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RSS Feed for Wolfram Community showing any discussions in tag Mathematics sorted by active[GIF] Bounce ((3,1)+(1,3) vibration mode of a square membrane)
https://community.wolfram.com/groups/-/m/t/1567736
![(3,1)+(1,3) vibration mode of square membrane][1]
**Bounce**
The vibration modes of a rectangular membrane of width $L_x$ and length $L_y$ are
$\Psi_{mn}(x,y) = \sin\left(\frac{\pi m}{L_x}x\right) \sin\left(\frac{\pi n}{L_y}y\right);$
in other words, each direction just consists of standing waves with wavelength given by the reciprocal of some integer multiple of the length of the side of the membrane.
Ψ[m_, n_, {x_, y_}] := Sin[m π x] Sin[n π y];
In general these modes all have different frequencies, but when, e.g., one side length is a multiple of the other, it is possible for two different modes to have the same frequency, and then linear combinations of modes of the same frequency will also be vibration modes of the membrane. See [Dan Russell's demo][2] for more.
This animation shows the combination of the $(1,3)$ mode and the $(3,1)$ mode of the square where each factor is equally weighted. (Compare previous vibration mode animations [_Square Up_][3], [_Drumbeat_][4], and [_Things That Go Bump in the Night_][5]).
Here's the code:
DynamicModule[{n = 25, a = 1.2, dots,
cols = RGBColor /@ {"#0098d8", "#f54123", "#0b3536"}},
Manipulate[
dots = Table[
{2 π (x - 1)/n, 2 π (y - 1)/n,
Cos[θ] (1/Sqrt[2] Ψ[3, 1, {(x - 1)/n, (y - 1)/n}] + 1/Sqrt[2] Ψ[1, 3, {(x - 1)/n, (y - 1)/n}])},
{x, 1, n + 1}, {y, 1, n + 1}];
Graphics3D[
{AbsoluteThickness[2],
Table[
Line[#[[i]], VertexColors -> (Blend[cols[[;; -2]], (# + a)/(2 a)] & /@ #[[i, ;; , 3]])],
{i, Length[#]}] & /@ {dots, Transpose[dots]}},
Boxed -> False, PlotRange -> {{0, 2 π}, {0, 2 π}, {-2, 2}},
ImageSize -> 540, ViewPoint -> {2, 0, 1/2},
SphericalRegion -> True, Background -> cols[[-1]]],
{θ, 0, 2 π}]
]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=rolling34Lr.gif&userId=610054
[2]: https://www.acs.psu.edu/drussell/Demos/rect-membrane/rect-mem.html
[3]: https://community.wolfram.com/groups/-/m/t/896369
[4]: https://community.wolfram.com/groups/-/m/t/899038
[5]: https://community.wolfram.com/groups/-/m/t/985795Clayton Shonkwiler2018-12-07T21:38:05ZA Prime Pencil
https://community.wolfram.com/groups/-/m/t/1569707
![a very prime pencil][1]
I just got a set of these pencils, from Mathsgear[2].
The number printed on it is prime, and will remain so as you sharpen the pencil from the left, all the way down to the last digit, 7.
Here is a recursive construction of all such *truncatable primes*.
TruncatablePrimes[p_Integer?PrimeQ] :=
With[{digits = IntegerDigits[p]},
{p, TruncatablePrimes /@ (FromDigits /@ (Prepend[digits, #] & /@ Range[9]))}
];
TruncatablePrimes[p_Integer] := {}
The one on the pencil is the largest one,
In[7]:= Take[Sort[Flatten[TruncatablePrimes /@ Range[9]]], -5]
Out[7]= {9918918997653319693967, 57686312646216567629137, 95918918997653319693967, 96686312646216567629137,
357686312646216567629137}
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=IMG_20181212_120939.jpg&userId=143131
[2]: https://mathsgear.co.uk/products/truncatable-prime-pencilRoman Maeder2018-12-12T12:01:36ZThe "Mathy" Arts of Coding Postcards
https://community.wolfram.com/groups/-/m/t/1569557
**[Open in Cloud][1]** | **Attachments for Desktop at the End** | *LARGE images, wait till they load*
----------
![enter image description here][2]
----------
And so, the holidays are upon us once more and celebrations are in order. Wolfram Language fans enjoy fun recreation and arts, because beautiful things can be made with beautiful code, concise and elegant. I wanted to find a few gems from the past to honor the holidays and the traditions our users have. Surprisingly those few gems combined into one Christmas postcard you can see above.
The story of this postcard begins six years ago, when our sister community Mathematica Stack Exchange sprang up a question about simulating a snow fall with Wolfram Language. One of Wolfram most creative users, [@Simon Woods][at0] gave a [wonderful answer][3] that was very popular. Then about five years ago I have run into a [viral Reddit discussion][4] dubbed
$$t * sin (t) ≈ Christmas tree$$
which showcased a beautiful minimalistic Christmas tree built with simple $t * sin (t)$ function and Java Script. I recreated the concept with Wolfram Language and our another wonderful user [@Silvia Hao][at1] ornamented it with [festoon lamps][5]. An idea came to me to combine them, because a Christmas Tree sparkling lights in a snowfall is the icon of winter holidays. But beware a few subtle tricks ;-) In depth those discussed at the original references I gave. Below are slightly changed code and a few comments.
## The Tree
Our Christmas Tree is indeed spun with $t * sin (t)$. But in 3D rather than 2D. This is basically a conical spiral whose amplitude increases, a 2D circle dragged along 3-rd axis like [this][6]:
![enter image description here][7]
but only with increasing radius. Density of lights and their motion is one subtlety to take care of with math. Another subtlety is increasing 3D depth perception by slightly dimming the lights that are further from the observer. This function defines the mathematics of the tree:
PD = .5;
s[t_, f_] := t^.6 - f
dt[cl_, ps_, sg_, hf_, dp_, f_, flag_] :=
Module[{sv, basePt},
{PointSize[ps],
sv = s[t, f];
Hue[cl (1 + Sin[.02 t])/2, 1, .8 + sg .2 Sin[hf sv]],
basePt = {-sg s[t, f] Sin[sv], -sg s[t, f] Cos[sv], dp + sv};
Point[basePt],
If[flag,
{Hue[cl (1 + Sin[.1 t])/2, 1, .8 + sg .2 Sin[hf sv]], PointSize[RandomReal[.01]],
Point[basePt + 1/2 RotationTransform[20 sv, {-Cos[sv], Sin[sv], 0}][{Sin[sv], Cos[sv], 0}]]},
{}]
}]
and this code uses the function to build 228 frames of the animated tree:
treeFrames = ParallelTable[
Graphics3D[Table[{
dt[1, .01, -1, 1, 0, f, True],
dt[.45, .01, 1, 1, 0, f, True],
dt[1, .005, -1, 4, .2, f, False],
dt[.45, .005, 1, 4, .2, f, False]},
{t, 0, 200, PD}],
ViewPoint -> Left, BoxRatios -> {1, 1, 1.3},
ViewVertical -> {0, 0, -1}, Boxed -> False,
ViewCenter -> {{0.5, 0.5, 0.5}, {0.5, 0.55}},
PlotRange -> {{-20, 20}, {-20, 20}, {0, 20}},
Background -> Black,ImageSize->350],
{f, 0, 1, .0044}];
Let's check a single frame of THe Tree:
First[treeFrames]
![enter image description here][8]
## The Snow
This function below builds a single random snowflake. They are of course six-fold symmetric polygons.
flake := Module[{arm},
arm = Accumulate[{{0, 0.1}}~Join~RandomReal[{-1, 1}, {5, 2}]];
arm = arm.Transpose@RotationMatrix[{arm[[-1]], {0, 1}}];
arm = arm~Join~Rest@Reverse[arm.{{-1, 0}, {0, 1}}];
Polygon[Flatten[arm.RotationMatrix[# \[Pi]/3] & /@ Range[6], 1]]];
Let's see a few random shapes, they are fun in black on white ;-)
Multicolumn[Table[Graphics[flake, ImageSize -> 50], 100], 10]
![enter image description here][9]
Now it's time to build the `snowfield` which has a few tricks. To simulate 3D perception 2 things need to be obsereved:
1. Real further snowflakes appear smaller
2. Real further snowflakes have slower perceived angular speeds
The 2nd observation is taken care of by the `size_` variable below.
snowfield[flakesize_, size_, num_] :=
Module[{x = 100/flakesize},
ImageData@
Image[Graphics[{White,Opacity[.8],
Table[Translate[
Rotate[flake, RandomReal[{0, \[Pi]/6}]], {RandomReal[{0, x}],
RandomReal[{0, x}]}], {num}]}, Background -> Black,
PlotRange -> {{0, x}, {0, x}}], ImageSize -> {size, size}]];
and by 3 different sizes given here:
size=455;
r=snowfield@@@{{.9,size,250},{1.2,size,30},{1.6,size,10}};
So we sort of have 3 different fields of vision reproaching the observer. The 1st observation is simulated with different speed with which different fields of vision are rotated, the closer one being the fastest. This simulates rotation of the fields of vision and builds the frames for the snowfall:
snowFrames=ParallelTable[Image[Total[(RotateRight[r[[#]],k #]&/@{1,2,3})[[All, ;;size]]]],{k,0,455,2}];
## The Postcard
Slight opacity is needed to to blend The Tree and The Snow appealingly. The opacity is given the snowflakes in the code above and `SetAlphaChannel` below is formally needed for image data to have the same dimensions (3 RGB + 1 Opacity channels) and to be able to combine. This builds the final frames
finalFrames=
Parallelize[MapThread[
ImageAdd[SetAlphaChannel[#1,1],#2]&,
{treeFrames,snowFrames}]];
and this exports the frames to the GIF you see at the top of the post:
Export["xmas.gif", finalFrames,"AnimationRepetitions"->Infinity]
I hope you had fun. Feel free to share your own crafts. Happy holidays!
[at0]: https://community.wolfram.com/web/swoods1
[at1]: https://community.wolfram.com/web/wyelen
[1]: https://www.wolframcloud.com/objects/wolfram-community/The-Mathy-Arts-of-Coding-Postcards-by-Vitaliy-Kaurov
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=8701ezgif.com-optimize.gif&userId=11733
[3]: https://mathematica.stackexchange.com/a/16889/13
[4]: https://redd.it/1tswai
[5]: https://community.wolfram.com/groups/-/m/t/175891
[6]: https://en.wikipedia.org/wiki/File:ComplexSinInATimeAxe.gif
[7]: https://community.wolfram.com//c/portal/getImageAttachment?filename=ComplexSinInATimeAxe.gif&userId=11733
[8]: https://community.wolfram.com//c/portal/getImageAttachment?filename=534yrsgfdgbd.png&userId=11733
[9]: https://community.wolfram.com//c/portal/getImageAttachment?filename=435wyrhgsfdasaW.png&userId=11733Vitaliy Kaurov2018-12-12T00:56:35ZGet step by step solution of this differential equation with W|A?
https://community.wolfram.com/groups/-/m/t/1569498
When I enter the following ode y''''+8y''+16y=e^t+e^(2t), wolfram alpha doesn't load all steps for me (clicking the button 'show all steps'). Is it just me or is it a general problem.
EDIT: Single steps are working but not all of them are loading. It usually stops after step 14 for me.Agre agrem2018-12-11T17:03:23Z[✓] Use a Wolfram Language equivalent to Matlab "linspace" function?
https://community.wolfram.com/groups/-/m/t/1568743
Hello
Matlab has a built in function called `linspace`, that, according to the documentation, generates a linearly spaced vector, but "gives direct control over the number of points and always includes the endpoints".
Mathematica doesn't work with vectors, rather with lists, but I wish to generate a list of numbers that start from `x1` and ends at `x2`, with `n` points in between (such that the spacing between the points is (x2-x1)/(n-1).
I couldn't find help with `Table`. `Table` generates a list of numbers with a starting point and an end point, with an extra option for the intervals, but I want to have `n` numbers in between.
Is there any work around?Ehud Behar2018-12-10T17:48:09ZSearching for collaborators about image tiling
https://community.wolfram.com/groups/-/m/t/1568407
Symmetry and symmetry breaking is a central topic of my artistic work. I am fascinated by image tiling as a source of symmetry but until recently the situation was not different than 25+ years ago when my interest started with the Photoshop plugin Terrazzo: There are thousands of known euclidean tilings ([Tiling Database][1]) but only the basic 17 wallpapergroups were used for image tiling.
I am working on a general approach to change this: Every image tiling or image pattern in general can be made by one or more proto-tiles (rectangle or masked polygon shaped images with transparency) and a list of clone-, rotate-, mirror- (flip,flop), and translate-commands collected in a CRMT command list. A CRMT interpreter would take such a list and a set of proto-tiles and generate an image tiling, ornament or pattern by step-by-step processing the commands.
For example the following CRMT command list is coding the 14 processing steps to generate a p3m1 tile from a given equilateral triangle proto-tile image (see my p3m1-examples using this CRMT approach on [p3m1-CRMT album 1][2] and [p3m1-CRMT album 2][3]):
C0, x0, y0, C0, Fo, R-60, x-1/2*t_w, y0, C0, Fo, R60, x1/2*t_w, y0, C0, Fo, R-60, Fi, xt_w, y0, C0, Fi, x3/2*t_w, y0, C0, Fo, R60, Fi, x2*t_w, y0, C0, Fo, R-60, x5/2*t_w, y0, C0, Fi, x0, yt_h, C0, Fo, R-60, Fi, x-1/2*t_w, yt_h, C0, Fo, R60, Fi, x1/2*t_w, yt_h, C0, Fo, R-60, xt_w, yt_h, C0, x3/2*t_w, yt_h, C0, Fo, R60, x2*t_w, yt_h, C0, Fo, R-60, Fi, x5/2*t_w, yt_h
The operation sequence for one proto-tile processing like "C0, Fo, R-60, x-1/2*t_w, y0" is interpreted as: clone the first (starting with 0) element in the proto-tile list, flop it, rotate it 60 degrees counterclockwise (+ trim), make a translation in x-direction with floor(-1/2*t_w) where t_w is the width of the equilateral triangle, make a translation in y-direction with 0 and then compose the proto-tile over the tile background which is in the p3m1 case a black image with an (2*t_h, 3*t_w) area were t_h = 1/2*sqrt(3)*t_w.
![p3m1 CRMT example from my art][4]
Additionally the CRMT interpreter also needs a list with coordinates for one or more masks that must be draw because the masked proto-tiles must come somewhere; in the p3m1 case the x and y coordinates of the triangle are: x_coord = [0 t_w t_w/2]; y_coord = [0 0 t_h];
Such an approach is not fast but universal and because of the no-overlap condition of tilings the processing steps for one proto-tile are independent and therefore the 14 steps in the p3m1 case could be made in parallel. And there is always the option to optimize some specific tiling by using some knowledge about its structure. In the p3m1 case run-time can be saved by the knowledge that the lower half is a flipped version of the upper half.
To further develop the CRMT approach I am searching for
**1) programmer implementing CRMT interpreter in different environments like Mathematica**
Programming a CRMT interpreter seems neither a difficult nor a too costly task because the core is some string processing combined with calling some image processing functions. And if the used environment has RGBA abilities and boundary methods like "-virtual-pixel mirror" (ImageMagick function) or simple inpainting functions it makes everything much easier because composing polygon shaped images results in artifacts at the edges if non-90 degree polygon-masks were used.
There is active development in other environments underway: I have programmed a batch processing prototype in Matlab and a first prototype for a CRMT filter in G'MIC was published this week ([G'MIC][5], [discussion about the G'MIC CRMT filter][6]).
**2) people writing their own CRMT lists**
Every high school kid with basic trigonometry knowledge can determine the angles, lengths and distances in a given tiling image from the Tiling Database to write a CRMT-list without any knowledge of a specific programming language or image processing, so there is potential a huge community. It is only a question of persistence to code even the most complex periodic tilings like [Islamic patterns][7].
Math teachers are often looking for something motivational. Real world applications and aesthetics in math (see [Bridges conferences][8]) are mostly the areas they come up with and I think that image tiling is an excellent example for the later. Writing a CRMT command list for a tiling would be a nice assignment if intermediate states could immediately be checked for feedback with a CRMT interpreter. And there is the general question about meaning and sustainability of assignments: Making the first command list for a tile is a meaningful and sustainable activity because such a list can generate aesthetic images even decades later independent of whatever programming languages will be used then for a CRMT interpreter.
**Future perspectives**
Using trigonometry to extract information from given tilings, ornaments and patterns will restrict the audience mostly to academia and math enthusiasts. A much wider audience would be reached with a GUI based system where proto-tiles are placed with drag&drop. Combined with an interaction log that records the relevant actions and a CRMT optimizer that combines all the actions related to one proto-tile to one command set a CRMT-list for a pattern should be reconstructible. I am thinking about modifying this [Mathematica demonstration][9]. If in a long-term perspective AI-agents with a learned sense for symmetry and aesthetic can play such a system an endless stream of interesting pattern descriptions would be accessible that can directly be used in every environment with a CRMT interpreter.
The CRMT approach is also extendible in many directions for example by simply adding a new command type like "S" for scaling which makes image types like euclidean [Fractal Tiling][10], Iterated Function Systems and [orbit trapping][11] accessible.
[1]: http://www.tilingsearch.org/
[2]: https://www.flickr.com/photos/gbachelier/albums/72157673790864417
[3]: https://www.flickr.com/photos/gbachelier/albums/72157674278683877
[4]: https://community.wolfram.com//c/portal/getImageAttachment?filename=32230404598_168f19be34_k.jpg&userId=753358
[5]: https://gmic.eu
[6]: https://discuss.pixls.us/t/collaborators-for-image-tiling/9966/24
[7]: https://patterninislamicart.com/drawings-diagrams-analyses/1/elements-art-arabe
[8]: http://bridgesmathart.org/
[9]: http://demonstrations.wolfram.com/TilingConstructorTileDraggingVariant/
[10]: https://www.mathartfun.com/encyclopedia/encyclopedia.html
[11]: http://2008.sub.blue/projects/fractal_explorer.htmlGuenter Bachelier2018-12-09T13:46:51ZFind all roots of the following equation?
https://community.wolfram.com/groups/-/m/t/1568356
I'm currently doing some Mathematica exercises, and I'm stuck on this one task where you're supposed to plot the functions h(t)= |3-t^2|+|t-1|-t^2 , g(t)=3sin(t) in the same grap, and then find all the roots. This is what I've got so far:
![enter image description here][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=UHvxo.png&userId=1540567
The instructions say that I should use FindRoot to exactly decide all the roots, but I don't think I've done it right. What should I change with the function in order to make it find all of the roots?
Thanks in advance.Jhn Snd2018-12-09T11:50:45ZImprove code for finding the coordinates of a triangular mesh?
https://community.wolfram.com/groups/-/m/t/1567948
Hello, The problem of finding the coordinates of a triangular (equilateral) mesh discussed earlier is solved. It also counts the number of equilateral triangles formed by the intersecting parallel lines. But the problem is that the code takes more time for larger values of n i.e. the size of the side of the triangle. Can the code be improved? The code is given here. Thanks for any suggestion.
n = 4;
Print["Number of lines/size of triangle = ", n]
h = Sqrt[3] /2;
Array[x, n];
Array[s, n];
x[0] = {{n/2, n h}};
For[i = 1, i <= n, i++,
x[i] = Table[{x[0][[1, 1]] - i/2 + j, n h - i h}, {j, 0, i}]];
set = Apply[Union, Table[x[i], {i, 0, n}]];
Print["Number of vertices = ", Length[set]]
cond := (EuclideanDistance[#[[1]], #[[2]]] ==
EuclideanDistance[#[[2]], #[[3]]] ==
EuclideanDistance[#[[1]], #[[3]]] && #[[1]] != #[[2]] != #[[
3]] && #[[1, 1]] < #[[2, 1]] < #[[3,
1]] && (#[[1, 2]] == #[[2, 2]] || #[[2, 2]] == #[[3, 2]] || #[[
3, 2]] == #[[1, 2]]) &)
tr0 = Tuples[set, 3];
tr1 = Select[tr0, cond];
Print["Number of Triangles = ", Length[tr1]]jagannath debata2018-12-09T07:39:12ZPlot Poincare map in order to analyze chaos?
https://community.wolfram.com/groups/-/m/t/1567675
Consider the following code:
U[t] + 3 U[t]^2 + 6 V[t] + 3 V[t]^2 + 5 W[t] + 2 W[t]^2 + 4 U[t]*V[t] == 2 U'[t] ;
6 U[t] + 3 U[t]^2 + 3 V[t] + 4 V[t]^2 + 8 W[t] + 4 W[t]^2 + 3 U[t]*V[t] == V'[t];
5 U[t] + 3 U[t]^2 + 5 V[t] + 3 V[t]^2 + 8 W[t] + 4 W[t]^2 + 8 U[t]*V[t]+ Q*Sin[100*t] == W'[t] + 2 W''[t];
U[0] == V[0] == W[0], U'[0] == V'[0] == W'[0]==0.0001
Q=const
I need to plot a Poincare map with W [t], W '[t]. I am having trouble. I thank everyone.Vũ Ngọc Việt Hoàng2018-12-08T07:33:18Z[✓] Use NMinimize calling an own defined function?
https://community.wolfram.com/groups/-/m/t/1567463
I want to call an own defined function through NMinimize. For the sake of simplicity, let us define the problem as follow:
radpatt[x1_] := (Print[x1]; x1 )
NMinimize[{radpatt[xx], 0.1 <= xx <= 1.1}, {xx}]
If you run the above-listed instructions, you will notice that x1 is not number as soon as the function radpatt is called by NMinimize, but it is equal to xx. This causes me an issue because in my original problem x1 needs to be a number just at the beginning of my own function. Any ideas?
Many many thanks in advance.Mario Junior Mencagli2018-12-08T06:23:38ZSolve the differential equation by Runge Kutta method?
https://community.wolfram.com/groups/-/m/t/1563269
I need to solve the equation as follows by numerical methods such as Runge Kutta, Newton Raphson, ... Hope everyone help me. I thank everyone.
![enter image description here][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Untitled.png&userId=1563397Vũ Ngọc Việt Hoàng2018-12-01T15:10:42ZGet formula to calculate text rotation angle (clock angle problem)?
https://community.wolfram.com/groups/-/m/t/1567764
Hello.
Let's say we have circle. We cut it to 4 'triangles' but longest hand is perfect arc. I want to write some letters on this arc but rotate it to be horizontal always.
In other words, how to make clock face with horizontal numbers. It can be any hour, so can be 12-3-6-9, but can be 1-4-7-11 etc.
Thanks for suggestions.John Cosmic2018-12-08T00:16:55ZUse NMinimize calling a own defined function with FindRoot?
https://community.wolfram.com/groups/-/m/t/1567276
I defined my own function which solves a transcendental equation by using a findroot, and it works. But, when I call my own defined function with NMinimize, it gives me an error. I do not understand where it comes from. Attached please find my code. Please let me know if you know where the problem might be.Mario Junior Mencagli2018-12-07T05:10:37ZAvoid problem using "Nested" NMaximize?
https://community.wolfram.com/groups/-/m/t/1566000
I'm having trouble with NMaximize in the following code (this is an example of my real problem, I know that this example has an exact analytical solution).
This part works just fine, is the clasical consumer maximization problem :
Util[x_, y_, a_] := x^a* y^(1 - a)
UtilMax[a_, px_, py_, P_] := Module[{temp, UtilMax, XYOptim},
temp = NMaximize[{Util[x, y, a], px*x + py*y <= P, x > 0,
y > 0}, {x, y}];
UtilMax = temp[[1]];
XYOptim = {x, y} /. Last[temp];
Flatten[{XYOptim, UtilMax}]]
Sales[vecAlfa_, vecP_, px_, py_] := Module[{temp},
temp = Table[
UtilMax [vecAlfa[[i]], px, py, vecP[[i]]], {i, 1,
Length[vecAlfa]}];
Sum[temp[[i, 1]] + temp[[i, 2]], {i, 1, Length[vecAlfa]}]
]
vecAlfa = {0.1, 0.9};
vecP = {10, 20};
Test that the functions are working ok:
In[6]:= UtilMax[vecAlfa[[1]], 5, 5, vecP[[1]]]
UtilMax[vecAlfa[[2]], 5, 5, vecP[[2]]]
Sales[vecAlfa, vecP, 5, 5]
Out[6]= {0.2, 1.8, 1.44493}
Out[7]= {3.6, 0.4, 2.88987}
Out[8]= 6.
The problem arises with the followin part:
NMaximize[{Sales[vecAlfa, vecP, px, py], 0 < px < 100, 0 < py < 50}, {px, py}]
During evaluation of In[10]:= NMaximize::bcons: The following constraints are not valid: {x>0,y>0,px x+py y<=10}. Constraints should be equalities, inequalities, or domain specifications involving the variables. >>
During evaluation of In[10]:= ReplaceAll::reps: {x,y} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>
During evaluation of In[10]:= NMaximize::bcons: The following constraints are not valid: {x>0,y>0,px x+py y<=20}. Constraints should be equalities, inequalities, or domain specifications involving the variables. >>
During evaluation of In[10]:= ReplaceAll::reps: {x,y} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>
During evaluation of In[10]:= NMaximize::nnum: The function value -x^0.9 y^0.1-x^0.1 y^0.9-2 ({x,y}/. {x,y}) is not a number at {px,py} = {1.91862,1.66351}. >>
Out[10]= NMaximize[{x^0.9 y^0.1 + x^0.1 y^0.9 + 2 ({x, y} /. {x, y}),
0 < px < 100, 0 < py < 50}, {px, py}]
Any idea what could I've been doing wrong?Augusto Umaña2018-12-06T05:47:13ZFind the Coordinates of a triangular mesh?
https://community.wolfram.com/groups/-/m/t/1565820
Hello , I have an equilateral triangle with vertices at {0,0},{4,0} and {2,2 Sqrt[3]}. Three points on each side are taken dividing the side into four equal parts .These points are joined so as to draw lines parallel to the sides. How do we get the coordinates of the intersecting points of the triangular mesh obtained? For example {{1/2,sqrt[3]/2},{1,0},{0,0}} is a triangle. Thanks for any help.jagannath debata2018-12-04T16:04:49ZFind the curvature of an object from an image
https://community.wolfram.com/groups/-/m/t/1562821
`curvatureMeasure.m` is a Mathematica script for calculating curvature along the boundary of an image object. It might be useful to people working in the computer vision community. This simple script can be easily extended to even track the curvature of an object as it deforms (will add soon) across several images.
Here is our input image:
![enter image description here][1]
boundary is discretized into equidistant points
![enter image description here][2]
circle fits to a given point point Pi, Pi + N-left, Pi + N-right, where N is the neighbour to the point Pi.
![enter image description here][3]
curvatures (found as 1/radius) is the colormap:
![enter image description here][4]
**code**
(* ::Package:: *)
BeginPackage["curvatureMeasure`"];
curvatureMeasure::usage = "measures the curvature along the image object";
Begin["`Private`"];
shiftPairs[perimeter_,shift_]:=Module[{newls},
newls=perimeter[[-shift;;]]~Join~perimeter~Join~perimeter[[;;shift]];
Table[{newls[[i-shift]],newls[[i]],newls[[i+shift]]},{i,1+shift,Length[newls]-shift}]
];
(* we can either use the suppressed code below to fit circles or the Built-In Circumsphere to find the fits
(* from Mathematica StackExchange: courtesy ubpdqn *)
circfit[pts_]:=Module[{reg,lm,bf,exp,center,rad},
reg={2 #1,2 #2,#2^2+#1^2}&@@@pts;
lm=LinearModelFit[reg,{1,x,y},{x,y}];
bf=lm["BestFitParameters"];
exp=(x-#2)^2+(y-#3)^2-#1-#2^2-#3^2&@@bf;
{center,rad}={{#2,#3},Sqrt[#2^2+#3^2+#1]}&@@bf;
circlefit[{"expression"->exp,"center"->center,"radius"->rad}]
];
circlefit[list_][field_]:=field/.list;
circlefit[list_]["Properties"]:=list/.Rule[field_,_]:>field;
circlefit/:ReplaceAll[fields_,circlefit[list_]]:=fields/.list;
Format[circlefit[list_],StandardForm]:=HoldForm[circlefit]["<"<>ToString@Length@list<>">"]
*)
curvatureMeasure[img_Image,div_Integer,shift_Integer]:=Module[{\[ScriptCapitalR],polygon,t,interp,sub,sampledPts,
pairedPts,circles,\[Kappa],midpts,regMem,col,g,fn},
\[ScriptCapitalR] = ImageMesh[img, Method -> "Exact"];
polygon = Append[#,#[[1]]]&@MeshCoordinates[\[ScriptCapitalR]][[ MeshCells[\[ScriptCapitalR],2][[1,1]] ]];
t = Prepend[Accumulate[Norm/@Differences[polygon]],0.];
interp = Interpolation[Transpose[{t,polygon}],InterpolationOrder -> 1,
PeriodicInterpolation->True];
sub = Subdivide[interp[[1,1,1]],interp[[1,1,2]],div];
sampledPts = interp[sub];
Print[Show[\[ScriptCapitalR],Graphics@Point@sampledPts,ImageSize-> 250]];
pairedPts = shiftPairs[sampledPts, shift];
circles = (Circumsphere/@pairedPts)/. Sphere -> Circle;
(*circles = (fn=circfit[#]; Circle[fn["center"],fn["radius"]])&/@pairedPts;*)
Print[Graphics[{{Red,Point@sampledPts},{XYZColor[0,0,0,0.1],circles}}]];
\[Kappa] = 1/Cases[circles,x_Circle:> Last@x];
midpts = Midpoint/@pairedPts[[All,{1,-1}]];
regMem = RegionMember[\[ScriptCapitalR],midpts]/.{True-> 1,False-> -1};
\[Kappa] *= regMem;
col = ColorData["Rainbow"]/@Rescale[\[Kappa], MinMax[\[Kappa]],{0,1}];
g = Graphics[{PointSize[0.018],MapThread[Point[#1,VertexColors->#2]&,{sampledPts,col}]}];
Print[Show[HighlightMesh[\[ScriptCapitalR],{Style[1, Black],Style[2,White]}],g,ImageSize->Medium]];
\[Kappa]
]
End[];
EndPackage[];
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=inputImage.png&userId=942204
[2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=mesh.png&userId=942204
[3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=curvatureFitstoShape_cropped.png&userId=942204
[4]: https://community.wolfram.com//c/portal/getImageAttachment?filename=curvatureFinal.png&userId=942204Ali Hashmi2018-11-30T07:26:13ZEdit specific parts of vector fields?
https://community.wolfram.com/groups/-/m/t/1565312
I would like to take a specific part of a given vector field, and make all the vectors bold.
For example, I want to take the {x, y} vector field, and make all the vectors from x=0 to x=10 to be
bold, and the rest of the field to be regular. I would appreciate it if anyone could help me figure this out!Hafez Rais2018-12-04T05:24:35ZGet 3358th digit of Pi, E and Phi using W|A?
https://community.wolfram.com/groups/-/m/t/1564133
According to subidiom.com the first occurrence of 1984 on Pi is at the 3358th decimal digit.
http://www.subidiom.com/pi/pi.asp
![enter image description here][1]
I can use the command "3359 digit of pi" to get a similar output on WolframAlpha including nearby digits as you can see below.
http://m.wolframalpha.com/input/?i=3359+digit+of+pi
Unfortunately it doesn't seem to work with e or phi. Does someone have a workaround for that ?
Thanks,
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=F22D299C-8F24-4047-9F52-CAA9DC8A2271.png&userId=1562578Renan José2018-12-02T23:51:41Z