How does the following evaluate?

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We have a recurrence as follows:




$$T(n) = 2Tleft(sqrtnright) + log n$$




Renaming $m = log n$ yields




$$T(2^m) = 2T(2^fracm2) + m$$




Renaming $S(m) = T(2^m)$ new recurrence becomes:




$$S(m) = 2Sleft(fracm2right) + m$$




How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?







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    We have a recurrence as follows:




    $$T(n) = 2Tleft(sqrtnright) + log n$$




    Renaming $m = log n$ yields




    $$T(2^m) = 2T(2^fracm2) + m$$




    Renaming $S(m) = T(2^m)$ new recurrence becomes:




    $$S(m) = 2Sleft(fracm2right) + m$$




    How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      We have a recurrence as follows:




      $$T(n) = 2Tleft(sqrtnright) + log n$$




      Renaming $m = log n$ yields




      $$T(2^m) = 2T(2^fracm2) + m$$




      Renaming $S(m) = T(2^m)$ new recurrence becomes:




      $$S(m) = 2Sleft(fracm2right) + m$$




      How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?







      share|cite|improve this question














      We have a recurrence as follows:




      $$T(n) = 2Tleft(sqrtnright) + log n$$




      Renaming $m = log n$ yields




      $$T(2^m) = 2T(2^fracm2) + m$$




      Renaming $S(m) = T(2^m)$ new recurrence becomes:




      $$S(m) = 2Sleft(fracm2right) + m$$




      How does this $T(2^fracm2)$ become $Sleft(fracm2right)$?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 15 at 12:40









      James

      2,002619




      2,002619










      asked Aug 15 at 12:31









      Jot Waraich

      104




      104




















          1 Answer
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          you agree that $$S(m) = T(2^m)$$
          now put $fracm2$ inplace of $m$.



          $$S(fracm2)=T(2^fracm2)$$
          so what is $T(2^fracm2)$ now it is $S(fracm2)$
          thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
          hope it clears your doubt!






          share|cite|improve this answer




















          • Never thought it is going to be this easy : |
            – Jot Waraich
            Aug 15 at 12:44










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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes








          up vote
          0
          down vote



          accepted










          you agree that $$S(m) = T(2^m)$$
          now put $fracm2$ inplace of $m$.



          $$S(fracm2)=T(2^fracm2)$$
          so what is $T(2^fracm2)$ now it is $S(fracm2)$
          thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
          hope it clears your doubt!






          share|cite|improve this answer




















          • Never thought it is going to be this easy : |
            – Jot Waraich
            Aug 15 at 12:44














          up vote
          0
          down vote



          accepted










          you agree that $$S(m) = T(2^m)$$
          now put $fracm2$ inplace of $m$.



          $$S(fracm2)=T(2^fracm2)$$
          so what is $T(2^fracm2)$ now it is $S(fracm2)$
          thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
          hope it clears your doubt!






          share|cite|improve this answer




















          • Never thought it is going to be this easy : |
            – Jot Waraich
            Aug 15 at 12:44












          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          you agree that $$S(m) = T(2^m)$$
          now put $fracm2$ inplace of $m$.



          $$S(fracm2)=T(2^fracm2)$$
          so what is $T(2^fracm2)$ now it is $S(fracm2)$
          thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
          hope it clears your doubt!






          share|cite|improve this answer












          you agree that $$S(m) = T(2^m)$$
          now put $fracm2$ inplace of $m$.



          $$S(fracm2)=T(2^fracm2)$$
          so what is $T(2^fracm2)$ now it is $S(fracm2)$
          thus we can replace $T(2^fracm2)$ with $S(fracm2)$ right?
          hope it clears your doubt!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 15 at 12:42









          James

          2,002619




          2,002619











          • Never thought it is going to be this easy : |
            – Jot Waraich
            Aug 15 at 12:44
















          • Never thought it is going to be this easy : |
            – Jot Waraich
            Aug 15 at 12:44















          Never thought it is going to be this easy : |
          – Jot Waraich
          Aug 15 at 12:44




          Never thought it is going to be this easy : |
          – Jot Waraich
          Aug 15 at 12:44












           

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