24,936 reputation
43778
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location Seattle, WA
age 44
visits member for 4 years, 3 months
seen 1 min ago

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


2h
comment Help finding the residue of $1/(z^8+1)$
Your first factorization looks incorrect - e.g., it suggests that $i^{1/8}$ is a root of $z^8+1=0$ (i.e. $z^8=-1$) when clearly $(i^{1/8})^8=i$.
22h
comment Limit of $\lim_{x\to\infty} (1+\frac{f(x)}{x})^x$ for $f(x) = o(x)$
You shouldn't need an image for this - all of this is writeable in MathJax.
1d
comment Surprising identities / equations
I presume there's some relatively clean bijection based around a canonical cycle structure of a permutation that explains this?
Nov
20
comment How to find a general sum formula for the series: 5+55+555+5555+…?
Agreed - this is one of the finest 'simple' answers I've ever seen on this site.
Nov
20
comment Poisson Distribution finding $E(x^4)$
What have you tried, what are your thoughts on the problem? Your 'given' formula looks incorrect to me; where did you obtain it from?
Nov
19
comment What does “The closure of the shift-orbit of the Fibonacci word” mean?
This is a fine answer but I'm worried that it might be over OP's head yet. An example of a word that's in the closure but not in the shift-orbit proper would go a long way, but at least at first glance I can't see any reasonably explicit way of defining such a word (IIRC this is coupled pretty closely to some of Connes' noncommutative geometry and specifically the examples on Penrose tilings)
Nov
19
comment Solve $\lim_{n\to\infty}\frac{1}{{2n+1}}\sqrt{(25-a^2)n^2+2n+3}=2$ for $a$
The poor 5-12-13 triangle never gets any love... :(
Nov
19
comment Prove this congruence
The LHS should be the set of all quadratic residues $\bmod p$ (or closely related); you can consider attacking the problem from that angle.
Nov
17
comment prove that either $[K_{n-1}[\alpha]:K_{n-1}]=1$ or $[K_{n-1}[\alpha]:K_{n-1}]=2$
What do you mean by 'constructible points generated by' specifically? Under my usual understanding of constructibility, a point constructible from constructible points is itself constructible, so you have $\mathscr{C}_n=\mathscr{C}_1$ for all $n$...
Nov
16
comment If $(A\times B)\cup (B\times A) = C\times C$ then $A = B = C$
Do you need to prove it for all $C$, or that there exists a particular $C$? If it's the latter, then the simplest $C$ you can imagine certainly satisfies it...
Nov
16
revised How to differentiate $y=x^{y^{\sin x}}$
Title fix
Nov
15
comment Why must a map from $X$ to $S^{n}$ not be onto for it to be null homotopic
Going the other direction, look at the identity map $\iota:S^n\mapsto S^n$ and see if you can understand why that map isn't null-homotopic; that's at the heart of it.
Nov
15
comment Lesser known derivations of well-known formulas and theorems
I love this derivation, but it's worth noting that it's basically completing the square in disguise: shifting your extremum to zero is exactly equivalent to putting the quadratic in the form $s^2-t$, where $s=x+(b/2a)$.
Nov
14
comment Algorithm to find a proof of every provable theorem.
@KevinCarlson But that's sort of the point - this is a field where technical precision is essential to make sure that the things one's stating are actually correct. I'm frankly more than a little surprised to see this sloppiness, given the instructor (who has a generally excellent reputation).
Nov
14
comment Need help determining the pairs of quaternions that anticommute
What's more, even in the purely imaginary case you haven't shown that each product $x_iy_i=0$; instead, all you have is that their sum - the dot product - is zero.
Nov
14
comment Need help determining the pairs of quaternions that anticommute
One small problem: you don't know that $x_iy_i=0$ for $i=2,3,4$ - you've shown that that must be so in the pure-imaginary case, but that doesn't mean that you can use the result you derived for the purely-imaginary case in the case where $x$ and $y$ aren't purely imaginary.
Nov
14
comment Sum $\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$
Ahhh, of course - I should have been able to see that from the definition (it's of the form $f(s)-f(-s)$, so reverses sign when the sign of $s$ reverses), but somehow in actually trying to do the subtraction I got tangled. Mea culpa!
Nov
14
comment Sum $\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$
Something seems wrong in that first term - won't $\ln(1+\sqrt{t}/1-\sqrt{t})$ have an even expansion in terms of $\sqrt{t}$ (and thus be expressible in terms of integral powers of $t$)? That's a little problematic since it's being multiplied by $t^{3/2}$...
Nov
13
answered Is there official name for “Manhattan visibility” measure?
Nov
13
comment Is there official name for “Manhattan visibility” measure?
@greenoldman I can fix that... :-) Hang on a bit!