Computational complexity sum of digits square root












1














I know that the sum of digits formula is:



$frac{k(k+1)}{2}$



I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:



$frac{k(k+1)}{2}>n$



At the time when the function increases over $n$, the while loop will exit.



Expanding the formula and solving for $k$:



$frac{k^2+k}{2} > n$



This next line I do not understand, can someone explain how this line is achieved?



$k = Oleft(sqrt{n}right)$



Regards










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    1














    I know that the sum of digits formula is:



    $frac{k(k+1)}{2}$



    I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:



    $frac{k(k+1)}{2}>n$



    At the time when the function increases over $n$, the while loop will exit.



    Expanding the formula and solving for $k$:



    $frac{k^2+k}{2} > n$



    This next line I do not understand, can someone explain how this line is achieved?



    $k = Oleft(sqrt{n}right)$



    Regards










    share|cite|improve this question



























      1












      1








      1







      I know that the sum of digits formula is:



      $frac{k(k+1)}{2}$



      I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:



      $frac{k(k+1)}{2}>n$



      At the time when the function increases over $n$, the while loop will exit.



      Expanding the formula and solving for $k$:



      $frac{k^2+k}{2} > n$



      This next line I do not understand, can someone explain how this line is achieved?



      $k = Oleft(sqrt{n}right)$



      Regards










      share|cite|improve this question















      I know that the sum of digits formula is:



      $frac{k(k+1)}{2}$



      I am calculating the computation complexity of an algorithm whose while loop is increasing at this factor, hence:



      $frac{k(k+1)}{2}>n$



      At the time when the function increases over $n$, the while loop will exit.



      Expanding the formula and solving for $k$:



      $frac{k^2+k}{2} > n$



      This next line I do not understand, can someone explain how this line is achieved?



      $k = Oleft(sqrt{n}right)$



      Regards







      computational-complexity






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 at 10:15









      bames

      1,9451315




      1,9451315










      asked Dec 26 at 9:50









      fynx

      83




      83



























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