Zach Langley
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 Jan23 awarded Popular Question Nov12 awarded Enlightened Nov12 awarded Nice Answer Feb27 awarded Yearling Feb6 comment How is $n^{1.001} + n\log n = \Theta (n^{1.001})$? Moreover, $n^k$ dominates $\log^c n$ for $k, c > 0$. Feb5 revised How to show that $ALL_{DFA}$ is in P deleted 13 characters in body Feb27 awarded Yearling Feb11 answered Bound on Stirling numbers of the first kind? Nov24 awarded Citizen Patrol Nov23 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 27 characters in body Nov22 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 3 characters in body Nov22 revised How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ added 35 characters in body Nov22 answered How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ Nov21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ @dreamcrash No. $(n^{0.5})^i = n^{0.5i} \ne n^{{0.5}^i}.$ Nov21 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.) Nov21 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)? Nov21 comment How to solve this recurrence relation: $T(n) = 4\cdot T(\sqrt{n}) + n$ The exponent of 4 should be just $i$, not $i+1$. Nov16 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. Nov16 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. Nov15 comment How to understand why $x^0 = 1$, where $x$ is any real number? Although, I would interpret "5 times itself 0 times" as the empty product, which is defined to be 1.