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loky
user (0)
Pe: 2010-02-14T12:34:09+02:00 In: In:

o problema frumusica':|:|..pls help

1. Demonstraţi că pentru orice număr real x are loc relaţia:

unde desemnează partea întreagă a numărului real a.

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3 raspunsuri

  1. user (0)

    am folosit identitatea lui hermite
    iata rezolvarea

        \[ \begin{array}{l} \left[ {\frac{x}{6}} \right] + \left[ {\frac{x}{6} + \frac{1}{6}} \right] + \left[ {\frac{x}{6} + \frac{2}{6}} \right] + \left[ {\frac{x}{6} + \frac{3}{6}} \right] + \left[ {\frac{x}{6} + \frac{4}{6}} \right] + \left[ {\frac{x}{6} + \frac{5}{6}} \right] - \left[ {\frac{x}{6}} \right] - \left[ {\frac{x}{6} + \frac{1}{6} + } \right] - \left[ {\frac{x}{6} + \frac{2}{6}} \right] - 2\left[ {\frac{x}{6} + \frac{4}{6}} \right] = \left[ {\frac{x}{2} + \frac{1}{2}} \right] + \left[ {\frac{x}{2}} \right] - \left[ {\frac{x}{2}} \right] - \left[ {\frac{{x + 1}}{3}} \right] \\ \left[ {\frac{{6x}}{6}} \right] - \left[ {\frac{x}{6}} \right] - \left[ {\frac{x}{6} + \frac{1}{6}} \right] - \left[ {\frac{x}{6} + \frac{2}{6}} \right] - 2\left[ {\frac{x}{6} + \frac{4}{6}} \right] = \left[ {\frac{{2x}}{2}} \right] - \left[ {\frac{x}{2}} \right] - \left[ {\frac{{x + 1}}{3}} \right] \\ \left[ x \right] - (\left[ {\frac{x}{6}} \right] + \left[ {\frac{x}{6} + \frac{1}{3}} \right] + \left[ {\frac{x}{6} + \frac{2}{3}} \right]) - \left[ {\frac{x}{6} + \frac{2}{3}} \right] - \left[ {\frac{x}{6} + \frac{1}{6}} \right] = \left[ x \right] - \left[ {\frac{x}{2}} \right] - \left[ {\frac{{x + 1}}{3}} \right] \\ [x] - \left[ {\frac{{3x}}{6}} \right] - \left[ {\frac{x}{6} + \frac{2}{3}} \right] - \left[ {\frac{x}{6} + \frac{1}{6}} \right] = \left[ x \right] - \left[ {\frac{x}{2}} \right] - \left[ {\frac{{x + 1}}{3}} \right]\left| { - [x]} \right. \\ - \left[ {\frac{x}{2}} \right] - \left[ {\frac{x}{6} + \frac{2}{3}} \right] - \left[ {\frac{x}{6} + \frac{1}{6}} \right] = - \left[ {\frac{x}{2}} \right] - \left[ {\frac{{x + 1}}{3}} \right]\left| + \right.\left[ {\frac{x}{2}} \right] \\ - \left[ {\frac{x}{6} + \frac{2}{3}} \right] - \left[ {\frac{x}{6} + \frac{1}{6}} \right] = - \left[ {\frac{{x + 1}}{3}} \right] \\ \left[ {\frac{{x + 1}}{3}} \right] - \left[ {\frac{{x + 1}}{6}} \right] - \left[ {\frac{{x + 1}}{6} + \frac{3}{6}} \right] = 0 \\ \left[ {\frac{{x + 1}}{3}} \right] - (\left[ {\frac{{x + 1}}{6}} \right] + \left[ {\frac{{x + 1}}{6} + \frac{1}{2}} \right]) = 0 \\ \left[ {\frac{{x + 1}}{3}} \right] - \left[ {\frac{{2(x + 1)}}{6}} \right] = 0 \\ \left[ {\frac{{x + 1}}{3}} \right] - \left[ {\frac{{x + 1}}{3}} \right] = 0 \\ \end{array} \]

    • 0
  2. user (0)

    mama…stai nu poti sa explicic putin mai….pe larg
    stiu cum e ident lui hermite dar nu imi dau seama cum ai folosit’o:(

    • 0
  4. user (0)

    [x]+[x+1/n]+[x+2/n]+…[x+(n-1)/n=[nx]
    am aplicat-o invers considerand [x]=[6*x/6] pt n=6 si x fiind x/6
    apoi n=3 x fiind x/6
    cam asta e tot
    pt detalii mai scrie …
    succes!!!

    • 0
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