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seen Oct 18 at 2:47

Don't have much time these days...


Oct
29
answered Adding integers to an infinite continued fraction expansion doesn't change the value?
Oct
28
revised Lower bound on product of distances from points on a circle
added 6 characters in body
Oct
28
comment How effective is this alternative to integration?
Maybe this will help regarding your rational function: en.wikipedia.org/wiki/Partial_fraction.
Oct
28
revised Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$
edited title
Oct
28
revised Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$
edited title
Oct
28
revised An inequality like Riemann sum involving $\sqrt{1-x^2}$
edited title
Oct
28
revised Existence of lines not containing given points in general position
edited title
Oct
28
revised Lower bound on product of distances from points on a circle
edited title
Oct
28
revised Lower bound on product of distances from points on a circle
added 95 characters in body; added 4 characters in body
Oct
28
revised Lower bound on product of distances from points on a circle
added 178 characters in body
Oct
28
answered Lower bound on product of distances from points on a circle
Oct
28
comment Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$
I have a feeling it is pretty standard (also supported by the paper stating it as a fact), but I am unable to place where I might have come across it. I had to derive it myself too, though.
Oct
28
comment Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$
+1: This is what I was about to add!
Oct
28
awarded  combinatorics
Oct
27
comment Iterated polynomial problem
What is the source of this problem? Do you know it, Jaska?
Oct
27
comment Iterated polynomial problem
@svenkat: No, in that case all the P(P(...P(1)...)) are odd.
Oct
27
comment Euler numbers formula
There must be better algorithms than using the formula you state. Perhaps this will have a better algorithm: emis.ams.org/journals/JIS/VOL4/CHEN/AlgBE2.pdf
Oct
27
comment Solving $2x \equiv 1 \pmod{p}$ where $p$ is an odd prime
When p=3, x=2. p=5,x=3. p=11,x=6. p=17,x=9. Do you see any pattern?
Oct
27
comment Do the equations used in Stargate make sense or are they gibberish?
Do you have a link/image?
Oct
27
answered Identity involving Euler's totient function: $\sum \limits_{k=1}^n \left\lfloor \frac{n}{k} \right\rfloor \varphi(k) = \frac{n(n+1)}{2}$