you put

√ {(√89+ 2√1193+ √4097)(√89+ 2√1193+ √4097-2√89)(√89+ 2√1193+ √4097-4√1193)(√89+ 2√1193+ √4097-2√4097)}

into wolfram alpha it will give you the correct result. Theresa Knott | token threats 22:16, 2 April 2010 (UTC)[reply]

As mentioned on the triangle article, you can also find the area directly from the coordinates using the formula:

Area={\frac {1}{2}}\left|\det {\begin{pmatrix}x_{A}&x_{B}&x_{C}\\y_{A}&y_{B}&y_{C}\\1&1&1\end{pmatrix}}\right|

Rckrone (talk) 18:37, 7 April 2010 (UTC)[reply]

A question for any "soccer" enthusiasts out there.

You are probably acquainted with the Premier League, the highest level of soccer in England. The Premier League consists of 20 teams, Tand each team plays 38 games in a season. For winning a match, a team is granted three points; for drawing (tying) a match, a team is granted one point; and for losing a match, a team is granted zero points.

At the end of each season, the three teams with the lowest number of points are "relegated" (in the next season they will be replaced by new teams).

I have been trying for a while to calculate the number of points needed for a team to be "safe" from relegation - the minimum amount of points needed to avoid finishing in the bottom three.

--T.M.M. Dowd (talk) 22:04, 2 April 2010 (UTC)[reply]

Well, it could be as low as 1 point, if there are at least 3 teams which lose every game, or as high as 58 points, if pretty much every team wins half and loses half, with no ties. So, before the season starts, 58 is the answer. However, once there have been games played, the number would likely go down. After half the season has been played, you'd just take the 3rd worst team's point total, and add 3×(remaining number of games)+1 to get the current "safe" value. This is, of course, assuming that those bad teams might win every game from here on, which is extremely unlikely, but technically possible. A more useful approach might be to just look at the highest number of points in each team to be relegated in each previous season, and take one more than the highest of those, as a reasonably safe score. StuRat (talk) 22:08, 2 April 2010 (UTC)[reply]

It is not possible for the three bottom teams to each have lost every game, since each plays each of the others twice. Each match among the bottom four teams contributes at least two points to their total, so the minimum number of points the bottom four teams can have between them is 24. Thus the fewest points a team can have and avoid relegation is 6, if the four bottom teams draw every game among themselves, and lose everything else. This of course ignores the possibility of a team being docked points for going bankrupt or somesuch. Algebraist 22:31, 2 April 2010 (UTC)[reply]

Good point about not losing every game, but I get 5, not 6. That is, if each of the worst 3 teams loses every game except when playing each other, they would each have 4 points. Thus, any team with 5 or more points would not be relegated, under this scenario. There also seems to be an inherent guarantee that any team with 3 or fewer points would be relegated. StuRat (talk) 22:42, 2 April 2010 (UTC)[reply]

If all the worst 3 teams lose every game against better teams, then all better teams have at least 9 points; there's no way they can have as few as 5. Thus 6 is the smallest number of points a team can have and stay up, as I claimed. This means a team with at most 5 points is guaranteed to be relegated. Algebraist 22:45, 2 April 2010 (UTC)[reply]

Similarly, the maximum number of points that can be scored among all teams in total is if no match is drawn, in which case 1140 points are scored in total, and the best the bottom teams can do is if these points are distributed evenly (e.g. by every match being won by the team playing at home), i.e. with every team scoring 57 points. Thus 57 points is the most a team can score and yet be relegated. Algebraist 22:36, 2 April 2010 (UTC)[reply]

Agreed, but they would need 58 to guarantee that they won't be relegated, which is how I interpret the last line of the original Q. StuRat (talk) 22:42, 2 April 2010 (UTC)[reply]

That analysis is of course a load of rubbish. We don't want the bottom teams to do as well as possible; we want the third-from-bottom team to do as well as possible. This requires the two bottom teams to do badly. The optimum situation is when the bottom two teams lose every match (except against each other) and wins are divided evenly otherwise. This leads to all the top 18 teams getting 63 points, so that's the most a team can have and be relegated. Algebraist 18:03, 3 April 2010 (UTC)[reply]

Thank you all for your answers. --T.M.M. Dowd (talk) 20:22, 3 April 2010 (UTC)[reply]

Leaving mathematics aside don't call it "soccer" with speech marks around it. Call it football damn you! It's a game played with a thing shaped like a ball that you kick with your foot. Just because some foolish people across the pond call a game played with a thing not shaped like a ball, that they pick up with their hands and throw "football" doesn't mean that we need to start using "soccer" ;-) Theresa Knott | token threats 23:47, 2 April 2010 (UTC)[reply]

The term soccer, of course, comes from England — it's derived from association football, which is the code of football propounded by the Football Association. In England I believe it's an informal term, but not an incorrect one. In the States, we use it even in formal contexts, to avoid confusion with the more interesting sort of football. --Trovatore (talk) 00:08, 3 April 2010 (UTC)[reply]

:-) Theresa Knott | token threats 00:27, 3 April 2010 (UTC)[reply]

If I wasn't familiar with either game, I would have assumed "football" was a version of kick-boxing which lacks the "no hitting below the belt" rule. :-) StuRat (talk) 03:16, 3 April 2010 (UTC) [reply]

The word "tangent" springs to mind. --T.M.M. Dowd (talk) 20:22, 3 April 2010 (UTC)[reply]