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.

SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM

This post continues with the work on some special sets of integers related series:

The square pyramidal numbers expression can be found on [1] and it is:

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

The Square pyramidal numbers reciprocal finite serie is:

$latex \displaystyle S(n)=\sum_{k=1}^n{\frac{6}{k(k+1)(2k+1)}}$

Splitting the main fraction into others:

$latex \displaystyle \frac{S(n)}{6}=\sum_{k=1}^n{\frac{1}{k(k+1)(2k+1)}}=\sum_{k=1}^n{\frac{A}{k}+\frac{B}{k+1}+\frac{C}{2k+1}}$

Solving the equations gives:

$latex \displaystyle A=1 ;\; B=1;\; C=-4;$

Substituing the series (ii), and the expression for the Harmonic Numbers (i):

$latex \displaystyle \frac{S(n)}{6}= \sum_{k=1}^n {\frac{1}{k}}+\sum_{k=1}^n{\frac{1}{k+1}}-4\sum_{k=1}^n{\frac{1}{2k+1}}=2H_n-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$

Taking into account the expressions: (i),(iv) and (vi), for the digamma function:

$latex \displaystyle \frac{S(n)}{6}= 2\gamma+2\psi_{0}(n+1)-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$

$latex \displaystyle S(n)=6[ 2\gamma+2\psi_{0}(n+1) -\frac{n}{n+1}-2[\psi_{0}(n+\frac{3}{2})+\gamma+2\log{2}-2]]$

$latex \displaystyle S(n)=6[2\psi_{0}(n+1)-2\psi_{0}(n+\frac{3}{2})-\frac{n}{n+1} -4\log{2}+4]$

Calculating the limit, and using (vii):

$latex \displaystyle S(\infty)= \lim_{x \to{+}\infty}{S(n)}= 6(3-4\log{2})=18-24\log{2}\approx 1.3644676665...$

(see reference [4])

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NOTES:

(i) Harmonic Numbers and Digamma Function at integer values:

$latex \displaystyle H_{n}=\sum_{k=1}^n {\frac{1}{k}}=\gamma+\psi_{0}(n+1)$

(ii) Changing series into Harmonic Numbers:

$latex \displaystyle \sum_{k=1}^n{\frac{1}{k+1}} =\sum_{k=1}^n{\frac{1}{k}}-1+\frac{1}{n+1}=H_{n}-\frac{n}{n+1}$

(iii) Changing one series into another:

$latex \displaystyle \sum_{k=1}^n{\frac{1}{2k-1}} -\sum_{k=1}^n{\frac{1}{2k+1}}=1-\frac{1}{2n+1}$

(iv) Digamma function at half-integer values:

$latex \displaystyle \psi_{0}(n+\frac{1}{2})= -\gamma-2\log{2}+2\sum_{k=1}^n{\frac{1}{2k-1}}$

(v) A Digamma function property:

$latex \displaystyle \psi_{0}(z+1)= \psi_{0}(z)+\frac{1}{z}$

(vi) Another expression for Digamma function at half-integer values:

Using (iii), (iv) and (v):

$latex \displaystyle \psi_{0}(n+\frac{3}{2}) = \psi_{0}(n+\frac{1}{2}) +\frac{1}{n+\frac{1}{2}}=\psi_{0}(n+\frac{1}{2})+\frac{2}{2n+1}=$
$latex = -\gamma-2\log{2} +2(1-\frac{1}{2n+1}+\sum_{k=1}^n{\frac{1}{2k+1}} )= -\gamma-2\log{2}+2+2\sum_{k=1}^n{\frac{1}{2k+1}} $

(vii) Limit for Digamma function at half-integer values:

$latex \displaystyle L_{1}(\infty)=\lim_{x \to{+}\infty}{L_{1}(n)}= \lim_{x \to{+}\infty}{[\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})] }$

$latex \displaystyle L_{1}(n)=\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})=2\log{2} + \sum_{k=1}^n {\frac{1}{k}}-2 \sum_{k=1}^n {\frac{1}{2k-1}}$

$latex \displaystyle L_{1}(n)=2\log{2}-\sum_{k=1}^n {\frac{1}{k(2k-1)}}$

The last serie limit can be derived from the Mercator-Mengoli infinite series for $latex \displaystyle \log{2}\;$. See [3].
This proof is interesting enough for another entry on the blog.

$latex \displaystyle \log{2}=\sum_{k=1}^\infty {\frac{1}{2k(2k-1)}}$

So:

$latex \displaystyle L_{1}(\infty)=0$

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References:
[1]-Square Pyramidal Numbers @ Wikipedia Square Pyramidal Numbers
[2]-Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Collection of formulae for log 2-Numbers Constants and Computation-Xavier Gourdon and Pascal Sebah.
[4]-A159354-Decimal expansion of 18-24*log(2) The On-Line Encyclopedia of Integer Sequences!
[5]-A000330-Square pyramidal numbers The On-Line Encyclopedia of Integer Sequences!

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Archives:
[a]-041109-SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM.nb

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