Përdoruesi:Armend/nënfaqe

Compart

Gjatësia valore e një vale sinusoidale është perioda hapësinore e saj dmth distanca për të cilën forma e valës përsëritet. Zakonisht konsiderohet largësia në mes dy pikave korresponduese të njëpasnjëshme me fazë të njejtë, siç janë majat e valës, pika më e ulët e valës etj. Ajo shënohet me simbolin llambda (λ) i cili përdoret edhe për shënimin e valëve josinusoidale.

Numri i thjeshtë

Ky artikull në encikloopedinë ruse është zgjedhur ndër artikujt më të mirë prandaj unë vendosa që ta përkthej dhe besoj se përkthimi është i kualitetit të lartë ka disa probleme të vogla me referencat. Unë mendoj se edhe në vikipedinë shqipe meriton të citohet si ndër artikujt më të mirë. Do të kisha dashur që këtë artikull d.m.th përkthimin e tij ta vlerësonte një matematikan profesionist por për fat të keq momentalisht nuk i kemi.--Armend 8 Mars 2009 00:57 (CET)

Merre spell-checker, se po më duket ka disa gabime. Une po e fusë por prap se prap ti kontrollo.--Hipi Zhdripi 8 Mars 2009 02:04 (CET)

polinomiale, pseudorastësishëm - nuk e kap spell çekiqi--Hipi Zhdripi 8 Mars 2009 02:13 (CET)

Ka më se një vit që këtë artikull e kam kandiduar për artikull perfekt por ka kaluar pa u vërejtnga përdoruesit tjerë. Ju lutem votoni dhe komentoni.--Armend 5 Qershor 2010 13:19 (CEST)

IQ testi

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3SmartCubes.com - <a title="IQ Test" href="http://www.3smartcubes.com?iq-score">IQ Test</a> ==S==tampa

Ky përdorues është adhurues i Leonhard Eulerit.

Koordinata polare

Le të jetë dhënë një gjysmëdrejtëzë e orientuar të cilën e quajmë edhe bosht në një rrafsh të caktuar. Origjinën e gjysmëdrejtëzës do ta quajmë pol atëherë pozita e çdo pike të këtij rrafshi mund të përcaktohet me largësinë e saj nga poli dhe me këndin të cilin e formon rrezja e kësaj pike me boshtin polar.

Poli është analog me origjinën e sistemit koordinativ këndrejt, dhe rrezja nga poli me drejtim të fiksuar quhet bosht polar. Largësia r nga poli quhet koordinatë radiale ose rreze, ndërsa këndi $phi$ që rrezja e pikës e formon me boshtin polar quhet koordinatë këndore

Numrat

Në fillim ishin numrat natyral pastaj lindën numrat tjerë.

$1,\;2,\;\ldots$ Numrat natyral $0,\;1,\;-1,\;\ldots$ Numrat e plotë $1,\;-1,\;{\frac {1}{2}},\;{\frac {2}{3}},\;0{,}12,\;\ldots$ Numrat racional $1,\;-1,\;{\frac {1}{2}},\;0{,}12,\;\pi ,\;{\sqrt {2}},\;\ldots$ Numrat real
$-1,\;{\frac {1}{2}},\;0{,}12,\;\pi ,\;3i+2,\;e^{i\pi /3},\;\ldots$	$1,\;i,\;j,\;k,\;\pi j-{\frac {1}{2}}k,\;\dots$
Numrat kompleks	Kuaternionet

Compositions of natural numbers over arithmetic progressions

Denote by

$A_{s}^{p}=\{x:x=si+p,i\in N,p\in I_{s}\}\,$

the set Each m-sequence of natural numbers that fulfill the conditions.

$c_{0}+c_{1}+...c_{m-1}=k\,$, $c_{i}\in A_{s}^{p}\,$

is called composition of natural number k in m parts over $A_{s}^{p}\,$
Denote by $C_{m}(k,A_{s}^{p})\,$ the set of compositions of natural number k in m parts over $A_{s}^{p}\,$

generating function    

      $\sum _{k=0}^{\infty }c_{m}(k,A_{s}^{0}-\{0\})x^{k}=\sum _{c_{i}\in A_{s}^{p}-\{0\},i\in I_{m}}x^{c_{0}+c_{1}+...+c_{m-1}}=\,$

                                         $=\sum _{c_{0}\in A_{s}^{0}-\{0\}}x^{c_{0}}\sum _{c_{1}\in A_{s}^{0}-\{0\}}x^{c_{1}}...\sum _{c_{m-1}\in A_{s}^{0}-\{0\}}x^{c_{m-1}}=\,$

                                         $=\left(\sum _{i=1}^{\infty }x^{si}\right)^{m}=\left({\frac {x^{s+1}}{1-x}}\right)^{m}\,$

now from binomial formula we get

       $\left({\frac {x^{p}}{1-x^{s}}}\right)^{m}=x^{pm}(x^{s})^{-m}=x^{pm}\sum _{i=0}^{\infty }{\binom {-m}{i}}(-x^{s})^{i}=\,$

                                $=\sum _{i=0}^{\infty }{\binom {m+i-1}{i}}x^{pm+si}$

if now substitute $pm+si=k\,$  thent  $i=(k-pm)/s\,$  taking in account that $i\in N\,$ follow

that $k\in \{pm,pm+s,pm+2s,...\},$

trace of sequences

Trace of sequence

Denote by $N=\{0,1,2,...\}\,$ the set of natural numbers and by $I_{m}=\{0,1,...,m-1\}\,$ the set of natural numbers lesser than given natural number m. Lets $c=(c_{0},c_{1},...,c_{m-1})\,$ a m-sequence of natural numbers and $p=max\{c_{0},c_{1},...,c_{m-1}\}\,$ the greatest term of sequence c then the sequence

$t(c)=(t_{0},t_{1},...,t_{p})\,$

where $t_{j},j\in I_{p+1}\,$ denote number of terms of sequence c thats are equal at j, is called trace of c. Is clear that terms of trace fulfills the conditions

$t_{0}+t_{1}+...+t_{p}=m\,$

$t_{1}+2t_{2}+...+pt_{p}=c_{0}+c_{1}+...+c_{m-1}\,$

Denote by

$t^{0}(c)=c\,$

$t^{n}(c)=t(t^{n-1}(c))\,$

1.The set of sequences

$B=\{(1,0,0,1),(2,2),(0,0,2),(2,0,1),(1,1,1),(0,3)\}\,$

that is cycle of length 6 is called bracelet of sequences because for each sequence c from B holds

$t^{6}(c)=c\,$

2.The set of sequences

$R=\{(0,1,1),(1,2)\}\,$

that is cycle of length 2 is called ring of sequences because for each sequence c from R holds

$t^{2}(c)=c\,$

The set

$H=B\ cupR\,$ is called black hole of sequences

Reasons for that name are because I suppose that:

Claim:For each finite sequence $a\,$ of natural numbers exists natural number n such that $t^{n}(a)\in H\,$ in other words each sequence converges to H.

Sequence $a\,$ is of type $B\,$ if its converge to H from B for example sequence (2,3) is of type B because

$t^{3}((2,3))=(0,0,2)\in B\,$

And sequences that converges to H from R are of type R for example the sequence (0) is of type R because

$t^{6}((0))=(1,2)\in R\,$

My question is.Is my assumption true and if it is true how to decide of which type is any given finite sequence of natural numbers, can be done any programme or algorithme. Thanks

This is not a complete solution, but it seems to reduce the problem to an analysis of the cases $m\le2$. I'll write $t^k$ for the $k$-fold composition of $t$ with itself, so that $t^1=t$, $t^2=t\circ t$, etc. I'll also write $|{c}|$ for the number of distinct elements of the $m$-tuple $c=(c_0,c_1,\dots,c_{m-1})\in\mathbb{N}^m$. And I'll write things like $(n,m^k,p)$ as shorthand for $(n,m,m,\dots,m,p)$ where $m$ is repeated $k$ times. A first observation is that $t(c)=t(c')$ for any permutation $c'$ of $c$. Lemma: $|t^2(c)|\geq |c|$ if and only if either $|{c}|=1$, or $|{c}|=2$ and $c=(a,b^{|c|-1})$ (up to permutation) where $a\ne b$. Proof: We have $|t^2(c)|=\max{t_0,\dots,t_p}+1$, so $|t^2(c)|\ge |c|$ if and only if $|t^2(c)|\in {|c|,|c|+1}\iff {|c|-1,|c|}\cap {t_0,\dots,t_p}\ne \emptyset$, which is equivalent to the conditions above. Claim: If $m=|c|\ge 3$ then $|t^n(c)|<|c|$ for some $n\ge1$. Remark: This allows us to reduce to the case $m\in {1,2}$ to establish the claim in the question, if it's true. Proof of the claim: this is a case-by-case analysis. If $|{c}|>2$ then $n=2$ will do by the lemma. If $|{c}|=1$, either $c=(0^m)\implies t(c)=(m)$, or $c=(1^m)\implies t(c)=(0,m)$, or $c=(2^3)\implies t^6(c)=(0,3)$, or $c=(2^m)$ where $m\ge4$, so that $t(c)=(0,0,m)$, or $c=(x^m)$ where $x\ge3$. In this case, $t(c)=(0^x,m)$, $t^2(c)=x,0^{m-1},1), t^3(c)=(m-1,1,0^{x-2},1)$, $t^4(c)=(x-2,2,0^{m-3},1)$. If $m\ge4$ or $x\ge5$ then $|{t^4(c)}|>2$, so $|t^4(c)|=m$ and $|t^6(c)| If $|{c}|=2$ and $c$ is not of the form $(x,y^{m-1})$ (up to permutation) then $|t^2(c)|<|c|$ by the lemma. If $|{c}|=2$ and $c=(x,y^{m-1})$, suppose first that $x 3$ then $|t^4(c)|=\max{m-3,2}+1=\max{m-2,3} y$ then $t(c)$ is a permutation of $t(y,x^{m-1})$, so the previous argument applies. -- mac 12 hours ago Following on from mac’s partial answer, the assumption is true for $m \in {1,2}$, so it appears to be true in general. If $m=1$, $c=(x)$ for some $x \in \mathbb{N}$. If $x>0$, $t(c) = (0^x,1)$, and either $x=1$, in which case $t^2(c) = (0,2), t^3(c) = (1,0,1)$, and $t^4(c) = (1,2) \in R$. If $x=0$, $t(c) = (1)$, and we’re in the first case. Now let $m=2$ and $c = (x,y)$. Observation: If $n>2$, $t^2((n,2)) = (n-1,2)$, while $t((2,2)) = (0,0,2) \in B$, so $t^{1+2(n-2)}((n,2)) = t^{2n-3}((n,2)) \in B$ whenever $n \ge 2$. If $x=y \in {0,1}$, $t(c) = (0^x,2),t^2(c) = (x,0,1)$, and $t^4(c) = (0,1,1) \in R$. If $x=y>1$, $t(c) = (0^x,2),t^2(c) = (x,0,1),t^3(c) = (1,1,0^{x-2},1)$, and $t^4(c) = (x-2,3)$. If $x=y=2$, $t^4(c) \in B$; if $x=y \in {3,4}$, $t^6(c) = (2,2) \in B$; and if $x=y=5$, $t^{10}(c) = (2,2) \in B$. If $x=y>5$, $t^5(c) = (0,0,0,1,0^{x-6},1)$, and $t^6(c) = (x-3,2)$, where $x-3>2$, whence $t^{6+2(x-3)-3}(c) = t^{2x-3}(c) \in B$ by the Observation. Now assume without loss of generality that $x Observation ensures that $t^{2+2(y-1)-3}(c) = t^{2y-3}(c) \in B$. -- Brian M. Scott 8 hours ago