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1h
comment the golden ratio and a certain simple continued fraction
As $k$ increases without bound, $q^k\to 0$, so the continued fraction just tends to $\frac1{1-q+(\phi-1)}$ (since we recognize the continued fraction for $\phi$).
1h
comment Elementary geometry question involving quadrilateral and bisector given with picture.
Can you elaborate on the very last sentence?
21h
comment Elementary geometry question involving quadrilateral and bisector given with picture.
I agree; I think there are two degrees of freedom still. Are you sure there weren't other assumptions in the problem, like line $BD$ bisecting angle $D$ also, or $ABCD$ being a special type of quadrilateral?
2d
revised How many ways can an integer $i$ appearing in a sequence with multiplicty at least $j$, be minimal
edited body
2d
comment How many ways can an integer $i$ appearing in a sequence with multiplicty at least $j$, be minimal
That's what I wrote - wait. What? That's what I should have written! Now it's what I wrote.
2d
answered How many ways can an integer $i$ appearing in a sequence with multiplicty at least $j$, be minimal
2d
comment Can one construct any n-gon if angle trisection is also allowed?
@QiaochuYuan: you're right - it follows from the prime number theorem in arithmetic progressions and from considering degrees of number fields.
Aug
27
answered Number of integer functions satisfying three constraints
Aug
27
comment bounding a sum using a definite integral
Yep, that's me!
Aug
26
answered bounding a sum using a definite integral
Aug
26
comment LCM of randomly selected integers
What kind of answer do you want? Exact number? Algorithm to determine it? Asymptotic size as 10000000 goes to infinity?
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Perhaps unhelpful comment: if $F(t) = \sum_{n=1}^\infty (-1)^n \log \Gamma(n t)$, then $S_m + \gamma\log 2 = F'(\frac1m)$.
Aug
19
comment Quadratic field extensions and complex conjugation
Do you intend that $K$ should be a subfield of $\Bbb R$ as well? If so, then the answer is yes: any non-real element of the quadratic extension of $K$ is a root of a quadratic polynomial with coefficients in $K$; the quadratic formula then shows that the two roots are complex conjugates of each other and their sum is in $K$, which implies that the complex conjugate is in the field generated by $K$ and the first root.
Aug
19
comment Prove that having 6 points in the interior of a square
Your rectangle solution works great! Note that indeed circles can be placed at the centers of those rectangles and that works too.
Aug
19
comment Prove that having 6 points in the interior of a square
Hmmm. What about the six subsquares consisting of the four along the main diagonal, together with the other two corners? Any subset of size 5 fails to cover all 16 subsquares.
Aug
19
comment Prove that having 6 points in the interior of a square
One way to rephrase part of this argument is in terms of the graph whose vertices correspond to the smaller squares, and edges correspond to vertical/horizontal adjacency; one needs to show that this graph has no independent vertex set of size 6.
Aug
19
comment Prove that having 6 points in the interior of a square
I prefer not to, since you have spent less than 3 minutes thinking about it.
Aug
19
answered Prove that having 6 points in the interior of a square
Aug
19
comment Showing an analytic function is exactly 0 or never 0 on a domain
Argument Principle!
Aug
18
comment Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
Can you give an example of a bijection on $\Bbb R^n$ that is connected but not continuous?