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Dec
21
answered A little question about convergence of sequence
Dec
8
comment How did mathematicians decide on the axioms of linear algebra
"(...) the early history of vectors was complicated by people wanting to have a way to multiply vectors in analogy with multiplying real numbers or complex numbers (...)" This is very reasonable, but is there a written example?
Nov
2
awarded  Nice Question
Oct
21
revised Find the sum of all $x$ such that $y = 2 - \frac{9}{x+1}$ has an integer solution $(x, y)$
deleted 13 characters in body; edited title
Oct
16
awarded  Yearling
Oct
7
revised Finding $y$ so that $y^2 + x^2 \equiv 1 (\mod x + y)$
added 5 characters in body
Oct
7
answered Finding $y$ so that $y^2 + x^2 \equiv 1 (\mod x + y)$
Oct
1
answered Is $f$ uniformly continuous?
Sep
30
comment On the number of integers whose prime factors are congruent to 1 modulo some given number.
The last formula should be $(2n^2H_{2n}N^{(2n+1)/2n})/((2n+1) \log^2N)$, sorry. We should expect, then, $$C_m(N)\sim\frac{N}{\log^2 N}\sum_{n\geq 1} \frac{2n^2}{2n+1}H_n \frac{N^{1/2n}}{\varphi^n(m)}.$$
Sep
30
comment On the number of integers whose prime factors are congruent to 1 modulo some given number.
In general, the $n$th sum is asymptotic to $$\frac12\sum_{k=1}^n\pi(N^{k/2n})\pi(N^{(2n+1-k)/2n}),$$ which is like $$\frac12 \sum_{k=1}^n \frac{2nN^{k/2n}}{k\log(N)} \frac{N^{1-(k/2n)+(1/2n)}}{\left(1-(k/2n) +(1/2n)\right)\log(N)} =\frac{4n^2H_{2n}N^{(2n+1)/2n}}{(2n+1) \log^2N}$$ by the prime number theorem.
Sep
30
comment On the number of integers whose prime factors are congruent to 1 modulo some given number.
I think this possible in principle. Write $[1,N] = [1, N^{1/2}]\cup[N^{1/4}, N^{1/2}]\cup[N^{1/2}, N^{3/4}][N^{3/4}, N]$. Then the second sum is approximately $$\frac{1}{2}\left[\pi(N^{1/2})^2 + (\pi(N^{1/2}) - \pi(N^{1/4}))(\pi(N^{3/4}) - \pi(N^{1/2}))+ \pi(N^{1/4})(\pi(N) - \pi(N^{3/4}))\right].$$ Keeping only the relevant powers and using the prime number theorem, this should be asymptotic to $$\frac{10}{3}\frac{N^{5/4}}{\log^2(N)}.$$
Sep
30
comment On the number of integers whose prime factors are congruent to 1 modulo some given number.
Heuristically, I would expect $C_m(N)$ to be like $$\frac{1}{\varphi(m)}\sum_{p<N} 1+ \frac{1}{\varphi^2(m)}\sum_{pq<N}1+\cdots+\frac{1}{\varphi^k(m)}\sum_{pqr\cdots z < N} 1+\cdots,$$ by Dirichlet's theorem. The inner sums are probably well approximated by some manipulations involving $\pi(N^{1/k})$, which in turn might be combined with the prime number theorem to yield an asymptotic. Disclaimer: this is a stretch, I'm not familiar with such problems.
Sep
30
comment How to extend a function $f: S \rightarrow N$ to $\bar{f}: \bar{S} \rightarrow N$
The last part doesn't seem right. That it is continuous on a closed set does not suffice to claim uniform continuity.
Sep
30
answered Radius convergence of $\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n} z^{n(n+1)}$
Sep
28
comment Tricky trig question from GRE
Heh, it is almost as if the testmaker made $\cos^2(\arccos(\pi/12)) \color{Red}{=} \cos(\pi/12)=\sqrt{\frac{1+\cos(\pi/6)}2}$...
Sep
11
comment Euler's errors?
The link is broken.
Sep
7
comment Mental Calculations
@Deepak This is very, very nice. I wonder what both the painting (photograph? Possibly religious?) and the paper in the wall mean.
Sep
1
reviewed Leave Open How many ways can a quadratic form represent a prime?
Sep
1
reviewed Leave Open Maximum of $xy^3z^7$ in the plane $x+y+z=1$
Aug
21
comment In the real domain, are there any theorems or definitions that state all functions are differentiable?
"the most famous mathematicians who lived two centuries ago did believe that every function had a derivative." This is a bit strange. Maybe "was differentiable almost everywhere" would be better? I'm thinking about examples like $f(x) = |x|$, as I don't know whether they were thought as functions back then.