text:
1- What causes ozone molecules to break down? 2- What is the benefit of the ozone layer in the atmosphere for living organisms (humans, animals, and plants)?

Question:

1- What causes ozone molecules to break down?
2- What is the benefit of the ozone layer in the atmosphere for living organisms (humans, animals, and plants)?

Which type of fats insulate the body, protect organs, and provide energy, are necessary for good health? Group of answer choices unhealthy essential unsaturated saturated

Which type of fats insulate the body, protect organs, and provide energy, are necessary for good health? Group of answer choices unhealthy essential unsaturated saturated...

What do the gods tell Creon the people of Thebes must do?

What do the gods tell Creon the people of Thebes must do?...

Playing certain sports increases your muscular fitness

playing certain sports increases your muscular fitness...

How do you start a Prologue for a story.

How do you start a Prologue for a story....

Tom is 13 years old Brad is 10 years old which number sentence can be used to find how much older Tom is than Brad

Tom is 13 years old Brad is 10 years old which number sentence can be used to find how much older Tom is than Brad...

What is the product of -4/7 and -5/6

What is the product of -4/7 and -5/6...

WHY DO SEED PLANTS NEED TO EVOLVE WAYS FOR DISPERSING THIS SEED?

WHY DO SEED PLANTS NEED TO EVOLVE WAYS FOR DISPERSING THIS SEED?...

A solution for y = x + 2

a solution for y = x + 2...

Find the measure of AB¯¯¯¯¯¯¯¯. A. 14 B. 22 C. 21 D. 20

Find the measure of AB¯¯¯¯¯¯¯¯. A. 14 B. 22 C. 21 D. 20...

How many miles did native americans march on the trail of tears?

How many miles did native americans march on the trail of tears?...

The sum of two numbers is 8. their difference is 12. find the numbers.

the sum of two numbers is 8. their difference is 12. find the numbers....

Hi, im mexican xd

hi, im mexican xd             ...

I have two bags of counters. The first bag contains two red counters and one blue counter. The second bag contains one red counter, one blue counter and two yellow counters. I take a counter at random from both bags. A). What is the probability that the two counters will be the same colour? b). What is the probability that exactly one of the counters will be red?

I have two bags of counters. The first bag contains two red counters and one blue counter. The second bag contains one red counter, one blue counter and two yellow counters. I take a counter at random from both bags. A). What is the probability that the two counters will be the same colour? b). ...

( NO LINKS ) How many edges does this prism have? * A. 4 B. 8 C. 12

( NO LINKS ) How many edges does this prism have? * A. 4 B. 8 C. 12...

SONO CHI NO ____________ finish

SONO CHI NO ____________ finish...

THIS IS MUSICAL THEATER WEST SIDE STORY

THIS IS MUSICAL THEATER WEST SIDE STORY...


text:
A complex

is a chain of abelian groups $C_k$ and homomorphisms $\partial_k$ such that $\partial_k\circ\partial_{k+1}=0$ for all $k$, that is, $\text{im }\partial_{k+1}\subset \text{ker }\partial_k$

Associated with the complex $C_\ast$ we define the cohomology groups

[Cohomological version]

A complex

is a chain of abelian groups $C^k$ and homomorphisms $\mathrm{d}^k$ such that $\mathrm{d}^k\circ\mathrm{d}^{k-1}=0$ for all $k$, that is, $\text{im }\mathrm{d}^{k-1}\subset \text{ker }\mathrm{d}^k$

Associated with the complex $C^\ast$ we define the cohomology groups

format: plain


text:
This book is in Open Review. I want your feedback to make the book better for you and other readers. To add your annotation, select some text and then click the on the pop-up menu. To see the annotations of others, click the button in the upper right hand corner of the page

10.4 Stability and forecastability conditions of ADAMX

It can be shown that any ADAMX{S} is not stable (as it was defined for pure additive ETS models), meaning that the weights of such model do not decline exponentially to zero. This becomes apparent, when we compare the explanatory part of any ADAMX with a deterministic model, discussed in the context of pure additive models. For example, we have already discussed that when $$\alpha=0$$ in ETS(A,N,N), then the model becomes equivalent to the global level, looses the stability condition, but still can be forecastable. Similarly, the X part of ADAMX{S} will always be unstable, but can be forecastable. However, this is unreasonable from the model building point of view: if for example, the ETS part of the model is stable but the model does not pass stability check just because of the X part, then the check itself is incorrect. Furthremore, there are no issues constructing ARIMAX models, but claim that there are issues with stability of ETSX. This only means that the stability / forecastability conditions should be checked for the dynamic part of the model (ETS or ARIMA) separately, ignoring the X part. Technically, this implies creating separate transition matrix, persistence and measurement vectors and calculating the discount matrix for the ETS part in order to check already discussed stability and forecastability conditions.

When it comes to the dynamic ADAMX, then the situation changes, because now the smoothing parameters for the coefficients of the model determine, how weights decline over time. It can be shown based on (5.9) that the values of the state vector on the observation $$t$$ can be calculated via the recursion (here we provide formula for the non-seasonal case, keeping in mind that in case of the seasonal one, the derivation and the main message will be similar): $$$\mathbf{v}_{t} = \prod_{j=1}^{t-1}\mathbf{D}_{t-j} \mathbf{v}_{0} + \sum_{j=0}^{t-1} \prod_{i=0}^{j} \mathbf{D}_{t-i} y_{t-j}, \tag{10.23}$$$ where $$\mathbf{D}_t=\mathbf{F} - \mathrm{diag}\left(\mathbf{w}_{t}\right)^{-1} \mathbf{g} \mathbf{w}_{t}'$$ is the time varying discount matrix. The main issue in the case of dynamic ADAMX is that the stability condition varies over time together with the values of explanatory variables. So, it is not possible to derive it for the general case. In order to make sure that the model is stable, we need for all eigenvalues of each $$\mathbf{D}_{j}$$ for all $$j=\{1,\dots,t\}$$ to lie in the unit circle.

Alternatively, we can introduce a new condition. We say that the model is stable on average if the eigenvalues of $$\mathbf{\bar{D}}=\frac{1}{t}\sum_{j=1}^t\mathbf{D}_t$$ all lie in the unit circle. This way, some of observations might have a higher impact on the final value, but they will be canceled out by those that have much lower weights. This condition can be checked during the model estimation, similar to how the conventional stability condition is checked.

As for the forecastability condition, for the ADAMX{D} it should be (based on and the same logic as in section 6.4): $$$\lim\limits_{t\rightarrow\infty}\left(\mathbf{w}'_{t}\prod_{j=1}^{t-1}\mathbf{D}_{t-j} \mathbf{v}_{0}\right) = \text{const} . \tag{10.24}$$$ However, for the reasons discussed earlier in this subsection, this condition will always be violated for the ADAMX models, just because the explanatory variables in $$\mathbf{w}_{t}$$ have their own variability and typically do not converge to a stable value with the increase of the sample size. So, if a forecastability condition needs to be checked for either ADAMX{D} or ADAMX{S}, we recommend checking it separately for the dynamic part of the model.

References

• Hyndman, R.J., Koehler, A.B., Ord, J.K., Snyder, R.D., 2008. Forecasting with Exponential Smoothing. Springer Berlin Heidelberg.


text:
How do I download the paper for study and future use? Will I be vio...
### Hard Real Time vs Soft Real Time A real time system is a sys...
CHILL is a procedural and object oriented programming language. Its...
The Ericsson MD110 is a PBX system (Private Branch Exchange), which...
### Declarative vs Imperative programming These are two programm...
### PROLOG PROLOG is a logic programming language (as opposed to ...
Even tho Ericsson's market share in the mobile phone market decline...
The AXE10 is part of the AXE telephone exchange product line of cir...
PLEX (Programming Language for EXchanges) is a special-purpose, pro...
PABX stands for *Private Automatic Branch Exchange* which is a tech...
It's important to underline that even tho Erlang is a language that...
### Agner Krarup Erlang Agner Erlang was a Danish mathematician, s...
BEAM is the virtual machine at the core of the Erlang Open Telecom ...
Strand is a parallel programming language. The citation goes to a b...
*JAM* stands for Joe's Abstract Machine
[Warren Abstract Machine (WAM)](https://en.wikipedia.org/wiki/Warre...
I wonder if this is still true? I would say that most software engi...
Further notes on the basic syntax of Erlang: ` -module(math)....
### SNMP Simple Network Management Protocol (SNMP) is an applicati...
I think it is interesting that [H|T] and {Key1, Val, S,B} (above) b...
You can read more about the effort to create a type system for Erla...
The development of Erlang
Joe Armstrong
Computer Science Laboratory
Ericsson Telecom AB
joeQcslab.ericsson.se
Abstract
This paper describes the development of the programming
language Erlang during the period 1985-1997.
Erlang is a concurrent programming language designed
for programming large-scale distributed soft real-time con-
trol applications.
The design of Erlang was heavily influenced by ideas from
the logic and functional programming communities. Other
sources of inspiration came from languages such as
Chill and
Ada which are used in industry for programming control
systems.
1 Introduction
This paper describes the development of the Erlang pro-
gramming language. Erlang is a language which draws heav-
ily from various traditions in the logic, functional and real-
time control programming communities.
Our goal was to make a language which could
be used
for building large soft real-time control systems. By large I
mean systems with possibly millions of lines of code. By a
soft real-time system I mean a system which does not faif
catastrophically if a real-time deadline is missed.
Erlang was developed by the author and his colleagues
at the Ericsson Computer Science Laboratory. Ericsson is
currently the world’s largest supplier of telecoms systems
and the world’s leading manufacturer of mobile phones.
Ericsson’s flagship product is a telephone exchange called
the AXE lO which is programmed in a proprietary language
called PLEX. A typicaf AXEl O system has over 5 million
Iines of PLEX. The AXEIO is specified to have a “down
time” of less than 3 minutes per year - such requirements
are common in the telecoms industry.
The programming languages which are used for building
such systems have to deal with a range of problems not usu-
ally encountered in traditional batch or interactive computer
systems. For example, one requirement is that the system
should be in “continuous” operation - this means we must
do software upgrades without stopping the system. It also
implies the use of fault-tolerant processors and a software
architecture which protects the integrity of the system from
various classes of programmer error,
Erlang grew out of a series of experiments which we per-
formed to see if we could find better ways of programming
telecoms systems.
Our criterion for success was that we
could program telecoms systems with less effort and fewer
errors than could be done using conventional technology.
Permission to
make digital/hard copy of part or all this work for
personal or classroom use is granted without fee provided that
tage, the copyright notice, the title of the publication and its date
appear, and notice is given that copying is by permission of ACM,
Inc. To copy otherwise, to republish, to post on servers, or to
redistribute to lists, requires prior specific permission and/or a fee
This paper starts with a brief discussion of the various
milestones which the language went through. This is fol-
lowed by sections describhrg the language, finally future di-
rections of research are mentioned.
2
Milestones
s
b
Early experiments
Being a telecoms laboratory we were in the fortunate
position of always having hardware to play with. The
lab had a Ericsson MD110 (which is a smafl telephone
exchange) which had been modified so it could be
controlled by a conventional Unix machine. We pro-
grammed POTS1 in as many programming languages
as we could lay our hands on.
The main conclusion [5] was that declarative language
programs for POTS were a lot shorter and easier to
understand than imperative language programs. Un-
fortunately the declarative languages lacked features
for concurrency control and had poor error handling
facilities.
We concluded that we would like something like Pro
log with added facilities for concurrency and improved
error handling. No such language existed at the time.
1986 - 1989
Having decided that we liked Prolog for programming
POTS we started a series of experiments call LOTS
(Lots of POTS) to program not only
POTS but aleo
an extremely large number of different telephony fea-
tures in Prolog. Before long I had programmed a large
subset of PABX features in Prolog. To do so I wrote
a Prolog meta-interpretor which added the notion of
vanced error detection and recovery. This work is de-
scribed in [1].
The Bollmora group
ter we were able to attract the curiosity and interest of
a group of engineers who
were responsible for a future
soft ware architecture of the Ericsson MD 110.
POTS stands for Plain Old Telephony Service, that is, subscriber
A calls subscriber B and they chat for a while
‘Who worked in Bollmora, Stockholm.
ICFP ’97 Amsterdam, ND
0 1997 ACM 0-89791 -918 -1/97 /0006 .,.\$3.50
196
This group thought that they could US( our results as
a vehicle for investigate ing their own software architec-
tures. This began a period of collaboration which led
to the development of Erlang as a fully fledged lan-
guage and to a new range of Ericsson products.
During the period 1986-1988 members of the computer
science lab worked closely with the Bollmora group,
meeting once or twice a week. During this time the
changed rapidly, The language grew and evolved and
somewhere along the line acquired a name, Erlang,
named in honor of the Danish mathematician Agner
Krarup Erlang (1878- 1929) whose name is associated
with the telecoms industry.
The jam machine
By about 1988 is was clear that Erlang as it was now
called wss a good language for prototyping telephone
exchanges [10]. It was a strange mixture, with declar-
ative features (inherited from Prolog), multi-tasking
and concurrency (inherited from EriPascaJ and Ada)
and an original combination of error handling mecha-
nisms.
Erlang was, however, tar too slow to be used for prod-
uct development, The Bolhnora group did some mea-
surements and decideil t}lat for product development
they needed a system that was 40 times faster than
This requirement lead to the development of a number
of different abstract lnachines and compilation tech-
niques used for implementing Erlang. A cross compiler
from Erlang to Strand [7] was developed. After a num-
ber of experiments tht JAM [4] machine was invented.
This was based on the Warren Abstract Machine with
added primitives for concurrency and exception han-
dling.
By this time the Erlang effort had grown to three peo-
ple. Mike Wifliams wrote a byte code emulator for
JAM code, I wrote the compiler and Robert Virding
wrote the support libraries.
While we only ever distributed one version of the sys-
tem to externaf user:., internally there were severaf
competing Erlang “engines” where we experimented
with different implementation techniques.
The resulting JAM implementation was delivered to
the Bollmora group in 1989. Fortunately it was 70
fortunately they had now revised their original esti-
mates and wanted a machine that was 280 times faster
than the original.
During the period 1989-1994 the Erlang “group” ex-
panded (there has ne~,er been a formal group). Claes
Vikstrom joined the group in 1990 and added distribu-
tion to the language. The original language had been
designed with “hooks’ for adding distribution - but
this was Iiever actually done until Claes implemented
it.
and build a new PABX in Erlang. Erlang wasn’t fast
9
enough but it was generally thought that the perfor-
mance problems could be solved. In 1992 they started
development of a product called the “Mobility Server”,
In 199.5 this was launched and became part of the
“Consono” product range. At the time of writing the
Mobility Server is marketed in 12 countries. Among
other things it is used to control the DECT mobile
phone in the European Parliament in Strasbourg.
Fight for acceptance
During the period 1992-1996 it was by no means clear
that Erlang was suitable for programming large-scale
industrial products. While we were very successful in
a number of smalf projects, Erlang programming and
programmers stilf represented a very small percentage
of the programming effort at Ericsson.
The Mobility server project continued and we attracted
several new smalf projects. During this time we were
involved in maasive “guerilla marketing” activities. I
don’t think there was a single person in Ericsson who
ever came out of Bjarne’s3 office without a copy of the
Erlang manual under their arm!
The beam
Performance has always been a major problem. In
1992 Bogumil Hausman started work on the BEAM.4
The BEAM compiles Erlang to C which can then be
compiled with a conventional C compiler. The BEAM
can afso compife to threaded code which can be freely
intermixed with compiled code.
Compiled code is faster but takes up more space than
threaded code. In a large embedded system with mil-
lions of lines of code, the volume of object code can be
a major problem,
Infrequently used and non-performance critical parts
of the system can be compiled to threaded code while
performance critical parts of the system can be com-
piled to native code. There are performance tools
which we can use to analyse the system and find out
which parts of the system need to be native code com-
piled.
The BEAM is described in [6]. In many applications
the BEAM is comparable with C in performance terms.
The BEAM replaced the JAM as the principle system
for new product development in 1997.
Erlang Systems
Up to 1995 most users were “enthusiasts” and Erlang
spread internally within Ericsson by “word of mouth”.
We didn’t find new users, they found us.
As the language spread we needed to produce training
material and hold courses to train new users. Initially,
all courses were held at the Computer Science Lab by
members of the Erlang “group”. Whife holding such
courses was great fun, demand soon far outstripped
supply.
In April 1993 a new company “Erlang Systems” was
formed to handle sales, marketing, education and con-
sulting of Erlang. We had been in the fortunate posi-
tion of having supervised a number of Master degree
3
Boss of the computer science lab.
4 Bogdans’s Erlang Abstract machine.
197


text:
# Tag Info

4

I will assume you generally understand the mechanics of the padding oracle attack. Say we have some ciphertext block $c_i = c_{i-1} \oplus E(k,m_i)$ and we want to find the last byte of $m_i$. Let $b$ denote the last byte of $m_i$, which we are trying to find. Consider ciphertexts of the form $(c_{i-1} \oplus \texttt{00} \cdots \texttt{00} g,c_i)$, where $g$ ...

2

If I know the set of possible characters in the plaintext will only be lowercase English letters, then is there any way in which I can speed up the deciphering process? If so, then specifically what would this look like? Yes, it helps. First, remember that the padding oracle works on the fact that the server sends back only one information at all; the ...

1

It is a block cipher with block size 3. Therefore for a given known-plaintext, we can build the current key's action table that has $2^3=8$ values. We will use the decryption part $$P_i = D(C_i) \oplus C_{i-1} \oplus P_{i-1}$$ with $C_{0} \oplus P_{0} = IV$ with conversion $$D(C_i) = P_i \oplus C_{i-1} \oplus P_{i-1}$$ P_i 000 001 010 011 100 101 ...

2

Am I thinking correctly and will an malicious user (which has access to encryption keys) not (or very hard) be able to generate a collision in my IV-ciphertext combination so that the server is grouping unequal messages together? If the server checks all the bytes in the IV-ciphertext (and declares a collision only every byte for two encrypted messages are ...

Top 50 recent answers are included


text:
# Larger sets of infinity

This seems to be a widely accepted theory on larger sets of infinite numbers, originally shown by Cantor.

After watching the video, I am trying to grasp in layman's terms why this is true. My understanding is the following, and I wondered if someone could confirm if this is correct;

The reason there are more real numbers between 0 and 1, than all the natural numbers, is because in this example, each real number can have a length of infinity.

Initially I thought that there is an infinite number of real numbers between 0 and 1, and an infinite number of natural numbers. This would allow for (to use the videos metaphore) a line to be drawn between every real and natural number. But if the real numbers are also infinite in length, there are "infinity to the power of infinity" real numbers, and just infinity natural numbers.

Have I understood this correctly? If not, could someone spell it out for me please?

-
I have not watched the video, but there as as many integers as there are fractions (i.e. the number of fractions are countable). However, there are uncountably many real numbers. –  picakhu May 30 '12 at 17:23
There are "countably infinite" natural numbers. There are "countably infinite" rational numbers. What does "countably infinite mean? A set A is said to be "countably infinite" if there is a bijection from N to A. The set of real numbers R is not countably infinite (we say that R is uncountably infinite). When we try to show that R is countably infinite, we fail due to a contradiction. You can use google and/or wiki for proofs of these facts. They are standard and accepted by most mathematicians. –  Adam Rubinson May 30 '12 at 17:33
Also, what do you mean by length of infinity ? Do you mean, infinite decimal places after the decimal point? If so, then yes, each real number can be written as an infinite string of numbers after the decimal point (but not uniquely, e.g. 0.99999... = 1) –  Adam Rubinson May 30 '12 at 17:36
If you look at all infinite sequences $x_1,x_2, x_3,\dots$ of real numbers, then it turns out that there are no more of these (in the sense of cardinality) than there are real numbers. So even though the explanation given is not wrong, there is detail to fill in before one can come up with a tight argument. –  André Nicolas May 30 '12 at 18:08

## 3 Answers

You seem to get the gist of the idea by stating that there are "infinity to the power of infinity" real numbers, and "just infinity" natural numbers. The problem is that you haven't clarified "which" infinity you are talking about, nor have you shown that raising "infinity to the power of infinity" (whatever that means) gets you a "bigger" infinity. Speaking loosely again, you know that there are infinitely many even numbers, and infinitely many integers, and that they are the same "size" of infinity, even though one is "infinity" and the other is "two times infinity". So, it is important that we are precise about defining what we mean by "infinity".

The cardinality (size of infinity) of the natural numbers is $\omega$. This is also the cardinality of the rational numbers, algebraic numbers, etc. In set theory, we can think of the real number line $\mathbb{R}$ as $\omega^\omega$, which is just the set of functions from $\omega$ to itself (this is the same thing as $\omega$ to the power of $\omega$). Each function from $\omega$ can be thought of as a function on the natual numbers, and hence is just a sequence. Thus each real number is just a sequence of natural numbers.

If you look at Cantor's Diagonalization Argument, he uses the fact that each real number is a sequence of $\omega$-many natural numbers to inductively construct a real number that could not have been enumerated. The crux of the argument relies on being able to demonstrate the existence of a real number that "defies" the enumeration up until the $n$th step, where $n$ is made arbitrarily large. This, in turn, requires that the real numbers can contain an "infinite amount of information," which I'm guessing is what you intuitively mean by stating that real numbers are "infinite in length."

-

The reason there are more real numbers between 0 and 1, than all the natural numbers, is because in this example, each real number can have a length of infinity.

That's pretty much right, but you must be a little bit careful about what you mean by length. If you just look at the real numbers whose decimal expansions terminate, like 23.5043958495, or 0.000000000004 or whatever, there are only as many of those as there are natural numbers. The "lines" thing will work, though it would be a bit tricky to say exactly which line goes where. But when you allow decimal expansions to go on forever to the right, that is when you look at the set of all real numbers, there are strictly more of these than there are naturals.

there are "infinity to the power of infinity" real numbers, and just infinity natural numbers.

Isaac talked about this, so I won't. Have a look at http://en.wikipedia.org/wiki/Cardinal_number to find out more about how people measure the relative sizes of infinite sets.

-
In saying "when you allow decimal expansions to go on forever to the right" you have oversimplified. Nonterminating repeating decimal expansions "go on forever to the right", but the set of those is merely countable. Austin Mohr's answer makes this distinction correctly. –  MJD May 30 '12 at 19:12

To supplement the answers already given, I would add that one should take care to distinguish between sets in which decimal representations use a finite, but unbounded number of digits (such as the naturals) and those that use an infinite number of digits (such as the irrationals).

The former means that the decimal representation of any particular natural number has finite length, but I may choose as large a finite length as I want. Considering numbers like this will only get you up to a countable set.

On the other hand, the decimal representation of any particular irrational number truly uses a (countably) infinite number of digits, and so may be said to be of (countably) infinite length. It is these kinds of numbers that give the "infinity to the power infinity" that you need to construct an uncountable set.

-


text:
# What's up in the Milky Way? The orientation of the disc relative to the triaxial halo

Debattista, V P; Roskar, R; Valluri, M; Quinn, T; Moore, B; Wadsley, J (2013). What's up in the Milky Way? The orientation of the disc relative to the triaxial halo. Monthly Notices of the Royal Astronomical Society, 434(4):2971-2981.

## Abstract

Models of the Sagittarius stream have consistently found that the Milky Way disc is oriented such that its short axis is along the intermediate axis of the triaxial dark matter halo. We attempt to build models of disc galaxies in such an intermediate-axis orientation'. We do this with three models. In the first two cases we simply rigidly grow a disc in a triaxial halo such that the disc ends up perpendicular to the global intermediate axis. We also attempt to coax a disc to form in an intermediate-axis orientation by producing a gas+dark matter triaxial system with gas angular momentum about the intermediate axis. In all cases we fail to produce systems which remain with stellar angular momentum aligned with the halo's intermediate axis, even when the disc's potential flattens the inner halo such that the disc is everywhere perpendicular to the halo's local minor axis. For one of these unstable simulations we show that the potential is even rounder than the models of the Milky Way potential in the region probed by the Sagittarius stream. We conclude that the Milky Way's disc is very unlikely to be in an intermediate-axis orientation. However we find that a disc can persist off one of the principal planes of the potential. We propose that the disc of the Milky Way must be tilted relative to the principal axes of the dark matter halo. Direct confirmation of this prediction would constitute a critical test of Modified Newtonian Dynamics.

Models of the Sagittarius stream have consistently found that the Milky Way disc is oriented such that its short axis is along the intermediate axis of the triaxial dark matter halo. We attempt to build models of disc galaxies in such an intermediate-axis orientation'. We do this with three models. In the first two cases we simply rigidly grow a disc in a triaxial halo such that the disc ends up perpendicular to the global intermediate axis. We also attempt to coax a disc to form in an intermediate-axis orientation by producing a gas+dark matter triaxial system with gas angular momentum about the intermediate axis. In all cases we fail to produce systems which remain with stellar angular momentum aligned with the halo's intermediate axis, even when the disc's potential flattens the inner halo such that the disc is everywhere perpendicular to the halo's local minor axis. For one of these unstable simulations we show that the potential is even rounder than the models of the Milky Way potential in the region probed by the Sagittarius stream. We conclude that the Milky Way's disc is very unlikely to be in an intermediate-axis orientation. However we find that a disc can persist off one of the principal planes of the potential. We propose that the disc of the Milky Way must be tilted relative to the principal axes of the dark matter halo. Direct confirmation of this prediction would constitute a critical test of Modified Newtonian Dynamics.

## Citations

25 citations in Web of Science®
23 citations in Scopus®

## Altmetrics

Detailed statistics

Item Type: Journal Article, refereed, original work 07 Faculty of Science > Institute for Computational Science 530 Physics English October 2013 12 Sep 2013 15:05 05 Apr 2016 16:59 Wiley-Blackwell 0035-8711 Publisher DOI. An embargo period may apply. https://doi.org/10.1093/mnras/stt1217
Permanent URL: https://doi.org/10.5167/uzh-80942


text:
## 1 Test description

This test uses a simple blade with few aerodynamical blade stations in steady conditions and compares the aerodynamical loads computed by Ashes to an analytical solution.

The following load cases are tested

## 2 Model

The model used for this test is shown in the figure below:

The chordlength accross the blade is
$$c = 1\text{ m}$$
$$L = 5\text{ m}$$
$$r_h = 0.5\text{ m}$$
.
The rotor is fixed, the blade is therefore not rotating. For the models that do not use the Stiff rotor approximation, the blade and the hub conections are infinitely stiff, therefore there are no deflections.

The blade used for this model has three Blade aerodynamical station . The innermost station is at the root and has a cylindrical airfoil with no drag and no lift. This station will therefore not produce any aerodynamic loads. The two other blade aerodynamical stations have the NACA64-618 airfoil, which polar can be exported from Ashes and are shown in the figure below:

The table below gives a summary of the relevant blade characteristics:

 Blade aerodynamical station Distance to blade root Influence length Airfoil 1 $$r_1 = 0$$ $$L_{I1} = 1.25\text{ m}$$ Cylinder with no lift and drag 2 $$r_2 = 2.5\text{ m}$$ $$L_{I2} = 2.5\text{ m}$$ NACA64-618 3 $$r_3 = 5\text{ m}$$ $$L_{I3}=1.25\text{ m}$$ NACA64-618

The air density is
$$\rho = 1.225\text{ kg}\cdot{m}^{-3}$$
. No gravity forces are applied, and tip and hub corrections are set to 0.

## 3 Analytical solution

The wind speed speed is denoted
$$V$$
. For the current analysis we use
$$V = 10\text{ m}\cdot\text{s}^{-1}$$
Since there is no rotational velocity, the angle of attack at all blade aerodynamical stations is 90 degrees. For the two outermost blade aerodynamical stations, this gives a lift and drag coefficient of
$$C_L = 0.053$$
and
$$C_D = 1.4565$$
(the innermost blade aerodynamical blade station experiences no lift or drag, and is therefore not considered in the rest of the analysis).

$$i$$
, the distributed lift and srag forces are, respectively:
$$F_{L,i}=\frac{1}{2}\rho c C_L V^2$$
and
$$F_{D,i}=\frac{1}{2}\rho c C_D V^2$$

Note: this is only valid because the rotational speed is zero, which implies that there is no induced velocities

For each blade aerodynamical station, since the blade is not rotating, the lift is in the direction of the torque force and the drag is in the direction of the torque force.

The aerodynamic thrust for the whole rotor is thus
$$F_T = \sum_{i=2}^3F_{D,i}\cdot L_{Ii}=334.5\text{ N}$$
and the aerodynamic torque for the whole rotor is
$$T=\sum_{i=2}^3 F_{L,i}\cdot L_{Ii}\cdot (r_i+r_h)=46.67\text{ Nm}$$
. These two output can be found in the Rotor sensor

The next three output are part of the Blade [Time] sensor sensor. The root force is the sum of the drag and the lift forces from both stations, which can be expressed as
$$F_r = \sum_{i=2}^3\sqrt{(F_{D,i}\cdot L_{I,i})^2+(F_{L,i}\cdot L_{I,i})^2}=334.8\text{ N}$$
. The in-plane bending moment is
$$M_{ip}=\sum_{i=2}^3 F_{L,i}\cdot L_{Ii}\cdot r_i=40.58\text{ Nm}$$
and the out-of-plane bending moment is
$$M_{oop}=\sum_{i=2}^3F_{D,i}\cdot L_{Ii}\cdot r_i=1115\text{ Nm}$$

Flexible blades, as opposed to stiff blades, are modelled structurally by dividing the blade into a number of finite elements, each with its own structural properties (which can be different for each element or the same accross the blade span, as is the case in this model). In the present test, flexible blades are modelled with elements that have a very large stiffness, which means that they will experience no deflection.

Because Flexible blades are divided into structural elements, the aerodynamic loads have to be lumped and applied to the nodes at both ends of the element. The lumping is done as follows: if part of an element lies within the Influence length of an Blade aerodynamical station, its nodes will experience aerodynamic loads. Each of the two loads (one on each node) will be equal to the linear aerodynamic loads multiplied the portion of the element that lies within the Influence length divided by two.

Below we show several examples of how the derivation is carried out for the present test.

A two-element blade will have three structural nodes, each located at a Blade aerodynamical station. The Influence length of each Blade aerodynamical station will vary depending on its position along the blade (a detailed explanation of how the influence length is obtained can be found in the Influence length document). This is illustrated in the figure below:

Note: there are two more nodes in this model, that belong to the Blade connection. These have been hidden in the picture of the left

To exemplify how the loads apply on each node, we first focus on the drag force and how it contributes to the Aerodynamic thrust.
As explained above, the drag force from station 3 will be
$$F_{D,3}\cdot L_{i,3}=111.5\text{ N}$$
. The influence length of this station is entirely within Element 2, so this load will only be applied to Element 2.
The drag force from station 2 will be
$$F_{D,2}\cdot L_{i,2}=223.0\text{ N}$$

Half on its influence length is on Element 2 and the other half on Element 1, which means that half of this load will be applied on Element 2 and the other half on Element 1. This means that Element 2 will experience a total load (from two stations) of
$$111.5+223.0/2=223.0\text{ N}$$
In Ashes, the loads experienced by an element are divided by two and lumped to its nodes. This means that, from Element 2 (namely Node 1 and Node 2 from Blade 1) will experience a load (more precisely, a BEM nodal load) in the direction of the drag of
$$111.5\text{ N}$$
. This is illustrated inthe figure below:

Note: in the above picture, not all possilbe fields from the Loads on node sensor are displayed. You can customise which fields are displayed in the Customise sensor fields document.

The figure above shows that Node 1 experiences two BEM forces (namely, Blade nodal BEM force 4 and Blade nodal BEM force 1). This is because this node belongs both to Element 1 and Element 2 and will therefore experience loads originating from both elements. It is possible to see where a load originates from by right clicking on the relevant node and then selecting the load. This is illustrated in the figure below:

Similarly, since half of the influence length of station 2 is on Element 1, half of the drag load from station 2 will be applied to Element 1. This will be the only drag load applied to Element 1 since station 1 does not experience aerodynamic loads (because its lift and drag coefficients are both 0), so it will have a value of
$$F_{D,2}\cdot (L_{i,2}/2)=111.5\text{ N}$$
. The load on Element 1 will be split into two equal components applied to both of its nodes, of value
$$55.75\text{ N}$$
. This is illustrated in the figure below:

The sum of all the BEM forces in the drag directions amounts to
$$334.5\text{ N}$$
, which is the value of the thrust force on the rotor

It is important to note that the way the aerodynamical loads are lumped onto structural nodes has implications on the bending moments at the blade root. For example, part of the drag from station 2 is applied to the node at the root of the blade (or more precisely, Blade nodal BEM force (0) is applied on Node 2 - Blade connection 1). This means that it will not contribute to the Out-of-plane bending moment. If we use the notation
$$BEM_{i,y}$$
to mean the y-component Blade nodal force (i) (i.e. the component in the drag direction) and by noticing that both elements have a length
$$E_L = 2.5\text{ m}$$
, we find that the Out-of plane bending moment will be
$$M_{oop,f}=BEM_{1,y}\cdot E_L+BEM_{4,y}\cdot E_L+BEM_{5,y}\cdot 2E_L=976\text{ Nm}$$

As expected , this value is lower than the Out-of-plane bendig moment with stiff blades, since part of the aerodynamic loads are applied at the root.

This lumping will also influence the aerodynamic torque. By following the same procedure as above including the hub radius, we obtain an aerodynamic torque in the Flexible blades case of

$$T = BEM_{0,x}\cdot r_h+BEM_{1,x}\cdot (E_L+r_h)+BEM_{4,x}\cdot (E_L+r_h)+BEM_{5,x}\cdot (2E_L+r_h)=41.59\text{ Nm}$$

Again, this is lower than with Stiff blade case.

Note: to simplify the equations, we have ommitted the negative signs in front of the components of the Blade nodal forces. A detailed explanation of the signs for the bending moments can be found in the Out-of-plane and In-plane documents. The aerodynamic torque is positive when it spins the rotor clockwise.

We now reproduce the derivation explained in the previous paragraph for a blade divided into three elements, shown in the image below:

As derived above, the drag loads from the three aerodynamical stations are
- for Station 3,
$$F_{D,3}\cdot L_{I,3}=111.5\text{ N}$$
- for Station 2,
$$F_{D,2}\cdot L_{D,2} = 223.0\text{ N}$$
- for Station 3,
$$0\text{ N}$$
Element 3 covers the whole Influence length 3 and one sixth of Influence length 2, therefore it will experience a drag force of
$$111.5+223.0/6=148.7\text{ N}$$
. This force is then split in two and the resulting force of
$$74.33\text{ N}$$
is applied to both nodes of element 3 (namely Node 2 and Node 3 on Blade 1).
Element 2 covers two thirds of Influence lenght 2, therefore it will experience a drag force of
$$223\cdot2/3=148.7\text{ N}$$
. Its nodes (namely Node 2 and Node 1 from Blade 1) will thus experience a force of
$$74.33\text{ N}$$
.
Element 1 covers one sixth of influence length 2 and the whole Influence length 1, but the latter does not produce any aerodynamic load. It will therefore experience a drag force of
$$223/6 = 31.27\text{ N}$$
and its nodes (namely Node 1 from blade 1 and Node 2 from Blade connection) will experience a force of
$$18.59\text{ N}$$
.
If we now calculate the out-of-plane bending moment as we did for the two-element blade case, we find that
$$M_{oop,3}=1022\text{ Nm}$$
.
Similarly, the aerodynamic torque is now
$$T=42.95\text{ Nm}$$
.
These values are again lower than the case of stiff blades, but are larger than the case with two elements. This is expected since the load applied at the root of the blade (Node 2 of Blade connection) is lower than in the two-element case.

#### 3.2.3 Analytical solution

If we consider the case of a cantilever beam with a partial distributed load, we can find the analytical solution for the bending moment at the root of the beam. This case is illustrated in the figure below:

The bending moment at the root of the beam is
$$M = q\frac{\left(l^2-d^2\right)}{2}$$

Applying this formula to the blade bending moments and aerodynamic torque, we find
$$M_{oop,an}=F_{D}\frac{L^2-L_{I,1}^2}{2}=1045\text{ Nm}$$

$$M_{ip,an}=F_{L}\frac{L^2-L_{I,1}^2}{2}=38.04\text{ Nm}$$

$$T_{an}=F_{L}\frac{(L+r_h)^2-(L_{I,1}+r_h)^2}{2}=44.13\text{ Nm}$$

A blade with an inifinite number of elements should produce these results. In practice, we expect that 20 elements should be enough to get with 0.05% of the analytical solution.

### 3.3 Summary of expected results

The following table sums up the results expected in Ashes, as calculated above:

 Test name Aerodynamic thrust $$[\text{N}]$$ Aerodynamic torque $$[\text{Nm}]$$ Root force $$[\text{N}]$$ In plane bending moment $$[\text{Nm}]$$ Out of plane bending moment $$[\text{Nm}]$$ Stiff blade 334.5 46.67 334.8 40.58 1115 two-element blade 334.5 41.59 334.8 33.50 975.4 three-element blade 334.5 43.28 334.8 37.20 1022 20-element blade 334.5 44.13 334.8 38.04 1045

## 4 Results

A simulation of ten seconds is run. The test is considered passed if the last two seconds of the results produced by Ashes lie within 0.05% of the expected solution.

The report with the results can be found here:


text:
# Can someone explain the proof of the following linear differential equation

Mathematics Asked by Lucas G on January 7, 2022

I was working through a robotics book, and came across the following equation, solution, and proof where $$A$$ is a $$n times n$$ constant matrix, and $$x_0 = x(0)$$
$$dot x (t)=Ax(t)$$
whose solution is :
$$x(t)=e^{At}x_0$$
Proof:
begin{align} dot{x}(t)&=left(frac{d}{dt}e^{At}right)x_0 \&=frac{d}{dt}left(I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dotsright)x_0 \&=left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0 \&=Ae^{At}x_0 \&=Ax(t) end{align}

I’m not super familiar with linear differential equations and operations, and am kind of confused as to what happens around the third line of the proof and onward. If anyone understands this well, an explanation would be appreciated

I'm going to go ahead and put in my two cents, which does not answer your actual question (but perhaps can shed some light for you). A linear differential equation is of the form:

$$A_nx^{(n)} + A_{n - 1}x^{(n - 1)} + A_{n - 2}x^{(n - 2)} + dots A_2ddot{x} + A_1dot{x} + A_0x = 0$$

where $$x^{(n)}$$ is the $$text{n}^text{th}$$ derivative of the function $$x$$. The general solution to this type of differential equation, called a homogeneous differential equation--because the RHS is equal to 0--is an exponential, of the form: $$x(t) = x_0e^{lambda t}$$ (note that $$x_0 neq lambda$$ or maybe it is--point is that there is no relation between the two).

So in your case, we have a first-order, homogeneous, linear differential equation (first order because it only involves a first derivative):

$$dot{x} = Ax$$

The typical way to solve is that $$x(t) = x_0x^{lambda t}$$ and the procedure is to find $$x_0$$ and $$lambda$$:

begin{align*} &x(t) = x_0e^{lambda t} \ &dot{x}(t) = x_0lambda e^{lambda t} \ &dot{x} = Ax leadsto x_0lambda e^{at} = Ax_0e^{lambda t} end{align*}

To solve we set this equal to zero and factor out the common $$x_0e^{lambda t}$$:

$$x_0e^{lambda t} left(lambda - Aright) = 0$$

Regardless of the value of $$e^{lambda t}$$, this equation is satisfied so long as $$lambda = A$$ (since then $$lambda - A = A - lambda = 0$$). This is similar to when solving a quadratic, we factor and set each factor to zero to solve, e.g. $$3x^2 + 5x + 2= (3x + 2)(x + 1) = 0 leadsto x = -frac{2}{3}, -1$$.

So we now have the general solution: $$x(t) = x_0e^{At}$$. Now we must find $$x_0$$. Unfortunately, regardless of $$lambda$$, $$x(0) = x_0$$--usually there is an equation to be solved to find the initial conditions.

So there it is--this is the first week of a differential equations class.

Answered by Jared on January 7, 2022

The proof is

begin{align} dot{x}(t)&=left(frac{d}{dt}e^{At}right)x_0tag{1} \&=frac{d}{dt}left(I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dotsright)x_0tag{2} \&=left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0tag{3} \&=Ae^{At}x_0tag{4} \&=Ax(t)tag{5}. end{align}

Explanation.

$$text{(1)}$$: Follows from the definition of $$x(t)$$. We have $$dot{x}(t)=frac{d}{dt}left(x(t)right)=frac{d}{dt}left(e^{At}x_0right)=frac{d}{dt}left(e^{At}right)x_0.$$ $$text{(2)}$$: The Taylor series of $$e^x$$ is $$e^x=1+frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+ldots=sum_{n=0}^{infty}frac{x^n}{n!}.$$ If $$A$$ is a $$n times n$$ matrix with real entries, define $$e^{At}=I+At+frac{A^2t^2}{2!}+frac{A^3t^3}{3!}+dots=sum_{n=0}^{infty}frac{t^nA^n}{n!},$$ where $$e^{0 ,cdot A}=I$$.

$$text{(3)}$$: Differentiate the series term-by-term. A more direct way is begin{align}frac{d}{dt}(e^{At})&=sum_{n=0}^{infty}nfrac{t^{n-1}A^n}{n!}\&=sum_{n=1}^{infty}frac{t^{n-1}A^n}{(n-1)!}\&=Asum_{n=1}^{infty}frac{t^{n-1}A^{n-1}}{(n-1)!}\&=Asum_{m=0}^{infty}frac{t^mA^m}{m!}\&=Ae^{At}. end{align} $$text{(4)}$$: This follows from $$left(A+A^2t+frac{A^3t^2}{2!}+dotsright)x_0=Aleft(I+At+frac{A^2t^2}{2!}+dotsright)x_0=Ae^{At}x_0,$$ because $$e^{At}=left(I+At+frac{A^2t^2}{2!}+dotsright).$$ $$text{(5)}$$: Directly from the definition $$x(t), ~x(t)=e^{At}x_0$$.

Answered by Axion004 on January 7, 2022

The proof you've attached shows why the given answer is the correct one. It assumes you are aware of the Taylor expansion of $$e^x$$, which is given in the proof.

With some experience in calculus, you might be aware of the fact that $$e^x$$ is its own derivative, and see the similarity to that statement with the first equation. If you know how to differentiate $$e^{Ax}$$, you can easily confirm the solution without the added machinery of Taylor expansion.

In general, solving differential equations like these come from practice and experience. It's about trying different functions with certain analytical properties which might work in the given context. You will only get better at them as you keep exploring differential equations like these.

Answered by Subhasish Mukherjee on January 7, 2022

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text:
MathSciNet bibliographic data MR1191610 17B67 (17B55) Cox, Ben \${\germ F}\$${\germ F}$-categories and \${\germ F}\$${\germ F}$-functors in the representation theory of Lie algebras. Trans. Amer. Math. Soc. 343 (1994), no. 1, 433–453. Article

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


text:
NEET Foundation Material for Class-VII ,NEET Physics units Measuremnt

# NEET Physics units and measurements Test-7

NEET Physics units and measurements Test-7

1. The liquid drop of density ρ, radius r and surface tension a oscillates with time period T. Which of the following expression for T2 is correct.

2. The volume of a liquid (v) flowing per second through a cylindrical tube depends upon the pressure gradient (p/l) radius of the tube (r) coefficient of viscosity (η) of the liquid by dimensional method the correct formula is

3. The dimensions of resistivity in terms of M, L, T and Q, where Q stands for the dimensions of charge is

4. The distance travelled by a particle in nth second is Sn = u + $latex frac { a }{ 2 } (2n – 1)$ where u is the velocity and a is the acceleration. The equation is

5. The position of a particle at time ‘t’ is given by the equation : x(t) = $latex frac { V0 }{A } (1 – eAt)$; V0= constant and A > dimensions of V0 and A respectively are ;

6. If J is the angular momentum and E is the kinetic energy, then J2/E has the dimensions

7. If L is the inductance, C is the capacitance and R is the resistance, then $latex R sqrt { frac { C }{ L } }$ has the dimension

8. If h is Planck’s constant and, &lamda;is the wavelength, h/&lamda; has the dimensions of

9. The power of a motor is 200W. If the unit of length is halved, that of mass is doubled and that of time is Ac doubled, then the power of the motor in the new system is

10. If the unit of mass is α kg, the unit of length is β metre and the unit of time is ‘ γ ‘ second, The magnitude of calorie in the new system is (1 Cal = 4.2 J)

11. If the mass of the electron (9x 10-31kg) is taken as unit of mass, the radius of the first Bohr orbit 0.5×10-10m) as unit of length and 500 newton as the unit of force, then the unit of time in the new system would be

12. If pressure P, velocity of light C and acceleration due to gravity g are chosen as fundamental units, then dimensional formula of mass is

13. If young’s modulus y, surface tension s and time t are the fundamental quantities then the dimensional formula of density is

14. If P represents radiation pressure, C speed of light, and Q radiation energy striking unit area per second and x,y,z are non zero integers, then Px Qy Cz is dimensionless. The values of x,y and z are respective

15. If the time period (T) of vibration and liquid drop depends on surface tension (S), radius (r) of the drop and density (ρ) of the liquid, then the expression of T is

16. Suppose, the torque acting on a body is given by T=KL+xI/(ω) . Where L = angular- momentum, I= moment of inertia, omega= angular speed. The dimensional formula for Kx is same as that for

17. The number of particles crossing unit area perpendicular to x-axis in unit time is given by n=-[-D(n2-n1)]/[x2-x1]where n1 and n2 are number of particles per unit volume for the value of x meant to x1and x2. The dimensions of D are

18. Turpentine oil is flowing through a capillary tube of length 1 and radius r. The pressure difference between the two ends of the tube is P. The viscosity of oil is given by : η=[P(r2-x2)]/[4vl]fd. Here v is velocity of oil at a distance x from the axis of the tube. From this relation, the dimensional formula of η is

19. Dimensional formula of the product of the two physical quantities P and Q is ML2T-2, the dimension formula of (P/Q) is MT -2. P and Q respectively are

20. The potential energy of a particle varies with distance x from a fixed origin as $latex V = frac { Asqrt { x } }{ x+B }$ where A and B are constants. The dimensions of AB are

21. If h is the Planck’s constant, m = mass of die electron, e – charge of the electron and ε0 = permittivity of vacuum, then [(h2ε0)/(me2)] has the unit

22. ii the displacement y of a particle is y = A sin(pt + qx), then dimensional formula of ‘ Apq ‘ is

23. A circular railway track of radius r is banked at angle θ so that a train moving with speed u can safely go round the track. A student writes : tan θ = rg/v2 Why this relation is not correct ?
Equality of dimensions does not guarantee correctness of the relation.
Dimensionally correct relation may not be numerically correct.
The relation is dimensionally incorrect.

24. The parameter $latex frac { m{ Q }^{ 4 } }{ { { varepsilon }_{ 0 } }^{ 2 }{ h }^{ 2 } }$ has the dimensions of (m = mass, Q = charge, epesolonat 0 Permittivity and h = Planck’s constant)

25. The dimensional formula of( 1/2) ε0 E2 is(ε0 is permittivity of free space and E is electric field)


text:
Multivariable calculus

Multivariable calculus (also known as multivariate calculus) is the extension of calculus in one variable to calculus in more than one variable: the differentiation and integration of functions involving multiple variables, rather than just one.

Typical operations

Limits and continuity

A study of limits and continuity in multivariable calculus yields many counter-intuitive results not demonstrated by single-variable functions. For example, there are scalar functions of two variables with points in their domain which give a particular limit when approached along any arbitrary line, yet give a different limit when approached along a parabola. For example, the function

$f(x,y) = \frac{x^2y}{x^4+y^2}$

approaches zero along any line through the origin. However, when the origin is approached along a parabola $y=x^2$, it has a limit of 0.5. Since taking different paths toward the same point yields different values for the limit, the limit does not exist.

Continuity in each argument does not imply multivariate continuity: For instance, in the case of a real-valued function with two real-valued parameters, $f(x,y)$, continuity of $f$ in $x$ for fixed $y$ and continuity of $f$ in $y$ for fixed $x$ does not imply continuity of $f$. As an example, consider

$f(x,y)= \begin{cases} \frac{y}{x}-y & \text{if } 1 \geq x > y \geq 0 \\ \frac{x}{y}-x & \text{if } 1 \geq y > x \geq 0 \\ 1-x & \text{if } x=y>0 \\ 0 & \text{else}. \end{cases}$

It is easy to check that all real-valued functions (with one real-valued argument) that are given by $f_y(x):= f(x,y)$ are continuous in $x$ (for any fixed $y$). Similarly, all $f_x$ are continuous as $f$ is symmetric with regards to $x$ and $y$. However, $f$ itself is not continuous as can be seen by considering the sequence $f(\frac{1}{n},\frac{1}{n})$ (for natural $n$) which should converge to $f(0,0)=0$ if $f$ was continuous. However, $\lim_{n \to \infty} f(\frac{1}{n},\frac{1}{n}) = 1.$

Partial differentiation

The partial derivative generalizes the notion of the derivative to higher dimensions. A partial derivative of a multivariable function is a derivative with respect to one variable with all other variables held constant.

Partial derivatives may be combined in interesting ways to create more complicated expressions of the derivative. In vector calculus, the del operator ($\nabla$) is used to define the concepts of gradient, divergence, and curl in terms of partial derivatives. A matrix of partial derivatives, the Jacobian matrix, may be used to represent the derivative of a function between two spaces of arbitrary dimension. The derivative can thus be understood as a linear transformation which directly varies from point to point in the domain of the function.

Differential equations containing partial derivatives are called partial differential equations or PDEs. These equations are generally more difficult to solve than ordinary differential equations, which contain derivatives with respect to only one variable.

Multiple integration

The multiple integral expands the concept of the integral to functions of any variable. Double and triple integrals may be used to calculate areas and volumes of regions in the plane and in space. Fubini's theorem guarantees that a multiple integral may be evaluated as a repeated integral.

The surface integral and the line integral are used to integrate over curved manifolds such as surfaces and curves.

Fundamental theorem of calculus in multiple dimensions

In single-variable calculus, the fundamental theorem of calculus establishes a link between the derivative and the integral. The link between the derivative and the integral in multivariable calculus is embodied by the famous integral theorems of vector calculus:

In a more advanced study of multivariable calculus, it is seen that these four theorems are specific incarnations of a more general theorem, the generalized Stokes' theorem, which applies to the integration of differential forms over manifolds.

Applications and uses

Techniques of multivariable calculus are used to study many objects of interest in the material world. In particular,

Domain/Codomain Applicable techniques
Curves $f: \mathbb{R} \to \mathbb{R}^n$ Lengths of curves, line integrals, and curvature.
Surfaces $f: \mathbb{R}^{2} \to \mathbb{R}^n$ Areas of surfaces, surface integrals, flux through surfaces, and curvature.
Scalar fields $f: \mathbb{R}^n \to \mathbb{R}$ Maxima and minima, Lagrange multipliers, directional derivatives.
Vector fields $f: \mathbb{R}^m \to \mathbb{R}^n$ Any of the operations of vector calculus including gradient, divergence, and curl.

Multivariable calculus can be applied to analyze deterministic systems that have multiple degrees of freedom. Functions with independent variables corresponding to each of the degrees of freedom are often used to model these systems, and multivariable calculus provides tools for characterizing the system dynamics.

Multivariable calculus is used in many fields of natural and social science and engineering to model and study high-dimensional systems that exhibit deterministic behavior. Non-deterministic, or stochastic systems can be studied using a different kind of mathematics, such as stochastic calculus. Quantitative analysts in finance also often use multivariate calculus to predict future trends in the stock market.


text:
Products

#### What are the Odour Neutralizers?

These products are the result of the study of natural plant substances (flowers, fruits, herbs, roots, etc.) and the development of processes of industrial biosynthesis in order to design complex mixtures of fragrances and essential oils which can neutralize any odour and gas emission coming from industrial manufacturing processes, wastewater treatment plants and zootechnics.

Unlike the traditional air fresheners, the odour neutralizers of the MasterDeo line do not mask the smells in the air but neutralize them completely.

#### Activities

The neutralization relies on the chemical or physical destruction of the molecules that build the odours.

###### We know how to eliminate the odours, without masking them.

Analytical methods: gas chromatography, mass spectrometry valid for specific substances dissolved in gaseous fluids. They do not reveal the origin of olfactory harassment .
Olfactometric method: human noses, appropriately selected assessors trained to recognize the intensity and the pleasantness of odours.
Sensory methods: electronic noses, biosensors that combine the action of olfactory receptors and physical-chemical transducers of the signal.
Olfactometric Unit (OU) expresses a value that combines the odour concentration threshold and its air dilution.

vegetable extracts of plants, herbs, flowers, fruits and essential oils, properly mixed to act specifically against any odour molecule.
common excipients (propylene glycol) and surfactants (which make them soluble in water).

How does the neutralization of odours work against some specific mixtures that contain these functional groups?

Es. 1) ${H}_{2}S$ : the process of chemical decomposition of ${H}_{2}S$ occurs by chemical interaction of the amine functional group $R\text{-}N{H}_{3}$ that is: $R\text{-}N{H}_{3}+{H}_{2}S$ $R\text{-}N{H}_{3}+SH$ $R\text{-}N{H}_{3}+SH+{O}_{2}+{H}_{2}O$ $R\text{-}N{H}_{3}+S{O}_{4}+OH$ $R\text{-}N{H}_{3}+S{O}_{4}+OH$ $R\text{-}N{H}_{3}+S{O}_{4}+{H}_{2}O$ in that reaction, the hydrogen sulphide is captured by the amino functional group. With oxygen, a non-volatile compound containing sulphate is then formed.

Es. 2) $N{H}_{3}$ : the process of chemical decomposition of $N{H}_{3}$ is made per condensation by the functional acid group $R\text{-}COOH$ namely: $R\text{-}COOH+N{H}_{3}$ $R\text{-}COON{H}_{4}$ the ammonia is captured by the functional acid group $COOH$ and a no more volatile compound containing ammonium salt is formed.

The odour neutralisation occurs with different chemical reactions for each substance which has to be treated:

Construction the foul-smelling gas reacts chemically with the neutralizing molecule and forms an odourless compound (ammonia).
Combination the foul-smelling gas is combined with a pleasant odour forming a third lightly scented or unscented product (chlorine derivatives).
Counteraction when you add to a bad smelling substance another one that opposes to it, interferes with it and finally eliminates the odour (derivatives of sulphur).
Absorption a foul-smelling substance is absorbed by the neutralizer and forms an odourless product (styrene).
Chemical reaction all the ingredients of the neutralizer interact with the bad odours and form an odourless product (mercaptans).
Collection the micelles of some ingredients capture the odour molecules in the air, they create then a heavy drop that falls to the ground and holds the odours.

Neutralization in the air: indoor application, wastewater treatment plants, sludge accumulation tanks, porous floors, toilets, urinals, scrubbers and fireplaces.
Contact neutralization: landfills, composting plants, waste disposal, sludge tanks, wastewater treatment plants, septic tanks, stables, farms, chemical toilets and well drains

#### More details

The odour neutralization against the smell pollution

The problem of the odours from industrial and composting plants, landfills, wastewater treatment plants etc. is becoming more and more important. Nowadays at a national level there are no specific regulations and no limit values for the odour emissions. The current legislation about the air quality, the waste and the wastewater treatment refers however to the prevention and the reduction of odour annoyance. Some Regions have already introduced some laws and regulations regarding the reduction of the odour emissions and have set the upper limits of emissions in the atmosphere before the forthcoming adaptation to European guidelines. The “odour annoyance” is part of the regulations concerning the emissions of pollutants in the atmosphere, even though there are no benchmarks to verify the intensity of the odour emissions. The bad smell is considered as a true form of environmental pollution.


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# Markov chain: how to estimate the transition matrix? I don't have the underlying observations, just the sum by state and time

I have a matrix for data that (supposedly) follows a Markov process with an absorbing state; I have 3 possible states and 50 periods (discrete states, discrete time). Element [t,s] of the matrix tells me how much of my population is in state s at time t. Something like:

90 10 0
80 10 10

The question is: how can I estimate the transition matrix?

I use Python but might use R or Julia for this - or I'd be happy to consider converting an algorithm to Python if not too complex.

Note that I only have this matrix as described - I do not have the underlying individual observations; in other words, I do not know which item went from which state to which state - only the total number of items in each state at each time.

The answers/comments I have found refer to cases where you know the underlying observations, which I don't in this case. I understand there is a package for R: https://cran.mtu.edu/web/packages/markovchain/ But, if I understood it correctly, it requires the actual observations, and cannot really be used with my kind of data - is that right?

I have put together a few lines of Python to simulate the kind of data I want to estimate.

Thoughts / ideas / suggestions? Thanks!

import numpy as np
import pandas as pd

# the initial state
s=[0.9,0.1,0]

# One matrix is applied in 15% of the cases; I do this to introduce
# some 'noise'
m1=np.array([[0.91,0.09,0 ], [0.3,0.4,0.3],[0,0,1] ])
m2=np.array([[0.9,0.1,0 ], [0.32,0.4,0.28],[0,0,1] ])

periods = 50
ran = np.random.rand(periods)

evol= np.zeros((periods,3))
evol[0,:]=s

for t in np.arange(1, periods):
if ran[t]<0.15:
trans = m1
else:
trans = m2
evol[t,:]=np.matmul( evol[t-1,:], trans )

• Can you better explain the first few lines of your question. What are you trying to achieve ? – rgk Mar 14 at 17:06
• Is the structure of the data I have clear? Like I said, I am trying to estimate the transition matrix. Let me try to rephrase. Let's suppose I have data on the medical status of some patients; there are 3 states: healthy, sick and dead. Dead is an absorbing state (obviously!). I do not have the individual observations that tell me that, say, patient 1 went from HHHHHHSSSSD while patient 2 went from SSHHHHHHSSHHHD All I have is, for each period, the total count of my population broken down among these 3 states. So I know that, at time 0, I have 90 healthy, 10 sick and 0 dead, etc. – Pythonista anonymous Mar 14 at 17:58
• In light of all this, how can one estimate the transition matrix? – Pythonista anonymous Mar 14 at 18:00
• Thank you for your comments. As far as I understand, since there is an absorbing state in your problem, the markov chain is not ergodic which means there is no n-step transition probability matrix. – rgk Mar 14 at 22:01
• I'm not sure I am following. Yes, the chain is not ergodic, but how does this affect whether/how the transition matrix can be estimated? – Pythonista anonymous Mar 14 at 23:47


text:
# Combinatorics homework question

The answer is 54912.

This is what I've tried so far: So first you have to pick a rank to occur 3 times so thats 13, now you gotta pick a suit that that rank has, which is now 13 * 4. Now you need to pick that same rank 2 more times. The second time it will be 13 * 3, then the third time it will be 13 * 2. So the expression so far is (13^3)*(4*3*2) but that number is already too high when I try to add the other two cards. What am I doing wrong?

The key to this question is the terms 'combination' and 'subset'. You want to model this problem with combinations for each subset/property and then multiply these together.

That is: rank $\times$ suit $\times$ other_rank $\times$ other_suits.

There are 13 ranks so we choose one with $\binom{13}{1}$, choose our suits with $\binom{4}{3}$, chose our other two ranks $\binom{12}{2}$ and then the suits for these two $\binom{4}{1}^2$.
So the answer is $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2=54912$.
$13 \cdot {4 \choose 3} \cdot {12 \choose 2} \cdot 4 \cdot 4 = 54912$.


text:
# Technical Details for Choice Modelling Utility Calculations

### Input data

This QScript uses data from individual-level parameter means from a choice modeling experiment. The variables should be combined into a Number - Multi question that has the string Individual-Level Parameter Means in the name. The individual-level parameters question can be generated in Q from a segmentation of an experiment question (by right-clicking a node of the tree and selecting Save Individual-Level Parameter Means and Standard Deviations).

Alternatively, the individual-level parameter means can be computed externally and imported into Q as part of a larger data set. In this case:

• There should be one variable for each attribute level.
• Each variable should have a Label in the form: <Attribute Name>: <Level Label>
• The variables should be combined as a Number - Multi question.
• The text Individual-Level Parameter Means should be included in the question name.

### Numeric attributes

Numeric attributes, for example Price, will be treated as categorical attributes for the purpose of this calculation. When numeric attributes are present the QScript will prompt you to specify the levels that you want to display. You should enter two or more numbers separated by commas. For example, if your attribute is Price and you want to display levels corresponding to $1,$1.50, and \$2, you should enter 1, 1.5, 2 into the prompt.

### Outputs

A new Number - Multi question is generated in the project, and a table is added to the report to show the average of the computed values.