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seen Jun 27 at 7:01

Jun
27
comment Geometric interpretation of a Taylor series like identity
Are you aware of the derivation of the formula by repeated integration by parts? That yields what might be called a geometric interpretation, though it's probably not quite as geometric as you'd like it to be: The area under $f$ is the area under $(xf)'$ minus the area under $xf'$, which in turn is the area under $(\frac12x^2f')'$ minus the area under $\frac12x^2f''$, and so on.
Jun
26
comment Can we qualitatively predict the strategy of the German and US teams in today's World Cup soccer match?
@Hayden: Yes, that's why I put it in the answer and not in the question -- perhaps someone can come up with a more comprehensive analysis.
Jun
12
comment Multiple integral over a disc
Due to the radial symmetry, you can perform the outermost angular integral to get a factor $2\pi$. I wouldn't be surprised if it turns out to be impossible to make any progress beyond that.
Sep
23
comment Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
@lanzariel: I multiplied out the three parentheses; that yields eight terms; the two terms $\pm1$ cancel, and I grouped the remaining six terms conventiently for further manipulation.
Sep
19
comment What's up with Plouffe's inverter? Is there an alternative?
Why the downvote?
Aug
19
comment Left or right edge in cubic planar graph
@draks: I don't understand what you mean by "group the edges".
Aug
8
comment Sequence of powers of Gaussian integers capturing all positive integers?
@Nemes: That isn't possible, either. Consider the residues mod $2$. If $z=0+0\mathrm i\bmod2$ or $z=1+1\mathrm i\bmod2$, then $z^2=0\bmod2$, and the exponentially accumulating factors of $2$ prevent you from hitting all even integers. On the other hand, if $z=0+1\mathrm i\bmod2$ or $z=1+0\mathrm i\bmod2$, then each power has one even part and one odd part, so due to the period $4$ in the residues mod $10$ you miss at least one of the five even residues mod $10$.
Jul
27
comment Volume of a region on the sphere
@Tomás: You're right, sorry about that. I made them all $x$.
Jul
15
comment Minimizing the time to produce $T$ items with machines that run less efficiently over time
@Mirror: Hmm -- now I don't understand your calculation. From the first term, it seems that there's a single person setting up these blenders, so the time it takes to set them up is $Q$ times the time it takes to set one of them up. But then shouldn't the blenders start producing at different times, each one as soon as you're done setting it up? And why is $f$ being multiplied by $R/Q$?
Jul
15
comment Expected Value of a Negative Binomial Random Variable
@Alex: There's no assignment here; this is an equation.
Jul
10
comment A continuous random walk of length 1
There shouldn't be an $N$ in the distribution for one step?
Jul
9
comment Nodes in spherical equations and graph matching
You've written an expression, not an equation. Perhaps you intended to set this expression equal to zero?
Jul
7
comment Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$
@Thomas: Thanks -- indeed, it should; and that $x_1$ has remainder $2$ mod $3$ was even more blatantly wrong.
Jul
6
comment Probability Combinatorial related; choosing couples
By contrast, you are not counting e.g. $(B_3,G_1)(G_2,G_3)(G_4,G_5),(G_6,G_7)(B_1,B_2)$; this doesn't occur because you always choose the boys couple as the first pair. The factor $5!$ would only be correct if all $5!$ permutations of each pairing would occur in the way you choose pairs, but they don't.
Jul
6
comment Probability Combinatorial related; choosing couples
@rbm: I'm not sure how to explain further. Your first choice is a choice of two boys. The boys couple can only be chosen in this choice; it cannot be chosen in any other choice. By contrast, when you then make four choices of $2$ out of $8$, the mixed couple, for instance, can be chosen in any of those four choices. You're counting as different pairings e.g. $(B_1,B_2)(B_3,G_1)(G_2,G_3)(G_4,G_5),(G_6,G_7)$ and $(B_1,B_2)(G_2,G_3)(G_4,G_5)(G_6,G_7)(B_3,G_1)$, and there are $4!$ such equivalent pairings, so you have to divide by $4!$ to correct for that.
Jul
6
comment Probability Combinatorial related; choosing couples
Imagine a slot for each of the five binomial coefficients you write down. Each slot can be filled with a pair. The first slot, corresponding to $\binom32$, is always filled with two boys, whereas if you had all $5!$ different permutations of the pairs, this slot would sometimes have to contain other pairs, and the boys pair would sometimes have to occur in a different slot.
Jul
6
comment Probability Combinatorial related; choosing couples
In your calculation, the boys couple can again only occur as the first choice, but then the mixed couple and the girls couples can all occur as any of the remaining four choices, yielding a factor $4!$ for the orders in which they can occur.
Jul
6
comment Probability Combinatorial related; choosing couples
@rbm: If I understand correctly what you're doing, you seem to be making a similar mistake as Ofek. You only have to divide through for duplicates that occur in different orders if they actually do so occur. In Ofek's calculation, the boys couple could only occur as first choice, the mixed couple could only occur as second choice, and only the three girls couples could appear as any of the remaining three choices, yielding a factor $3!$ for the orders in which they can occur.
Jul
5
comment Intuition about whether to switch in box problem
That doesn't change the fact that this is a duplicate of that question. To make it a new question, you could edit it to point out what you're missing in the answer to that question -- as specifically as possible, but if the problem is that you don't understand that answer well enough to be able to describe what you're missing in it, you could just ask for an explanation in simpler terms -- in that case it would help if you describe as specifically as possible what sort of level of explanation you're looking for.
Jul
5
comment Card doubling paradox
@Ittay: How do you mean, "There is a uniform distribution on the real numbers"? There isn't in the usual sense of the term (see e.g. math.stackexchange.com/questions/14777).