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Feb
12
comment Number of pairs $(a, b)$ with $gcd(a, b) = 1$
Have you tested your answer for small values of $n$? Where does the 2 come from?
Feb
9
comment The globe, spherical disks and spherical straight lines
@sphere, the "angle excess" is proportional to the area of the triangle.
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
No, but it's still convenient for the reader to tell them why a statement isn't just pulled from thin air. And no, but I haven't had any coffee today. Aargh.
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
Ok, if you add a justification for assuming $f$ and $g$ to be positive then that covers the bases. BTW I think you need to change the bound on $c$ to be $c > 1$ or your example of $f(x) = x^{-1}$ can be changed to $f(x) = x^{-1} + c$ for small $c$ and the conclusion again fails to hold.
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
You're trying to have your cake and eat it. Either the absolute values are part of the definition, so you should include them when you state the definition, or they aren't, so you can't rely on them in the second paragraph.
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
Because not everyone who comes along here needs to read the whole chain. Besides, I was expecting an edit I made to add the absolute values from the definition you link to be accepted, and then it would have made no sense. Why did you reject it?
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
Are you working with a different definition of $O(g(n))$ than the one given in the first paragraph, then? Because it's certainly the case that $\forall n \ge 1 : -\log n < 1$
Feb
8
comment If $f(n) \in O(g(n))$, and $g(n) \geq 2$ then $\log{f(n)} \in O(\log{g(n)})$?
If this is homework, please tag it as such. Also, how far have you got by yourself? And what does $g(n) \ge 2$ mean? $\forall n : g(n) \ge 2$? $\forall n > 0: g(n) \ge 2$? $\exists N : \forall n > N : g(n) \ge 2$? Something else?
Feb
7
comment Optimization Puzzle
@user946850, it's actually ok (although, as I said, I've only given a sketch rather than a full formal proof). Big-O notation is about asymptotic behaviour, and asymptotically we can get tighter bounds on the distribution of primes. See the section "Bounds on the prime counting function" in the linked Wikipedia page on the prime number theorem.
Feb
7
comment Optimization Puzzle
@user946850, things like that are the reason that the primes need to be roughly equal. Note that having the bound $k$ from using the decision form of the problem is critical because it allows us to determine an objective criterion for what is "close enough".
Feb
7
comment Optimization Puzzle
@user946850, done, and thanks, respectively.
Feb
7
comment Optimization Puzzle
@ZevChonoles, I'd already made the edit to my own post following user946850's comment above. I was trying to let him get the rep for any upvotes because he came up with the key idea, but oh well.
Feb
6
comment Sine Approximation of Bhaskara
Analogous to taking $\pi \approx 16/5$ to get the correct gradient at $x=0$, you can take $\pi \approx \sqrt{10}$ to get the correct second differential at $x=\pi/2$.
Feb
6
comment Sine Approximation of Bhaskara
You can compute the Taylor series of the difference between this function and sine around $\frac{\pi}{2}$. I think the leading term is $\left(\frac{1}{2}-\frac{5}{\pi^2}\right) x^2 \approx -0.0066 x^2$. (At least, that's what I got by converting the approximation into an approximation for cos and then expanding the difference.
Feb
5
comment Combinatorics - pigeonhole principle question
Looks fine to me.
Feb
4
comment Solution of a system with exponentials
Smarter than subtracting $1$ from both sides and taking logs?
Feb
4
comment Optimization Puzzle
@Ross, yes, the interesting question is which GCDs to build when you have several targets.
Feb
3
comment Optimization Puzzle
So you have a directed graph with vertices corresponding to $\mathbb{Z}^+$ and edges $n \to kn$ with weight $k$ for all $n, k$. The task is to find the connected subgraph containing $1$ and the specified targets which has least total weight. As an optimisation, I think Patrick87's answer shows that you only need consider prime values of $k$.
Feb
2
comment How to compute Shannon information?
@Yrogirg, if you define "good" tightly enough and have a sufficiently precise value of $H_n$ then you might be able to recover the frequencies and calculate the true probabilities; otherwise, don't expect much joy.
Feb
2
comment How to solve quadratic over square root of quartic equals constant?
Can't you square and rearrange to a quartic, solve the quartic with Sturm sequences and Newton's method, and then check the solutions against the original equation?