Wikipedia:Reference desk/Archives/Mathematics/2016 May 8

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May 8

statistical difference

Hi all,
It says here (page 99) that if we have a 2-universal hash functions family $H$, mapping $\{0,1\}^{n}$ to $\{0,1\}^{m}$ and $T\subseteq \{0,1\}^{n}$, then the following distributions have statistical difference of $(2^{m}/|T|)^{\Omega (1)}$:

1. Choose $h$ uniformly from $H$, and then choose $y$ uniformly from $T\cap h^{-1}(0)$. Output $(h,y)$.
2. Choose $y$ uniformly from $T$, and then choose $h$ uniformly from $\{h\in H|h(y)=0\}$. Output $(h,y)$.

I couldn't find any explanation why is it guaranteed to have this statistical difference. Can someone please explain it to me?
Thanks a lot! — Preceding unsigned comment added by 109.186.49.82 (talk) 21:41, 8 May 2016 (UTC)

Rotation

I have a 2-sphere and I want to define 2 rectangular areas on it. They are "spherical rectangles" with borders that are some meridians and parallels. Sorry for not using Greek letters all the way through-they refused to take subscripts, superscripts. Let angle θ <t> be the inclination (polar) angle, and angle φ <f> the azimuthal angle. I will use Greek and Latin letters interchangeably. I need three angles t: $0<t_{1}<t_{2}<t_{3}<90^{o}$ and two angles f: $0<f_{1}<f_{2}<360^{o}$ It is clear they define two contiguous (adjacent) spherical rectangles. The rectangles touch each other on $t_{2}$ parallel. A function f(θ,φ) is defined on the area that is the sum of both rectangles. The portions of the function f(θ,φ) on both rectangles are different. Let $f_{1}$(θ,φ) be the portion of the function f(θ,φ) on one rectangle and $f_{2}$(θ,φ) the corresponding portion on the other rectangle. A basis of fully normalized Spherical Functions

$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$

(1)

is defined on the whole 2-sphere but I will consider only the portion that is covered by the above two rectangles. Each function $f_{1}$(θ,φ), $f_{2}$(θ,φ) and $f$(θ,φ) will be expressed as

${\mathfrak {f}}_{1}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(2)

${\mathfrak {f}}_{2}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(3)

and

${\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(4)

I then fix one particular index ${l}$ which defines a subspace in the functional Hilbert space and compute these expressions (asterisk marks complex conjugate):

${\mathfrak {I}}_{1}={\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}f_{l}^{m*}$

(5)

${\mathfrak {I}}_{2}={\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}g_{l}^{m*}$

(6)

${\mathfrak {I}}={\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}q_{l}^{m*}$

(7)

It is very important for me to know if additivity is preserved and

${\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}$

(8)

Thank you. --AboutFace 22 (talk) 22:54, 8 May 2016 :(UTC)

In order to expand your function into spherical harmonics you need to know it on the whole sphere. Ruslik_Zero 03:10, 9 May 2016 (UTC)

NO, I don't need to know if it is on the whole sphere. It is NOT on the whole sphere. It is a locally defined function and when the coefficients are computed integration has local limits. It can be assumed that the function is defined on the two rectangles but on the rest of the sphere the amplitude of the function is zero. --AboutFace 22 (talk) 12:25, 9 May 2016 (UTC)

I think Ruslik's point is valid, but you can get around this by assuming the function is 0 outside the domain you're interested in. Otherwise, while the expansion may exist, it won't be unique. As a simple example using Fourier series instead of spherical harmonics, you can expand the function f(x)=1 on the interval [0, π] as a sin series and get a square wave, or you can expand it as simply 1. If you are assuming 0 outside the two rectangles then yes, the expansions of the two separately will match what you get for their union. But the two functions you're talking about have to match on the common boundary, otherwise it the function f isn't defined on the whole rectangle. Presumably if the functions did not agree then the series would converge to the average of the two values, I know this is true at least for Fourier series. It should be noted that, unless your functions are 0 on the boundaries, convergence will be slow and not uniform, so the series may not be practical. --RDBury (talk) 18:29, 9 May 2016 (UTC)

Expansion in spherical Bessel functions would be more appropriate for domains of this kind. Alternatively, you can conformally map to the disk via a Schwarz-Christoffel map and take a Bessel expansion in the disk. Sławomir Biały (talk) 18:34, 9 May 2016 (UTC)

Thank you, @RDBury and @Sławomir Biały. Yes it is a physical situation, function $f$ is a well behaved function with no singularities anywhere and it is smooth and the two parts $f_{1}$ and $f_{2}$ match on their common boundary. It is possible that there is an "abrupt drop" at the outside boundary of the combination of two rectangles $f$. I am sorry, I don't understand your (@RDBury) sentence "yes, the expansions of the two separately will match what you get for their union." I am sorry, I would appreciate if you confirm the expression (8). I am sorry about it. I need yes or no. Otherwise I feel your sentence is ambiguous.

I just meant to say additivity is preserved, since you're assuming the function 0 outside the given domains. Not sure about ${\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}$, but f₁ and f₂ are orthogonal so I would think so. --RDBury (talk) 22:36, 9 May 2016 (UTC)

@Sławomir Biały, Thank you for the suggestion about Bessel Functions. I have thought about it before but now I need what I posted about. --AboutFace 22 (talk) 21:56, 9 May 2016 (UTC)

@RDBury, thank you for the confirmation. --AboutFace 22 (talk) 22:48, 9 May 2016 (UTC)