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Solve $T(n)=T(n-2)+\frac{1}{\log(n)}$ for $T(n)$.

I am getting the answer as $O(n)$ by treating $1/\log(n)$ as $O(1)$. The recursive call tree of this is a lop-sided tree of height $n$. Hence, considering the amount of work done in each step, the answer comes out to be $O(n)$. Please verify my answer, and tell me if I am correct.

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If $T(n)=T(n-2)+\frac{1}{\log(n)}$, then $$ T(n)=\left\{\begin{array}{}T(0)+\sum_{k=1}^{n/2}\frac{1}{\log(2k)}&\text{if }n\text{ is even}\\T(1)+\sum_{k=1}^{(n-1)/2}\frac{1}{\log(2k+1)}&\text{if }n\text{ is odd}\end{array}\right.\tag{1} $$ Note that $$ \begin{align} \sum_{k=1}^{(n-1)/2}\frac{1}{\log(2k+1)} &\le\frac{1}{\log(3)}+\int_1^{(n-1)/2}\frac{\mathrm{d}x}{\log(2x+1)}\\ &=\color{#C00000}{\frac{1}{\log(3)}+\frac12\int_3^8\frac{\mathrm{d}x}{\log(x)}}+\frac12\int_8^n\frac{\mathrm{d}x}{\log(x)}\tag{2} \end{align} $$ and $$ \begin{align} \sum_{k=1}^{n/2}\frac{1}{\log(2k)} &\le\frac{1}{\log(2)}+\int_1^{n/2}\frac{\mathrm{d}x}{\log(2x)}\\ &=\color{#C00000}{\frac{1}{\log(2)}+\frac12\int_2^8\frac{\mathrm{d}x}{\log(x)}}+\frac12\int_8^n\frac{\mathrm{d}x}{\log(x)}\tag{3} \end{align} $$ For $n\ge8$, we have that $\log(x)\le2(\log(x)-1)$, thus $$ \begin{align} \frac12\int_8^n\frac{\mathrm{d}x}{\log(x)} &\le\int_8^n\frac{(\log(x)-1)\mathrm{d}x}{\log(x)^2}\\ &=\frac{n}{\log(n)}\color{#C00000}{-\frac{8}{\log(8)}}\tag{4} \end{align} $$ Combining $(1)$ though $(4)$, we get that $$ T(n)\le\color{#C00000}{C}+\frac{n}{\log(n)}=O\left(\frac{n}{\log(n)}\right)\tag{5} $$

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  • $\begingroup$ Excellent! this is really nice $\endgroup$ – pritam Jul 8 '12 at 9:03
  • $\begingroup$ Should not it be as shown below in eq 4. Notice the power 2 is on the whole log. $$ \begin{align} \frac12\int_8^n\frac{\mathrm{d}x}{\log(x)} &\le\int_8^n\frac{(\log(x)-1)\mathrm{d}x}{\log^2(x)}\\ &=\frac{n}{\log(n)}\color{#C00000}{-\frac{8}{\log(8)}} \end{align} $$ And if that is the case then the inequality is wrong. Also, I did not understand that how you arrived at this equation. $\endgroup$ – V K Jun 1 at 4:51
  • $\begingroup$ @VK: I am not sure what you are claiming. Are you saying that you disagree with equation $(4)$, or that you think equation $(4)$ somehow invalidates the inequality? Which equation do you not see how I arrived at? You've simply copied equation $(4)$. $\endgroup$ – robjohn Jun 1 at 11:59
  • $\begingroup$ I was saying that in eq 4 on the right hand side denominator, x is squared. But for the right hand side to result in n/log(n) after integration, instead of x being squared full log(x) term should be squared in denominator. $\endgroup$ – V K Jun 1 at 12:12
  • $\begingroup$ @VK: No. $\log(x)^2=(\log(x))^2$ not $\log\left(x^2\right)$, which would actually be $2\log(x)$. Operator precedence dictates that the parentheses around the argument of a function go before the exponent $2$. $\endgroup$ – robjohn Jun 1 at 12:16
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$$T(n)=T(n-2)+1/\mbox{log }n=T(n-4)+1/\mbox{log }(n-2)+1/\mbox{log }n=\cdots $$ So $$T(n)=a+\sum_{i=1}^{\lfloor n/2\rfloor}\frac{1}{\mbox{log}(n-2i)}\leq a+\sum_{i=1}^{\lfloor n/2\rfloor}\frac{1}{b}=a+bn/2=O(n) $$ Where $a=T(0),T(1)$ and $b=\mbox{log }2,\mbox{log }3$ depending on even or odd $n$. This shows that O(n) is an upper bound

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$T(n)=T(n-2)+\frac{1}{\log(n)}$

$\implies T(n-2)=T(n-4)+\frac{1}{log(n)}+\frac{1}{\log(n-2)}$

So, $T(n) = T(n-2r)+\sum_{0≤2s<2r}\frac{1}{\log(n-2s)}$

If $n$ is even, $2r\leq n-2$, $T(n) = T(2)+\sum_{0≤2s<n-2}\frac{1}{\log(n-2s)}$

If $n$ is even, $2r\leq n-1$, $T(n) = T(1)+\sum_{0≤2s<n-1}\frac{1}{\log(n-2s)}$

In either cases, $O(n) = \frac{n}{\log(n)}$

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  • $\begingroup$ I don't think the bound is that tight. $\endgroup$ – user17794 Jul 6 '12 at 9:39
  • $\begingroup$ @lab bhattacharjee: Can you explain how did you get O(n/log n) from that sum ? $\endgroup$ – pritam Jul 6 '12 at 9:48
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    $\begingroup$ @Pritam, each term in the summation is $O(log(n))$. Now there are 1+$\frac{(n-2)}{2} =\frac{n}{2}$ or 1+ $\frac{(n-1)}{2}= \frac{n+1}{2}$ terms respectively. Don't they result in $\frac{n}{log(n)}$ ? $\endgroup$ – lab bhattacharjee Jul 8 '12 at 4:28
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    $\begingroup$ @labbhattacharjee: You mean each term of the summation is $O(1/log(n))$ ? Thats not correct, in fact all the terms in the summation are bigger than $O(1/log(n))$ $\endgroup$ – pritam Jul 8 '12 at 9:01

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