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-(II)

SQUARE FRACTIONAL PART DEFINITE INTEGRAL:

For any $latex \displaystyle x\geq 0$:

$latex \displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx = \int_0^{\lfloor x \rfloor} \left\{{x}\right\}^2 dx+ \int_{{\lfloor x \rfloor}}^ {x} \left\{{x}\right\}^2 dx $

The function $latex \displaystyle \left\{{x}\right\}^2$ has a periodical behaviour, so the limits can be reduced to the first period:

$latex \displaystyle I_1= \lfloor x \rfloor \int_0^{1} {\left\{{x}\right\}^2 dx}+ \int_0^{\left\{{x}\right\}}{ \left\{{x}\right\}^2 dx} = \lfloor x \rfloor \int_0^1 x^2 dx+ \int_0^{\left\{{x}\right\}} {x^2 dx}$

$latex \displaystyle I_1= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^3}{3} $

$latex \displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx= \frac{x+ \left\{{x}\right\}^{3}-\left\{{x}\right\}}{3}$

POWER FRACTIONAL PART DEFINITE INTEGRAL:

The same formula can be extended to any given positive integer power:

$latex \displaystyle I_2= \int_0^x \left\{{x}\right\}^n dx= \frac{x+ \left\{{x}\right\}^{n+1}-\left\{{x}\right\}}{n+1}$

$latex \displaystyle I_2= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^{n+1}}{n+1} $

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-Notes

The final result, in the preceeding post, can not be derived from a telescoping series [3], if $latex \displaystyle k$ is not integer (See comments at reference [1]).

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

This lack of generality, can be avoided, if we consider a more general definition for the Harmonic Numbers [4], extended to the complex plane, using the function:

$latex \displaystyle H_z=\gamma+\psi_0(z+1) $


Where $latex \displaystyle \psi_0 \;$ is the so called digamma function, and $latex \displaystyle \;\gamma\;$ is the Euler-Mascheroni constant.

If you take a look at the expresion (15), in the reference [2] : We can find that one asymptotic expansion for the digamma function is:

$latex \displaystyle \psi_0(k+1)=-\gamma+\sum_{n=1}^\infty{\frac{k}{n(n+k)} $

This is why the Polygonal Numbers Series sum is working:

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

$latex \displaystyle \frac{H_k}{k}=\sum_{n=1}^\infty{\frac{1}{n(n+k)}=\frac{\gamma+\psi_0(k+1)}{k} $

And the polygonal numbers infinite sum, can be expressed (if $latex \displaystyle \;s\neq4\;$) as:

$latex \displaystyle S_{\infty}(s)=\frac{2}{4-s}*(\gamma+\psi_{0}\left(\frac{2}{s-2}\right)) $

This expresion works for all $latex \displaystyle s>2$, as well as for all nonreal $latex \displaystyle s$, It also works for all $latex \displaystyle s<2$, except if $latex \displaystyle s<2$, and $latex \displaystyle s$ is $latex \displaystyle \;\;0, 1, 4/3, 6/4, 8/5, 10/6, ... \;$, because $latex \displaystyle \;\psi_0\;$ is not defined for negative integers (See reference) [1]

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References:

[1]-Charles R Greathouse IV - Comments @ My Math Forum Inverse Polygonal Series
[2]-Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Sondow, Jonathan and Weisstein, Eric W. "Harmonic Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HarmonicNumber.html
[5]-Photo Martin Gardner, Mathematical Games, Scientific American, 211(5):126-133, taken from http://bit-player.org/2007/hung-over

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