INTEGRATING ROUNDING FUNCTIONS-(III)

SQUARE FLOOR FUNCTION DEFINITE INTEGRAL:

$latex \displaystyle I_3= \int_0^x \lfloor x \rfloor^2 dx = \int_0^{\lfloor x \rfloor} \lfloor x \rfloor^2 dx+ \int_{{\lfloor x \rfloor}}^ {x} \lfloor x \rfloor^2 dx $

$latex \displaystyle I_3=\sum_{k=1}^{ \lfloor x \rfloor-1}{k^2} + \left\{{x}\right\}\lfloor x \rfloor^2 $

$latex \displaystyle I_3=P(\lfloor x \rfloor-1) + \left\{{x}\right\}\lfloor x \rfloor^2 $

Where: $latex \displaystyle P(n)$ gives the n-th Square Pyramidal Number.

$latex \displaystyle P(n) =\frac{n(n+1)(2n+1)}{6} $

POWER FLOOR FUNCTION DEFINITE INTEGRAL:

$latex \displaystyle I_4= \int_0^x \lfloor x \rfloor^n dx = \sum_{k=1}^{ \lfloor x \rfloor-1}{k^n} + \left\{{x}\right\}\lfloor x \rfloor^n$

$latex \displaystyle S(n,m)=\sum_{k=1}^{m}{k^n} \;\;\;\;$ is the Faulhaber's formula.

If $latex \displaystyle n=1$ , the formula gives the Triangular Numbers.

And if $latex \displaystyle n=2$ , the formula gives the Square Pyramidal Numbers.

INTEGRATING ROUNDING FUNCTIONS-(I)

Integer Rounding Functions can be found in many Number Theory texts, but I wasn´t able to find something about its integrals.

The following expressions can be derived just from their plots, adding and subtracting areas. They hold if $latex \displaystyle \;x\geq 0$

The Triangular Numbers function is used to get shorter expressions.

$latex \displaystyle T(n)=\frac{n^{2}+n}{2} $


FLOOR FUNCTION DEFINITE INTEGRAL:

$latex \displaystyle \int_0^x \lfloor x \rfloor dx = \left\{{x}\right\}\lfloor x \rfloor +T(\lfloor x \rfloor-1)$

FRACTIONAL PART FUNCTION DEFINITE INTEGRAL:

$latex \displaystyle \int_0^x \left\{{x}\right\} dx = \frac{x}{2}- \left\{{x}\right\} +T( \left\{{x}\right\} ) $

$latex \displaystyle \int_0^x \left\{{x}\right\} dx = \frac{x}{2} +T( \left\{{x}\right\}-1 ) $

$latex \displaystyle \int_0^x \left\{{x}\right\} dx =\frac{x+ \left\{{x}\right\}^{2}-\left\{{x}\right\}}{2}$

CEILING FUNCTION DEFINITE INTEGRAL:

$latex \displaystyle \int_0^x \lceil x \rceil dx = \lceil x \rceil (x-\lceil x \rceil)+T(\lceil x \rceil)$

All these formulas can be changed using the relations between them:

$latex \displaystyle \left\{{x}\right\}=x- \lfloor x \rfloor $

This topic doesn't finish here it's going to be used on incoming posts.

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Archives:

[a]-032209-INTEGRATING INTEGER ROUNDING FUNCTIONS-(I).nb
[b]-032309-CheckIntegratingIntegerRoundingFunctions.pdf

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References:

[1]-Štefan Porubský:Retrieved 2009/3/22 from Interactive Information Portal for Algorithmic Mathematics, Institute of Computer Science of the Czech Academy of Sciences, Prague, Czech Republic http://www.cs.cas.cz/portal/AlgoMath/NumberTheory/ArithmeticFunctions/IntegerRoundingFunctions.htm
[2]-Greg Gamble: The University of Western Australia SCHOOL OF MATHEMATICS & STATISTICS The Floor or Integer Part function

INVERSE POLYGONAL NUMBERS SERIES

Let there be $latex \displaystyle P(n,s)$ the n-th Polygonal Number [4] of s sides, then:

$latex \displaystyle P(n,s)=\frac{ (s-2)*n^{2}-(s-4)*n }{2}$

For s=3, we get the formula for Triangular Numbers

$latex \displaystyle P(n,3)=\frac{n^{2}+n}{2}=T_n$

and with s=4 then we get the Squares:

$latex \displaystyle P(n,4)=n^{2}$

Polygonal numbers hold for the next identity:

$latex \displaystyle P(n,s+1)=P(n,s)+ P(n-1,3)=P(n,s)+ T_{n-1}$

The sum of the inverse of the first k polygonal numbers with side s, is:

$latex \displaystyle S_{k}(s)=\sum_{n=1}^k{} \frac{1}{P(n,s)}$

And its infinite series sum:
$latex \displaystyle S_{\infty}(s)=\sum_{n=1}^{\infty{}} \frac{1}{P(n,s)}$

Than this infinite series is convergent, can be proved, using some convergence test:

Raabe's convergence Test:

$latex \displaystyle \rho=\displaystyle\lim_{n \to{+}\infty}{n*\left(\frac{P(n+1,s)}{P(n,s)}-1\right) }=2>1$

A series with a lower number of sides, upper bounds, series with higher sides: So the convergence in triangular numbers, implies the convergence of the remaining polygonal numbers series:

If $latex \displaystyle s_1>s_2 \longrightarrow{} \frac{1}{P(n,s_1)}<\frac{1}{P(n,s_2)} \longrightarrow{} S_{\infty}(s_1)$

The inverse series with Triangular numbers is a telescoping series [3]:

$latex \displaystyle S_{\infty}(3)=2\cdot\sum_{n=1}^{\infty{}} \frac{1}{n\cdot(n+1)}= 2$

The squares sum is the so called Basel Problem, [1] [2], related with the Riemann Zeta Function:

$latex \displaystyle S_{\infty}(4)=\sum_{n=1}^{\infty{}} \frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$

If $latex \displaystyle s\neq4$, then:

The series is a more general case of a telescoping series [3], related with the Harmonic Numbers.

$latex \displaystyle S_{\infty}(s)=2\cdot \sum_{n=1}^{\infty{}} \frac{1}{(s-2)\cdot n^{2}-(s-4)*n}=\frac{2}{s-2}\cdot\sum_{n=1}^{\infty{}}\frac{1}{n\cdot(n+\frac{4-s}{s-2})}$

$latex \displaystyle \sum_{n=1}^{\infty{}} \frac{1}{n*(n+k)}=\frac{H_k}{k} $

$latex \displaystyle S_{\infty}(s)= \frac{2}{4-s}\cdot H_{\frac{4-s}{s-2}}$

Example: In the hexagonal numbers case:

$latex \displaystyle S_{\infty}(6)= -H_{-\frac{1}{2}} \approx 1.38629$

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Archives:

[a]-031009-INVERSE POLYGONAL NUMBERS SERIES.nb

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References:

[1]-Estimating Basel Problem@ MAA Online How Euler did it, Ed.Sandifer
[2]-Basel Problem @ Wikipedia Basel Problem
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Polygonal Numbers @ Wikipedia Polygonal Number

SUMS INSIDE POWER SET

Let's consider the set of integers less or equal than a given one:

$latex \displaystyle A=\{1,2,3,...,n\}\:\:n\in\mathbb{N} $

If we add all the elements in this set, we have:

$latex \displaystyle S=\sum_{i\in A}{i}=\sum_{i=1}^n{i}=\frac{n^2+n}{2} $

If we look, at the figure, we can see how the sum ,is equal to the area below the "ladder" plotted:

$latex \displaystyle S=S_{(i)}+S_{(ii)} $

Where $latex S_{(i)} $ is the area formed for $latex n$ small triangles, with a $latex \displaystyle \frac{1}{2}$ area, each one; And $latex S_{(ii)} $ is the area of an isosceles triangle with an equal base and height of $latex n$.

$latex \displaystyle S_{(i)}=\frac{1}{2}\cdot n;\quad S_{(ii)}=\frac{1}{2}\cdot n \cdot n;$

And if the aim, of this blog, where to be rigurous instead of, to give simple ideas, an induction proof, should fit here, perfectly [1].
To extend this result, to the sum of all the elements, in the subsets included, in the Power Set of integers less or equal to a given one:

$latex \displaystyle P(A) = \{X:X\subseteq A=\{1,2,3,...,n\}\}$

There are $latex 2^n$ subsets in $latex \displaystyle P(A) $, (see [3] and [4]):

$latex \displaystyle |P(A)| = 2^{|A|} =2^n$

If $latex \displaystyle i\in A$, then this element is in half of the subsets of $latex \displaystyle P(A)$ (observe the relation between the binary digits and the power set [3]): $latex \displaystyle 2^{n-1}$
To get the final result, it is only necessary to multiply both expresions [2]:

$latex \displaystyle h(n)=\sum_{i\in X\subseteq P(A)}{i}= \frac{n^2+n}{2} \cdot 2^{n-1}=(n^2+n) \cdot 2^{n-2}$

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NOTE:

$latex \displaystyle h(n)=A001788(n)$

$latex \displaystyle\lim_{x \to{+}\infty}{\frac{h(n+1)}{h(n)}}\displaystyle\lim_{x \to{+}\infty}{2+\frac{4}{x}}=2 $

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Archives:

[a]-020609-SUMSINSIDEPOWERSET.nb

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References:

[1]-A000217 The On-Line Encyclopedia of Integer Sequences!
[2]-A001788 The On-Line Encyclopedia of Integer Sequences!
[3]-Powerset Construction Algorithm Shriphani Palakodety.
[4]-Course Notes 8: Size of the Power Set. Chris Nowlin.