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Sep
11
answered Estimating $\sum n^{-1/2}$
Sep
11
revised Moebius function log sum
edited tags
Sep
11
comment Moebius function log sum
Are you asking for a proof of that? Where did you find that identity?
Sep
10
comment Recognizable vs Decidable
@EricLeschinski: I am not sure what you are trying to say. We are talking about Turing machines. Talking about programming languages seems irrelevant.
Sep
10
awarded  Enlightened
Sep
10
awarded  Nice Answer
Sep
9
comment How can you prove that a function has no closed form integral?
@Mauris: Thanks.
Aug
12
awarded  Yearling
Jul
21
revised For any integer $n>1$ exist integers $a$ and $b$ so that $\tau(a)+\tau(b)=n$
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Jul
19
comment Solving recurrence $T(n) = T(\lceil n/2 \rceil) + T(\lfloor n/2 \rfloor) + \Theta(n)$
@IntrepidTraveller: We are considering a mathematical recurrence, it could be space/time complexity or even something else (like number of comparisons). Besides, we can replace $T$ in the above post with $G$ where $G(n) = T(n) - T(1)$ and for that $G$, $G(1)$ is indeed $0$. $T(1) = 0$ is taken to simplify the math, and does not change the end result.
Jul
8
revised What is $x$ if $\{x\}+\{\frac{1}{x} \}=1$ ? ({} - fractional part)?
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Jul
8
revised What is $x$ if $\{x\}+\{\frac{1}{x} \}=1$ ? ({} - fractional part)?
edited tags; edited tags
Jul
4
revised If, $x+y=1, x^2+y^2=2$ Find $x^7+y^7=??$
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Jun
13
awarded  Favorite Question
Jun
11
awarded  Nice Answer
Jun
9
comment How do I solve inequalities of the form $\left|\frac{f(x)}{g(x)}\right| \geq 1$?
@StevenGregory: You can. Notice the absolute values...
Jun
9
comment If $A,B,C,D$ are complex numbers on the unit circle with $A+B+C+D=0$, then they form a rectangle
@YotasTrejos: It is not a square. Basically, given $A$, you just reflect it along x and y axis, and then take the negative to get the four corners of the quadrilateral. Now take $A$ close to the y axis and far far away from the x axis. Do you still get a square?
May
7
revised First number $\ge n$ that is divisible by $k$?
edited tags
May
4
comment $g^q-q$ and $g^q-gq$ are primitive roots modulo $q^2$
Using $g$ and $q$ is confusing!
Apr
26
comment Reduction from Hamiltonian cycle to Hamiltonian path
@graphtheory92: Seems valid to me. What are your concerns?