Wikipedia:Reference desk/Archives/Mathematics/2009 April 22

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April 22

The Floor Function

I am trying to evaluate

$\int _{0}^{1}\int _{0}^{1}\cdots \int _{0}^{1}[x_{1}+\cdots +x_{n}]dx_{n}\cdots dx_{1}$

which is basically the floor function in a n-fold integral over the unit cube. I tried to start with a simple case and go step by step.

$\int _{0}^{1}[x]dx=0$ that is easy to see.

$\int _{0}^{1}\int _{0}^{1}[x+y]dydx$ can be rewritten as

$0\cdot R_{0}+1\cdot R_{1}$ where

$R_{1}=\{(x,y)\in \mathbb {R} ^{2}|0<x,y<1,x+y>1\}$

which has an area of 1/2 so the entire integral is 1/2. And similarly for three dimensions, I got

$\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}[x+y+z]dzdydx$ as

$1\cdot R_{1}+2\cdot R_{2}$ where

$R_{1}=\{(x,y,z)\in \mathbb {R} ^{3}|0<x,y,z<1,2>x+y+z>1\}$

$R_{2}=\{(x,y,z)\in \mathbb {R} ^{3}|0<x,y,z<1,3>x+y+z>1\}.$

My questions is, I can't find the volume of these regions correctly (because I can't set up the triple integral correctly). Can someone please shed some light on this on how to find the volume of both of these regions and then how to generalize this to n-dimensions? Thanks!69.224.116.142 (talk) 18:08, 22 April 2009 (UTC)

I guess it is $I_{n}=(n-1)/2$. Change variable in the integral putting: $y_{i}:=1-x_{i}$ and sum the two, so you get twice $I_{n}$ is the integral on the n-cube of $\scriptstyle [x_{1}+\cdots +x_{n}]+[n-(x_{1}+\cdots +x_{n})]$; then use the identities $[n+t]=n+[t]$ and $[t]+[-t]=-1$ (a.e.) --pma (talk) 20:33, 22 April 2009 (UTC)

As to the volume of the sets $R_{n,k}$ here they are [1] pma (talk) 22:22, 22 April 2009 (UTC)

By the way, "a.e" as in pma's post stands for "almost everywhere" in case you were wondering... --PST 03:19, 23 April 2009 (UTC)

This is great my now my question is how can I show that making that change of variables in the integral still equals $I_{n}$? It is pretty easy to show it for the n=1 case but for higher n's, it doesn't seem to work out.130.166.159.98 (talk) 02:09, 24 April 2009 (UTC)

It's just the change of variables formula. But here you need a very particular case of it, for the change of variable map φ(x):=(1,1..1)-x is quite an elementary isometry and the integrand is a simple function. So, if you prefer, write your integral as you did,

$\textstyle I_{n}=\sum _{k=0}^{n-1}k|R_{k}|$,

where

$R_{k}:=\{x\in [0,1]^{n}:[x_{1}+..+x_{n}]=k\}$

and observe that $\textstyle R_{k}$ and $\textstyle R_{n-1-k}$ have the same measure, because they are obtained from each other (up to a null set) with a simmetry (the change of sign) and a translation. For instance in your computation above for $n=3$, $\textstyle |R_{0}|=|R_{2}|$ so $\textstyle I_{3}=|R_{1}|+2|R_{2}|=|R_{0}|+|R_{1}|+|R_{2}|=1$. pma (talk) 10:45, 24 April 2009 (UTC)

Factoring a cubic

Does the cubic $20u^{3}-65\pi u^{2}+50\pi ^{2}u-16\pi ^{3}$ factor nicely? If it does, what is the factorization? Lucas Brown 42 (talk) 19:23, 22 April 2009 (UTC)

The first step would be to get rid of the pi in the equation. Introduce a substitution $u=a\pi$ and the cubic reduces to

$20a^{3}-65a^{2}+50a-16=0$. Readro (talk) 19:48, 22 April 2009 (UTC)

Which polynomial is irreducible over the rationals. Algebraist 19:58, 22 April 2009 (UTC)

What about the irrationals? 72.197.202.36 (talk) 04:35, 23 April 2009 (UTC)

$\textstyle 20a^{3}-65a^{2}+50a-16=20(a-a_{1})(a-a_{2})(a-a_{3})\,$ ,

$a_{1}={\frac {R}{60}}+{\frac {245}{12R}}+{\frac {13}{12}}\,$ ,

$a_{2}=-{\frac {R}{120}}-{\frac {245}{24R}}+{\frac {13}{12}}+i{\sqrt {3}}\left({\frac {R}{120}}-{\frac {245}{24R}}\right)\,$ ,

$a_{3}=-{\frac {R}{120}}-{\frac {245}{24R}}+{\frac {13}{12}}-i{\sqrt {3}}\left({\frac {R}{120}}-{\frac {245}{24R}}\right)\,$ ,

$R:={\sqrt[{3}]{68525+300{\sqrt {31749}}}}\,$ .

--78.13.138.117 (talk) 06:53, 23 April 2009 (UTC)

In other words: No, it doesn't factorise nicely! --Tango (talk) 16:50, 23 April 2009 (UTC)

You can see right away that it has at least one positive root. If you find such a root, then u minus that root is a factor of the polynomial. If you divide the polynomial by that factor, you get a quadratic polynomial, and then it's just a matter of solving a quadratic equation. But whether the positive root you find can be expressed "nicely" is another question. If "Tango" has the details right, then it's no nicer than the messiest you could expect under the circumstances. Michael Hardy (talk) 22:07, 23 April 2009 (UTC)

The anon did the calculation (using the cubic formula, by the looks of it), I just concluded that that wasn't "nice" (it clearly doesn't simplify significantly). --Tango (talk) 10:24, 24 April 2009 (UTC)

apparent error in pages on kernel smoothing

Hi, there appears to be an inconsistency in the pages Kernel smoother and Kernel (statistics). The second of these gives the requirement for a K function that $\int _{-\infty }^{+\infty }K(u)du=1\,;$ whereas the Kernel smooth article gives the following as a K function: $D(t)=\left\{{\begin{aligned}&1,\,\,\,if\,\,\left|t\right|\leq 1\\&0,\,\,otherwise\\\end{aligned}}\right.$

I add that the full notation for kernel smoothers, using the K function, is: $K_{h_{\lambda }}(X_{0},X)=D\left({\frac {\left\|X-X_{0}\right\|}{h_{\lambda }(X_{0})}}\right)$

This, however, doesn't seem to make any difference, since I assume the K funtion is to be interpreted as a function of X, not X-nought. In this case, it looks the integral of the D function is 2, which is incompatible with the requirement that it be 1. Am I making a fundamental oversight, and if not, what is the resolution for the inconsistency? Regards, It's been emotional (talk) 23:57, 22 April 2009 (UTC)

I suspect it should have said 1/2 if |t| ≤ 1. Michael Hardy (talk) 04:26, 23 April 2009 (UTC)