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andrei1397
Pe: 2015-09-02T09:15:52+03:00 In: In:

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  1. ex-admin
    profesor

    Indicatie: (\omega - 1)(\omega ^{2} + \omega + 1)=\omega ^{3} - 1\Rightarrow \omega ^{3}=1

  2. adrianS

    Buna ziua
    Am o idee:
    Rezolvand ecuatiax^2+x+1=0\ gasim\ solutiile\ x_{1,2}=\dfrac{-1\pm 3i}{2} Mergem\ pe\ x_1=\dfrac{-1+3i}{2} In complex aceasta inseamna :
    \omega^1=cos120^0+isin120^0\ apoi\\ \omega^2=cos240^0+isin240^0\\ \omega^3=cos360^0+isin360^0\\ \omega^4=cos120^0+isin120^0\\ \omega^5=cos240^0+isin240^0\\ \omega^6=cos360^0+isin360^0\\ \omega^7=cos120^0+isin120^0\\ \omega^8=cos240^0+isin240^0\\ \omega^9=cos360^0+isin360^0\\ \dots\dots\dots\dots\dots\dots\\ \omega^1+\omega^2+\omega^3=cos120^0+cos240^0+cos360^0+isin120^0+isin240^0+isin360^0\\ =-cos60^0-cos60^0+1+isin60^0-isin60^0=-2cos60^0+1=1-2\dfrac{1}{2}=0\\ \omega^2+\omega^4+\omega^6=cos120^0+cos240^0+cos360^0+isin120^0+isin240^0+isin360^0=0\\ \omega^3+\omega^6+\omega^9=cos360^0+isin360^0+cos720^0+isin720^0+cos1080^0+isin1080^0=1+1+1=3
    Deci in final facand suma matricelor obtinem:
    \begin{pmatrix}\0&0&2013\\ 0&0&0\\ \end{pmatrix}
    Am mers cu grade pentru a fi mai usor de urmarit.

  4. ex-admin
    profesor

    Suma oricaror trei puteri consecutive ale lui \small \omega este nula pentru ca \small {\omega ^k} + {\omega ^{k + 1}} + {\omega ^{k + 2}} = {\omega ^k} \cdot (1 + \omega + {\omega ^2}) = 0.
    Cum 2013 este multiplu de 3, rezulta ca \small \sum\limits_{k = 1}^{2013} {{\omega ^k}} = 0.

    Apoi, tinand seama de indicatia din postul anterior obtinem imediat ca \small \sum\limits_{k = 1}^{2013} {{\omega ^{3k}}} = \sum\limits_{k = 1}^{2013} {{{({\omega ^3})}^k} = \sum\limits_{k = 1}^{2013} 1 = 2013} ,

    singura suma mai interesanta parand a fi \small \sum\limits_{k = 1}^{2013} {{\omega ^{2k}}} .

    Dar \small 1 + {\omega ^2} + {({\omega ^2})^2} = 1 + {\omega ^2} + {\omega ^4} = 1 + {\omega ^2} + \omega = 0, deci \small \omega ^2 are aceeasi proprietate ca si \small \omega, asadar \small \sum\limits_{k = 1}^{2013} {{\omega ^{2k}}} = 0 .

  5. andrei1397

    admin wrote: Suma oricaror trei puteri consecutive ale lui \small \omega este nula pentru ca \small {\omega ^k} + {\omega ^{k + 1}} + {\omega ^{k + 2}} = {\omega ^k} \cdot (1 + \omega + {\omega ^2}) = 0.
    Cum 2013 este multiplu de 3, rezulta ca \small \sum\limits_{k = 1}^{2013} {{\omega ^k}} = 0.

    Apoi, tinand seama de indicatia din postul anterior obtinem imediat ca \small \sum\limits_{k = 1}^{2013} {{\omega ^{3k}}} = \sum\limits_{k = 1}^{2013} {{{({\omega ^3})}^k} = \sum\limits_{k = 1}^{2013} 1 = 2013} ,

    singura suma mai interesanta parand a fi \small \sum\limits_{k = 1}^{2013} {{\omega ^{2k}}} .

    Dar \small 1 + {\omega ^2} + {({\omega ^2})^2} = 1 + {\omega ^2} + {\omega ^4} = 1 + {\omega ^2} + \omega = 0, deci \small \omega ^2 are aceeasi proprietate ca si \small \omega, asadar \small \sum\limits_{k = 1}^{2013} {{\omega ^{2k}}} = 0 .

    Cred ca e o metoda mai usoara decat cea cu sinus si cosinus. As prefera sa o utilizez pe aceasta, cu toate ca am stat putin pe ganduri cum ai rezolvat {{\omega ^{2k}}}. Multumesc amandurora pentru ajutorul acordat.

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