sum of Time complexities












1












$begingroup$


Question: Say we have $f(n) = O(n)$ and $g(n) = O(n)$, Show (or not) that $f(n) + c g(k) = O(n+k)$



Solution:



We have : $f(n) = O(n) Leftrightarrow exists a textrm{ and } n_0 textrm{ s.t. } f(n) leq a cdot n forall n geq n_0$



and similarly: $g(k) = O(k) Leftrightarrow exists b textrm{ and } k_0 textrm{ s.t. } g(k) leq b cdot k forall n geq k_0$



so
$$
f(n) + cg(k) leq acdot O(n) + c cdot b cdot O(k)
$$



But at this point I start thinking maybe the statement does not make sense, how could we have one algorithm running on two different input sizes.
Any Hints would be much appreciated! Thank you!






proof-verification asymptotics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Question: Say we have $f(n) = O(n)$ and $g(n) = O(n)$, Show (or not) that $f(n) + c g(k) = O(n+k)$



    Solution:



    We have : $f(n) = O(n) Leftrightarrow exists a textrm{ and } n_0 textrm{ s.t. } f(n) leq a cdot n forall n geq n_0$



    and similarly: $g(k) = O(k) Leftrightarrow exists b textrm{ and } k_0 textrm{ s.t. } g(k) leq b cdot k forall n geq k_0$



    so
    $$
    f(n) + cg(k) leq acdot O(n) + c cdot b cdot O(k)
    $$



    But at this point I start thinking maybe the statement does not make sense, how could we have one algorithm running on two different input sizes.
    Any Hints would be much appreciated! Thank you!






    proof-verification asymptotics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Question: Say we have $f(n) = O(n)$ and $g(n) = O(n)$, Show (or not) that $f(n) + c g(k) = O(n+k)$



      Solution:



      We have : $f(n) = O(n) Leftrightarrow exists a textrm{ and } n_0 textrm{ s.t. } f(n) leq a cdot n forall n geq n_0$



      and similarly: $g(k) = O(k) Leftrightarrow exists b textrm{ and } k_0 textrm{ s.t. } g(k) leq b cdot k forall n geq k_0$



      so
      $$
      f(n) + cg(k) leq acdot O(n) + c cdot b cdot O(k)
      $$



      But at this point I start thinking maybe the statement does not make sense, how could we have one algorithm running on two different input sizes.
      Any Hints would be much appreciated! Thank you!






      proof-verification asymptotics







      share|cite|improve this question











      $endgroup$




      Question: Say we have $f(n) = O(n)$ and $g(n) = O(n)$, Show (or not) that $f(n) + c g(k) = O(n+k)$



      Solution:



      We have : $f(n) = O(n) Leftrightarrow exists a textrm{ and } n_0 textrm{ s.t. } f(n) leq a cdot n forall n geq n_0$



      and similarly: $g(k) = O(k) Leftrightarrow exists b textrm{ and } k_0 textrm{ s.t. } g(k) leq b cdot k forall n geq k_0$



      so
      $$
      f(n) + cg(k) leq acdot O(n) + c cdot b cdot O(k)
      $$



      But at this point I start thinking maybe the statement does not make sense, how could we have one algorithm running on two different input sizes.
      Any Hints would be much appreciated! Thank you!






      proof-verification asymptotics




      proof-verification asymptotics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 1 at 16:39









      YuiTo Cheng

      2,3694937




      2,3694937










      asked Feb 1 at 15:45









      rannoudanamesrannoudanames

      570717




      570717






















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          $begingroup$

          For $n+k ge n_0 + k_0$, you have



          $$vert f(n)+ c g(k) vert le vert f(n) vert + c vert g(k) vert le an+cbk le max(a,cb)(n+k)$$



          So indeed saying that $f(n) + c g(k) = O(n+k)$ makes sense using Big O notation.






          share|cite|improve this answer









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            $begingroup$

            For $n+k ge n_0 + k_0$, you have



            $$vert f(n)+ c g(k) vert le vert f(n) vert + c vert g(k) vert le an+cbk le max(a,cb)(n+k)$$



            So indeed saying that $f(n) + c g(k) = O(n+k)$ makes sense using Big O notation.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For $n+k ge n_0 + k_0$, you have



              $$vert f(n)+ c g(k) vert le vert f(n) vert + c vert g(k) vert le an+cbk le max(a,cb)(n+k)$$



              So indeed saying that $f(n) + c g(k) = O(n+k)$ makes sense using Big O notation.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For $n+k ge n_0 + k_0$, you have



                $$vert f(n)+ c g(k) vert le vert f(n) vert + c vert g(k) vert le an+cbk le max(a,cb)(n+k)$$



                So indeed saying that $f(n) + c g(k) = O(n+k)$ makes sense using Big O notation.






                share|cite|improve this answer









                $endgroup$



                For $n+k ge n_0 + k_0$, you have



                $$vert f(n)+ c g(k) vert le vert f(n) vert + c vert g(k) vert le an+cbk le max(a,cb)(n+k)$$



                So indeed saying that $f(n) + c g(k) = O(n+k)$ makes sense using Big O notation.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 1 at 16:40









                mathcounterexamples.netmathcounterexamples.net

                26.9k22158




                26.9k22158






























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