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# API¶

This page details the methods and classes provided by the isochrones package. The ModelGrid object handles the book-keeping aspects of loading a given grid into a pandas.DataFrame object. However, ModelGrid works behind the scenes; you should usually only work directly through the Isochrone object, which handles the grid interpolation. Finally, the StarModel object is used to fit the stellar models constrained by observed data.

## ModelGrid¶

class isochrones.grid.ModelGrid(bands=None, **kwargs)[source]

Base class for Model Grids.

Subclasses must implement the following (shown below is the Dartmouth example):

name = 'dartmouth'
common_columns = ('EEP', 'MMo', 'LogTeff', 'LogG', 'LogLLo', 'age', 'feh')
phot_systems = ('SDSSugriz','UBVRIJHKsKp','WISE','LSST','UKIDSS')
phot_bands = dict(SDSSugriz=['sdss_z', 'sdss_i', 'sdss_r', 'sdss_u', 'sdss_g'],
UBVRIJHKsKp=['B', 'I', 'H', 'J', 'Ks', 'R', 'U', 'V', 'D51', 'Kp'],
WISE=['W4', 'W3', 'W2', 'W1'],
LSST=['LSST_r', 'LSST_u', 'LSST_y', 'LSST_z', 'LSST_g', 'LSST_i'],
UKIDSS=['Y', 'H', 'K', 'J', 'Z'])

default_kwargs = {'afe':'afep0', 'y':''}
zenodo_record = 159426  # if you want to store data here
zenodo_files = ('dartmouth.tgz', 'dartmouth.tri') # again, if desired
master_tarball_file = 'dartmouth.tgz'

Subclasses also must implement the following methods:

get_band, phot_tarball_file, get_filenames, get_feh, to_df, hdf_filename. See DartmouthModelGrid and MISTModelGrid for details.

df_all(phot)[source]

Subclasses may want to sort this

classmethod extract_master_tarball()[source]

Unpack tarball of tarballs

classmethod get_band(b)[source]

Must defines what a “shortcut” band name refers to.

Param: b (string) Band name. phot_system, band b maps to the band defined by phot_system as band.
classmethod get_feh(filename)[source]

Parse [Fe/H] from filename (returns float)

get_filenames(phot, **kwargs)[source]

Returns list of all filenames corresponding to phot system and kwargs.

hdf_filename(phot)[source]

Returns HDF filename of parsed/stored phot system

phot_tarball_file(phot, **kwargs)[source]

Returns name of tarball file for given phot system and kwargs

classmethod to_df(filename)[source]

Parses specific file to a pandas DataFrame

## Isochrone¶

Any usage of isochrones involves instantiating an Isochrone, usually through a specific subclass such as mist.MIST_Isochrone.

class isochrones.Isochrone(m_ini, age, feh, m_act, logL, Teff, logg, mags, tri=None, minage=None, maxage=None, ext_table=False)[source]

Basic isochrone class. Everything is a function of mass, log(age), Fe/H.

Can be instantiated directly, but will typically be used with a pre-defined subclass, such as dartmouth.Dartmouth_Isochrone. All parameters must be array-like objects of the same length, with the exception of mags, which is a dictionary of such array-like objects.

Parameters: m_ini (array-like) – Array of initial mass values [msun]. age – log10(age) [yr] feh – Metallicity [dex] m_act – Actual mass; same as m_ini if mass loss not implemented [msun] logL – log10(luminosity) [solar units] Teff – Effective temperature [K] logg – log10(surface gravity) [cgs] mags (dict) – Dictionary of absolute magnitudes in different bands tri (scipy.spatial.qhull.Delaunay, optional) – Triangulation object used to initialize the interpolation functions. If pre-computed triangulation not provided, then the constructor will calculate one. This might take several minutes, so be patient. Much better to use pre-computed ones, as provided in, e.g., dartmouth.Dartmouth_Isochrone. minage,maxage – If desired, a minimum or maximum age can be manually entered.
agerange(m, feh=0.0)[source]

For a given mass and feh, returns the min and max allowed ages.

delta_nu(*args)[source]

Returns asteroseismic delta_nu in uHz

reference: https://arxiv.org/pdf/1312.3853v1.pdf, Eq (2)

density(*args)[source]

Mean density in g/cc

evtrack(m, feh=0.0, minage=None, maxage=None, dage=0.02, return_df=True)[source]

Returns evolution track for a single initial mass and feh.

Parameters: m – Initial mass of desired evolution track. feh – (optional) Metallicity of desired track. Default = 0.0 (solar) maxage (minage,) – (optional) Minimum and maximum log(age) of desired track. Will default to min and max age of model isochrones. dage – (optional) Spacing in log(age) at which to evaluate models. Default = 0.02 return_df – (optional) Whether to return a DataFrame or dicionary. Default is True. Either a pandas.DataFrame or dictionary representing the evolution track—fixed mass, sampled at chosen range of ages.
isochrone(age, feh=0.0, minm=None, maxm=None, dm=0.02, return_df=True, distance=None, AV=0.0)[source]

Returns stellar models at constant age and feh, for a range of masses

Parameters: age – log10(age) of desired isochrone. feh – (optional) Metallicity of desired isochrone (default = 0.0) maxm (minm,) – (optional) Mass range of desired isochrone (will default to max and min available) dm – (optional) Spacing in mass of desired isochrone. Default = 0.02 Msun. return_df – (optional) Whether to return a :class:pandas.DataFrame or dictionary. Default is True. distance – Distance in pc. If passed, then mags will be converted to apparent mags based on distance (and AV). AV – V-band extinction (magnitudes). pandas.DataFrame or dictionary containing results.
nu_max(*args)[source]

Returns asteroseismic nu_max in uHz

reference: https://arxiv.org/pdf/1312.3853v1.pdf, Eq (3)

random_points(n, minmass=None, maxmass=None, minage=None, maxage=None, minfeh=None, maxfeh=None)[source]

Returns n random mass, age, feh points, none of which are out of range.

Parameters: n – Number of desired points. maxmass (minmass,) – (optional) Desired allowed range. Default is mass range of self. maxage (minage,) – (optional) Desired allowed range. Default is log10(age) range of self. maxfeh (minfehs,) – (optional) Desired allowed range. Default is feh range of self. np.ndarray arrays of randomly selected mass, log10(age), and feh values within allowed ranges. Used, e.g., to initialize random walkers for StarModel fits.

## MIST¶

Stellar model grids from the MESA Isochrones and Stellar Tracks. These grids cover a larger range in mass, age, and [Fe/H], compared to the Dartmouth grids, and are thus preferred. Because this is a larger grid, interpolation here is implemented via the FastIsochrone implementation, rather than the standard Delaunay triangulation-based Isochrone implementation.

class isochrones.mist.MIST_Isochrone(*args, **kwargs)[source]

MESA Isochrones and Stellar Tracks

Parameters: bands – (optional) List of desired photometric bands. Default list of bands is ['G','B','V','J','H','K','W1','W2','W3','g','r','i','z','Kepler']. Here B and V are Tycho-2 mags, griz are SDSS, and G is Gaia G-band.

Details of models are here.

modelgrid

alias of isochrones.mist.grid.MISTModelGrid

## Dartmouth¶

Stellar model grids from the Dartmouth Stellar Evolution Database. Because these grids contain fewer points, both interpolation options are available: Dartmouth_Isochrone and Dartmouth_FastIsochrone.

class isochrones.dartmouth.Dartmouth_Isochrone(bands=None, afe='afep0', y='', **kwargs)[source]

Dotter (2008) Stellar Models, at solar a/Fe and He abundances.

Parameters: bands – (optional) List of desired photometric bands. Default list of bands is ['B','V','g','r','i','z','J','H','K','W1','W2','W3','Kepler'].

Model grids are obtained from here

agerange(m, feh=0.0)[source]

For a given mass and feh, returns the min and max allowed ages.

## StarModel¶

Estimating a star’s physical properties based on a set of observed properties involves instantiating a StarModel.

class isochrones.StarModel(ic, obs=None, N=1, index=0, name='', use_emcee=False, RA=None, dec=None, coords=None, **kwargs)[source]
Parameters: ic – Isochrone object used to model star. obs – (optional) ObservationTree object containing photometry information. If not provided, then one will be constructed from the provided keyword arguments (which must include at least one photometric bandpass). This should only happen in the simplest case of a single star system—if multiple stars are detected in any of the observations being used, an ObservationTree should be passed. If obs is a string, then it is assumed to be a filename of an obs summary DataFrame. N – Number of model stars to assign to each “leaf node” of the ObservationTree. If you want to model a binary star, provide N=2. **kwargs – Keyword arguments must be properties of given isochrone, e.g., logg, feh, Teff, and/or magnitudes. The values represent measurements of the star, and must be in (value,error) format. All such keyword arguments will be held in self.properties. parallax is also a valid property, and should be provided in miliarcseconds, as is density [g/cc], and nu_max and delta_nu (asteroseismic parameters in uHz.)
corner_observed(**kwargs)[source]

Makes corner plot for each observed node magnitude

evidence

Log(evidence) from multinest fit

fit_mcmc(nwalkers=300, nburn=200, niter=100, p0=None, initial_burn=None, ninitial=50, loglike_kwargs=None, **kwargs)[source]

Fits stellar model using MCMC.

Parameters: nwalkers – (optional) Number of walkers to pass to emcee.EnsembleSampler. Default is 200. nburn – (optional) Number of iterations for “burn-in.” Default is 100. niter – (optional) Number of for-keeps iterations for MCMC chain. Default is 200. p0 – (optional) Initial parameters for emcee. If not provided, then chains will behave according to whether inital_burn is set. initial_burn – (optional) If True, then initialize walkers first with a random initialization, then cull the walkers, keeping only those with > 15% acceptance rate, then reinitialize sampling. If False, then just do normal burn-in. Default is None, which will be set to True if fitting for distance (i.e., if there are apparent magnitudes as properties of the model), and False if not. ninitial – (optional) Number of iterations to test walkers for acceptance rate before re-initializing. loglike_args – Any arguments to pass to StarModel.loglike(), such as what priors to use. **kwargs – Additional keyword arguments passed to emcee.EnsembleSampler constructor. emcee.EnsembleSampler object.
fit_multinest(n_live_points=1000, basename=None, verbose=True, refit=False, overwrite=False, test=False, **kwargs)[source]

Fits model using MultiNest, via pymultinest.

Parameters: n_live_points – Number of live points to use for MultiNest fit. basename – Where the MulitNest-generated files will live. By default this will be in a folder named chains in the current working directory. Calling this will define a _mnest_basename attribute for this object. verbose – Whether you want MultiNest to talk to you. overwrite (refit,) – Set either of these to true if you want to delete the MultiNest files associated with the given basename and start over. **kwargs – Additional keyword arguments will be passed to pymultinest.run().
classmethod from_ini(ic, folder='.', ini_file='star.ini', **kwargs)[source]

Initialize a StarModel from a .ini file

The “classic” format (version <= 0.9) should still work for a single star, where all properties are just listed in the file; e.g.,

J = 10, 0.05 H = 9.5, 0.05 K = 9.0, 0.05 Teff = 5000, 150

If there are multiple stars observed, you can either define them in the ini file, or use the obsfile keyword, pointing to a file with the summarized photometric observations. In this case, spectroscopic/parallax info should still be included in the .ini file; e.g.,

obsfile = obs.csv Teff = 5000, 150

The obsfile should be a comma-separated table with the following columns: [name, band, resolution, mag, e_mag, separation, pa, relative].

• name is the name of instrument
• band is the photometric bandpass
• resolution is the approximate spatial resolution of instrument
• mag, e_mag describe magnitude of source (absolute or relative)
• separation, pa describe position of source
• relative: single-bit flag; if 1 then magnitudes taken with this instrument are assumed to be relative rather than absolute.

If an obsfile is not provided, you can also define all the same information in the ini file, following these rules:

• Every instrument/survey gets its own [section]. Sections are only

created for different photometric observations.

• if photometry relates to all stars in aperture, there is no extra info in the section, just the photometry. In this case, it is also assumed that the photometry is absolute. (relative=False)
• If ‘resolution’ is an attribute under a particular survey section (and

‘relative’ is not explicitly stated), then the survey is assumed to have relative photometry, and to be listing information about companion stars. In this case, there must be “separation” and “PA” included for each companion. If there is more than one companion star, they must be identifed by tag, e.g., separation_1, PA_1, Ks_1, J_1, etc. The tag can be anything alphanumeric, but it must be consistent within a particular section (instrument). If there is no tag, there is assumed to be only one companion detected.

• If there are no sections, then bands will be interpreted at face value

and will all be assumed to apply to all stars modeled.

• Default is to model each star in the highest-resolution observation as a

single star, at the same distance/age/feh/AV.

The N and index parameters may also be provided, to specify the relations between the model stars. If these are not provided, then N will default to 1 (one model star per star observed in highest-resolution observation) and index will default to all 0 (all stars physically associated).

classmethod load_hdf(filename, path='', name=None)[source]

A class method to load a saved StarModel from an HDF5 file.

File must have been created by a call to StarModel.save_hdf().

Parameters: filename – H5 file to load. path – (optional) Path within HDF file. StarModel object.
maxlike(p0, **kwargs)[source]

Finds (local) optimum in parameter space.

mnest_analyzer

PyMultiNest Analyzer object associated with fit.

See PyMultiNest documentation for more.

mnest_basename

Full path to basename

mnest_loglike(cube, ndim, nparams)[source]

loglikelihood function for multinest

print_ascii()[source]

Prints an ascii representation of the observation tree structure.

random_samples(n)[source]

Returns a random sampling of given size from the existing samples.

Parameters: n – Number of samples pandas.DataFrame of length n with random samples.
sampler

Sampler object from MCMC run.

samples

Dataframe with samples drawn from isochrone according to posterior

Columns include both the sampling parameters from the MCMC fit (mass, age, Fe/H, [distance, A_V]), and also evaluation of the Isochrone at each of these sample points—this is how chains of physical/observable parameters get produced.

save_hdf(filename, path='', overwrite=False, append=False)[source]

Saves object data to HDF file (only works if MCMC is run)

Samples are saved to /samples location under given path, ObservationTree is saved to /obs location under given path.

Parameters: filename – Name of file to save to. Should be .h5 file. path – (optional) Path within HDF file structure to save to. overwrite – (optional) If True, delete any existing file by the same name before writing. append – (optional) If True, then if a file exists, then just the path within the file will be updated.


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# Quantifying if two datasets are from the same distribution, if I only have distances

Let's say that I have two datasets. The target dataset is 1K instances, the "predicted" dataset is 1M instances (but I could downsample).

I can compute the distance between instances.

How do I quantify how much the two datasets are "similar"? e.g. drawn from the same distribution?

As an example, if the predicted dataset oversamples a particular instance (or near a particular instance) in the target dataset, that would be bad.

Additionally, if the predicted dataset has instances that are very far away from all the target instances, that would also be bad.

What are appropriate well-motivated measures for this task? Particularly with good Python implementations, but not required. I have looked for similar questions, but most seem to be based around comparing individual samples to a particular dataset or 1d datasets.

I am aware of the Ben David Discepancy approach but would want to use distances (between target and predicted dataset instances) to solve my problem.

• Do you have the points $x_i,y_j$ from the two datasets, or just the distances $d_{ij} = d(x_i, x_j)$ ? – ArnoV Mar 5 at 8:51

format: plain


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# Problem: The mass of a sample of gas is 827 mg. Its volume is 0.270 L at a temperature of 88°C and a pressure of 975 mmHg. Find its molar mass.

###### FREE Expert Solution

Molar mass:

First, we have to calculate the amount of gas in moles using the ideal gas equation.

$\overline{){\mathbf{P}}{\mathbf{V}}{\mathbf{=}}{\mathbf{n}}{\mathbf{R}}{\mathbf{T}}}$

P = pressure, atm
V = volume, L
n = moles, mol
R = gas constant = 0.08206 (L·atm)/(mol·K)
T = temperature, K

Isolate n (number of moles of gas):

$\frac{\mathbf{P}\mathbf{V}}{\mathbf{R}\mathbf{T}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{R}\mathbf{T}}}{\overline{)\mathbf{R}\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{n}}{\mathbf{=}}\frac{\mathbf{P}\mathbf{V}}{\mathbf{R}\mathbf{T}}}$

Given:

P = 1.2829 atm

V = 0.270 L
T = 88°C + 273.15 = 361.15 K
R = 0.08206 (L·atm)/(mol·K)

Calculate for n:

###### Problem Details

The mass of a sample of gas is 827 mg. Its volume is 0.270 L at a temperature of 88°C and a pressure of 975 mmHg. Find its molar mass.


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## What was the political motive behind United States Osama Bin Laden's killing? Why not capture him?

52

8

On May 2, 2011, Al Qaeda leader Osama bin Laden was killed by US Special Forces during an early morning (approximately mid-afternoon on May 1 in the United States) raid in Abbottabad, Pakistan.

The body of Osama Bin Laden was buried at sea, according to U.S. officials.

What was the reason for the non-recovery of the body? Also, why was the option of capture off the table? Wouldn't this just cause more doubt?

5Can you make it a little clearer what the question is? "Why would don't we parade corpses?", "Why do we kill our enemies?", "Is there any evidence Bin Laden might not be dead?", but I'm also not clear that any of those are really on topic. – None – 2018-01-04T05:18:39.793

22I think the question is "why kill and not capture?" @notstoreboughtdirt. – None – 2018-01-04T05:20:48.483

9Isn't the premise of this question wrong? The mission was officially to "kill or capture" Bin Laden. Did most of the people involved in the manhunt want him dead? Probably. Did the Navy Seal who shot him have the option to capture him, or was he forced by the situation to shoot? We'll never know. – ArrowCase – 2018-01-04T17:35:08.940

What do you mean by "non-recovery of the body"? It was recovered and transported to a US Navy ship, then buried. There may have been some massaging of details, i.e. "confirmed dead" on the ship, so that way burial at sea was halal. – Nick T – 2018-01-04T19:05:18.057

2@notstoreboughtdirt, it's pretty clear. Why not capture him and if not, why dump it at sea? It would create doubt and controversy. – Noah4343 – 2018-01-04T19:17:38.087

8Who says capture was off the table? AFAIK, the White House has maintained that if bin Laden had surrendered, they would have taken him alive. – mikeazo – 2018-01-05T15:32:32.270

1I fail to understand why "unclear what you're asking" stigma was applied to this question. I agree with Carpetsmoker about its meaning. Also, it has a high score, a fair amount of stars and a high quality answer. I am voting to reopen it. – Alexei – 2018-01-13T20:41:19.080

@Alexei, yeah this site has a very grey area of what questions can be asked. – Noah4343 – 2018-01-14T02:00:01.813

You have to have evidence to convict. – dan-klasson – 2020-03-24T10:30:56.293

88

### Kill versus capture

1. If they had captured Osama bin Laden, what would they have done with him? Barack Obama opposed the Guantanamo Bay facility, so they would have had to keep him on United States soil. Where?

2. If captured, he would have been tried, convicted, and executed. So not kill versus capture but kill then versus later.

3. His trial would have been an obvious target and excuse for retaliation.

4. His trial would have kept him in the news, encouraging more attacks.

5. He was perceived as dangerous at the moment that he was shot. Source

When the commandos reach Bin Laden’s room on the third floor, an AK-47 and a Makarov pistol are seen in arm’s reach of Bin Laden.

6. While there have been times and circumstances when the US military has taken prisoners intentionally, that's not really their job.

### Burial at sea

If he had been buried on land, his grave would be a landmark. Supporters might use it as a rallying point or symbol. Opponents might vandalize it.

No grave makes it harder to focus on a location.

"It was a joint decision. We thought it was important to think through ahead of time how we would dispose of the body if he were killed in the compound," says the president.

Note that this also indicates that it was not guaranteed that bin Laden be killed at the compound. Capture was on the table.

### Published photos

1. Ick. Who wants to see that?
2. Another thing to make people mad.

Obama said in the interview:

It is important for us to make sure that very graphic photos of somebody who was shot in the head are not floating around as an incitement to additional violence. As a propaganda tool. You know, that's not who we are. You know, we don't trot out this stuff as trophies. You know, the fact of the matter is this was somebody who was deserving of the justice that he received. And I think-- Americans and people around the world are glad that he's gone. But we don't need to spike the football. And I think that given the graphic nature of these photos, it would create some national security risk.

### Summary

Apparently they felt that the risks of a capture were worse than the risks of a death. So the mission parameters were such that they shot him rather than attempt a capture. Although apparently there were some circumstances in which he would have been captured, the reaction to the actual circumstances erred on the side of safety for the raiding party.

Once he was dead, they seem to have focused on not making him any more of a martyr than he already was. That was more important than proving his death.

Obama, from the same interview:

You know, the truth is that and we -- we're monitoring worldwide reaction. There's no doubt that Bin Laden is dead. Certainly there's no doubt among al Qaeda members that he is dead. And so we don't think that a photograph in and of itself is going to make any difference. There are going be some folks who deny it. The fact of the matter is, you will not see bin Laden walking on this Earth again.

52+1. Minor criticisms: (1) IMO it was within the resources of the US government to detain bin Laden in a secure fashion on US soil, most likely on a military base; (2) "that's not really their job" On the contrary, under the Geneva Conventions there are conditions when the US military is obliged to take prisoners. – Royal Canadian Bandit – 2018-01-04T08:32:43.293

3+1 small nit to pick,If he had been buried on land, his grave would be a landmark., the Salafist sect which Bin Laden and his followers adhered to oppose making burial grounds a landmark or any kind of land mark really which is why KSA bulldozed the shrines on graves of Prophet Muhammad's companions along with rest of the historically important buildings in Mecca and Medina. It was not going to become a pilgrimage ground – NSNoob – 2018-01-04T09:09:57.993

39"Published photos - 1. Ick. Who wants to see that?" I think you're giving humanity much more credit than it deserves here... – Guy G – 2018-01-04T09:33:36.927

11@GuyG, And they did published photos of Gaddafi's corpse later that year. And Che Guevara's too years before. Why not publish Bin Laden's too? – user28434 – 2018-01-04T09:51:07.160

8@NSNoob While his own faction might feel that way, that doesn't mean other factions might not use him as a martyr to further their cause, potentially including his grave. – Cronax – 2018-01-04T10:27:29.473

52@user28434: Who exactly is "they"? Neither Gaddafi nor Guevara was killed by US Special Forces. The USA is certainly not bound to follow precedent established by the Bolivian government in the 1960s. – Royal Canadian Bandit – 2018-01-04T10:38:49.610

3@Royal Canadian Bandit: Re "obliged to take prisoners", that's IF they're surrendering, Bin Laden wasn't surrendering. But the force did take a number of other prisoners, who did surrender. And that's putting aside the question of whether the Geneva Conventions even apply in a situation like this. – jamesqf – 2018-01-04T17:44:12.487

8I have reservations about the "they'd have to keep him on US soil. Where?" part of the answer. Except for the people who mostly did nothing wrong, or ones they tortured, or both, this has never been an obstacle for anyone (except Bush/Cheney). Terrorists have, historically, been tried in our federal system, convicted and sentenced to a lifetime in the federal prison system, to fade into obscurity. I don't see that what to do with him was a consideration, as much as "we're happy to take him alive, but we're going to invest zero additional energy or risk to make sure that happens." – PoloHoleSet – 2018-01-04T19:32:48.543

5@NSNoob Besides Cronax's sound observation that non-Salafists aren't obliged to respect Salafist custom or doctrine, the notion that bin Laden personally and Islamoterrorists in general don't act hypocritically with regard to religious doctrine is laughable. – Beanluc – 2018-01-04T20:02:06.883

1Your point 2 does not sound legit to me - or had there already been a trial in absence with a death sentence? – Hagen von Eitzen – 2018-01-05T19:18:22.847

1@GuyG I think the claim that being repulsed enough by a dead body to not want to see it is somehow "credit" is... a bit silly. – AmagicalFishy – 2018-01-05T19:19:09.667

@Hagen von Eitzen: Do you really think there would have been the slightest doubt about the verdict in any trial, had bin Laden actually surrendered? – jamesqf – 2018-01-05T20:34:51.127

@jamesqf Whether Hagen von Eitzen has doubt about a verdict is irrelevant. Or are you saying that, generally speaking, people who you're "really sure" are guilty don't need a trial? Because that is super duper scary. – Beska – 2018-01-05T22:01:06.100

@Beska: No, what I'm saying is that if bin Laden HAD surrendered, and a trial had been held, then given the volume of evidence - including his own public statements - there would have been no real doubt about the outcome. If he'd wanted a trial, he had years to surrender, up until the last second before the trigger was pulled. He chose not to do so. – jamesqf – 2018-01-06T00:02:24.957

A trial might raise awkward questions, such as how the Saudis could have been recruited without the knowledge of the regime. Bin Laden might name names. – Keith McClary – 2018-01-06T01:13:27.400

1@Keith McClary: But we have the nearly parallel case of Saddam Hussein. He was captured and given a trial, regardless of any political repercussions there might have been. The only real differences, IIRC, were that the capturing forces had much greater control of the situation, and Saddam came out with his hands up. – jamesqf – 2018-01-06T04:17:53.003

@jamesqf He was not put on trial for the gas attacks where the provenance of the chemicals and the aircraft used to drop them might have come out (because he was executed for something else first). ObL would presumably be tried in a US court where that would not happen. Of course, they could procrastinate forever, as with Khalid Sheikh Mohammed.

– Keith McClary – 2018-01-06T04:47:45.560

Also note that legally proving that he was behind the terrorist attacks would probably be tricky, he would also need a defender. The worst that could happen would be a long court filling the media and then letting him go. – Sulthan – 2018-01-06T18:47:23.803

@Keith McClary: But the decision on what grounds to try Saddam was made long after his capture, and probably not entirely by the US. Regardless of that, he WAS captured, because he chose to surrender. Bin Laden did not choose to do so. – jamesqf – 2018-01-07T03:23:06.460

7

To cop a phrase, "intellectual and social insecurity". From that, there are more logical steps that could be drawn in the lead-up, however.

Terrorists are killed overseas on a daily/weekly basis by drone strikes, but virtually none have committed any high-profile attacks on the US or other Western countries. Bin Laden did coordinate such an attack, so he had perverse celebrity status. Thus, a unique effort was expended (boots on the ground vs. yet another drone strike) to confirm his capture/killing and set people's minds at ease. Knowing where he was, and following it up with anything short of an unconfirmed drone kill would be political suicide.

Furthermore, because he was in Pakistan, a drone strike there has the potential to cause a tremendous diplomatic incident. If a bomb levels the building and Pakistani personnel are first on the scene, they could frame the incident basically however they want. Maybe bin Laden wasn't there and the US just senselessly attacked an ally. It still caused a kerfuffle, but Pakistan was in a tight spot trying to claim support for the US while simultaneously harboring its #1 Most Wanted.

Thus, the team on the ground was fated and Brythan's answer explains it from there.

4He was in a fairly suburban, upscale neighborhood in Pakistan, as well, with a lot of high ranking military types living there. A drone strike could have been a real, real mess. Nice perspective on what wasn't already touched upon. +1. Liked the term "perverse celebrity status," as well. – PoloHoleSet – 2018-01-04T20:27:46.637

1Re "virtually none have committed any high-profile attacks...": 1) Should low-profile attacks be acceptable? 2) Surely the point is to prevent attacks, not simply avenge them? 3) Since most attacks are suicide ones, it's fundamentally impossibe to do anything about the people who actually commit the attacks. So why should those who plan, organize, and supply materiel for the attacks be able to work with impunity? – jamesqf – 2018-01-05T04:01:52.657

1>

• No, but realpolitiks. 2. Yes, prevention is best, but to quote Iron Man, "if we can't protect the [country], you can be damned sure we'll avenge it." 3. I kinda glossed over that, but yeah, bin Laden didn't pilot the planes or bomb the towers. I think legally if you're a party to the crime by material support it's kinda the same thing?
• < – Nick T – 2018-01-05T07:01:05.563

The "fairly suburban, upscale neighborhood in Pakistan" was in Pakistan's equivalent to Garrison, NY (West Point) – pojo-guy – 2018-01-06T03:12:52.610

6

Your question has a false premise: that the goal was to kill him. The goal would have been to put an end to his terrorist activities.

He probably could have been more useful alive, it just didn't happen.

While war has been called politics by other means, it does tend to end up with more people dead than sitting around a table and talking. And terrorists aren't exactly known for their tendency to surrender peacefully.

2

Because Osama didn't want to be captured

If Osama had tried to surrender the SEALs were under instructions to allow it. He didn't try to surrender however, which mean the SEALs used lethal force. See Wikipedia article on the Death of Osama bin Laden.

The Associated Press reported at the time two U.S. officials as stating the operation was "a kill-or-capture mission, since the U.S. doesn't kill unarmed people trying to surrender", but that "it was clear from the beginning that whoever was behind those walls had no intention of surrendering". White House counterterrorism advisor John O. Brennan said after the raid: "If we had the opportunity to take bin Laden alive, if he didn't present any threat, the individuals involved were able and prepared to do that." CIA Director Leon Panetta said on PBS NewsHour: "The authority here was to kill bin Laden. ... Obviously under the rules of engagement, if he in fact had thrown up his hands, surrendered and didn't appear to be representing any kind of threat, then they were to capture him. But, they had full authority to kill him."

1

I'll see if I can find a link to back this up but I remember reading an article about this at the time. The mission wasn't approved by Pakistan, it was a top secret incursion into a nation that might have responded with force had they known. That meant they were on the clock. A fast in and out mission was essential to it's success.

One of the choppers went down, making getting in and out fast all the more difficult.

The orders were "Take him alive if he'd come quietly", the Marine, when he saw Bin Laden said "surrender" or "lay down", presumably in Arabic. Bin Laden didn't comply, so he shot him right then and there. If he'd laid down, they might have taken him alive.

Given the necessity for speed, capture alive wasn't the priority, but it was the plan if he'd not resisted. That's what I read anyway shortly after the event. It's possible this was just words to make it sound better and not how it actually went down, but that's what I read.


text:
Preprint D42/2007
On the moduli space of quasihomogeneous foliations on $(\Bb{C}^2,0)$
Leonardo Camara
Keywords: holomorphic foliations | holonomy
We study the relationship between singular holomorphic foliations on $(\mathbb{C}^{2},0)$} and their separatrices. Under mild conditions we shall describe a complete set of analytic invariants characterizing foliations with quasihomogeneous separatrices.

Anexos:


text:
# What is a googol?

##### 1 Answer
Mar 5, 2018

One googol is ${10}^{100}$.

#### Explanation:

Googol is a very large number, in which one hundred zeros follow one. In scientific notation, you can write it as ${10}^{100}$.


text:
5

# Write the expression as the sine cosine, or tangent of an angle. Then find Ihe exact value of the expression sin 75" cos 159 cos 75" sin 150Write the expr...

## Question

###### Write the expression as the sine cosine, or tangent of an angle. Then find Ihe exact value of the expression sin 75" cos 159 cos 75" sin 150Write the expression as the sine cosinetangent of an anglesin 750 cos 150 cos 758 sin 158 (Type an exact answer; using as needed Use integers or fractions for any numbers in the expression )FInd tho exact value of the expression_sin 75 cos 150 cos 759 sin 150 (Simplily your answer: Type an exact answer; using radicals as needed. Use integers or fra

Write the expression as the sine cosine, or tangent of an angle. Then find Ihe exact value of the expression sin 75" cos 159 cos 75" sin 150 Write the expression as the sine cosine tangent of an angle sin 750 cos 150 cos 758 sin 158 (Type an exact answer; using as needed Use integers or fractions for any numbers in the expression ) FInd tho exact value of the expression_ sin 75 cos 150 cos 759 sin 150 (Simplily your answer: Type an exact answer; using radicals as needed. Use integers or fractions for any numbers the expresslon: Ratlonallze all denominators, Enter - your arswer In each of the answer boxes.

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text:
# How to find the base point given public and private key and EC parameters except the base point

I'm working on the curve secp256k1 and I've private key $$k$$ and public key $$P = [k]G$$. I look for solutions to the following problems;

• What is the inverse of a scalar $$k$$ in ECC?

• If I'm only missing $$G$$ but know all parameters of the curve can I find $$G$$ from $$k$$ and $$P$$?

• $G$ is publicly known so there is no need to calculate that. Do you want to check that given a public key is belongs to a private key? then check with $key_{pub} = [key_{private}]G$ – kelalaka Dec 4 '20 at 11:04
• @kelalaka Sure, I know that G is known and I know both public and private key. This is more of a theoretical exercise. Can a calculate G back from publickey and privatkey? – Alexander Weps Dec 4 '20 at 11:19

Most standardized curves has a base point $$G$$ of prime order $$q$$ so computing the inverse of $$k$$ (the private key) modulo $$q$$ can be easily done as $$k_{\text{inv}} = k^{p-2} \bmod q,$$ thanks to Fermat's Little Theorem since $$k_{\text{inv}}\times k \equiv k^{p-1} \equiv 1 \pmod{q}$$

In the case the order is a composite integer $$n$$, then Euclid Extended algorithm is a good way to find it.

In your problem you have a public point $$P_{\text{pub}}$$ such that $$P_{\text{pub}} = [k]G,$$ and you only know $$k$$ and $$P_{\text{pub}}$$. To find $$G$$, use of the above algorithm (if the order of $$G$$ is known) to find $$k_{\text{inv}}$$ and we have $$G = [k_{\text{inv}}] P_{\text{pub}}.$$

In ECC, the private key $$k$$ is a scalar and a public key is $$[k]G$$, this is scalar multiplication defined as

\begin{align} [k]:& E \to E\\ &P\mapsto [k]G=\underbrace{G+G+\cdots+G}_{\text{k times}}.\end{align}

As we can see, it is a short definition of multiple additions of the same element. Since in ECC we work in finite fields $$\mathbf{F}$$, the elements of the curve are finite and actually forms an abelian group with an identity $$\mathcal{O}$$, this can be the point of infinity or not, depending on the curve definition. If the order of the point $$G$$ is $$t$$ ( the $$t$$ can be equal to the order of the EC group or a divisor of it, i.e. $$ord(G)\mid \#|E(\mathbf{F})|$$) then

$$\underbrace{G+G+\cdots+G}_{\text{t times}} = [t]G= \mathcal{O}$$

Inverse of private key

Now if you want to the inverse of $$k^{-1}$$ it is just $$t-k$$. You can see this from this

$$[k]G + [t-k]G = [k+t-k]G = [t]G = \mathcal{O}$$

Finding the base point only with private and public key

The question has little meaning since the base point comes with a standard and it is publicly known for all parties. To able to calculate any arithmetic, you already must know the Elliptic curve construction parameters.

let's consider you all know but the basepoint $$G$$ and formalizing ;

find $$x$$ such that $$[x]P = G$$ where $$P = [k]G$$ given $$P$$ and $$k$$.

We will use this identity on the scalars;

$$[k]([a]P) = [ak]P$$

$$[k]([a]P)=\underbrace{ \underbrace{P+P+\cdots+P}_{\text{a times}} + \cdots + \underbrace{P+P+\cdots+P}_{\text{a times}} }_{ \text{k times}}$$

Since the scalars are working a modulus, i.e.

$$[k]P = [k \bmod t]P$$ where $$t$$ was the order of the $$G$$. So, as given by the corspini's answer if the multiplicative inverse of $$k$$ over the group of order $$t$$ then we can use it to find $$G$$. In ECC, we use usually use prime order subgroups so inverses exist in them.

Let $$k'\cdot k \equiv 1 \bmod t$$ and thake $$P = [k]G$$ now add $$k'$$ times

$$[k' \bmod t]P = [k'([k]]G) = [k' \cdot k \bmod t]G = G$$


text:
# How do you solve 5x+15=8x-40?

Dec 30, 2016

$x = 18 \frac{1}{3}$

#### Explanation:

$5 x + 15 = 8 x - 40$

Add $40$ to each side.

$5 x + 55 = 8 x$

Subtract $5 x$ from each side.

$55 = 3 x$

Divide both sides by $3$.

$\frac{55}{3} = \frac{3 x}{3}$

$18 \frac{1}{3} = \frac{\cancel{3} x}{\cancel{3}}$

$18 \frac{1}{3} = x$ or $x = 18 \frac{1}{3}$


text:
# The decay width of the $Z_c(3900)$ as an axialvector tetraquark state in solid quark–hadron duality

The European Physical Journal C, Jan 2018

In this article, we tentatively assign the $Z_c^\pm (3900)$ to be the diquark–antidiquark type axialvector tetraquark state, study the hadronic coupling constants $G_{Z_cJ/\psi \pi }, G_{Z_c\eta _c\rho }, G_{Z_cD \bar{D}^{*}}$ with the QCD sum rules in details. We take into account both the connected and disconnected Feynman diagrams in carrying out the operator product expansion, as the connected Feynman diagrams alone cannot do the work. Special attentions are paid to matching the hadron side of the correlation functions with the QCD side of the correlation functions to obtain solid duality, the routine can be applied to study other hadronic couplings directly. We study the two-body strong decays $Z_c^+(3900)\rightarrow J/\psi \pi ^+, \eta _c\rho ^+, D^+ \bar{D}^{*0}, \bar{D}^0 D^{*+}$ and obtain the total width of the $Z_c^\pm (3900)$. The numerical results support assigning the $Z_c^\pm (3900)$ to be the diquark–antidiquark type axialvector tetraquark state, and assigning the $Z_c^\pm (3885)$ to be the meson–meson type axialvector molecular state.

This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-017-5514-1.pdf

Zhi-Gang Wang, Jun-Xia Zhang. The decay width of the $Z_c(3900)$ as an axialvector tetraquark state in solid quark–hadron duality, The European Physical Journal C, 2018, 14, DOI: 10.1140/epjc/s10052-017-5514-1


text:
# How do you solve 10+|m-6|=7?

The absolute value gives always a positive number. So we have that $10$ plus a positive number has to give $7$. This is clearly impossible because $10$ is bigger than $7$ and when we add a positive number we obtain a bigger number.


text:
Homework Help: Gravitation Exercise

1. Aug 25, 2011

carlosbgois

1. The problem statement, all variables and given/known data

What is the intensity of the force $F_{g}$ between the ring and a mass 'm', which is at a distance 'x' from the center of the ring?

http://img52.imageshack.us/img52/6859/ringre.th.jpg [Broken]

3. The attempt at a solution

I have got to my own answer, but it is different from the given one. Here's what I did:

There's a potential energy between 'm' and a dM from the ring, which is given by $dU=-G*m*dM/d$ and $d=\sqrt{r^{2}+x^{2}}$.

From this, I can find the total potential energy by integrating dU from 0 to M, which gives me $U=-\int^{M}_{0}\frac{G*m*dM}{d}=\frac{-G*m*M}{\sqrt{r^{2}+x^{2}}}$

As the variation in the potential energy is equal to the negative of the work done, I did $-\int^{0}_{d}F(d)*dd=\frac{-G*m*M}{\sqrt{r^{2}+x^{2}}}$$\Rightarrow$$F(d)=\frac{2*G*m*M}{(r^{2}+x^{2})^{3/2}}$

So, what's wrong? Thanks

Last edited by a moderator: May 5, 2017
2. Aug 25, 2011

carlosbgois

I think $-\int^{0}_{d}F(d)*dd$ should be $[itex]-\int^{0}_{d}F(d)*cos\theta*dd$, but even this way I didn't get the right answer.

3. Aug 26, 2011

carlosbgois

Sorry, forgot to post the given answer for the exercise: $F = \frac{GMmx}{(r^2+x^2)^{3/2}}$


text:
## Algebra 2 (1st Edition)

The error is in the second step because it should be: $4(x^2+6x)=11\\4(x^2+6x+9)=11+4\cdot9=11+36$


text:
12:23 AM
@ACuriousMind I made it out ;)

1:22 AM
@dmckee @ChrisWhite Either of you around?

Sorta. What's up?

@dmckee Do you know memory bug debugging in C++?

I'm out of practice and might not know your particular environment, but I've done it in the past.
Do you have a leak, or something more concrete?
Can you make it do it on command or does it pop up unexpectedly?

*** Error in ./ackermann.o': double free or corruption (fasttop): 0x0000000000c
as soon as it starts
I know where the bug is
I just don't know why it happens

Know how to run with the debugger?

1:25 AM
@dmckee I know how to run it, just not what to do with it

First question: does the bug occur with the debugger running?
If no, you probably have an uninitialized variable.
If so, then when you get the crash ask the debugger what line it occurs on.

That's not it I think. The bug only happens when I try to clear memory of variables I no longer need. Let me try this into gdb

All those commented out mpz_clearss?

Yep
Even if I use only the first one (that clears m and n) it breaks
are function arguments constants in C?
no right

No, though changes to them are in local execution context only.

1:32 AM
Yeah, so I thought

You don't have a lot of code and aren't calling new/delete (or alloc/free) yourself. SO the problem is likely one of understanding how the library is suppose to work. I've never used GMP so I can't help much.

It doesn't crash on Valgrind .-.
AH
AAAAH

Oh. Uhm. Your m and n are local to main and are loop variables.

Got it :)
you shouldn't call clear in the C++ API

Mmmm. Good. But I'd also think carefully about what you should be doing to those variables; their automatic and shouldn't be delete(free)ed.

1:37 AM
@dmckee Yeah, I'm still getting the hang of C++. It's a funny language

Well, it's built on a C foundation, and C is a very raw language. On 1970s hardware there was very nearly a 1-to-1 correspondence between language constructs and features of the chips.
Since then the chips have gained more features, but C still presents almost the same virtual machine.

C is cool, but I just can't get around to using it more, it's too much work to do simple things at times
and string management is dreadful

A lot of pros don't use the stdlib.h and string.h string interfaces for real projects. They bring in safer and more featureful libraries to manage strings for them.
If you do use them you use getline for IO and always use the n` variants of the string functions.

What I like about C is that it's so kick in the balls
it takes 0 bullshit home
that makes you better

It will teach you discipline or break your spirit in the attempt.

1:46 AM
Lol

@BernardMeurer The core language, and any reasonable library, will always have malloc/new-ish and free/delete-ish commands in pairs. Whenever you use one of the former, immediately write exactly one of the latter.

2:24 AM
Have you guys ever written a toy code to simulate and calculate diffusion coefficients?

2:40 AM
In double slit interference, will we see interference pattern, if we send protons from one slit and neutrons from other?

Why not? They both have a de Broglie wavelength.

@Mikhail In feynmanlectures.caltech.edu/III_01.html , they had problem in identifying which slit did electron came from. If they pass proton from one slit and neutron from 2nd slit, can't problem be solved. They know which slit did particles come from and also they would see interference pattern?

The experiment was done already with nuetrons, indeed it is not clear if there were some stray protons in the group.
The real problem with using protons it that I expect them to anti-bunch due to the charge.

What is anti bunch?
Was experiment tried actually for electrons?

3:29 AM
@ChrisWhite Got it. Thanks for the tip :)

3:49 AM
@AnubhavGoel electron interference is the mechanism for electronic (SEM) image formation. Actually people have been able to interfere larger multi-atom structures.
Also the famous bose-einstein condensation interference

4:14 AM
@JohnRennie hello

Morning
Or I guess it isn't morning in your part of the world :-)

4:53 AM
@JohnRennie Today I learned of a cool way of integrating over a nonorientable Riemannian manifold

I probably don't know enough about the subject to be suitably impressed :-)
When you say integrating over a manifold do you mean an integral like $\int something dx^n$ ?

@JohnRennie Yes.

Why are non-orientable manifolds different from orientable ones in that respect?

@JohnRennie $\mathrm d^nx$ encodes orientation so it's only defined for orientable manifolds.

Aaaaaaaaaaaaah, I get it.

4:57 AM
There's now two solutions to the problem that I know of

I remember last year, I got an A- in physics 'cause the teacher hated me

OK so what do you with a non-orientable manifold?

Said I asked too many questions above the scope of the class

@JohnRennie It turns out you can define a scalar density known as the Riemannian density
In the orientable case it's just the volume form $\sqrt g dx^1\wedge\cdots\wedge dx^n$

Hello, Anyone here worked on VASP ?

4:59 AM
@JohnRennie you basically construct it like this:
you pick a bunch of open sets that cover your manifold, and each one of these open sets is an orientable Riemannian manifold
then you take the volume forms on each of these pieces and glue the whole thing together

@Gowtham I've seen few comments about VASP, but right now is a bad time as most of the site members are asleep. Around 17:00 UTC is the liveliest time.

this isn't really well-defined but it turns out if you integrate it, it is
so you get a density $\mu$ and then you can integrate functions like $f\mapsto\int_M f\mu$.
But there's a really neat way that I learned today

@JohnRennie thanks

The key concept is something called the "cut locus." It's a set of the manifold that basically describes where geodesics fail to minimize length.
The cut locus of a point $p$ is denoted $C(p)$
Now, one can show that (a) $C(p)$ has measure zero in $M$ (b) $M-C(p)$ is homeomorphic to an open ball

5:05 AM
This means that $M-C(p)$ is contractible
By a theorem of topology, any fiber bundle defined on $M-C(p)$ is trivial
In particular, $T(M-C(p))=(M-C(p))\times\Bbb R^n$
By another theorem of topology, this implies that $M-C(p)$ is orientable

$M-C(p)$ is what's left if you snip out the troublesome bits?

@JohnRennie I guess so! I'd have to explicitly calculate it for some nonorientable manifold to be sure!
But it's often said that $C(p)$ carries all topological information of $M$
Oh I forgot
If you give $M$ a Riemannian metric you need a theorem by Nomizu and Ozeki that says there is a complete metric globally conformal to $g$
And we use this conformal metric to calculate $C(p)$
cut locus stuff only works with complete metrics.

So you do the integration on $M-C(p)$ then do some clever trick with the remaining bit?

So, because $C(p)$ has measure zero, $\int_{M-C(p)}\equiv \int_M$
@JohnRennie no, that's the cool part

@0celo7 Aaaaaaaaaaaaahhh (again)

5:10 AM
it's like integrating over $\Bbb R^2$ vs. $\Bbb R^2-\{0\}$
there's no difference

That feels somehow like cheating ....

But on ${M-C(p)}$ you can define an orientation

But I guess it works.

So you just restrict your original metric to this set
and find the volume form
and ta-da you used a bunch of high-powered theorems to define the integral in a really cool way

I'd be cautious about anyone's views on the consequences of Brexit. The truth is that no-one knows what's going to happen.
My guess is that things will change less than we think. It's in everyone's interests to find a compromise solution that keeps things much as they are.

5:50 AM
@JohnRennie I just want to say, there's a certain satisfaction to knowing I've been doing the right thing all along (even if my motivation was really one of laziness).

@ChrisWhite :-)
Soap is dreadful stuff. It destroys cell membranes. You'll hear people say that soap is very good at killing bacteria. Well yes, that's because it's very good at killing everything including your skin cells. The reason it stings when it gets in your eyes is because it's killing the cells in your eyes!

Is that just from its alkalinity? Or is there a more nuanced chemical reason?

Cell membranes are basically surfactant bilayers, where the surfactants are phospholipids. Foreign surfactants disrupt this bilayer structure and rupture the cell membrane, which kills the cell.
Soap is a mixture of sodium salts of carboxylic acids, and these happen to be particularly good at disrupting the phospholipid bilayers.
By contrast sodium lauryl ether sulphate doesn't get into the lipid bilayers so it doesn't damage them. Well, that's not strictly true, but it damages them a lot less.
See, colloid science is just awesome! I can't understand why everyone doesn't want to study it :-)

6:06 AM
::pours another glass of milk::

@JohnRennie Because colloid sounds like something nasty
"Yeah he couldn't stop seizing so he got his colloids removed"


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# Sign of the real part of the roots of a particular polynomial of degree 4

I am working with the polynomial $x^4+ax^3+bx^2+cx+d$, with $a>0, b>0, c>0, d>0$. I would like to understand under which conditions, on the coefficients, this polynomial has only complex roots with negative real part or real negative roots. Does someone knows this conditions or someone has some literature to suggest?

Given a polynomial $P=a_0 x^n + a_1 x^{n-1} + \cdots + a_n$, set up the $(n+1)\times (n+1)$ Hurwitz matrix $$H(P)= \begin{matrix} a_0 & a_2 & \ldots &\\ 0& a_1 & a_3 & \ldots\\ 0& a_0 & a_2 & \ldots\\ 0 & 0 & a_1 & \ldots \\ 0& 0 & \ldots & a_{n+1} \end{matrix}$$

$H(P)$ is constructed as follows: on the first row we place $a$'s with even indexes. Then, in each column, the indexes decrease by $1$. On the diagonal we have the coefficients $a_0$, $\ldots$, $a_{n+1}$. Note that $a_{\mu}$ is $0$ for $\mu>n$, or for $\mu<0$.

Assume that $a_0>0$. Then $P$ has all roots with real part $<0$ if and only if all the leading minors of the matrix $H(P)$ are $>0$. Note that this is equivalent to $a_0>0$, $a_n>0$ and the principal minors $(2,\ldots,k)$ positive for all $2\le k \le n$, that is, $a_0$, $a_1$, $a_n>0$ and $n-2$ other algebraic conditions.

In the case $n=4$, appearing also in Hurwitz original paper, we have $$P= a_0 x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4$$ the Hurwitz matrix is $$H(P) = \left(\begin{matrix} a_0 & a_2 & a_4 & 0 & 0\\ 0& a_1 & a_3 & 0 & 0\\ 0& a_0& a_2 & a_4 & 0\\ 0& 0& a_1 & a_3 & 0\\ 0& 0& a_0& a_2 & a_4\end{matrix} \right)$$ and the conditions, on top of $a_i>0$, are $$\left| \begin{matrix} a_1 & a_3 \\ a_0 & a_2 \end{matrix} \right |>0 \\ \left | \begin{matrix} a_1 & a_3 & 0\\ a_0 & a_2 & a_4\\ 0& a_1 & a_3 \end{matrix} \right |>0$$

What you are after is called the Routh-Hurwitz criterion.

As you guessed, it will require some reading, but the bottom line is that there is an algorithm to draw up a table, and the condition is that all entries from the first column should be of the same sign.

See this, for instance.


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Hong Kong
Stage 1 - Stage 3

# Volume of Cylinders II

Lesson

We have already seen how the volume of prisms can be calculated using the rule $\text{Volume }=\text{Area of Base }\times\text{Height of Prism }$Volume =Area of Base ×Height of Prism

(If you would like to refresh these topics try these pages, here and here)

A cylinder is very similar to a prism (except for the curved face), but the volume can be found using the same process we have already learnt.

$\text{Volume of Cylinder }=\text{Area of Base }\times\text{Height of Prism }$Volume of Cylinder =Area of Base ×Height of Prism

$\text{Volume of Cylinder }=\pi r^2\times h$Volume of Cylinder =πr2×h

$\text{Volume of Cylinder }=\pi r^2h$Volume of Cylinder =πr2h

#### Example

You are at the local hardware store to buy a can of paint. After settling on one product, the salesman offers to sell you a can that is either double the height or double the radius (your choice) of the one you had decided on for double the price. Assuming all cans of paint are filled to the brim, is it worth taking up his offer?

If so, would you get more paint for each dollar if you chose the can that was double the radius or the can that was double the height?

Since the volume of a cylinder is given by the formula $\pi r^2h$πr2h, if the height doubles, the volume becomes $\pi r^2\times2h=2\pi r^2h$πr2×2h=2πr2h (ie the volume increases two-fold).

Whereas if the radius doubles, the volume becomes $\pi\left(2r\right)^2h=4\pi r^2h$π(2r)2h=4πr2h (ie the volume increases four-fold).

That is, the volume increases by double the amount when the radius is doubled compared to when the height is doubled.

To see how changes in height and radius affect the volume of a can to different extents, try the following interactive. You can vary the height and radius by moving the sliders around.

#### Worked Examples

##### question 1

Find the volume of a cylinder correct to one decimal place if its diameter is $2$2 cm and its height is $19$19 cm.

##### question 2

Find the volume of a cylinder with radius $7$7 cm and height $15$15 cm, correct to two decimal places.


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What's up in

# logic

## Mathematicians Bridge Finite-Infinite Divide

A surprising new proof is helping to connect the mathematics of infinity to the physical world.


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# Independent component analysis with nonnegative mixing matrix

In independent component(s?) analysis, I have the observed signal, $O$, the mixing matrix, $A$, and the source matrix, $S$, with $O ≈ AS$

I've found some literature on ICA with the sources assumed to be nonnegative, but haven't been able to find anything on ICA with the mixing matrix assumed nonnegative. Does anyone know of whether algorithms for this exist?


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# Speech¶

## Introduction¶

Up to now, we have assumed that all sounds are equal. In this chapter, I finally single out speech as being the most equal, though I may refer to music every now and then.

One of the goals of this chapter is for you to become familiar with acoustic representations, such as that of a recording of me pronouncing the vowels [ieaou]:

I open it in Praat to the following screen; you can download it yourself from here if you want to follow along at home:

This image stacks two graphs one on top of the other.

Whenever you are confronted with a graph, the first thing that you should do is look at the labels on its axes. The graphs above have two of them, a horizontal one which is usually called “x” and a vertical one which is usually called “y”.

Questions

What is the x axis? What is its range?

I tried to articulate each vowel for the same length of time. How long is this?

Praat has set the cursor (red line) at the middle of the recording by default. How long is each half?

What are the two y axes? What is the range of the first? What about the second?

Why does the last vowel in the top graph look different from the others?

You should be able to figure out most of the answers. I’ll give you a hand with the two hardest ones.

The top graph is extremely unhelpful, since its three measurements, -0.1416, 0, and 0.145 are not labeled with any unit. They are measures of a sound’s amplitude. In theory, this should be pressure, whose unit is the pascal, though I have not found any mention of it in the Praat documentation. Praat normalizes this amplitude tracing to its highest and lowest measures – here nearly 1 and -1. These extremes indicate the levels at which the equipment used to record the sound and the equipment used to play it back reach their limits of fidelity, beyond which the sound will be distorted.

The last vowel has a lesser ‘footprint’ because we tend to reduce our speech towards the end of an utterance, presumably to save effort. Thus the amplitude tracing at the top gradually shrinks, while the frequency trace underneath gets gradually fainter.

By the way, a graph with two axes is often called two dimensional, under the assumption that each axis measures a different dimension.

## Articulatory phonetics: like having a guitar string in your throat¶

So how do the vowels in Fig. 15 come to be? They come from my favorite sound-production device, the human voice. At the largest scale, it is composed of the three parts in Fig. 16:

The respiratory system produces a column of air that makes the vocal cords vibrate in the laryngeal system which is filtered by the supralaryngeal system to produce the sounds that we recognize as speech.

### The laryngeal system¶

The air pushed up the trachea by the lungs does not in itself have any interesting linguistic properties, but once it passes through the vocal cords or folds, it certainly does. If you cut off my head and looked straight down my trachea, you would see something like this:

When the vocal folds are closed, as in the left side of the image, no expiratory airflow can enter the supralaryngeal system, so there is silence, and expiratory air is compressed, raising its pressure. When the vocal folds are open, as in the right side of the image, an opening appears called the glottis which allows expiratory airflow to enter the supralaryngeal system and create speech sounds, or phonation, and expiratory airflow is decompressed or rarefied and its pressure falls.

The vocal folds are more of a mucous membrane than a muscle and so tend to snap back to their closed position after being open. Their opening and closing compresses and releases the expiratory airflow as in next image, producing a cycle which you should now be familiar with:

Increasing darkness symbolizes increasing compression of the airflow. The vertical scale represents the pressure of the airflow through the vocal folds as a single quantity between a minimum and a maximum. A single cycle of closing and opening is defined by the distance between two peaks, marked by dotted white lines.

The outcome is that the each vocal fold vibrates periodically, like a guitar string. And like in Fig. 8, the image above illustrates the fundamental frequency of the vocal folds.

Modern recording technology is fast enough to catch the vocal folds in action. The image below zooms in on the beginning of the first vowel in [ieaou]:

Question

How much of the original recording does this one capture?

The peak of each pressure wave corresponds with the darkest band of the frequency spectrum, as I tried to point out by placing the cursor over one.

### The supralaryngeal system¶

But just like a plucked guitar string, the vocal folds vibrate at a variety of frequencies that are multiples of the fundamental. If we could hear just this pulse, it would sound, as Loritz (1999:93) says, “more like a quick, dull thud than a ringing bell”.

Yet the human voice does not sound like a quick, dull thud; it sounds, well, it sounds like a human voice. This is because the supralaryngeal system sits on top of the larynx and filters the glottal pulse just like different fingerings of a guitar string filters its vibrations:

In particular, the oral and nasal cavities resonate at certain frequencies, thereby exaggerating some harmonics while muting others.

#### Vowels¶

For a vowel, the mouth reaches its maximum opening.

Question

How many vowels are there in (American) English?

If you said five, i.e. ‘i, e, a, o, u’, I invite you pronounce the following words:

‘beat, bit, bait, bet, bat, but, cot, bought, boat, book, boot’

Now how many vowels do you think there are in (American) English?

Eleven is a much better answer than five. But of course, the English alphabet only has symbols for five of them, so to represent the others consistently, we will have to draw on a broader inventory. This is the purpose of the International Phonetic Alphabet.

Before seeing the additional symbols, let me say a few words about how vowels are described.

Put your hand under your jaw and say the vowel of ‘beat, bit, bait, bet, bat, cot’. What does your jaw do?

Now try the sequence in reverse. Put your hand under your jaw and say the vowel of ‘cot, bat, bet, bait, bit, beat’. What does your jaw do?

For the first sequence, you should feel your jaw going down. This is because your mouth is opening to allow your tongue to go down. Conversely, for the first sequence, you should feel your jaw going up. This is because your mouth is closing to allow your tongue to go down.

The parallel motion happens for the second half of the set of English vowels:

Put your hand under your jaw and say the vowel of ‘boot, book, boat, bought, cot’. What does your jaw do?

Now try the sequence in reverse. Put your hand under your jaw and say the vowel of ‘cot, bought, boat, book, boot’. What does your jaw do?

As before, your jaw goes down, then up, because your tongue goes down, then up.

The result is that vowels can be classified by height (of the tongue). The vowel of ‘cot’ is the lowest, and the vowels of ‘beat’ and ‘boot’ are the highest, with the rest falling in between.

There is a perpendicular motion of the tongue, too – tongue advancement – but you can’t feel it the way you can feel your jaw move. The vowel of ‘beat’ occupies the front-most position, while the vowel of ‘boot’ occupies the rear-most position. This leads to the following cross-classification:

The only bit that may need some additional explanation is the difference between [ʌ] “caret” and [ə] “schwa”. The former is the choice for stressed syllables and the latter for unstressed ones, such as the last syllable of ‘comma’, though I think that the two sound identical.

For a detailed overview of how difficult it is to describe English vowels, see the Vowel heading of Wikipedia’s English phonology page.

As for the full IPA chart of vowels, Wikipedia has a nice rendition with sample pronunciations at IPA vowel chart with audio.

With a little bit of imagination, you can see the vowels arrayed in a triangular fashion in Fig. 21, with the three points at [i], [a], and [u].

Question

All languages have at least [i, a, u]. Can you think of a reason why?

There is one vowel descriptor that I have skipped over.

Question

Say [u]. What are your lips doing? Now say [i]. Do your lips stay in the same position? How does [a] compare to the other two?

The bottom line of Fig. 21 summarizes the shape of the lips. For the articulation of a back vowel, the lips are rounded, which you can easily feel. A front vowel is articulated in the opposite manner, with the corners of the mouth drawn away so that the lips take on a slit-like shape, labeled “retracted”. A central vowel is articulated in between the two extremes, with a neutral shape.

Given how easy it is for you to feel what your lips are doing, lip shape would be a much more intuitive property of vowels than tongue advancement. Unfortunately, the tradition in phonetics relies on tongue advancement. Moreover, in English, lip shape can be predicted from tongue advancement – just read down from tongue advancement to find lip shape – so it adds no new information. Lip shape is therefore redundant, and I won’t have much to say about it in the upcoming text.

Lip shape is not redundant in all languages, though.

Say [i] and round your lips for [u] at the same time. Do you know this vowel? Likewise, say [e] and round your lips for [o]. How about this one?

The following two images illustrate how position in the vowel triangle translates into different configurations of buccal cavities:

The oral cavity itself sits in a channel between two smaller cavities whose size varies according to the position of the tongue and lips. Counting from the back, there is:

1. a pharyngeal cavity,
2. an oral cavity properly speaking, and
3. a labiodental cavity, between the teeth and the lips.

Notice how the difference in tongue position for [i] and [a] changes the size of the oral and pharyngeal cavities. For [i], the tongue is high and forward, shrinking the oral cavity while expanding the pharyngeal one. For [a], the tongue is low and centered, expanding the the oral cavity greatly while shrinking the pharyngeal one. In the next section, you will learn how such cavity shapes produce particular constellations of frequency.

But let’s fix in your mind the arbitrary nature of English spelling by transcribing a few vowels to IPA.

Practice. Indicate the IPA symbol for the vowel in the following words

 beef breast brat brew bra goose cook Brit brunt

#### Consonants¶

Have I defined what a vowel is? Well, imagine that you spoke like this:

Source: http://soundbible.com/ 1188-Chimpanzee by Mike Koenig.

Question

Does this sound like a vowel to you?

As you will see below, it certainly looks like a vowel, repeated three times. So imagine that you could only speak in vowels. How many messages could you convey? Unless you start mixing vowels, only about eleven. And even if you did mix vowels to make different messages, the transitions between vowels would be hard to perceive.

Humans have solved this problem by creating a whole new category of articulations, called consonants, which close the mouth around a vowel. Thus we propose the definitions of a vowel as an articulation in which the mouth is maximally open, in contrast to a consonant, which closes the mouth, partially or completely.

Consonants are distinguished by three characteristics, which you will figure out with a little prodding from me.

Task. Pronounce the consonant at the beginning of these words

‘paw, fee, thigh, sea, shy, key, hay’

Except for the last two, you should be able to feel distinctly what your mouth is doing.

In sequence, the articulators and the phonetic term used to describe them are:

Table 12 Place of articulation
example articulators term
[p] both lips bilabial
[f] the bottom teeth and top lip labiodental
[θ] between the teeth interdental
[s] the alveolar ridge on the gums right above the top teeth alveolar
[ʃ] the hard palate or roof of the mouth, where peanut butter sticks palatal
[k] the soft palate velar
[h] the back of the mouth behind the tongue glottal

This characteristic of consonants is known as place of articulation.

You can figure out the second characteristic of consonants by …

Task. Pronounce the consonant at the beginning of these words

‘paw, moo, Joe, Lou/roux, so, yo’

Again, you should have clear intuitions about what your mouth does. The following table puts them into prose:

Table 13 Manner of articulation
example articulators term
[p] the expiratory airflow is momentarily halted stop
[m] the mouth is closed but the expiratory airflow is diverted through the nose nasal
[ʤ] the expiratory airflow pauses but then resumes with turbulence affricate
[l, ɹ] the expiratory airflow exits through one side of the mouth or over the curled tongue liquid: lateral or rhotic
[s] the expiratory airflow exits with turbulence palatal
[j] the mouth is open enough that the expiratory airflow approximates a vowel glide or semiconsonant

This characteristic of consonants is known as their manner of articulation.

The third and final characteristic of consonants becomes apparent by contrasting pairs …

Task. Place the palm of your hand on your throat and pronounce the consonant at the beginning of these words. Do you notice a vibration or not?

‘fee/bee, thin/then, sue/zoo, bash/badge, hey’

In the first member of each pair, plus ‘hey’, you should feel nothing in your throat, while in the second you should feel a distinct buzzing. This is called voicing. A consonant that lacks it is unvoiced or voiceless; a consonant that has it is voiced.

The three features cross-classify the consonants as depicted in this table:

Here are some examples of the symbols that you may not be familiar with:

Table 14 Obscure IPA consonant symbols
symbol example
[ʔ] Uh-oh!
[ŋ] Ang
[ʧ] chew
[ʤ] Joe
[θ] thigh
[ð] thy
[ʃ] shy
[ʒ] edge

One closing comment is in order. [w] is listed twice because it has two closures, a bilabial one and a velar one. It is known as a labiovelar, which doesn’t fit comfortably into the table.

Question

What is the voicing of the nasals, liquids, and glides?

Let’s continue fixing in your mind the arbitrary nature of English spelling by forcing you to focus on the actual pronunciation of a handful of words.

Practice. Transcribe the following words into the IPA.

 sting queen unique gin button enough rose emotion finger

A consonant can be identified by its three features, usually in the sequence “voicing manner place”. For instance, [b] is a voiceless bilabial stop, while [ʤ] is a voiced palatal affricate.

Practice. Identify the following symbols by their features

 [s] [d] [h] [l] [m] [j] [ʧ] [ʔ] [ð]

From these articulatory characteristics of the mouth, we can jump to their acoustic consequences, to get you started with actually understanding how the brain looks at speech.

## Acoustic phonetics¶

Phonetics (and phonology) makes an initial distinction between those aspects of speech that tend to be captured by an alphabet and those that are captured by punctuation, called segmental and supra-segmental or prosodic acoustic phonetics, respectively.

### Segmental acoustic phonetics¶

A segment is the basic unit of phonetic analysis. Each segment in a language can by written by a symbol drawn from the International Phonetic Alphabet.

#### Vowels¶

Here is a spectrogram of my pronunciation of the vowels [i, a]:

The frequency tracing at the bottom is marked by distinctive horizontal bands, called formants. They are named by counting them from the bottom up. In theory, the initial one is the fundamental frequency, $$F_0$$, but it is too low to be readable in such a broad range of frequencies. The next one, the first formant $$F_1$$ is the lowest for both vowels but is at different heights. They continue upwards, but it is usually just the bottom two that are most informative. This image picks them out for you:

Question

What do the colors mean?

Rob Hagiwara has a wonderful series of spectrograms of the non-schwa English vowels, over which I have superimposed colored bars to highlight how the formants rise or fall with vowel quality:

In the red bars, you should be able to discern that as $$F_1$$ rises, tongue height falls. That is, the higher $$F_1$$, the lower the vowel is in The vowel triangle of American English.. Thus $$F_1$$ is inversely correlated to vowel height. As for the orange bars, you should be able to make out that as $$F_2$$ rises, tongue advancement increases. Which is to say that, the higher $$F_2$$, the further forward the vowel is in The vowel triangle of American English.. Thus $$F_1$$ is (directly) correlated to tongue advancement.

Questions

• Should [i] have the largest value for $$F_1$$ or $$F_2$$?
• Should [i] and [u] have the smallest values for $$F_1$$ or $$F_2$$?
• Should [a] have the smallest or greatest value for $$F_1$$?

#### Consonants¶

The acoustic repercussions of consonant closures are varied.

Most consonants must be articulated along with a vowel. The image below analyzes [abadaga], from the files speech-abadaga.wav and speech-abadaga.TextGrid:

Question

Looking just at the two waveforms and knowing that a consonant is a constriction of the airflow, where do you think [b, d, g] are?

Underneath the spectrogram, Praat opens up a tier for me to label the waveform. I divided it into its segments and left the second [a] selected. The constriction for the three voiced stops is where the pressure drops abruptly and the spectrogram goes cloudy. My closure on [b] was not very crisp – you can see how the vowel almost continues straight through it, but I succeeded a bit better on [d] and [g] – especially [d], which clearly interrupts the acoustic signature of the surrounding vowels.

Turning to the fricatives, the image below displays [fθsʃh], from the files speech-fthsshh.wav and speech-fthsshh.TextGrid:

Question

What is the range of frequencies?

While I have maintained that fricatives sound like wind – or random frequencies – and so can’t bear any information, not all random bunches of frequencies are created equal. [ʃ] has a band of energy starting at about 2205 Hz, [s] has one at about 3550 Hz and [θ] has one at about 5220 Hz. [f] might have one in the vicinity of 7500 Hz, it’s hard to tell. In contrast, [h] appears to have no energy at all, though it does suggest a ghostly background of formants. Looking at it from front to back, fricative energy appears to step downwards as the place of articulation moves back. [h] is said to have no constriction at all, so it is just air passing through the mouth and may reflect the basic architecture of the vocal tract.

#### Spectral vs. temporal information¶

The notion of spectrum can be generalized to the continuum of frequencies of sound, and in particular to those of speech. The figure below superimposes the spectrum of the rainbow onto the first four formants of “BP”:

The spectral information is fundamental to identifying the sounds in Fig. 29, but the temporal information should not be overlooked. The vowels are much longer than the consonants, while the transitions between them are fairly crisp, usually taking only between two and four vibrations of the vocal folds. Fig. 30 overlays the previous spectrogram with long and short windows of time that highlight representative steady-state frequencies of the vowels and several frequency transitions:

Question

Why is the long window red and the short window blue?

The long window measures about 180 ms, while the short one measures about 30 ms. Dividing 1000 ms by each one, 1000/180 and 1000/30 rounds out to 6 and 33 Hz, respectively. That is to say, six red windows and thirty-three blue windows fit into a second, so each window can be understood as a cycle and thus be measured in Hertz, as a frequency. As a low-frequency entity, the long window is labelled red; conversely, as a high(er)-frequency entity, the short window is labelled blue.

### Supra-segmental or prosodic acoustic phonetics¶

Prosody is the quality of spoken language that provides its melodic contour and rhythm, features which help the hearer to decode syntactic and lexical meaning as well as emotional content. Prosody differentiates, say, the neutral statement of fact “It’s my fault” from the sarcastic question-like rejoinder “It’s MY fault?”.

Such distinctions are produced by variation in three parameters, which are borne in turn by three qualities of sound waves, respectively:

Table 15 Physical origin of prosodic parameters
Sound waves Prosody
fundamental frequency pitch
intensity stress
timing duration

Question

Which one of these parameters should already be able to elaborate a hypothesis for?

I invite you to contemplate the graph of three quantities in Fig. 31, measured from its source:

The top panel displays the spectrogram with its scale on the left, along with a plot of the loudness or intensity of the sound in deciBels, with its scale on the right. The high-frequency spectrogram contrasts with the low-frequency pitch of the speaker’s voice beneath it in red. Her voice barely touches the 90 Hz average of my own. The lower panel splits the sound wave into phrases.

The first phrase ends with a falling pitch, which signals declarative phrases in English. The second phrase also ends falling pitch, typical of questions introduced with an interrogative pronoun in English. The drop is much more abrupt, which is heard as the curious change in intonation of the word “on”. The third phrase repeats the contour of the second, but its force is declarative. The final one ends with a plunge, which signals the end of the entire complex sentence.

Here are three different ways of visualizing my pronunciation of “the first thousand”:

The phones are named across the bottom of the image. The graph above them labeled “Amplitude” displays the sound as a wave oscillating up and down, whose displacement tracks the oscillation of the vocal folds. It is curious to see how uninformative it is. Could you use it to distinguish, say, [f] from [n]?

It is the graph above it, labeled “Frequency (Hz)” on both sides that contains the information which allows us to distinguish [f] from [n], and hopefully from all the other speech sounds, too. This graph analyzes the sound wave into its component frequencies, in two ways. Most of the graph depicts the broad range of frequencies labeled on the left scale. This is what we hear as speech. The three clumps of dots, in contrast, are set on the narrow range of low frequencies labeled on the right scale. This depicts the low intonation of my male voice – note that it averages about 89 Hz. It is such a tiny bit of the full range that it would be smashed down into a featureless line at the very bottom of the graph, which is why it is promoted to its own scale on the right.

The goal of speech perception is to transform the continuous waveforms graphed in the top of the figure into the neat parcellation of symbols in the bottom. The goal of speech comprehension is to figure out what this string of symbols means.

The former may look easy, especially since I have already done it for you. Yet if you look at the frequency plot up close, you will see a transition at each edge that is a mixture of the two abutting segments, neither fully one nor the other. And if you listen to the transitional areas in the recording yourself, you will not hear any such sharply drawn distinctions.

### Voice onset time (VOT)¶

Question

Say the name BP. Do you notice any difference between the first and second consonants?

Here is my pronunciation, just in case its not clear.

Below is its rendering into Praat. You can download the recording and its TextGrid, too in order to reproduce Fig. 33.

The ‘b’ and the ‘p’ are articulated rather differently. The ‘p’ is followed by a puff of air called aspiration and transcribed as a superscript ‘h’, [ʰ], so that the entire segment would be [pʰ].

Aspiration is not just a quirk of ‘p’.

Question

Pronounce the pairs ‘Dan/tan’ and ‘goon/coon’. Which is aspirated? Refer back to Fig. 23 and try to come up with a succinct statement of which consonants have the potential to be aspirated.

Returning to the bilabial stops,

### Lexical prosody¶

Noun vs. verb in English (±15)

CONvert vs. conVERT

Thai, a tone language

• naa with a rising pitch tone means “thick”
• naa with a falling pitch tone means “face”

### Phrasal prosody¶

Compound noun rule noun phrase: hot DOG (a dog that is hot) adjective+noun: HOTdog (a frankfurter) noun+noun: SHEEPdog (a breed of dogs) Stress retraction After eating fourTEEN, CAKES did not tempt him. After eating FOURteen CAKES, he threw up.

### Clausal prosody¶

Contrastive (or emphatic or focal) stress

Examples

The horses were racing from the BARN. The HORSES were racing from the barn.

Sentence type

Types declarative: fall in pitch at end I eat chocolate. interrogative: rise for yes-no question (a); fall for interrogative pronoun (b) Do you eat chocolate? What do you eat? imperative: even pitch throughout; rise in intensity at end Eat chocolate!

## The next topic¶

Come to class having read Perception and answered the questions.

Last edited Aug 28, 2019