double sigma in complex numberIf complex no. ($z$) satisfying $frac12leq |z|leq 4;,$ Then Max. and Min. of $left|z+frac1zright|$complex no. $z$ such that $ |z|<frac13$ and $ sum^n_r=1a_rz^r = 1;,$ Where $|a_r|<2$Product of one minus the tenth roots of unityComplex number (cube roots of unity)Summation over roots of unityMaximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.Proving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityEvaluating the complex summationvalue of $|b_1|+|b_2|+cdots cdots +|b_k|$ isProve that $|z_1+z_2|^2leq (1+c)|z_1|^2+bigg(1+frac1cbigg)|z_2|^2$

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double sigma in complex number


If complex no. ($z$) satisfying $frac12leq |z|leq 4;,$ Then Max. and Min. of $left|z+frac1zright|$complex no. $z$ such that $ |z|<frac13$ and $ sum^n_r=1a_rz^r = 1;,$ Where $|a_r|<2$Product of one minus the tenth roots of unityComplex number (cube roots of unity)Summation over roots of unityMaximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.Proving that $sum cos p theta = 0$ where $theta$ is argument of the primitive roots of unityEvaluating the complex summationvalue of $|b_1|+|b_2|+cdots cdots +|b_k|$ isProve that $|z_1+z_2|^2leq (1+c)|z_1|^2+bigg(1+frac1cbigg)|z_2|^2$













1












$begingroup$


if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is



what i try



assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots



$x=beta_1,beta_2,cdots,beta_100$



$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$



how do i solve it help me please






complex-numbers







share|cite|improve this question









$endgroup$











  • $begingroup$
    This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
    $endgroup$
    – enedil
    Mar 7 at 14:51










  • $begingroup$
    @enedil dod not understand please explain me
    $endgroup$
    – jacky
    Mar 7 at 18:01






  • 1




    $begingroup$
    What did you not understand?
    $endgroup$
    – enedil
    Mar 7 at 19:50










  • $begingroup$
    @enedil first line
    $endgroup$
    – jacky
    Mar 7 at 20:20










  • $begingroup$
    First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
    $endgroup$
    – enedil
    Mar 7 at 20:32















1












$begingroup$


if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is



what i try



assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots



$x=beta_1,beta_2,cdots,beta_100$



$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$



how do i solve it help me please






complex-numbers







share|cite|improve this question









$endgroup$











  • $begingroup$
    This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
    $endgroup$
    – enedil
    Mar 7 at 14:51










  • $begingroup$
    @enedil dod not understand please explain me
    $endgroup$
    – jacky
    Mar 7 at 18:01






  • 1




    $begingroup$
    What did you not understand?
    $endgroup$
    – enedil
    Mar 7 at 19:50










  • $begingroup$
    @enedil first line
    $endgroup$
    – jacky
    Mar 7 at 20:20










  • $begingroup$
    First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
    $endgroup$
    – enedil
    Mar 7 at 20:32













1












1








1


1



$begingroup$


if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is



what i try



assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots



$x=beta_1,beta_2,cdots,beta_100$



$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$



how do i solve it help me please






complex-numbers







share|cite|improve this question









$endgroup$




if $beta_1,beta_2,cdots ,beta_100$ are $100$ th roots of unity. The numerical value of $displaystyle mathopsum sum_1leq ileq jleq 100bigg(beta_ibeta_jbigg)^5$ is



what i try



assuming $x=(1)^frac1100,$ then $x^100-1=0$ has roots



$x=beta_1,beta_2,cdots,beta_100$



$displaystyle x^100-1=(x-beta_1)(x-beta_2)cdots cdots (x-beta_100)$



how do i solve it help me please






complex-numbers




complex-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 7 at 14:36









jackyjacky

1,028715




1,028715











  • $begingroup$
    This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
    $endgroup$
    – enedil
    Mar 7 at 14:51










  • $begingroup$
    @enedil dod not understand please explain me
    $endgroup$
    – jacky
    Mar 7 at 18:01






  • 1




    $begingroup$
    What did you not understand?
    $endgroup$
    – enedil
    Mar 7 at 19:50










  • $begingroup$
    @enedil first line
    $endgroup$
    – jacky
    Mar 7 at 20:20










  • $begingroup$
    First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
    $endgroup$
    – enedil
    Mar 7 at 20:32
















  • $begingroup$
    This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
    $endgroup$
    – enedil
    Mar 7 at 14:51










  • $begingroup$
    @enedil dod not understand please explain me
    $endgroup$
    – jacky
    Mar 7 at 18:01






  • 1




    $begingroup$
    What did you not understand?
    $endgroup$
    – enedil
    Mar 7 at 19:50










  • $begingroup$
    @enedil first line
    $endgroup$
    – jacky
    Mar 7 at 20:20










  • $begingroup$
    First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
    $endgroup$
    – enedil
    Mar 7 at 20:32















$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51




$begingroup$
This sum equals $1/2(sum_1leq i, j leq nbeta_i^5 beta_j^5 + sum_1 leq i leq nbeta_i ^ 5)$. Then use the lemma that $sum_i in S_1sum_jin S_2f(i)g(j) = (sum_i in S_1f(i)) cdot (sum_j in S_2 f(j))$
$endgroup$
– enedil
Mar 7 at 14:51












$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01




$begingroup$
@enedil dod not understand please explain me
$endgroup$
– jacky
Mar 7 at 18:01




1




1




$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50




$begingroup$
What did you not understand?
$endgroup$
– enedil
Mar 7 at 19:50












$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20




$begingroup$
@enedil first line
$endgroup$
– jacky
Mar 7 at 20:20












$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32




$begingroup$
First you double count elements, without checking the condition that $i leq j$, then add elements with indices $i, i$. Every element was doubly counted, so you just divide by two.
$endgroup$
– enedil
Mar 7 at 20:32










1 Answer
1






active

oldest

votes


















1












$begingroup$


We obtain
beginalign*
colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
&=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
&=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
&=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
&=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
&=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
&,,colorblue=0
endalign*




Comment:



  • In (1) we rewrite the sum as double sum and factor out $beta_j$.


  • In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.


  • In (3) we apply the finite geometric series formula.


  • In (4) we factor out common terms and multiply out the sums.


  • In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.


Note:



Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$


    We obtain
    beginalign*
    colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
    &=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
    &=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
    &=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
    &=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
    &=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
    &,,colorblue=0
    endalign*




    Comment:



    • In (1) we rewrite the sum as double sum and factor out $beta_j$.


    • In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.


    • In (3) we apply the finite geometric series formula.


    • In (4) we factor out common terms and multiply out the sums.


    • In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.


    Note:



    Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$


      We obtain
      beginalign*
      colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
      &=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
      &=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
      &=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
      &=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
      &=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
      &,,colorblue=0
      endalign*




      Comment:



      • In (1) we rewrite the sum as double sum and factor out $beta_j$.


      • In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.


      • In (3) we apply the finite geometric series formula.


      • In (4) we factor out common terms and multiply out the sums.


      • In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.


      Note:



      Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$


        We obtain
        beginalign*
        colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
        &=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
        &=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
        &=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
        &=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
        &=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
        &,,colorblue=0
        endalign*




        Comment:



        • In (1) we rewrite the sum as double sum and factor out $beta_j$.


        • In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.


        • In (3) we apply the finite geometric series formula.


        • In (4) we factor out common terms and multiply out the sums.


        • In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.


        Note:



        Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).






        share|cite|improve this answer









        $endgroup$




        We obtain
        beginalign*
        colorbluesum_1leq ileq jleq 100colorblueleft(beta_ibeta_jright)^5
        &=sum_j=1^100beta_j^5sum_i=1^jbeta_i^5tag1\
        &=sum_j=1^100beta_1^5jsum_i=1^jleft(beta_1^5right)^itag2\
        &=sum_j=1^100beta_1^5jfracbeta_1^5-left(beta_1^5right)^j+11-beta_1tag3\
        &=fracbeta_1^51-beta_1left(sum_j=1^100beta_1^5j-sum_j=1^100beta_1^10jright)tag4\
        &=fracbeta_1^51-beta_1left(fracbeta_1^5-beta_1^5051-beta_1^5-fracbeta_1^10-beta_1^10101-beta_1^10right)tag5\
        &,,colorblue=0
        endalign*




        Comment:



        • In (1) we rewrite the sum as double sum and factor out $beta_j$.


        • In (2) we use $beta_j=expleft(frac2pi i j100right)=expleft(frac2pi i100right)^j=beta_1^j$.


        • In (3) we apply the finite geometric series formula.


        • In (4) we factor out common terms and multiply out the sums.


        • In (5) we again apply the finite geometric series formula. We observe, since $beta_1^100=1$ that $beta_1^505=beta_1^5$ and $beta_1^10=beta_1^1010$.


        Note:



        Since we have $beta_1^100-1=(beta_1-1)left(1+beta_1+cdots+beta_1^99right)=0$ we can deduce the result zero already from (4).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Markus ScheuerMarkus Scheuer

        62.6k459149




        62.6k459149



























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            Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee

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            Parai AttamNatarajaTandavaRasa lilaLasyaBollywood song and danceHindi dance songsNautchAndhra PradeshAssamHimachal PradeshOdishaPunjabTamil NaduMizoramKerala KalamandalamKerala Lalitakala AcademyKerala Sangeetha Nadaka Academy, ThrissurKerala Folklore AcademyKshetram vadyamKuzhal PattuMappila PaattuPanchari melamPanchavadyamPandi MelamParisha VadyamPulluvan PaattuSopanamThayambakaKerala mural paintingKalaripayattuSarpa KavuTharavadTemplatesPortalCategoryWikiProject KathakaliTheatre in IndiaArts of KeralaClassical theatre of india Malayalamclassical Indian danceMalayalamKeralaCEHindu templesPuranasNatya ShastraNatya ShastraBharataShivaMalayalamKeralaShivaKrishnaKrishnaJayadevaoperaticKrishnaVishnuRamaShivaSuryaYudhishthiraArjunaNalaRavanaDushasanaHiranyakashipuSitaHanumanGarudaHamsaGuṇaSamkhyaHindu philosophysign languagemudrasUnnayi VariyarplaywrightsNalaDamayantimusical modeChristianityMiguel de CervantesJohann Wolfgang von GoetheWilliam ShakespeareGurukulTravancorePalakkadBrahmi...
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            Method to test if a number is a perfect power? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Detecting perfect squares faster than by extracting square rooteffective way to get the integer sequence A181392 from oeisA rarely mentioned fact about perfect powersHow many numbers such $n$ are there that $n<100,lfloorsqrtn rfloor mid n$Check perfect squareness by modulo division against multiple basesFor what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.Do there exist any positive integers $n$ such that $lfloore^nrfloor$ is a perfect power? What is the probability that one exists?finding perfect power factors of an integerProve that the sequence contains a perfect square for any natural number $m $ in the domain of $f$ .Counting Perfect Powers

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            Proof of work - 51% attack Would it be possible to dictate a bech32 address as a list of English words? How do I find out the mythology and history of my Fortress? Sentence order: Where to put もう Is there any word for a place full of confusion? Do wooden building fires get hotter than 600°C? What does it mean that physics no longer uses mechanical models to describe phenomena? Project Euler #1 in C++ How fail-safe is nr as stop bytes? Why does it sometimes sound good to play a grace note as a lead in to a note in a melody? How to draw/optimize this graph with tikz How much damage would a cupful of neutron star matter do to the Earth? What would you call this weird metallic apparatus that allows you to lift people? Importance of からだ in this sentence "Lost his faith in humanity in the trenches of Verdun" — last line of an SF story A term for a woman complaining about things/begging in a cute/childish way How to ...
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