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


1h
awarded  Nice Answer
2h
answered How to prove $(1-\frac1{36})^{25}\lt\frac12$?
3h
comment Speeding up the convergence of a series
As far as convergence speedup goes, you may be able to use Euler's transform ( en.wikipedia.org/wiki/Series_acceleration#Euler.27s_transform ) to convert the series; it would take finding a good formula for the $k$'th difference of terms, but that may be feasible.
3h
comment Speeding up the convergence of a series
Out of curiosity, where does the sequence come from? (Normally it would be written with 'n' instead of 'x', FWIW, but that's a tiny thing.) The value of this sequence can probably be found explicitly using digamma functions, but presumably you're not particularly interested in that?
4h
comment Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?
@robjohn Sorry, yes - that's a very easy conflation to make. I almost never bump into the latter so I always tend to think of the former meaning, but you're definitely correct.
23h
comment Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?
@anonymous To be fair, the only case in which $c-b$ can divide coprime $b$ and $c$ is when $c-b=(\pm)1$.
23h
comment Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?
@miket The statement that $\sqrt{3}\sqrt{4a^3-1}$ isn't an integer is equivalent to the statement that there's no $b$ with $4a^3-1=3b^2$. This is an elliptic equation, and while there's a rich and developed theory behind them, at that point you're veering back to completely standard proofs of the theorem.
23h
comment Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?
To hit your edit: you haven't shown that of necessity there's no integer solution. You're right that the algebraic expression $\frac16(\sqrt{3}\sqrt{4a^3-1}+3)$ isn't a polynomial in $a$, but you still haven't shown any reason why it can't be an integer. (For a simpler example, $\sqrt{2x^2-1}$ isn't a polynomial expression, but that doesn't mean that it can't be an integer - take $x=12$, for instance.)
1d
revised Why is $f(n) =\frac{n(n+1)(n+2)}{(n+3)}$ in $O(n^2)$?
edited title
1d
comment Deciding whether a number is rational (2 examples)
@Just_a_fool The reason that the trick works here is that the 'inner' roots are the same, so after two squaring steps you're left with a single square root. OTOH, the method that you used for 1) (with the rational root theorem) can be used as an algorithm to determine the rationality or irrationality of an arbitrary algebraic expression (since the equation gives you a finite number of possible values to test against and all roots can be computed to sufficient precision). OTOH, there's a lot of subtlety in this; look up the 'sum-of-square-roots' problem in computational geometry for details..
2d
comment Is $\{1,1,2,3,4,5,\cdots,i,\cdots \} $ the simple continued fraction algebraic or transcendental?
@WillJagy As noted there, the expectation would be that almost all algebraic numbers should have unbounded CF coefficients, but just like with normal numbers I unfortunately don't know of any specific examples. I wonder if something can be said about the continued fraction expansion of Pisot numbers; hrm...
2d
answered Is $\{1,1,2,3,4,5,\cdots,i,\cdots \} $ the simple continued fraction algebraic or transcendental?
2d
comment Tiling a $23 \times 23$ square by $1 \times 1$, $2 \times 2$, and $3 \times 3$ tiles
@Jebediah Since neither of these answers has quite made it explicit: for either of these numberings, any 2x2 or 3x3 tile covers a set of numbers that adds to a multiple of 3. So if you could tile with only 2x2 and 3x3 tiles, then your total sum of numbers over the grid would have to be a multiple of 3. But the total sum of all numbers on the grid is one more than a multiple of 3, so...
2d
comment There is no polynomial $q$ such that $\int_0^1 p(x)q(x)\,dx=p(0)$ for each polynomial $p$.
Look ma, no topology! This is a really nice clean algebraic proof. (Not that there's anything wrong with the other proofs - they use appropriate machinery! I just like how this one could be explained to a gifted high-schooler, albeit with a little handwaving around showing that $\int_0^1p(x)dx = 0$ and $p(x)\geq 0$ on $[0,1]$ $\implies p(x)\equiv 0$ for polynomial $p()$.)
2d
comment Expected time to roll all 1 through 6 on a die
@DouglasZare Performing a little algebromancy... (a) trying to find the point at which a probability becomes 50% and presuming that's the expectation (i.e., confusing mean for median); (b) Assuming that the probabilities each number is rolled is independent? If you assume independence, that gives the LHS: $1-\left(\frac56\right)^n$ gives the probability that e.g. 1 has been rolled after $n$ rolls; now include one of those factors for each of the six faces...
Jul
28
comment Unit circle is divided into $n$ equal pieces, what is the least value of the perimeters of the $n$ parts?
@JackD'Aurizio The advantage with the triangular tiling was that it chopped up so cleanly into manageable pieces, but I agree that you can do much better. I don't have a ton of time to work on this, sadly (preparations for some awesome life events in the near future), but I have thoughts about an annulus sequence - definitely far-from-optimal - that would guarantee the $\Theta(\sqrt{n})$ bound, and I'll see if I can't get a specific constant out of it. If so, I'll amend this answer with a quick description of it?
Jul
28
comment The sum of two irrational square roots
@just1question Well, you know that $1+\frac nx$ is rational, and $\sqrt{x}$ is irrational, so...
Jul
28
comment Unit circle is divided into $n$ equal pieces, what is the least value of the perimeters of the $n$ parts?
(This construction depends on being able to split the wedges appropriately, but I believe that you can prove that a convex shape with diameter to 'minimum diameter' ratio less than 2 can be split transversely into two other convex shapes that also have diameter-to-minimum-diameter ratio less than 2.)
Jul
28
comment Unit circle is divided into $n$ equal pieces, what is the least value of the perimeters of the $n$ parts?
I think the $\Theta(\sqrt{n})$ estimate is almost certain to hold for exactly the reason you suggest; while the constant won't be best, I think you can prove the conjecture with a relatively explicit construction - just take a 'nearly-inscribed' triangle of area a rational multiple of pi with its vertices near the circumference, then chop the triangle into $k^2$ pieces for suitably-chosen $k$ and split the three 'wedges' surrounding it into $n-k^2$ pieces to fit. This won't necessarily work for all $n$, but you should be able to get infinitely many $n$ this way.
Jul
28
comment A sequence defined as $a(n)=n-a(a(n-1))$ $n\geq 1,\ a(0)=0$, how to prove that $a(n)=⌊(n+1)(-1+√5)/2⌋$
What have you tried so far? This is related to the so-called Wythoff sequence (the 'simplest' example of a Beatty Sequence: en.wikipedia.org/wiki/Beatty_sequence ) and you may be able to find some good hints with those terms.