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I'm not sure if I'm correct, but could you please verify if this is right:

$$\begin{align} T(n) &= T\left(\frac{n}{2}\right) + log_{2}(n)\\ T(n) &= T\left(\frac{n}{2^{i}}\right) + n\sum_{i-1}^{k=0} \left ( \frac{1}{2} \right )^{i}\\ T(n) &= 1 + n\left[2 - \frac{1}{2^{log(n)-1}}\right] \\ T(n) &= 2n - 1\\ \end{align}$$

I would be very grateful. Thanks.

TooTone
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Paolo Bernasconi
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    man forget about 4 spaces, you are so close to latex nirvana, you should put those double dollars in and achieve equation enlightenment :) – TooTone Feb 27 '14 at 00:20
  • this looks a lot like http://math.stackexchange.com/questions/689108/how-to-solve-this-recurrence-tn-2tn-2n-log-n#comment1444949_689108 – TooTone Feb 27 '14 at 00:31

1 Answers1

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No - for example if $$T(4)=2 \times 4-1=7$$ then $$T(8)=T(4)+\log_2(8)= 7+3=10 \not = 2 \times 8-1.$$

Just looking at powers of 2:

$$T(2)=T(1)+1$$ $$T(4)=T(2)+2=T(1)+3$$ $$T(8)=T(4)+3=T(1)+6$$ $$T(2^n)=T(1)+\frac{n(n+1)}{2}$$

Henry
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