1

I need to find the upper asymptotic bound for the recursion:

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

I was able to determine:

  1. The height of this tree is $k-1$.

  2. The cost of each level is $\displaystyle{\frac{2^i}{k-i}}$.

So then I am using the following formula to get an upper bound estimate: $$ T(n) = 2^h*T(1) + \sum \frac{2^i}{k-i}, $$ where the sum term is from $i=0$ to $h-1$. By plugging in the fact that the height, $h = k-1$, this formula can be expanded to $$ T(n) = 2^{k-1}*T(1) + \sum\frac{2^i}{k-i}, $$ where the sum term is from $i=0$ to $k-2$. However, I am stuck at reducing the summation term to produce an estimated Big(O) of just the first time, or $O(2^k)$.

Thanks.

user84756
  • 275

1 Answers1

0

I tried to solve this from scratch: $$ \\ T(n) = 2T(n - 1) + \frac{1}{n} \\ T(n) = 2(2T(n - 2) + \frac{1}{n - 1}) + \frac{1}{n} \\ T(n) = 2(2(2T(n - 3) + \frac{1}{n - 2}) + \frac{1}{n - 1}) + \frac{1}{n} = 2^3T(n - 3) +\frac{2^2}{n - 2}+\frac{2^1}{n - 1} + \frac{2^0}{n} \\ T(n) = 2^kT(n - k) + \sum_{i = 0}^{k - 1} \frac{2^i}{n - i} \\ \text{When } n - k = 0 \text{(instead of n - k = 1, to simplify)} \Rightarrow k = n \\ T(n) = 2^nT(0) + \sum_{i = 0}^{n - 1} \frac{2^i}{n - i} \\ T(n) = 2^n -\frac{\frac{1}{2}^{n + 1}\ { }_2F_1(1, n + 1;n+2; \frac{1}{2})}{n + 1} - \frac{1}{2} - \log(-\frac{1}{2}) $$ With a helpful answer here and here.

Intuitively, however, I think that (proof here): $$ { }_2F_1(1, n + 1;n+2; \frac{1}{2})\ \in\ O(2^n).$$

Community
  • 1
Mohamed Ennahdi El Idrissi
  • 315