Recurrence relation for the sequence $sqrt{n}, sqrt{sqrt{n}}, sqrt{sqrt{sqrt{sqrt{n}}}}cdots$












0












$begingroup$


To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:



$$
f(n) =
begin{cases}
1 & text{if $n=2$}\
fleft(sqrt{n}right) & text{otherwise}
end{cases}
$$



For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).



I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$



$Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.



To visualize, the sequences are:



$A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$



and



$B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$






recurrence-relations recursion







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$endgroup$

















    0












    $begingroup$


    To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:



    $$
    f(n) =
    begin{cases}
    1 & text{if $n=2$}\
    fleft(sqrt{n}right) & text{otherwise}
    end{cases}
    $$



    For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).



    I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$



    $Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.



    To visualize, the sequences are:



    $A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$



    and



    $B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$






    recurrence-relations recursion







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      2



      $begingroup$


      To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:



      $$
      f(n) =
      begin{cases}
      1 & text{if $n=2$}\
      fleft(sqrt{n}right) & text{otherwise}
      end{cases}
      $$



      For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).



      I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$



      $Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.



      To visualize, the sequences are:



      $A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$



      and



      $B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$






      recurrence-relations recursion







      share|cite|improve this question











      $endgroup$




      To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:



      $$
      f(n) =
      begin{cases}
      1 & text{if $n=2$}\
      fleft(sqrt{n}right) & text{otherwise}
      end{cases}
      $$



      For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).



      I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$



      $Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.



      To visualize, the sequences are:



      $A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$



      and



      $B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$






      recurrence-relations recursion




      recurrence-relations recursion






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 2 '18 at 16:31







      dibyendu

















      asked Dec 2 '18 at 16:06









      dibyendudibyendu

      356318




      356318






















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
          $f(k) =
          begin{cases}
          n & text{if }k = 0 \
          sqrt{f(k-1)} & text{if }k > 0
          end{cases}$



          Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.



          For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence



          $f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.



          We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
            $endgroup$
            – dibyendu
            Dec 4 '18 at 22:30





















          0












          $begingroup$

          I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].



          For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:



          $$
          f(n)=f_1(n, 1)
          $$
          $$
          f_1left(n, iright) =
          begin{cases}
          f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
          1 & text{if, $n=2$}\
          f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
          end{cases}
          $$



          Any other thought/suggestion is appreciated.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
            $f(k) =
            begin{cases}
            n & text{if }k = 0 \
            sqrt{f(k-1)} & text{if }k > 0
            end{cases}$



            Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.



            For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence



            $f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.



            We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
              $endgroup$
              – dibyendu
              Dec 4 '18 at 22:30


















            0












            $begingroup$

            I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
            $f(k) =
            begin{cases}
            n & text{if }k = 0 \
            sqrt{f(k-1)} & text{if }k > 0
            end{cases}$



            Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.



            For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence



            $f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.



            We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
              $endgroup$
              – dibyendu
              Dec 4 '18 at 22:30
















            0












            0








            0





            $begingroup$

            I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
            $f(k) =
            begin{cases}
            n & text{if }k = 0 \
            sqrt{f(k-1)} & text{if }k > 0
            end{cases}$



            Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.



            For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence



            $f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.



            We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$






            share|cite|improve this answer









            $endgroup$



            I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
            $f(k) =
            begin{cases}
            n & text{if }k = 0 \
            sqrt{f(k-1)} & text{if }k > 0
            end{cases}$



            Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.



            For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence



            $f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.



            We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 2:49









            AkashnilAkashnil

            1




            1












            • $begingroup$
              I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
              $endgroup$
              – dibyendu
              Dec 4 '18 at 22:30




















            • $begingroup$
              I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
              $endgroup$
              – dibyendu
              Dec 4 '18 at 22:30


















            $begingroup$
            I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
            $endgroup$
            – dibyendu
            Dec 4 '18 at 22:30






            $begingroup$
            I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
            $endgroup$
            – dibyendu
            Dec 4 '18 at 22:30













            0












            $begingroup$

            I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].



            For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:



            $$
            f(n)=f_1(n, 1)
            $$
            $$
            f_1left(n, iright) =
            begin{cases}
            f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
            1 & text{if, $n=2$}\
            f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
            end{cases}
            $$



            Any other thought/suggestion is appreciated.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].



              For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:



              $$
              f(n)=f_1(n, 1)
              $$
              $$
              f_1left(n, iright) =
              begin{cases}
              f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
              1 & text{if, $n=2$}\
              f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
              end{cases}
              $$



              Any other thought/suggestion is appreciated.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].



                For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:



                $$
                f(n)=f_1(n, 1)
                $$
                $$
                f_1left(n, iright) =
                begin{cases}
                f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
                1 & text{if, $n=2$}\
                f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
                end{cases}
                $$



                Any other thought/suggestion is appreciated.






                share|cite|improve this answer









                $endgroup$



                I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].



                For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:



                $$
                f(n)=f_1(n, 1)
                $$
                $$
                f_1left(n, iright) =
                begin{cases}
                f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
                1 & text{if, $n=2$}\
                f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
                end{cases}
                $$



                Any other thought/suggestion is appreciated.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 22:54









                dibyendudibyendu

                356318




                356318






























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