text:
# Simulator-specific optimizations¶

This page explains how to optimize your code, in pyNN, for particular simulators. Those are specific options/code instructions that will make your pyNN code tuned to a particular backend, sometimes breaking compatibility with the others. However, if you would like to speed up your code and if at a particular time, here are some tricks you should know

## General observations¶

### Assemblies¶

When recording an Assembly object, i.e the concatenation of several populations, you should notice that saving the recorded data to file is taking some time. In fact, even if the user is allowed to do that, this is something you could maybe avoid, in order to speed up the writing times of large data files. Merging all the indices and the recorded data from the populations within the Assembly involves sorting large arrays, and this is not optimal. So instead of doing:

>>> exc = Population(10000, IF_curr_exp)
>>> inh = Population(2500, IF_curr_exp)
>>> all_cells = exc + inh
>>> all_cells.record()
>>> run(simtime)
>>> all_cells.printSpikes(file)

You should do:

>>> exc = Population(10000, IF_curr_exp)
>>> inh = Population(2500, IF_curr_exp)
>>> all_cells = exc + inh
>>> all_cells.record()
>>> run(simtime)
>>> exc.printSpikes(file_exc)
>>> inh.printSpikes(file_inh)

This is more convenient in a sense that data will be sorted by populations in two files at the end, and it will be faster.

### Recording to memory¶

If your simulations are short enough, and if you know that the size of the recorded data will fit in the memory, you should better record data to memory, instead of a file. This is not the default option, but then, at the end of the simulations, accessing the raw recorded data will be faster, because they won’t have to be read from the disk. This is only tru for pyNN.nest, because NEST is the only one able to record to file while running. To so do, you should do:

>>> population.record(to_file=False)

### Projections¶

If you want to speed up the building times of your networks, you have to be sure that you are using all the options in the Connector object. Basically, you should avoid as much as possible the use of the function setDelays() or setWeights(). Those functions will loop again over all the synapses that have been created to set their weights/delays, but except if you have a good reason to do so, you should provide those values when synapses are created, during the projection. Instead of doing:

>>> prj = Projection(population, population, AllToAllConnector())
>>> prj.setDelays(0.1)
>>> prj.setWeights(0.3)

You should do:

>>> prj = Projection(population, population, AllToAllConnector(weights=0.3, delays=0.1))

## Optimizations for NEST¶

### Fast Connectors¶

In NEST, the pyNN.connectors objects are not taking advantages of the mpirun command, because they are based on the function nest.ConvergentConnect which does not scale well with the number of nodes. Therefore, if you want to reduce the building times of your network according to the number of nodes used by your parallel simulations, you should use the following Connectors:

• FastFixedProbabilityConnector
• FastDistanceDependentProbabilityConnector
• FastAllToAllConnector
• FastSmallWorldConnector
• FastFromListConnector
• FixedNumberPreConnector

However, note that those connectors (based on nest.DivergentConnect function) will not lead to the same networks.

### SpikeSourcePoisson¶

The SpikeSourcePoisson in pyNN is meant to be an object that can be connected to various targets, and send the same spike train realisation to all its targets. However, quite often, you may end up by building a large network of cells that you would like to be able to bombard with external noise. Creating as many SpikeSourcePoisson as cells in your network is very inefficient in NEST, because those Stimulating Devices are not meant to be numerous. Moreover, in NEST, a single poisson_generator object is able to send independent realizations of the spike trains to all its targets. So instead of doing:

>>> population = Population(10000, IF_curr_exp)
>>> noise      = Population(10000, SpikeSourcePoisson, {'rate' : 1000.})
>>> prj        = Projection(noise, population, OneToOneConnector(weights=w))

You should use the native cell type poisson_generator from NEST and write:

>>> population = Population(10000, IF_curr_exp)
>>> noise      = Population(1, native_cell_type("poisson_generator"), {'rate' : 1000.})
>>> prj        = Projection(noise, population, AllToAllConnector(weights=w))

This will be much more efficient. However, you won’t be able anymore to record the noise population, because those devices can not be recorded

## Optimizations for BRIAN¶

### setup¶

If you are launching Brian from a machine with a gcc compiler installed (i.e tested with Linux, should also work from windows, see Brian website for that), you could take advantage of the fact that Brian can pre-compile parts of the code in C++ and speed up the resolution of the differential equations governing the dynamics. To do that, you just have to pass extra arguments in the setup() function:

>>> setup(timestep=0.1, min_delay=0.1, extra={'useweave':True, 'usecodegen':True}

### Fast Connectors¶

In BRIAN, the pyNN.connectors objects are not taking advantages of the homogeneous delays, because they are made such that a user is always able to change the delays, even after connections have been created. However, BRIAN is much more efficient when using homogeneous delays in a projection. Building and simulations times are faster, and memory usage more efficient. Therefore, if you know in advance that you’re creating a projection with homogeneous delay that will not be changed afterwards, then you should use the following connectors:

- FastFixedProbabilityConnector
- FastDistanceDependentProbabilityConnector
- FastAllToAllConnector
- FastSmallWorldConnector
- FastOneToOneConnector

>>> population = Population(10000, IF_curr_exp)

>>> population = Population(10000, IF_curr_exp)


text:
## Wednesday, May 14, 2014

### New BICEP rumours: nothing to see here

This week there has been a minor kerfuffle about some rumours, originally posted on Adam Falkowski's Resonaances blog, regarding the claimed gravitational wave detection by BICEP. The rumours asserted that Planck had proven BICEP had made a mistake, BICEP had admitted the mistake, and that this might mean that all the excitement about the detection of gravitational waves was misplaced and all that BICEP had seen was some foreground dust emission contaminating their maps. (Since then there has been a strong public denial of this by the BICEP team.)

Now, with the greatest respect to Resonaances, which is an excellent particle physics blog, this is really a non-issue, and certainly not worth offending lots of people for (see for instance Martin Bucher's comment here). I really do not see what substantial information these rumours have provided us with that was not already known in March, and therefore why we should alter assessments of the data  made at that time.

Let me explain a bit more. One of the important limitations of the BICEP2 experiment is that it essentially measured the sky at only one frequency (150 GHz) — the data from BICEP1, which was at 100 GHz, was not good enough to see a signal, and the data from the Keck Array at 100 GHz has not yet been analysed. When you only have one frequency it is much harder to rule out the possibility that the "signal" seen is not due to primordial gravitational waves at all but due to intervening dust or other contamination from our own Galaxy.

The way that BICEP addressed this difficulty was to use a set of different models for the dust distribution in that part of the sky, and to show that all of them predict that the possible level of dust contamination is an order of magnitude too small to account for the signal that they see. Now, some of these models may not be correct. In fact none of them are likely to be exactly right, because they may be based on old and likely less accurate measurements of the dust distribution or rely on a bit of extrapolation, wishful thinking, whatever. But the point is that they all roughly agree about the order of magnitude of dust contamination. This does not mean that we know there is or isn't any foreground contamination; this is merely a plausibility argument from BICEP (that is supported by and supports some other plausibility arguments in the paper).

Now the "new" rumour is based on the fact that it turns out that one of the dust models was based on BICEP's interpretation of preliminary Planck data, and that this data was not officially sanctioned but digitally extracted from a pdf of a slide shown at a talk somewhere. This is not exactly news, since the slide in question is in fact referenced in the BICEP paper. What's new is that now somebody unnamed is suggesting that the slide was in fact misinterpreted, and therefore this one dust model is more wrong than we thought, though we already accepted it was probably somewhat wrong. This is not the same as proving that the BICEP signal has been definitively shown to be caused by dust contamination! In fact I don't see how it changes the current picture we have at all. Ultimately the only way we can be sure about whether the observed signal is truly primordial or due to dust is to have measurements that combine several different frequencies. For that we have to wait a bit for other experiments — and that's the same as we were saying in March.

It's worth noting that when BICEP quote their result in terms of the tensor-to-scalar ratio r, the headline number $r=0.2$ assumes that there is literally zero foreground contamination. This was always an unrealistic assumption, but that hasn't stopped some 300 theorists from writing papers on the arXiv that take the number as face value and use it to rule out or support their favourite theories. The foreground uncertainty means that while we can be reasonably confident that the gravitational wave signal does exist (see here), model comparisons that strongly depend on the precise value of r are probably going to need some revision in the future.

So what new information have we gained since March? Well, Planck released some more data, this time a map of the polarized dust emission close to the Galactic plane.

 The polarization fraction at 353 GHz observed by Planck. From arXiv:1405.0871.

Since these maps do not include the part of the sky that BICEP looked at (which is mostly in the grey region at the bottom), they don't tell us a huge amount about whether that part of the sky is or is not contaminated by polarized dust emission! Some people have speculated that this is something to do with the rivalry between Planck and BICEP, which is a bit over-the-top. Instead the reason is more scientific: the mask excludes areas where the error in determining the polarisation fraction is high, or the overall dust signal itself is too small. So the fact that the BICEP patch is in the masked region indicates that the dust emission does not dominate the total emission there, at least at 353 GHz (dust emission increases with frequency). This means there is not a whole lot of dust showing up in the BICEP region — if anything, this is good news! But even this interpretation should be treated with caution: dust doesn't contribute too much to the total intensity in that region, but it may well still contribute a large fraction of whatever B-mode polarization is seen. Based on my understanding and things I have learned from conversations with colleagues, I don't think Planck is going to be sensitive enough to make definitive statements about the dust in that specific region of the sky.

Another interesting paper that has come out since March has been this one, which claims evidence for some contamination in the CMB arising from the "radio loops" of our Galaxy. It also has the great benefit of being an actual scientific paper rather than a rumour on somebody's blog. (Full disclaimer: one of the authors of this paper was my PhD advisor, and another is a friend who was a fellow student when I was at Oxford.)

The radio loops are believed to be due to ejected material from past supernovae explosions; the idea is that if this dust contains ferrimagnetic molecules or iron, it would contribute polarized emission that might be mistaken for true CMB when it is in fact more local. What this paper argues is that does appear to be some evidence that one of the CMB maps produced by the WMAP satellite (which operated before Planck) does show some correlation between map temperature and the position of one of these radio loops ("Loop I"). In particular, synchrotron emission from Loop I appears to be correlated with the temperature in the WMAP Internal Linear Combination (or ILC) map. I'm not going to comment on the strength of the statistical evidence for this claim; doubtless someone more expert than I will thoroughly check the paper before it is published. For the time being let us treat it as proven.

The relevance of this to BICEP is somewhat intricate, and proceeds like this: given our physical understanding of how the radio loops formed, it seems likely that they produce both synchrotron and dust emission which follow the same pattern on the sky. Therefore perhaps the correlation of the synchrotron emission from Loop I with the ILC map is because both are correlated with dust emission from the loop. If the correlation is because of dust emission, this might be polarized because of the postulated ferrimagnetic molecules etc., leading to a correlation between the WMAP polarization and Loop I. And if Loop I is contaminating the WMAP ILC map, it is perhaps plausible that a different radio loop, called the "New Loop", is also contaminating other CMB maps, in particular those of BICEP. Whereas Loop I doesn't get very close to the BICEP region, the New Loop goes right through the centre of it (see the figure below), so it is possible that there is some polarized contamination appearing in the BICEP data because of the New Loop. At any rate, the foreground dust models that BICEP used didn't account for any radio loops, so likely underestimate the true contamination.

 Position of some Galactic radio loops and the BICEP window. "Loop I" is large one in the upper centre, that only skims the BICEP window; the "New Loop" is the one in the lower centre that passes through the centre of it. Figure from Philipp Mertsch.

So far so good, but this is quite a long chain of reasoning and it doesn't prove that it is actually dust contamination that accounts for any part of the BICEP observation. Instead it makes a plausible argument that it might be important; further investigation is required.

At the end of the day then, we are left in pretty much the same position we were in back in March. The BICEP result is exciting, but because it is only at one frequency, it cannot rule out foreground contamination. Other observations at other frequencies are required to confirm whether the signal is indeed cosmological. One scenario is that Planck, operating on the whole sky at many frequencies but with a lower sensitivity than BICEP, confirms a gravitational wave signal, in which case pop the champagne corks and prepare for Stockholm. The other scenario is that Planck can't confirm a detection, but also can't definitively say that BICEP's detection was due to foregrounds (this is still reasonably likely!), in which case we wait for other very sensitive ground-based telescopes pointed at that same region of sky but operating at different frequencies to confirm whether or not dust foregrounds are actually important in that region, and if so, how much they change the inferred value of r.

Until then I would say ignore the rumours.

1. Sesh, the new Planck paper says "The data are not shown in the grey areas where the dust emission is not dominant or where residuals were identified comparing individual surveys." The term residuals refers to uncertainties. Later they say, "we only show the Planck polarization data and derived quantities, where the systematic uncertainties are small, and where the dust signal dominates total emission." So there is no reason for assuming that this is "good news" as you claim. It could be that they are not showing the BICEP2 region because the systematic uncertainties are high.

In fact, far from being "good news" for BICEP2, I think the paper you cite is actually bad news, for reasons I explain in this blog post:

http://futureandcosmos.blogspot.com/2014/05/why-new-planck-paper-casts-grave-doubts.html

The new paper is bad news for BICEP2 because the new, revised polarized fraction graph (with lots of yellow regions now showing as red, with a higher 20 percent polarization fraction) now shows a lot higher polarization fraction than the previous version used by BICEP2.

1. Hello and thanks for the comment. I read your blog post, but I don't really agree with you. Firstly, there's no justification for assuming the polarization fraction is the same, or even roughly the same, across the sky. The same goes for the total intensity of dust emission. Therefore extrapolating from the areas Planck did not mask to the areas they did is likely to give you a number that has nothing much to do with reality. The only numbers that really matter to this discussion are (a) how much dust emission is there in the BICEP window, and (b) what is the polarization fraction of that dust?

Since that window lies in the region masked by Planck, the answer to both of those questions is "we don't know". What we do know is that since Planck chose to mask the region, either (i) there isn't a lot of dust emission in that windowl, even at 353 GHz (and therefore even less at 150 GHz), or (ii) the error in determining the polarization fraction is large, or (iii) both.

It seems likely to me that option (iii) is correct (if (i) is true, (ii) probably follows as well). This is not bad news for BICEP; it may even be good news. The option that would be bad news for BICEP would be if (i) were not true but somehow (ii) were - there is a lot of dust, we're just not sure how polarized it is - I don't think that is likely given how measurements work, but only Planck people would know for sure.

Incidentally, the "revision" to the old polarization fraction map is basically due to subtracting out the (unpolarized) cosmic infrared background (CIB), which isn't to do with dust. This is quite an elementary step, so people think it is unlikely that the BICEP team forgot to do something so straightforward themselves! (For instance, see Peter Coles' blog.)

2. Thanks for the discussion of Liu et al, and especially for the figure overlaying the BICEP2 area. If not Planck, who will clean up the dust, and when?

1. This is a tricky question, the answer to which depends a little on whether there is any primordial signal to be seen, or whether it is all due to dust. So if there is a large primordial $r$ out there (say with a value 0.1 or larger) and Planck eventually manage to sort out all the systematics issues that having been holding back their polarization maps, then it will be Planck that should confirm the BICEP result, within a few months. (They have a couple of conferences scheduled in December to present their results, we should know one way or another by then.)

If Planck either cannot resolve their systematics, or does not have the sensitivity to unambiguously confirm a primordial signal, then we are in a difficult grey zone, because it seems they are also not going to provide a clear statement about dust in the BICEP window. In this case I think the Keck Array data at 100 GHz will be the first information we get - this data is already being collected, so we should not need to wait too long. Having two frequencies is better than one, and will give some indication of possible dust levels, but may still not be conclusive. Eventually I believe Keck will also work at 220 GHz.

There are also several other polarization experiments currently in operation or just about to start. SPTPol, POLARBEAR and ACTPol are sensitive to much smaller, arcminute angular scales than the degree scales (or larger) on which the GW signal shows up; they're much better for looking for B-modes from lensing. I also don't think they look at the same patch of sky as BICEP, so not sure whether they can say anything useful about the dust there. There's the ABS experiment in Chile, which should be starting about now, targeting primordial B-modes, but they're also only one frequency (145 GHz). There's a balloon-borne experiment called SPIDER which should provide multi-frequency data in something like 2-3 years (I believe it has been launched already, though their website isn't clear!), and another one due to launch soon called PIPER.

But for the short term, let's hope Planck provides the answers quickly!

2. By the way, thanks for asking that question: it made me go and look up the answer, and in the process I learned some things I didn't know!

3. Thank you. Re a third frequency, there's an amusing back-and-forth in Flauger's presentation (at 1:08) where he tries and fails to get a date for 220 GHz data from BICEP3 (?) (later that 2015?) from someone in the audience. I think the audience member says "no 220 GHz receivers exist right now"...

format: plain


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# modular arithmetic question

know some rules for modular arithmetic expressions, for example,

1. $A+B=C\implies ((A\bmod M) + (B\bmod M))\bmod M = C\bmod M$.

-


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# PyTorch: optim¶

A fully-connected ReLU network with one hidden layer, trained to predict y from x by minimizing squared Euclidean distance.

This implementation uses the nn package from PyTorch to build the network.

Rather than manually updating the weights of the model as we have been doing, we use the optim package to define an Optimizer that will update the weights for us. The optim package defines many optimization algorithms that are commonly used for deep learning, including SGD+momentum, RMSProp, Adam, etc.

import torch

# N is batch size; D_in is input dimension;
# H is hidden dimension; D_out is output dimension.
N, D_in, H, D_out = 64, 1000, 100, 10

# Create random Tensors to hold inputs and outputs, and wrap them in Variables.
x = Variable(torch.randn(N, D_in))

# Use the nn package to define our model and loss function.
model = torch.nn.Sequential(
torch.nn.Linear(D_in, H),
torch.nn.ReLU(),
torch.nn.Linear(H, D_out),
)
loss_fn = torch.nn.MSELoss(size_average=False)

# Use the optim package to define an Optimizer that will update the weights of
# the model for us. Here we will use Adam; the optim package contains many other
# optimization algoriths. The first argument to the Adam constructor tells the
# optimizer which Variables it should update.
learning_rate = 1e-4
for t in range(500):
# Forward pass: compute predicted y by passing x to the model.
y_pred = model(x)

# Compute and print loss.
loss = loss_fn(y_pred, y)
print(t, loss.data[0])

# Before the backward pass, use the optimizer object to zero all of the
# gradients for the variables it will update (which are the learnable weights
# of the model)


text:
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text:
# Create a “perspective animation”

I would like to create my new experiment/stimuli using Mathematica. The idea is to study how people move their eyes in a dynamic scene.

To do so I need to build some animations (1 minute long to start) where we feel like going forward in a perspective grid like the one below.

I will need to make some object appear/disappear/change within that scene. The grid itself could just be made out of point, and i will add some geometric colored shapes.

While I know how to draw perspective I dont know how i could "program" it and then animate it.

-
Should it be done in 2D (like FJRA´s solution) or is a native 3D scene an option? –  Yves Klett Mar 5 '12 at 7:10
native 3D would be great I think. As long as I could map what is seen in 2D. That is I will correlate the scene properties with the gaze observed when it is displayed on the screen. –  500 Mar 5 '12 at 17:05

It could be something like this?

outersquare = {{-1, -.8}, {1, .8}};

innersquare = outersquare/10. + {-.1, 0};

corners[sq_] := {sq[[1]], {sq[[1, 1]], sq[[2, 2]]},
sq[[2]], {sq[[2, 1]], sq[[1, 2]]}};

lines[sq_] := Partition[corners[sq], 2, 1, 1];

pointOfSquare[sq_, side_, i_, n_] :=
With[{line = lines[sq][[side]]},
line[[1]] + (n - i)/n (line[[2]] - line[[1]])];

alllines :=
With[{n = 20},
Flatten@Table[
Line[{pointOfSquare[innersquare, side, i, n],
pointOfSquare[outersquare, side, i, n]}], {side, 4}, {i, 0,
n}]];

rectangle[i_, n_] :=
With[{factor = Log[i]/Log[n]},
Rectangle @@ (factor innersquare + (1 - factor) outersquare)];

Animate[Graphics[{Red, alllines, EdgeForm[Red], FaceForm[Transparent],
Table[rectangle[i, 10], {i, 1 + ministep, 10, 1}]}], {ministep,
.99, 0, -.01}]

alllines2[shift_] :=
With[{n = 20},
Flatten@Table[
Line[{pointOfSquare[innersquare, side, i, n],
pointOfSquare[outersquare, side, i, n]}], {side, 4}, {i, shift,
n, 1}]];

Animate[Graphics[{Red, alllines2[ministep], EdgeForm[Red],
FaceForm[Transparent],
Table[rectangle[i, 10], {i, 1 + ministep, 10,
1}]}], {ministep, .99, 0, -.01}]

EDIT2: Adding moving points (if you want to take out the red lines, get rid of alllines).

pointInLine[linepoints_, i_, shift_, n_] :=
If[i + shift < n,
With[{factor = Log[i + shift]/Log[n]},
Point[linepoints[[1,
1]] + (1 - factor) (linepoints[[1, 2]] -
linepoints[[1, 1]])]], {}]

allPoints[linesperside_, shift_, n_] :=
With[{lines =
Flatten@Table[
Line[{pointOfSquare[innersquare, side, i, linesperside],
pointOfSquare[outersquare, side, i, linesperside]}], {side,
4}, {i, 0, linesperside}]},
Flatten[
Table[pointInLine[line, i, shift, n], {line, lines}, {i, n}]]];

Animate[Graphics[{Red, alllines, Blue, PointSize[Medium],
allPoints[4, ministep, 10], EdgeForm[Red], FaceForm[Transparent],
Table[rectangle[i, 10], {i, 1 + ministep, 10,
1}]}], {ministep, .99, 0, -.01}]

In conclusion, I use as "move" factor Log[i+shift]/Log[n], so for each frame, the change from the previous one is that shift` value (between 0 and 1).

-
very cool, could the line that converge in the bcd be moving points as well ? That is instead of lines to have points close to one another ? –  500 Mar 3 '12 at 15:28
I edited and added a moving background... that's what you wanted? –  FJRA Mar 3 '12 at 15:36
wow, what a weird feeling, here is a ppt mockup of what i had uin mind, the points coming toward you, not rotating. Thank You very much for your time and attention. : dl.dropbox.com/u/20775208/Sparrow/… –  500 Mar 3 '12 at 17:58
Thank you once agin ! –  500 Mar 4 '12 at 4:09
Cool! I really missed the exhaust port from the end. –  István Zachar Mar 4 '12 at 20:14


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# 17-29-076

floor cavity ratio

Kf; K

quotient of five times the sum of the area of the walls bounded by the floor and the working plane and the sum of the areas of the floor and the working plane, used in the calculation of the utilization factor or utilance

Note 1 to entry: The floor cavity ratio is given by the formula

where a and b are the dimensions of the sides of the room and hf is the distance from the floor to the utilization plane.

Note 2 to entry: See also "ceiling cavity ratio" and "room cavity ratio".

Note 3 to entry: The floor cavity ratio has unit one.

Note 4 to entry: This entry was numbered 17-452 in CIE S 017:2011.

Publication date: 2020-12


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Oscillator types

Status
Not open for further replies.

ElectronicTy

New Member
if the required frequency of the local oscillator is 150MHz, what type of oscillator would you recommend?

JimB

Super Moderator
A bit short on detail.

Is this a local oscillator in a superhet receiver?

Tunable?
If so, what frequency range?

Fixed frequency?
Crystal control or synthesiser?

Is this college homework?

JimB

Status
Not open for further replies.


text:
# Worksheet on Sets in Roster Form

Practice the worksheet on sets in roster form to write a set using the roster or tabular method.  We know, to express the set in roster form the elements of a set are listed within the curly brackets and are separated by commas.

Now, let's practice the questions given in the worksheet on sets in Roster form.

1. Write each of the following sets in the Roster form:

(i) The set of first seven natural numbers.

(ii) The set of whole numbers less than 5.

(iii) The set of five numbers each of which is divisible by 3.

(iv) The set of whole numbers less than 20 and divisible by 3.

(v) The set of integers greater than -2 and less than 4.

(vi) The set of integers between -4 and 4.

(vii) The set of letters in the word ‘mathematics’.

(viii) The set of consonants in the word ‘possession’.

(ix) The set of first three letters in the word ‘booklet’.

(x) {x : x is a letter in the word ‘SCHOOL’}.

(xi) {x : x is an odd natural number between 10 and 20}.

(xii) {x | x = 1 – p, p ∈ M}.

(xiii) {x | x = p$$^{2}$$, p ∈ W}.

(xiv) {Vowels used in the word ‘AMERICA’}

(xv) {Consonants used in the word ‘MADRAS’}.

(xvi) Set of prime factors of 36.

2. Express each given set in the Roster Form:

(i) All prime numbers between one and twenty.

(ii) The squares of first four natural numbers.

(iii) Even numbers between 1 and 9.

(iv) First eight letters of the English alphabets.

(v) The letters of the word ‘BASKET’.

(vi) Names of months in a year.

(vii) Four cities of America whose names start with the letter J.

(viii) Any four closed geometrical figures names.

(ix) Vowels used in the word ‘MONDAY’

(x) Single digit numbers which are perfect squares also.

3. Write in Roster Form the set of:

(i) The first four odd natural numbers each divisible by 3.

(ii) The first four odd natural numbers each divisible by 5.

(iii) The counting numbers between 15 and 35, each of which is divisible by 6.

(iv) The names of the last three days of a week.

(v) The names of the last four months of a year.

Answers for the worksheet on sets in Roster form are given below to check the exact answers of the above questions on sets in Roster form or Tabular form.

1. (i) {1, 2, 3, 4, 5, 6, 7}

(ii) {0, 1, 2, 3, 4}

(iii) {3, 6, 9, 12, 15}

(iv) {3, 6, 9, 12, 15, 18}

(v) {-1, 0, 1, 2, 3}

(vi) {-3, -2, -1, 0, 1, 2, 3}

(vii) {m, a, t, h, e, i, c, s}

(viii) {p, s, n}

(ix) {b, o, k}

(x) {s, c, h, o, l}

(xi) {11, 13, 15, 17, 19}.

(xii) {0, -1, -2, -3, ….}

(xiii) {0, 1, 4, 9, 16, ….}

(xiv) {a, e, i}

(xv) {m, d, r, s}

(xvi) {2, 3}

2. (i) {2, 3, 5, 7, 11, 13, 17, 19}

(ii) {1$$^{2}$$, 2$$^{2}$$, 3$$^{2}$$, 4$$^{2}$$} = {1, 4, 9, 16}

(iii) {2, 4, 6, 8}

(iv) {a, b, c, d, e, f, g, h}

(v) {b, a, s, k, e, t}

(vi) {January, February, March, April, May, June, July, August, September, October, November, December}

(vii) {Janesville, Jasper, Jeannette, Johnston}

(viii) {triangle, circle, square, pentagon}

(ix) {o, a}

(x) {0, 1, 4, 9}

3. (i) {3, 9, 15, 21}

(ii) {5, 15, 25, 35}

(iii) {18, 24, 3}

(iv) {Friday, Saturday, Sunday}

(v) {September, October, November, December}

Sets and Venn-diagrams Worksheets

Worksheet on Set

Worksheet on Elements Form a Set

Worksheet to Find the Elements of Sets

Worksheet on Properties of a Set

Worksheet on Sets in Roster Form

Worksheet on Sets in Set-builder Form

Worksheet on Finite and Infinite Sets

Worksheet on Equal Sets and Equivalent Sets

Worksheet on Empty Sets

Worksheet on Subsets

Worksheet on Union and Intersection of Sets

Worksheet on Disjoint Sets and Overlapping Sets

Worksheet on Difference of Two Sets

Worksheet on Operation on Sets

Worksheet on Cardinal Number of a Set

Worksheet on Venn Diagrams


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I have noticed that there are quite a few links to Blender documentation that have gone down since the Manual URL was changed recently.

Since this change, it seems that some of the older documentation links are not redirecting, and instead give a 404 error like below.

Some examples:

What is the difference between a UV Sphere and an Icosphere?

Creating a Black Hole in Blender

Make a gold reflective object

How to change to spherical Axis

While I have read through this question and agree with the points made, since these links are the documentation for Blender, it seems to me that it would make more sense to fix them if possible rather than to just remove them.

My main question is, is there a way to search the site specifically for the wiki links/ older documentation links? Then one could update as appropriate if so inclined to ensure that links go to the correct place, and that there is current and correct information. Or would it be better to just update them as they are found?

• For old wiki links, you can view their original content by inserting?title= into the URL after index.php and before Doc:2.6. For example, https://wiki.blender.org/index.php/Doc:2.6/Manual becomes https://wiki.blender.org/index.php?title=Doc:2.6/Manual. Can be useful for determining which manual page would be most appropriate. – gandalf3 Mar 18 '17 at 2:03
• That's a useful tip, thanks. I always read through the URL and do my best to judge which page is the correct one based on the content for the ones I have changed. Sometimes, not so easy though to find the right page though. – Timaroberts Mar 18 '17 at 2:56


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Thread: Factors and Simplest Form: Concept Check

1. Factors and Simplest Form: Concept Check

From the book
"Why is the following way to simplify the fraction 17/75 incorrect?"

$\frac{1\not7}{\not75}=\frac{1}{5}$

This is an incorrect way to simplify the fraction because 17/75 cannot be simplified any further by dividing out digits within whole numbers in the numerator and denominator.

Do you think this clear and correct?

2. Originally Posted by Euclid Alexandria
From the book
"Why is the following way to simplify the fraction 17/75 incorrect?"

$\frac{1\not7}{\not75}=\frac{1}{5}$

This is an incorrect way to simplify the fraction because 17/75 cannot be simplified any further by dividing out digits within whole numbers in the numerator and denominator.

Do you think this clear and correct?
Perhaps something like:

When cancelling you are removing common factors of the numerator and
the denominator. In this case the 7's are not factors of the 17 or 75 but
are digits in their decimal representation. As 7 is not a common factor of
17 and 75 the 7's cannot be cancelled

RonL.

3. Hello, Euclid Alexandria!

Why is the following way to simplify the fraction $\frac{17}{75}$ incorrect?

$\frac{1\!\!\!\not7} {\not75}=\frac{1}{5}$

But it works for these . . . LOL!

. . $\frac{16}{64}\;=\;\frac{1\!\!\!\not6}{\not64}\;=\; \frac{1}{4}$

. . $\frac{19}{95}\;=\;\frac{1\!\!\!\not9}{\not95} \;=\;\frac{1}{5}$

. . $\frac{26}{65}\;=\;\frac{2\!\!\!\not6}{\not65} \;=\;\frac{2}{5}$

. . $\frac{49}{98}\;=\;\frac{4\!\!\!\not9}{\not98} \;=\;\frac{4}{8}\:=\:\frac{1}{2}$

4. Thanks Captain, that explanation is right on the mark (well, except for the first sentence, if you consider Soroban's reply).

5. Originally Posted by Euclid Alexandria
From the book
"Why is the following way to simplify the fraction 17/75 incorrect?"

$\frac{1\not7}{\not75}=\frac{1}{5}$


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However, I have read this amazing blog about binning data this whereby the author claims that adaptive binning is better than fixed-width binning. I understand the idea behind this because the some of the bins that we define in the fixed-width approach may have too little data distribution in comparison to other bins, which won't be a fair game to play whereby the adaptive approach, motivated by the idea of quantiles is better. Are there any more arguments, or more in depth analysis, of this hypothesis ?
I took a look at that link. It's pretty informative. As you already know, The cut function is used to specifically define the bin edges. There is no guarantee about the distribution of items in each bin. In fact, you can define bins in such a way that no items are included in a bin or nearly all items are in a single bin. The qcut function is slightly different. The qcut function tries to divide up the underlying data into equal sized bins. The function defines the bins using percentiles based on the distribution of the data, not the actual numeric edges of the bins. In conclusion, if you want equal distribution of the items in your bins, use qcut . If you want to define your own numeric bin ranges, then use cut.


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shannon wiener index calculator

2021年1月17日

(May 1975)[2]. Charles. The calculator uses the following formula to calculate the Shannon-Wiener diversity index: H = - S p i ln ( p i) . This calculator uses The Shannon-Wiener species diversity index per sample plot was calculated. In the Shannon index, p is the proportion (n/N) of individuals of one particular species found (n) divided by the total number of individuals found (N), ln is the natural log, Σ is the sum of the calculations, and s is the number of species. This tutorial explains how to calculate the Shannon Wiener diversity index and Evenness. Simpsons diversity index. Simpson's reciprocal index - abbreviated 1/D in the literature. Diversity calculator online (bpmsg). In previous descriptions, the Shannon Wiener index was mostly calculated using multiple degrees of data, and in some papers coverage was used. Using a joint dataset retrieved from Foursquare API, … Ecology how can i calculate shannon-wiener diversity index in. Photo and Microscopy for VIRTUE-s projects, Understanding what we see. How to calculate shannon index for plankton diversity? Shannon wiener species diversity index calculator â. Diversity index wikipedia. In ecological studies, this order is characterized by the number of individuals observed for each species in the sample plot (e.g., biofilm on a plexiglass disc). , 2007; Pandey and Kulkarni, 2006; Price, 1975), where it is known as Shannon’s diversity index, the Shannon -Wiener index, the Shannon- Weaver index and the Shannon entropy It measures the rarity and commonness of species in a community. Here p i is the proportion of total number of species made up of the ith species. Moreover, inverse Simpson is asymptotically equal to rarefied species richness in sample of two individuals, and Fisher's $\alpha$ is … the diversity of landscape factors. 150  0    2 I'm trying to find out how to perform Hutcheson's t-test for significance on Shannon-Wiener Indices. It measures the rarity and commonness of species in a community. Reply. Reply. where c = INT(n/k) and d = MOD(n, k).. 1    15   1 Biodiversität) eingesetzt wird.Er beschreibt die Vielfalt in betrachteten Daten und berücksichtigt dabei sowohl die Anzahl unterschiedlicher Datenkategorien (z. If abundance is primarily concentrated into one species, the index will be close to zero. sum = (60+10+25+1+4) = 100 Shannon index - abbreviated H in the literature. Additionally it also provides Species Richness and Evenness in a single button click. First, let us calculate the sum of the given values. The Shannon equitability index is simply the Shannon diversity index divided by the maximum diversity $$E_{H} = \frac{H}{\log(k)}$$ This normalizes the Shannon diversity index to a value between 0 and 1. the following definition for Shannon's entropy: Here pi is the probability for the i-th event. Shannon index calculation (in google sheets or excel) youtube. where n i is the number of observations from the sample in the i th of k (non-empty) categories and n = is the sample size. 35. Let’s compute the Shannon-Weiner diversity index for the same hypothetical community in the previous example. 11   1    1 Shannon-Weiner index (H) - This diversity measure came from information theory and measures the order (or disorder) observed within a particular system. The calculation is performed using a natural logarithm. Also known as the Shannon-Wiener or Shannon-Weaver index. The calculation is performed using a natural logarithm. Temperature impact on the shannon-wiener biodiversity index (bdi. Shannon-Wiener Diversity Index. Shannon-Wiener Diversity Index. The calculation is performed using the natural logarithm. The user only have to enter the number of individuals recorded for each species present. Here each column is one community, each row is one species, and the numbers are either probabilities or numbers found for each species. The Shannon index has been a popular diversity index in the ecological literature (Tandon et al. Also known as the Shannon-Wiener or Shannon-Weaver index. ii) Shannon maximum . Please let us know your suggestions and comments. Shannon-Wiener Index (H’) Most commonly used index of diversity in ecological studies; Values range from 0 to 5, usually ranging from 1.5 to 3.5 ; Calculated: Where: n i = number of individuals or amount (e.g., biomass or density) of each species (the i th species) N = total number of individuals (or amount) for the site, and ln = the natural log of the number. After testing for normalization using the Kolmogorov–Smironov test, the Shannon - Wiener index was used as the dependent variable, and spectral values from original and synthetic bands from different processing on the ETM + data were used as the independent variables. Laformula dell'indice di Shannon-Wienerè la seguente: In genere è indicata con la lettera H o H'. First, enter the number of species, and then enter the name you wish to give the species, if available, and the given populations for each of the species—in any given order. Thanks for using our software! About image and VIRTUE, Short video lessons on biofilms and biodiversity. Sea water in constant motion. The calculation is performed using a natural logarithm. How to calculate a biodiversity index | amnh. In plainer (but longer) language, here’s the Shannon-Wiener index: To practice, calculate the Shannon-Wiener diversity index (H) for one of the five samples from Activity 1 (Sample _____). The calculator uses the following formula to calculate the Shannon-Wiener diversity index: Here pi is the proportion of total number of species made up of the ith species. Using cover data for vegetation. It measures the rarity and commonness of species in a community. … This calculator is free to use and is designed for biologists, ecologists, teachers, and students needing to quickly calculate the biodiversity indexes of an ecosystem. The samples of 5 species are 60,10,25,1,4. First, enter the number of species, and then enter the name you wish to give the species, if available, and the given populations for each of the species—in any given order. Shannon wiener diversity index calculation | birds | biodiversity. To understand the basic concept of diversity, you might watch my video here; it explains how diversity can be characterized using diversity indices – like the Simpson index– taking into account richness and evenness. Sugar Maple. Simpson Diversity Index (D): D = S pi2. Shannon Entropy (Information Content) Calculator. The BPMSG diversity online calculater allows you to calculate diversity indices from your sample input data. Diversity indices like the Shannon entropy ("Shannon-Wiener index") and the Gini-Simpson index are not themselves diversities.They are just indices of diversity, in the same way that the diameter of a sphere is an index of its volume but is not itself the volume. View Shannon-Wiener Excel sheet_AMK.pdf from BIOL 220W at Pennsylvania State University. The Shannon index has been a popular diversity index in the ecological literature, where it is also known as Shannon's diversity index, the Shannon–Wiener index, the Shannon–Weaver index and the Shannon entropy. Note that lower values indicate more diversity while … Stack Exchange Network. These notes will show you how to conduct the Hutcheson t-test and so get a statistical significance of the difference in Shannon diversity between two samples. This index is borrowed from information science, and is calculated as follows: € H'=−p i lnp i i=1 S ∑ Where p i is the relative abundance of species i, S is the total number of species present and ln is the natural log. Simpson's reciprocal index - abbreviated 1/D in the literature. Evenness (E): E = H / ln( S ) . In particular, the exponent of the Shannon index is linearly related to inverse Simpson (Hill 1973) although the former may be more sensitive to rare species. The more unequal the abundance of species, the larger the weighted geometric mean of the pi values, the smaller the index. Species richness - abbreviated S in the literature. Charles says: March 13, 2018 at 9:59 am Mary, Perhaps I don’t understand your question, but Shannon’s index doesn’t assume that numbers need to be equal. Shannon-Wiener Diversity Index is a term in biology/ecology. A diversity index (also called phylogenetic or Simpson's Diversity Index) is a quantitative measure that reflects how many different types (such as species) there are in a dataset (a community) and that can simultaneously take into account the phylogenetic relations among the individuals distributed among those types, such as richness, divergence or evenness. Shannon index - abbreviated H in the literature. Simpson's reciprocal index - abbreviated 1/D in the literature. Formula: H = -SUM[(pi) * ln(pi)] E=H/H max Where, SUM = Summation pi= Numbe of individuals of species i/total number of samples S = Number of species or species richness H max = Maximum diversity possible E= Eveness=H/H max In general the concept of diversity can be formulated using the power mean. SDR-IV Help Measuring and understanding biodiversity by R M H Seaby and P A Henderson SDR is designed to help you measure and understand the variation in … Shannon index - abbreviated H in the literature. An equivalent formula is. Biodiversity calculator for the simpson and shannon indexes. Shannon Wiener Species Diversity Index Formula - Data Analysis. The maximum value of H is. Shannon-Wiener Index (H’) Most commonly used index of diversity in ecological studies; Values range from 0 to 5, usually ranging from 1.5 to 3.5; Calculated: Where: n i = number of individuals or amount (e.g., biomass or density) of each species (the i th species) How many categories? I have 11 locations (paired fields) for which I already defined (with QGIS) and calculated the proportions of 5 landscape factors (crops, grassland, semi-natural, forest and others) within a … Effective number of species . Bpmsg diversity online calculator – bpmsg. Shannon–Wiener index of diversity (information index) A measure used by ecologists when a system contains too many individuals for each to be identified and examined.A small sample is used; the index (D) is the ratio of the number of species to their importance values (e.g. Simpson's index of diversity - abbreviated 1-D in the literature. Effects of currents, tides and waves. The Simpson index is based on the arithmetic mean, in the general concept of diversity it corresponds to a “true” diversity of order two. use the Shannon-Wiener diversity index, or H'. The key is the formula that determines the variance of the Shannon index. The Shannon-Wiener Diversity Index, often symbolized by H' (or, H-prime), is a measure of species diversity that takes into consideration not only the number of species present, but includes their relative abundance in the population. H = -SUM [ (pi) * ln (pi)] E=H/H max Where, SUM = Summation pi= Numbe of individuals of species i/total number of samples S = Number of species or species richness H max = Maximum diversity possible E= Evenness=H/H max. The user only have to enter the number of individuals recorded for each species present. This calculator is free to use and is designed for biologists, ecologists, teachers, and students needing to quickly calculate the biodiversity indexes of an ecosystem. For a population (or a non-random sample), we can use Brillouin’s index of diversity, instead of Shannon’s index.Brillouin’s index is defined as. Calculate the Shannon diversity index and Evenness for these sample values. Species richness - abbreviated S in the literature. 香農多樣性指數（Shannon's diversity index）用來估算群落多樣性的高低，亦稱為(Shannon-Weiner diversity index)。公式如下： ′ = − ∑ = 其中S表示總的物種數，pi表示第i個種占總數的比例(Pielou 1975)。 Shannon-Wiener diversity index is commonly used in measuring the community species diversity or assessing the ecological status of marine environments. Mannigfaltigkeit der Arten wird häufig der Shannon-Wiener-Index Hs verwendet, welcher oft fälschlicherweise als Shannon-Weaver-Index bezeichnet wird, da Shannon seinen Beitrag zusammen mit einem weiteren Artikel von Weaver in Buchform publizierte. No. In the Shannon-Wiener diversity index, you have to add one pi lnpi term for EACH species that you counted. Simpson's reciprocal index - abbreviated 1/D in the literature. Simpson's index of diversity - abbreviated 1-D in the literature. Select the number of categories or classes (between 2 and 20) and input your sample data (e.g. There are two methods to compute the Shannon-Wiener diversity index in the surveyed region, which is calculated based on each sample data (Method 1) or based on the pooled data of the survey region (Method 2). What I can't understand, is the math behind the index; for example: why is it multiplying n by n-1? Shannon wiener index (Shannon-Weiner index) is used to describe the disorder and uncertainty of individual species. Biodiversity Calculations made simple! Where pi = the proportion of individuals in the species. Simpson's index of diversity - abbreviated 1-D in the literature. For the environment (B) demonstrated in table 4 the diversity index calculated by Shannon-Wiener diversity index model w as 1.6094 with evenness of 1. Simpson's index of diversity - abbreviated 1-D in the literature. Shannon index. Calculating racial/ethnic diversity using the shannon-wiener index. In particular, for a random sample, we can use Shannon’s index of diversity (aka as Shannon-Weiner’s index), which is defined as. Also known as the Shannon-Wiener or Shannon-Weaver index. Identification plates by feeding strategies, Identification plates by taxonomic groupings, Counting organisms in photographs with ImageJ (Fiji), Water sampling and handling of water samples. The free online Shannon's diversity index calculator. ある．前者ではShannon-Wiener 指数，後者ではSimpson の多様度指数がよく用いられる．この2つは個体数に一定 の分布型を仮定しないノンパラメトリックな指数である． Shannon-Wiener 指数H’は， H' pilnpi i 1 s で表される．Sは群集内の種数，pi は群集内の全種の個体 Der Shannon-Index (häufig auch Shannon-Wiener-oder Shannon-Weaver-Index) ist eine mathematische Größe, die in der Biometrie für die Beschreibung der Diversität (vgl. Shannon-Wiener Diversity Index is a term in biology/ecology. where n i is the number of observations from the sample in the i th of k (non-empty) categories and n is the sample size. Rejestracja i składanie ofert jest darmowe. Shannon-Wiener Diversity Index is a term in biology/ecology. I would like to calculate Shannon's diversity index for habitat diversity, i.e. Read more on aquatic biodiversity and calculations and download instructional documents. Here p i is the proportion of total number of species made up of the ith species. Also known as the Shannon-Wiener or Shannon-Weaver index. The calculator uses the following formula to calculate the Shannon-Wiener diversity index: H = - S p i ln ( p i) . Species. shannon wiener diversity index explanation, From what I have read, the Index calculates the probability that any 2 randomly selected organisms belong to the same species, and it is thought that the closer the result is to zero, the more diverse the region is. This app is designed to simplify the calculation process of species diversity to get Shannon-Wiener Diversity Index. Example $$\PageIndex{3}$$:Calculating Shannon-Weiner Index. This app is designed to simplify the calculation process of species diversity to get Shannon-Wiener Diversity Index. The Shannon-Wiener diversity index is a heterogeneity measure that incorporates species richness and evenness (Hollenbeck and Ripple, 2007). The higher the uncertainty, the higher the diversity. The Simpson index is a dominance index because it gives more weight to common or dominant species. Relative to other diversity indices, such as Simpson’s index, it is considered sensitive to the addition of rare species (Krebs, 1989). The Shannon-Weiner index is most sensitive to the number of species in a sample, so it is usually considered to be biased toward measuring species richness. When all species in the data set are equally common, all pi values = 1/R and the Shannon-Weiner index equals ln(R). 2    0    33. 33   44   11 Szukaj projektów powiązanych z Shannon diversity index calculator lub zatrudnij na największym na świecie rynku freelancingu z ponad 19 milionami projektów. What biodiversity measure(s) would you like to have computed? The value of the Shannon-Wiener Index usually lays between 1.5 and 3.5 for ecological data and rarely exceeds 4.0. shannon weiner index for meadowa Posted on 2013-10-21 2017-09-04 Author bpmsg Categories Online Tools , Other Articles Tags Berger Parker , calculation , Excel , Shannon Entropy , Shannon Weaver Index , Shannon-Wiener Index , Simpson Index Shannon-Wiener Index This diversity measure is based on information theory; simply, the measure of order (or disorder) within a particular system. Als ein Maß für die Biodiversität, bzw. Richness (S): Total number of species in the community Additionally it also provides Species Richness and Evenness in a single button click. of individuals. Copyright © 2020 University of Gothenburg, Reports and project work about / with VIRTUE, Cooperation partners and financial support. There is also a spreadsheet calculator, that you can download and use for … Species diversity index to calculate, one of: Shannon index - abbreviated H in the literature. When calculating Shannon’s index, how do you standardize by the number of quadrats sampled when unequal sample unit size occurs among sites? Shannon's entropy or information content is an important concept that bridges physical entropy and information theory.


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# Lecture 7: Step functions and exponential modalities

Teacher: Paul-André Melliès

## Step functions

$D_{A ⊸ B} = \underbrace{D_A ⊸ D_B}_{\text{domain of linear functions}}$

NB: this theory is heavily inspired by functional analysis, and in particular step functions.

$A ⊸ B = (A ⊗ B^⊥)^⊥ = A^⊥ ⅋ B$

Every pair $(a,b) ∈ \vert A \vert × \vert B \vert$ defines a linear function:

$δ_{(a,b)} = u ⟼ \begin{cases} \lbrace b \rbrace &&\text{ if } a ∈ u \\ ∅ &&\text{ else} \end{cases}$ $\Tr δ_{(a,b)} = \lbrace (\lbrace a \rbrace, b)\rbrace$

Think of $δ_{(a,b)}$ as a “step function”.

Analogously, for $u ∈ !A, \, b ∈ B$:

$δ_{(u,b)} = v ⟼ \begin{cases} \lbrace b \rbrace &&\text{ if } v ⊆ u \\ ∅ &&\text{ else} \end{cases}$

Exercise: Show that $δ_{(u,b)}: D_A ⟶ D_B$ is a stable function.

$\Tr δ_{(u,b)} = \lbrace (u, b)\rbrace$

Fact: $f$ is the union of $δ_{(u,b)}$ for $(u,b) ∈ \Tr f$ whenever $f: D_A ⟶ D_B$ is a stable function:

$f: x ⟼ \bigcup \lbrace δ_{(u,b)} \; \mid \; (u,b) ∈ \Tr f\rbrace (x) \\ = \bigcup \lbrace δ_{(u,b)}(x) \; \mid \; (u,b) ∈ \Tr f\rbrace\\ = \lbrace b \; \mid \; ∃ u ∈ !A, u ⊆ x; (u,b) ∈ \Tr f\rbrace$

In the case of linear functions $D_A ⟶ D_B$: when can we define the union of $δ_{(a,b)}$ and $δ_{(a’,b’)}$? You want to map $\lbrace a, a’\rbrace$ to $\lbrace b, b’\rbrace$, if $a \sim_A b$ and $b \sim_B b’$: this precisely defines $A ⊸ B$!

## Exponential modalities

Linear decomposition of the intuitionistic implication:

$A ⇒ B \; = \; !A ⊸ B$ $u \sim_{!A} v ⟺ \text{the union } u ∪ v \text{ is a finite clique }\\ ⟺ u ↑ v \text{ (compatible)}$ $?A = (!A^⊥)^⊥$

If $f: D_A ⟶ Σ = D_1 = D_⊥$ is a stable function determines and is determined by a set of pairwise incompatible elements of $!A$

So a stable function $D_A ⟶ Σ = D_1 = D_⊥$ is the same thing as an anticlique of $!A$, hence the same as a clique of $!A ⊸ ⊥$

More generally, a stable function $D_A ⟶ D_B$ is the same thing as a clique of $A ⊸ B$ (every stable function is the union of step functions $δ_{(u,b)}$ compatible with one another!)

$!(A \& B) \; ≅ \; !A ⊗ !B \qquad !⊤ = 1$

Categorical properties of the exponential:

$Δ_A : A ⊸ A \& A\\ Δ_A = \lbrace (a, \texttt{inl } a) \; \mid \; a ∈ \vert A \vert\rbrace ∪ \lbrace (a, \texttt{inr } a) \; \mid \; a ∈ \vert A \vert\rbrace$ $\underbrace{!A ⟶ !(A \& A) \overset{≅}{⟶} !A ⊗ !A}_{\text{comonoid structure → contraction}}$

Using the $!$, we can turn the cartesian product (which is not closed!) into some kind of diagonal for the tensor

### Monoids (category theory)

A monoid in $Set$:

$M × M \overset{m}{⟶} M$

that is associative (the evident rectangle commutes)

Tags:

Updated:


text:
# Contour Plot not taking previously defined expressions?

I am new to Mathematica so my question might sound a bit silly, but I hope you help me.

While learning ContourPlot, I have learned that when I use the previously defined expressions as inputs onto it, it doesn't seem to work, and only putting them in manually seems to work.

For example, defining

f1 = x^2/9 + y^2/4 == 1
f2 = x^2 - 1 == y

and then evaluating

ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]

results in empty plot, while putting the expressions manually inside such as

ContourPlot[{x^2/9 + y^2/4 == 1, x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]

results in what I want...

What's more is, this also seems to occur when using NSolve... Can anyone tell me what's going on?

• I think it has something to do with the HoldAll attribute of ContourPlot... It works with ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]. – MelaGo Oct 15 '19 at 5:44
• Thank you! Could you specify what you mean by that? – Danny Han Oct 15 '19 at 7:57
• @DannyHan As @MelaGo noted, ContourPlot has attribute HoldAll. Before evaluating anything (in particular, before inserting the definitions of f1 and f2), ContourPlot will look at the first argument and decide what type of plot you want - whether you're giving it a single curve or a list and whether you are giving it a function or an equation. (continued...) – Lukas Lang Oct 15 '19 at 8:25
• (...continued) Here, it sees {f1, f2} and decides to plot two functions. When it starts evaluating f1 and f2 for given x and y, it doesn't get a number, but True and False, so it doesn't plot anything (non-numerical results are simply ignored by all/most plotting functions). Evaluate forces evaluation to occur before ContourPlot, so ContourPlot sees a list of equations, which it can then correctly plot. – Lukas Lang Oct 15 '19 at 8:25
• @LukasLang Great explanation! Worth posting as an answer. – Alexey Popkov Oct 15 '19 at 10:24

As suggested, I'm expanding my comment into an answer

The problem here is the HoldAll attribute of ContourPlot. Like Plot and similar functions the process goes something like this:

• Look at the first argument, and decide what form it has:
• If it's a list, the user wants to plot multiple functions
• If it's an equation with ==, the user wants to plot the solution to that equation
• Start the evaluation at different points (this part is done recursively on many points)
• Set the values of the variables (second and third argument) to the correct values (similar to Block, as noted in the details section of ContourPlot)
• Evaluate the first argument and use the result

Now we see what the problem in your case is:

ContourPlot[{f1, f2}, {x, -3, 3}, {y, -3, 8}]

When ContourPlot decides on the type of plot you want, it decides on "plot contours for two functions", since all it sees at that point is {f1, f2}. Now, values like e.g. x=0,y=0 are assigned and f1,f2 are evaluated. The problem is that this results in e.g.

          {f1, f2}
(* --> *) {x^2/9 + y^2/4 == 1,x^2 - 1 == y}
(* --> *) {0^2/9 + 0^2/4 == 1,0^2 - 1 == 0}
(* --> *) {0 == 1,- 1 == 0}
(* --> *) {False, False}

And like most plotting functions, non-numeric results (like the False above, are simply discarded).

It is now also clear why Evaluate fixes the issue: It forces the first argument of ContourPlot to be evaluated before ContourPlot has a chance to look at it. So now the evaluation sequence goes like this:

          ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]
(* --> *) ContourPlot[{x^2/9 + y^2/4 == 1,x^2 - 1 == y}, {x, -3, 3}, {y, -3, 8}]

At this point ContourPlot examines the first argument and decides on "plot the solutions of two equations", which is what we want.

It should be noted that the only thing that ContourPlot needs to see are the lists (when multiple things are to be plotted) and the equations - everything else can be evaluated later. This means the following will also work:

g1 = x^2/9 + y^2/4
g2 = x^2 - 1

ContourPlot[{g1 == 1, g2 == y}, {x, -3, 3}, {y, -3, 8}]

## TL;DR;

To summarise, force evaluation of the first argument of ContourPlot using Evaluate to ensure that the right type of plot is chosen:

ContourPlot[Evaluate[{f1, f2}], {x, -3, 3}, {y, -3, 8}]

• OMG thank you so much! Now I understand :) – Danny Han Oct 15 '19 at 12:29


text:
+0

# triangle similarity

0
44
1

Ramona wants to find the height of the giraffe statue outside the library.

Mar 18, 2021

#1
+141
+1

i believe the two triangles are similar, because of the mirror.

so we can get this ratio. let's have g represent the giraffe's height:

$$\frac{15}{4} = \frac{g}{5}$$

$$75 = 4g$$ (cross multiply)

$$g = 18.75$$

18.75 ~ 19, so our answer is 19 feet.

hope this helped! please let me know if you have any questions :)

Mar 18, 2021

#1
+141
+1

i believe the two triangles are similar, because of the mirror.

so we can get this ratio. let's have g represent the giraffe's height:

$$\frac{15}{4} = \frac{g}{5}$$

$$75 = 4g$$ (cross multiply)

$$g = 18.75$$

18.75 ~ 19, so our answer is 19 feet.

hope this helped! please let me know if you have any questions :)

idyllic Mar 18, 2021


text:
# 4 ways to solve ±1 RMQ

You can test @ BNUOJ 4363 RE&MLE Quiz

Source code can be found @ Wiki

Test package(include data maker and source code) can be found @ Box

RMQ(Range Minimum/Maximum Query) is a kind of problem that asks you the minimum(for convenience) value in given intervals of a given array.

±1 RMQ is a special kind of RMQ that the difference between adjacent values is exactly 1.

For normal RMQ, we have ST(sparse table) algorithm.

Let f[i][j] be the minimum value of interval $$[i,i + {2^{j}})$$, which is

$f[i][j] = \min \{ {A_k}|i \le k < i + {2^j}\}$

so we have

$f[i][j] = \left\{ \begin{array}{l}{A_i},j = 0\\\min \{ f[i][j - 1],f[i + {2^{j - 1}}][j - 1]\} ,j > 0\end{array} \right.$

Hera we say

$f = RMQ(A)$

For each query [x, y], the answer is

$\min \{ f[x][k],f[y - {2^k} + 1][k]\}$ where $k = \min \{ j|{2^{j + 1}} \ge y - x + 1\}$

Apply to the problem I mentioned above, where there are at most N(1<=N<=1,000,000) numbers in the array and M(1<=M<=1,000,000) queries in total, we’ll need  f[MAXN][MAXE] where

$MAXN = N = 1,000,000$

$MAXE = {\log _2}^N \approx 20$

and another copy for maximum values which result in about 160 MB of memory, obviously beyond the limit.

If we can take doubling measure, why not triple or more? Similarly, we let p[i][j] denote the minimal value in range $$[i,i + {2^{5j}})$$, correspondingly we have

$f[i][j] = \left\{ \begin{array}{l}{A_i},j = 0\\\min \{ f[i + k \cdot {2^{5(j - 1)}}][j - 1]|0 \le k < {2^5}\} ,j > 0\end{array} \right.$

Queries can be similarly handled.

As Rujia Liu introduced in his book, we can divide the array into several blocks, each of

$l = {\log _2}^N/2$

numbers.

So there are

$N' = \left\lceil {\frac{N}{l}} \right\rceil = \left\lceil {\frac{{2N}}{{{{\log }_2}^N}}} \right\rceil \approx 100,000$

parts in total. Let A’ be the minimum value of each block

${A_i}' = \min \{ {A_j}|j \in [(i - 1)l + 1,il]$

Here we calculate

$f' = RMQ(A')$

Now for each query [x, y], first we find out where x, y is, say

x is the q1-th number in the p1-th block

y is the q2-th number in the p2-th block

the minimum value is

$answer = \min \left\{ \begin{array}{l}In - RMQ(p1,q1,l)\\In - RMQ(p2,1,q2)\\f'(p1 + 1,p2 - 2)\end{array} \right.$

here In-RMQ(i,j,k) means the minimum value from the j-th number to the k-th in block i.

And now we finally come to use the feature of ±1, the length of each block is l, so there are at most

${2^{l - 1}} = 512$

different blocks in essence. Calculate all the possible In-RMQ(i,j,k) queries in advance, save them in g[i][j][k].

Now all the queries can be done in O(1), and the preparation can be done in

$O(N'\log N' + {2^{l - 1}}{l^2}) = O(\frac{{2N}}{{\log N}}(\log N - \log \log N + 1) + \sqrt n {\log ^2}N) = O(N)$

So the complexity of the whole algorithm is O(N).

Actually, we can run faster if we use the most simple way to do the In-RMQ, enumerate, since l<10. Personally, I think it’s because of the addressing time for array g[][][], calculate g[][][] doesn’t take much time as I tested.

Here is the Test Result.

Corresponding to the order of the red text above, it’s

Sparse Table

Tricky1

±1RMQ & ±1RMQ + IO Optimize

Tricky2 & Tricky2 + IO Optimize

Well, the better ones get better opportunities.


text:
# k-vertex-connected graph

In graph theory, a connected graph G is said to be k-vertex-connected (or k-connected) if it has more than k vertices and remains connected whenever fewer than k vertices are removed.

The vertex-connectivity, or just connectivity, of a graph is the largest k for which the graph is k-vertex-connected.

## Definitions

A graph with connectivity 4.

A graph (other than a complete graph) has connectivity k if k is the size of the smallest subset of vertices such that the graph becomes disconnected if you delete them.[1] Complete graphs are not included in this version of the definition since they cannot be disconnected by deleting vertices. The complete graph with n vertices has connectivity n − 1, as implied by the first definition.

An equivalent definition is that a graph with at least two vertices is k-connected if, for every pair of its vertices, it is possible to find k vertex-independent paths connecting these vertices; see Menger's theorem (Diestel 2005, p. 55). This definition produces the same answer, n − 1, for the connectivity of the complete graph Kn.[1]

A 1-connected graph is called connected; a 2-connected graph is called biconnected. A 3-connected graph is called triconnected.

## Applications

### Polyhedral combinatorics

The 1-skeleton of any k-dimensional convex polytope forms a k-vertex-connected graph (Balinski's theorem, Balinski 1961). As a partial converse, Steinitz's theorem states that any 3-vertex-connected planar graph forms the skeleton of a convex polyhedron.

More generally, the 3-sphere regular cellulation conjecture claims that every 2-connected graph is the one-dimensional skeleton of a regular CW-complex on the three-dimensional sphere (http://twiki.di.uniroma1.it/pub/Users/SergioDeAgostino/DeAgostino.pdf).

## Computational complexity

The vertex-connectivity of an input graph G can be computed in polynomial time in the following way[2] consider all possible pairs ${\displaystyle (s,t)}$  of nonadjacent nodes to disconnect, using Menger's theorem to justify that the minimal-size separator for ${\displaystyle (s,t)}$  is the number of pairwise vertex-independent paths between them, encode the input by doubling each vertex as an edge to reduce to a computation of the number of pairwise edge-independent paths, and compute the maximum number of such paths by computing the maximum flow in the graph between ${\displaystyle s}$  and ${\displaystyle t}$  with capacity 1 to each edge, noting that a flow of ${\displaystyle k}$  in this graph corresponds, by the integral flow theorem, to ${\displaystyle k}$  pairwise edge-independent paths from ${\displaystyle s}$  to ${\displaystyle t}$ .


text:
# Phase space Lagrangian?

Reading out of this lecture series we define a phase space Lagrangian $\mathcal L$ to be a function of $4n+1$ variables namely $q,\dot q,p,\dot p,t$. My question is, what space is this function defined on? (I know that the $\dot p$ is there for names sake only).

My stab at an answer is it is a product space between $\mathcal L:TQ\times T^*Q\times \mathbb R\to \mathbb R$ since we are dealing with both velocity and momentum (where $Q$ is a configuration manifold). However this makes zero intuitive sense to me. If indeed it is phase space, my understanding was that taking the fibre derivative with respect to a function on the tangent bundle changed the velocity coordinates to momentum coordinates - the Legendre transform?

## 2 Answers

Let $X$ be the phase space. Then $L_\text{ph}(q,p,\dot{q},\dot{p},t)$ is a function on $TX\times \mathbb{R}$1, since the coordinates of $TX\times\mathbb{R}$ are precisely the coordinates of $X$, i.e. $(q,p)$ and their derivatives $(\dot{q},\dot{p})$ (and time $t$).

If Hamilton's equations are fulfilled, there are relations among $q,\dot{q},p,\dot{p}$ (the defining relations of the Legendre transformation) that reduce $L_\text{ph}(q,\dot{q},p,\dot{p},t)$ to the usual Lagrangian $L(q,\dot{q},t)$.

1In full analogy to the usual Lagrangian $L(q,\dot{q},t)$ being a function on $TQ\times\mathbb{R}$, where $Q$ is the $q$-space ("configuration space").

1. If $Q$ is configuration space, then the Lagrangian is a function $L: TQ\times \mathbb{R}\to \mathbb{R}$.

2. Let the cotangent bundle $M:=T^{\ast}Q$ be the corresponding phase space.

3. The Hamiltonian/phase space Lagrangian is a function $L_H: TM\times \mathbb{R}\to \mathbb{R}$.


text:
A colorful metapuzzle (of a sort)

edit: realized two puzzles ended up being variants. Replaced subpuzzle 5 completely, and swapped subpuzzle 2 to the original version. Either version has the same solution.

edit 2: my apologies. this was originally meant for a test solve, and I pushed it live far too early it seems. Added a couple more givens to puzzles 2 and 6 to bring their difficulty more in line with the rest of the subgrids. Also, had to replace the Minesweeper because I am made of fail.

Hint:

• It looks to me like the second and fifth puzzles are variants, and I haven't identified the eighth yet...
– Deusovi
Mar 24 '20 at 1:53
• The eighth looks similar to rot13(uggcf://ybtvpchmmyr.wxyngrk.qr/fgnefnaqneebjf) but not sure what the black circle is. perhaps an rot13(nyernql cynprq fgne)? Mar 24 '20 at 2:07
• Update: solved all puzzles except second, fifth, and sixth so far. (The sixth just because I'm bad at that genre; the other two seem underconstrained, but may also just be me being bad.)
– Deusovi
Mar 24 '20 at 3:09

Each of these grids is

a common logic puzzle genre. These genres are, in order:

Hidato; Str8ts; Battleships; Sudoku
Kakurasu; Skyscrapers; Minesweeper; Stars and Arrows
Tapa; Light and Shadow; Nurikabe; Fillomino

Solutions are below:

(Puzzles 2, 5, 6, and 7 were solved by Reinier.)

For the next step:

Each of the grids is a different color. (The grids on the right look black, but they are very dark shades of red, green, and blue.)

These colors are hexadecimal 88, 44, 22, and 11 in one of the channels, and 0 in the other two.

The text says "get to the heart of everything... all that is left is a single color". In all of the grids, the center cell is unclued; we can use the top right guide to convert the solutions to each puzzle to "shaded" or "unshaded", and then "add" the shaded colors together.

#000000 + #440000 + #220000 + #000000
+ #000000 + #000000 + #002200 + #001100
+ #000011 + #000044 + #000022 + #000011
... = 663399

This is a color that has the name "rebeccapurple" in the CSS color list; its "deeper meaning" is that it was named after Rebecca Meyer, who was a daughter of web designer Eric Meyer and who passed away at a young age due to a brain tumor.

• That is not the answer, but your logic seems to make some form of sense... Mar 24 '20 at 4:32
• Your last row is a bit off... I'm assuming you mean #000088 + #000000 + #000000 + #000001 Mar 25 '20 at 20:19

Continuing from Deusovi's work, here are the solutions to the remaining subpuzzles
(the second from the first row, and the first, second and third from the second row. The rest is copied from Deusovi)

• Now you just need to make the squares with even numbers black and....uh...figure out what to do next. :)
– Jens
Mar 24 '20 at 21:07


text:
# 13.1 Convergence and the central limit theorem  (Page 3/4)

 Page 3 / 4

Absolutely continuous examples

By use of the discrete approximation, we may get approximations to the sums of absolutely continuous random variables. The results on discrete variables indicatethat the more values the more quickly the conversion seems to occur. In our next example, we start with a random variable uniform on $\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ .

## Sum of three iid, uniform random variables.

Suppose $X\sim$ uniform $\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ . Then $E\left[X\right]=0.5$ and $\mathrm{Var}\phantom{\rule{0.166667em}{0ex}}\left[X\right]=1/12$ .

tappr Enter matrix [a b]of x-range endpoints [0 1] Enter number of x approximation points 100Enter density as a function of t t<=1 Use row matrices X and PX as in the simple caseEX = 0.5; VX = 1/12;[z,pz] = diidsum(X,PX,3);F = cumsum(pz); FG = gaussian(3*EX,3*VX,z);length(z) ans = 298a = 1:5:296; % Plot every fifth point plot(z(a),F(a),z(a),FG(a),'o')% Plotting details (see [link] )

For the sum of only three random variables, the fit is remarkably good. This is not entirely surprising, since the sum of two gives a symmetric triangulardistribution on $\left(0,\phantom{\rule{0.166667em}{0ex}}2\right)$ . Other distributions may take many more terms to get a good fit. Consider the following example.

## Sum of eight iid random variables

Suppose the density is one on the intervals $\left(-1,-0.5\right)$ and $\left(0.5,1\right)$ . Although the density is symmetric, it has two separate regions of probability. From symmetry, $E\left[X\right]=0$ . Calculations show $\mathrm{Var}\left[X\right]=E\left[{X}^{2}\right]=7/12$ . The MATLAB computations are:

tappr Enter matrix [a b]of x-range endpoints [-1 1] Enter number of x approximation points 200Enter density as a function of t (t<=-0.5)|(t>=0.5) Use row matrices X and PX as in the simple case[z,pz] = diidsum(X,PX,8);VX = 7/12; F = cumsum(pz);FG = gaussian(0,8*VX,z); plot(z,F,z,FG)% Plottting details (see [link] )

Although the sum of eight random variables is used, the fit to the gaussian is not as good as that for the sum of three in Example 4 . In either case, the convergence is remarkable fast—only a few terms are needed for good approximation.

## Convergence phenomena in probability theory

The central limit theorem exhibits one of several kinds of convergence important in probability theory, namely convergence in distribution (sometimes called weak convergence). The increasing concentration of values of the sample average random variable A n with increasing n illustrates convergence in probability . The convergence of the sample average is a form of the so-called weak law of large numbers . For large enough n the probability that A n lies within a given distance of the population mean can be made as near one as desired. The fact that the variance of A n becomes small for large n illustrates convergence in the mean (of order 2).

$E\left[|{A}_{n}-\mu {|}^{2}\right]\to 0\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{as}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}n\to \infty$

In the calculus, we deal with sequences of numbers. If $\left\{{a}_{n}:1\le n\right\}$ is a sequence of real numbers, we say the sequence converges iff for N sufficiently large a n approximates arbitrarily closely some number L for all $n\ge N$ . This unique number L is called the limit of the sequence. Convergent sequences are characterized by the fact that for largeenough N , the distance $|{a}_{n}-{a}_{m}|$ between any two terms is arbitrarily small for all $n,\phantom{\rule{0.277778em}{0ex}}m\ge N$ . Such a sequence is said to be fundamental (or Cauchy ). To be precise, if we let $ϵ>0$ be the error of approximation, then the sequence is

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text:
What is the equation of the line normal to f(x)=- tan(2pix -2) at x=3?

Jul 21, 2016

$y - \left(- \tan \left(6 \pi - 2\right)\right) = \frac{x - 3}{2 \pi \cdot {\sec}^{2} \left(6 \pi - 2\right)}$

Explanation:

f(x)=−tan(2πx−2)
f(3) =−tan(6π−2)

Point: (3, −tan(6π−2))

To find the slope, we need to take the derivative of $f \left(x\right)$.

$f ' \left(x\right) = - 2 \pi \cdot {\sec}^{2} \left(2 \pi x - 2\right)$

The slope normal/perpendicular would be 1/(2pi*sec^2(2pix-2).

Therefore at x=3, the perpendicular slope would be 1/(2pi*sec^2(6pi-2).

Now, we just need to plugin everything in the point-slope form.

$y - \left(- \tan \left(6 \pi - 2\right)\right) = \frac{x - 3}{2 \pi \cdot {\sec}^{2} \left(6 \pi - 2\right)}$


text:
Where parallels cross

Interesting bits of life

Moldable Emacs: query your Org Roam notes by tag

I display the Org Roam notes that intersect over a selection of tags (i.e., Org Roam notes titled with a single word).

The problem

Lately I am working on tagging. I recently reworked the way I maintain my agenda and yes, Org Mode tags are amazing! Tags are better than hierarchy because you can interpret tags as hierarchy or something else according to your needs. For example, having a hierarchy like king -> lord -> worker is less flexible than labelling/tagging something king/lord/worker. With hierarchy you enforce the order with structure, while with tags you need an interpreter to do the same. But you can also create other interpreters for the same tags.

Anyway, I got side tracked here... I just meant to say: I found out I have been creating tags consistently in my Org Roam notes. And these tags are notes themselves. For instance, I have a book note. When I add notes I took from a book I read, I link these notes to book. I also have a note management that explains what I believe management is. Moreover I linked to it some notes I collected over time. Now the question: what if I want to find all the notes that link to both book and management?

It is a problem indeed

I thought my enthusiasm for Org Roam would eventually fade. It didn't. Also I simply failed the principle from the book How to take smart notes:

The idea is not to collect, but to develop ideas, arguments and
discussions. Does the new information contradict, correct, support or
you combine ideas to generate something new? What questions are
triggered by them?

I got 6k-ish notes at the moment. I hoarded a bit. I can find most of what I need, but better tools for searching are necessary. And the funny thing is that tagging is also a concept in Org Mode. It sounds like whatever we devise, we could reuse for our TODO lists!

And there is a solution

Let me show you where I got so far.

The idea is simple: collect the notes that link to the notes you query. How it works is naive: find the notes containing each tags and return the intersection of these. Then I show them with org-transclusion.

You can find the mold definition in the repository here.

The definition of what is an Org Roam tag is a bit custom to my setup. Probably others use a different approach to it. Showing this I want to both share how helpful is tagging your notes for search and how molds can be also an interface to search your notes. Once you have the notes you can compose other molds to learn more about your notes. Using the Org Roam Elisp API you can gather the notes they link to or when you accessed them last time.

Conclusion

Grab moldable-emacs and start investigating your Org Roam notes! And try out using tags over hierarchy ;)

Happy searching!


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# I keep getting x1 wrong... t least one of the answers above is NOT correct. (4...

###### Question:

i keep getting x1 wrong...
t least one of the answers above is NOT correct. (4 points) Find the set of solutions for the linear system 2x + - 5x2 2x2 - x - + 8x3 - 2x3 + x 12x4 x = = = 18 5 - 12 If your answer includes any arbitrary parameters name them 51, 52, etc., as needed. (x1, x2, x3, xx) = ( VII - 265-881 | = (-12-8112 61 5 ) Note: You can earn partial credit on this problem. Preview My Answers Submit Answers Your score was recorded You have attempted this problem 29 times. You received a score of 75% for this attempt Your overall recorded score is 75%. You have unlimited attempts remaining. 00

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## Maximizing the Net Benefits of Pollution

Read this section about how to apply the Coase Theorem to solve environmental problems.

### 7. The Marginal Cost of Emissions

Pollutants harm people and the resources they value. The marginal cost curve for a pollutant shows the additional cost imposed by each unit of the pollutant. As we saw in Figure 18.1 "Determining the Efficient Level of Pollution", the marginal cost curves for all the individuals harmed by a particular pollutant are added vertically to obtain the marginal cost curve for the pollutant.

Like the marginal benefit curve for emissions, the marginal cost curve can be interpreted in two ways, as suggested in Figure 18.3 "The Marginal Cost of Emissions and the Marginal Benefit of Abatement". When read from left to right, the curve measures the marginal cost of additional units of emissions (MCE). If increasing the motorists' emissions from four pounds of carbon monoxide per week to five pounds of carbon monoxide per week imposes an external cost of $2, though, the marginal benefit of not being exposed to that unit of pollutant must be$2. The marginal cost curve can thus be read from right to left as a marginal benefit curve for abating emissions (MBA). This marginal benefit curve is, in effect, the demand curve for cleaner air.

Figure 18.3 The Marginal Cost of Emissions and the Marginal Benefit of Abatement

The marginal cost of the first few units of emissions is zero and then rises once emissions begin to harm people. That is the point at which the air becomes a scarce resource. Read from left to right the curve gives the marginal cost of emissions (MCE). Read from right to left, the curve gives the marginal benefit of abatement (MBA).

Economists estimate the marginal cost curve of pollution in several ways. One is to infer it from the demand for goods for which environmental quality is a complement. Another is to survey people, asking them what pollution costs - or what they would pay to reduce it. Still another is to determine the costs of damages created by pollution directly.

For example, environmental quality is a complement of housing. The demand for houses in areas with cleaner air is greater than the demand for houses in areas that are more polluted. By observing the relationship between house prices and air quality, economists can learn the value people place on cleaner air - and thus the cost of dirtier air. Studies have been conducted in cities all over the world to determine the relationship between air quality and house prices so that a measure of the demand for cleaner air can be made. They show that increased pollution levels result in lower house values.

Surveys are also used to assess the marginal cost of emissions. The fact that the marginal cost of an additional unit of emissions is the marginal benefit of avoiding the emissions suggests that surveys can be designed in two ways. Respondents can be asked how much they would be harmed by an increase in emissions, or they can be asked what they would pay for a reduction in emissions. Economists often use both kinds of questions in surveys designed to determine marginal costs.

A third kind of cost estimate is based on objects damaged by pollution. Increases in pollution, for example, require buildings to be painted more often; the increased cost of painting is one measure of the cost of added pollution.

While all of the attempts to measure cost are imperfect, the alternative is to not try to quantify cost at all. To economists, such an ostrich-like approach of sticking one's head in the sand would be unacceptable.