## SEO Keyword Density Issue – Webmasters Stack Exchange

I have a website that has a keyword density of 8% and 4% for my keywords, but I only used the keyword once. The website doesn’t have a lot of actual text. Does this high keyword density hurt my site’s SEO even though I only used it once? I checked my keyword density using the SEO Review Tools density checker.

## mp.mathematical physics – Diagonalization of the generalized 1-particle density matrix

Let $$mathscr{H}$$ be a complex separable Hilbert space and $$mathscr{F}$$ be the corresponding fermionic Fock space generated by $$mathscr{H}$$. Let $$rho: mathscr{L}(mathscr{F}) to mathbb{C}$$ be a bounded linear functional on all bounded operators of $$mathscr{F}$$ with $$rho(I)=1$$ and $$rho(A^*)=rho(A)^*$$, and define the 1-particle density matrix (1-pdm) by the unique bounded self-adjoint $$Gamma: mathscr{H}oplus mathscr{H}^* to mathscr{H}oplus mathscr{H}^*$$ such that

$$langle x|Gamma yrangle = rho((c^*(y_1)+c(y_2))(c(x_1)+c^*(x_2)))$$

where $$x=x_1 oplus bar{x}_2$$ and $$y=y_1 oplus bar{y}_2$$ (I use the notation $$bar{x} (cdot) = langle x|cdotrangle$$) and $$c,c^*$$ are the annihilation/creation operators.

In references V. Bach (Generalized Hartree-Fock theory and the Hubbard model)(Theorem 2.3) and J.P Solovej (Many Body Quantum Mechanics)(9.6 Lemma and 9.9 Theorem), the authors claim that (under suitable conditions) $$Gamma$$ is diagonalizable by a Bogoliubov transform $$W:mathscr{H}oplus mathscr{H}^* to mathscr{H}oplus mathscr{H}^*$$ so that $$W^* Gamma W = operatorname{diag}{(lambda_1,…,1-lambda_1,…)}$$. The main idea of the proof is that $$Gamma$$ is diagonalizable by an orthonormal basis, and that if $$xoplus bar{y}$$ is an eigenvector with eigenvalue $$lambda$$, then $$yoplus bar{x}$$ is an eigenvector with eigenvalue $$1-lambda$$. The proof is fine when $$lambdane 1/2$$, since the 2 eigenvectors are orthonormal to each other. However, if $$lambda=1/2$$, then things become a little more difficult. J.P Solove solves this in the case where the eigenspace of $$lambda =1/2$$ is even-dimensional, but as far as I know, I can’t understand why would it be.

Question. Is there something I’m forgetting? If not, is there a way or are there references that complete the proof?

## simplifying expressions – Calculate density of states for 1D system

The density of states for a 1D system can be written as:

$$Dleft(omegaright)=frac{L}{pi}frac{1}{domegaleft(kright)/dk}$$

I have some expressions for $$omegaleft(kright)$$:

``````w1(k_) = Sqrt((k1 (m1 + m2)/(m1*m2))*(1 -
Sqrt(1 - (2*(1 - Cos(k*a)) m1*m2)/((m1 + m2)^2))));
w2(k_) = Sqrt((k1 (m1 + m2)/(m1*m2))*(1 +
Sqrt(1 - (2*(1 - Cos(k*a)) m1*m2)/((m1 + m2)^2))));
``````

I need to calculate $$Dleft(omegaright)$$ analytically eliminating $$k$$ but I can only do that if I simplify the expression for w1,2(k) by assigning values to k1,m1,m2,a:

``````k1 = 1; a = 1; m1 = 1; m2 = 2;
w1(k_) = Sqrt((k1 (m1 + m2)/(m1*m2))*(1 -
Sqrt(1 - (2*(1 - Cos(k*a)) m1*m2)/((m1 + m2)^2))));
w2(k_) = Sqrt((k1 (m1 + m2)/(m1*m2))*(1 +
Sqrt(1 - (2*(1 - Cos(k*a)) m1*m2)/((m1 + m2)^2))));
sol = Solve(p == D(w2(k), k) && w == w2(k), {p}, {k});
Simplify((p /. sol((2)))^-1)
``````

Output:

``````(2 Sqrt((-3 + 2 w^2)^2))/Sqrt(6 - 11 w^2 + 6 w^4 - w^6)
``````

Is it possible to do it for the general case?

## mg.metric geometry – Fast way to generate random points in 2D according to a density function

I’m looking for a fast way to generate random points in 2D according to a given 2D density function.
For instance something like this:

Right now I’m using a modified version of “Poisson disc” method I found here:
https://www.jasondavies.com/poisson-disc/

To make this fast, it relies on a grid on fixed interval.
In my case, I don’t know the right size of the grid cells since the density changes over the plane.
(To generate this image I used no grid at all and it’s pretty slow)
Is there a “right way” to generalize this algorithm for this need?
Maybe a different approach altogether?

## calculus – Some question about proving \$displaystylelimsup_{ntoinfty}|cos{n}|=1\$ by using density of \${a+bpi|a,binmathbb{Z}}\$

I have seen Proving \$displaystylelimsup_{ntoinfty}cos{n}=1\$ using \${a+bpi|a,binmathbb{Z}}\$ is dense and got this question.
Hagen von Eitzen gave the solution as following:

Pick an integer $$n$$. By density of $$Bbb Z+piBbb Z$$, there exist $$a_n,b_ninBbb Z$$ with $$frac 1{n+1}. If $$a_m=a_n$$, then $$|b_npi-b_mpi|<1$$, which implies $$b_n=b_m$$ and ultimately $$n=m$$. We conclude that $$|a_n|to infty$$.
As $$cos|2a_n|=cos 2a_n=cos(2a_n+2pi b_n)>cosfrac 2nto 1,$$
the desired result follows.

I was wondering why $$|a_n|to infty.$$ Could someone give more details about it? –Moreover, is $$|a_n|$$ increasing to $$infty?$$

## unity – Controlling noise density

I’m trying to fake terrain blending by using an opacity mask on my road mesh to reveal the grass underneath. I’m currently multiplying some Perlin noise by a Rectangle node displaying my texture at 80% width which is giving me the following:

Obviously this doesn’t look great for reasons that should be apparent. The blotches are unnaturally spaced and they abruptly end with a sharp line at the end of the texture. What I would like to do is generate noise that looks more along the lines of this:

Is there a way to apply some sort of gradient to my noise to achieve this? I know I can just use the texture itself but I would like to be able to adjust properties like the extent of the noise during runtime.

## neutral density – Do I still need filters for photographing landscape under an eclipse, if I’m not zooming in on the sun itself?

Context: I find myself due to be in the path of the upcoming annular solar eclipse. This was unplanned and I am unable to get a proper filter delivered on time.

The general consensus on the internet appears to be that photographing an eclipse requires a solar filter, or at least a 16 stop, ND 100,000 filter, e.g.: https://www.bhphotovideo.com/explora/photography/tips-and-solutions/how-to-photograph-a-solar-eclipse

But it seems people frequently take photographs under a normal sun without apparently needing such extreme levels of filtering. Perhaps such advice is geared towards people trying to fill the frame with the sun using a super telephoto lens? It feels intuitive that aiming a telephoto lens directly at the sun would be more dangerous, like starting fires with a magnifying class.

So what I am wondering is, if I were to take a landscape photo with a mild telephoto lens (e.g. a 85mm), with an eclipsing sun in a corner, do I still need the recommended protection? Or would a 10 stop, ND 1000 filter be sufficient?

Would this be different with 35mm or wider lens?

## neutral density – What is the best option for fitting an ND filter to an 82mm lens?

I’d recommend the square filters you’ve found. A great many filter types are made in the 100mm size, which is the size I’d recommend for this.

The most popular system for this is the Cokin system, but there are competing systems. HiTech makes them, too, for instance, as does Lee, as you’ve found.

The basic idea is that you buy a filter holder and then as many adapter rings as you need for the lens filter sizes you want to use the filter system with.

One nice thing about such a system is that you can get graduated ND filters which you can then slide up and down in the holder to position the boundary line where you want it. The system also allows easy rotation.

The Cokin system is much broader than this, offering such things as filters that fade from one color to another, but these sort of filters aren’t very useful in the digital world, IMHO, where a gradient overlay in a photo editor can achieve the same effect. Graduated ND is still quite useful today, though.

## physics – Simulating Gas Density and Pressure in a 2D World

I’m building a small spaceship simulation app that looks a lot like a game for an upcoming talk I’m giving where I use this sample app to teach the F# programming language.

This small app is something like FTL meets Oxygen Not Included where you have a top-down 2D grid of tiles (similar to an old RPG) where each tile has its own mixture of gasses – right now oxygen and carbon dioxide, but potentially others.

I’ve got a few things I’m trying to simulate:

1. When new gasses are added to a tile by something like a vent or a life support system, that gas should expand to neighboring tiles if possible
2. When a pressure changes (e.g. opening a door to another area of the ship or a hull breach), air should flow from the high pressure tile to the low pressure tile next to it.

Given this, and given that some gasses naturally sift to the top of others, I’m trying to figure out a small set of simple rules to govern this behavior.

Previously I had all gasses equalizing with their neighbors and no concept of pressure, but that made it very difficult to treat scenarios like hull ruptures, so I’m looking for something a bit more realistic without getting complex or hyper-accurate.

For example, given tile A with 15g oxygen and 6g CO2 and neighboring tile B of 3g oxygen and 1g CO2, some air should clearly flow from A to B. However, what flows? Is it the lightest gasses? The heaviest gasses? A random or representative sampling of gasses in A? Are there any relevant physics principles I should be aware of?

Note: I posted here instead of in physics because I don’t care extremely about nuanced accuracy, just something simple and believable

## Finding the cumulative distribution function and the probability density function of a set X which is not a set

A material point M moves at a constant angular velocity around a circle with the centre at (0,0) and
radius 1 (uniform angle distribution). Let X be the distance of point M from point (1,0). Find the cumulative
distribution function and the probability density function of X.

What I know so far

With some help of geometry you’ll get $$X = d(x)=sqrt{2-2x}$$ as the function which describes the distance between (1,0) and M

Then we got

$$F_Xleft(tright)=Pleft(Xle tright)=left{ begin{array}{lr} 0 & t<0\ frac{t^2}{4} & 0≤t≤2\ 1 & t>1\ end{array} right.$$

It’s very easy to check this, simply $$sqrt{2-2x} ≤ t$$ and you’ll see that $$frac{2-t^2}{2} = x$$

But that’s the strange point here: P is defined by $$Pleft(xright)=frac{1-x}{2}$$ (that’s why for t=0 P is equal to 0, for t=2 P is equal to 1, and if you go below 0 or above 1 the value cannot change to something lower than 0/higher than 1 because P is a probability measure). But why is P(x) defined in such a way?

And also, why do we calculate here with$$frac{2-t^2}{2} = x$$ and not $$frac{2-t^2}{2}le x$$?