Zach Langley
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 Nov 2 awarded Popular Question Jan 23 awarded Popular Question Nov 12 awarded Enlightened Nov 12 awarded Nice Answer Feb 27 awarded Yearling Feb 6 comment How is $n^{1.001} + n\log n = \Theta (n^{1.001})$? Moreover, $n^k$ dominates $\log^c n$ for $k, c > 0$. Feb 5 revised How to show that $ALL_{DFA}$ is in P deleted 13 characters in body Feb 27 awarded Yearling Feb 11 answered Bound on Stirling numbers of the first kind? Nov 24 awarded Citizen Patrol Nov 23 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 27 characters in body Nov 22 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 3 characters in body Nov 22 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 35 characters in body Nov 22 answered How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ Nov 21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ @dreamcrash No. $(n^{0.5})^i = n^{0.5i} \ne n^{{0.5}^i}.$ Nov 21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ It looks like $T(n) = O(n)$. (Look at the last term in the summation.) Nov 21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ @dreamcrash Are you looking for an exact solution or just the asymptotic growth (Big-Oh notation)? Nov 21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ The exponent of 4 should be just $i$, not $i+1$. Nov 16 comment How to understand why $x^0 = 1$, where $x$ is any real number? Are you assuming $x^0 = 1$? You should reverse your argument. Nov 16 comment How to understand why $x^0 = 1$, where $x$ is any real number? @Rhys It doesn't help. I was just suggesting that there is nothing wrong with thinking about $5^0$ as $5$ times itself $0$ times.