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7h
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14h
comment Derivative of sum of powers
@user258692 $p^k(1-p)^{n-k} = (1-p)^n\frac{p^k}{(1-p)^k}$, so your sum as a whole is $(1-p)^n\sum_{k=0}^n\left(\frac{p}{1-p}\right)^k$; now you can use the usual (partial) geometric formula to compute the sum explicitly.
15h
comment Derivative of sum of powers
Why not evaluate the sum first using the geometric theorem and then take the derivative of that?
18h
comment Is $\int_{x=0}^1\int_{y=0}^1\int_{z=0}^1 \frac{1}{(x-y)^2 (y-z)} dx dy dz$ finite?
Actually there's a perfectly reasonable way of assigning a value (specifically, if I've done my math right, $-\ln(2\min(y, 1-y))$) to your non-absolute value integral, but that only works because the singularity is isolated.
19h
comment Is $\int_{x=0}^1\int_{y=0}^1\int_{z=0}^1 \frac{1}{(x-y)^2 (y-z)} dx dy dz$ finite?
There is the broad notion of Cauchy Principal Value for one-dimensional integrals, but that doesn't work when the singularity isn't isolated; in this case, your singularity is the union of the planes $x=y$ and $y=z$, and it's not clear how any meaningful value can be assigned with a non-isolated singularity like this.
2d
comment Why this $\sigma \pi \sigma^{-1}$ keeps apearing in my group theory book? (cycle decomposition)
You very likely have the first two swapped around. Note that you can in fact find a permutation $\sigma$ with $\sigma(123)\sigma^{-1}=(456)$ using your third statement.
Jul
30
comment $2^{m+1}-2^n\geq(m-n)^2$
A very broad hint: since you know that $m\geq n$, consider setting $m=k+n$ and rephrasing the problem in terms of $k=m-n$ and $n$. You should be able to factor the LHS into two pieces each of which you have a good size for...
Jul
28
comment Finding all real numbers x such that $x \lceil x \lceil x \lceil x \rceil \rceil \rceil = 88$
A tiny gap here: $f(x)$ is not an increasing function for negative $x$...
Jul
28
awarded  Nice Answer
Jul
28
comment Is the Green-Tao theorem valid for arithmetic progressions of numbers whose Möbius value $\mu(n)=-1$?
(1) is trivially implied, as you suggest, but (2) is not trivially implied. That said, it would be very surprising if the proof of Green-Tao could not be extended to handle your case - in fact, probably even the special case (which implies your question) which covers only integers that are the products of three distinct primes.
Jul
26
comment $x^p -x-1$ irreducible over $\mathbb{F}_{p}$
A big hint: Fermat's little theorem shows that it can't have any linear factors - why?
Jul
26
comment Describe a fast (polynomial time)algorithm who takes as input the elements $g^a,g^b$ and gives as output the element $g^{a \cdot b}$
A loose hint: $(a+b)^2-a^2-b^2=2ab$...
Jul
24
comment Wondering how to rotate a normal vector in 4 dimensions?
That said, this is missing details: It's still not clear what you're trying to accomplish; clearly you want to rotate a point or figure in 4-space, but you still need to figure out how to specify the rotation you want. You may want to start with en.wikipedia.org/wiki/…, which gives (among other things) a representation of 4-dimensional rotations by pairs of quaternions (note: this is not the same thing as a biquaternion!).
Jul
24
comment Wondering how to rotate a normal vector in 4 dimensions?
There are six planes because a plane represents a pair of axes to rotate around, and there are $4\cdot3/2=6$ possible (unordered) pairs of axes - but you don't need a $6\times 6$ matrix to represent a rotation just because there are 6 possible planes. Instead, 6 represents the number of degrees of freedom available for a rotation. (This can also be proved more carefully through a sort of dimensional analysis).
Jul
24
answered Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$
Jul
24
comment Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$
@BolzWeir I vehemently disagree. After all, you could say that as $x\to\infty$, $\frac{x}{2x}\to\frac{\infty}{\infty}=1$. The way the notion can be made rigorous here is to divide numerator and denominator by $\tan(A)$so that none of the terms actually tend to infinite values.
Jul
24
reviewed Edit Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$
Jul
24
revised Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$
improved formatting
Jul
23
comment Is the alias method “stable”?
So if I understand correctly you're looking for a way of updating your background data structures in time linear (at worst) in $d$, where $d$ is some suitable measure of the 'distance' between the two probability distributions; is that a fair characterization?
Jul
23
comment Prove that a number is NOT periodic
@Semiclassical A terminating expansion is also a periodic expansion; just one in which every digit after a certain point is zero. (Or alternately, $b-1$ - e.g., $9$ in decimal expansions - but let's not get into that...)