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# Please, what are the products ? I thought it was an elimination happening but there is...

###### Question:

please, what are the products ? I thought it was an elimination happening but there is no halogen, so I was confused
A OH H₂SO4 "CH₃ 120°C, distill

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## Poincare inequality from different approach

$\mathbf{Problem:}$ Let $\Omega$ be convex and $u\in C^1(\Omega)$. Suppose $S\subset\Omega$ with $\displaystyle \frac{|S|}{|\Omega|}>0$,

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy$

where $d$ is the diameter of $\Omega$ and $\displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy$.

$\mathbf{Proof\,1: }$

$\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt$

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy$

Set $z=ty+(1-t)x$, then $t|x-y|=|z-x|$. Since $|x-y|\leq d$, $\displaystyle t\geq \frac{|z-x|}{d}$. We also have $t^ndy=dz$

Then

$\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$.

$\mathbf{Proof\,2: }$ Using

$\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt$

where $\displaystyle w=\frac{x-y}{|x-y|}$. Then

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy$

$\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy$

$\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt$

$\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt$

$\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$

where $\xi(w)$ is the intersection of ray $x+tw$ with $\partial \Omega$ (it is unique because $\Omega$ is convex).

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

format: plain


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fixdiv - Maple Programming Help

# Online Help

###### All Products    Maple    MapleSim

fixdiv

compute the fixed divisor of a polynomial

 Calling Sequence fixdiv(a, x)

Parameters

 a, x - a is polynomial in x over the integers

Description

 • Compute the fixed divisor of $a\left(x\right)$ in ${Z}_{x}$.  The fixed divisor is the largest integer that divides $a\left(k\right)$ for all k in $Z$.  For primitive polynomials, the fixed divisor is no greater than $n!$ where $n=\mathrm{degree}\left(a,x\right)$.
 • The probability that the fixed divisor is greater than one for random polynomials of degree n is approximately 0.278 as n goes to infinity.

Examples

 > $\mathrm{fixdiv}\left(x\left(x+1\right)\left(x+2\right)\left(x+3\right),x\right)$
 ${24}$ (1)

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# Prove the time evolution operator is unitary

## Main Question or Discussion Point

How is (5.240b) derived? I get ${U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)=I$ instead.

My steps:
\begin{align}<\psi(t_0)\,|\,\psi(t_0)>&=\,<U(t_0, t)\,\psi(t)\,|\,U(t_0, t)\,\psi(t)>\\ &=\,<U^{-1}(t, t_0)\,\psi(t)\,|\,U^{-1}(t, t_0)\,\psi(t)>\\ &=\,<\psi(t)\,|\,{U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)\,|\,\psi(t)>\end{align}

Also, to get (5.240a), do we use the fact that $<\psi(t_0)\,|\,\psi(t_0)>\,=\,<\psi(t_0)\,|\,U^\dagger(t, t_0)\,U(t, t_0)\,|\,\psi(t_0)>$is true for any $\psi(t_0)$?

How is (5.240b) derived? I get ${U^{-1}}^\dagger(t, t_0)\,U^{-1}(t, t_0)=I$ instead.
Also, to get (5.240a), do we use the fact that $<\psi(t_0)\,|\,\psi(t_0)>\,=\,<\psi(t_0)\,|\,U^\dagger(t, t_0)\,U(t, t_0)\,|\,\psi(t_0)>$is true for any $\psi(t_0)$?


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# The quantity of oxygen that can dissolve in water depends on the temperature of the water. (So thermal pollution influences the oxygen content of water.) The graph shows how oxygen solubility $S$ varies as a function of the water temperature $T$. (a) What is the meaning of the derivative $S'(T)$? What are its units?(b) Estimate the value of $S'(16)$ and interpret it.

## a) oxygen solubility ,per liter per degree Celsius.b. -0.25 milligrams

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David Base G.

October 26, 2020

Finally, now I'm done with my homework

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Catherine A.

October 27, 2020

CA

Catherine A.

October 27, 2020

That was not easy, glad this was able to help

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### Video Transcript

So in this problem were given the soluble bility of oxygen in water and it is a various with temperature we were given a graph T. In degrees C. And yeah, sigh ability? S in milligrams per leader. And we have a graph that looks something like this, right? Or this is 16 up here and on down. Okay. And the first question we're asked is, what does S prime at T. Was the meaning of the derivative. And what are its units? Well, this property means the rate at which the oxygen, sorry, ability yeah, changes with respect two temperature, specifically the temperature of water in this case. Okay. And the units will be milligrams for leader degrees C. Okay, then, the next thing we're asked to do is to estimate the value Value of s. prime at 16 and interpret it. All right part. So from the graph we can see that? S prime at 16. Yeah. Is s of 16 S. Of a. Yeah, over 16 minus eight. Which coming off the graph Is approximately 10 -12 over eight, which is going to be -2/8. Or negative 0.25. Okay. So what does that mean? That means is the temperature temperature past 16 degrees C. Yeah, the oxygen, sorry, ability is decreasing ah a rate 0.25 milligrams for leader degrees C.

DM
Oklahoma State University

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# Chemical Equations

Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken

Nie wieder prokastinieren mit unseren Lernerinnerungen.

Reactions are fundamental to the study of chemistry, as you will know. Chemical equations are how we show every chemical reaction simply and easily.

• We are going to learn about the two different kinds of chemical equations you might find.
• We will consider what state symbols are.
• We'll also learn how to balance chemical equations.

## What is a chemical equation?

Before we delve into topics like balancing chemical equations, we need to know what they are.

A chemical equation is a symbolic representation of a chemical reaction in the form of symbols and formulae, showing the reactants and products of the reaction.

Chemical equations serve to represent reactions in a standardised and simple way that allows us to get the important information at a glance, that is, the reactants and products and their state, and the stoichiometry and direction of the reaction.

### Law of conservation of mass

The Law of conservation of mass is one of the most important laws in chemistry. It states that atoms cannot be created or destroyed in a chemical reaction.

In chemical equations, this means that there must be the same number of every type of atom on the reactant (left) side of the equation as on the product (right) side of the equation.

When this is the case, we say that the equation is balanced. Given that this law can never be broken, we have to make sure that every chemical equation we write and find must be balanced!

Another thing we need to know about this law is that it also applies to mass. Remember that everything has a mass, even gases! So, if the total mass of all the reactants put together is 10g, the total mass of the products put together must also equal 10g. You might find some reactions that seem to break this rule. But remember, that isn’t 'possible'; there must always be an explanation. In most cases, it’s because there is a gas involved.

Some examples:

• One of the products is a gas and the reaction is done in an open container. It will seem like the mass has decreased.

• This is only because a gas has been produced and has left the test tube and gone into the air, so the mass of the gas hasn't been registered.

• Heating in air - it might seem like suddenly the mass of the substance you were heating has increased.

• This only appears so because the substance has reacted with the oxygen in the air. The scale is measuring not only the mass of the original substance, but also the mass of the extra oxygen.

### State symbols

Sometimes, you will find state symbols in a balanced symbol equation. They tell us some specific information about the substance they are next to.

Table 1. State symbols

State symbol

Meaning

(s)

solid

(l)

liquid

(g)

gas

(aq)

aqueous solution/dissolved in water

In an equation, they go immediately after the substance they are describing and so look like this:

$$Fe_{(s)} + CuSO_{4\ (aq)} \rightarrow FeSO4_{(aq)}+ Cu_{(s)}$$

In this example, by looking at the state symbols, we know that iron (Fe) is in its solid state, copper sulfate (CuSO4) is in aqueous solution, iron sulfate (FeSO4) is also in aqueous solution, and copper (Cu) is in its solid state.

### Why do we need chemical equations?

You can’t escape reactions in chemistry. They’re integral to understanding how things work and finding out the properties of a substance. We use chemical equations to represent every single reaction, so it’s pretty important that you understand them!

## Chemical equations formula

In any reaction, you start off with substances that, after some time and changes, become different substances. These substances can be molecules of elements e.g. H2, or of compounds e.g. NH3.

The substances at the beginning of a chemical reaction are called reactants, and the ones at the end are called products.

$$N_{2}+3H_{2}\rightarrow 2NH_{3}$$

To represent the changes during the reaction, the simplified formula above is usually used. This is a chemical equation formula. It just gives information about the products and reactants, so any intermediaries that are consumed during the chemical reaction will not appear in a basic chemical equation.

A little number after an element shows how many atoms of the element are found in one unit of that substance.

For example, next to the N for nitrogen on the reactant side here, there is a little ‘’ after the element; this means there are 2 nitrogen atoms!

### What are word equations?

One way we can represent a reaction is by using a word equation. That is to say by writing out the names of all the substances, instead of using symbols.

E.g. the chemical word equation for photosynthesis is:

$$Carbon\ dioxide + water \rightarrow glucose + oxygen$$

The balanced chemical formula for photosynthesis, though, is:

$$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$$

Or,

$$Hydrogen+ oxygen \rightarrow water$$

This equation tells us that if we add hydrogen and oxygen together, we make water.

As you can see, the reactants go on the left, and the products go on the right. If there are multiple substances, we separate them using a ‘+’ sign.

It is exactly the same as a sum in maths, e.g. 2+3 = 5, just replace the ‘=’ into a ‘→’!

### What are symbol equations?

Word equations are great, as they tell us what substances are being used and what they become in a reaction. But what if you wanted to know how much of everything is reacting? As in the example of hydrogen and oxygen making water, is there an equal amount of hydrogen and oxygen reacting, or is there more of one or the other? Using symbol equations can help us understand that!

The symbol equation for the same reaction is:

$$2H_{2} + O_{2}\rightarrow 2H_{2}O$$

You will have noticed that there is a number ‘2’ in front of some of the substances. If there isn’t a number in front of a substance, we always assume that it is ‘1’. We put numbers in front of every substance involved in a reaction to show the ratio of units of each substance. If there isn't a number there, we assume there is a '1'.

In this example, 2 units of hydrogen react with one unit of oxygen to produce 2 units of water. It doesn't matter how big each unit is, only that we have twice as many hydrogen units as oxygen units.

This means that in this reaction, there are twice as many hydrogen and water molecules as there are oxygen molecules.

Don’t worry if you don’t know where these numbers came from or why they are 2s and not 3s or 4s! We will learn all about that later when we balance chemical equations.

## Chemical equation examples

Here is a list of chemical equations to help you recognise when you have one!

$$N_{2\ (g)} + 3H_{2\ (g)} \rightarrow 2NH_{3\ (g)}$$

Here, we have 1 unit of nitrogen and 3 units of hydrogen, which react to form 2 units of ammonia.

Here are some others. Can you understand what each equation is showing?

$$Mg(OH)_{2\ (s)} + SO_{2\ (g)}\rightarrow MgSO_{3\ (s)} + H$$

$$C_{6}H_{12}O_{6} \rightarrow 2 C_{2}H_{5}OH + 2 CO_{2}$$

## Balancing chemical equations

We have mentioned balancing chemical equations a lot through this article. Now is the time for you to learn how to do it!

Learning to do this is important because you will have to do this hundreds of times during the entire course. There is always a question asking you to do this in the GCSE Chemistry exam.

To balance a chemical equation, we need to have the same number of atoms of each element involved on either side of the equation. If we don't, we would be breaking the Law of the Conservation of Mass! We do this by changing the quantities of each substance involved. However, we can't add in any different substances, because this would create an entirely new reaction.

The steps to balancing an equation are:

1. Write out the reaction as a symbol equation, and count the number of atoms of each element on the reactant side, and the number of atoms of each element on the product side.
2. Multiply the quantity of one substance, to have the same number of atoms of one element on both sides of the equation.
3. Recount the number of atoms of each element on the reactant and product sides of the equation.
4. Repeat steps 2 and 3 until there are equal numbers of each element atom on both sides of the equation. Then, the equation is balanced.

Let’s go through these steps together with an example - hydrogen reacting with oxygen to make water.

### 1. Write out the reaction as a symbol equation

Hydrogen is usually represented as H, oxygen as O and water as HO. As we know what the reactants and products are, let’s demonstrate this as a symbol equation which we are going to balance. This step might already be done for you.

$$H_{2}+O_{2}\rightarrow H_{2}O$$

### 2. Below the reactant and product side of the equation, write a list of all the elements involved in the reaction

Now count the number of atoms of each element on the reactant side, and then on the product side.

This is to keep track of how many of each element there are on either side of the equation. in this case, the elements involved are hydrogen (H), and oxygen (O).

Make sure to count every single atom, not forgetting about the little numbers - like a little ‘’ after an element! Therefore ‘H’ means there are 2 H atoms in that molecule, so we can write 2 beside the H in the element list we wrote in step 2.

Check hydrogen (H) doesn’t appear in any of the other substances on that side!

Once you’re sure, repeat this counting for all of the other elements.

Remember if there isn’t a little number after an element, we always assume there is only 1 - as in the case of O in HO.

$$H_{2}+O_{2}\rightarrow H_{2}O$$

H: 2 H: 2

O: 2 O: 1

### 3. Multiply the quantity of one substance, to have the same number of atoms of one element on both sides of the equation

Look at each row individually in the list you wrote for step 2 which shows the number of atoms of each element on both sides of the equation. You can see that there are some elements with different numbers of atoms on either side of the equation. In this case, there are two oxygen atoms on the left-hand side, but only one on the right-hand side.

Remember, according to the law of conservation of mass, we can't create or destroy atoms; there has to be the same number of atoms of each element on both sides of the equation. We need to add more oxygen atoms to the left-hand side of the equation, as right now there aren't enough.

Looking at oxygen (O) though, we see that on the reactant side there are 2 oxygen atoms, and only 1 on the product side. So, this means we need to add one more oxygen atom to the product side.

So how can we add an oxygen atom? We can’t add a different substance or change the formula of water as that would change the equation completely - it would be a whole new reaction! So we add one more water molecule, and have 2 water molecules in total in the equation (1 originally + 1 added now). We represent this by writing a 2 in front of the water.

So we have:

H + O 2HO

### 4. Recount the number of atoms of each element on the reactant and product sides of the equation. Update your list!

Because we have added another molecule, the number of atoms of one, some, or all the elements will change. So, we need to update our list to check!

$$H_{2}+O_{2}\rightarrow 2H_{2}O$$

H: 2 H: 4

O: 2 O: 2

Previously, we had 1 water molecule. This gave us 2 hydrogen atoms and 1 oxygen atom. We now have 2 water molecules, so we have twice as many hydrogen and oxygen atoms.

### 5. Repeat steps 3 and 4 until the equation is balanced!

We can see that by adding that water molecule, the number of oxygen atoms on both sides of the equation is the same, but the number of hydrogen atoms is now not the same!

Thankfully, we can do the same thing that we did last time and look for a reactant that contains hydrogen atoms to try and make the number of hydrogen atoms the same on both sides. In this case, we need 2 more hydrogen atoms on the reactant side, as the difference between 4 (product side) and 2 (reactant side) is 2!

We can add 1 more H molecule, which contains 2 hydrogen atoms to make the number of hydrogen atoms the same on both sides.

$$2H_{2}+O_{2}\rightarrow 2H_{2}O$$

And repeat step 4, updating the list again:

$$2H_{2}+O_{2}\rightarrow 2H_{2}O$$

H: 4 H: 4

O: 2 O: 2

We can see that we have finally balanced this chemical equation!

### 6. Count the atoms on each side of the equation for the last time, like in step 2, just to check!

$$2H_{2}+O_{2}\rightarrow 2H_{2}O$$

H: 4 H: 4

O: 2 O: 2

## Chemical Equations - Key takeaways

• In a reaction, the substances you have at the beginning are called reactants, and the ones you have at the end are called products.
• In a word equation, you write out the names of each substance. For example, hydrogen + oxygen → water.
• Symbol equations let us know how much of each substance is actually reacting.
• The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction. So, the total mass of reactants always equals the total mass of products
• State symbols give us extra information about substances: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous solution.
• In a balanced chemical equation, there is the same number of each type of atom on either side of the equation.

We can combine several equations by placing the reactants on the left side of the equation, and the products that are formed over the reaction time, on the right side of the equation.

A chemical equation is a way we have of theoretically representing how products are formed from reactants. Normally, the products are placed on the right side of the equation and the reactants are on the left side, separated by an arrow pointing to the products.

To write a chemical equation, the products are placed on the right side of the equation and the reactants are on the left side, separated by an arrow pointing to the products.

1. Write out the reaction as a symbol equation.

2. Below the reactant and product side of the equation, write a list of all the elements involved in the reaction.

3. Multiply the quantity of one substance, to have the same number of atoms of one element on both sides of the equation.

4. Recount the number of atoms of each element on the reactant and product sides of the equation. Update your list.

5. Repeat steps 3 and 4 until the equation is balanced.

6. Count the atoms on each side of the equation for the last time, like in step 2, just to check.

An example of chemical equation is the formation of water:

2H2+O2-->2H2O

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text:
# The Unapologetic Mathematician

## Characteristic Functions as Idempotents

I just talked about characteristic functions as masks on other functions. Given a function $f:X\rightarrow\mathbb{R}$ and a subset $S\subseteq X$, we can mask the function $f$ to the subset $S$ by multiplying it by the characteristic function $\chi_S$. I want to talk a little more about these functions and how they relate to set theory.

First of all, it’s easy to recognize a characteristic function when we see one: they’re exactly the idempotent functions. That is, ${\chi_S}^2=\chi_S$, and if $f^2=f$ then $f$ must be $\chi_S$ for some set $S$. Indeed, given a real number $a$, we can only have $a^2=a$ if $a=0$ or $a=1$. That is, $f(x)=0$ or $f(x)=1$ for every $x$. So we can define $S$ to be the set of $x\in X$ for which $f(x)=1$, and then $f(x)=\chi_S(x)$ for every $x\in X$. Thus the idempotents in the algebra of real-valued functions on $X$ correspond exactly to the subsets of $X$.

We can define two operations on such idempotent functions to make them into a lattice. The easier to define is the meet. Given idempotents $\chi_S$ and $\chi_T$ we define the meet to be their product:

$\displaystyle\left[\chi_S\wedge\chi_T\right](x)=\chi_S(x)\chi_T(x)$

This function will take the value ${1}$ at a point $x$ if and only if both $\chi_S$ and $\chi_T$ do, so this is the characteristic function of the intersection

$\displaystyle\chi_S\wedge\chi_T=\chi_{S\cap T}$

We might hope that the join would be the sum of two idempotents, but in general this will not be another idempotent. Indeed, we can check:

$\displaystyle(\chi_S+\chi_T)^2={\chi_S}^2+{\chi_T}^2+2\chi_S\chi_T=(\chi_S+\chi_T)+2\chi_{S\cap T}$

We have a problem exactly when the corresponding sets have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it off to define

$\displaystyle\left[\chi_S\vee\chi_T\right](x)=\chi_S(x)+\chi_T(x)-\chi_S(x)\chi_T(x)$

We can multiply this out to check its idempotence, or we could consider its values. If $x$ is not in $T$, then $\chi_T(x)=0$, and we find $\chi_S\vee\chi_T=\chi_S$ — it takes the value ${1}$ if $x\in S$ and ${0}$ otherwise. A similar calculation holds if $x\notin S$, which leaves only the case when $x\in S\cap T$. But now $\chi_S(x)$ and $\chi_T(x)$ both take the value ${1}$, and a quick calculation shows that $\chi_S\vee\chi_T$ does as well. This establishes that

$\displaystyle\chi_S\vee\chi_T=\chi_{S\cup T}$

We can push further and make this into an orthocomplemented lattice. We define the orthocomplement of an idempotent by

$\displaystyle\left[\neg\chi_S\right](x)=1-\chi_S(x)$

This function is ${1}$ wherever $\chi_S$ is ${0}$, and vice-versa. That is, it’s the characteristic function of the complement

$\displaystyle\neg\chi_S=\chi_{X\setminus S}$

So we can take the lattice of subsets of $X$ and realize it in the nice, concrete algebra of real-valued functions on $X$. The objects of the lattice are exactly the idempotents of this algebra, and we can build the meet and join from the algebraic operations of addition and multiplication. In fact, we could turn around and do this for any commutative algebra to create a lattice, which would mimic the “lattice of subsets” of some “set”, which emerges from the algebra. This sort of trick is a key insight to quite a lot of modern geometry.

December 23, 2009 Posted by | Algebra, Fundamentals, Lattices | 2 Comments

## Galois Connections

I want to mention a topic I thought I’d hit back when we talked about adjoint functors. We know that every poset is a category, with the elements as objects and a single arrow from $a$ to $b$ if $a\leq b$. Functors between such categories are monotone functions, preserving the order. Contravariant functors are so-called “antitone” functions, which reverse the order, but the same abstract nonsense as usual tells us this is just a monotone function to the “opposite” poset with the order reversed.

So let’s consider an adjoint pair $F\dashv G$ of such functors. This means there is a natural isomorphism between $\hom(F(a),b)$ and $\hom(a,G(b))$. But each of these hom-sets is either empty (if $a\not\leq b$) or a singleton (if $a\leq b$). So the adjunction between $F$ and $G$ means that $F(a)\leq b$ if and only if $a\leq G(b)$. The analogous condition for an antitone adjoint pair is that $b\leq F(a)$ if and only if $a\leq G(b)$.

There are some immediate consequences to having a Galois connection, which are connected to properties of adjoints. First off, we know that $a\leq G(F(a))$ and $F(G(b))\leq b$. This essentially expresses the unit and counit of the adjunction. For the antitone version, let’s show the analogous statement more directly: we know that $F(a)\leq F(a)$, so the adjoint condition says that $a\leq G(F(a))$. Similarly, $b\leq F(G(b))$. This second condition is backwards because we’re reversing the order on one of the posets.

Using the unit and the counit of an adjunction, we found a certain quasi-inverse relation between some natural transformations on functors. For our purposes, we observe that since $a\leq G(F(a))$ we have the special case $G(b)\leq G(F(G(b)))$. But $F(G(b))\leq b$, and $G$ preserves the order. Thus $G(F(G(b)))\leq G(b)$. So $G(b)=G(F(G(b)))$. Similarly, we find that $F(G(F(a)))=F(a)$, which holds for both monotone and antitone Galois connections.

Chasing special cases further, we find that $G(F(G(F(a))))=G(F(a))$, and that $F(G(F(G(b))))=F(G(b))$ for either kind of Galois connection. That is, $F\circ G$ and $G\circ F$ are idempotent functions. In general categories, the composition of two adjoint functors gives a monad, and this idempotence is just the analogue in our particular categories. In particular, these functions behave like closure operators, but for the fact that general posets don’t have joins or bottom elements to preserve in the third and fourth Kuratowski axioms.

And so elements left fixed by $G\circ F$ (or $F\circ G$) are called “closed” elements of the poset. The images of $F$ and $G$ consist of such closed elements

May 18, 2009

## DeMorgan’s Laws

And here’s the post I wrote today:

Today, I want to prove two equations that hold in any orthocomplemented lattice. They are the famous DeMorgan’s laws:

$\displaystyle\neg(x\vee y)=\neg x\wedge\neg y$
$\displaystyle\neg(x\wedge y)=\neg x\vee\neg y$

First, we note that $x\leq x\vee y$ by definition. Since our complementation reverses order, we find $\neg(x\vee y)\leq\neg x$. Similarly, $\neg(x\vee y)\leq\neg y$. And thus we conclude that $\neg(x\vee y)\leq\neg x\wedge\neg y$.

On the other hand, $\neg x\wedge\neg y\leq\neg x$ by definition. Then we find $x=\neg\neg x\leq\neg(\neg x\wedge\neg y)$ by invoking the involutive property of our complement. Similarly, $y\leq\neg(\neg x\wedge\neg y)$, and so $x\vee y\leq\neg(\neg x\wedge\neg y)$. And thus we conclude $\neg x\wedge\neg y\leq\neg(x\vee y)$. Putting this together with the other inequality, we get the first of DeMorgan’s laws.

To get the other, just invoke the first law on the objects $\neg x$ and $\neg y$. We find

\displaystyle\begin{aligned}\neg x\vee\neg y&=\neg\neg(\neg x\vee\neg y)\\&=\neg(\neg\neg x\wedge\neg\neg y)\\&=\neg(x\wedge y)\end{aligned}

Similarly, the first of DeMorgan’s laws follows from the second.

Interestingly, DeMorgan’s laws aren’t just a consequence of order-reversal. It turns out that they’re equivalent to order-reversal. Now if $x\leq y$ then $x=x\wedge y$. So $\neg x=\neg(x\wedge y)=\neg x\vee\neg y$. And thus $\neg y\leq\neg x$.

May 8, 2009 Posted by | Algebra, Lattices | 8 Comments

## Orthogonal Complements and the Lattice of Subspaces

We know that the poset of subspaces of a vector space $V$ is a lattice. Now we can define complementary subspaces in a way that doesn’t depend on any choice of basis at all. So what does this look like in terms of the lattice?

First off, remember that the “meet” of two subspaces is their intersection, which is again a subspace. On the other hand their “join” is their sum as subspaces. But now we have a new operation called the “complement”. In general lattice-theory terms, a complement of an element $x$ in a bounded lattice $L$ (one that has a top element ${1}$ and a bottom element ${0}$) is an element $\neg x\in L$ so that $x\vee\neg x=1$ and $x\wedge\neg x=0$.

In particular, since the top subspace is $V$ itself, and the bottom subspace is $\mathbf{0}$ we can see that the orthogonal complement $U^\perp$ satisfies these properties. The intersection $U\cap U^\perp$ is trivial, since the inner product is positive-definite as a bilinear form, and the sum $U+U^\perp$ is all of $V$, as we’ve seen.

Even more is true. The orthogonal complement is involutive (when $V$ is finite-dimensional), and order-reversing, which makes it an “orthocomplement”. In lattice-theory terms, this means that $\neg\neg x=x$, and that if $x\leq y$ then $\neg y\leq\neg x$.

First, let’s say we’ve got two subspaces $U\subseteq W$ of $V$. I say that $W^\perp\subseteq U^\perp$. Indeed, if $p$ is a vector in $W^\perp$ then it $\langle w,p\rangle=0$ for all $w\in W$. But since any $u\in U$ is also a vector in $W$, we can see that $\langle u,p\rangle=0$, and so $p\in U^\perp$ as well. Thus orthogonal complementation is

Now let’s take a single subspace $U$ of $V$, and let $u$ be a vector in $U$. If $v$ is any vector in $U^\perp$, then $\langle v,u\rangle=\overline{\langle u,v\rangle}=0$ by the (conjugate) symmetry of the inner product and the definition of $U^\perp$. Thus $u$ is a vector in $\left(U^\perp\right)^\perp$, and so $U\subseteq U^{\perp\perp}$. Note that this much holds whether $V$ is finite-dimensional or not.

On the other hand, if $V$ is finite-dimensional we can take an orthonormal basis $\left\{e_i\right\}_{i=1}^n$ of $U$ and expand it into an orthonormal basis $\left\{e_i\right\}_{i=1}^d$ of all of $V$. Then the new vectors $\left\{e_i\right\}_{i=n+1}^d$ form a basis of $U^\perp$, so that $V=U\oplus U^\perp$. A vector in $V$ is orthogonal to every vector in $U^\perp$ exactly when it can be written using only the first $n$ basis vectors, and thus lies in $U$. That is, $U^{\perp\perp}=U$ when $V$ is finite-dimensional.

May 7, 2009

## Lattices

A poset which has both least upper bounds and greatest lower bounds is called a lattice. In more detail, let’s say we have a poset $(P,\preceq)$ and give it two operations: meet (written $\wedge$) and join (written $\vee$). These satisfy the requirements that

• $x\preceq x\vee y$ and $y\preceq x\vee y$.
• If $x\preceq z$ and $y\preceq z$ then $x\wedge y\preceq z$.
• $x\wedge y\preceq x$ and $x\wedge y\preceq y$.
• If $z\preceq x$ and $z\preceq y$ then $z\preceq x\wedge y$.

Not every poset has a meet and a join operation, but if these operations do exist they are uniquely specified by these requirements. In fact, we can see this sort of like how we saw that direct products of groups are unique up to isomorphism: if we have two least upper bounds for a pair of elements then they must each be below or equal to the other, so they must be the same.

We can derive the following properties of the operations:

• $a\vee(b\vee c)=(a\vee b)\vee c$ and $a\wedge(b\wedge c)=(a\wedge b)\wedge c$.
• $a\vee b=b\vee a$ and $a\wedge b=b\wedge a$.
• $a\vee(a\wedge b)=a$ and $a\wedge(a\vee b)=a$.

from these we see that there’s a sort of duality between the two operations. In fact, we can see that these provide two commutative semigroup structures that happen to interact in a certain nice way.

Actually, it gets even better. If we have two operations on any set satisfying these properties then we can define a partial order: $x\preceq y$ if and only if $x=x\wedge y$. So we can define a lattice either by the order property and get the algebraic properties, or we can define it by the algebraic properties and get the order property from them.

In many cases, a lattice also satisfies $x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z)$, or equivalently $x\wedge(y\vee z)=(x\wedge y)\vee(x\wedge z)$. In this case we call it “distributive”. A bit weaker is to require that if $x\preceq z$ then $x\vee(y\wedge z)=(x\vee y)\wedge z$ for all $y$. In this case we call the lattice “modular”.

A lattice may have elements above everything else or below everything else. We call a greatest element of a lattice $1$ and a least element ${0}$. In this case we can define “complements”: $x$ and $y$ are complements if $x\vee y=1$ and $x\wedge y=0$. If the lattice is distributive, then the complement of $x$ is unique if it exists. A distributive lattice where every element has a complement is called “Boolean”.

May 14, 2007 Posted by | Algebra, Lattices, Orders | 10 Comments


text:
Python Program to Convert Miles to Kilometers and vice versa

In this example, we will write a simple program to convert units of measurement from miles to kilometers and vice versa in Python. The conversion formula plays an important role in writing the program. Let us see how the formula looks:-

To Convert Miles to Kilometers
1 Mile = 1.60934 Kilometers
1 Kilometer = 0.621371 Miles

To better understand this example, make sure you have knowledge of the following tutorials:-

Python Program to Convert Miles to Kilometers:

miles = float(input("Enter the value in miles: "))
conversion_factor = 1.60934
kilometers = miles * conversion_factor
print('%.4f miles = %0.4f kilometers' %(miles, kilometers))

The output of the above program is:-

Enter the value in miles: 4
4.0000 miles = 6.4374 kilometers

Python Program to Convert Miles to Kilometers:

kilometers = float(input("Enter the value in kilometers: "))
conversion_factor = 0.621371
miles = kilometers * conversion_factor
print('%0.4f kilometers = %0.4f miles' %(kilometers, miles))

The output of the above program is:-

Enter the value in kilometers: 2
2.0000 kilometers = 1.2427 miles


text: EN|RU Volume 15, No 6, 2008, P. 63-89 UDC 519.718 V. V. Chugunova About realizations of boolean functions by asymptotically optimal reliable circuits Abstract: The following two problems are solved in this work:  1. It is shown that if each of irreducible complete base of functions of one or two variables is supplemented by one nonconstant Boolean function $\varphi(x_1,x_2)$ noncongruent to the basis functions, in most bases the estimation of reliability of functional elements circuits, subject to inverse malfunctions on the input gates of elements, reduces for almost all functions. 2. It is shown that if each of irreducible complete base of functions of one or two variables is supplemented by $k (k\ge3)$ nonconstant Boolean functions noncongruent to the basis functions, then in all obtained bases the estimation of reliability of functional elements circuits, subject to inverse malfunctions on the input gates of elements, is asymptotically (as $\varepsilon\to0$) $2\varepsilon$ (i.e., trivial) for all functions $f(x_1,x_2,\dots,x_n)$ except constants 0 and 1 and functions $x_i$ and $\overline x_i$, where $\varepsilon$ is the failure probability at each input of functional element, $i=\overline{1,n}$. It is also shown that the circuits constructed while solving the last two problems are asymptotically optimal reliable circuits as $\varepsilon\to0$.  Illustr. 3, tabl. 6, bibl. 5. Keywords: boolean functions, asymptotically optimal reliable circuits. Chugunova Varvara Valerevna 1 1. Penza State University, 40 Krasnaya str., 440026 Penza, Russia e-mail: burchug@sura.ru © Sobolev Institute of Mathematics, 2015


text:
The virtualspecies package embeds a function to randomly generate virtual species generateRandomSp, with many customisable options to constrain the randomisation process.

Let’s take a look:

library(virtualspecies)
## Loading required package: raster
## Loading required package: sp
worldclim <- getData("worldclim", var = "bio", res = 10)

my.stack <- worldclim[[c("bio2", "bio5", "bio6", "bio12", "bio13", "bio14")]]
random.sp <- generateRandomSp(my.stack)
##  - Perfoming the pca
##  - Defining the response of the species along PCA axes
##  - Calculating suitability values
##    The final environmental suitability was rescaled between 0 and 1.
##                   To disable, set argument rescale = FALSE
##  - Converting into Presence - Absence
##    --- Generating a random value of beta for the logistic conversion
##    --- Determing species.prevalence automatically according to alpha and beta
##    Logistic conversion finished:
##
## - beta = 0.585585585585586
## - alpha = -0.1
## - species prevalence =0.174187069412391
random.sp
## Virtual species generated from 6 variables:
##  bio2, bio5, bio6, bio12, bio13, bio14
##
## - Approach used: Response to axes of a PCA
## - Axes:  1, 2 ;  83.24 % explained by these axes
## - Responses to axes:
##    .Axis 1  [min=-19; max=2.6] : dnorm (mean=1.136441; sd=0.3143425)
##    .Axis 2  [min=-3.44; max=10.95] : dnorm (mean=0.2830632; sd=4.88196)
## - Environmental suitability was rescaled between 0 and 1
##
## - Converted into presence-absence:
##    .Method = probability
##    .probabilistic method    = logistic
##    .alpha (slope)           = -0.1
##    .beta  (inflexion point) = 0.585585585585586
##    .species prevalence      = 0.174187069412391

We can see that the species was generated using a PCA approach. Indeed, as explained in the PCA section, when you have a lot of variables, it becomes very difficult to generate a species with realistic environmental requirements. Hence, by default the function generateRandomSp uses a PCA approach if you have 6 or more variables, and a ‘response functions’ approach if you have less than 6 variables.

In the next sections I detail the many possible customisations for the function generateRandomSp.

## 5.1. General parameters

### 5.1.1. Choosing the approach

You can choose approach =:

• "automatic": a ‘response’ approach is used if you have less than 6 variables and a ‘PCA’ approach is used if you have 6 or more variables
• "random": a random approach is chosen (response or PCA)
• "response": to use a response approach
• "pca": to use a pca approach

### 5.1.2. Rescaling of the environmental suitability

By default, rescale = TRUE, which means that the environmental suitability is rescaled between 0 and 1.

### 5.1.3. Conversion to presence-absence

You can choose to enable the conversion of environmental suitability to presence-absence. To do this, set convert.to.PA = TRUE. You can customise the conversion:

• choose the conversion method with PA.method. You should leave it to probability unless you have a very particular reason to use the threshold approach.
• define the parameters alpha, beta and species.prevalence exactly as explained in the convert to presence-absence section, or leave them as they are to create a random conversion into presence-absence

## 5.2. Parameters specific to a ‘response’ approach

### 5.2.1. Define the possible response functions

At the moment, four relations are possible for a random generation of virtual species: Gaussian (gaussian), linear (linear), logistic (logistic) and quadratic (quadratic) relations. By default, all the relation types are used. You can choose to use any combination with the argument relations:

# A species with gaussian and logistic response functions
random.sp1 <- generateRandomSp(worldclim[[1:3]],
relations = c("gaussian", "logistic"),
convert.to.PA = FALSE)
##  - Determining species' response to predictor variables
##  - Calculating species suitability
## Generating virtual species environmental suitability...
##  - The response to each variable was rescaled between 0 and 1. To
##             disable, set argument rescale.each.response = FALSE
##  - The final environmental suitability was rescaled between 0 and 1.
##             To disable, set argument rescale = FALSE
random.sp1
## Virtual species generated from 3 variables:
##  bio2, bio1, bio3
##
## - Approach used: Responses to each variable
## - Response functions:
##    .bio2  [min=9; max=211] : logisticFun   (alpha=8.99818181818182; beta=81.4302024302024)
##    .bio1  [min=-269; max=314] : dnorm   (mean=202.375273752738; sd=157.190031900319)
##    .bio3  [min=8; max=95] : logisticFun   (alpha=-295.8; beta=29.2549042549043)
## - Each response function was rescaled between 0 and 1
## - Environmental suitability formula = bio1 * bio2 * bio3
## - Environmental suitability was rescaled between 0 and 1
plotResponse(random.sp1)
# A purely gaussian species
random.sp2 <- generateRandomSp(worldclim[[1:3]],
relations = "gaussian",
convert.to.PA = FALSE)
##  - Determining species' response to predictor variables
##  - Calculating species suitability
## Generating virtual species environmental suitability...
##  - The response to each variable was rescaled between 0 and 1. To
##             disable, set argument rescale.each.response = FALSE
##  - The final environmental suitability was rescaled between 0 and 1.
##             To disable, set argument rescale = FALSE
random.sp2
## Virtual species generated from 3 variables:
##  bio1, bio3, bio2
##
## - Approach used: Responses to each variable
## - Response functions:
##    .bio1  [min=-269; max=314] : dnorm   (mean=-170.437034370344; sd=100.247852478525)
##    .bio3  [min=8; max=95] : dnorm   (mean=63.5163351633516; sd=29.9352593525935)
##    .bio2  [min=9; max=211] : dnorm   (mean=56.4954449544495; sd=119.932639326393)
## - Each response function was rescaled between 0 and 1
## - Environmental suitability formula = bio1 * bio2 * bio3
## - Environmental suitability was rescaled between 0 and 1
plotResponse(random.sp2)

### 5.2.2. Rescale individual response functions

As explained in the section on the ‘response’ approach, you can choose to rescale or not each individual response function, with the argument rescale.each.response. TRUE by default.

### 5.2.3. Try to find a realistic species

An important issue with the generation of random responses to the environment, is that you can obtain a species willing to live in summer temperatures of 35Â°C and winter temperature of -50Â°C. This may interesting for generating species from another planet, but you are probably more interested in generating species that can actually live on Earth. There is therefore an option to do that, activated by default: realistic.sp.

When activating this argument, the function will proceed step-by-step to try defining a realistic species. At step one, one of the variable is chosen, and the program randomly determines a response function for this variable. Then, it will compute the environmental suitability of this species. At step two, the program will pick a second variable, and will constrain its random generation depending on the environmental suitability obtained at step one. For example, if at step 2 a gaussian response is picked, then the mean will be chosen in areas where the species had a high environmental suitability. Then, the environmental suitability is recalculated on the basis of the first two response functions. At step 3, another variable is picked, a response function randomly generated with respect to areas where the species already has a high suitability, and so on until there are no variables left. While this process can help, it does not always work, and can provide completely unrealistic results also. In this case, you should try different runs, or switch to a ‘PCA’ approach.

Let’s see an example :

realistic.sp <- generateRandomSp(worldclim[[c(1, 5)]],
realistic.sp = TRUE,
convert.to.PA = FALSE)
##  - Determining species' response to predictor variables
##  - Calculating species suitability
## Generating virtual species environmental suitability...
##  - The response to each variable was rescaled between 0 and 1. To
##             disable, set argument rescale.each.response = FALSE
##  - The final environmental suitability was rescaled between 0 and 1.
##             To disable, set argument rescale = FALSE
unrealistic.sp <- generateRandomSp(worldclim[[c(1, 5)]],
realistic.sp = FALSE,
convert.to.PA = FALSE)
##  - Determining species' response to predictor variables
##  - Calculating species suitability
## Generating virtual species environmental suitability...
##  - The response to each variable was rescaled between 0 and 1. To
##             disable, set argument rescale.each.response = FALSE
##  - The final environmental suitability was rescaled between 0 and 1.
##             To disable, set argument rescale = FALSE

Note that you can always let the function randomly determine the conversion to presence-absence by changing the argument convert.to.PA = TRUE (be careful: since parameters are defined randomly, the results can be bizarre), or you can do it by yourself later:

realistic.sp <- convertToPA(realistic.sp, beta = 0.5)
##    --- Determing species.prevalence automatically according to alpha and beta
##    Logistic conversion finished:
##
## - beta = 0.5
## - alpha = -0.05
## - species prevalence =0.491698330769981

If, in spite of all your attempts, you are struggling to obtain satisfactory species, then perhaps you should try the ‘PCA’ approach.

### 5.2.4. Define how response functions are combined to compute the environmental suitability

You can define how response functions are combined with the argument sp.type, exactly as explained in the ‘response’ approach section:

• species.type = "additive": the response functions are added.
• species.type = "multiplicative": the response functions are multiplied (default value).
• species.type = "mixed": there is a mix of additions and products to calculate the environmental suitability, randomly generated.

## 5.3. Parameters specific to a ‘PCA’ approach

### 5.3.1. Generate a random species with a PCA approach on a low-memory computer

As explained in the PCA section, if you have low-memory computer, or are working with very large raster files (large extent, fine resolution), then you can still perform the PCA by using a sample of the pixels of your raster.

To do this, set sample.points = TRUE, and choose the number of pixels to sample, depending on your computer’s memory, with nb.points.

# A safe run for a low memory computer
safe.run.sp <- generateRandomSp(worldclim[[c(1, 5, 6)]],
sample.points = TRUE,
nb.points = 1000)
##  - Determining species' response to predictor variables
##  - Calculating species suitability
## Generating virtual species environmental suitability...
##  - The response to each variable was rescaled between 0 and 1. To
##             disable, set argument rescale.each.response = FALSE
##  - The final environmental suitability was rescaled between 0 and 1.
##             To disable, set argument rescale = FALSE
##  - Converting into Presence - Absence
##    --- Generating a random value of beta for the logistic conversion
##    --- Determing species.prevalence automatically according to alpha and beta
##    Logistic conversion finished:
##
## - beta = 0.0940940940940941
## - alpha = -0.1
## - species prevalence =0.904424306397888

### 5.3.2. Generate stenotopic or eurytopic species

You can modify the niche.breadth argument to generate stenotopic (niche.breadth = 'narrow'), eurytopic (niche.breadth = 'wide'), or any type of species (niche.breadth = 'any').

-----------------
Do not hesitate if you have a question, find a bug, or would like to add a feature in virtualspecies: mail me!


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# UK Electronics industry needs re-structuring.

#### Flyback

##### Well-Known Member
Please find here advice on how the UK electronics industry can easily improve profitability....by adopting a "visible workplace" ethos, where engineers can be seen at all times. This stops non-work-related activity and encourages progress. Other countries already operate like this, and reap the benefits. This is why I direct it at UK, because other countries are already doing this.
I am writing this from my experience of having worked in over 27 different electronics engineering companies/departments in UK.
The idea involves electronics engineering companies/departments, simply adopting a “visible workplace” policy, -a working environment where engineers are visible at all times to their managers, and indeed to preferably most of the company’s other employees.
This creates a more hard-working environment, where engineers can develop better their skills. Also, such an environment makes it more difficult for engineers to waste time in less constructive activities such as messing about, arguing continuously with each other, forming “benefit cliques” (or gangs) so as to put pressure on engineers outside the “clique”, or even sabotaging one another’s work.
I have found that the best electronics companies have such a “visible workplace”. The Dyson company is a good example. The main hall at Dyson is totally open, everybody can be seen and heard at all times, by a number of other people. There are also cameras which is a great idea if not essential. (I believe microphones are also a great idea) At Dyson, there is little scope for engineers to engage in ‘non-engineering activities’ such as ganging up on a bright new engineer who has recently arrived at the company, etc etc. Even the mezzanine lab at Dyson is very open. Indeed, whilst working at Dyson, a young engineer who did not feel that he was being given enough suitable work came up to the mezz lab and continuously moaned at me about how he wanted me to teach him about a certain aspect of electronics. I didn’t have the time to do this. What impressed me was the speed with which a manager came up to the mezzanine lab and took charge of this lad, taking him away and finding something for him to do. This without me saying anything.
The “open” workplace in Dyson helps this to happen.
Unfortunately, I have often seen how engineers, instead of striving to better themselves technically, strive instead to simply try & make some of their other engineering colleagues look worse. –Continuously distracting engineers who are busy working, stealing or hiding their equipment, spreading stories about them, sabotaging their work, putting virus’s on their PC when they are not looking, etc etc. The “visible workplace” makes it so much harder for such less edifying activity to take place.
Sadly, there are few other UK engineering companies with the “open workplace” that Dyson has. Most of the company’s have a multitude of closed rooms, in which engineers can get up to all kinds of shenanigans behind the bosses back…
Example 1:
*****************************
In one UK based electronics company, I worked in a “closed lab” with three electronics engineers and one assembly girl. One of the engineers continually distracted the other two younger engineers, engaging them daily in hours & hours of non-work-related chat about just about anything. Very often, he would get them to join him in foul jokes directed at the assembly girl. The two younger engineers were of a disposition to 'get-on-with-their-work', however, the older engineer continuously distracted them, making them feel ‘anitsocial’ or ‘geekish’ if they did not join him in messing about instead of working. The manager of that site new that the work was going too slowly, however, that manager was seated in an office in a building across the road, and he simply could not see or hear what was going on in the lab…..and indeed, no other staff could see inside there either. Every month or so, the manager would insist that these engineers, due to their slow progress, should come and work a whole weekend, full-time, unpayed, to try and catch-up….the engineers would agree to this, but come the Monday, the larking about would just start all over again.
(I was in this lab as a contractor, I reported the foul language directed at the girl to her friend in the upstairs offices, but this girl told me not to report it further, as the girl might then be taken off her soldering exams, which would be useful to her in her imminent job application to a different company.)
One interesting point about this company, was that the older engineer discussed above was quite close to the Chief Design Engineer. The Chief Design Engineer gave me the impression of not wanting the body of engineers to become too technically adept, as indeed, perhaps this would simply act as a ‘threat’ to his technical superiority within the company. On one occasion, a bright new engineer had joined the company, and I overheard the Chief Design Engineer telling the older engineer “if he gets too keen, punch him one!”.
*****************************
There are multiple examples like the above, I wont bore yourself with them. These things happen far, far less in “open type” workplaces, such as Dyson.
Altogether, only five of the 27 UK workplaces that I have attended benefit at all from the “open workplace” principle.
Tridonic benefits from virtually all internal walls being see-through, thus managers always have good visibility over staff, and multiple different staff members can see each other. –Engineers can’t so easily get up to shenanigans. The engineering office and lab is actually ‘glass-walled’ to the entire production unit. Almost total visibility.
At a certain amplifier company, the entire staffing of the company, including the owners/directors were seated in a large, open room. It was simply not possible for anybody to loaf about or lark about without being in the sight of others who may well object to it/report it. There was a meeting room, but it had a huge glass window into the main room, so when the bosses were in the meeting room, they still had total visibility over all staff.
I have often noticed electronics companies where a number of engineering graduates are simply not given work, or are just given “token” jobs which are monitored loosely , if at all. Where this happens in non “open-workplace” type places, it often just descends into a shambles, with these engineers constantly chattering all day, -typically, at least one engineer will try to distract & stop his/her fellow junior colleagues from working & learning, -not wanting to be ‘left-behind’.
I have been to companies where they are paying hugely for consultancies to come in and present workplace efficiency ideologies such as the “Agile” working principle. The problem has a far simpler solution than that…..as discussed here, the “open workplace” principle. Get people working and stop the shenanigans/time-wasting.

#### JimB

##### Super Moderator
I bet that you are a real barrel of laughs to your co-workers. I wonder why that group that you worked with ganged up on you recently.

Cameras, microphones, what next a spot bonus for "fitting up" your colleagues.
You would be a good fit for a post with the STASI, but they were disbanded with the fall of the DDR.

Just to add a little flourish to your ideal workplace, you could put a sign above the front gate, Arbeit Macht Frei has a nice ring to it in this context.

JimB

#### Flyback

##### Well-Known Member
Seriously these “open workplaces” are the happier , more enjoyable places to work.. as well as being more productive. Think of a premier league football match….that’s effectively an “open workplace”, and nobody complains there.
Many of the UK electronics industries are surviving on Government grants derived from North Sea Oil. The UK pound currently being too high to allow a vibrant UK electronics/engineering industry. (North Sea Oil is almost run out though). Surviving on grants, and not having customer deadlines to meet, can make for slack type activity, so shaping things up a bit is no bad thing. If UK cannot make Engineering work when North Sea Oil runs out, then UK is a gonner.

#### picbits

##### Well-Known Member
I work in a department with 7 other engineers - it is an enclosed room and we have a party in there every day. We also get on with out work, hit deadlines and come up with innovative new designs to help the business sometimes through banter, mick taking and random conversation.

Considering the level of intellect in there, personality clashes are non existent and we've never fallen out with each other.

#### Flyback

##### Well-Known Member
OK thanks, you didn’t mention whether you seven are the only people who work there, and/or whether or not any supervisors , or the company owner actually sits in that room with you.
Another point is that you say you meet deadlines. But many UK companies are meeting deadlines, but these are just “invented deadlines”, for “invented products” , for “Invented markets”, and the whole shebang is simply being funded by UK government grants which derive from North sea oil.
If you are making it work then all the best to you , but I have many many stories of situations of companies where there are closeted little rooms, and productivity certainly is not high at all.
Another point is that the “open workplace” gets more important, the bigger is the company…..a little consultancy staffed by a group of hardworking mates with similar goals can often work without being in an “open workplace”, as such.

Also, supposing that an eigth engineer was introduced into that room of yours. Suppose that they were someone unknown to any of the clique.
Now suppose that this new engineer happens to be better at some job than an engineer that’s already in there…..and suppose that the new guy actually ends up taking all the nice juicy, technically interesting work away from the engineer who had been there for a while…..your boss demands this. In some situations, the seven may “gang” together in order to try and oust the new guy, after all, they don’t want to see their old mate getting pushed away from his beloved job, even if the new guy is better at it than their mate.

In UK , it is necessary for all to think about these things now, as the UK is drastically falling way behind Mainland Europe and the Far East, and North Sea Oil is about to run out.

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#### picbits

##### Well-Known Member
The group has over 60,000 employees, we have around 200 in our part of the business. Our manager alternates with one week in our office and one week in a different office.

Deadlines are pretty vital - it is a fast turnaround business and hourly costs for missing a deadline for a repair or new installation run into thousands.

A new engineer recently was assigned a project/product I'd come up with and had spent a lot of time researching in my own time - not a problem for us as ultimately it benefits the company and we support each other 100% on anything we do. New employees are carefully chosen both on their skills and personality - it is a well run department with plenty of cross skill matching and a great atmosphere.

Not all roses though - maybe 80%+ of my work is general maintenance and pretty tedious stuff but your job is what you make it. A lot of staff in this industry get poached by customers.

#### Flyback

##### Well-Known Member
Thanks PICbits, and it sounds like you are a software engineer, of course, that’s slightly easier to manage because your always at your laptop terminal, and each laptop has a camera and microphone, and so your boss can look at you and the room through that at any time. Also, the boss can tell on the network, from your computer strokes if you are working on what you should be. The boss may in fact have a hidden camera and microphone in your office of course.

#### picbits

##### Well-Known Member
Nope - we have CCTV in the office as we have many £100k's worth of kit - it is only used if we need to see who has pinched a bit of kit. I do both software programming and hardware systems - mainly hardware. I also walk several miles a day doing everything from changing filters in industrial computers to testing fire systems and redesigning data acquisition equipment.

#### Flyback

##### Well-Known Member
Well you are lucky, but as you can see from "Example 1" in the top post, the poor assembly woman wasn't quite so fortunate to have worked in such a civilised place as your own. Ditto many many other places with similar situations.
You sound like you are doing "real" work, and of course, many of the UK owned companys simply don't have that, and can "drift".

#### Flyback

##### Well-Known Member
The group has over 60,000 employees
..so we can tell that this company isn't UK owned...this is my point.
It was Nicolas Sarkozy who said "The British cannot do industry"

#### KeepItSimpleStupid

##### Well-Known Member
The only guy that gets ahead is the one that "kisses the bosses A". That's just the way it is.

One guy I worked for would interact with you "one on one". Basically a mentor. But for my area I was on my own.

Another guy I worked for basically let me interact with anybody as long as the time wasn't excessive. For that, I needed approval. The neat part is that he knew that a little bit of my time saved lots of effort. Two Phd's were trying to figure out why a piece of equipment sparked. They spent an entire weekend and got no where. 5 minutes of my time on Monday morning the problem was identified.

The last guy I worked for wanted to account for every minute of my time. He owned me. A genuine micro-manager.

At one point of my employment, we had a "matrix management" scheme. One guy had control of your vacation, time sheet and could divide your time percentage wise. The other person had control of what you had to do. I was pretty autonomous in that situation.

#### Flyback

##### Well-Known Member
Nicolas Sarkozy was exactly right when he said “Britain has no industry”
http://www.dailymail.co.uk/news/article-2093546/Nicolas-Sarkozy-claims-UK-industry-cheap-shot-economy.html
…Sarkozy said this because of his deep concern that the most densely populated country in Europe is not capable of doing industry for itself. This does not bode well, especially when our one lifeline, North Sea Oil, runs out.
We all know that UK industrial output figures are massaged to include overseas industrial sites which have been invited to UK, -UK using North Sea Oil wealth to literally pay them to set up divisions in UK. Also, the profits from north sea oil are also massaged into uk industrial output figures.
Companies like Jaguar-landrover are great, but its profits go overseas as its owned by Tata. Also, Nissan sunderland is great, but its profits don't funnel into the UK tax system, albeit a good wage earner for the guys working there.
Britain is well on the way down, and needs MASSIVE RESTRUCTURING IN its electronics/engineering industry….this includes a stopping of dependency on north sea oil (because soon it will run out), a depreciation of sterling, and a massive re-think on how British Electronics companies are managed…as in the top post
Nicolas Sarkozy was right, he was speaking out of a genuine concern for Britain, a close neighbour of france.

#### Flyback

##### Well-Known Member
OK nice one, not so nice for the poor woman of "Example 1" in the top post though eh?

#### large_ghostman

##### Well-Known Member
OK nice one, not so nice for the poor woman of "Example 1" in the top post though eh?
poor women was going for another job anyway, and had it bothered her that much she could have taken it further. Harassment in the work place has some stiff penalties. You seem to have a huge chip on your shoulder?? You might like to look at altering how you interact with people, also making people feel like they are constantly watched etc may well make them do there £20 hour job upto the value of £20, people who feel valued and trusted tend to work harder and have greater loyalty.

We have enough BB in the UK, I dont see why you would want more. I can imagine your fairly hard to work with and not is generally thought of as a team player. As a kid even I realize it matters to get on with your peers unless you want to impress management, but who wants to impress the kind of people who like people like that? Stop bashing the UK you have the benefit of hard won freedoms so dont be in too much of a rush to give that freedom back. There is something not right with an employee thinking the way you do, I could understand it if you owned your own company and run it so poorly people felt they could or should take the pea.

#### Flyback

##### Well-Known Member
I am not "uk bashing", and neither was Nicolas Sarkozy. Sarkozy spoke with his genuine concern for UK, a country with whom France may like to shimmy up with, to form an alliance with uk, but the state uk is in, that's not very likely. However, it can all change so easily.
Seriously, the "open workplace" places are really really enjoyable to work in.....its a thriving and buzzing atmosphere....seriously, no one complains at Dyson..everybody absolutely loves it...same with tridonic.
There is not a single UK owned car designer and maker...even morgans use BMW engines.
TVR haven't got any products out at the minute.
Being concerned for the poor woman of ‘example 1’ of the top post is hardly having a chip on the shoulder.
They were giving her absolutely foul, awful language, and if she had complained , she would maybe have been taken off her soldering exams…also, she would have found herself complaining against the senior engineer, who is best buddies with the chief design engineer, which would have put her in a perilous position.
Sorry good friend but objecting to foul language against woman by a group of blokes is not having a chip on the shoulder

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# 7.3 Plotting FRF data

To plot FRF data, right-click on the data of interest in the FRF Data card and select Plot. An overview of the dataset will appear in the Matrix Viewer, and selecting individual entries in the Matrix Viewer will plot them in the Graphs card, as illustrated in the following two figures. If the data has only been imported but not yet loaded, then plotting the data will automatically load it as well.

## Using the Matrix Viewer

The Matrix Viewer is used to provide a visualization of the complete FRF matrix. By default, the average magnitude of the dataset is visualized. Drop-down menus are also available just below the Matrix Viewer, where the mode can be set to magnitude or coherence, and the averaging can be set to mean, minimum or maximum. Note that coherence may be empty or hidden in case this is not provided in the FRF dataset.

Clicking within the individual boxes of the Matrix Viewer will display the corresponding plot of that specific FRF within the Graphs card.

I: Using the Matrix Viewer (left) to plot a single FRF (right)

II: Using the Matrix Viewer (left) to plot a CMIF and condition number (right)

## Plotting a single FRF

As mentioned, the Graphs card will display the FRF (including magnitude, phase and coherence) for whichever entry is selected in the Matrix Viewer, as shown in figure I. For VP-transformed data ($$\mathbf{Y_{um}}$$, $$\mathbf{Y_{qm}}$$), the coherence plot will be empty, as coherence is only valid for raw FRF measurements. For additional information and tips on plotting, see 4 Plotting.

To view another FRF, there are two options:

1. Select another one from the Matrix card;
2. In the plot in the Graphs card, hover the mouse over the channels in the title and scroll with the mouse wheel. Specific channels can also be selected individually from the dropdown menus in the title, as explained in 4.3 Data selection.

## Showing a singular value breakdown (CMIF)

In SOURCE, it is also possible to show the singular value breakdown, often referred to as Complex Mode Indicator Function (CMIF) of any subset of the FRF matrix. Any time you try to plot more than one row along with more than one column of the FRF matrix, the CMIF will automatically be displayed, rather than the individual FRFs. In this mode, the CMIF is displayed in the upper subplot, and the condition number is displayed in the lower subplot, as shown in figure II.

The number of curves shown in the CMIF will be the minimum amount of the number of rows or columns. Reviewing the CMIF is useful for setting the matrix inverse settings in the Analyze module and can be plotted in two ways:

1. Select the desired entries for the CMIF directly in the Matrix Viewer. Create a box around the FRF data of interest by clicking on one entry, holding SHIFT, and selecting another entry. The two entries should be selected to form the opposite corners of the subset of interest.
2. In the plot in the Graphs card, select the drop-down menus in the title and select multiple channels and reference channels by holding SHIFT or CTRL while clicking.


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# Justifying the "Physicist's method" for ODEs using differential forms

I need some help in untangling and solving the following exercise:

Let the curve $$c:[a,b] \to \mathbb{R}^2, t \mapsto (t, y(t))$$ be a solution for the ODE $$y'(x) = f(x, y(x)).$$ Justify the "Physicist's method" (no offense intended) of rearranging the equation $$\frac{dy}{dx} = f(x,y)$$ through formal multiplication of $$dx$$ to $$dy = f(x,y)dx,$$ by showing that both differential forms agree in every point $$c(t)$$ on the tangent space.

As far as I understand the situation, we are considering the two differential forms $$$$dy: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\ f(x,y)dx: \mathbb{R}^2 \to {\bigwedge}^1(T_p\mathbb{R}^2)\\$$$$ Here $$dy$$ is a constant differential form, in the sense that $$dy(p)(x,y) = y$$ independent of $$p=(v,w) \in \mathbb{R}^2$$. However, in general $$f(x,y)dx$$ is not a constant differential form. Now, if we choose a point on the curve $$c$$, say $$p = c(t_0) \in c([a,b])$$ then we have $$f(p)dx = f(c(t_0))dx = y'(t_0)dx,$$ since $$c$$ is a solution to the ODE given above. I don't know how to proceed from this point. How can we argue that this equals $$dy$$?

The equality in $$dy = f\, dx$$ is very misleading, because strictly speaking it's not true. To see why, note that $$dy$$ is a differential $$1$$-form defined on $$\Bbb{R}^2$$, which means for each $$p \in \Bbb{R}^2$$, $$dy_p : T_p \Bbb{R}^2 \to \Bbb{R}$$ is linear. Similarly, $$dx$$ is also a differential $$1$$-form on $$\Bbb{R}^2$$. Let's for the sake of concreteness say $$f: \Bbb{R}^2 \to \Bbb{R}$$ is defined on all of $$\Bbb{R}^2$$, so that $$f \, dx$$ is still a $$1$$-form on $$\Bbb{R}^2$$.

So, if we just write $$dy = f \, dx$$, this means that the $$1$$-form on the LHS must equal the $$1$$-form on the RHS. But this is just not the case, because it amounts to saying that $$dy$$ and $$dx$$ are linearly-dependent over the module $$C^{\infty}(\Bbb{R}^2)$$. Just to really drive this point home, let's fix a point $$p \in \Bbb{R}^2$$, then, if that equality were true, it would mean $$dy_p = f(p)\, dx_p$$, where the equality is as elements in $$T_p^*(\Bbb{R}^2)$$ (the dual of the tangent space; i.e the cotangent space). But this is of course absurd, because if you evaluate both sides on the tangent vector $$\dfrac{\partial}{\partial y}\bigg|_{p} \in T_p\Bbb{R}^2$$, you'll get the absurd equality $$1 = 0$$. Yet again, the statement $$dy = f \, dx$$ is kind of like saying the row vector $$(0 , 1)$$ equals $$\lambda \cdot (1,0)$$ for some $$\lambda \in \Bbb{R}$$... which is plain wrong.

Now that I've hopefully convinced you that the equation taken literally is false, how do we interpret it? Well, the last sentence of your question gives a clue it says

"... by showing that both differential forms agree in every point c(t) on the tangent space."

But the tangent space of what? $$\Bbb{R}^2$$? Clearly not, as I've just shown above. What is actually meant is that these two differential forms agree at every point $$c(t) \in \Bbb{R}^2$$, when restricted to the (one-dimensional) subspace $$T_{c(t)}\left(\text{image}(c) \right) \subset T_{c(t)} \Bbb{R}^2$$. But what is the tangent space to the image of $$c$$? It shouldn't be too hard to convince yourself that if you write $$c(t) = (t, c_2(t))$$ then the tangent space to the image equals the linear span of the (non-zero) vector \begin{align} \xi_{c(t)} :=\dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)} \in T_{c(t)} \Bbb{R}^2 \end{align} (i.e $$c(t) = (t, c_2(t))$$ implies $$c'(t) = (1, c_2'(t))$$ so the tangent space is just the span of this vector).

So, we have to show that for all $$t \in [a,b]$$ and for all $$\zeta_{c(t)} \in T_{c(t)} \left( \text{image}(c)\right)$$, \begin{align} dy_{c(t)}(\zeta_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\zeta_{c(t)}) \end{align} But notice that since the tangent space to the image is one-dimensional, it suffices to verify equality when evaluated on the basis vector $$\xi_{c(t)}$$ defined above; i.e it's enough to prove \begin{align} dy_{c(t)}(\xi_{c(t)}) &= f(c(t)) \cdot dx_{c(t)}(\xi_{c(t)}). \end{align} This is straight forward: \begin{align} dy_{c(t)}(\xi_{c(t)}) &= dy_{c(t)}\left( \dfrac{\partial}{\partial x}\bigg|_{c(t)} + c_2'(t) \dfrac{\partial}{\partial y}\bigg|_{c(t)}\right) \\ &= c_2'(t) \\ &= f(c(t)) \tag{c solves the ODE} \\ &= f(c(t)) \cdot 1 \\ &= f(c(t))\cdot dx_{c(t)}(\xi_{c(t)}). \end{align}

So, this completes the proof.

Note that another way of stating the equality is that $$c^*(dy) = c^*(f \, dx)$$; i.e when you pull-back the two $$1$$-forms on $$\Bbb{R}^2$$ via the curve $$c$$, you get two $$1$$-forms , but now defined on $$[a,b]$$; and it is these forms which are equal.

• Very illuminating, thank you! There's one thing I don't quite understand. Why is $dy$ a $2$-Form? Up until now I thought a $2$-Form over $\mathbb{R}^2$ has to be of the form $\alpha(x,y) dx \wedge dy$, but in $dy$ there's no exterior product to be seen? Mar 28 '20 at 13:35
• @jazzinsilhouette you're right, it's a 1-form on $\Bbb{R}^2$... lol I got confused because there were two of them, $dy$ and $dx$ (and they were both defined on $\Bbb{R}^2$ so my brain registered $2$ everywhere ;) ). I'll edit now Mar 28 '20 at 13:36

I think I may be the 'physicist' in question, but I'll give it a go.

$$dy$$ is a one-form on $$\mathbb{R}^2$$, so: $$dy_p: T_p \mathbb{R}^2\to \mathbb{R}$$, i.e. it takes 1-simplexes defined in space isomorphic to $$T_p \mathbb{R}^2$$ and gives real values (via integration). In what is to follow $$p\in\mathcal{D}$$ is a fixed point and $$\mathcal{D}\subseteq\mathbb{R}^2$$ is the subspace where the solution of $$y'=f$$ is defined. Same defnitions apply to $$\left(fdx\right)_p$$

Next, since linear functionals can be linearly combined, we can define $$dq_p=\left(dy-fdx\right)_p$$. We want to prove that this new functional is zero.

Now, consider integral along the sufficiently short line-segment, 1-simplex, $$\sigma_p=(p_0,p_1)$$, that contains $$p$$. Let $$\phi_p:[0,1]\to\mathcal{D}$$ be the push-forward such that (with slight abuse of notation) $$\phi_p\left([0,1]\right)=\sigma_p$$. Then:

$$\int_{\sigma_p} dq_p=\int_{\phi_p[0,1]} dq_p=\int_0^1\, \phi_p^*dq_p=\int_0^1 \left(\frac{d\bar{y}}{ds}-f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}\right)ds$$

where $$\bar{y}\left(s\right)=y\left(\phi\left(s\right)\right)$$ and the same for $$\bar{x}$$. The good thing is that now we are no longer dealing with forms, so the simple chain rule applies:

$$f\left(\phi\left(s\right)\right)\frac{d\bar{x}}{ds}=\frac{dy}{dx}\bigg|_{\phi\left(s\right)}\frac{d}{ds}x\left(\phi\left(s\right)\right)=\frac{d}{ds}\bar{y}$$

so:

$$\int_{\sigma_p} dq_p=0$$

Thus we have a functional that gives zero for any 'vector' (simplex) we apply it to (via integration). It must be that $$dq_p=0$$ (how else would you define a zero-functional), which proves what you wanted.


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# The solutions of system of equations are not continuous

I'm solving the system of equations that is equivalent to polynomial equation and I'm trying to plot the solutions as a function of one parameter. The problem is that instead of continuous functions I have switching between two solutions at some point. Below you can see as two solutions plotted in red and orange flips in the middle. Still it is clear that there are tree distinct branches. Further I will choose the only one branch that is always positive. But to do so I need to define the function which describe only this branch.

Here is the code to solve the system of equations:

Manipulate[
m = n + 1;
solution = Solve[{r*M0 ==
P + M0*Sum[
i*Product[Evaluate[Symbol["K" <> ToString[j]]]*1, {j, i}]*
P^i, {i, n}]/
Sum[Product[Evaluate[Symbol["K" <> ToString[j]]]*1, {j, i}]*
P^i, {i, 0, n}],
M == M0/Sum[
Product[Evaluate[Symbol["K" <> ToString[j]]]*1, {j, i}]*
P^i, {i, 0, n}],
Table[Evaluate[Symbol["MP" <> ToString[j]]]*1 ==
Times @@ (Table[
Evaluate[Symbol["K" <> ToString[i]]]*1, {i, j}]) M P^j, {j,
n}]} //
Flatten, {P, M,
Table[Evaluate[Symbol["MP" <> ToString[j]]]*1, {j, n}]} //
Flatten];, {n, 3, 1, 1}]

And here is the code to plot the solutions as a functions of parameter 'r':

valP = P /. Take[solution, {1, 4}];

molP[rVal_, K1Val_, K2Val_, K3Val_, M0Val_, DifInd_: 0] :=
Re@With[{r = rVal, K1 = K1Val, K2 = K2Val, K3 = K3Val, M0 = M0Val, índex = DifInd}, (Evaluate@D[valP, {r, índex}])];

Manipulate[
Plot[Evaluate[{molP[r, 10^k1, 10^k2, 10^k3, 10^p0, 0]}],
{r, 0.1, rrmax}, ImageSize -> 500, Frame -> True],
{{rrmax, 10, "r"}, 1, 10},
{{k1, 6, "k1"}, 2, 9},
{{k2, 3, "k2"}, 2, 9},
{{k3, 6, "k3"}, 2, 9},
{{p0, -3.5, "10^P0"}, -1, -10},
ControlPlacement -> Left]

Parameter n in Manipulate define the system of equations and at n=1 or n=2 there is no problems with branches switching:

The problems starts only when n=3. So, the question is - How I can mix the solutions in such way, so I can define all branches as continuous functions ?

Other example is below

solution = Solve[{
K1*P*L == A,
K2*P*L*L == B,
K2*A*L*L == F,
P0 == P + A + B + F,
r*P0 == L + A + 2*B + 3*F
}, {P, L, A, B, F}];

valF = F /. Take[solution, {1, 4}];

FF[rVal_, K1Val_, K2Val_, P0Val_, DifInd_: 0] :=
Re@With[{r = rVal, K1 = K1Val, K2 = K2Val, P0 = P0Val,
índex = DifInd}, (Evaluate@D[valF, {r, índex}])];

Plot[Evaluate[FF[r, 10^6, 10^7, 10^-3.8, 0]],
{r, 0.05, 2},
PlotStyle -> {Red, Blue, Green, Magenta},
PlotRange -> {{0, 1.5}, {-0.00005, 0.0001}},
ImageSize -> Large
]

I need to define 4 continuous functions instead of 4 current not continuous functions!

• Try using Solve with Quartics->False. – Carl Woll May 16 '17 at 22:49
• Can you please fix your code so that it indeed gives an output? Just call Quit and try the Manipulate block yourself. You that there are several undefined variables including P, M0, and so on. – halirutan May 16 '17 at 22:50
• @CarlWoll So easy! Thank you! – Филипп Цветков May 16 '17 at 22:51
• @CarlWoll It is not universal solution ;( if I plug another parameters into molP function the solutions changes the branches even with Quartics->False option. molP[r, 10^5.34, 1000, 10^6.1, 10^-3.5, 0] – Филипп Цветков May 17 '17 at 19:34
• Interesting problem. If you set all of the parameters = 1 so r is the remaining variable, then you can see internally there is a rational polynomial inside a square root where the numerator and denominator both go to zero simultaneously. At that point, the left limit is negative and the right limit is positive. Taking the absolute value gives you a smooth curve that is positive, minus the discontinuity. When it goes negative, the square root is complex. this gives the break in the curve. – MikeY May 19 '17 at 22:15

For the sake of plotting, NDSolve is more likely to track a evolution of a root continuously, since it uses the derivative to predict its next value. The large number of parameters in the OP's problem lead to a correspondingly large system here. The parameters could be incorporated into a ParametricNDSolve code, but for sake of developing and debugging a working solution, it seemed convenient to have access to the components in global variables.

The following shows how to do it for the OP's "other example." Note it uses the solution from Solve to set up the initial conditions. The system algsys consists of equations for all four roots. These equations are identical except for the names of the variables. They are differentiated before being passed to NDSolve. The root each subsystem represents is determined by the initial conditions algics.

varsP = {P1, P2, P3, P4};     (* one variable for each solution *)
varsL = {L1, L2, L3, L4};
varsA = {A1, A2, A3, A4};
varsB = {B1, B2, B3, B4};
varsF = {F1, F2, F3, F4};
allvars = {{varsP, varsL, varsA, varsB, varsF}, {P, L, A, B, F}};
algsys = {K1*P*L == A, K2*P*L*L == B, K2*A*L*L == F,
P0 == P + A + B + F, r*P0 == L + A + 2*B + 3*F} /.
Equal -> Subtract /. MapThread[#2 -> Through[#1[r]] &, allvars] // Flatten;
algics = MapThread[Through[#1[r]] == (#2 /. solution) &, allvars] /. r -> 0.05;
Clear[FF2];
FF2[K1Val_, K2Val_, P0Val_, DifInd_: 0] :=
Block[{K1 = K1Val, K2 = K2Val, P0 = P0Val, índex = DifInd},
Through[varsF[r]], {r, 0.05, 2}]
];

The real and imaginary parts (green and magenta have the same real parts, blue and red have the same imaginary parts):

Plot[Evaluate[Re@FF2[10.^6, 10.^7, 10^-3.8, 0],
{r, 0.05, 2}, PlotStyle -> {Red, Blue, Green, Magenta}]

Plot[Evaluate[Im@FF2[10.^6, 10.^7, 10^-3.8, 0]],
{r, 0.05, 2}, PlotStyle -> {Red, Blue, Green, Magenta}]

• Is it possible to use FF2 as a model for NonlinearModelFit ? – Филипп Цветков Jul 10 '17 at 8:37
• One could use Indexes[] to fit a particular Part[] perhaps. One should be able to fit the vector of roots – Michael E2 Jul 10 '17 at 12:45
• @ФилиппЦветков Sorry, on a phone. Hit Send accidentally -- I won't be able to check all day, so that's really just a guess at this point. (And it's Indexed[], not Indexes[].) – Michael E2 Jul 10 '17 at 12:49
• For instance it's continuous function of r, but what about the other parameters? Need to check – Michael E2 Jul 10 '17 at 12:52
• Sorry! Me too.. on vocations. Cannot check (( – Филипп Цветков Jul 12 '17 at 7:49

This answer uses my result from another thread, Is there a way to sort NSolve solution (roots) automatically? It doesn't use any information about the equations, it just looks to identify smooth curves from the data.

solution =
Solve[{K1*P*L == A, K2*P*L*L == B, K2*A*L*L == F,
P0 == P + A + B + F, r*P0 == L + A + 2*B + 3*F}, {P, L, A, B, F}];

valF = F /. Take[solution, {1, 4}];

FF[rVal_, K1Val_, K2Val_, P0Val_, DifInd_: 0] :=
Re@With[{r = rVal, K1 = K1Val, K2 = K2Val, P0 = P0Val,
índex = DifInd}, (Evaluate@D[valF, {r, índex}])];

Instead of immediately plotting, first generate a list of points, then manipulate them, and plot that. Here's the data.

data = Table[
Evaluate[FF[r, 10^6, 10^7, 10^-3.8, 0]], {r, 0.05, 2, .01}];

ListPlot[data // Transpose, Joined -> True]

Below is the kernel of a FoldList call. You may need to fiddle with the WEIGHT constant to get the results to come out just right. It controls how much the derivative information is used in discriminating the curves.

WEIGHT = 100;
permMatchWeight[{vals_, dels_}, l2_] :=
Module[{l1 = {vals, dels}, ps, bestPerm},
ps = Map[({#, WEIGHT (# - vals)}) &, Permutations[l2]];
bestPerm = Sort[ps, Norm[l1 - #1] < Norm[l1 - #2] &] // First
];

Here's where the data gets manipulated.

res = FoldList[permMatchWeight, {{0, 0, 0, 0}, {0, 0, 0, 0}}, data] // Rest;
orderedCurves = Map[(# // First) &, res] // Transpose;

The orderedCurvesis now a list of 4 sets of points, each tied to a single curve. On examination, you find the first two curves are identical.

ListLinePlot[orderedCurves , PlotStyle -> {Red, Blue, Green, Magenta}, ImageSize -> Large ]

In generating the data, I ignored tracking the independent variable, so you'd need to massage the result to get it back in. You could do an interpolating fit at this point to get a function for each of the curves.


text:
Browse Questions

The molal freezing point depression constant of benzene ($C_6H_6$) is 4.90 K kg $mol^{−1}$. Selenium exists as a polymer of the type ***. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is $0.112^{\large\circ}$C lower than pure benzene. Deduce the molecular formula of selenium. (Atomic mass of Se = 78.8 g $mol^{−1}$)

Can you answer this question?

$Se_8$


text:
Free Version
Moderate

# Multiple Math Methods in a Statement

APCSA-VWVKGG

What range of values could num possibly have after this statement is executed?

double num = Math.pow(Math.random()*10,Math.sqrt(Math.pow(2, 0))*2);
System.out.println(num);

A

Any decimal number between 0 (inclusive) and 100 (exclusive): [0, 100).

B

Any decimal number between 0 (inclusive) and 100 (inclusive): [0, 100].

C

Any decimal number between 0 (exclusive) and 100 (exclusive): (0, 100).

D

Any decimal number between 0 (inclusive) and 10 (exclusive): [0, 10).

E

Any decimal number between 0 (inclusive) and 10 (inclusive): [0, 10].