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location Seattle, WA
age 43
visits member for 3 years, 11 months
seen 24 mins ago

Software engineer and long-time dabbler in mathematics; 11th in the Putnams forever ago but I've long since atrophied.


1h
comment sequence $x_{n+1} = x_{n} + \sin x_{n}$
There's something wrong with your definition - $\frac\pi2$ isn't a fixed point of your recurrence. That is, if $|x_n-\frac\pi2|\lt\epsilon$ for some small $\epsilon$, $\sin x_n\approx 1$ so $|x_{n+1}-\frac\pi2|\gt\frac12$. Are you sure you don't want $\cos$ there, for instance?
17h
answered Differentiating $\frac{te^{\tan t}}{ln(3t+1)}$?
22h
comment About amicable numbers
Trivially there are fewer Mersenne primes than there are powers of 2, and the density of powers of 2 itself is exponentially small, so the density of Mersenne primes must be exponentially small (i.e., there can't be more than $O(\log n)$ Mersenne primes less than $n$). But remarkably, I can't actually find anything that implies that odd perfect numbers couldn't have asymptotic density much greater than even perfect numbers - even though the empirical evidence is compelling, it's even possible that there are only finitely many even perfect numbers, but infinitely many odd ones!
22h
comment About amicable numbers
Your asymptotic estimate doesn't work - you only have bounds, not actual values - and only upper bounds, at that - so you can't divide out to get an estimate. (For instance, if $a\lt 1$ and $b\lt 3$, you can't say anything about the relation between $\frac ab$ and $\frac13$.)
22h
comment About amicable numbers
@azimut It's certainly the case that the density of even perfect numbers is 0, and I wouldn't be surprised if we know enough about odd perfect numbers to say the same about them.
1d
comment Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$
@AsdrubalBeltran By their definitions - as I said, $a_1\lt a_3$ implies that $a_2\gt a_4$, and that implies that $a_3\lt a_5$, etc. In terms of the $b$ and $c$ sequences, this means that $c_1\lt c_2$, $b_1\gt b_2$, $c_2\lt c_3$, etc.
1d
comment Solve for numbers to appear on two six-sided dice
@vkaul11 Another way of thinking about it: fix the 3 to be on die A. Now we have to choose two out of the remaining five numbers to be on die A, and there are ${5\choose 2}=10$ ways of doing so.
1d
comment Solve for numbers to appear on two six-sided dice
@vkaul11 Because the other three faces on the dice (012) are indistinguishable, so for instance there's no difference between putting 356 on A and 478 on B vs. putting 478 on A and 356 on B. This is where the factor of 2 in my denominator comes from.
1d
revised Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$
added 195 characters in body
1d
revised Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$
Added a few more details.
1d
comment Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$
@AsdrubalBeltran Yes, that would be a better word for it. :-)
1d
answered Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$
1d
comment The ambiguity of set theory language
This is a fine answer all around, but I'm not sure about the second paragraph - there's a canonical notion of mapping a function onto a set of arguments, but a priori I wouldn't say that $\cos(\emptyset)$ is defined, nor would I say that $\cos(\{0, \frac\pi2\})=\{0,1\}$.
2d
comment Is chess Turing-complete?
The traditional definition of a 'position' does encode some information about past game data (e.g., whether either side can castle, what if any en passant captures are legal) but otherwise doesn't include any information as to how the pieces got to where they are; this is what makes e.g. retrograde analysis problems interesting.
2d
comment Is chess Turing-complete?
(And nitpickily, there's a small problem with your encoding - if a queen is on e.g. b8 and she wants to move in a NW direction, your encoding would try to move her to c9 and 'wrap' that around to c5 - but b8-c5 isn't a legal queen move.)
2d
comment Is chess Turing-complete?
The question is asking for a translation into a position, not into a move sequence - otherwise the answer is relatively trivial (there are plenty of non-interacting move sequences that you could use to encode just about anything).
2d
awarded  Nice Answer
2d
comment I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?
@Matt It's mostly pattern recognition (and I'll edit this into my answer later): seeing that the leading terms in the fraction are the same ($3x$ in both numerator and denominator) suggests that the fraction can be written as $1+f(x)$ for some $f()$ that goes to zero, and so the overall limit can be written colloquially as $(1+0)^\infty$, which is a good candidate for trying to bring into the form of the classical exponential limit.
2d
answered I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?
Jul
7
comment Why do people lose in chess?
@TonyK I think if I had to choose between the two, I would say 'Chess is a first-player win' is orders of magnitude more likely than Goldbach's conjecture being false. (Contrariwise, I think a case could be made that Goldbach being false is marginally more likely than chess being a win for black!)