BETA FUNCTION MATRIX DETERMINANT

If $latex A_n$ is an square matrix of order $latex n$, whose elements are defined as:

$latex \displaystyle a_{i,j}=\frac{1}{\beta(i,j)}$, where $latex \beta$ is the beta function. Then:

$latex |A_{n}|=n!$

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Proof:

If we use Cholesky method, we can decompose this matrix as:

$latex A_{n}=U_{n}^{T}*U_n$

Where $latex U_n$ is an upper triangular matrix.

But instead of applying the algorithm to a generic case, we are going to propose a factorization, and after we will check that this decomposition generates the appropriate matrix. This mean to use software to speed up the proof, like:

$latex \displaystyle A_{5}=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 6 & 12 & 20 & 30 \\ 3 & 12 & 30 & 60 & 105 \\ 4 & 20 & 60 & 140 & 280 \\ 5 & 30 & 105 & 280 & 630 \end{array} \right)= U_{n}^{T}* \left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 0 & \sqrt{2} & 3 \sqrt{2} & 6 \sqrt{2} & 10 \sqrt{2} \\ 0 & 0 & \sqrt{3} & 4 \sqrt{3} & 10 \sqrt{3} \\ 0 & 0 & 0 & 2 & 10 \\ 0 & 0 & 0 & 0 & \sqrt{5} \end{array} \right)$

The elements of $latex U_n$, seems to be:

$latex \displaystyle u_{i,j}=\sqrt{i}\cdot \binom{j}{i}$

$latex U_{n}^T$ is the transpose of $latex U_n$, so:

$latex \displaystyle u_{i,j}^{*}=u_{j,i}=\sqrt{j} \cdot \binom{i}{j}$

If we multiply both triangular matrices:

$latex \displaystyle a_{i,j}=\sum_{r=1}^n{u_{i,r}^{*} \cdot u_{r,j}} =\sum_{r=1}^n{r \cdot \binom{i}{r} \binom{j}{r}}=\sum_{r=1}^{min(i,j)}{r \cdot \binom{i}{r} \binom{j}{r}}$

This last binomial expression, can be added to a closed form, equal to the reciprocal of beta function, (See proof on reference [1])

$latex \displaystyle a_{i,j}=i \cdot \binom{i+j-1}{j-1}=\frac{1}{\beta(i,j)}$

Now, that we have found this decomposition, for $latex A_n$, then it is unique and $latex A_n$ is Hermitian and positive definite.

$latex \displaystyle |A_{n}|=|U_{n}^{T}*U_{n}|=|U_{n}|^{2}=\left(\prod_{i=1}^{n}{u_{i,i}}\right)^{2}=\prod_{i=1}^{n}{u_{i,i}^2}=\prod_{i=1}^{n}{\left(\sqrt{i}\cdot\binom{i}{i}\right)^2}=\prod_{i=1}^{n}{i}=n!$

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Archives:[a]-092109-Beta Determinant.nb

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References:
[1]-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - page 181 Problem 5
[2]-A000142-Factorial numbers: n! The On-Line Encyclopedia of Integer Sequences!

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CURIOUS SERIES-002

This is the Dirichlet convolution of $latex \displaystyle \omega*1$:

$latex \displaystyle a(n)=\sum_{d|n}{\omega(d)}$

Where $latex \displaystyle \omega$ is the number of distinct prime factors function.

This function, $latex \displaystyle \omega(n)$ is an additive function:

$latex \displaystyle \omega(n)=\omega(d \cdot n/d ) \leq \omega(d)+\omega(n/d)$

Where the equal symbol holds iff $latex GCD(n/d,d)=1$

If we sum over all divisors:

$latex \displaystyle \sum_{d|n}{\omega(n)} \leq \sum_{d|n}{\bigg(\omega(d)+\omega(n/d)\bigg)}=\sum_{d|n}{\omega(d)}+\sum_{d|n}{\omega(n/d)}=2 \sum_{d|n}{\omega(d)} $

$latex \displaystyle \omega(n)\cdot \sum_{d|n}{1}=\omega(n)\cdot\tau_{2}(n) \leq 2 \sum_{d|n}{\omega(d)}=2\cdot a(n)$

If $latex s$ is a squarefree number and if $latex d|s$ then $latex GCD(d,d/s)=1$ and, the number of divisors , $latex \displaystyle \tau_{2}(s)=2^{\omega(s)}$, then:

$latex \displaystyle a(s)=\omega(s)\cdot 2^{\omega(s)-1}$

For every number $latex n>1$:

$latex \displaystyle a(n)=\sum_{d|n}{\omega(d)} \leq \sum_{d|n \; d\neq 1}{\omega(n)}= \omega(n) \cdot \sum_{d|n \; d\neq 1}{1}= \omega(n)\cdot(\tau_{2}(n)-1) $

$latex \displaystyle \omega(n)\cdot(\tau_{2}(n)-1) \geq a(n) \geq \bigg\lceil \frac{\omega(n)\cdot \tau_{2}(n)}{2} \bigg\rceil$

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NOTE: Sequences in OEIS:

$latex a(n)=A062799(n)$

$latex \omega(n)=A001221(n)$

$latex \tau_{2}(n)=A000005(n)$

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References:

[1]-A062799-Inverse Moebius transform of A001221, the number of distinct prime factors of n The On-Line Encyclopedia of Integer Sequences!
[2]-A001221-Number of distinct primes dividing n (also called omega(n)). The On-Line Encyclopedia of Integer Sequences!
[3]-A000005-d(n) (also called tau(n) or sigma_0(n)), the number of divisors of n.. The On-Line Encyclopedia of Integer Sequences!

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SPLITTING FTA FUNCTIONS (I)

The Fundamental Theorem of Arithmetic (FTA) grants every natural number, $latex n>1$, a unique factorization of the form:

$latex \displaystyle n=p_1^{\alpha_1}p_2^{\alpha_2}...p_{\omega(n)}^{\alpha_{\omega(n)}} = \prod_{i=1}^{\omega(n)} p_i^{\alpha_i}$

Where $latex \omega(n)$ is the number of distinct prime factors of n.

The arithmetical functions can be evaluated once the factorization of $latex n$ is known, (although there are many of them that can be calculated without factorization)

In fact, the only way to "express some arithmetical property of $latex n$" is that the function, must be dependant on the primes, $latex p_i$ and (or) on the coefficients, $latex \alpha_i$

So the arithmetical functions can be classified, in a psychedelic and unorthodox way of course, in:

1) Functions that depend only on coefficients.

2) Functions that depend only on primes.

3) Functions that depend both on primes and coefficients.

This classification, mathematically speaking, seems to be useless, but it is only an alternative to the alphabetical order, when it comes to deal with this topic.

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References:

[1]-D. Joyner, R. Kreminski, J. Turisco @ Applied Abstract Algebra The Fundamental Theorem of Arithmetic
[2]-Arithmetic Function @ Wikipedia Arithmetic Function
[3]-Prime Factor @ Wikipedia Prime Factor

COROLLARY TO EULER´S COROLLARY

It is sure that, after a very short fraction of a second, the first time Euler saw (despite he was one-eyed and partially blind) Gauss´s Gamma Function Multiplication Formula:

$latex \displaystyle \prod _{k=0}^{n-1} \Gamma \left(\frac{k}{n}+z\right)=(2 \pi )^{\frac{n-1}{2}} n^{\frac{1}{2}-n z}\Gamma (n z)$

Euler tested the expresion with $latex \displaystyle z=\frac{1}{n}$ to get his corollary:

$latex \displaystyle \prod _{k=1}^{n-1} \Gamma \left(\frac{k}{n}\right)=\frac{(2 \pi )^{\frac{n-1}{2}}}{\sqrt{n}}$

Or maybe was Gauss who generalized, Legendre´s Gamma Duplication Formula with Euler´s ideas, I haven´t found anything about the real history.

Anyway, if we multiply Euler´s corollary by the Gamma Formula with $latex \displaystyle z=1$, and if we practice the "good habit" of multiplying things by $latex 1$:

$latex \displaystyle \Gamma \left(\frac{n}{n}\right)=\Gamma(1)=\Gamma \left(\frac{2n}{n}\right)=\Gamma(2)=1$

Then we get:

$latex \displaystyle \prod_{k=1}^{2n} \Gamma \left(\frac{k}{n}\right)=\frac{(2 \pi )^{n-1}}{n^n}\Gamma(n)$

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References:

[1]-Xavier Gourdon and Pascal Sebah, Introduction to the Gamma Function
[2]-Gamma Function @ Wikipedia Gamma Function

CURIOUS SERIES-001

There´s a very common finite series, that use to be, at the begining on every book:

$latex \displaystyle S_{n}(z)=\sum _{k=0}^n z^k =\frac{z^{n+1}-1}{z-1}$

Where $latex z$ can be real or complex.

There is a, very well known, particular case of this series where $latex z=2$:

$latex \displaystyle S_{n-1}(2)=\sum _{k=0}^{n-1} 2^{k} =2^{n}-1=M_n$

$latex \displaystyle M_n$ are the Mersenne Numbers, and due to this sum, is easy to see that the Mersenne numbers consist of all 1s in base-2 (they are base 2 repunits)

But this entry is about another particular case of this finite sum:

$latex \displaystyle S_{n}(\textbf{i})=\sum _{k=0}^{n} \textbf{i}^k$

Where: $latex \textbf{i}=\sqrt{-1}$, is the complex unit.

$latex \displaystyle S_{n}( \textbf{i} ) =\frac{1}{2}(1+\textbf{i}) \left(1-\textbf{i}^{n+1}\right)$

This sum shows periodical behaviour with a period of $latex 4$, and its values changes from one vertex to another in a square of side equal to $latex 1$, if we plot them in the complex plane:

$latex \displaystyle S_{n}( \textbf{i} )=\{1,1+\textbf{i},\textbf{i},0,1,1+\textbf{i},\textbf{i},...\}$

$latex \displaystyle S_{n}(\textbf{i})=1\;$ if $latex \;n\equiv 0\;mod\;4$

$latex \displaystyle S_{n}(\textbf{i})=1+\textbf{i}\;$ if $latex \;n\equiv 1\;mod\;4$

$latex \displaystyle S_{n}(\textbf{i})=\textbf{i}\;$ if $latex \;n\equiv 2\;mod\;4$

$latex \displaystyle S_{n}(\textbf{i})=0\;$ if $latex \;n\equiv 3\;mod\;4$

If we take a look at the real part of the complex number $latex S_{n}(\textbf{i})$:

$latex \displaystyle Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)=\{1,1,0,0,1,1,0,0,...\}$

Then we had just found the sequence A133872 from OEIS, and then we can construct another expressions for this sequence, and also for the problem series:

$latex \displaystyle A133872(n)=Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)$

$latex \displaystyle A133872(n)=\frac{1}{2}\bigg(\sum _{k=0}^{n} \textbf{i}^k + \sum _{k=0}^{n} \textbf{i}^{-k}\bigg)$

$latex \displaystyle A133872(n)=\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)$

Then, if we expand to trigonometrical functions $latex S_{n}( \textbf{i} )$:

$latex \displaystyle S_{n}( \textbf{i} ) =\left(\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right) + \textbf{i} \left( \frac{1}{2}-\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right)$

And finally using the information inside OEIS:

$latex \displaystyle S_{n}( \textbf{i} )= \text{mod}\left(\bigg\lfloor\frac{n+2}{2}\bigg\rfloor,2\right)+ \textbf{i}\cdot \text{mod}\left(\bigg\lfloor\frac{n+1}{2}\bigg\rfloor,2\right)\cdot \textbf{i}$

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References:[1]-A133872-Period 4: repeat 1,1,0,0. The On-Line Encyclopedia of Integer Sequences!

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SUMMERTIME CHANGES

Now that the Fish are jumpin' and the cotton is high: the $latex \LaTeX$ code renderer this blog was using, is not working in the same way it was.

I´ve discovered today, that some older posts with tables and floor function symbols, are not been displayed properly. As far as I know, the server from http://yourequations.com/ has exceeded its files quota, and they have made some changes that are causing this kind of problems.

So I´ve added a new script that provides very good quality $latex \LaTeX$ rendering:

http://www.watchmath.com

with the added advantage that the code is easier to be included inside any page.

I´ve also updated my Mathematica´s version from 5.2 to 6.0 with some files I´ve found "somewhere" on the internet.

INTEGRATING ROUNDING FUNCTIONS (IV)

FLOOR AND INTEGER PART PRODUCT DEFINITE INTEGRAL:
$latex \displaystyle I_4= \int_0^x \lfloor x \rfloor \left\{x\right\} dx $ $latex \displaystyle x^2=(\lfloor x \rfloor + \left\{x\right\})^2 = {\lfloor x \rfloor}^{2} +2 \lfloor x \rfloor \left\{x\right\} + \left\{x\right\}^2 $
$latex \displaystyle \lfloor x \rfloor \left\{x\right\}=\frac{1}{2}(x^2 - \left\{x\right\}^2 - {\lfloor x \rfloor}^{2} )$
$latex \displaystyle I_4=\frac{1}{2} ( \frac{x^3}{3} - I_3 - I_1)$
$latex \displaystyle I_4=\frac{1}{2} ( \lfloor x \rfloor \left\{x\right\}^2 +\frac{{\lfloor x \rfloor}^{2}-\lfloor x \rfloor}{2}) $

The same result can be derived just adding the areas under the curve.