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  • Filipe | planetmath.org
    ergodic theorem Kolmogorov s extension theorem Landau Ramanujan Constant Maximal ergodic theorem Multiple Recurrence Theorem multiplicative cocycle Oseledets multiplicative ergodic theorem Proof of Dulac s Criteria Proof of Fekete s subadditive lemma Proof that every absolutely convergent series is unconditionally convergent Structural stability theorem Von Neumann s ergodic theorem Buddy list actions Create Buddy List Designate Existing Team Member of World Writable Forename Filipe City Porto State Porto Country Portugal

    Original URL path: http://www.planetmath.org/users/filipe (2016-04-25)
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  • Multiple Recurrence Theorem | planetmath.org
    there exist a certain time n n n such that the subset of E E E for which all elements return to E E E simultaneously for all transformations T i subscript T i T i is a subset of E E E with positive measure Observe that the theorem may be applied again to the set G E T 1 n E T q n E G E superscript subscript T 1 n E normal superscript subscript T q n E G E cap T 1 n E cap cdots cap T q n E obtaining the existence of m ℕ m ℕ m in mathbb N such that μ G T 1 m G T q m G 0 μ G superscript subscript T 1 m G normal superscript subscript T q m G 0 mu G cap T 1 m G cap cdots cap T q m G 0 so that μ E T 1 m n E T q m n E μ G T 1 m G T q m G 0 μ E superscript subscript T 1 m n E normal superscript subscript T q m n E μ G superscript subscript T 1 m G normal superscript subscript T q m G 0 mu E cap T 1 m n E cap cdots cap T q m n E geq mu G cap T 1 m G cap cdots cap T q m G 0 So we may conclude that when E E E has positive measure there are infinite times for which there is a simultaneous return for a subset of E E E with positive measure As a corollary since the powers T T 2 T q T superscript T 2 normal superscript T q T T 2 cdots T q

    Original URL path: http://www.planetmath.org/multiplerecurrencetheorem (2016-04-25)
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  • convergence in the mean | planetmath.org
    n a k 1 a n absent A 1 n subscript a 1 normal subscript a k 1 n subscript a k 1 normal subscript a n displaystyle A frac 1 n a 1 ldots a k frac 1 n a k 1 ldots a n 1 n A a 1 A a k 1 n A a k 1 A a n absent 1 n delimited A subscript a 1 normal A subscript a k 1 n delimited A subscript a k 1 normal A subscript a n displaystyle frac 1 n A a 1 ldots A a k frac 1 n A a k 1 ldots A a n A a 1 A a k n A a k 1 A a n n absent A subscript a 1 normal A subscript a k n A subscript a k 1 normal A subscript a n n displaystyle leqq frac A a 1 ldots A a k n frac A a k 1 ldots A a n n The supposition implies that there is a positive integer k k k such that A a i ε 2 for all i k formulae sequence A subscript a i ε 2 for all i k A a i frac varepsilon 2 quad mbox for all i k Let s fix the integer k k k Choose the number l l l so great that A a 1 A a k n ε 2 for n l formulae sequence A subscript a 1 normal A subscript a k n ε 2 for n l frac A a 1 ldots A a k n frac varepsilon 2 quad mbox for n l Let now n max k l n k l n max k l The three above inequalities yield A b n ε 2 1 n n k ε 2 ε 2 ε 2 ε A subscript b n ε 2 1 n n k ε 2 ε 2 ε 2 ε A b n frac varepsilon 2 frac 1 n n k frac varepsilon 2 frac varepsilon 2 frac varepsilon 2 varepsilon whence we have lim n b n A subscript normal n subscript b n A lim n to infty b n A Note The converse of the theorem is not true For example if a n 1 1 n 2 assign subscript a n 1 superscript 1 n 2 a n frac 1 1 n 2 i e if the sequence 1 has the form 0 1 0 1 0 1 0 1 0 1 0 1 normal 0 1 0 1 0 1 ldots then it is divergent but converges in the mean to the limit 1 2 1 2 frac 1 2 the corresponding sequence 2 is 0 1 2 1 3 2 4 2 5 3 6 3 7 4 8 4 9 0 1 2 1 3 2 4 2 5 3 6 3 7 4 8 4 9 normal 0 frac 1 2 frac

    Original URL path: http://www.planetmath.org/convergenceinthemean (2016-04-25)
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  • definition of vector space needs no commutativity | planetmath.org
    Tue 09 29 2015 13 10 Briefly the sketch proof consists of the following steps 1 Iteration Let p 0 be the largest known prime with shape x 2 x 1 Let x 0 be the relevant value of x i e x 0 2 x 0 1 p 0 Consider the interval x 0 x 0 p 0 this interval is chosen because f x 0 k f x 0 is congruent to 0 mod f x 0 In each iteration there is a percentage of values of x not covered by the relevant failure functions these are such that the relevant f x s are prime which need not be tested for primality Do the relevant failure functions cover the whole interval If so the iteration has come to an end which means p 0 is the largest prime of the shape x 2 x 1 and that there are only a finite number of primes of this shape If not we go to the second iteration i 2 let x 1 be the largest value of x not covered by the relevant failure functions f x 1 is prime let this be p 1 Consider the interval x 1 x 1 p 1 Do the relevant failure functions cover the this interval completely If so iteration has come to an end and p 1 is the largest prime of this shape this also means there are only a finite number of primes of this shape If not we go to iteration i 3 My conjecture the percentage of xs not covered by the relevant failure functions will decrease from iteration to iteration progressively however the decrease of percentage is asymptotic to 3 i e it never reaches 3 Hence the iteration is perpetural Therefore the infinitude of primes of shape x 2 x 1 is proved to be continued Log in to post comments Conjecture sketch proof concluding message Permalink Submitted by akdevaraj on Wed 09 30 2015 03 18 Note 1 p 0 less than p 1 less than p 2 Hence the intervals get progressively larger even a small percentage of xs not covered by the relevant failure functions means a sizable number of candidates for initiating the next iteration However we need only one x not covered by any failure function to go to iteration i 1 This is the largest of the xs not covered 2 Only a competent programmer can program the failure functions and perform the iteration 3 For the sake of demonstration I have selected x 12 and the interval x 12 12 157 12 2 12 1 157 The iteration is done as follows I make a list of xs starting with 12 and ending with 169 Which are the relevant failure functions 1 2k 2 7k 3 13k 4 7k 5 31k 6 43k ending with 47 61k These are relevant because these cover the interval chosen Next I circle the xs covered by each failure function For example 1 3k generates 4 7 10 After performing the same procedure with all the relevant failure functions I found I was left with 37 unmarked xs The percentage of unmarked xs is 24The list of 37 unmarked xs 14 15 17 20 24 27 33 38 41 50 ending with 167 3 What are the implications of the iteration coming to an end after the ith iteration p i 1 is the largest prime with with shape x 2 x 1 This means x i 1 is the largest uncovered x all subsequent xs are such that f x are composite having shape 3 q7 m13 n x i 1 z here q m n z are exponents belonging to N This is highly improbable perhaps impossible Log in to post comments conjecture three points Permalink Submitted by akdevaraj on Sat 10 03 2015 10 58 1 Carl Pomerance has just pointed out an exception to my generalisation However I stand by my message on x 2 x 1 2 A minor correction second last line shape 3 q p i 1 z 3 I ran the program p n factorint n 2 n 1 from n 1 to n 1000 I could not find a single exponent greater than 2 Log in to post comments Conjecture clarification Permalink Submitted by akdevaraj on Sat 10 03 2015 01 53 Sorry I misunderstood Pomerance he has clarified that it has not yet been proved Log in to post comments conjecture a fewpoints Permalink Submitted by akdevaraj on Mon 10 05 2015 04 44 1 I double checked my computations I am convinced that a value of x say x if generated by one or more of the relevant failure functions then f x is a failure composite If not f x is prime and it need not be tested for primality 2 in the demonstration sample the largest uncovered value of x in the interval chosen is 167 f 167 28057 and the second interval to be considered is 167 28224 Obviously this cannnot be dealt with manually Log in to post comments conjecture a fewpoints contd Permalink Submitted by akdevaraj on Mon 10 05 2015 10 58 3 Bonus when a value of x say x is generated by a failure function we not only know that f x is composite we also know what f x is a multiple of For example let us take the failure function x 14 211k when k 1 x 225 f 14 211 is a composite and it is a multiple 211 Needless to say this bonus aspect is not important for application of failure functions in proving the infinitude of primes os shape f x Log in to post comments Random thoughts on proofs Permalink Submitted by akdevaraj on Tue 10 06 2015 04 29 1 Perhaps some conjectures cannot be proved without collaboration with a programmer unless the mathematician himself happens to be a gifted programmer 2 What is the

    Original URL path: http://www.planetmath.org/definitionofvectorspaceneedsnocommutativity (2016-04-25)
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  • redundancy of two-sidedness in definition of group | planetmath.org
    a right inverse x 1 superscript x 1 x 1 We have to show that the right identity e e e is also a left identity and that any right inverse is also a left inverse Let the above assumptions on G G G be true If a 1 superscript a 1 a 1 is the right inverse of an arbitrary element a a a of G G G the calculation a 1 a a 1 a e a 1 a a 1 a 1 1 a 1 e a 1 1 a 1 a 1 1 e superscript a 1 a superscript a 1 a e superscript a 1 a superscript a 1 superscript superscript a 1 1 superscript a 1 e superscript superscript a 1 1 superscript a 1 superscript superscript a 1 1 e a 1 a a 1 ae a 1 aa 1 a 1 1 a 1 e a 1 1 a 1 a 1 1 e shows that it is also the left inverse of a a a Using this result we then can write e a a a 1 a a a 1 a a e a e a a superscript a 1

    Original URL path: http://www.planetmath.org/redundancyoftwosidednessindefinitionofgroup (2016-04-25)
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  • Nucleus | planetmath.org
    p F q F p wedge q F p wedge F q Usually the term nucleus is used in frames and locales theory when the semilattice mathfrak A is a frame 1 Some well known results about nuclei Proposition If F F F is a nucleus on a frame mathfrak A then the poset Fix F Fix F operatorname Fix F of fixed points of F F F with order

    Original URL path: http://www.planetmath.org/nucleus-0 (2016-04-25)
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  • porton | planetmath.org
    org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A06B99 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module History Member for 10 years 11 months My articles amnestic functor Background of several open problems I stumbled upon closed sublattice collection complete lattice homomorphism Conjecture Every regular paratopological group is Tychonoff difference of lattice elements Direct products in a category of funcoids disjunction property of Wallman endomorphism Filtrator Intersecting two staroidal products Intersection with a staroid product through its upper sets Meta singular numbers Multifuncoid has atomic arguments Nucleus order preserving map proximity continuous proximity generated by uniformity pseudodifference sections and retractions selector Stone functor uniformly continuous is proximity continuous Buddy list actions Create Buddy List Designate Existing Team Member of World Writable My questions Arrow function canonical isomorphism closed complex plane closed sublattice Completely distributive lattice Countably complete upper lower semilattice decideable Lattice of filters Lattices as algebraic structures Lax algebras and lax monads Product of uniform spaces Pseudodifference quasi inverse functor Quasicategory Remark and note Semicategory Separable lattice surjective functor Forename Victor Surname Porton City State Country Homepage http www mathematics21 org Preamble this is the default PlanetMath preamble as your knowledge of TeX increases you will probably want to edit this but it should be fine as is

    Original URL path: http://www.planetmath.org/users/porton (2016-04-25)
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  • Articles | planetmath.org
    on Monday July 13 2015 04 36 Carmichal number 561 Last updated by akdevaraj on Sunday July 12 2015 06 12 Carmichal number 561 Last updated by akdevaraj on Sunday July 12 2015 06 12 Carmichal number 561 Last updated by akdevaraj on Sunday July 12 2015 06 12 Failure functions Last updated by akdevaraj on Wednesday May 27 2015 05 00 induction proof of fundamental theorem of arithmetic Last updated by pahio on Wednesday April 8 2015 07 32 Birkhoff Recurrence Theorem Last updated by Filipe on Friday March 20 2015 00 56 Multiple Recurrence Theorem Last updated by Filipe on Friday March 20 2015 00 29 convergence in the mean Last updated by pahio on Wednesday April 8 2015 07 29 definition of vector space needs no commutativity Last updated by pahio on Sunday January 25 2015 12 26 redundancy of two sidedness in definition of group Last updated by pahio on Tuesday January 20 2015 17 28 Nucleus Last updated by porton on Thursday December 18 2014 15 34 spam Last updated by vinayets10 on Wednesday December 31 2014 16 13 sequence of bounded variation Last updated by pahio on Friday November 28 2014 21 01 Numerical

    Original URL path: http://www.planetmath.org/articles (2016-04-25)
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