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# Thread: Hessian

1. ## Hessian

Hello there I am investigating catastrophe theory although I have discovered something in a book I am reading that I dont understand

we consider a smooth function of two variables with $f(0,0) = f_x(0,0) = f_y(0,0) = 0$, i.e. there is a critical point at the origin, so the Taylor series is

$f(x,y) = ax^2 + 2h x y + b y^2)$+ higher order terms

where $a = f_{xx}, b = f_{yy}, h = f_{xy}$

We consider the case when the hessian $ab - h^2 = 0$, and that not all the derivatives are zero

So the book says that this means $|ax^2 + 2h x y + b y^2 |$ is a perfect square and allows us to write

$|f(x,y)| = \frac{1}{2}(\sqrt{|a|}x + \sqrt{|b|}y)^2$ + higher order terms

BUT!

if we do $f(x,y) = \frac{1}{2}(2x^2 - 4 x y + 2 y^2)$, all the conditions are satisfied but

$|f(x,y)| \not=\frac{1}{2}((\sqrt{|a|}x + \sqrt{|b|}y)^2 = \frac{1}{2}(2x^2 + 4 x y + 2 y^2)$

could anyone explain? thanks very much

2. Just replace y with -y. Wouldn't that fix the problem? It seems to me that the issue is whether you allow negative y's in the picture. Why wouldn't you allow negative y's? It doesn't disprove what the author is saying.


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## Sunday, April 4, 2010

### ## More GHCN results.

I've put up version v1.1 of my least squares GHCN code. It now makes it possible to easily analyse subsets using any of the information on the inventory file. I've used it to calculate a number of much discussed examples. The code now runs very fast. On my fairly old PC, the biggest case (global) takes less than 100 secs. So although each special case is calculated pretty much from scratch, there's not much penalty there. The code is here, and also on the document repository.

## New code features.

The program has now been split into two. The first is a preprocessor which reads v2.mean, does the merging of duplicates, and prints a condensed file called v2mean.txt. Unlike ver 1, it doesn't require prior editing of v2,mean. The idea is that once you have merged the duplicates, you don't need to do it again un til you want to deal with a new version of v2,mean.

The second part is somewhat like V1, but much more clearly laid out. It's even shorter (with the preprocessing gone) - about 170 lines. There's a lot of infrastructure with printed tables, plots with trends etc.

The selection process works thus. The 18 parameters from the .inv file have been given names, like tv.$country, listed in colnames. You declare a list of names that you'll want to use and have appear on graphs etc: title = c("Globe","NH","SH","ConUS","Airport","Rural","Urban"); choices = 1:7 #c(4,5,6,7); In the second line, you declare the ones you want to run. The logic of the choices is expressed in the list: if (tc == "NH") i01 = i01[cellnums>1296]; if (tc == "SH") i01 = i01[cellnums<1297]; if (tc == "Airport") i01 = i01[tv$airpt == "A"];
if (tc == "Rural")   i01 = i01[tv$urban == "A"]; if (tc == "Urban") i01 = i01[tv$urban != "A"];
That's my current list, but you can change it.

## Some results

Firstly, a table of trends, in deg C/decade.
 Trends Globe.... NH........ SH......... Airport.. Rural... Urban No. Stations.. 7280 6056 1224 2390 2712 4568 1900-2009 0.0760 0.0824 0.0690 0.0596 0.0606 0.0752 1978-2009 0.2717 0.3144 0.1106 0.2628 0.2561 0.2631

No real surprises. The main outlier is the SH, but remember, this is land stations only. The classifications of airport, rural and urban make little difference. The urban/rural differentiation is by night lighting, and I've included small towns with urban.

Remember that this is based on raw GHCN data, with no corrections made by me.

Update - there was a bug in V 1.1, which had a minor effect on the results, but which could cause a crash. It didn't affect global calcs - only subregions. Here are the recomputed trends:
 Trends Globe NH SH Airport Rural Urban No. Stations 7280 6056 1224 2390 2712 4568 1900-2009 0.0760 0.0833 0.0479 0.0661 0.0678 0.0761 1978-2009 0.2717 0.3297 0.1199 0.2438 0.2599 0.2721
I'll fix the posted code in all locations, calling it v1.11, and the results file.

I've posted the results file on the document repository.

## Some plots

 NH 1900-2009 SH 1900-2009 NH 1978-2009 SH 1978-2009 Rural 1900-2009 Urban 1900-2009 Rural 1978-2009 Urban 1978-2009 Airport 1900 Airport 1978

#### 1 comment:

1. Nice work Nick... something else you may want to do: add uncertainties to your series and trend fits. Part of doing the trend fit is remembering that your uncertainty generally is increasing as your number of stations drop (especially before 1950), i.e., so really one also needs to weight by year, if you fit to an interval before that.

Of course, this is already done in CRUTEM3, so if you haven't seen it, here is a plot including 95% uncertainty bounds

format: plain


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# Math Help - identity

1. ## identity

((sin x)^2 - (cos x)^2) / (sin x- cos x) = sin x + cos x

I want to change the LHS only.

2. Originally Posted by Stuck Man
((sin x)^2 - (cos x)^2) / (sin x- cos x) = sin x + cos x

I want to change the LHS only.
Remember $a^2-b^2=(a+b)(a-b)$? If so, apply that to the LHS, then cancel the common factor $a-b$ of numerator and denominator.


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# How would the night sky look from inside a globular cluster?

When the weather is clear, we can look at the stars. And we normally would see several thousands of them, they all being more than a $\textrm{pc}$ away from us.

Now, there are globular clusters, which consist of some $10^5$ stars concentrated is a few $\textrm{pc}$ area. From oustside they look like this:

Now, how do they (and the whole sky) look from inside? Imagine that solar system is inside such a cluster. Would it be a big change? Would there be a significant difference between night and day? Would it be easier or harder to study astronomy there?

P.S. The credits for the question go to Ross Church.

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I'm not sure, but I would think that there is no stable habitable zone within a globular cluster (GC), except perhaps in its outer parts. The stellar density is so high that Solar-system-like objects will be destroyed/destablised by stellar encounters. Hence life as we know it may never form there. Moreover, the fraction $Z$ of non-primordial elements ("metals" in astronomy slang) is very low, at least in the old GCs of the Milky Way. Low $Z$ stars are much less likely to host planets. Thus, there may be nobody so observe the GC night sky. –  Walter Dec 25 '13 at 13:51
Dear @Walter, yes, this is correct. Solar system would not be stable for more than some $10^3$ years and would not even have formed there. However, I am still curious to know, how would the sky look like if one managed to take the Sun and the Earth and put them into such a cluster. –  Alexey Bobrick Dec 25 '13 at 14:05
The Asimov story Nightfall describes this very situation. –  dotancohen Dec 27 '13 at 9:05

Globular clusters occupy an interesting place in the spectrum of composite stellar systems. As you point out, they are highly concentrated populations of stars, and seem to lack any dark matter component, unlike more massive dwarf galaxies.

Binary interactions become very important in simulating globular clusters, and interestingly enough (maybe unsurprisingly), the one example of a discovery of a planet found in a globular cluster has been around a binary star system (see: PSR B1620-26 b; this circumbinary planet was found orbiting a pulsar and a white dwarf.). This is not to say there are not other examples, however, this was the easiest for me to come across. I would be interested to know how common this situation is, and in addition, how stable it is given the potentially highly chaotic environment it lives in. These speculations don't answer your question, but I thought it interesting enough to bring up as evidence in favor of your question not being an unreasonable one to ask.

From the wiki page:

Globular clusters can contain a high density of stars; on average about 0.4 stars per cubic parsec, increasing to 100 or 1000 stars per cubic parsec in the core of the cluster.[26] The typical distance between stars in a globular cluster is about 1 light year,[27] but at its core, the separation is comparable to the size of the Solar System (100 to 1000 times closer than stars near the Solar System).[28]

This seems to indicate to me that location within the globular cluster would matter quite a bit. If at the core the average distance between stars is about three thousand times closer than our nearest neighbor is to our sun (my estimate to give some perspective: a few lightyears to Proxima Centauri divided by 100 is about 3000AU (about 100 times further than Pluto from the sun)), then stable orbits may be shifted inward, or simply may not exist due to two-body interactions.

However, if life were to exist (an assumption we're going to make for the purposes of your question), one would see a very different night sky. According to this paper, the number density profile of stars within the globular cluster M92 follows a Wilson Profile fairly well, which has the form:

$$f_{W} = A\{ e^{-aE} - e^{-aE_{0}} [1 - a(E-E_{0}) ] \}$$ where $E \le E_{0}$. E is the specific energy of the star:

$$E = v^{2}/2 + \Phi(r)$$

and where $\Phi(r)$ is the mean-field gravitational potential, determined from the Poisson equation. For each family of models, the constants A, $E_{0}$, and a in the above distribution function define two dimensional scales (a typical radius and a typical mass or velocity) and one dimensionless parameter, the central depth of the potential well (related to the concentration parameter) (all of the information is taken from the paper I've linked).

It seems to be the case that globular clusters are not "simple stellar populations", in that they are usually made of multiple generations (sources: 1,2). However, globular clusters generally consist of population II stars and are older stellar systems when compared to other star clusters. I bring all of this up because in addition to the number density of stars, the stellar distribution of stellar types would certainly be an important factor in how the night sky would look. If you lived a thousandth of a lightyear from a blue supergiant, you could imagine that that would make a huge difference in what you would see on a day to day basis. At the same distance, a supergiant star is on the order of $10^{5}$ times the luminosity of our sun (and therefore is $10^{5}$ the flux, since $L \propto f$ holding $D_{L}$ constant). At the same distance as our sun, the magnitude of a star with $10^{5}$ times the flux would have an apparent magnitude of about -38 (I used Rigel as my test case; this produces a star in our sky which is 12 magnitudes brighter than our sun). Moving this to the average distance between stars at the center of a globular cluster we would get an apparent magnitude of:

• $M = -6.43$ from $m=-38$ at distance of $d=1$AU.
• $m=-26.43$ from $M=-6.43$ at a distance of $d = .00326 ly$ (the new average distance between stars at the center of a globular cluster)

In other words, a blue supergiant at an average distance between stars within a globular cluster would appear to be as bright as our sun is to us! This is absolutely crazy. Depending on where it is in relation to the sun, it could effective cause two days, or potentially one day which is greater than half the time it takes your planet to rotate once. I would image that this would certainly interfere with observing in optical (and shorter wavelengths).

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Great answer, thank you very much! I think you can easily expand it a bit to make it complete. I am not sure if supergiants are important in cluster cores. In any case, I don't think they dominate in total luminosity. A relevant one-liner here would be just to get the total flux from all the stars within half-mass radius as seen from the core and compare it to that of the Sun. Then one other small thing, which seems confusing is the usage of $m$ and $M$, former is typically used for apparent and latter for absolute magnitudes. Thank you very much again, it was very interesting and informative! –  Alexey Bobrick Dec 26 '13 at 19:01
Sure - supergiants are probably not the most common stellar type, but I wanted to point out that there could potentially be a drastic change in what you might see at the core of a globular cluster. I am also using M for absolute magnitude and m for apparent magnitude. I just presented it in reverse order. Also, I couldn't find values of A, $E_{0}$, and a for a typical globular cluster, though I will look for them. Once I do find them I can do exactly as you say - finish off the calculation regarding how many stars and of what types within some volume from the test observer's point of view. –  astromax Dec 26 '13 at 20:21
I guess, it would be "hard" to do, because you will need then to get the grav.potential, which you don't yet know. At least, it wouldn't be a one-liner. A simpler approach would be to take a gaussian cluster (in velocities and positions), or take an isothermal model. In any case you will find that the relevant flux will be coming from a sunlike star located at $R_{core}/N_{*,core}^{1/3}$. –  Alexey Bobrick Dec 26 '13 at 21:00
Just as a sidenote, another interesting question would be whether anything would change if the cluster were or were not core-collapsed. Also, you are right, stars leaving post-MS would make a difference now and then, when viewed from within the cluster. –  Alexey Bobrick Dec 26 '13 at 21:02

In a typical position in a globular cluster (maybe halfway between center and edge), there'd be many more bright stars in the sky due to the star density. These would be distributed unevenly in the sky, with more light coming from the center of the globular cluster.

Depending on the globular cluster's orbit, we might be able to see the Milky Way face-on. This would be beautiful and would take up a large fraction of the sky. However, it wouldn't be particularly bright--the apparent magnitude would be $\approx -3$, the brightness of Jupiter.

If our alien eyes were good at discerning wavelengths, the sky would seem redder. Globular clusters have many old stars (more red), and very few massive stars (fewer blue) since they are not sites of active star formation.

-

Let us assume the data for a globular cluster to be equivalent to that of M13.

Given 300,000 stars and a radius of 1 ly, let us assume uniform density.

Another assumption is to consider all stars to be Sun-like.

The number density can be calculated as $$\rho = \frac{300000*3}{4\pi(1\ ly)^3} \approx 9 \times10^{-44}\ m^{-3}$$

Now, using the formula for mean free path $$\lambda = \frac{1}{\sqrt{2}\pi D^2 \rho}$$, and using $D = 14 \times 10^{8}\ m$, we have $$\lambda \approx 10^{24}\ m \equiv 10^{8}\ ly$$.

Now you don't have to be a genius to see how the sky will look if the nearest stars to you are this far. It will be almost like our own sky but just a lot of stars in every direction. There will be no special hike in flux received.

The number density is much higher at the centre. But even if you assume the density at centre to be orders higher than the average value, the uninteresting view exists! As Zack pointed out, we'll have a lot of long wavelength light due to abundance of old stars.

It is mentioned that the optical view won't be too attractive, but there is a certain thrill in being at the centre of a globular cluster. It is very difficult to stay there for a long time avoiding collisions or surviving the radiation and stellar winds from collisions and novae which are frequent for globular cluster.

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Many thanks! As a comment, I don't see why would mean free path estimate would be relevant here. Perhaps you mention it for estimating the maximum possible lifetime of a planetary system. But then you need to account for gravitational attraction, which will decrease $\lambda$ significantly. Then also, the average distance between the stars will be in fact of order $pc/N^{1/3}\approx 10^{-2} pc$, which is a few thousands of AU, which is not too much, but not too little, given that you have quite many stars here. –  Alexey Bobrick Dec 26 '13 at 14:35
No, I use it as a rough estimation of the order of mean distance between any two stars in the globular cluster. –  Cheeku Dec 26 '13 at 14:38
but then it gets really brainexploding: if the cluster is a parsec in size, how can you imagine the stars been separated by $10^8$ parsec in it? –  Alexey Bobrick Dec 26 '13 at 14:48
Yes, this particular estimate is highly dependent on number density which I have assumed to be uniform. I searched through lots of articles and have lengthy calculations on my notebook but then you needed an idea of how it would "look" so... –  Cheeku Dec 26 '13 at 15:05
@AlexeyBobrick Now, I see. I got something to learn from my own answer. Good! –  Cheeku Dec 26 '13 at 23:47

In fact, some people had looked at it more seriously recently and conducted a computer simulation to visualize the night sky as seen from within a globular cluster.

The article has appeared recently in Astronomy journal.

This is just one example of a typical image inside a globular cluster:

Some more discussion can be found here: http://io9.com/what-the-night-sky-would-look-like-from-inside-a-globul-1589324556

And the journal issue containing the article by William Harris and Jeremy Webb can be found here: http://www.astronomy.com/magazine/press-releases/2014/05/july-2014

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# Math Help - stupid test question.

1. ## stupid test question.

i DO understand how to do this, i just cant remember how to get all of the numbers in the equation in order to solve it.
"The population of a town is 10000 people. it grows at a rate of 1.4% per year for four years, what is the growth factor, b, for the equation y=a*bx (x being the small integer over the b)"

i get that a = 10000
i get that b = 1.4
i get that x = 4
but none of the answers i come out with is in the multiple choice.

2. Originally Posted by autumn
i DO understand how to do this, i just cant remember how to get all of the numbers in the equation in order to solve it.
"The population of a town is 10000 people. it grows at a rate of 1.4% per year for four years, what is the growth factor, b, for the equation y=a*bx (x being the small integer over the b)"

i get that a = 10000
i get that b = 1.4 <<<<<<<< where did you get this number from?
i get that x = 4
but none of the answers i come out with is in the multiple choice.
1. If x = 1 then the population is:

$p = 10000 + \frac{1.4}{100} \cdot 10000 = 10140$

2. Plug in these values and calculate the value of b.

3. Hello, autumn!

Your explanation could be clearer . . .

The population of a town is 10,000 people.
It grows at a rate of 1.4% per year for four years.
Find the growth factor $b$ for the equation: . $y\:=\:a\cdot b^x$

When $x = 0,\;y = 10,\!000$

We have: . $10,\!000 \:=\:a\cdot b^0 \quad\Rightarrow\quad a \,=\,10,\!000$

. . The function is: . $y \:=\:10,\!000\cdot b^x$

A little baby-talk:

After one year $(t = 1)$, the population increased by 1.4%.
. . The increase is: . $0.014 \times 10,\!000 \,=\,140$
Hence, the new population is: $10,\!140$

We have: . $10,\!140 \:=\:10,\!000\cdot b^1 \quad\Rightarrow\quad b \:=\:\frac{10,\!140}{10,\!000}$

Therefore, the growth factor is: . $\boxed{b \:=\:1.014}$

. . The population function is: . $y \;=\;10,\!000\cdot 1.014^x$

4. ## hmmm. thanks.

thank you. these answers will realy help me in the long run if i just sit and mess with that formula.
i feel stupid asking such a simple question though.

5. ## explination

i got the 1.4 because it is the population growth rate. population times growth rate to the power of the number of years. am i wrong?

6. Originally Posted by autumn
i DO understand how to do this, i just cant remember how to get all of the numbers in the equation in order to solve it.
"The population of a town is 10000 people. it grows at a rate of 1.4% per year for four years, what is the growth factor, b, for the equation y=a*bx (x being the small integer over the b)"

i get that a = 10000
i get that b = 1.4
i get that x = 4
but none of the answers i come out with is in the multiple choice.

Hi Autuum,

So what did you get for the growth factor.Did you divide the % increase by 100? How many people will there be after one year? Can you see how to expand for 4 years?

bjh

7. Originally Posted by autumn
i got the 1.4 because it is the population growth rate. population times growth rate to the power of the number of years. am i wrong?
You got 1.4 for what? The rate of change was given as 1.4% or 0.014 so the population will increase by 0.14 of the original amount. Added to the original amount, the population will be 1.014 of the original amount which means that in one year the population is multiplied by 1.014. That, as Soroban told you, is b.

8. ## i get it now

at first i couldnt remember to divide the percentage (or 0.14) by the original number. i knew i was missing a step, but i couldnt find it. but now it makes a lot of sence. im dont have an algabra teacher. im on my own in this subject, hence my ignorence. your help is appreceated.
(also... please disregard my spelling. or lack there of)


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# Remarks of Body.IsSolid Property

Valued Contributor

Hello everyone,

In the Remarks of Solid Edge Geometry Type Library > Body Object > IsSolid in programming reference, It states:

"This does not necessarily imply that each Face, for example, is connected at all its edges. The Body may still consist of standalone Faces that collectively define a finite volume."

So, what does it mean "standalone Faces" and what kind of model it is? Can anyone give me an example of this kind of body consisting of "standalone Faces"? (I mean how to create this kind of body in Solid Edge manually?)

Thanks,

Guan

# Re: Remarks of Body.IsSolid Property

Valued Contributor

Hi, all

Body can hold standalone faces. The case exists. Standalone faces cannot be created by SE manually (I couldn't), but may be imported somehow.


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# Find reaction heat of a reaction [closed]

1) A(s)+O2(g) →Ao2; ΔH^0=-394kj/mol

2)AO(g)+ $\frac{1}{2}$O2(g) →AO2; ΔH^0=-283kj/mol

3)A(s)+ $\frac{1}{2}$O2(g) →AO(g); ΔH^0=-Xkj/mol

Find the value of X? I want to know How to find this value.

## closed as off-topic by airhuff, Martin - マーチン♦May 16 '17 at 6:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

You need to reverse the 2nd reaction. This leaves you with Ao2 ---> frac{1}{2}O2(g) + AO(g). And since you're reversing the reaction, you have to switch the sign of delta H, so it becomes 283 kj/mol (endothermic). Once you do this, you can cancel out the AO2 and frac{1}{2}O2 with the 1st reaction. This leaves you with your desired reaction, that is, A(s) + frac{1}{2}O2 ---> AO(g). That is the reaction you want. To find delta H for that reaction, simply add the delta H's from the previous reactions. So, -394+283 = -111 kj/mol. So your X-value is 111. This process works because of Hess's Law. In fact, it is a classic example of how Hess's law would be implemented in chemistry.

To wrap up: The reason we reversed the 2nd reaction was because doing that allows us to cancel out some stuff with the 1st reaction in order to get the 3rd reaction. Our goal is to get the third reaction you described. The reason we switched the sign of delta H (from -283 to 283) is because if a reaction is exothermic (negative delta H) in one direction, it is endothermic (positive delta H) in the reverse direction. That is just a rule of thumb; you can obtain further understanding of such concepts by learning about bond energies, activation energies and the like.

You can get further understanding of why Hess's Law works, as well as practice, here.


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# PITCH

The PITCH function computes the second angle, in radians, of a body-3 Yaw-Pitch-Roll rotation sequence.

PITCH(I, J)

## Description

Given two coordinate systems, I and J, which are initially having the same orientation, this rotation sequence is defined as follows:
• First, rotate in a counterclockwise direction about the z-axis of the I coordinate system [YAW].
• Second, rotate counterclockwise about the negative y-axis of the I coordinate system [PITCH].
• Third, rotate counterclockwise about the x-axis of the I coordinate system [ROLL].

The first argument, marker I, must be specified. The second argument, marker J, is optional.

## Arguments

I
The marker whose rotational displacement is to be computed.
J
The marker with respect to which the rotational displacement is to be computed. This argument is optional. If omitted, it defaults to the ground coordinate system.

## Example

<Post_Request
comment   = "Vehicle Orientation (degress)"
id        = "70000256"
type      = "EXPRESSION"
expr1     = "NULL"
expr2     = "YAW(30301010,30101010)*RTOD"
expr3     = "PITCH(30301010,30101010)*RTOD"
expr4     = "ROLL(30301010,30101010)*RTOD"
/>

## Comments

The figure below shows the physical meaning of the Yaw-Pitch-Roll rotation sequence.


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# A Neighborhood of Infinity

## Saturday, October 10, 2009

### Vectors, Invariance, and Math APIs

Many software libraries, especially those for physics or 3D graphics, are equipped with tools for working with vectors. I'd like to point out how in these libraries the functions for manipulating vectors sometimes have special and useful properties that make it worthwhile to separate them out into their own interface.

### Two types of Function

Suppose an object of mass m is moving with velocity v and we apply force f to it for time t. What is the final velocity? This is given by g:

g(m,t,f,v) = v+(t/m)*f

Now suppose that R is a rotation operation, typically represented by a matrix. What happens if we apply it to both of the vector arguments of g?

g(m,t,Rf,Rv) = Rv+(t/m)*Rf = Rg(m,t,f,v)

In other words, rotating the vector arguments is the same as rotating the vector result.

Another example: Consider the function that gives the force on an electric charge as a function of its velocity and the magnetic field:

f(e,v,B) = ev×B

It's essentially just the cross product. If you rotate both of the arguments to the cross product then the result is rotated too. The result is that

f(e,Rv,RB) = Rf(e,v,B)

On the other hand, many 3D APIs come with a function to perform componentwise multiplication of vectors. Write vectors x as triples (x0,x1,x2), and so on, we can write such a function as:

f(x,y) = (x0y0,x1y1,x2y2)

You can show that this doesn't have a similar property.

### Rotational Invariance

To make things easy, let's restrict ourselves to functions of scalars and vectors. And when I say vector, I'm talking strictly about vectors representing magnitude and direction, but not positions. Examples of such vectors are velocities, accelerations, angular velocities, magnetic fields, and the difference between two positions. A function is said to be rotationally invariant if applying a rotation R to all of its vector arguments results in the same thing as applying R to all of the vectors in its value. This allows you to have a function that returns multiple vectors, like a tuple or array.

The first two functions I described above were rotationally invariant but the third wasn't. Notice how the first two examples also described physical processes. This is the important point: as far as we know, all of the laws of physics are rotationally invariant. If you write down an equation describing a physical process then replacing all of the vectors in it by their rotated counterparts must also result in a valid equation. So if you're describing a physical process with a computer program, and you end up with a function that isn't rotationally invariant, you've made a mistake somewhere.

### Vector APIs

Vector APIs frequently come with all manner of functions. Some have the invariance property and some don't. If you write code that you'd like to be rotationally invariant, but it turns out that it isn't, you usually have to examine the code to find the bug. But if you separate the invariant functions into their own interface, and then write code using just that interface, the code is guaranteed to be invariant. If your programming language has reasonably strict types then you may even be able to arrange things so that the type signature of the function alone is enough to tell you that the function is invariant. In effect you are able to make the compiler prove that your function is invariant.

(As an aside, this is an example of why a good type system does much more than you might at first have guessed. They don't just stop you making typos, they can do things like prove that your programs satisfy certain geometrical properties.)

So what functions would you have in such an API? Among the essential rotationally invariant functions are:

1. Multiplcation of a vector by a scalar
3. Dot product
4. Cross product

In terms of these you can build functions such as

1. Vector length
2. Vector normalization
3. Rotation of one vector around an axis specified by another vector
4. Linear interpolation between vectors

What kinds of functions would be excluded?

1. Constructing a vector from three scalars, ie. f(x,y,z) = (x,y,z).
2. Constructing a vector form a single scalar, ie. f(x) = (x,x,x).
3. Extracting the ith component of a vector, ie. f(i,(x0,x1,x2)) = xi.
4. Pointwise multiplication of vectors.
5. Computing the elementwise cosine of a vector.

On seeing the first excluded example above you might ask "how am I supposed to construct vectors?" The point is that you don't program exclusively with an invariant API, you simply use it whenever you need to prove invariance.

### Coordinate System Invariance

One purpose of writing to a particular interface is that it allows the API to hide implementation details from the user. Using a rotationally invariant API has a role to serve here. For example, many 3D renderers allow you to write shaders. These are essentially functions that compute the colour of a piece of geometry that needs rendering. You write a shader and the renderer then calls your shader as needed when a fragment of geometry passes through its pipeline. Frequently these are used for lighting calculations but there are all kinds of other things that may be computed in shaders.

In a 3D renderer different parts of the computation are often performed in different coordinate systems. For example it may be convenient to perform lighting calculations in a coordinate system oriented with the direction of the light. But the author of a renderer doesn't want to be committed to a particular choice. In order to do this, it is essential to be able to write shaders that are agnostic about which coordinate system is being used. If we work with rotationally invariant functions, our shaders are guaranteed to be agnostic in this way (assuming that the only kind of coordinate change that takes place is a rotation).

### Lots More Types

I've concentrated on just one type of invariance, rotational invariance. If we consider more types of invariance then more types of interface naturally emerge. It would take too long to cover all of the details here so I'm just going to briefly sketch the beginnings of the more general case. So just read this section as a list of pointers to further reading.

For example, some functions are invariant under translations. These can be thoght of as functions of points in space. If we allow more general linear transformations then we find that some common functions transform 'oppositely' to vectors. In particular, normals to surfaces transform in this way. In fact, Pixar's Renderman has three distinct types, vectors, points and normals that captures these different invariances.

If we go back to rotations again but now extend these by allowing reflections then we find an interesting new phenomenon. For example, consider the result of reflecting in the x-y-plane, followed by reflecting in the y-z-plane followed by reflecting in the x-z-plane. This simply multiplies vectors by -1. Dot product is invariant under this: (-x)·(-y)=x·y. But cross product isn't because (-x)×(-y)=x×y. Even though the cross product is apparently vector, it doesn't get multiplied by -1. When we start considering invariance under reflection we find that some vectors behave differently. These are the pseudovectors and in effect they have their own separate type and interface. Interestingly, nature likes to keep pseudovectors and vectors separate except in parity violating phenomena. There are even pseudoscalars.

Incidentally, if you consider invariance under scaling you're led to the idea of encoding dimensions in your types.

### Conclusion

If you're writing a vector API think about the invariance properties that your functions may have. If any are useful then it may be worth placing those in a separate interface. The more distinct types you have, the more properties you can make your compiler prove. Obviously this needs to be balanced against practicality, complexity for users and what you actually need. To some extent, many existing APIs make some of these distinctions with varying degrees of strictness. The main point I want to make clear is that these distinctions are based on invariance properties, something that not all developers of such APIs are aware of.

At some point I hope to return to this topic and enumerate all of the common vector-like types in a single framework. Unfortunately it's a big topic and I've only been able to scratch the surface here. In particular there are some subtle interplays between dimensions and types.

On a deeper level, I think there must be some type theoretical framework in which these invariance properties are free theorems.

Update: I believe some of this material is covered in Jimm Blinn's Vectors and Geometry and Objects, Oh My!, but I don't have access to that. I suspect that there is one big difference in my presentation: I'm not so interested here in vectors (or normals or whatever) in themselves but as defining interfaces to functions with invariance properties. Like the way category theorists shift the focus from objects to arrows. It makes a difference because it immediately gives theorems that our code is guaranteed to satisfy. It's the invariance property of the cross product (say) that is useful here, not the fact that the components of a vector transform a certain way when we change coordinates (because I might not even want to refer to coordinates in my code).

### Example Code

To show that I'm talking about something very simple, but still powerful, here's some Haskell code:

> data Triple = T Float Float Float deriving Show> class Vector v where>   (.+) :: v -> v -> v>   (.*) :: Float -> v -> v>   dot :: v -> v -> Float>   cross :: v -> v -> v> instance Vector Triple where>   T x y z .+ T u v w = T (x+u) (y+v) (z+w)>   a .* T x y z = T (a*x) (a*y) (a*z)>   dot (T x y z) (T u v w) = x*u+y*v+z*w>   cross (T x y z) (T u v w) = T>       (y*w-v*z)>       (z*u-x*w)>       (x*v-y*u)

You can freely apply the four primitive functions to elements of type Triple but if you have a function of, say, signature

> f :: Vector v => (v,v,Float) -> [(v,v)]

you are guaranteed it is invariant.

Derek Elkins said...

As I'm sure you well know, the geometric algebra and particularly the conformal geometric algebra makes it easier and clearer to recognize and maintain these invariants. For example, in any geometric algebra the distinction between a free vector and a normal vector is clear as normals are represented by tangent bivectors and which automatically transform appropriately under reflections. Similarly, the conformal geometric algebra distinguishes free vectors, tangent vectors, position vectors and normal "vectors" readily. Usually when I write my own 3D code, I make a Point type distinct from a Vector type, and also, Bivector, Spinor, and Pseudoscalar types.

Any rotationally invariant function is a natural transformation between (products of) the inclusion of the category of vectors and orthogonal transformations into the category of vectors and general linear transformations (or more or less general categories, e.g. it could be the identity functor).

Call the inclusion I, then g : I -> I means for a vector space V and an orthogonal transformation on it, R, we have g(IRv) = IRg(v).

Anonymous said...

Amazing post as usual! I'll soon learn Haskell just to try out your mind warping examples. :)

Andrew Sackville-West said...

Isn't this really just a distinction between linear versus non-linear transformations? I think as a matter of practicality, lots of people stick all sorts of vector operations into libraries but don't make the distinction that certain operations are most definitely *not* linear transformations.

And there is probably a lot of misunderstanding among users of libraries as to what rules apply to linear or non-linear transformations.

Rotation is a linear transformation and thus preserves vector addition and scalar multiplication. So you can do all sorts of stuff with it and the order of the operations doesn't really matter.

The component-wise multiplication of vectors is not a linear transformation and so none of those rules apply.

I think you are right, that the segregation of APIs makes some sense, but even more important is getting the user's of APIs to understand the distinctions between different types of transformations. (But maybe I'm missing your point and being too simplistic).

gelisam said...

This comment spells out the free theorem instantiation in which invariance properties are free theorems.

First, I do not think that the rotational invariance of (.+), (.*), dot and cross are free theorems. Rather, for all other operations made out of those primitives, the fact that it too is rotation invariant is a free theorem.

With a class annotation like (Vector v =>), the free relation corresponding to the type v can be any relation as long as it satisfies the free theorems corresponding to the types of the class methods. In our case, we wish to use the relation (Rx == x'), which needs to satisfy the following requirements:

R (x + y) == Rx + Ry
R (a * x) == a * Rx
dot x y == dot Rx Ry
R (cross x y) == cross Rx Ry

Those requirements are specifically stating that the (.+), (.*), dot and cross primitives are rotation invariant.

Now if those requirements are satisfied (which they are), we can compute the free relation corresponding to the right-hand side of (Vector v =>) and we will obtain a free theorem. With the f example at the end of the post, for example, the free theorem is

map rotpair (f (x,y,a)) == f (Rx, Ry, a)
where
rotpair (s, t) = (Rs, Rt)

which, as desired, states that f is rotation invariant.


text:
I’m looking over Casella and Berger’s Statistical Inference and reviewing some of the concepts. For the record, this book is exactly what you want if you need to take a statistics qualifying exam as a graduate student and not at all what a generalist will want. I actually enjoyed the material for the relatively clean overview it gave. There’s depth, but not so much as to deter someone without a degree in mathematics. That said, I would not recommend this book for beginners. If you do want to slog through, get a supplementary book. I’m currently looking at a section on page 4 of the second edition. There is a section that begins:

The operations of union and intersection can be extended to infinite collections of sets as well If $$A_1, A_2, A_3, ...$$ is a collection of sets, all defined on a sample space $$S$$, then $$\quad\bigcup_{i=1}^\infty A_i = \lbrace x \in S : x \in A_i \text{ for some } i\rbrace$$
$$\quad\bigcap_{i=1}^\infty A_i = \lbrace x \in S : x \in A_i \text{ for all } i \rbrace$$ For example, let $$S = (0, 1]$$ and define $$A_i = [(1/i), 1]$$. Then $$\bigcup_{i=1}^\infty A_i = \bigcap_{i=1}^\infty [(1/i), 1] = \lbrace x \in (0,1] : x \in[(1/i), 1] \text{ for some } i \rbrace$$
$$\quad\quad\quad = \lbrace x \in (0, 1]\rbrace = (0,1];$$ $$\bigcap_{i=1}^\infty A_i = \bigcap_{i=1}^\infty[(1/i), 1] = \lbrace x \in (0, 1] : x \in [(1/i), 1] \text{ for all } i \rbrace$$
$$\quad\quad\quad = \lbrace x \in (0, 1] : x \in [1, 1]\rbrace = \lbrace 1 \rbrace$$ (the point 1)

The above occurred after a discussion about countable and uncountable sets and proving theorems about sets from first principles (rather than Venn diagrams). If your eyes kind of glazed over while reading the above, no worries – mine did, too. Actually, my reaction was worse than glazing over: I skimmed and thought I understood. However, as I started to tease apart what was written, I realized that there was much more going on here than I realized.

First of all, the preceding section describes reasoning about events that can be represented as finite sets, over finite sample spaces. This section builds on what we know to discuss the infinite case.

In our example, the sample space is the interval ((0, 1]), which is defined over the reals. We are defining a countably infinite set of intervals, each denoted by some (A_i). How do we know it’s countably infinite? This is implied by the notation: we start at 1 and go to infinity. Therefore, there is a 1-1 correspondence with the natural numbers and thus the set of intervals is infinite. Let’s take a look at some of the intervals (note that I can’t draw lines, so imagine that the dashed line is actually connecting the endpoints and they’re actually aligned):

$$A_1 : \underset{0}{\circ} \quad \quad \quad \quad \quad \quad \underset{1}{\bullet}$$
$$A_2 : \underset{0}{\circ} \quad \quad \quad \underset{\frac{1}{2}}{\bullet} \mbox{-------}\underset{1}{\bullet}$$
$$A_3 : \underset{0}{\circ} \quad \quad \underset{\frac{1}{3}}{\bullet} \mbox{-----------}\underset{1}{\bullet}$$
$$\vdots$$
$$A_\infty : \underset{0}{\bullet}\mbox{--------------------}\underset{1}{\bullet}$$

Now, the first statement says that the union of infinite subsets of a sample space should be equal to the sample space:

$$\lbrace x \in (0,1] : x \in[(1/i), 1] \text{ for some } i \rbrace$$.

The statement on the far left of the set notation, $$x\in (0,1]$$, gives us $$x$$’s domain: it is defined over $$S$$, which we have defined to be $$(0,1]$$. The right side of the “such that” states that every x in the domain $$S$$ that is in some partition $$A_i$$ is in this set. Recall that $$A_i$$ was defined as $$[(1/i), 1]$$.

The above statement seems self-evident for rational numbers: for any rational number

$$m/n, m > 0 \wedge n > 0$$,

we know that there is a coarse-grained bound such that $$1/n$$ is less than or equal to $$m/n$$ and therefore $$m/n$$ is in $$A_n$$. But what about irrational numbers? For this to work, we would need a theorem that gives rational bounds on irrational numbers. This seems like something that ought to be out there, but I’m not sure where to look. I suspect hard-core PL theory, such as work on PCF, would have something to say about this. Number theory and/or real analysis would also be good candidates.

In any case, we are trying to make the argument that for every number on the real interval $$(0,1]$$, there exists at least one sub-interval with rational endpoints, and that the union of these sub-intervals gives us back the interval $$(0, 1]$$ exactly. We don’t miss any numbers on the interval $$(0, 1]$$. We might make some argument about the compactness of the interval.

The second statement defines intersection in this context. Here we would make an argument about the uniqueness of the intervals. That is, if $$i\not = j$$, then $$A_i \not = A_j$$. Every sub-interval is unique by virtue of its lower bound. However, the upper bound (1) is included in every interval. Therefore, the only interval all sub-intervals could share is the point 1.

It seems the point of the example was to show that countably infinite sets do occur and that the principles of set theory still hold. Defining the sample space as the reals appeals to practical considerations not fully explored in the text: we often model phenomena we measure as having infinite precision, but we know that our instruments can only be finitely precise. A countably infinite partition seems like it would introduce less error into a calculation than a finite one, i.e., a histogram.


text:
parcolumns, Vertical Misalignment of Columns

The columns produced by parcolumns are not vertically aligned. Why, and how can this be fixed using the parcolumns package as intended?

I have found similar questions with answers that I find unsuitable here:

Question (2) is essentially the same; however, I do not consider the accepted answer satisfactory.

Thanks to the comments below, the following code produces vertically aligned columns.

\documentclass[graybox,envcountchap,sectrefs]{svmono}

\usepackage{parcolumns}

\begin{document}

\begin{parcolumns}{2}
\colchunk{%
\section{Introduction}\null}
\colchunk{%
\section{Einf{\"u}hrung}\null}
\end{parcolumns}

\end{document}

Add \null (or \mbox{}) after each section so they both have content following them, and they'll be aligned.


text:
My Math Forum Multiplying a Random Variable, poisson dists and variance

 November 9th, 2011, 04:56 PM #1 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 Multiplying a Random Variable, poisson dists and variance OK, I either have one simple question "How do you multiply a random variable by a number?", or utter confusion. I think it's all centered around the weird notations in Probability. A random variable, X, is a coin flip variable (that is, it's either a 0 or a 1), and the Poisson distribution's Probability Mass D(something) (PMD) is $p_X(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$. I have that Expected Value and Variance are $E[X]=\sum_{\forall x}{x \cdot p_{X}(x)} \\ var(X)=E[(X-E[x])^2].$ I'm also told that the expected value and variance of a Poisson Distribution is lambda (which makes sense). I have answers to my friend's quiz which says that, if lambda=.2 of Poisson variable X, then $E[X]=.2 \\ var(X)=.2 \\ E[3X+.2]=1.1 \\ var(3X)=1.8.$ She thinks that E[3X+2]=3(.2)+.2=1.1, which I know is wrong, but can't convince her otherwise. Neither of us understand why the var(3X)=1.8. But my question is not just how did they those answers. I really want to know how do you multiply 3 times a random variable, X (and then how do you add 2 to that)? The variable is not a number, it's an idea - it's the number of occurrences of an event (e.g. coin flips heads). After I get that, I need to apply it to finding the EV and variance of 3X, 3X+2, etc. I'm also trying to figure out how they calculate that E[poisson variable]=lambda and variance(poisson variable)=lambda. Thanks, everyone.
 November 9th, 2011, 08:39 PM #2 Joined: Oct 2011 Posts: 26 Thanks: 0 Re: Multiplying a Random Variable, poisson dists and varianc Rules: -Adding a constant to a random variable increases the expectation of the random variable by that constant $E(X+C)= E(X)+C$ -Multiplying a random variable by a constant means that the expected value of that random variable is multiplied by the constant $E(CX)= CE(X)$ -Adding a constant to a random variable does not change the variance of the random variable $Var(X+C)= Var(X)$ -Multiplying a random variable by a constant multiplies the variance of the random variable by the square of that constant $Var(CX)= C^2Var(X)$ Memorise these rules and you'll be able to answer exam questions. Do some googling if you want to see proof of these rules and really understand them. In your example, you have $E(3X+.2)= E(3X)+.2 = 3E(X)+.2 = 3(.2)+.2 = .8$ $Var(3X)= 3^2Var(x) = 9(.2) = 1.8$ It looks like our answer for the expectation doesn'td agree. Now I'm confused... Anyway, something else: you say that the it "makes sense" that the expectation and variance of a Poisson distribution are lambda, but as far as I know that's at least as complicated as the rules for the expectation and variance. You might want to look here for an explanation of why the expectation of a Poisson distribution is lambda:
November 9th, 2011, 09:11 PM   #3
Senior Member

Joined: Sep 2009

Posts: 251
Thanks: 0

Re: Multiplying a Random Variable, poisson dists and varianc

Quote:
 Originally Posted by newguy123 Rules: -Adding a constant to a random variable increases the expectation of the random variable by that constant $E(X+C)= E(X)+C$ -Multiplying a random variable by a constant means that the expected value of that random variable is multiplied by the constant $E(CX)= CE(X)$ -Adding a constant to a random variable does not change the variance of the random variable $Var(X+C)= Var(X)$ -Multiplying a random variable by a constant multiplies the variance of the random variable by the square of that constant $Var(CX)= C^2Var(X)$ Memorise these rules and you'll be able to answer exam questions. Do some googling if you want to see proof of these rules and really understand them. In your example, you have
Ahhh. That clears everything up. Good answer.
The first three will be easy to memorize. I will have to look up a proof of that variance one. (Edit: I retract that last sentence - it was easy to figure out.)

Quote:
 Originally Posted by newguy123 $E(3X+.2)= E(3X)+.2 = 3E(X)+.2 = 3(.2)+.2 = .8$ $Var(3X)= 3^2Var(x) = 9(.2) = 1.8$ It looks like our answer for the expectation doesn'td agree. Now I'm confused...
The confusion is that my OP is almost certainly copied wrong. My friend was late for her class and I had to drive her back to school. I copied the problems down in the hurry. Lambda, the probability that I made a write-o (that's a type-o made with pencil and paper) is somewhere north of .7 for each word I rushed to write (you can use a Poisson distribution to determine the likelihood that my original post was completely accurate ). So don't be confused.

Quote:
 Originally Posted by newguy123 Anyway, something else: you say that the it "makes sense" that the expectation and variance of a Poisson distribution are lambda, but as far as I know that's at least as complicated as the rules for the expectation and variance. You might want to look here for an explanation of why the expectation of a Poisson distribution is lambda:
I mean it "makes sense" in that it's an intuitive answer. If lambda is the probability of something happening, and they're all equal weighted, then it just "seems" like the right answer for expected. I don't mean that I can calculate it or genuinely understand it.

In fact, I first started to try and calculate the expected of a poisson using the integral
$E[X]=\int{x\ p_{X}(x)}\ dx=\int{x\ \frac{e^{-\lambda}\ \lambda^{x}}{x!}}\ dx$
but I could not figure out how to do it and I couldn't get Maple to do it (I can never get Maple to do anything useful). Instead, I looked in another textbook and found it.

Thanks again, newbie.

 Tags dists, multiplying, poisson, random, variable, variance

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# [OS X TeX] How does Texhshop count words?

Themis Matsoukas tmatsoukas at icloud.com
Wed Dec 20 16:59:05 CET 2017

Hi Herb,

Thanks for the scripts. I used DropTeXCount and got the following:

• Encoding: utf8
• Words in text: 2866
• Words outside text (captions, etc.): 0
• Number of floats/tables/figures: 0
• Number of math inlines: 236
• Number of math displayed: 41

The word count is in the ballpark of TeXShop, which gave about 3000.

Themis

> On Dec 20, 2017, at 10:50 AM, Herbert Schulz <herbs at wideopenwest.com> wrote:
>
>> On Dec 20, 2017, at 9:29 AM, Themis Matsoukas <tmatsoukas at icloud.com> wrote:
>>
>> I used Edit/ Statistics… to count words in my source file. With the entire file selected (from first line to last, comments and all), I get a word count of 3045/20021 characters. If I select from my \begin{abstract} to the end of the file, I get 3381 words/22663 characters! I didn’t expect the count to be exact but I would have thought that a subset of the document would give a lower count that the whole. With some experimentation I discovered that when  the line \begin{document} is excluded,  the count increases by ~300 words. That’s with a revtex document, I am not sure if the same is seen with other classes.
>>
>> Not a big deal but since Dick is a mathematician I thought he’d want to know that there is a discontinuity that breaks monotonicity in the count :)
>>
>> Themis
>
> Howdy,
>
> If I had to guess I'd say it uses detex (to remove commands) and then wc to count words.
>
> For curiosities sake try downloading DropTeXCount.zip and use of of the two drop-scripts enclosed in the unzipped folder to do a count in the whole document. You can get that file from <https://herbs.github.io>. Those scripts use texcount (part of TeX Live) to do the word count.
>
> Good Luck,
>
> Herb Schulz
> (herbs at wideopenwest dot com)
>
>
>
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----------- Please Consult the Following Before Posting -----------
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# PoS(FPCP2015)017

Lattice QCD Calculation of direct CP violation and long distance effects in kaon mixing and rare decays

N.H. Christ

Contribution: pdf


text:
Applied Mathematics-II-Nagpur-Summer-2017 - Grad Plus

# Applied Mathematics-II-Nagpur-Summer-2017

## B.E. Second Semester All Branches (C.B.S.)

Time: 3 hours
Maximum marks: 80

Notes:
1. All questions carry marks as indicated.
2. Solve Question 1 OR Questions No. 2.
3. Solve Question 3 OR Questions No. 4.
4. Solve Question 5 OR Questions No. 6.
5. Solve Question 7 OR Questions No. 8.
6. Solve Question 9 OR Questions No. 10.
7. Solve Question 11 OR Questions No. 12.
8. Use of non programmable calculator is permitted.
9. Assume suitable data wherever necessary.

1. a) Prove that $latex \int_2^1x^{n-1}\left(\log\frac12\right)^{m-1}dx=\frac{\left|\overline m\right.}{n^m}$ [6M]

b) Evaluate $latex \int_0^1\frac{x^\alpha-1}{\log x}dx,\;\alpha\geq0$ by differentiating under integral sign. [6M]

OR

2. a) Evaluate $latex \int_0^1x^4\left(1-\sqrt x\right)^5dx$ [6M]

b) Find Root Mean square value of $latex \log_e^x$ over the range x = 1 to
x = e. [6M]

3. a) Trace the curve $latex y^2\left(2a-x\right)=x^3$. [6M]

b) Find the area enclosed by two parabolas y2=4x and $latex y^2=-4\left(x-2\right).$ [6M]

OR

4. a) Trace the curve $latex r=a\left(1+cos\theta\right)$ and find the perimeter of the curve. [6M]

b)Find the volume of the solid obtained by revolving the ellipse
$latex \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about x axis. [6M]

5. a) Evaluate $latex \iint\limits_Rydx\;dy$ where R is the region bounded by parabolas y2=4x and x2=4y. [6M]

b) Evaluate by changing order of integration. [6M]
$latex \int_0^4\int_y^4\frac x{x^2+y^2}dydx$

c) Evaluate $latex \int_{0}^{2}\int_{0}^{\sqrt{2x-x^{2}}}\frac{x}{\sqrt{x^{2}}+y^{2}}dy dx dz$ by changing into polar coordinates. [6M]

OR

6. a) Evaluate $latex \int_{-1}^{1}\int_{0}^{z}\int_{x-z}^{x+z}\left ( x+y+z \right )dy dx dz$ [6M]

b) Evaluate $latex \iint r^3drd\theta$ over the area bounded by circles $latex r=2cos\theta\;and\;r=4cos\theta.$ [6M]

c) Find the area lying between the parabola $latex y=4x-x^2$ and the line y = x. [6M]

7. a) Prove that [6M]
i) $latex \left[\overline b-\overline c\;\;\overline c-\overline a\;\;\overline a-\overline b\right]=0$
ii) $latex \bar{b}\times \bar{c}\cdot \left \{ \left ( \bar{c}\times \bar{a} \right )\times \left ( \bar{a}\times \bar{b} \right ) \right \}=\left \{ \left ( \bar{a}\times \bar{b} \right )\cdot \bar{c} \right \}^{2}$

b) A particle moves along a curve $latex x=t^3+1,\;y=t^2,\;z=2t+5,$ where t is the time. Find the component of its velocity and acceleration at t = 1 in the direction i + j + 3k. [6M]

c) Find the angle between the tangents to the curve $latex \overline r=t^2i-2tj+t^3k$ at the points t = 1 and t = 2. [6M]

OR

8.    a) Find the directional derivative of $latex \phi=x^2-y^2+2z^2$ at the point P(1, 2, 3) in the direction of line PQ where Q is the point (5, 0, 4). In what direction will it be maximum. [6M]

b) A vector field is given by -$latex \overline A=\left(x^2+xy^2\right)i+\left(y^2+x^2y\right)j$ Show that field is irrotational and find its scalar potential. [6M]

c) If $latex \overline r=xi+yj+zk$ show that [6M]
i) grad $latex r=\frac{\overrightarrow r}r$
ii) $latex \nabla r^n=n\;r^{n-2}\overrightarrow r$

9. Find the total work done in moving a particle in a field of force given by $latex \overline F=3xyi-5zj+10xk$ along the curve $latex x=t^2+1,\;y=2t^2,\;z=t^3$ from t=1 and t=2. [7M]

OR

10. Verify Greens’ theorem in the plane for $latex \int_C\left(3x^2-8y^2\right)dx+\left(4y-6xy\right)dy$ where C is the boundary of the region defined by $latex y=\sqrt x,\;y=x^2$. [7M]

11. a) Fit a curve $latex y=a+bx^2$ for the following data : [7M]

b) Using Lagrange’s interpolation formula, find the value of y when x = 10 from the
following table. [6M]

OR

12. a) The two lines of regressions are $latex 8x-10y+66=0\;:\;40x-18y=214$ If $latex \sigma_x^2=9$ Find : [7M]
i) Mean values of x and y
ii) Coefficient of correlation, and
iii) $latex \sigma_y$, the standard deviation of y.

b) Solve $latex u_{n+2}-2u_{n+1}+u_n=n^2\cdot2^n$ [6M]

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text:
## Trigonometry (11th Edition) Clone

The graph of all the complex numbers, $z$, will be a horizontal line passing through the imaginary axis at $i$.
For $z = x + yi$, The imaginary part of $z$ is 1, shall mean, $y = 1$ or $z = x + i$ The graph of all the complex numbers, $z$, satisfying the captioned condition will be a horizontal line passing through the imaginary axis at $i$.


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# Environment.EvaluatorSparseMultiPlayers module¶

EvaluatorSparseMultiPlayers class to wrap and run the simulations, for the multi-players case with sparse activated players. Lots of plotting methods, to have various visualizations. See documentation.

Warning

FIXME this environment is not as up-to-date as Environment.EvaluatorMultiPlayers.

Environment.EvaluatorSparseMultiPlayers.uniform_in_zero_one()

random() -> x in the interval [0, 1).

Environment.EvaluatorSparseMultiPlayers.REPETITIONS = 1

Default nb of repetitions

Environment.EvaluatorSparseMultiPlayers.ACTIVATION = 1

Default probability of activation

Environment.EvaluatorSparseMultiPlayers.DELTA_T_PLOT = 50

Default sampling rate for plotting

Environment.EvaluatorSparseMultiPlayers.MORE_ACCURATE = True

Use the count of selections instead of rewards for a more accurate mean/std reward measure.

Environment.EvaluatorSparseMultiPlayers.FINAL_RANKS_ON_AVERAGE = True

Default value for finalRanksOnAverage

Environment.EvaluatorSparseMultiPlayers.USE_JOBLIB_FOR_POLICIES = False

Default value for useJoblibForPolicies. Does not speed up to use it (too much overhead in using too much threads); so it should really be disabled.

Environment.EvaluatorSparseMultiPlayers.PICKLE_IT = True

Default value for pickleit for saving the figures. If True, then all plt.figure object are saved (in pickle format).

class Environment.EvaluatorSparseMultiPlayers.EvaluatorSparseMultiPlayers(configuration, moreAccurate=True)[source]

Evaluator class to run the simulations, for the multi-players case.

__init__(configuration, moreAccurate=True)[source]

Initialize self. See help(type(self)) for accurate signature.

activations = None

Probability of activations

collisionModel = None

Which collision model should be used

full_lost_if_collision = None

Is there a full loss of rewards if collision ? To compute the correct decomposition of regret

startOneEnv(envId, env)[source]

Simulate that env.

getCentralizedRegret_LessAccurate(envId=0)[source]

Compute the empirical centralized regret: cumsum on time of the mean rewards of the M best arms - cumsum on time of the empirical rewards obtained by the players, based on accumulated rewards.

getFirstRegretTerm(envId=0)[source]

Extract and compute the first term $$(a)$$ in the centralized regret: losses due to pulling suboptimal arms.

getSecondRegretTerm(envId=0)[source]

Extract and compute the second term $$(b)$$ in the centralized regret: losses due to not pulling optimal arms.

getThirdRegretTerm(envId=0)[source]

Extract and compute the third term $$(c)$$ in the centralized regret: losses due to collisions.

getCentralizedRegret_MoreAccurate(envId=0)[source]

Compute the empirical centralized regret, based on counts of selections and not actual rewards.

getCentralizedRegret(envId=0, moreAccurate=None)[source]

Using either the more accurate or the less accurate regret count.

getLastRegrets_LessAccurate(envId=0)[source]

Extract last regrets, based on accumulated rewards.

getAllLastWeightedSelections(envId=0)[source]

Extract weighted count of selections.

getLastRegrets_MoreAccurate(envId=0)[source]

Extract last regrets, based on counts of selections and not actual rewards.

getLastRegrets(envId=0, moreAccurate=None)[source]

Using either the more accurate or the less accurate regret count.

strPlayers(short=False, latex=True)[source]

Get a string of the players and their activations probability for this environment.

__module__ = 'Environment.EvaluatorSparseMultiPlayers'
Environment.EvaluatorSparseMultiPlayers.delayed_play(env, players, horizon, collisionModel, activations, seed=None, repeatId=0)[source]

Helper function for the parallelization.

Environment.EvaluatorSparseMultiPlayers.with_proba(proba)[source]

True with probability = proba, False with probability = 1 - proba.

Examples:

>>> import random; random.seed(0)
>>> tosses = [with_proba(0.6) for _ in range(10000)]; sum(tosses)
5977
>>> tosses = [with_proba(0.111) for _ in range(100000)]; sum(tosses)
11158


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Home > Calculus calculators > Find maximum and minimum value of y calculator

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
 Solution Find maximum and minimum value of y=4x^3+19x^2-14x+3Solution:Your problem -> maximum and minimum value of y=4x^3+19x^2-14x+3Here, y=4x^3+19x^2-14x+3:. (dy)/(dx)==(4x^(3)+19x^(2)-14x+3)=4x^(3)+19x^(2)-14x+3d/(dx)(4x^(3)+19x^(2)-14x+3)=d/(dx)(4x^(3))+d/(dx)(19x^(2))-d/(dx)(14x)+d/(dx)(3)=12x^(2)+38x-14+0=12x^(2)+38x-14=2(6x^(2)+19x-7)=2((3x-1)(2x+7))=2(3x-1)(2x+7)For stationary values, (dy)/(dx)=0=>2(3x-1)(2x+7)=0=2(6x^(2)+19x-7)=>x=0:.At x=0 we get stationary values.Now, (d^2y)/(dx^2)=d/(dx)(2(3x-1)(2x+7))=2(d/(dx)(3x-1))(2x+7)+2(3x-1)(d/(dx)(2x+7))d/(dx)(3x-1)=3d/(dx)(3x-1)=d/(dx)(3x)-d/(dx)(1)=3-0=3d/(dx)(2x+7)=2d/(dx)(2x+7)=d/(dx)(2x)+d/(dx)(7)=2+0=2=2(3)(2x+7)+2(3x-1)(2)=6(2x+7)+4(3x-1)=12x+42+12x-4=24x+38=2x(12+19/x)=2(12x+19)=24x+38((d^2y)/(dx^2))_(x=0)=24(0)+38=0+38=38 (positive):. At x=0 the function is minimumNow, y=4x^3+19x^2-14x+3"putting " x=0y_(min)=4(0)^(3)+19(0)^(2)-14(0)+3=0+0+0+3=3 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me


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On the angle condition for the perturbation of elliptic systems
Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 4, Volume 29 (2000) no. 2, p. 253-268
@article{ASNSP_2000_4_29_2_253_0,
author = {Landes, R\"udiger},
title = {On the angle condition for the perturbation of elliptic systems},
journal = {Annali della Scuola Normale Superiore di Pisa - Classe di Scienze},
publisher = {Scuola normale superiore},
volume = {Ser. 4, 29},
number = {2},
year = {2000},
pages = {253-268},
zbl = {0961.35042},
mrnumber = {1784175},
language = {en},
url = {http://www.numdam.org/item/ASNSP_2000_4_29_2_253_0}
}

Landes, Rüdiger. On the angle condition for the perturbation of elliptic systems. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 4, Volume 29 (2000) no. 2, pp. 253-268. http://www.numdam.org/item/ASNSP_2000_4_29_2_253_0/

[F] J. Frehse, On a class of nonlinear diagonal elliptic systems with critical growth and Cα- regularity, In: "Partial Differential Equations and the Calculus of Variations", F. Colombini et al. (eds.), Birkhäuser, Boston, 1989. | Zbl 0685.35027

[E] L.C. Evans, "Weak Convergence Methods for Nonlinerar Partial Differential Equations", AMS CBMS Regional conference series in Mathematics, Providence, Rhode Island, 1990. | MR 1034481 | Zbl 0698.35004

[Ga] M. Giaquinta, "Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic Systems", Princeton University Press, Princeton, NJ, 1983. | MR 717034 | Zbl 0516.49003

[Gi] E. Giusti, "Metodi Diretti nel Calcolo delle Variazioni ", Unione Matematica Italiana, Bologna, 1994. | MR 1707291 | Zbl 0942.49002

[GT] D. Gilbarg - N.S. Trudinger, "Elliptic Partial Differential Equations of Second Order", Springer-Verlag, New York-Berlin-Heidelberg, 1977. | MR 473443 | Zbl 0361.35003

[H] St. Hildebrandt, "Nonlinear elliptic systems and harmonic mappings" , Proceedings of Beijing Symp. on Diff. Geom. and Diff. Equ. Sci. Sience Press, Beijing, 1980. | MR 714341 | Zbl 0515.58012

[LU] O.A. Ladyshenskaya - N.N. Ural'Ceva, "Linear and Quasilinear Elliptic Equations ", Academic Press (English translation from Russian), New York and London, 1968. | MR 244627 | Zbl 0164.13002

[L1] R. Landes, Testfunctions for elliptic systems and maximum principles, To appear in Forum Mathematicum. | MR 1736096 | Zbl 0942.35065

[L2] R. Landes, On the existence of weak solutions ofperturbed systems with critical growth, J. Reine Angew. Math. 393 (1989), 21-38. | MR 972359 | Zbl 0664.35027

[L3] R. Landes, Some remarks on bounded and unbounded solutions of elliptic systems, Manuscripta Math. 64 (1989), 227-234. | MR 998488 | Zbl 0699.35093

[LM] R. Landes - V. Mustonen, On parabolic initial-boundary value problems with critical growth for the gradient, Ann. Inst. H. Poincaré Anal. Non Linéaire 11 (1994), 135-158. | Numdam | MR 1267364 | Zbl 0836.35078

[Z] K.-W. Zhang, Remarks on perturbated systems with critical growth, Nonlinear Anal. 18 (1992),1167-1179. | MR 1171604 | Zbl 0786.35061


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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# Ceva's Theorem

Ceva’s theorem is a theorem regarding triangles in Euclidean Plane Geometry. Consider a triangle ABC. Let CE, BG and AF be a cevians that forms a concurrent point i.e. D.

## Ceva’s Theorem Statement

Then according to Ceva’s theorem,

$$\begin{array}{l}\large\frac{AG}{GC} \times \frac{CF}{FB}\times\frac{BE}{EA}=1\end{array}$$

Also, the converse of the above is true, i.e. if

$$\begin{array}{l}\large\frac{AG}{GC} \times \frac{CF}{FB}\times\frac{BE}{EA}=1\end{array}$$
, then the line AF, BG, CE forms a concurrent point at D.

## Ceva’s Theorem Proof

Let h1 and h2 be the altitudes of triangles ABG, BGC and ADG, GDC, respectively. Let the area of the triangle be denoted using closed square brackets such as [ABG], [BGC], and so on.

When h1 and h2 are constructed, [BGC] is equal to 0.5(GC)(h1) and [ABG] is equal to 0.5(AG)(h1). [DGC] is equal to 0.5 (GC)(h2) and [ADG] is equal to 0.5 (AG)(h2).
That means,

$$\begin{array}{l}\frac{0.5\times AG\times h_1}{05\times GC\times h_1}=\frac{[ABG]}{[BGC]}\end{array}$$

And
$$\begin{array}{l}\frac{0.5\times AG\times h_1}{05\times GC\times h_1}=\frac{[ADG]}{[DGC]}\end{array}$$

On simplification we have,
$$\begin{array}{l}\frac{AG}{GC}=\frac{[ABG]}{[BGC]}….(i)\\ \frac{AG}{GC}=\frac{[ADG]}{[DGC]}….(ii)\end{array}$$

From (i) and (ii),
$$\begin{array}{l}\frac{AG}{GC}=\frac{[ABG]}{[BGC]}=\frac{[ADG]}{[DGC]}\end{array}$$

Similarly we can get,
$$\begin{array}{l}\frac{AG}{GC}=\frac{[BDA]}{[BDC]}\\\frac{CF}{FB}=\frac{[ADC]}{[BDA]}\\\frac{BE}{EA}=\frac{[BDC]}{[ADC]}\end{array}$$

Multiplying all these three equations we get,

$$\begin{array}{l}\frac{AG}{GC}\times \frac{CF}{FB}\times \frac{BE}{EA}=\frac{[BDA]}{[BDC]}\times \frac{[ADC]}{[BDA]}\times \frac{[BDC]}{[ADC]}\end{array}$$

On simplification,
$$\begin{array}{l}\frac{AG}{GC}\times \frac{CF}{FB}\times \frac{BE}{EA}=1\end{array}$$

### Converse of Ceva’s Theorem

We have,

$$\begin{array}{l}\frac{(AG)}{(GC)}\frac{(CF)}{(AB)}\frac{(BE)}{(EA)}=1\end{array}$$

Here CE, BG, and AF Cevians are concurrent.

Estimate that Cevians CE and AF intersect at D and assume that the Cevians passing through D is BH. So according to Cevians Theorem we have,

$$\begin{array}{l}\frac{AH}{HC}\frac{CF}{FB}\frac{BE}{EA}=1\end{array}$$

As assumed

$$\begin{array}{l}\frac{AG}{GC}\frac{CF}{FB}\frac{BE}{EA}=1\end{array}$$

According to Transitive Property, we have

$$\begin{array}{l}\frac{AH}{HC}\frac{CF}{FB}\frac{BE}{EA}=\frac{AG}{GC}\frac{CF}{FB}\frac{BE}{EA}\end{array}$$

By simplifying

$$\begin{array}{l}\frac{AH}{HC}=\frac{AG}{GC}\end{array}$$
<

It holds true when H and G illustrate the same point. Hence BG, CE, and AF should be concurrent.


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## A fixed-point theorem for integral type inequality depending on another function.(English)Zbl 1216.54015

Let $$(X,d)$$ be a complete metric space, $$0 \leq \alpha < 1$$, and $$S, T$$ be selfmaps of $$X$$ such that $$T$$ is injective, continuous, and subsequentially convergent. Suppose that there exists $$x \in X$$ such that $$\int_0^{d(Tsy,TS^2y)}\varphi(t)\,dt \leq \alpha\int_0^{d(T,TSy)}\varphi(t)\,dt$$ for each $$y$$ in the orbit of $$x$$, where $$\varphi := [0, +\infty) \to [0, +\infty)$$ is a Lebesgue integrable mapping which is summable, nonnegative, and such that $$\int_0^{\varepsilon}\varphi(t)\,dt > 0$$ for each $$\varepsilon > 0$$. Then the authors show that
(i) $$\lim_nTS^nx = Tq$$,
(ii) $$\int_0^{d(Tq,TS^nx)}\varphi(t)\,dt \leq \alpha_n\int_0^{d(Tq,Tx)}\varphi(t)\,dt$$, and
(iii) $$q$$ is a fixed point of $$S$$ if and only if $$G(x) := d(TSx, Tx)$$ is $$S_T$$-orbitally lower semicontinuous at $$q$$.

### MSC:

 54H25 Fixed-point and coincidence theorems (topological aspects)
Full Text:


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# A large open tank has two holes in the wall. One is a square hole of side $'L'$ at a depth $'y'$ from the top and the other is a circular hole of radius R at a depth $'4y'$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then value of R is :

$\begin {array} {1 1} (a)\;\frac{L}{\sqrt {2 \pi}} & \quad (b)\;2 \pi L \\ (c)\;L \sqrt {\frac{2}{\pi}} & \quad (d)\;\frac{L}{2 \pi} \end {array}$

$(a)\;\frac{L}{\sqrt {2 \pi}}$


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5

# 5) Suppose C caalom Va /alle has CDE Functi 04 FC) = 38 fr OLXLz Find Hbe eX Pected value Of...

## Question

###### 5) Suppose C caalom Va /alle has CDE Functi 04 FC) = 38 fr OLXLz Find Hbe eX Pected value Of

5) Suppose C caalom Va /alle has CDE Functi 04 FC) = 38 fr OLXLz Find Hbe eX Pected value Of

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text: Speaker : Professor Sheng-Chi Liu (Washington State University) Title : The $L^2$ restriction norm of a $GL(n)$ Maass form Time : 2018-07-06 (Fri) 10:00 - Place : Seminar Room 722, Institute of Mathematics (NTU Campus) Abstract: In this talk, I will discuss the $L^2$ restriction norm of a Maass form on $GL(n+1)$ restricted to $GL(n)$. Using the spectral decomposition on $GL(n)$ and the Rankin-Selberg method, we are led naturally to a family of $GL(n+1)XGL(n)$ $L$-functions. We give sharp upper and lower bounds for the restriction norm under certain assumptions (i.e., the Lindelof hypothesis, the weighted local Weyl law and others). This is joint work with Xiaoqing Li and Matthew Young.


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derivative¶

dolfin.fem.formmanipulations.derivative(form, u, du=None)

UFL form operator: Given any form, compute the linearization of the form with respect to the given Coefficient. The resulting form has one additional Argument in the same finite element space as the coefficient. A tuple of Coefficients may be provided in place of a single Coefficient, in which case the new Argument argument is based on a MixedElement created from this tuple.


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# Least-Squares Density Ratio Estimation¶

Implementation of Least-Squares Density Ratio Estimation (LS-CDE) method introduced in [SUG2010] with some extra features.

This approach estimates the conditional density of multi-dimensional inputs/outputs by expressing the conditional density in terms of the ratio of unconditional densities r(x,y):

$p(y|x) = \frac{p(x,y)}{p(x)} = r(x,y)$

Instead of estimating both unconditional densities separately, the density ratio function r(x,y) is directly estimated from samples. The density ratio function is modelled by the following linear model:

$\widehat{r_{\alpha}}(x,y) := \alpha^T \phi(x,y)$

where $$\alpha=(\alpha_1, \alpha_2,...,\alpha_b)^T$$ are the parameters learned from samples and $$\phi(x,y) = (\phi_{1}(x, y),\phi_{2}(x,y),...,\phi_{b}(x,y))^T$$ are kernel functions such that $$\phi_{l}(x,y) \geq 0$$ for all $$(x,y)\in D_{X} \times D_{Y}$$ and $$l = 1, ..., b$$.

The parameters $$\alpha$$ are learned by minimizing the a integrated squared error.

$J(\alpha) = \int\int ( \widehat{r_{\alpha}}(x,y) - r(x,y))^2 p(x)dxdy.$

After having obtained $$\widehat{\alpha} = argmin_{\alpha} ~ J(\alpha)$$ through training, the conditional density can be computed as follows:

(1)$\widehat{p}(y|x=\tilde{x}) = \frac{\widehat{\alpha}^T\phi(\tilde{x},y)}{\int\widehat{\alpha}^T\phi(\tilde{x},y)dy}$

[SUG2010] propose to use a Gaussian kernel with width $$\sigma$$ (bandwidth parameter), which is also the choice for this implementation:

$\phi_{l}(x,y) = exp \left( \frac{||x-u_{l}||^2}{2 \sigma^2} \right) exp \left( \frac{||y-v_{l}||^2}{2 \sigma^2} \right)$

where $$\{(u_{l},v_{l})\}_{l=1}^b$$ are center points that are chosen from the training data set. By using Gaussian kernels the optimization problem $$argmin_{\alpha} ~ J(\alpha)$$ can be solved analytically. Also, the denominator in (1) is traceable and can be computed analytically. The fact that training does not require numerical optimization and the solution can be computed fully analytically is the key advantage of LS-CDE.

While [SUG2010] propose to select center points for the kernel functions randomly from the training set, our implementation offers further center sampling methods:

• all: use all data points in the train set as kernel centers

• random: randomly selects k points as kernel centers

• k_means: uses k-means clustering to determine k kernel centers

• agglomorative: uses agglomorative clustering to determine k kernel centers

class cde.density_estimator.LSConditionalDensityEstimation(name='LSCDE', ndim_x=None, ndim_y=None, center_sampling_method='k_means', bandwidth=0.5, n_centers=500, regularization=1.0, keep_edges=True, n_jobs=-1, random_seed=None)[source]

Least-Squares Density Ratio Estimator

http://proceedings.mlr.press/v9/sugiyama10a.html

Parameters
• name – (str) name / identifier of estimator

• ndim_x – (int) dimensionality of x variable

• ndim_y – (int) dimensionality of y variable

• center_sampling_method – String that describes the method to use for finding kernel centers. Allowed values [all, random, distance, k_means, agglomerative]

• bandwidth – scale / bandwith of the gaussian kernels

• n_centers – Number of kernels to use in the output

• regularization – regularization / damping parameter for solving the least-squares problem

• keep_edges – if set to True, the extreme y values as centers are kept (for expressiveness)

• n_jobs – (int) number of jobs to launch for calls with large batch sizes

• random_seed – (optional) seed (int) of the random number generators used

conditional_value_at_risk(x_cond, alpha=0.01, n_samples=1000000)

Computes the Conditional Value-at-Risk (CVaR) / Expected Shortfall of the fitted distribution. Only if ndim_y = 1

Parameters
• x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

• alpha – quantile percentage of the distribution

Returns

CVaR values for each x to condition on - numpy array of shape (n_values)

covariance(x_cond, n_samples=1000000)

Covariance of the fitted distribution conditioned on x_cond

Parameters

x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Covariances Cov[y|x] corresponding to x_cond - numpy array of shape (n_values, ndim_y, ndim_y)

eval_by_cv(X, Y, n_splits=5, verbose=True)

Fits the conditional density model with cross-validation by using the score function of the BaseDensityEstimator for scoring the various splits.

Parameters
• X – numpy array to be conditioned on - shape: (n_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_samples, n_dim_y)

• n_splits – number of cross-validation folds (positive integer)

• verbose – the verbosity level

fit(X, Y, **kwargs)[source]

Fits the conditional density model with provided data

Parameters
• X – numpy array to be conditioned on - shape: (n_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_samples, n_dim_y)

fit_by_cv(X, Y, n_folds=3, param_grid=None, verbose=True, n_jobs=-1)

Fits the conditional density model with hyperparameter search and cross-validation. - Determines the best hyperparameter configuration from a pre-defined set using cross-validation. Thereby,

the conditional log-likelihood is used for simulation_eval.

• Fits the model with the previously selected hyperparameter configuration

Parameters
• X – numpy array to be conditioned on - shape: (n_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_samples, n_dim_y)

• n_folds – number of cross-validation folds (positive integer)

• param_grid

(optional) a dictionary with the hyperparameters of the model as key and and a list of respective parametrizations as value. The hyperparameter search is performed over the cartesian product of the provided lists. Example: {“n_centers”: [20, 50, 100, 200],

”center_sampling_method”: [“agglomerative”, “k_means”, “random”], “keep_edges”: [True, False]

}

get_configuration(deep=True)

Get parameter configuration for this estimator.

Parameters

deep – boolean, optional If True, will return the parameters for this estimator and contained subobjects that are estimators.

Returns

params - mapping of string to any Parameter names mapped to their values.

get_params(deep=True)

Get parameters for this estimator.

Parameters

deep (boolean, optional) – If True, will return the parameters for this estimator and contained subobjects that are estimators.

Returns

params – Parameter names mapped to their values.

Return type

mapping of string to any

kurtosis(x_cond, n_samples=1000000)

Kurtosis of the fitted distribution conditioned on x_cond

Parameters

x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Kurtosis Kurt[y|x] corresponding to x_cond - numpy array of shape (n_values, ndim_y, ndim_y)

log_pdf(X, Y)[source]

Predicts the conditional log-probability log p(y|x). Requires the model to be fitted.

Parameters
• X – numpy array to be conditioned on - shape: (n_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_samples, n_dim_y)

Returns

conditional log-probability density log p(y|x) - numpy array of shape (n_query_samples, )

mean_(x_cond, n_samples=1000000)

Mean of the fitted distribution conditioned on x_cond :param x_cond: different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Means E[y|x] corresponding to x_cond - numpy array of shape (n_values, ndim_y)

mean_std(x_cond, n_samples=1000000)
Computes Mean and Covariance of the fitted distribution conditioned on x_cond.

Computationally more efficient than calling mean and covariance computatio separately

Parameters

x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Means E[y|x] and Covariances Cov[y|x]

pdf(X, Y)[source]

Predicts the conditional density p(y|x). Requires the model to be fitted.

Parameters
• X – numpy array to be conditioned on - shape: (n_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_samples, n_dim_y)

Returns

conditional probability density p(y|x) - numpy array of shape (n_query_samples, )

plot2d(x_cond=[0, 1, 2], ylim=(-8, 8), resolution=100, mode='pdf', show=True, prefix='', numpyfig=False)

Generates a 3d surface plot of the fitted conditional distribution if x and y are 1-dimensional each

Parameters
• xlim – 2-tuple specifying the x axis limits

• ylim – 2-tuple specifying the y axis limits

• resolution – integer specifying the resolution of plot

plot3d(xlim=(-5, 5), ylim=(-8, 8), resolution=100, show=False, numpyfig=False)

Generates a 3d surface plot of the fitted conditional distribution if x and y are 1-dimensional each

Parameters
• xlim – 2-tuple specifying the x axis limits

• ylim – 2-tuple specifying the y axis limits

• resolution – integer specifying the resolution of plot

predict_density(X, Y=None, resolution=50)

Computes conditional density p(y|x) over a predefined grid of y target values

Parameters
• X – values/vectors to be conditioned on - shape: (n_instances, n_dim_x)

• Y – (optional) y values to be evaluated from p(y|x) - if not set, Y will be a grid with with specified resolution

• resulution – integer specifying the resolution of simulation_eval grid

Returns: tuple (P, Y)
• P - density p(y|x) - shape (n_instances, resolution**n_dim_y)

• Y - grid with with specified resolution - shape (resolution**n_dim_y, n_dim_y) or a copy of Y in case it was provided as argument

sample(X)[source]

sample from the conditional mixture distributions - requires the model to be fitted

Parameters

X – values to be conditioned on when sampling - numpy array of shape (n_instances, n_dim_x)

Returns: tuple (X, Y)
• X - the values to conditioned on that were provided as argument - numpy array of shape (n_samples, ndim_x)

• Y - conditional samples from the model p(y|x) - numpy array of shape (n_samples, ndim_y)

score(X, Y)

Computes the mean conditional log-likelihood of the provided data (X, Y)

Parameters
• X – numpy array to be conditioned on - shape: (n_query_samples, n_dim_x)

• Y – numpy array of y targets - shape: (n_query_samples, n_dim_y)

Returns

average log likelihood of data

set_params(**params)

Set the parameters of this estimator.

The method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.

Returns

Return type

self

skewness(x_cond, n_samples=1000000)

Skewness of the fitted distribution conditioned on x_cond

Parameters

x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Skewness Skew[y|x] corresponding to x_cond - numpy array of shape (n_values, ndim_y, ndim_y)

std_(x_cond, n_samples=1000000)

Standard deviation of the fitted distribution conditioned on x_cond

Parameters

x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

Returns

Standard deviations sqrt(Var[y|x]) corresponding to x_cond - numpy array of shape (n_values, ndim_y)

tail_risk_measures(x_cond, alpha=0.01, n_samples=1000000)

Computes the Value-at-Risk (VaR) and Conditional Value-at-Risk (CVaR)

Parameters
• x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

• alpha – quantile percentage of the distribution

• n_samples – number of samples for monte carlo model_fitting

Returns

• VaR values for each x to condition on - numpy array of shape (n_values)

• CVaR values for each x to condition on - numpy array of shape (n_values)

value_at_risk(x_cond, alpha=0.01, n_samples=1000000)

Computes the Value-at-Risk (VaR) of the fitted distribution. Only if ndim_y = 1

Parameters
• x_cond – different x values to condition on - numpy array of shape (n_values, ndim_x)

• alpha – quantile percentage of the distribution

Returns

VaR values for each x to condition on - numpy array of shape (n_values)

SUG2010(1,2,3)

Sugiyama et al. (2010). Conditional Density Estimation via Least-Squares Density Ratio Estimation, in PMLR 9:781-788 (http://proceedings.mlr.press/v9/sugiyama10a.html)