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# Highlight integer points in RegionPlot

Is there any way to mark integer points after plotting region in Mathematica? For example, if I:

RegionPlot[x >= 4 y && x <= 4 y + 3 , {x, 0, 63}, {y, 0, 15}]

then it highlights the region but I want to see Y value corresponding to integer X value. Can I do it with Mathematica?

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I added a graphic to your post to help illustrate. Could you describe in more detail what you expect to see, please? –  Mr.Wizard Dec 11 '12 at 16:39
Like in above image I want to mark (x,y) where both x and y are integers. –  username_4567 Dec 11 '12 at 16:44
You might want to wait a bit with the accept, as others may provide better answers. –  István Zachar Dec 11 '12 at 17:00
Yes I can do that .. –  username_4567 Dec 11 '12 at 17:01

Use the inequality to sow all integer coordinates that are inside the boundary of the region when iterating through all integer pairs of the full range:

pts = First@Last@Reap@Do[If[x >= 4 y && x <= 4 y + 3, Sow@{x, y}], {x, 0, 63}, {y, 0, 15}]
RegionPlot[x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}, Epilog -> {Red, Point@pts}]

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Amazing..Thanks a lot.. –  username_4567 Dec 11 '12 at 16:58

Another way to generate all the points is by using Reduce:

points = {x, y} /.
List@ToRules@
Reduce[x >= 4 y && x <= 4 y + 3 && 0 < x < 63 && 0 < y < 15, {x, y}, Integers]

If you give bounds (and thus constrain the possible solutions to a finite set), Reduce will typically list all solutions.

Then just plot them with Point or ListPlot:

ListPlot[points]

Show them together with the RegionPlot:

Show[ListPlot[points], RegionPlot[...]]

Thanks to Mr.Wizard to pointing me to the following relevant note in the documentation:

Mathematica enumerates the solutions explicitly only if the number of integer solutions of the system does not exceed the maximum of the $p^{\text{th}}$ power of the value of the system option DiscreteSolutionBound, where $p$ is the dimension of the solution lattice of the equations, and the second element of the value of the system option ExhaustiveSearchMaxPoints.

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Might want to mention ExhaustiveSearchMaxPoints. +1 –  Mr.Wizard Dec 11 '12 at 18:29
@Mr.Wizard Thanks for the pointer. I didn't know about this option. –  Szabolcs Dec 11 '12 at 23:42
I changed your documentation link to an anchor I thought was more appropriate. If you disagree change it back, or perhaps add a second one. –  Mr.Wizard Dec 12 '12 at 19:30
@Mr.Wizard I agree, but I am usually too lazy to dig out the precise anchor (if it is not linked to and I can't just copy the link). Is there an easier way than using the dev tools of the browser or looking at the web page source? –  Szabolcs Dec 13 '12 at 0:23
There's a Meta post about that. I still use View Selection Source myself; it really doesn't take long. If your browser only lets you view the source of the entire page I can see the problem. –  Mr.Wizard Dec 13 '12 at 1:36

Alternatively you can generate just the points you want and then plot them :

data = DeleteCases[Flatten[Outer[Boole[4 #2 <= #1 <= 4 #2 + 3] {#1, #2} &,
Range[0, 63], Range[0, 15]], 1], {0, 0}];

Show[RegionPlot[x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}],  ListPlot[data]]

-

Another, using smart and fast functions like Array and Tuples, thus a bit more recommended way :

RegionPlot[ x >= 4 y && x <= 4 y + 3, {x, 0, 63}, {y, 0, 15}, Epilog -> {
Red, PointSize[0.005],
Point[ Join @@ Tuples /@ Array[ {Range[4 #, 4 # + 3], {#}} &, {16}, 0]]},
AspectRatio -> 15/63 ]

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By the way, you can just do AspectRatio -> Automatic... –  Rahul Dec 12 '12 at 2:30
@RahulNarain That's all ? –  Artes Dec 12 '12 at 19:04
I mean, instead of manually specifying 15/63. –  Rahul Dec 12 '12 at 21:49


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### 4.7.1 Design / Operating principle

The principle of operation of single-stage Roots pumps corresponds to the operating principle of multi-stage pumps as described in Chapter 4.5. In the Roots vacuum pump, two synchronously counter-rotating rotors (4) rotate contactlessly in a housing (Figure 4.16). The rotors have a figure-eight configuration and are separated from one another and from the stator by a narrow gap. Their operating principle is analogous to that of a gear pump having one two-tooth gear each that pumps the gas from the inlet port (3) to the outlet port (12). One shaft is driven by a motor (1). The other shaft is synchronized by means of a pair of gears (6) in the gear chamber. Lubrication is limited to the two bearing and gear chambers, which are sealed off from the suction chamber (8) by labyrinth seals (5) with compression rings. Because there is no friction in the suction chamber, a Roots vacuum pump can be operated at high rotation speeds (1,500 – 3,000 rpm ). The absence of reciprocating masses also affords trouble-free dynamic balancing, which means that Roots vacuum pumps operate extremely quietly in spite of their high speeds.

### Design

The rotor shaft bearings are arranged in the two side covers. They are designed as fixed bearings on one side and as movable (loose) bearings on the other to enable unequal thermal expansion between housing and rotor. The bearings are lubricated with oil that is displaced to the bearings and gears by splash disks. The driveshaft feedthrough to the outside on standard versions is sealed with radial shaft seal rings made of FPM that are immersed in sealing oil. To protect the shaft, the sealing rings run on a protective sleeve that can be replaced when worn. If a hermetic seal to the outside is required, the pump can also be driven by means of a permanent-magnet coupling with a can. This design affords leakage rates $Q_I$ of less than 10-6 Pa m3 s-1.

### Pump properties, heat-up

Since Roots pumps do not have internal compression or an outlet valve, when the suction chamber is opened its gas volume surges back into the suction chamber and must then be re-discharged against the outlet pressure. As a result of this effect, particularly in the presence of a high pressure differential between inlet and outlet, a high level of energy dissipation is generated, which results in significant heat-up of the pump at low gas flows that only transport low quantities of heat. The rotating Roots pistons are relatively difficult to cool compared to the housing, as they are practically vacuum-insulated. Consequently, they expand more than the housing. To prevent contact or seizure, the maximum possible pressure differential, and so also the dissipated energy, is limited by an overflow valve (7). It is connected to the inlet side and the pressure side of the pump-through channels. A weight-loaded valve plate opens when the maximum pressure differential is exceeded and allows a greater or lesser portion of the intake gas to flow back from the pressure side to the inlet side, depending on the throughput. Due to the limited pressure differential, standard Roots pumps cannot discharge against atmospheric pressure and require a backing pump. However Roots vacuum pumps with overflow valves can be switched on together with the backing pump even at atmospheric pressure, thus increasing their pumping speed right from the start. This shortens evacuation times.

Figure 4.16: Operating principle of a Roots pump

### Backing pumps

Single-stage or two-stage rotary vane pumps or external vane pumps are used as oil-lubricated backing pumps. Screw pumps or multi-stage Roots pumps can be used as dry backing pumps. Pump combinations such as these can be used for all applications with a high pumping speed in the low and medium vacuum range. Liquid ring pumps can also be used as backing pumps.

### Gas-circulation-cooled Roots pumps

To allow Roots vacuum pumps to work against atmospheric pressure, some models are gas-cooled and do not have overflow valves (Figure 4.17). In this case, the gas that flows from the outlet flange (6) through a cooler (7) is re-admitted into the middle of the suction chamber (4). This artificially generated gas flow cools the pump, enabling it to compress against atmospheric pressure. Gas entry is controlled by the Roots pistons, thus eliminating the need for any additional valves. There is no possibility of thermal overload, even when operating at ultimate pressure.

Figure 4.17: Operating principle of a gas-cooled Roots pump

Figure 4.17 shows a cross-section of a gas-circulation-cooled Roots vacuum pump. The direction of gas flow is vertical from top to bottom, enabling the liquid or solid particles entrained in the inlet stream to flow off downward. In phase I, the chamber (3) is opened by the rotation of the pistons (1) and (2). Gas flows into the chamber through the inlet flange (5) at pressure $p_1$. In phase II, the chamber (3) is sealed off against both the inlet flange and the pressure flange. The inlet opening (4) for the cooling gas is opened by the rotation of the pistons in phase III. The chamber (3) is filled to the outlet pressure $p_2$, and the gas is advanced toward the pressure flange. Initially, the suction volume does not change with the rotary movement of the Roots pistons. The gas is compressed by the inflowing cooling gas. The Roots piston now continues to rotate (phase IV), and this movement pushes the now compressed gas over the cooler (7) to the discharge side (Phase V) at pressure $p_2$.

Gas-cooled Roots pumps can be used in the inlet pressure range of 130 to 1,013 hPa. Because there is no lubricant in the suction chamber, they do not discharge any mist or contaminate the medium that is being pumped. Connecting two of these pumps in series enables the ultimate pressure to be reduced to 20 to 30 hPa. In combination with additional Roots vacuum pumps, the ultimate pressure can be reduced to the medium vacuum range.

### Pumping speed and compression ratio

The characteristic performance data of Roots pumps are the pumping speed and compression ratio. The theoretical pumping speed $S_{th}=S_0$ is the volume flow rate which the pump displaces without counterpressure. The compression ratio $K_0$ when operated without gas displacement (inlet flange closed) depends on the outlet pressure $p_2$. Pumping speeds range from 200 m3 · h-1 to several thousand m3 · h-1. Typical $K_0$ values are between 10 and 75.

Figure 4.18: No-load compression ratio for air for Roots pumps

The compression ratio is negatively impacted by two effects:

• By the backflow into the gaps between the piston and housing
• By the gas that is deposited by adsorbtion on the surfaces of the piston on the outlet side and re-desorbs after rotating toward the suction side.

In the case of outlet pressures of 10-2 to 1 hPa, molecular flow prevails in the seal gaps,which results in less backflow due to their low conductivities. However the volume of gas that is pumped back through adsorption, which is relatively high by comparison with the pumped gas volume, reduces the compression ratio.

$K_0$ is highest in the 1 to 10 hPa range, since molecular flow still prevails due to the low inlet pressure in the pump’s sealing gaps, and backflow is therefore low. Since gas transport through adsorption is not a function of pressure, it is less important than the pressure-proportional gas flow that is transported by the pumping speed.

At pressures in excess of 10 hPa, laminar flow occurs in the gaps and the conductivities of the gaps increase significantly, which results in declining compression ratios. This effect is particularly noticeable in gas-cooled Roots pumps that achieve a compression ratio of only approximately $K_0$ = 10.

The gap widths have a major influence on the compression ratio. Due to the different thermal expansion of the pistons and the housing, they must not, however, fall below certain minimum values in order to avoid rotor-stator-contact.

format: plain


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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp018049g818h
 Title: Investigating the Interaction of PqsE and RhlR in Pseudomonas aeruginosa Authors: Pousont, Dorian Advisors: Bassler, Bonnie Department: Molecular Biology Class Year: 2021 Abstract: Pseudomonas aeruginosa is an opportunistic pathogen, responsible for causing multi-drug resistant infections in immunocompromised patients. P. aeruginosa pathogenicity is largely controlled through quorum sensing (QS), the process of bacterial cell-cell communication mediated by the release and detection of small molecule signals, called autoinducers. One major branch of QS in P. aeruginosa, the rhl branch, controls the release of several virulence factors. Interestingly, a protein, PqsE, that is annotated to be part of a separate QS system, has been implicated in activation of the rhl system. Preliminary work in the Bassler group has shown that PqsE makes a direct protein-protein interaction with the QS receptor of the rhl system, RhlR, in vitro. This study attempts to prove the hypothesis that PqsE and RhlR engage in a protein-protein interaction and that they regulate similar genes, potentially as a result of this interaction. Isolation and sequencing of RNA from ∆pqsE and rhl pathway mutant P. aeruginosa strains revealed a number of genes regulated by PqsE, RhlR, and RhlI, the rhl system auto-inducer synthase. Importantly, many genes were regulated by both PqsE and RhlR as predicted, suggesting a link between PqsE and the rhl system. Further, immunoprecipitation and mass spectrometry provided evidence that PqsE engages in a direct protein-protein interaction with RhlR. Data from the immunoprecipitation experiments also showed that PqsE is potentially involved in a number of other systems in P. aeruginosa such as the production of autoinducers, the regulation of drug efflux pumps and membrane porins, and the production of acetyl-CoA. Together, the results presented herein provide evidence of the PqsE-RhlR interaction and suggest that it may be a major driver of virulence and a much-needed potential drug target for treating P. aeruginosa infections. URI: http://arks.princeton.edu/ark:/88435/dsp018049g818h Type of Material: Princeton University Senior Theses Language: en Appears in Collections: Molecular Biology, 1954-2021

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# Countabilty

Printable View

• September 11th 2011, 09:43 AM
slevvio
Countabilty [SOLVED]
Hello, I just had a quick question about countability.

Let $A$ be a countable set, so there exists a map $f: A \rightarrow \mathbb{N}$ such that $f$ is injective. How can I show that there is a bijection between these sets?

Any help would be appreciated.
• September 11th 2011, 09:56 AM
slevvio
Re: Countabilty [SOLVED]


text:
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# Tag Info

## Hot answers tagged code-request

94

Amusingly enough, the images above actually arose as an accidental by-product of browsing inane YouTube conspiracy theory videos. I happened across a rather beautiful video of a "mirror cube" device produced by a man in Germany named Ben Palmer, who apparently produced it in an attempt to bring recognition to a philosopher named Walter Russell (the first ...

89

TL;DR: A package (Mathematica v10) can be found at the very bottom of this post. UPDATES 6: Tiny update: Import can now use the ".bvh" extension to determine the import type. The code that does this is ugly, but I don't see any other way at the moment. out = Import["C:\\Female1_C03_Run.bvh"] 5: Added error checking and registered the package as an ...

45

EDIT: I found a version of the source image with fewer jpeg artifacts. My idea would be to extract pixels along the border of the shadow, inside and outside of the shadow. Then I have a list of pairs of RGB values, and I can find a suitable transformation from "shadowed" pixels to "non-shadowed" pixels. But first, I must find the shadow area. I would ...

45

Edit Here is a different approach using Graphics to actually draw some brush strokes. I run a GradientOrientationFilter on a smaller version of the image to estimate the local image gradient, and use that information to create a collection of randomly shaded lines: img = Import["http://i.stack.imgur.com/XwYg7.jpg"]; im = img ~ImageResize~ 200 ~...

44

First, define the dimensions and colors associated with our matrix: {mheight, mwidth} = mdim = {12, 20}; mdepth = 20; mcolors = Reverse@Array[ Blend[{{0, Darker[Green, 0.9]}, {0.4, Darker[Green]}, {0.6, Darker[Green]}, {0.90, Lighter[Green]}, {1, Lighter[Green, 0.8]}}, #/(mdepth - 1)] &, mdepth, 0]; Next, define some useful ...

41

There are two aspects of this question that distinguish it from previous questions: The request for a general template, as opposed to just a single example. The fact that the given example is a polynomial of degree three, whereas as opposed to the quadratic examples which appear in many places, including Stan's book. To deal with the first issue, in part, ...

37

I am not sure, if one can separate one curve out of the image. There is, however, another possibility to do what you want. It is possible to get the curve points out of the image. I am not sure, if I have already published here this answer. I checked but did not find it. Hence, I am publishing it. The task is fulfilled by a copyCurve function first described ...

34

Forward Mapping One way to do it is to create the texture for one tile and then transform repeated copies of it in a way that resembles the original illusion. First we create the tile: tile = Module[{KeyHole}, KeyHole[base_] := Sequence[ Disk[{0, 1/3} + base, 1/10], Rectangle[{-1/30, 1/15} + base, {1/30, 1/3} + base] ]; Image@Rasterize@...

28

You'll be interested in the (undocumented!) functions GraphicsMeshIntersectQ[] (for checking the intersections) and GraphicsMeshFindIntersections[] (for actually finding them). As a sample: BlockRandom[SeedRandom[42, Method -> "MersenneTwister"]; (* for reproducibility *) lins = Table[{Line[RandomReal[1, {2, 2}]]}, {42}];] GraphicsMesh...

28

I got my CIE color matching functions from here. These are the CIE 1931 2-deg, XYZ CMFs modified by Judd (1951) and Vos (1978). {λ, x, y, z} = Import["http://www.cvrl.org/database/data/cmfs/ciexyzjv.csv"]\[Transpose]; ListLinePlot[{{λ, x}\[Transpose], {λ,y}\[Transpose], {λ, z}\[Transpose]}, PlotLegends -> {"X", "Y", "Z"}] Conversion of color ...

25

The kite-domino tiling is based the pinwheel tiling which is falls out of a particular decomposition of a right triangle with legs of length 1 and 2. In the code that follows, rt[{a,b,c}] represents such a right triangle and dissect indicates how such a triangle should be decomposed into smaller copies of itself. We simply iterate the dissect function on ...

25

Hyperbolic lines in the Poincare disk are arcs of circles which, if extended, would meet the unit circle boundary orthogonally. Tiling the disk requires reflection of a tile's vertices and circular edges in those same circular edges. This corresponds to inversion in a circle. Begin with a central polygon having p edges, with q such polygons meeting at each ...

24

Found this somewhere: ϕ = GoldenRatio; s = 1.75; ContourPlot3D[ -(4*(ϕ^2*x^2 - y^2)*(ϕ^2*y^2 - z^2)*(ϕ^2*z^2 - x^2) - (1 + 2 ϕ)*(x^2 + y^2 + z^2 - 1)^2) == 1.1, {x, -s, s}, {y, -s, s}, {z, -s, s}, ContourStyle -> White, Boxed -> False, Axes -> False, SphericalRegion -> True, Mesh -> 5, BoundaryStyle -> None, PlotPoints -> 45, ...

23

This small app allows you to select data points and put them in a bin. In that way you can manually select several curves. I imagine this might be useful to someone who has complex plots where lines might intersect, making it hard or impossible to extract the data programmatically. You first have to set the image at the first line. After that when you want ...

23

Usually viruses have icosahedron symmetry. So I propose to generate a random chain of balls and translate it appropriately n = 2000; f = GaussianFilter[#, 5] &; p = f@RandomReal[{3.0, 4.0}, n] #/Sqrt@Total[#^2, {2}] &@ Accumulate@Prepend[0.08 f@RandomReal[NormalDistribution[], {n - 1, 3}], Normalize@RandomReal[NormalDistribution[], 3]]; r = f@...

22

You can use the LCE package by Dr Sandri Marco. It is updated for version 7 and I just tried in on your system for V9 and it worked. Download lcm.zip and use the package as instructed Here is the result of running your system on it on my PC << lce.m ?LCEsC These are the 3 Lyapunov Exponents for the Rossler system: rossler[{x_, y_, z_}] := {-y - z,...

22

Show it: RegionPlot[ region = RegionUnion[ Sequence @@ RegionIntersection @@@ Subsets[{Disk[{-1, 0}], Disk[{0, -1}], Disk[{1, 0}], Disk[{0, 1}]}, {2}], Fold[RegionDifference, {Disk[{0, 0}, 2], Disk[{-1, 0}], Disk[{0, -1}], Disk[{1, 0}], Disk[{0, 1}]}]], Frame -> False] Area Area[region] 4 (-2 + π) Perimeter ...

21

I thought I'd share my attempt at this, even though it doesn't seem to have worked properly. The CIE color matching functions are tabulated in the Image`ColorOperationsDump context which is used by ChromaticityPlot. The context can be loaded by calling ChromaticityPlot and then we can interpolate the data to obtain functions: ChromaticityPlot["RGB"]; {x, ...

21

The Pitsianis-Van Loan algorithm turns out to be surprisingly easy to implement in Mathematica: nearestKroneckerProductSum[mat_?MatrixQ, dim1_?VectorQ, dim2_?VectorQ, k_Integer?Positive, opts___] /; TrueQ[Dimensions[mat] == dim1 dim2] := Module[{tmp}, Check[tmp = SingularValueDecomposition[Flatten[...

20

Definition GaussCurvature[f_] := With[{dfu = D[f, u], dfv = D[f, v]}, Simplify[(Det[{D[dfu, u], dfu, dfv}] Det[{D[dfv, v], dfu, dfv}] - Det[{D[f, u, v], dfu, dfv}]^2) / (dfu.dfu dfv.dfv - (dfu.dfv)^2)^2]]; Sphere As @ ubpdqn already remarked GaussCurvature[{Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]}] 1 Ellipsoid ellipsoid = {2 Cos[u] Sin[v], ...

19

Edit Old answer follows below The features of your new set are very different from the first image. For example: There are components "touching" the borders. They have "hidden" structures like this: So it needs a different approach: l = {"http://i.stack.imgur.com/OZEvk.png", "http://i.stack.imgur.com/Tl1Pm.png", "http://i.stack.imgur.com/8enYZ.png"}...

19

Catmull-Clark Subdivision Indeed, I have some code for Catmull-Clark subdivision and I planned to post it here for quite some time. This seems to be a good opportunity. The code is optimized for performance, so it involves a lot of CompiledFunction and SparseArray hacks. I am sorry if you find it somewhat unidiomatic. CatmullClarkSubdivisionMatrix creates ...

18

Update: If you want 4-neighborhood, you can use MorphologicalComponents to do most of the work, which is fast and easy to implement (that was my original attempt, see below). But I don't think this can be adapted for 8-neighborhood. For 8-neighborhood, I would implement the standard 2-pass connected component labeling algorithm (this might be what ...

18

You may use Interpretation. You will need some fancy formatting function that I will proxy here with Style. The following will create a column of the code snippet and its annotation that can be directly used in input cells. Interpretation[ Column[ { HoldForm[RandomVariate[NormalDistribution[], 20]], Style["Twenty random numbers", Blue, "Output"] ...

17

You could start to scrape allowed characters in that font, e.g.: mcolors = Blend[{{0.00, Darker[Green, 0.7]}, {0.90, Darker@Green}, {0.90, Lighter[Green]}, {1.00, Lighter[Green, 0.9]}}, #] &; $unic :=$unic = Join[{"\[Wolf]", "\[MathematicaIcon]", "\[HappySmiley]"}, Drop[ToExpression[ "\"" <> StringReplace[#, "U+" -> "...

17

A simple numerical maximization using NMaximize as suggested by b.gatessucks: pts = Array[{x[#], y[#]} &, 10]; mindist2 = Min[#.# & /@ Subtract @@@ Subsets[pts, {2}]]; vars = Flatten[pts]; constraints = Thread[0 <= vars <= 1]; {md2, rules} = NMaximize[{mindist2, constraints}, vars]; minimaldistance = Sqrt[md2] (* 0.381759 *) Graphics[{Yellow,...

17

I have seen your question on math.SE and I know how it is when you don't find the right place to ask a question. Consider this a one-time present. If you truly want to learn Mathematica, you need to put in some effort in understanding what's going on here. Basically, your iteration can be written down almost as you have it: s[x_] := x^Log[x]; t[x_] := 1/81*...

17

Using CirclePoints, Mod, Throughand Range More than enough lines Multicolumn@With[ { n = 500, m = 17 }, Table[ Graphics[ {Opacity[0.1], Through@{Point, Line}[ Part[CirclePoints[n], Mod[{1, k } #, n] + 1]] & /@ Range[n] } ] , {k, 2, m}] ] Simpler Multicolumn@With[ { n = 100, m = 7 }, Table[ ...

15

Update Nov 6, 2018: Tweaked to work on nonautonomous systems, to fix problem reported here. Here's an updated implementation (works with MMA v11) that generalizes Housam Binous & Nasri Zakia's update to Marco Sandri's package and incorporates ideas from @bbgodfrey. First, define @halirutan's GramSchmidt: GramSchmidt[w_?MatrixQ] := Module[{v = ...

15

I decided to take a slightly different approach. Instead of transforming an image, I thought of constructing a function that will look like the illusory figure in the OP after performing the log-polar transform. Here's what I came up with: checkerboard[x_, y_] := Boole[EvenQ[Floor[x] - Floor[y]]] keyholes[x_, y_] := Boole[(Mod[x - 1/2, 1] - 1/2)^2 + (Mod[y, ...

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text:
# Simple algebra Problem

Algebra Level 5

Given that $$x$$ and $$y$$ are positive real such that $$x+y=1$$, find the value of $$k$$ such that the maximum value of $${ x }^{ 4 }y+x{ y }^{ 4 }$$ is $$\dfrac { 1 }{ k }.$$

• This is not an original problem.
• For more problems try my set.
×


text:
MathSciNet bibliographic data MR2322678 (2008k:11055) 11F72 (11B75 11R58 22E40) Sarveniazi, Alireza Explicit construction of a Ramanujan \$(n\sb 1,n\sb 2,\dots,n\sb {d-1})\$$(n\sb 1,n\sb 2,\dots,n\sb {d-1})$-regular hypergraph. Duke Math. J. 139 (2007), no. 1, 141–171. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
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The Thom-Pontryagin construction is ingenious and unexpected, and turns out to be tremendously important. To construct an oriented cobordism group, consider the set of oriented compact $n$-manifolds, modulo the equivalence relation that two such manifolds are considered equivalent if they together bound an oriented compact $(n+1)$-manifold. This is seen to have the structure of an abelian group, where the inverse of a manifold is that manifold with the opposite orientation. The oriented cobordism group of points is isomorphic to $\mathbb{Z}$, where a pair of points oriented "+" and "-" cobound an oriented interval. The oriented cobordism group of circles is ${0}$, because any set of circles together cobound a surface, and the oriented cobordism group of compact surfaces is also ${0}$ because any oriented surface bounds a handlebody.
Cobordism groups had been in the air for a long time, with Poincaré having considered them as a first (failed) attempt to define homology. In general, there are an uncountable number of $n$--manifolds, so one would have expected oriented cobordism groups to be hopelessly untractable for large $n$. But, by a simple ingenious construction for which he won the Field's medal, René Thom showed that the oriented cobordism group $N^{+}{d-1}$ \mathcal{N}^+_{d-1}$ is isomorphic to$\pi{d-1}\Omega^{\infty} \pi_{d-1}\Omega^\infty MSO$, which are homotopy groups of the infinite loop space of the Thom spectrum. This is a very large space, but calculating its homotopy turns out to be quite simple, and it turns out to be the case that $N^{+}{d-1}$, \mathcal{N}^+_{d-1}$, modulo torsion is nothing more than a polynomial algebra $\mathbb{Q}[y{4i}]{i\geq \mathbb{Q}[y_{4i}]_{i\geq 1}$ where $y{4i}$ y_{4i}$ is a formal variable which can be represented by the complex projective space$\mathbb{CP}^{2i}$. The Thom-Pontryagin construction itself is explained very nicely in an MO answer of Greg Kuperberg. A short lucid introduction to this story, on which this answer is loosely based, is given in the first few minutes of Ulrike Tillman's talk at the recent IMA conference in honour of John Milnor. 2 latex fixed The Thom-Pontryagin construction is ingenious and unexpected, and turns out to be tremendously important. To construct an oriented cobordism group, consider the set of oriented compact$n$-manifolds, modulo the equivalence relation that two such manifolds are considered equivalent if they together bound an oriented compact$(n+1)$-manifold. This is seen to have the structure of an abelian group, where the inverse of a manifold is that manifold with the opposite orientation. The oriented cobordism group of points is isomorphic to$\mathbb{Z}$, where a pair of points oriented "+" and "-" cobound an oriented interval. The oriented cobordism group of circles is${0}$, because any set of circles together cobound a surface, and the oriented cobordism group of compact surfaces is also${0}$because any oriented surface bounds a handlebody. Cobordism groups had been in the air for a long time, with Poincaré having considered them as a first (failed) attempt to define homology. In general, there are an uncountable number of$n$--manifolds, so one would have expected oriented cobordism groups to be hopelessly untractable for large$n$. But, by a simple ingenious construction for which he won the Field's medal, René Thom showed that the oriented cobordism group$\mathcal{N}^+N^{+}{d-1}$is isomorphic to$\pi{d-1}\Omega^\infty d-1}\Omega^{\infty} MSO$, which are homotopy groups of the infinite loop space of the Thom spectrum. This is a very large space, but calculating its homotopy turns out to be quite simple, and it turns out to be the case that$\mathcal{N}^+N^{+}{d-1}$, modulo torsion is nothing more than a polynomial algebra$\mathbb{Q}[y{4i}]{i\geq 1}$where$y{4i}$is a formal variable which can be represented by the complex projective space$\mathbb{CP}^{2i}$. The Thom-Pontryagin construction itself is explained very nicely in an MO answer of Greg Kuperberg. A short lucid introduction to this story, on which this answer is loosely based, is given in the first few minutes of Ulrike Tillman's talk at the recent IMA conference in honour of John Milnor. 1 [made Community Wiki] The Thom-Pontryagin construction is ingenious and unexpected, and turns out to be tremendously important. To construct an oriented cobordism group, consider the set of oriented compact$n$-manifolds, modulo the equivalence relation that two such manifolds are considered equivalent if they together bound an oriented compact$(n+1)$-manifold. This is seen to have the structure of an abelian group, where the inverse of a manifold is that manifold with the opposite orientation. The oriented cobordism group of points is isomorphic to$\mathbb{Z}$, where a pair of points oriented "+" and "-" cobound an oriented interval. The oriented cobordism group of circles is${0}$, because any set of circles together cobound a surface, and the oriented cobordism group of compact surfaces is also${0}$because any oriented surface bounds a handlebody. Cobordism groups had been in the air for a long time, with Poincaré having considered them as a first (failed) attempt to define homology. In general, there are an uncountable number of$n$--manifolds, so one would have expected oriented cobordism groups to be hopelessly untractable for large$n$. But, by a simple ingenious construction for which he won the Field's medal, René Thom showed that the oriented cobordism group$\mathcal{N}^+{d-1}$is isomorphic to$\pi{d-1}\Omega^\infty MSO$, which are homotopy groups of the infinite loop space of the Thom spectrum. This is a very large space, but calculating its homotopy turns out to be quite simple, and it turns out to be the case that$\mathcal{N}^+{d-1}$, modulo torsion is nothing more than a polynomial algebra$\mathbb{Q}[y{4i}]{i\geq 1}$where$y{4i}$is a formal variable which can be represented by the complex projective space$\mathbb{CP}^{2i}\$.


text:
# Functor Obandit.MakeParametrizableEpsilonGreedy

module MakeParametrizableEpsilonGreedy: functor (P : RateBanditParam) -> Bandit  with type bandit = banditEstimates
The $\epsilon$-Greedy Bandit with a parametrizable exploration rate.
Parameters:
 P : RateBanditParam

type bandit
The internal data structure of the bandit algorithm.
val initialBandit : bandit
The initial state of the bandit algorithm.
val step : bandit -> float -> int * bandit
step r advances the bandit game one step, where r is the reward for the last action. The result of this call is the next action, encoded as an integer in $\{ 0, \cdots , K-1 \}$, and the new state of the bandit. The reward range depends on the bandit algorithm in use and the first reward provided to the algorithm is discarded.


text:
# Substitution format string¶

## config.core.v3.SubstitutionFormatString¶

[config.core.v3.SubstitutionFormatString proto]

Configuration to use multiple command operators to generate a new string in either plain text or JSON format.

{
"text_format": "...",
"json_format": "{...}"
}

text_format

(string) Specify a format with command operators to form a text string. Its details is described in format string.

text_format: %LOCAL_REPLY_BODY%:%RESPONSE_CODE%:path=\$REQ(:path)%

The following plain text will be created:

upstream connect error:204:path=/foo

Precisely one of text_format, json_format must be set.

json_format

(Struct) Specify a format with command operators to form a JSON string. Its details is described in format dictionary. Values are rendered as strings, numbers, or boolean values as appropriate. Nested JSON objects may be produced by some command operators (e.g. FILTER_STATE or DYNAMIC_METADATA). See the documentation for a specific command operator for details.

json_format:
status: %RESPONSE_CODE%

The following JSON object would be created:

{
"status": 500,
"message": "My error message"
}

Precisely one of text_format, json_format must be set.


text:
# Error Analysis – 1 – 1

1. The steady state error due to a ramp input for a type two system is equal to :

(a) zero
(b) infinite
(c) non-zero number
(d)
constant

EXPLANATION

2. The impulse response of a system is

$\dpi{100} 5 e-10t.$

Its step response is equal to :

(a) 0.5e-10t
(b) 5(1 – e-10t)

(c) 0.5(l – e-10t)
(d) 10(1 – e-10t)

EXPLANATION

3. A unity feedback system has open-loop transfer function G(s). The steady-state error is zero for :

(a) step input and type-1 G( s)
(b) step input and type-0 G(s)
(c) ramp input and type-1 G(s)
(d)
ramp input and type-0 G(s)

EXPLANATION

4. In the type 1 system, the velocity error is :

(a) inversely proportional to bandwidth
(b) directly proportional to error constant
(c) inversely proportional to error constant
(d) independent of error constant


text:
12 class Maths Notes Chapter 9 Differential Equations  free PDF| Quick revision Differential Equations  Notes class 12 maths

CBSE Revision Notes for CBSE Class 12 Mathematics Differential Equations Definition, order and degree, general and particular solutions of a differential equation.Formation of differential equation whose general solution is given.Solution of differential equations by method of separation of variables solutions of homogeneous differential equations of first order and first degree.

Class 12 Maths Chapter-9 Differential Equations  Quick Revision Notes Free Pdf

🔷 Chapter - 9 🔷
👉  Differential Equations  👈

1. INTRODUCTION

Differential equation constitute a very important part of mathematics as it has many applications in real life. Various laws of physics are often in the form of equations involving rate of change of one quantity with respect to another. As the mathematical equivalent of a rate is a derivative, differential equation arise very naturally in real life and methods for solving them acquire paramount importance.

Definition

An equation involving the dependent variable and independent variable and also the derivatives of the dependable variable is known as differential equation.

For example:

Differential equations which involve only one independent variable are called ordinary differential equation.

2. ORDER AND DEGREE OF DIFFERENTIAL EQUATIONS

2.1 Order

The order of a differential equation is the order of the highest derivative involved in the differential equation

For example:

2.2 Degree

The degree of a differential equation is the degree of the highest differential coefficient when the equation has been made rational and integral as far as the differential coefficients are concerned.

For example:

Illustration 1: Find the order and degree of the following differential equations.

3. FORMATION OF ORDINARY DIFFERENTIAL EQUATION

An ordinary differential equation is formed in an attempt to eliminate certain arbitrary constants from a relation in the variables and constants. Consider an equation containing n arbitrary constants. Differentiating this equation n times we get n additional equations containing n arbitrary constants and derivatives. Eliminating n arbitrary constants from the above (n + 1) equations, we obtain differential equation involving nth derivative.

Thus if an equation contains n arbitrary constants, the resulting differential equation obtained by eliminating these constants will be a differential equation of nth order.
i.e.,      an       equation     of          the            form

Illustration 2 : Find the differential equation of the family of all circles which pass through the origin and whose centre lie on y-axis

Sol. Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0
If it passes through (0, 0), then c = 0
•       The equation of circle is x² + y² + 2gx + 2fy = 0
Since the centre of the circle lies on y-axis then g =          0
•       The equation of the circle is
x² + y² + 2fy = 0.                                                 ...(i)
This represents family of circles.
Differentiating, we get

4. SOLUTION OF A DIFFERENTIAL EQUATION

The solution of the differential equation is a relation is a relation between the independent and dependent variable free from derivatives satisfying the given differential equation.

Thus the solution of dy/dx = m could be obtained by simply integrating both sides i.e., y = mx + c, where c is arbitrary constant.

(a) General solution (or complete premitive)

The general solution of a differential equation is the relation between the variables (not involving the derivatives) which contain the same number of the arbitrary constants as the order of the differential equation.

Thus the general solution of the differential equation

(b) Particular solution or Integral

A solution which is obtained by giving particular values to the arbitrary constants in the general solution is called a particular solution.

5. METHOD OF SOLVING AN EQUATION OF THE FIRST ORDER AND FIRST DEGREE

A differential equation of the first order and first degree can be written in the form

Method – 1

(i) Variable Separation:

The general form of such an equation is
f(x)dx + f(y)dy = 0                                             ..(i)
Integrating, we get
∫f(x )dx +∫f(y)dy = c which is the solution of (i)

(ii) Solution of differential equation of the type

Method -3

(i) Linear differential equation:

A differential equation is said to be linear if the dependent variable y and its derivative occur in the first degree.

where P and Q are functions of y only. To get the general solution of the above equations we shall determine a function R of x called Integrating function (I.F). We shall multiply both sides of the given equation by R

(ii) Differential equation reducible to the linear form:

Sometimes equations which are not linear can be reduced to the linear form by suitable transformation.

Method 4

Exact differential equation:

A differential equation is said to be exact if it can be derived from its solution (primitive) directly by differentiation, without any elimination, multiplication etc.

For example, the differential equation x dy + y dx = 0 is an exact differential equation as it is derived by direct differentiation for its solution, the function xy = c

Illustration 11 : Solve (1 + xy) y dx + (1 - xy) x dy = 0

Sol. The given equation can be written as
y dx + xy²dx + x dy x²y dy = 0
or,  (y dx + x dy) + xy (y dx - x dy) = 0
or,  d (xy) + xy (y dx - x dy) = 0

Dividing by x²y², we get
Which is the required solution.

Application of differential equations :

In solving some geometrical problems, the following results are very helpful.

Illustration 12: If the length of the sub-normal at any point P on the curve is directly proportional to OP², where O is the origin, then form the differential equation of the family of curves and hence find the family of curves.


text:
# Lebesgue integral (existence and finiteness) of $\sin(1/x^2)$

I think I am getting a little better at these MCT, DCT-type exercises. The issue is to show/prove the existence and finiteness (if they apply) to the following function:

$$f(x)=\sin \left(\dfrac1{x^2} \right)$$

Where applicable I want to show as rigorously as possible the justification for existence, etc. For example, it is not enough to say simply, "this exists by MCT"; rather, I need to show that the conditions of measurability, monotone-increasing, etc. are met.

Work/attempt at a solution:

First, although I have a limited experience with this type of integral, I believe the function can be re-written as follows:

$$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}$$

If I have this correctly (I attempted to extrapolate from an integral table I found regarding $\sin\left(\dfrac1x \right)$, but please correct me if I am wrong), I can utilize Dominated Convergence to show the integral exists. That is,

If $f_1,f_2,...$ are measurable functions and $|f_n|\le g$ for $\mu$-integrable function, $g$, and if $f_n\to f$, $\mu$-a.e, then $\lim_{n\to\infty}\int f_n d\mu\to\int fd\mu$.

First, I need to show measurability. I believe that continuity on the interval in question, $(0,\infty)$, is sufficient to show measurability, correct?

Next, define

$$S_n(x):=\sum_{n=0}^{t}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}\to f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}$$

Then, define

$g(x)=\dfrac1x\ge |f_n|,\forall x>0$

and therefore conclude by Dominated Convergence Theorem, that $\displaystyle \int_{(0,\infty)}\sin \left(\dfrac1{x^2} \right)d\mu$ exists.

Please comment, add answers, critique, tell me where I made mistakes, etc. I am learning this on my own so any and all improvements are welcome.

From here I need to show whether this integral is finite or infinite. Is there a way to show this without an explicit calculation of the integral?

-

$\sin(1/x^2)$ is continuous in $(0,\infty)$ and hence it is also measurable.
Split your domain $(0,\infty)$ into $(0,1] \cup (1,\infty)$.
In the interval $(0,1]$, bound $\sin(1/x^2)$ by $1$ and in the interval $(1,\infty)$, bound $\sin(1/x^2)$ by $1/x^2$ and then argue out.


text:
# L3 Winter Build Thread - 3/4 Scale PAC-3 Patriot

### Help Support The Rocketry Forum:

#### mccordmw

##### Well-Known Member
I've been working on the specs, designs, and calculations on this for quite a while. The Patriot is a beautiful design, and I wanted to use that for my L3 build. I love the look of the MIM104F next generation version, so here it begins. I've got a ton of calculations to post before I get to the build portion.

So to kick it off, here's the summary pic and first release version of the ORK file.

I want to use is as a workhorse for (much much later) EX 75-98 motors once I've mastered formulations in the 24-29mm range. So this was built with a 98mm motor mount and an Aero Pack 98>75 adapter in the design to house my Loki 76mm M1969 sparky. At a 6.8:1 TTW ratio, it's a bit close. If I get too heavy, I'll bump up to an M3400 at a nice 10:1 ratio.

Stuff I want to cover before the build portion:
1. the airframe and fiberglassing strategy
3. fin flutter analyses to calculate safe speeds
4. fin strength to check safe AoA parameters
5. electronics
6. deployment scheme

--- the build ---

Then:

1. chute calculations
2. weight and motors
3. flight profiles

I think that covers most of it. Going to take a while... I guess that gives me time to save for the Loki engines.

Last edited:

#### Mustang67

I have often dreamed of building a full scale Patriot, going to be a plasure to watch your build come to fruition. Question what is the difference between a PAC-3 Patriot and the standard Patriot?

Subscribed!

#### mccordmw

##### Well-Known Member
Oops. I totally forgot to attach the ORK file. Here it is. You'll notice the fin tabs are quite extended. I'm lining up tabs with bulk plates to form a rigid, inline structural framework. That allows me to only worry about the outer airframe for compressive forces as it hits mach. I want to do something new for me and learn in the process.

You can see the structural support on the 3D unfinished view. Sure, that adds weight, but I plan on testing an adhesion primer on the tube before adding the drain sleeve and epoxy. That will massively cut down on epoxy absorption into the tube which will cut a lot of weight. I just need to test strength. Science!

View attachment 8 inch scratch MEADS PAC-3 L3.ork

#### mccordmw

##### Well-Known Member
I have often dreamed of building a full scale Patriot, going to be a plasure to watch your build come to fruition. Question what is the difference between a PAC-3 Patriot and the standard Patriot?
PAC-3 is the Patriot Advanced Capability-3 version. A good link for the version comparisons is here.

https://en.wikipedia.org/wiki/MIM-104_Patriot#Variants

The PAC-3 has rear fins that look more like Nike Smoke fins which I like. Also has forward control surfaces kind of like strakes (at least that's what it looks like to me). Several different looks are out. I modeled mine after the Lokheed Martin version shown here:

https://m.lockheedmartin.com/conten...ment-enhancement/mfc-pac-3-mse-photo-01-h.jpg

To save weight, I reduced the forward fin height to more resemble the German AMD variant:

#### Mustang67

Awesome and Thx. I found this as well for operational range.
PAC-1: 70 km
PAC-2: 96 km-160 km
PAC-3: 20 km against ballistic missile PAC-3 MSE: 35 km against ballistic missile

Wonder if we could try some of their solid fuel??? Wonder if there is a difference..lol

#### dixontj93060

##### Well-Known Member
That design is really sharp. I like the clean lines of the strikes in combination with those forward positioned square fins. I probably would have never gone there in a million sketches, but it looks great!

#### mccordmw

##### Well-Known Member
Data crunching and planning progress has been made.

First off, I wasn't so sure about the strength of the 8" Quikrete tubes, so I did a lot of research on the forces the tubes will see and how they can fail. The end result (as covered in another thread) was the publishing of a tube crush/buckle calculator on Google Sheets.

Plugging in my numbers, I determined that the tubes alone would buckle under a moderate flight if they weren't reinforced in some way.

In another thread, I've been testing ways to fiberglass the tube without adding crazy amounts of weight. After testing, I've settled on the following method:

1. skip peeling oro sanding away the outer paper wrap. It makes the tube too porous.
2. instead, spray the outside of the tube with adhesion promoter. I used Dupli-color adhesion promoter on a 1' test section.
3. I tested applying 2 layers of polyester drain sleeve with US Composites 3:1 laminating epoxy.

The end result was a pretty robust tube. However, paranoia took over, and I tested again with another section of tubing with 2 x 6oz wraps of fiberglass and one 2oz veil wrap. The weight gain was only a bit higher than the drain sleeve, but it was a bit stiffer.

I both cases, I could not peel off the drain sleeve or the fiberglass in the slightest bit. Using pliers, I could pull it off, but it came with a generous layer of the tube. I'd say I'm happy with how well the epoxy stuck to the adhesion promoter treated tube. Plus, the tube didn't soak up enormous amounts of epoxy; avoiding wasted money and weight.

The final weight for an 8" x 48" Quikrete tube sprayed and fiberglassed was 2.4 kg, or 5.3 lbs.

I used this modified density to recalculate the model.

View attachment 8 inch scratch MEADS PAC-3 L3.ork

#### mccordmw

##### Well-Known Member
Next up, I'm figuring out how much harness to use. I'll be using a pilot chute to pull the main from a bag while the nose cone separates with the pilot. I'm estimating this amount of harness. Sound about right?

30' from booster to upper airframe
8' from upper airframe to main chute
6' from nosecone to empty deployment bag

If those are about right, next up is calculating the right chute sizes for the weights.

#### rfjustin

##### Well-Known Member
Sound about right?...30' from booster to upper airframe...
IMO, way too short for a drogue recovery cord on an 8" bird. Others may disagree, but I think 50-70 feet would be better for an 8" bird.

:2:

#### UhClem

##### Well-Known Member
I tried this once before with no luck but here goes again. I started work on a dimensioned drawing fpr the PAC3-MSE some time ago and this is what I have.

The diameter is based on the known diameter of the base PAC3 of 255mm. While the motor diameter increased, the forward control section remained the same. You have to add 6mm to that for the ablative coating to get the true diameter of the finished vehicle. I didn't get to the fin spans but the tip-to-tip distance for the fixed forward fins should be very close to that of the PAC3 since they need to fit into the same space.

The aft control fins fold so that they can fit their extra span into the launch tube.

View attachment pac3mse.topsilk.pdf

#### CzTeacherMan

##### Well-Known Member
IMO, way too short for a drogue recovery cord on an 8" bird. Others may disagree, but I think 50-70 feet would be better for an 8" bird.

:2:
Agreed with Justin. For my 8"Gizmo, I used 2-30 foot harnesses. Might've been a 30 and a 25... In either case, at that size, the extra length can only help.

#### [email protected]

##### Well-Known Member
IMO, way too short for a drogue recovery cord on an 8" bird. Others may disagree, but I think 50-70 feet would be better for an 8" bird.

:2:
Double ditto. You ejection charges are going to be big to make dang sure that rocket separates and the will dissipate all that energy, too short and all those components are going to be stressed or you can decrease your ejection charge but that's not worth the risk

#### mccordmw

##### Well-Known Member
Thanks for the feedback. I bumped up the cord length to 60' for the drogue section. I found a nice deal on 2500# test milspec 5/8" tubular nylon.

https://www.amazon.com/dp/B00PHRUDMG/?tag=skimlinks_replacement-20 for $15 plus$8 shipping. That should fit the bill if I put a sleeve on it to guard from ejection charges.

I have 2 x 30" nomex sleeves already on hand I can use to guard them.

Last edited by a moderator:

#### mccordmw

##### Well-Known Member
I tried this once before with no luck but here goes again. I started work on a dimensioned drawing fpr the PAC3-MSE some time ago and this is what I have.

The diameter is based on the known diameter of the base PAC3 of 255mm. While the motor diameter increased, the forward control section remained the same. You have to add 6mm to that for the ablative coating to get the true diameter of the finished vehicle. I didn't get to the fin spans but the tip-to-tip distance for the fixed forward fins should be very close to that of the PAC3 since they need to fit into the same space.

The aft control fins fold so that they can fit their extra span into the launch tube.
Thanks, David for the specs. I tweaked my dimensions a bit to allow for stability when using larger 98mm motors later on. Here are my final dimensions.

Next up... Fin flutter analysis. I haven't built a model (and consequently fins) this big, so I'm checking everything.

#### mccordmw

##### Well-Known Member
I'm using Finsim for my divergence/flutter analysis to see if I need to do any tip-to-tip glassing for strength. For fin material, I'm using 1/2" 9-ply Baltic birch. Based on my previous post, here are my Finsim results.

Since I have a symmetrical fin, I'm assuming the elastic axis and the center of gravity are both the same at the root midpoint (0.5). Someone correct me if I made an incorrect assumption. I only use the following screen to input my CG and EA. The divergence and flutter speeds are more accurately calculated using the Torsion-Flexure analysis.

The Torsion-Flexure analysis tool says I'm safe from unstable flutter at mach 1.63. U vs g shows max speed for divergence as mach 1.59 (not shown here).

Looking at the hottest 98mm engines I can put in this, the theorhetical best speed I can attain is mach 1.27 using the N5800 from CTI. That's 80% of the safe maximum for divergence, so I don't believe any reinforcement is necessary. For my L3 cert, I'll use a 75mm Loki M1969 sparky which will give a max velocity of mach 0.55. Well within tolerance limits before ripping the fins off.

I'll skip the extra weight and not fiberglass the fins.

Last edited:

#### mccordmw

##### Well-Known Member
I forgot to add wild flight check...

Even if I take a 20 degree kick off the rod from winds, the fins won't rip off. They can survive an angle of attack of 20 degrees up to mach 0.3. Well below the speed after weathercocking has leveled off.

#### Handeman

##### Well-Known Member
TRF Supporter
IMO, way too short for a drogue recovery cord on an 8" bird. Others may disagree, but I think 50-70 feet would be better for an 8" bird.

:2:
I'm one that totally disagrees with having that long of drogue cord. I believe you only need enough drogue charge to reliably open the rocket, it doesn't have to extend to the ends. It only needs to get the drogue chute into the air stream and you will have a good air stream since the rocket will be falling pretty quick.

Don't over do the drogue charge! That causes all kinds of other problems like the main deploying when the payload section hits the end of the cord. My opinion is that you only need a long drogue cord if you have too large of drogue charge or fly drogueless. In either case, you are using a long cord to compensate for a poor recovery design.

The key is the drogue. It has to be big enough to keep the payload section above the fin can. If it doesn't do that, then all bets are off when the main deploys. You might get it to work fine the first 5 or 10 times, but if you deploy the main below the fin can, it will bite you sooner or later.

The other issue with long drogue chords when you are drogueless or have too small of drogue is that the main will open and since the payload is the only section supplying weight, it will almost stop in the air because of the light load when the fin can doesn't hit and foul it. Then after the fin can falls the full length of that too long of drogue cord, it hits the end and puts a huge shock load on the whole system. This is almost never a problem with L1 sized rockets because of material strengths. When you start getting into the larger L2 rockets, it can become more of an issue because the material strengths and larger shock loads give you less margin. At L3, you better design the whole recovery right because doing it wrong can leave you with a fin can falling by itself and the upper section drifting away on an under loaded chute.

BTW, the drogue cord on my 8ft 4" rocket is 15 ft long. It's got 37 flights on it without issue.
/rant off

#### mccordmw

##### Well-Known Member
... Then after the fin can falls the full length of that too long of drogue cord, it hits the end and puts a huge shock load on the whole system....
Question for those more experienced...

My model sims a drogue descent rate of 20 m/s (drogue plus airframe drag calculated).

Now, when the main deploys, the booster section will still be falling at an approximately constant 20 m/s regardless of whether the shock cord holding the booster was 10' or 100'? Plus, a longer shock cord would allow more elastic shock absorption. What am I missing here from a physics perspective? I'm not understanding how the booster would accelerate more on a longer shock cord. Then again, I'm possibly overlooking something.

With a properly sized drogue, and the attachment at a point that holds the upper rocket body over the booster, I see the cons of an overly long shock cord as:

1. weight
2. tangling potential

The pros are:

1. more shock absorption at apogee event
2. more shock absorption at main event

Assuming weight isn't an issue on a large rocket where I'm not min/maxing for speed or altitude, I can mitigate tangling with z-folds and tape/bands.

What else to consider?

#### mccordmw

##### Well-Known Member
Thinking out loud (is that possible when typing? :tongue: )

The main deploys.
Before snapping open, the upper starts to decelerate.
- short drogue cord tension pulls on the booster a few m/s deceleration before main snaps open
- long drogue cord allows the booster to continue falling at 20 m/s before main snaps open

So the question is, how long does it take for a main to usually snap open? And what's the deceleration rate of the upper body before then? If I can figure out the deceleration rate, the goal would be to size the drogue cord to become taught before the main snaps open while keeping it at a max length for shock absorption (plus more safety margin at apogee event).

Any parachute specs for deceleration before full deployment? And typical times to open fully from a bag assuming normal packing and not loose base jumper packing?

I have the itch to do some maths!

#### Handeman

##### Well-Known Member
TRF Supporter
Thinking out loud (is that possible when typing? :tongue: )

The main deploys.
Before snapping open, the upper starts to decelerate.
- short drogue cord tension pulls on the booster a few m/s deceleration before main snaps open
- long drogue cord allows the booster to continue falling at 20 m/s before main snaps open

So the question is, how long does it take for a main to usually snap open? And what's the deceleration rate of the upper body before then? If I can figure out the deceleration rate, the goal would be to size the drogue cord to become taught before the main snaps open while keeping it at a max length for shock absorption (plus more safety margin at apogee event).

Any parachute specs for deceleration before full deployment? And typical times to open fully from a bag assuming normal packing and not loose base jumper packing?

I have the itch to do some maths!
Question for those more experienced...

My model sims a drogue descent rate of 20 m/s (drogue plus airframe drag calculated).

Now, when the main deploys, the booster section will still be falling at an approximately constant 20 m/s regardless of whether the shock cord holding the booster was 10' or 100'? Plus, a longer shock cord would allow more elastic shock absorption. What am I missing here from a physics perspective? I'm not understanding how the booster would accelerate more on a longer shock cord. Then again, I'm possibly overlooking something.

With a properly sized drogue, and the attachment at a point that holds the upper rocket body over the booster, I see the cons of an overly long shock cord as:

1. weight
2. tangling potential

The pros are:

1. more shock absorption at apogee event
2. more shock absorption at main event

Assuming weight isn't an issue on a large rocket where I'm not min/maxing for speed or altitude, I can mitigate tangling with z-folds and tape/bands.

What else to consider?

That is also why I said the problem isn't usually apparent in L1 rockets. Their fin cans are lighter, the shock cords and attachment points are relatively stronger and the safety margins are a bigger. When that fin can goes from 3 lbs on a L1 to 30 lbs on a L3, your shock loads get much larger very quickly.

If you want to do math, calculate how much force various weight fin can impart at various speeds. You can calculate the drop weight of the main chute with just the weight of the payload section and use that as the difference in the fin can drop rate to calculate the impact speed.

#### mlrtime99

##### Well-Known Member
+1 for support. Looking forward to seeing this come together.

#### mccordmw

##### Well-Known Member
Thanks. TAP is triggering the project now. We'll see what changes will need to be made.

In the meantime, I'm turning all those 8" wood circles down by an 1/8th inch so they will fit in the tubes as centering rings and bulkheads . 10 down. 10 to go.

#### mccordmw

##### Well-Known Member
If you want to do math, calculate how much force various weight fin can impart at various speeds. You can calculate the drop weight of the main chute with just the weight of the payload section and use that as the difference in the fin can drop rate to calculate the impact speed.
I think so. If the lower half is dropping at the calculated speed of 20 m/s and snaps to a slower 7 m/s, the force will be about 41 lbs force. Still it too bad.

#### mccordmw

##### Well-Known Member
I got all my centering rings sanded to fit in the Quikrete tube, and I use a 98 mm hole saw to cut the hole for the motor mount. I had to sand the hole a tiny bit larger to get a tight fit on to the LOC motor mount tube.

The centering rings will all be glassed with a layer of drain sleeve and then doubled with a second ring to create a core of two layers of fiberglass. I did a test with a couple circles of this 1/8th inch burch plywood, and it came out like steel. The centering ring sandwich will be:

fiberglass
wood
fiberglass
fiberglass
wood
fiberglass

Here's a picture of the test fit with the centering rings and the axial strengthening strakes. The fins will go in the lower large gap. Shorter strakes (not shown) have been cut to fit between the thrust plate rings and the lower centering rings. A coupler will butt up against the topmost rings to create a continuous reinforcement from thrust plate up to the payload bay.

Next, I need to cut the fins from the 1/2 inch birch plywood and start the compound bevel. I have a template and guides for the primary and secondary bevels, so hopefully they'll all come out really close to the same shape. After sanding, those will get a layer of glass for plenty of flutter safety margin.

I don't want to get too far ahead until my TAP has a chance to provide feedback.

#### mccordmw

##### Well-Known Member
Yay. I got the green light to proceed from my first TAP who's local. Need to find a good second TAP to help advise now. The hunt begins... I reached out to all the local(ish) TAPs to see who might have the time to advise.

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#### mccordmw

##### Well-Known Member
I had time tonight to create the epoxied centering rings. The starting material was 3-ply 1/8th inch birch plywood. I sourced the 8" circles from https://www.craftparts.com/wooden-circles-sign-p-4028.html at \$13 for a bag of 10. Since they were 8", they matched the outer diameter and had to be sanded down. I did this by marking the center, drilling a small hole, and putting a bolt with large washers and nuts through to clamp them all together. The bolt was put in a drill and run across a clamped sanding block until they fit the ID of the Quikrete tubes. Worked surprisingly well.

The inner hole for the 98 mm motor mount was cut out using a 98 mm hole saw. https://www.amazon.com/gp/product/B000E8FP6M/?tag=skimlinks_replacement-20

I marked the grain direction on each ring since they will be layered in perpendicular directions when making the fiberglassed sandwich.

The rings were inserted into 8" diameter polyester drain sleeve for the reinforcing fabric; making sure the grain direction registration marks were perpendicular. The rings are in pairs since they will be folded over as a sandwich after the epoxy is added.

I used US Composites 3:1 medium laminating epoxy. The polyester drain sleeve turns moderately transparent when wetted out. Excess epoxy is scraped out using a teflon scraper since it only adds weight and not strength. Unfortunately, I didn't deglove and take an in-progress pic. All the epoxy work was done on top of a sheet of mylar to keep from sticking to the board that I've sanded smooth as a good base for epoxy work.

The folded over sandwich was covered on top with mylar to and weighted down with a board for compression.

Tomorrow, I'll rip off the mylar and trim the edges with a razor.

Last edited by a moderator:

#### Handeman

##### Well-Known Member
TRF Supporter
I think so. If the lower half is dropping at the calculated speed of 20 m/s and snaps to a slower 7 m/s, the force will be about 41 lbs force. Still it too bad.
I'm not sure I have all the figures right, but assuming a ~29 lbs, or 13kg fincan and it goes from 20 m/s to 6 m/s or 14m/s to zero in 0.1 seconds when hitting the end of the shock cord. The acceleration would be 14/0.1squared) or 1400, your formula would be F = 13 * (14/(0.1 *0.1)) = 13* (14/ 0.01) = 13 * 1400 = 18,200kg or 40124.132 lbs of force. It would be much more then 41 lbs.

That's why I say you almost never see the failures in L1 or smaller L2. Assuming a 5 lbs fincan or 2.27kg, the force would be 2.27 * 1400 or 3178kg or 7006.291 lbs, which is much easier to handle. a 2.5 lb. fincan would only impart about 3500 lbs of force.

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#### mlrtime99

##### Well-Known Member
40 THOUSAND lbs??? Seemed a little large so I also did the math, I also get 40.9 lbf.


text:
# Thevenin equivalent and RC circuit

Here is the schematic:

I got a little problem in electronics: it involves simple things but I can't figure it out (and found no other circuit that was the same configuration).

At t < 0 ms, both switches are open.

The problem is the following : when t = 0 ms, the switch I is closed and the switch II is open. Then I need to make an equivalent circuit when one DC source is in series with A and B (from UAB = UTH), to find the value V0 and calculate the exponential equation for the voltage of the capacitor.

Do we remove the branch that is open to calculate UTH or do we keep it? In short, is Uth (when t = 0ms) equal to 0V or 10V?

The question is here: if we close the switch I, and if V1 is equal to 10V, the equation for the capacitor being equal to $V_1 + (V_0 - V_1) \times e^{(-t/TAU)}$ will be equal to 10V constant, which doesn't seem logical since we added one resistance in the circuit!

But if V1 is equal to 0V, it would not be logical since the capacitor is in series with the 10V DC...

Can you please guide me in the right direction or tell me what I did wrong?

For t < 0 ms, the circuit can be re-drawn as shown.

simulate this circuit – Schematic created using CircuitLab

at t = 0 ms the $30\Omega$ resistance comes in parallel with the old resistance. Now the circuit becomes.

simulate this circuit

You can directly write $R_{th}$ and $U_{th}$ from this circuit.

The only thing happened here is that you made the capacitor's charging fast. But it won't have any effect on voltage across C if it has reached V1 volts before t = 0 ms.

• Okay, thanks you a lot. It means when we close the second switch, the voltage would be still 10V ? Then u(t) would be constant 10V from t<0ms to t->+infinite ? And current would be 0A from t<0ms to t->+infinite? – user2687718 May 21 '14 at 9:22
• Nop. When 2nd switch is closed, the capacitor will start to discharge because $U_{th}$ in that case will be less than 10V. Hence there will be non zero current whose value will saturate as $t\rightarrow\infty$ – nidhin May 21 '14 at 12:35
• I get Rth = 20V and Uth = -8.4V, am I right ? :) – user2687718 May 21 '14 at 12:48
• I think $R_{th} = 20\Omega$. But I am not sure about $U_{th}$. Are we closing both the switches? or only 2nd switch? – nidhin May 21 '14 at 12:54
• Both of them when t = 10ms. :) – user2687718 May 21 '14 at 12:54


text:
2021 CMS Winter Meeting

Ottawa, June 7 - 11, 2021

Arithmetic Statistics
Org: Chantal David (Concordia), Matilde Lalin (Universite de Montreal) and Wanlin Li (CRM)
[PDF]

SIEGFRED BALUYOT, American Institute of Mathematics
Sign changes of the error term in the Piltz divisor problem  [PDF]

For an integer $k\geq 3$, let $\Delta_k(x):= \sum_{n\leq x} d_k(n)-\mbox{Res}_{s=1} (\zeta^k(s)x^s/s)$, where $d_k(n)$ is the $k$-fold divisor function and $\zeta(s)$ is the Riemann zeta-function. In the 1950's, Tong showed for all large enough $X$ that $\Delta_k(x)$ changes sign at least once in the interval $[X,X+C_k X^{1-1/k}]$, where $C_k$ is some constant. Assuming the Lindelof Hypothesis, we show the existence of many subintervals of $[X,2X]$ of length $X^{1-1/k-\varepsilon}$ such that $\Delta_k(x)$ does not change sign in any of these subintervals. This is joint work with Cruz Castillo.

LEA BENEISH, University of California, Berkeley
On the proportion of everywhere locally soluble superelliptic curves  [PDF]

We investigate the proportion of superelliptic curves that have a $\mathbb{Q}_p$ point for every place $p$ of $\mathbb{Q}$. We show that this proportion is positive and given by the product of local densities, we provide lower bounds for this proportion in general, and for superelliptic curves of the form $y^3 = f(x,z)$ for an integral binary form $f$ of degree 6, we determine this proportion to be 96.94$\%$. More precisely, we give the local density as an explicit rational function in $p$. This is joint work with Christopher Keyes.

SERGIO CEBALLOS, Western University
Distribution of the p-Torsion of Jacobian Groups of Regular Matroids  [PDF]

Given a regular matroid $M$ on $E$ and a map $\lambda\colon E\to \mathbb{N}$, we can construct a regular matroid $M_\lambda$. In this talk, we discuss the distribution of the $p$-torsion of the Jacobian groups of the family $\{M_\lambda\}_{\lambda\in\mathbb{N}^E}$. We show that those Jacobian groups with nontrivial $p$-torsion can be parametrized by the $\mathbb{F}_p$-rational points of the configuration hypersurface associated to $M$. In this way, we reduce the problem to counting points over finite fields. As a result, we obtain a closed formula for the proportion of these groups. In addition, we show that the Jacobian groups with nontrivial $p$-torsion appear with frequency close to $1/p$, provided $M$ is irreducible.

MARTIN ČECH, Concordia University
Ratios Conjectures and Multiple Dirichlet Series  [PDF]

Conrey, Farmer and Zirnbauer formulated the Ratios Conjectures. These powerful conjectures give asymptotic formulas for the products of ratios of shifted L-functions averaged over a family, and have many applications to notoriously difficult problems in arithmetic statistics.

In this talk, we will use multiple Dirichlet series to conditionally under GRH prove the ratios conjectures in the family of real Dirichlet L-functions, with one shift in the numerator and denominator, in some range of the shifts.

STEPHANIE CHAN, University of Michigan
Integral points on the congruent number curves  [PDF]

Taking the quadratic twists family of an elliptic curve, we may ask how often a curve has an integral point, and how many integral points there are on average. We will discuss some results related to the distribution of integral points in the quadratic twists family of the congruent number curve $E_D:y^2=x^3-D^2x$.

PRANENDU DARBAR, Indian Statistical Institute
Correlation of L-functions over function fields  [PDF]

In this meeting, I shall speak about the correlation of shifted values of L-functions in the hyperelliptic ensemble. More precisely, I will present lower and upper bounds for the mean values of the quadratic Dirichlet L-functions associated with the hyperelliptic curves of genus $g$ over a fixed finite field $\mathbb{F}_q$ in the large genus limit.

Bias in cubic Gauss sums  [PDF]

We prove, in this joint work with Maksym Radziwill, a 1978 conjecture of S. Patterson (conditional on the Generalised Riemann hypothesis) concerning the bias of cubic Gauss sums. This explains a well-known numerical bias in the distribution of cubic Gauss sums first observed by Kummer in 1846.

There are two important byproducts of our proof. The first is an explicit level aspect Voronoi summation formula for cubic Gauss sums, extending computations of Patterson and Yoshimoto. Secondly, we show that Heath-Brown's cubic large sieve is sharp under GRH. This disproves the popular belief that the cubic large sieve can be improved.

An important ingredient in our proof is a dispersion estimate for cubic Gauss sums. It can be interpreted as a cubic large sieve with correction by a non-trivial asymptotic main term. This estimate relies on the Generalised Riemann Hypothesis, and is one of the fundamental reasons why our result is conditional.

ALEXANDRA FLOREA, UC Irvine
The Ratios Conjecture over function fields  [PDF]

I will talk about some recent results on the Ratios Conjecture for the family of quadratic L-functions over function fields. I will also discuss the closely related problem of obtaining upper bounds for negative moments of L-functions, which allows us to obtain partial results towards the Ratios Conjecture in the case of one over one, two over two and three over three L-functions. Part of the work is joint with H. Bui and J. Keating.

ALIA HAMIEH, UNBC
Value-Distribution of Logarithmic Derivatives of Quadratic Twists of Automorphic $L$-functions  [PDF]

In this talk, I report on an ongoing work with Amir Akbary in which we study the distribution of values of the logarithmic derivative of quadratic twists of the $L$-function of a fixed automorphic representation $\pi$ on $\mathrm{GL}_d$ for any $d\in\mathbb{N}$. We establish an upper bound on the discrepancy in the convergence of the family $\frac{L'}{L}(1+it,\pi\times\chi_D)$ (with fixed $\pi$ and $t$) to its limiting distribution.

VALERIYA KOVALEVA, University of Oxford
Correlations of Riemann Zeta on the critical line  [PDF]

In this talk we will cover the correlations of the Riemann Zeta in various ranges on the critical line. We will discuss Motohashi's formula for the fourth moment on average, and the connection of the so-called moments of moments of the Riemann Zeta with the its maximum in short intervals. We will also discuss and prove a conjecture of Bailey and Keating.

PETER KOYMANS, University of Michigan
The negative Pell equation and applications  [PDF]

In this talk we will study the negative Pell equation, which is the conic $C_D: x^2 - Dy^2 = -1$ to be solved in integers $x,y \in \mathbb{Z}$. We shall be concerned with the following question: as we vary over squarefree integers $D$, how often is $C_D$ soluble? Stevenhagen conjectured an asymptotic formula for such $D$. Fouvry and Kluners gave upper and lower bounds of the correct order of magnitude. We will discuss a proof of Stevenhagen's conjecture, and potential applications of the new proof techniques.

YU-RU LIU, U. of Waterloo
On the local solubility of multidimensional Hilbert-Kamke's problem  [PDF]

Waring's problem is about representations of integers as sums of fixed powers, and Hilbert-Kamke's problem is about a system of Diophantine equations of Waring's type. Motivated by the asymptotic estimates for multidimensional Waring's problem, we consider multidimensional analogues of Hilbert-Kamke's problem. We proved that the corresponding singular series is bounded below by an absolute positive constant without any nonsingular local solubility assumption. The number of variables we need is near-optimal. This is joint work with Wentang Kuo and Xiaomei Zhao.

ALLYSA LUMLEY, CRM
Selberg's Central Limit Theorem for Quadratic Dirichlet L-functions over Function Fields  [PDF]

In this talk, we will discuss the logarithm of the central value $L\left(\frac{1}{2}, \chi_D\right)$ in the symplectic family of Dirichlet $L$-functions associated with the hyperelliptic curve of genus $\delta$ over a fixed finite field $\mathbb{F}_q$ in the limit as $\delta\to \infty$. Unconditionally, we show that the distribution of $\log \big|L\left(\frac{1}{2}, \chi_D\right)\big|$ is asymptotically bounded above by the Gaussian distribution of mean $\frac{1}{2}\log \deg(D)$ and variance $\log \deg(D)$. Assuming a mild condition on the distribution of the low-lying zeros in this family, we obtain the full Gaussian distribution.

PATRICK MEISNER, Concordia University
Lower Order Terms in the Katz-Sarnak Philosophy  [PDF]

For a nice family of $L$-functions, $\mathcal{F}$, defined over $\mathbb{F}_q[T]$, the Katz-Sarnak philosophy states that as $q$ tends to infinity, the Frobenii $\Theta$ of the $L$-functions equidistribute in a compact matrix Lie group. More concretely, it predicts that for any continuous class function $f$, we have $$\lim_{q\to\infty} \mathbb{E}_{\mathcal{F}} (f(\Theta)) = \int_G f(U) dU$$ where $G$ is some compact matrix Lie group and $dU$ is the corresponding Haar measure. In this talk we will consider lower order terms which vanish with $q$ for certain families of $L$-functions defined over $\mathbb{F}_q[T]$.

NATHAN NG, University of Lethbridge
The eighth moment of the Riemann zeta function  [PDF]

In this talk I explain how the Riemann hypothesis and a conjecture for quaternary additive divisor sums implies the conjectured asymptotic for the eighth moment of the Riemann zeta function. This builds on earlier work on the sixth moment of the Riemann zeta function (Ng, Discrete Analysis, 2021). One key difference is that sharp bounds for shifted moments of the zeta function on the critical line are required. This is joint work with Quanli Shen and Peng-Jie Wong.

ROBERT LEMKE OLIVER, Tufts University
The average size of 3-torsion in class groups of 2-extensions  [PDF]

We determine the average size of the $3$-torsion in class groups of $G$-extensions of a number field when $G$ is any transitive $2$-group containing a transposition, for example $D_4$. It follows from the Cohen--Lenstra--Martinet heuristics that the average size of the $p$-torsion in class groups of $G$-extensions of a number field is conjecturally finite for any $G$ and most $p$ (including $p\nmid|G|$). Previously this conjecture had only been proven in the cases of $G=S_2$ with $p=3$ and $G=S_3$ with $p=2$. We also show that the average $3$-torsion in a certain relative class group for these $G$-extensions is as predicted by Cohen and Martinet, proving new cases of the Cohen--Lenstra--Martinet heuristics. Our new method also works for many other permutation groups $G$ that are not $2$-groups. (Joint with Jiuya Wang and Melanie Matchett Wood.)

ALVARO LOZANO ROBLEDO, University of Connecticut
ell-adic Galois representations attached to elliptic curves with CM  [PDF]

In a recent preprint, Rouse, Sutherland, and Zureick-Brown have given a (conjectural) explicit classification of all the $\ell$-adic Galois representations (up to conjugation) attached to elliptic curves over $\mathbb{Q}$ without complex multiplication. In this talk, we apply a recent classification (by the speaker) of Galois representations attached to curves with CM to give a complete and explicit classification of all the $\ell$-adic Galois representations in the CM case over $\mathbb{Q}$. In particular, we will describe how many different representations appear for each rational $j$-invariant with CM.

STELIOS SACHPAZIS, Université de Montréal
Prime values of divisor-bounded multiplicative functions with small partial sums  [PDF]

There are results of analytic number theory which require information about the prime values of a multiplicative function, in order to provide information about its averages. A characteristic example of such a result is the Landau-Selberg-Delange method. In this talk, we are interested in the opposite direction. In particular, we are going to see that if $f$ is a suitable divisor-bounded multiplicative function with small partial sums, then $f(p)\approx -p^{i\gamma_1}-\cdots -p^{i\gamma_m}$ on average, where the $\gamma_j$'s are the ordinates of the zeros of the Dirichet series corresponding to $f$. This extends an existing result of Koukoulopoulos and Soundararajan and it is built upon ideas coming from previous work of Koukoulopoulos for the case $|f|\leqslant 1$.

SOUMYA SANKAR, Ohio State University
Counting Elliptic curves with a rational N isogeny  [PDF]

The classical problem of counting rational Elliptic curves with a rational N isogeny can be phrased in terms of counting rational points on moduli stacks of elliptic curves. I will talk about how recent work of Ellenberg, Satriano and Zureick-Brown on heights on stacks can be used to answer this classical problem for certain values of N. This talk is based on joint work with Brandon Boggess.

FRANK THORNE, University of South Carolina
Fourier Analysis in Bhargava's Averaging Method  [PDF]

I will describe a version of Bhargava's averaging method that uses Fourier analysis in place of the geometry of numbers, with applications. Joint work with Theresa Anderson and Manjul Bhargava.

ILA VARMA, University of Toronto
Geometry of Numbers Methods for Counting in the Cusp  [PDF]

In joint work with Arul Shankar, Artane Siad, and Ashwin Swaminathan, we develop a new method for counting integral orbits having bounded invariants and satisfying congruence conditions that lie inside the cusps of fundamental domains for coregular representations — i.e., representations of semisimple groups for which the ring of invariants is a polynomial ring. During this talk, we will illustrate this method in the case of counting 3-torsion elements in class groups of quadratic orders, and time permitting, we will discuss the new applications of these methods, including to counting 2-torsion ideal classes of monogenized degree-n orders.

EZRA WAXMAN, University of Haifa
A Hardy Littlewood Conjecture for Artin Primes  [PDF]

We say that a prime $p \in \mathbb{N}$ is an Artin prime for $g$ if $g$ is a primitive root mod $p$. For appropriately chosen $g$, we present a conjecture for the asymptotic number of prime $k$-tuples $(p+d_{1},...,p+d_{k})$ such that $p+d_{i}$ is an Artin prime for $g$, for all $1 \leq i \leq k$. Our results suggest that the distribution of Artin prime $k$-tuples, amongst the ordinary prime $k$-tuples, is largely governed by a Poisson binomial distribution (Joint work in part with Magdaléna Tinková and Mikuláš Zindulka; and in part with August Liu).

PENG-JIE WONG, National Center for Theoretical Sciences
Square-free orders for elliptic curves modulo $p$  [PDF]

Let $E$ be an elliptic curve defined over $\Bbb{Q}$, and let $\bar{E}(\Bbb{F}_p)$ denote the mod $p$ reduction of $E$. There is a question of finding the number of primes $p\le x$ such that $|\bar{E}(\Bbb{F}_p)|$ is square-free, which appears as an intermediate problem between the cyclicity problem for $\bar{E}(\Bbb{F}_p)$ and Koblitz's conjecture on the primality of $|\bar{E}(\Bbb{F}_p)|$. In this talk, we will discuss Cojocaru's work and talk about some of the average results, improvements, and short interval variants for such a square-freeness problem.

PETER ZENZ, McGill University
Quantum variance restriction problem for holomorphic Hecke cusp forms  [PDF]

In this talk we explore a distribution result for holomorphic Hecke cusp forms on the vertical geodesic. More precisely, we show how to evaluate the quantum variance of holomorphic Hecke cusp forms on the vertical geodesic for smooth, compactly supported test functions. The variance is related to an averaged shifted-convolution problem that we evaluate asymptotically. We encounter an off-diagonal term that matches exactly with a certain diagonal term, a feature reminiscent of moments of $L$-functions.


text:
## The Annals of Applied Probability

### Inequalities for the Time Constant in First-Passage Percolation

#### Abstract

Consider first-passage percolation on $\mathbb{Z}^d$. A classical result says, roughly speaking, that the shortest travel time from $(0, 0,\ldots, 0)$ to $(n, 0, \ldots, 0)$ is asymptotically equal to $n \mu$, for some constant $\mu$, which is called the time constant, and which depends on the distribution of the time coordinates. Except for very special cases, the value of $\mu$ is not known. We show that certain changes of the time coordinate distribution lead to a decrease of $\mu$; usually $\mu$ will strictly decrease. Two examples of our results are: (i) If $F$ and $G$ are distribution functions with $F \leq G, F \not\equiv G$, then, under mild conditions, the time constant for $G$ is strictly smaller than that for $F$. (ii) For $0 < \varepsilon_1 < \varepsilon_2 \leq a < b$, the time constant for the uniform distribution on $\lbrack a - \varepsilon_2, b + \varepsilon_1 \rbrack$ is strictly smaller than for the uniform distribution on $\lbrack a, b\rbrack$. We assume throughout that all our distributions have finite first moments.

#### Article information

Source
Ann. Appl. Probab., Volume 3, Number 1 (1993), 56-80.

Dates
First available in Project Euclid: 19 April 2007

Permanent link to this document
https://projecteuclid.org/euclid.aoap/1177005507

Digital Object Identifier
doi:10.1214/aoap/1177005507

Mathematical Reviews number (MathSciNet)
MR1202515

Zentralblatt MATH identifier
0771.60092

JSTOR


text:
# Removing aliasing/shimmering from vertical objects?

1. I've already processed this clip through AvsPMod and deinterlaced but this weird aliasing problem still remains (see how it affects the lines on old man's shirt in attached video) It is inherent to the source.

And i'm unsure if that's shimmering on the floor or just some light flickering off it.

SetMTMode(5,4)
AVISource("F:\xxxxx.avi", audio=true)
SetMemoryMax(1000)
SetMTMode(2)
AssumeTFF()
AssumeFPS(25)
QTGMC(preset="slow", matchpreset="slow", matchpreset2="slow", sourcematch=3, Lossless=2, EdiThreads=2)
2. Better might be to provide a piece from the source. But if it's also like that I doubt much can be done.
3. Originally Posted by manono
Better might be to provide a piece from the source. But if it's also like that I doubt much can be done.
Have now attached an unrendered piece of the original file.

I don't know if it could be MPEG compression artefacts as it was captured via a lossy method (AverMedia from copy protected PVR) But captured at a bitrate of 15Mbits that far exceeds the broadcast bitrate, which i suspect isn't high enough to give a proper clean HD picture.
4. The artifacts on the yellow shirt are aliasing/moire from the thin vertical stripes. You always get that when the size of the stripes approaches the resolution of the video. In this case I think the video was shot at a lower resolution (probably 720p) and upscaled. The artifacts are baked into the video and there's probably not a lot you can do about it other than blurring them away. After deinterlacing, rotating the frame 90 degrees and using QTGMC(InputType=2) to reduce the shimmering might help a little (then rotating the frame back to normal, obviously).
5. You can try using Santiag(strv=0, nns=4, nsize=5) for aliasing.

I don't think you're going to be able to do much about the lines on the shirt. The material its made of is supposed to shine like that under light. When there's an up close shot of the guys clothing you can see it better. That looks like those silky shirts my grandpa used to wear. It's more likely that when far away it looks like aliasing because of the way it reflects light.
6. Yep doing QTGMC both ways didn't help, i think you're both right it's not a problem with the video, just the way that shirt is, and would be viewed the same under the lights if there for real. I didn't notice it myself when i went to see the show live, but then i was too far away.

How do you get Santiag to show up? (just for future reference when i do want to remove other aliasing that may occur) I have both MaskTools2 and NNEDI3 plugins in my folder, but Santiag is just highlighting in red in AvsPmod.
7. Import Santiag.avs in your script.

Code:
Import("Drive:\path\to\Santiag.avs")
Or put Santiag.avs in AviSynth's plugins folder and rename it Santiag.avsi.

QTGMC(InputType=2) does help but only a little. Not worth the extra processing time. I did try applying a large BinomialBlur() to just the yellow areas. That removed a lot of it but looked bad otherwise.
8. Originally Posted by jagabo

Code:
Import("Drive:\path\to\Santiag.avs")
Or put Santiag.avs in AviSynth's plugins folder and rename it Santiag.avsi.

QTGMC(InputType=2) does help but only a little. Not worth the extra processing time. I did try applying a large BinomialBlur() to just the yellow areas. That removed a lot of it but looked bad otherwise.
Thanks for that, working now

Things ground to a halt with my settings on top of QTGMC(InputType=2) so i tried just doing TurnRight().QTGMC(InputType=2).TurnLeft().QTGMC(In putType=2) and made the mistake of putting (InputType=2) twice.. which caused problems. But TurnRight().QTGMC(InputType=2).TurnLeft().QTGMC did help a little, cleaner picture.

Originally Posted by killerteengohan
You can try using Santiag(strv=0, nns=4, nsize=5) for aliasing.

I don't think you're going to be able to do much about the lines on the shirt. The material its made of is supposed to shine like that under light. When there's an up close shot of the guys clothing you can see it better. That looks like those silky shirts my grandpa used to wear. It's more likely that when far away it looks like aliasing because of the way it reflects light.
An error keeps showing up saying Santiag does not have a named argument "nns"
9. Originally Posted by Master Tape
An error keeps showing up saying Santiag does not have a named argument "nns"
You probably have an old version. Get v 1.6:

http://avisynth.nl/index.php/Santiag


text:
### Home > APCALC > Chapter 9 > Lesson 9.4.1 > Problem9-107

9-107.

The region under $y =\sqrt { x - 1 }$ over the interval $1 ≤ x ≤ a$ for some $a > 1$ is rotated about the line $y = −1$. If the volume of the resulting solid is $10π$ cubic units, what is the value of $a$?

$\int_1^a\Big(\pi(y-(-1))^2-\pi(1)^2\Big)dx=10\pi$


text:
Kaon

Composition K+: usK0: ds and sdK−: su Bosonic Strong, weak, electromagnetic, gravitational K+, K0, K− 1947 4 K±: 493.677±0.013 MeV/c2K0: 497.648±0.022 MeV/c2 K±: ±1 eK0: 0 e 0 K+: +1K0: ±1K−: -1
The decay of a kaon (
K+
) into three pions (2
π+
, 1
π
) is a process that involves both weak and strong interactions.

Weak interactions : The strange antiquark (
s
) of the kaon transmutes into an up antiquark (
u
) by the emission of a boson; the
W+
boson subsequently decays into a down antiquark  (
d
) and an up quark (
u
).

Strong interactions: An up quark (
u
) emits a gluon (
g
) which decays into a down quark (
d
) and a down antiquark (
d
).

In particle physics, a kaon n/, also called a K meson and denoted
K
,[a] is any of a group of four mesons distinguished by a quantum number called strangeness. In the quark model they are understood to be bound states of a strange quark (or antiquark) and an up or down antiquark (or quark).

Kaons have proved to be a copious source of information on the nature of fundamental interactions since their discovery in cosmic rays in 1947. They were essential in establishing the foundations of the Standard Model of particle physics, such as the quark model of hadrons and the theory of quark mixing (the latter was acknowledged by a Nobel Prize in Physics in 2008). Kaons have played a distinguished role in our understanding of fundamental conservation laws: CP violation, a phenomenon generating the observed matter–antimatter asymmetry of the universe, was discovered in the kaon system in 1964 (which was acknowledged by a Nobel Prize in 1980). Moreover, direct CP violation was discovered in the kaon decays in the early 2000s by the NA48 experiment at CERN and the KTeV experiment at Fermilab.

Basic properties

The four kaons are :

1. K
, negatively charged (containing a strange quark and an up antiquark) has mass 493.677±0.013 MeV and mean lifetime (1.2380±0.0020)×10−8 s.

2. K+
(antiparticle of above) positively charged (containing an up quark and a strange antiquark) must (by CPT invariance) have mass and lifetime equal to that of
K
. Experimentally, the mass difference is 0.032±0.090 MeV, consistent with zero; the difference in lifetimes is (0.11±0.09)×10−8 s, also consistent with zero.

3. K0
, neutrally charged (containing a down quark and a strange antiquark) has mass 497.648±0.022 MeV. It has mean squared charge radius of −0.076±0.01 fm2.

4. K0
, neutrally charged (antiparticle of above) (containing a strange quark and a down antiquark) has the same mass.

As the quark model shows, assignments that the kaons form two doublets of isospin; that is, they belong to the fundamental representation of SU(2) called the 2. One doublet of strangeness +1 contains the
K+
and the
K0
. The antiparticles form the other doublet (of strangeness −1).

Properties of kaons
Particle name Particle
symbol
Antiparticle
symbol
Quark
content
Rest mass (MeV/c2) IG JPC S C B' Mean lifetime (s) Commonly decays to
(>5% of decays)
Kaon[1]
K+

K
493.677±0.016 12 0 1 0 0 (1.2380±0.0021)×10−8 + μ or
+ or
+ + or
+ + e
Kaon[2]
K0

K0
497.611±0.013 12 0 1 0 0 [§] [§]
K-Short[3]
K0
S
Self ${\displaystyle \mathrm {\tfrac {d{\bar {s}}-s{\bar {d}}}{\sqrt {2}}} \,}$[†] 497.611±0.013[‡] 12 0 [⁎] 0 0 (8.954±0.004)×10−11 + or
+
K-Long[4]
K0
L
Self ${\displaystyle \mathrm {\tfrac {d{\bar {s}}+s{\bar {d}}}{\sqrt {2}}} \,}$[†] 497.611±0.013[‡] 12 0 [⁎] 0 0 (5.116±0.021)×10−8 + + e or
+ + μ or
+ + or
+ +
Quark structure of the kaon plus (K⁺).

[⁎] See Notes on neutral kaons in the article List of mesons, and neutral kaon mixing, below.
[§]^ Strong eigenstate. No definite lifetime (see neutral kaon mixing).
[†]^ Weak eigenstate. Makeup is missing small CP–violating term (see neutral kaon mixing).
[‡]^ The mass of the
K0
L
and
K0
S
are given as that of the
K0
. However, it is known that a relatively minute difference between the masses of the
K0
L
and
K0
S
on the order of 3.5×10−6 eV/c2 exists.[4]

Although the
K0
and its antiparticle
K0
are usually produced via the strong force, they decay weakly. Thus, once created the two are better thought of as superpositions of two weak eigenstates which have vastly different lifetimes:

• The long-lived neutral kaon is called the
K
L
("K-long"), decays primarily into three pions, and has a mean lifetime of 5.18×10−8 s.
• The short-lived neutral kaon is called the
K
S
("K-short"), decays primarily into two pions, and has a mean lifetime 8.958×10−11 s.
Quark structure of the kaon minus (K⁻).

(See discussion of neutral kaon mixing below.)

An experimental observation made in 1964 that K-longs rarely decay into two pions was the discovery of CP violation (see below).

Main decay modes for
K+
:

Quark structure of the neutral kaon (K⁰).
Results Mode Branching ratio

μ+

ν
μ
leptonic 63.55±0.11%

π+

π0

π+

π+

π

π+

π0

π0

π0

e+

ν
e
semileptonic 5.07±0.04%

π0

μ+

ν
μ
semileptonic 3.353±0.034%

Decay modes for the
K
are charge conjugates of the ones above.