635 reputation
614
bio website
location Rochester, NY
age 23
visits member for 3 years, 8 months
seen May 13 at 21:21

I am an undergraduate at RIT studying Computer Science and Mathematics.


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.
Nov
15
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.
Nov
15
revised Algorithmic Complexity of $i^2$
adding tex
Nov
15
suggested suggested edit on Algorithmic Complexity of $i^2$
Sep
15
comment Algorithmic Complexity of $i^2$
@MichaelGuantonio You won't be able to prove that because it's not true. $\sum_{i=1}^n i^k \ge n^k$ so the function is $\omega(n^{k-1})$. If you meant to switch the exponents, the exact same techniques I used above will work.