## html – A circle in a circle

Just experimenting some things with css, I made a centered circle div (inner) which is inside a circle div (outer).

``````<!DOCTYPE html>
<html lang="en">
<title>Document</title>
<style>
body {
margin: 0;
}
.outer {
background-color: skyblue;
height: 200px;
width: 200px;
padding: 25px 25px 25px 25px;
}
.inner {
background-color: lightgreen;
height: 150px;
width: 150px;
margin: 25px 25px 25px 25px;
}
</style>
<body>
<div class="outer">
<div class="inner"></div>
</div>
</body>
</html>
``````

The thing is I had to give padding in “outer” and also margin to “inner”. So I want to know if there is a better way of doing this.

## graphics3d – Quarter circle surface opening outward

I have a `Graphics3D` issue, I have a cylinder within a cylinder. I want to generate a surface from a quarter circle, but the quarter circle should be “curving outward”. The normal surface generated for a quarter circle “curving inward” is a hemisphere. I do not know if this surface has a specific name but it is similar to a wormhole, although the wormhole is somehow like a hyperboloid. The surface should start from the inner cylinder and ending perpendicularly on the outer cylinder, that is why I’m not using a hyperboloid for this surface. I have drawn a quarter circle curve below, if you rotate it by $$2 pi$$ then that is the surface I’m looking for.

``````rbelow = 0.5;
dbelow = {0, 0, -1};
rabove = 0.5;
dabove = {0, 0, 1};
HemisphereBelow = ParametricPlot3D[rbelow {Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]} + dbelow, {u, -[Pi]/2, [Pi]/2}, {v, -[Pi]/2, [Pi]/2}, PlotStyle -> Blue, Mesh -> None, Boxed -> False, Axes -> None][[1]];
HemisphereAbove = ParametricPlot3D[rabove {Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]} + dabove, {u, -[Pi]/2, [Pi]/2}, {v, [Pi]/2, 3 [Pi]/2}, PlotStyle -> Blue, Mesh -> None, Boxed -> False, Axes -> None][[1]];
Graphics3D[{Opacity[0.05, Black], Cylinder[], Opacity[0.2, Green], Cylinder[{{0, 0, -1}, {0, 0, 1}}, 0.5], Opacity[0.5, Blue], HemisphereBelow, Opacity[0.5, Blue], HemisphereAbove}, Boxed -> False]
``````

## dnd 5e – Do the Circle of the Land druid’s Circle Spells stack with your normal druid spells, or do you have to switch out?

The circle spells are, as you say, always prepared on top of your other spells.

There are a number of spells you have prepared from the basic Spellcasting trait common to all druids:

You prepare the list of druid spells that are available for you to cast, choosing from the druid spell list. When you do so, choose a
number of druid spells equal to your Wisdom modifier + your druid
level (minimum of one spell)

If you are a land druid, then once you reach certain levels you have other spells that are always prepared on top of what you have chosen (note the bolded section).

Once you gain access to a circle spell, you always have it prepared,
and it doesn’t count against the number of spells you can prepare each
day

Remember that druids can choose (change) their “normally” prepared spells every day. So if you already have a circle spell prepared as one of your normal spells, then you can simply prepare something else because being a land druid will mean you have it always prepared anyway.

E.g. If you are an arctic druid, then no need to prepare Spike Growth as one of your standard prepared spells, because as an arctic druid that spell is always prepared anyway on top of your other spells.

## Spells known and Spell slots

None of the above affects spell slots, which is the number of spells you cast per day.

You have spells known/prepared and you have spell slots.

1. Spells prepared are the spells you have available and ready.

E.g. a 3rd level druid may have, say, 6 spells prepared which can be 1st level or 2nd level spells. A 3rd-level druid can only choose 1st or 2nd level spells but you could prepare, say, four 1st level spells and two 2nd level spells, or 3 1st level spells and 3 2nd level spells etc.

1. The spell slots that you have are indicated on the class features table for druids. For a 6th level druid that would be four 1st level slots and two 2nd level slots. This is how many spells you can actually cast in a day.

When you cast a spell, you can pick any spell that you have prepared and that will use up a spell slot of at least the level of the spell (some spells have greater effect if you spend a higher level slot).

E.g. If you cast “Faerie Fire” – a 1st level spell – you must spend a 1st level spell slot or higher. That slot is now “gone” until you take a long rest or use some other ability that gets slots back.

Circle spells are additional spells prepared and available but do not affect number of slots. But note that land druids do get the “Natural Recovery” feature that allows them to recover some spent spell slots during a short rest, allowing you to cast more spells in a day.

## traveling salesman problem – TSP on a unitary circle

I am interested in knowing which is the expected length of a random path in a circle. That is, if there are $$n$$ random points located in the unitary circle $${(x,y): x^2+y^2leq 1}$$, what is the expected length of the shortest path the connects all of them starting from the origin? In Wikipedia there are some results regarding the unitary square, but I have found nothing for the unitary circle. Thanks!

## dnd 5e – Do circle of the land druid spells stack with your normal druid spells, or do you have to switch out?

The circle spells are, as you say, always prepared on top of your other spells.

There are a number of spells you have prepared from the basic Spellcasting trait common to all druids:

You prepare the list of druid spells that are available for you to cast, choosing from the druid spell list. When you do so, choose a
number of druid spells equal to your Wisdom modifier + your druid
level (minimum of one spell)

If you are a land druid, then once you reach certain levels you have other spells that are always prepared on top of what you have chosen (note the bolded section).

Once you gain access to a circle spell, you always have it prepared,
and it doesn’t count against the number of spells you can prepare each
day

Remember that druids can choose (change) their “normally” prepared spells every day. So if you already have a circle spell prepared as one of your normal spells, then you can simply prepare something else because being a land druid will mean you have it always prepared anyway.

E.g. If you are an arctic druid, then no need to prepare Spike Growth as one of your standard prepared spells, because as an arctic druid that spell is always prepared anyway on top of your other spells.

## Spells known and Spell slots

None of the above affects spell slots, which is the number of spells you cast per day.

You have spells known/prepared and you have spell slots.

1. Spells prepared are the spells you have available and ready.

E.g. a 3rd level druid may have, say, 6 spells prepared which can be 1st level or 2nd level spells. A 3rd-level druid can only choose 1st or 2nd level spells but you could prepare, say, four 1st level spells and two 2nd level spells, or 3 1st level spells and 3 2nd level spells etc.

1. The spell slots that you have are indicated on the class features table for druids. For a 6th level druid that would be four 1st level slots and two 2nd level slots. This is how many spells you can actually cast in a day.

When you cast a spell, you can pick any spell that you have prepared and that will use up a spell slot of at least the level of the spell (some spells have greater effect if you spend a higher level slot).

E.g. If you cast “Faerie Fire” – a 1st level spell – you must spend a 1st level spell slot or higher. That slot is now “gone” until you take a long rest or use some other ability that gets slots back.

Circle spells are additional spells prepared and available but do not affect number of slots. But note that land druids do get the “Natural Recovery” feature that allows them to recover some spent spell slots during a short rest, allowing you to cast more spells in a day.

## set theory – Bounds for a covering number of the circle group \$mathbb T\$ by some its small subgroups


Recall that a circle $$mathbb T={zinmathbb C:|z|=1}$$, endowed with the operation of multiplication of complex numbers and the topology inherited from $$mathbb C$$ is a topological group. We consider a cardinal $$cov(A(IT))$$ which is the smallest size of a family $$mathcal U$$ of strictly increasing sequences $$(u_n)_{ninomega}$$ of natural numbers such that for each $$zinIT$$ there exists $$(u_n)_{ninomega}inmathcal U$$ such that a sequence $$(z^{u_n})_{ninomega}$$ converges to $$1$$.
It would be ideally for us to find a known small cardinal equal to $$cov(A(IT))$$. While $$cov(A(IT))$$ remains unknown, we are interested in bounds for it by known small cardinals.

Our try.

Upper bounds.

Let $$(w)^w$$ denote the family of all infinite subsets of $$w$$. A subfamily $$mathcal Rsubseteq(w)^w$$ is called reaping if for any set $$Xin(w)^w$$ there is $$Rinmathcal R$$
such that one of sets $$Rcap X$$ and $$Rsetminus X$$ is finite. The reaping number $$mathfrak r$$ is the cardinality of the smallest reaping family. By Proposition 9.9 from (1),
$$mathfrak r$$ is the minimum cardinality of any ultrafilter pseudobase. Recall that a
pseudobase for a filter $$F$$ on $$omega$$ is a family $$mathcal P$$ of infinite subsets of
$$omega$$ such that every set in $$F$$ has a subset in $$mathcal P$$.

A family $$mathcal R$$ of infinite subsets of $$omega$$ is called $$sigma$$-reaping,
if for any countable family $$mathcal X$$ of infinite subsets of $$omega$$ there
is $$Rinmathcal R$$ such that for any $$Xin mathcal X$$ one of sets $$Rcap X$$ and $$Rsetminus X$$
is finite. The $$sigma$$-reaping number $$mathfrak r_sigma$$ is the cardinality
of the smallest $$sigma$$-reaping family. Clearly, $$mathfrak rlemathfrak r_sigma$$
and there is an old open problem whether $$mathfrak r is consistent,
see (4), (3), and (1, 3.6).
By cite(3), $$mathfrak r_sigmalemathfrak u_p$$,
where $$mathfrak u_p$$ is the smallest base of a $$P$$-point if a $$P$$-point exists and $$mathfrak u_p=mathfrak c$$ if no $$P$$-point exists. It is known that $$mathfrak u_p=mathfrak u$$ if $$mathfrak u. Let us recall that $$mathfrak u$$ is the smallest cardinality of a base of a free ultrafilter on $$omega$$.

By Theorem 3.7 from (1), $$mathfrak r_sigma$$ is equal to the smallest cardinality of a
family $$mathcal Rsubseteq(w)^w$$ such that for any bounded sequence of real numbers
$$(x_n)_{ninw}$$ there exists $$Rinmathcal R$$ such that the subsequence $$(x_n)_{nin R}$$ converges
in the real line. It easily follows that $$cov(A(IT))lemathfrak r_sigma.$$

Problem. Is $$cov(A(IT))lemathfrak r$$?

Lower bounds.

For any family $$I$$ of sets with $$bigcupInotinI$$ let $$cov(I)=min{|J|:JsubseteqI;wedge;bigcupJ=bigcupI}$$ and $$non(I)=min{|A|:AsubseteqbigcupI;wedge;AnotinI}$$. Let $$M$$ and $$N$$ be the ideals of meager and Lebesgue null subsets of the real line, respectively.

It is easy to show that $$cov(A(IT))gemax{cov(M),cov(N),x}$$, where $$x$$ is an auxiliary cardinal introduced as follows. An infinite set $$Rsubseteqomega$$ of natural numbers is called remote if there exists $$zinIT$$ such that $$inf_{nin R}|z^n-1|>0$$. Let $$x$$ be the smallest cardinality of a family $$F$$ of infinite subsets of $$omega$$ such that for any remote set $$R$$ there exists $$FinF$$ such that $$Fcap R$$ is finite. So it would be good for us to find a known small cardinal equal to $$x$$. While $$x$$ remains unknown, we are interested in bounds (especially lower) for it by known small cardinals.

Our try for $$x$$.

We can prove that $$cov(M)le x$$ and are interested whether this bound can be improved and whether $$cov(N)le x$$.

Lyubomyr Zdomskyy suggested that it is consistent that $$mathfrak d, where $$mathfrak d$$ is the cofinality of $$w^w$$ endowed with the natural partial order: $$(x_n)_{ninw}le (y_n)_{ninw}$$ iff
$$x_nle y_n$$ for all $$i$$.

We introduced an auxiliary cardinal $$x_{lac}$$, which is the smallest cardinality of a family $$F$$ of infinite subsets of $$w$$ such that for any lacunary set $$L$$ there exists $$FinF$$ such that $$Fcap L$$ is finite. Recall that an infinite set $$L$$ of natural numbers is called lacunary, if $$inf{b/a:a,bin L,;a1$$. We have $$x_laclex$$, because Pollington in (2) proved that any lacunary set is remote, as John Griesmer informed us. But it turned out that $$x_lac$$ is rather small. Namely, Will Brian showed that $$x_laclenon(N)$$ and the strict inequality here is consistent.

References

(1) A. Blass, Combinatorial Cardinal Characteristics of the Continuum, in: M. Foreman, A. Kanamori (eds.), Handbook of Set Theory, Springer Science+Business Media B.V. 2010, 395–489.

(2) Andrew D. Pollington, On the density of sequences $${n_kxi}$$, Ill. J. Math. 23* (1979) 511–515, ZBL0401.10059.

(3) J. Vaughan, Small uncountable cardinals and topology, Open problems in topology (J. van Mill and G. Reed, eds.), North-Holland, Amsterdam, 1990, 195–218.

(4) P. Vojtáš, Cardinalities of noncentered systems of subsets of $$omega$$, Discrete Mathematics 108 (1992) 125–129.

Thanks.

## algebraic topology – If \$A\$ is circle then why is \$A \$ contractible?

Show that there are no retractions $$r: X rightarrow A$$ in the following cases:

(c) $$X = S^1 times D^2$$ and $$A$$ the circle shown in the figure.

My attempt : I found the answer here But i didn’t undertstand the answer.If $$A$$ is circle then why is $$A$$ contractible ?

My thinking: I had two things in my mind.

case I: Here $$pi_1(A) cong mathbb{Z}$$ and $$pi_1(X) cong mathbb{Z} implies$$
Every subgroup of $$mathbb{Z}$$ is cyclic $$implies$$ the homomorphism $$i_* : mathbb{Z} to mathbb{Z}$$ is injective

Cases II :let $$i : A hookrightarrow X$$ be the inclusion , and fix $$x_0 in A$$.The induced map $$i_*:pi_1(A,x_0) to pi_1(X,x_0)$$ is defined by $$i_*((f))=(if)$$

we know that $$(f) in pi_1(A,x_0)$$ where $$f: S^1 to A$$ is a loop based at $$x_0$$

similarly $$(if) in pi_1(X,x_0)$$ where $$if: S^1 to X$$ is a loop based at $$x_0$$

Use the fact : circle is not contractible

Also, it is given that $$A$$ is circle $$implies pi_1(A,x_0) cong mathbb{Z}$$ which is non-trivial

similarly $$pi_1(S^1 times D^2) simeq pi_1(S^1) times pi_1(D^2) simeq mathbb{Z}$$ which is non-trivial

This implies the inclusion $$i: A hookrightarrow X$$ induces a non-trivial map $$i_* : pi_1(A) to pi_1(X)$$

Therefore $$i_*$$ is injective

Tell me where I’m wrong?

## python – Trying to graph half circle with matplotlib

My half circle doesn’t really look like how I expected.

Am I doing this right or am I missing something pretty big here ?

``````import math
import numpy as np
import matplotlib.pyplot as plt

coord_list = ()

h = 0
k = 0
r = 6

for x in range((1 + h - r), (h + r - 1), 1):
y1 = k + math.sqrt(r**2 - (x-h)**2)
coord_list.append((x, y1))

for each in coord_list:
print(each)

data = np.array((coord_list))

x, y = data.T

figure = plt.scatter(x, y)
figure = plt.grid(color = 'green', linestyle = '--', linewidth = 0.2)
figure = plt.show()
``````

## graphics – How to move around a circle and count the number of points inside it?

I have the following data:

``````data = {{223., 275.}, {212.5, 271.5}, {97.3889, 270.167}, {40., 269.}, {52.75, 266.875},{104.5, 265.5}, {241.7, 265.5}, {205.318, 263.318}, {217.045, 262.136}, {117.3, 257.5}, {69.2, 253.8}, {106.611, 253.833}, {198.389, 253.833}, {222., 254.}, {233.5, 254.}, {36.3889, 252.833},{210.214, 252.643}, {125.5, 250.5}, {92.5, 246.7}, {136.2, 246.8}, {115., 246.}, {147.682, 245.682}, {21.3, 244.5}, {104.5, 244.5}, {217.3, 244.5}, {244.3, 244.5}, {42.5, 242.3}, {72.8, 242.2}, {193.864, 241.955}, {31.8636, 240.955}, {123.1, 241.1}, {156., 241.}, {52., 240.}, {84.3889, 239.833}, {97.5, 237.5}, {142.786, 237.5}, {237.167, 237.611}, {64.5, 236.3}, {133.5, 236.5}, {109.045, 234.136}, {223.045, 234.136}, {34.4091, 232.864}, {77.3889, 232.833}, {117.318, 232.773}, {87.625, 231.625}, {151.3, 231.5}, {12.9545, 228.136}, {98.7, 228.}, {241.611, 225.833}, {139.5, 225.5}, {23.7, 224.5}, {81.1364, 223.409}, {171.3, 223.5}, {35.3889, 221.167}, {120.625, 221.375}, {158.2, 221.2}, {89.1364, 220.045}, {133.5, 217.5}, {147.7, 216.5}, {237.167, 216.611}, {110.167, 215.611}, {166.375, 215.375}, {177.167, 215.389}, {27.1364, 212.955}, {121.9, 211.9}, {38.8333, 210.611}, {92.5, 210.5}, {131.5, 208.3}, {158.056, 208.167}, {143., 207.3}, {19.0833, 204.75}, {72.5, 204.7}, {30.5, 203.5}, {171.167, 203.611}, {239.5, 203.5}, {252.3, 203.5}, {264.136, 203.591}, {84.7, 202.5}, {228.7, 202.5}, {42.3889, 199.833}, {123., 200.}, {51.6111, 198.833}, {76.3182, 196.682}, {64.5, 196.5}, {104.3, 195.5}, {144.167, 194.25}, {259.389, 194.167}, {34.1364, 191.955}, {235., 192.}, {114.833, 190.389}, {187.864, 190.591}, {56.5, 188.5}, {136.409, 188.318}, {158.3, 188.5}, {10.6111, 187.167}, {106.625, 183.625}, {139.409, 181.5}, {230.3, 182.5}, {47.2273, 181.318}, {90.3182, 181.318}, {59.8, 177.8}, {130., 178.}, {187.7, 178.}, {208.7, 175.5}, {81.1364, 174.409}, {139.5, 173.7}, {197.3, 173.5}, {241.833, 173.389}, {89.1154, 170.885}, {59.8, 167.2}, {175.625, 167.375}, {188., 166.}, {72.6111, 165.167}, {110.167, 165.056}, {159.611, 165.167}, {204., 165.}, {84.5, 161.5}, {143.136, 161.409}, {168.833, 161.389}, {262.7, 161.5}, {15.6111, 159.167}, {121.833, 159.389}, {102., 156.9}, {152.5, 157.}, {133.5, 154.5}, {75.0455, 153.136}, {113.625, 153.375}, {162.833, 153.389}, {85.5909, 152.136}, {241.611, 152.167}, {125.5, 149.7}, {96.3, 148.5}, {35.5, 146.5}, {105.5, 145.3}, {152.389, 145.167}, {197., 144.}, {85.7, 141.5}, {134.5, 141.5}, {162.833, 141.611}, {260.5, 141.5}, {175.833, 140.389}, {69.3889, 138.167}, {28.6429, 136.786}, {97.2, 136.8}, {129.167, 136.}, {39.625, 135.625}, {55.1667, 133.611}, {77.5, 133.5}, {164.3, 133.5}, {138.5, 132.5}, {213.5, 132.3}, {9.5, 131.}, {87.6111, 131.167}, {22.3889, 128.833}, {35.3889, 128.833}, {66.7, 128.5}, {177.136, 128.591}, {131.5, 125.3}, {189.722, 125.167}, {202., 125.}, {77.4091, 124.136}, {106.611, 123.833}, {160.611, 124.167}, {28.5, 122.7}, {142., 123.}, {214.5, 123.}, {226.5, 123.}, {15.5, 120.5}, {87.8333, 120.389}, {171.167, 120.611}, {5., 119.3}, {183., 117.}, {259.389, 117.167}, {73.1667, 115.583}, {98.6818, 115.682}, {131.318, 115.682}, {160.682, 114.682}, {34.1667, 114.611}, {193.7, 114.5}, {79.6667, 112.5}, {106.5, 112.5}, {203.786, 112.5}, {214.5, 112.5}, {13., 111.}, {59.3, 111.9}, {87.8636, 111.045}, {228.611, 111.167}, {69.3, 107.5}, {174.7, 107.5}, {151.3, 106.5}, {97.3889, 104.167}, {9.61111, 103.167}, {81., 103.}, {166.611, 103.167}, {61.2, 101.8}, {143., 102.}, {251.056, 101.833}, {71.5, 98.5}, {175.625, 98.375}, {238.167, 98.3889}, {90.3889, 95.8333}, {155.864, 95.9545}, {198.2, 94.9}, {138.375, 94.625}, {14.1667, 93.6111}, {59.8333, 93.6111}, {79.5, 93.5}, {112.682, 91.3182}, {33.875, 90.125}, {69.3182, 87.6818}, {150., 87.7}, {233.5, 87.5}, {180.5, 86.5}, {94., 85.3}, {190.167, 85.3889}, {242.833, 85.3889}, {19.1154, 82.8846}, {82., 83.}, {160.7, 83.}, {202., 83.}, {226.5, 79.7}, {129., 79.7}, {25.6818, 78.2273}, {39., 78.3}, {75.1364, 77.0455}, {48.1364, 74.9545}, {242.864, 74.9545}, {100.7, 73.5}, {120.5, 73.5}, {233.389, 73.1667}, {31.8333, 70.3889}, {69.2, 70.2}, {117.3, 65.5}, {226.5, 64.3}, {214.5, 62.}, {35.5, 60.7}, {205.3, 58.5}, {122., 57.3}, {180.375, 56.375}, {233.5, 56.3}, {43.5, 53.7}, {134.7, 54.}, {171.136, 53.9545}, {222.864, 53.9545}, {143.833, 49.0556}, {114.611, 47.8333}, {48.3, 45.5}, {81., 45.5}, {176., 45.5}, {40., 44.5}, {138.389, 40.8333}, {56.5, 39.7}, {127.239, 34.3261}, {67.0455, 32.8636}, {160.625, 32.625}, {118.5, 31.3}, {141.7, 31.5}, {55.0455, 29.1364}, {104.389, 28.1667}, {81., 26.7}, {93.7, 26.5}, {150., 26.7}, {110., 21.}};
``````

which looks like:

``````p1 = ListPlot(data, PlotStyle -> Black, AspectRatio -> 1, Axes -> False)
``````

Now, I need to draw a circle, let’s say, with a radius $$50$$, centered at each of the above points step by step, and count the number points inside the circle. For example, for one of the above points as a center of the circle, one has:

``````p2 = Graphics(Circle({79.66666666666667`, 112.5`}, 50));
Show(p1, p2)
``````

which looks like:

I can do the above repeatedly for other points as centers of the circle, and then count the number of points inside the circle at each step. However, this is very tedious and time-consuming.

I have the following two questions:

1. Is there an easier way to do the above procedure, and not manually do the counting and moving the circle around?

2. If the answer to the above question is yes, I need also the circle to be remained completely inside the region of the points, that is, the center-points which result in a circle which some parts of it go outside the region, to be excluded (when doing by hand, one can detect these points. For example, one set of these points are those ones at the edges of the picture).

## fractional calculus – Solid angle through a circle

I am trying to calculate the proportion of light emitted by a fluorophore that would be transmitted through a small hole. This is similar to calculating the solid angle defined by a circle of radius r and seen by a point.

I can calculate it for the case where the point is on top of the center of the hole, then solid angle is the solid angle defined by a circular cone : $$displaystyle Omega =2pi (1-cos {theta }$$ .
In the case where the point is misaligned it ‘sees’ the circle as en ellipse. i.e there are two different angles on the major and minor axis of the ellipse. I don’t really know how to handle this… i guess the solution lies in formulating some differential surface integration…