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Wednesday, September 21, 2022

# What Does Aq Mean In Chemistry

## What Is A Chemical Equation

What does “aqueous” mean?

nomenclature

A chemical equation is an expression of a chemical process. For example:

In this equation, AgNO3 is mixed with NaCl. The equation shows that the reactants react through some process to form the products . Since they undergo a chemical process, they are changed fundamentally.

Often chemical equations are written showing the state that each substance is in. The sign means that the compound is a solid. The sign means the substance is a liquid. The sign stands for aqueous in water and means the compound is dissolved in water. Finally, the sign means that the compound is a gas.

Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of molecules, or the relative number of moles . If no coefficient is shown, a one is assumed.

On some occasions, a variety of information will be written above or below the arrows. This information, such as a value for temperature, show what conditions need to be present for a reaction to occur. For example, in the graphic below, the notation above and below the arrows shows that we need a chemical Fe2O3, a temperature of 1000 degrees C, and a pressure of 500 atmospheres for this reaction to occur.

The graphic below works to capture most of the concepts described above:

## What Does Simp Mean In Slang

Urban Dictionarys top definition of a simp is someone who does way too much for a person they like. Other definitions on the crowdsourced online dictionary include a man who puts the hoes before the bros, and a guy that is overly desperate for women, especially if she is a bad person, or has expressed her

## What Is An Aqueous Solution

###### What Is an Aqueous Solution in Chemistry?

What is an aqueous solution in chemistry? Well, to start, its the 4th state of matter. To a chemist, any substance is found in one of the 4 states of matter. You already know about 3 of them: solid, liquid, and gas. And, in chemistry, an aqueous solution is the 4th state of matter. So, you might say that aqueous solutions matter a lot, as they are literally an entire state, or configuration, of this stuff we call matter.

First thing first, I know there are other states of matter. To your physics teacher, the 4th state of matter would be called plasma. To your chemistry teacher, the 4th state of matter would be called an aqueous solution. A child playing with corn starch might think that oobleck is the 4th state of matter. Oobleck definitely would win the contest if it were based on how fun the name was to say. Yet, at least in chemistry, the 4th state of matter is an aqueous solution. It turns out that an aqueous solution is an extremely practical concept, perhaps unlike plasma and oobleck, which we dont really encounter so much in our everyday lives.

###### What Is an Aqueous Solution: The Simple Answer

So what is an aqueous solution? It simply means dissolved. Yes, the everyday concept of dissolving something. Thats it. The aqueous solution definition means simply that something has been dissolved in water. The aqueous symbol is .

Aqueous solution definition: something is dissolved in water

Aqueous symbol:

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## Types Of Arrows In A Reaction:

1. Right Arrow: The right arrow is the most prevalent arrow in a chemical reaction formula. The direction indicates the reaction’s direction.

• A chemical equation is an empirical representation of a chemical process.
• Base materials are represented on the left-hand side of the equation as reactants.
• The right-hand side of the equation contains the products, which are the reaction results.
• A chemical or physical transition is represented by the arrow.
• In chemical reaction formulae, the right arrow is by far the most prevalent arrow. The direction indicates the reaction’s direction.

## What Is The Difference In And States In Reaction Mechanism

In these reactions between alkyl halide and KOH in different states the product is different only because of the change in state of KOH from aqueous and alcohol .

Can someone explain the mechanism which causes this drastic change?

Nucleophilic substitution always competes with Elimination reaction in such cases.

What is the difference between them?

In alcoholic solution the $\ce$ is basic enough to deprotonate a small amount of alcohol molecules ), thus forming alcohlate salts $\ce$. The alcoholate anions $\ce$ are not only more basic than pure $\ce$ but they are also bulkier . The higher bulkiness makes $\ce$ a worse nucleophile than $\ce$ and the higher basicity makes it better at $\mathrm2$ eliminations.

Mechanisms below:

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## Types Of Chemical Reactions

Two or more reactants combine to make 1 new product.Examples:

A single reactant breaks down to form 2 or more products.Examples:

A single element replaces a similar element of an adjacent reactant compound.Example:

Two ionic compounds exchange ions, producing 2 new ionic compounds.Examples:HCl + NaOH NaCl + H2O

## Balancing A Chemical Equation

Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure \.

Original molecule H2O: if the coefficient 2 is added in front, that makes 2 water molecules but if the subscript 2 is added to make H2O2, that’s hydrogen peroxide.

The simplest and most generally useful method for balancing chemical equations is inspection, better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method.

##### Steps in Balancing a Chemical Equation
• Identify the most complex substance.
• Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
• Balance polyatomic ions as a unit.
• Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
• Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.
• ##### Example \: Combustion of Heptane

Balance the chemical equation for the combustion of Heptane ).

##### Example \: Combustion of Isooctane

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## State Symbols And Phase Changes

Everything you need for better grades in university, high school and elementary.

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• Chemical phases and state symbols.
• Other key phase/state language.
• ###### Lessons
• Recall the different states of matter from descriptions of chemicals.Write “solid”, “liquid”, “gas”, or “aqueous”, next to each term below to show which state it is describing.
• Powder
• Precipitate
• Write full chemical equations using information from descriptions of laboratory experiments.Read the following experiment notes and write a balanced chemical equation, with state symbols, to describe what is happening.
• Strips of magnesium were weighed out and placed in a beaker. Shortly afterwards, a solution of hydrochloric acid was added to the beaker to produce magnesium chloride and gaseous hydrogen.
• In a reaction vessel, chlorine gas reacts with chunks of sodium metal to form sodium chloride in a reaction releasing a lot of heat energy.
• When hot steam is passed over iron filings, gaseous hydrogen and solid iron oxide is produced.
• Aqueous hydrochloric acid and sodium carbonate powder react to form sodium chloride in solution, water and bubbles of carbon dioxide.

Notes:

## Properties Of Aqueous Solutions

What Is Electrolysis | Reactions | Chemistry | FuseSchool

Aqueous solutions often conduct electricity. Solutions that contain strong electrolytes tend to be good electrical conductors , while solutions that contain weak electrolytes tend to be poor conductors . The reason is that strong electrolytes completely dissociate into ions in water, while weak electrolytes incompletely dissociate.

When chemical reactions occur between species in an aqueous solution, the reactions are usually double displacement reactions. In this type of reaction, the cation from one reactant takes the place for the cation in the other reactant, typically forming an ionic bond. Another way to think of it is that the reactant ions “switch partners”.

Reactions in aqueous solution may result in products that are soluble in water or they may produce a precipitate. A precipitate is a compound with a low solubility that often falls out of solution as a solid.

The terms acid, base, and pH only apply to aqueous solutions. For example, you can measure the pH of lemon juice or vinegar and they are weak acids, but you can’t obtain any meaningful information from testing vegetable oil with pH paper.

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## Definition Of Dissolve And The State Label In Chemical Equations

I am confused with the meaning of “” in chemical equations, and the meaning of “dissolved”. What is the actual meaning it is supposed to convey?

For instance, in the document “Preparing Bromine Water“, bromine molecules in bromine water are labelled $\ce}$.

• What does dissolving in water mean? Must it be that a hydration shell is formed around the molecule? See:Dissolving of non polar gases in water

• Even if no hydration shell is formed, small molecules just insert themselves into the space between solvent molecules, is it still called “dissolve”? It does seem more like being “dispersed”, as in a colloid.

• Must “” mean that there is a hydration shell surrounding the molecule/ion?

• For non-polar molecules dissolved in water, do we label it as “” in chemical equations? Must they form hydration shells?

• The abbreviation in chemical equations or symbols simply means that this material is present in water as a homogeneous phase. The state symbol has nothing to do with the actual and true “structure” in solution. In short forget about hydration shell.

From IUPAC’s recommendations “Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions” Pure & Appl.Chem., Vol.54, No.6, pp. 12391250, 1982.

$\ce$ would imply dissolved carbon dioxide in water .

$\ce$ would imply homogeneous solution of liquid bromine in water.

$\ce$: Salt dissolved in water.

A colloid is a heterogeneous phase.

## How To Identify States Of Matter In A Chemical Formula

Chemical equations and the formulas of substances that appear in them are powerful recipes for laboratory experiments, and information about the state of matter of each reactant and product is vital. If you are only told that H2O is a reactant, you will not know whether to add ice — the solid form — or liquid water or steam — the gaseous form. Even if you are not physically performing an experiment, knowledge of the state of matter of each substance is necessary to fully understand the equation, and it can even help you identify the type of experiment. Since identifying states of matter is so important, it is fortunate that the process is very simple.

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## Empirical Formula And Molecular Formula

One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of convertingfrom percentage to mass.

Example: Calculate the empirical formula for a compound that has 43.7 g P and 56.3 grams of oxygen.First we convert to moles:

Next we divide the moles to try to get a even ratio.

When we divide, we did not get whole numbers so we must multiply by two . The answer=P2O5

Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula.It is also possible to do this with one of the elements in the formula simply divide the mass of that element in one mole of compound by the massof that element in the empirical formula. The result should always be awhole number.

Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two . The answer is: H2C2N2.

## How To Write Balanced Chemical Equations

• Page ID
47502
• Explain the roles of subscripts and coefficients in chemical equations.
• Balance a chemical equation when given the unbalanced equation.
• Explain the role of the Law of Conservation of Mass in a chemical reaction.

Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the productsthey are merely reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.

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## What Does Chemistry Mean In Love

Simply put, the feeling of chemistry with another person is that of connection. Its a draw to someone else that makes you want more of them. That doesnt have to be in a romantic relationship context, though that is the way we most often use the word. A few common types of chemistry are outlined below.

## Use Of Arrow In Chemical Equations

When a reaction necessitates energy use, it is frequently shown above the arrow. If energy is provided to the reaction in the form of heat, a capital Greek letter delta is placed on top of the reaction arrow if energy is supplied in the form of light, hv is inscribed.

There are a wide variety of reactions possible in our everyday lives: Elements can form compounds , compounds can generate elements , or compounds can mix, split away, or reorganize to form new compounds.

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## Finding And Identifying States Of Matter

#### 1

Locate the parentheses after the chemical formula, whether or not it is within the context of an equation.

#### 2

Identify the parentheses as for solid, for liquid, for gas, or for aqueous solution. An aqueous solution is a substance dissolved in water.

#### 3

If no parentheses follow the chemical formula, look for key words or phrases. These are especially helpful in the context of chemical reactions. For example, a precipitate is a solid, a combustion reaction produces water and carbon dioxide in their gaseous forms, and in solubility experiments, ions are in aqueous solution.

References

## Aqueous Solution Definition In Chemistry

What does “aq” mean? (aqueous, dissolved in water)
• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

An aqueous solution is any solution in which water is the solvent. In a chemical equation, the symbol follows a species name to indicate that it is in aqueous solution. For example, dissolving salt in water has the chemical reaction:

NaCl Na+ + Cl-

Although water is often called the universal solvent, it dissolves only substances that are hydrophilic in nature. Examples of hydrophilic molecules include acids, bases, and many salts. Substances that are hydrophobic do not dissolve well in water and tend not to form aqueous solutions. Examples include many organic molecules, including fats and oils.

When electrolytessuch as NaCl and KCldissolve in water, the ions allow the solution to conduct electricity. Nonelectrolytes like sugar also dissolve in water, but the molecule remains intact and the solution is not conductive.

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## How Will U Know If A Guy Loves U

If your boyfriend loves you, he will treat you with respect. That means that he listens to you and cares about whats going on in your life. He notices the little things that you like and goes out of his way to give them to you. He values you as a person, and he genuinely listens to your opinions.

## Will It Dissolve

Whether or not a substance forms an aqueous solution depends on the nature of its chemical bonds and how attracted the parts of the molecule are to the hydrogen or oxygen atoms in water. Most organic molecules won’t dissolve, but there are solubility rules that can help identify whether or not an inorganic compound will produce an aqueous solution. In order for a compound to dissolve, the attractive force between a part of the molecule and hydrogen or oxygen has to be greater than the attractive force between water molecules. In other words, dissolution requires forces greater than those of hydrogen bonding.

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# A projectile is ejected with vertical speed v from the surface of a planet of mass M and radius R. [closed]

–1 vote
18 views

Show that it comes to rest at a distance r from the centre of the planet where $$r=R/(1−Rv^2/2GM)$$
If v = c in the above example what will be the condition between M and R such that the planet acts like a Newtonian black hole.

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# Kazhdan–Lusztig polynomials in terms of Ext groups

Let $$P_{x,w}$$ be the Kazhdan–Lusztig polynomial, $$\rho$$ be the half sum of positive roots in $$\Phi^+$$, $$M_x$$ be the Verma module with highest weight $$x\cdot(-2\rho)$$ and $$L_w$$ be the simple highest weight module with highest weight $$w\cdot(-2\rho)$$.

It is well-known that Kazhdan–Lusztig Conjecture is equivalent to $$P_{x,w}(q) = \sum_{i\ge 0}q^i \dim \mathrm{Ext}_\mathcal{O}^{\ell(x,w)−2i}(M_x, L_w)$$ for all $$x\le w$$. And it is also well-known that Kazhdan–Lusztig Conjecture is true.

My question: Does $$P_{x,w}(q) = \sum_{i\ge 0}q^i \dim \mathrm{Ext}_\mathcal{O}^{\ell(x,w)−2i}(M_x, L_w)$$ even for all $$x\not\le w$$?

If the answer is yes, any reference about that would be appreciated. If the answer is no, I would like to know why.

• Is it not the case that then both sides are zero? LHS is trivial by definition, while for the RHS this follows from BB localization (but probably there is a more direct way to see it). – dhy May 25 '19 at 17:36
• I agree with you. I would like to know how to show the RHS is zero for $x\not\le w$ by BB localization or any other methods. – James Cheung May 26 '19 at 8:34
• The BB localization argument goes as follows: $M_x$ is sent to $j_{x,!}\omega_x,$ where $j_x$ is the inclusion of the Bruhat cell corresponding to $x$. Taking Hom from this object is equivalent to taking the !-fiber on this cell. On the other hand, $L_w$ is supported on cells $\leq w$, so this !-fiber is zero if $x\not\leq w.$ – dhy May 26 '19 at 14:03

The answer is yes, for fairly elementary reasons, though it's not easy to give a reference. The point is partly that the polynomials are undefined for two elements of the Weyl group not related by the Bruhat partial orderijg. More precisely, the "linkage principle"(or "Harish-Chandra principle") ensures that the Hom functor is zero on these pairs of highest weight modules. Thus the derived functors Ext$$^n$$ vanish too.
(By the way, I meant to comment on the original version of the question, pointing out that for $$i=0$$ and $$q=0$$ you'd get 1 as constant term of the polynomial; $$i=0$$ was missing then and has been supplied in the sum apparently without editing. On the other hand, the tags have heen edited. But 'coxeter-groups', and 'co.combinatorics' which includes algebraic methods, are already available and are appropriate here.)


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My Math Forum Trig Interval Question

 Trigonometry Trigonometry Math Forum

 April 23rd, 2009, 10:30 AM #1 Newbie   Joined: Apr 2009 Posts: 3 Thanks: 0 Trig Interval Question I solved each problem, so if someone can check them I would appreciate it! 1. Solve the equation in the interval [0,2pi) tanxcosx-cosx=0 I found all solutions of the equation to be Cos: pi/2 +2npi ; 3pi/2 + 2npi The problem is both yield error when plugging into a calc so do I just exclude? n is any integer Tan: pi/4 +npi (this is just pi instead of 2pi because one period of tan is from (-pi/2, pi/2) If the question asks for the solutions in the interal [0,2pi) then the answers would just be pi/4? 2. Solve the equation in the interval [0,2pi) 2 cos^2x+sinx=1 For all solutions of the above I got 7pi/6 + 2npi; 11pi/6 + 2npi; pi/2 + 2npi So, for the interval [0,2pi] would it just be pi/2, 7pi/6, 11pi/6
 April 23rd, 2009, 10:42 AM #2 Newbie   Joined: Apr 2009 Posts: 3 Thanks: 0 Re: Trig Interval Question Here is some additional work for #1 and #2 1. tanxcosx-cosx=0 cosx(tanx-1)=0 cosx=0 thus x is pi/2 and 3pi/2 but when checked yields error tanx=1 @pi/4 2. 2cos^2x+sinx=1 2(1-sin^2x)+sinx-1=0 2-2sin^2x+sinx-1=0 -2sin^2x+sinx+1=0 -1(2sinx+1)(sinx-1)=0 2sinx+1=0 sin x=-1/2 @7pi/6 and 11pi/6 sinx-1=0 sin x =1 @pi/2

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# 1/sinx cosx interal

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April 23, 2009

I’ve been casually reading Field’s “Saving Truth from Paradox” for some time now. I think it’s a fantastic book, and I highly recommend it to anyone interested in the philosophy of logic, truth or vagueness.

I’ve just read Ch. 21 where he discusses a paradox presented in Restall 2006. The discussion was very enlightening for me, since I had often thought this paradox to be fatal to non-classical solutions to the liar. But although Fields discussion convinced me Restall’s argument wasn’t as watertight as I thought it was, I was still left a bit uneasy. (I think there is something wrong with Restall’s argument that Field doesn’t consider, but I’ll come to that.)

Before I continue, I should state the paradox. The problem is that if one has a strong negation in the language, $\neg$, one can generate a paradoxical liar sentence which says of itself that it’s strongly not true. Strong negation has the following properties which ensures that that last sentence is inconsistent:

1. $p, \neg p \models \bot$
2. If $\Gamma , p \models \bot$ then $\Gamma \models \neg p$

Roughly, the strong negation of p is the weakest proposition inconsistent with p – the first condition guarantees that it’s inconsistent with p, the second that it’s the weakest such proposition. It’s not too hard to see why having such a connective will cause havoc.

Restall’s insight (which was originally made to motivate a “strong” conditional, but it amounts to the same thing) was that one can get such a proposition by brute force: the weakest proposition inconsistent with p is equivalent to the disjunction of all propositions inconsistent with p. Thus, introducing infinitary disjunction into the language, we may just “define” $\neg p$ to be $\bigvee \{q \mid p \wedge q \models \bot \}$. Each disjunct is inconsistent with p so the whole disjunction must be inconsistent with p, giving us the first condition. If q is inconsistent with p, then q is one of the disjuncts in $\neg p$ so q entails $\neg p$, giving us (more or less) the second condition.

An initial problem Field points out is that this definition is horribly impredicative – $\neg p$ is inconsistent with p, so $\neg p$ must be one of it’s own disjuncts. Field complains that such non-well founded sentences give rise to paradoxes even without the truth predicate, for example, the sentence that is it’s own negation. (I personally don’t find these kinds of languages too bad, but maybe that’s best left for another post.) This problem is overcome since you can run a variant of the argument by only disjoining atomic formulae so long as you have a truth predicate.

The second point, Field’s supposed rebuttal of the argument, is that to specify a disjunction by a condition, F say, on the disjuncts, you must first show F isn’t vague or indeterminate, or else you’ll end up with sentences such that it is vague/indeterminate what their components are. Allowing such sentences means they can enter into vague/indeterminate relations of validity – for example, it is vague whether a sentence such that it is vague whether it has “snow is white” as a conjunct entails “snow is white”. But the property F, in this case, is the property of entailing a contradiction if conjoined with p. Thus to assess whether F is vague/indeterminate or not, we must ask if entailment can ever be vague. But to do this we must determine whether there are sentences in the language such that it is indeterminate what their components are. Since the language contains the disjunction of the F’s, this requires us to determine whether F is vague – so we have gone in a circle.

Clearly something weird is going on. That said, I don’t quite see how this observation refutes the argument. It’s perfectly consistent with what’s been said above that entailment for the expanded language with infinitary disjunction is precise, that there is a precise disjunction of the things inconsistent with p, and that Restall’s argument goes through unproblematically. It’s also consistent that there *are* vague cases of entailment – but that the two conditions for strong negation above determinately obtain (there are some subtle issues that must be decided here, e.g., is “p and q” determinately distinct from the sentence that has p as its first conjunct, but only has q as its second conjunct indeterminately.)

Even so, I think there are a couple of problems with Restall’s argument. The first is a minor problem. To define the relevant disjunction, we must talk about the property of “entailing a contradiction if conjoined with p”. But to do this we are treating “entails” like it was a connective in the language. However, one of Fields crucial insights is that “A entails B” is not an assertion of some kind of implication holding between A and B, but rather the conditional assertion of A on B. “entails” cannot be thought of like a connective. For one thing, connectives are embeddable, whereas it doesn’t make much sense to talk of embedded conditional assertions. Secondly, a point which I don’t think Field makes explicit, is that it is crucial that “entails” doesn’t work like an embeddable connective, otherwise one could run a form of Curry’s paradox using entailment instead of the conditional.

This not supposed to be a knockdown problem. After all, so what if you can’t *define* strong negation, there is, nonetheless, this disjunction whose disjuncts are just those propositions inconistent with p. We may not be able to define it or refer to it, but God knows which one it is all the same.

The real problem, I think, is the following. How are we construing $\neg p$? Is it a new connective in the language, stipulated to mean the same as “the disjunction of those things inconsistent with p”? If it is, how do we know it is a logical connective? (If $\neg$ weren’t logical neither (1) nor (2) would hold, since there would be no logical principles governing it.) Field objects to a similar argument from Wright, because “inconsistent with p” is not logical. Inconsistency is not logical: for a start it can only be had by sentences, so it is not topic neutral.

The way of construing $\neg p$ that makes it different from Wright’s argument, and allegedy problematic, is to construe $\neg p$ as schematic for a large disjunction. The symbol $\neg$ does not actually belong to the language at all – writing $\neg p$ is just a metalinguistic shorthand for a very long disjunction, a disjunction that will change, depending in each case, on p. Treating it as such guarantees that (1) and (2) hold, since when they are expanded out, are just truths about the logic of disjunction and don’t contain $\neg$ at all.

But treating $\neg p$ as schematic for a disjunction means it doesn’t behave like an ordinary connective. For one you can’t quantify into it’s scope. What sentence would $\exists x\neg Fx$ be schematic for? What we want it to mean is that there is some object, a, such that the disjunction of things inconsistent with Fa holds. But there’s no single sentence involved here.

Another crucial shortcoming is that it’s not clear that we can “put a dot” under $\neg$. That is, define a function which takes the Gödel number of p, to the Gödel number of the disjunction of things inconsistent with p. Firstly there might not be enough Gödel numbers to do this (since we have an uncountable language now!) But secondly, how do we know we can code “inconsistent with p” in arithmetic? Fields logic isn’t recursively axiomatizable (Welch, forthcoming) so it seems like we’re not going to be able to code “inconsistent with p” or the strong negation of p – and thus it seems we’re not going to be able to run the Gödel diagonalisation argument. (I was always asleep in Gödel class so maybe someone can check I’m not missing something here.)

So you can’t get a strongly negated liar sentence through Gödel diagonalisation, but what about indexical self reference? “This sentence is strongly not true” is schematic for a sentence not including “strongly not”, but with a large disjunction instead. However, which disjunction is it? We’re in the same pickle we were in when we tried to quantify into the scope of $\neg$. In both cases, the disjunction needed to vary depending on the value of the variable “x” or in this case, the indexical “this”.

I can’t say I’ve gotten to the bottom of this, but it’s no longer clear to me how problematic Restall’s argument is for the non classical logician.

I agree that Field has a way out, as he describes, but I think it comes at a significant cost. My initial statement of the problem wasn’t directed at Field, of course (I’d been working on this before Hartry was working on non-classical approaches to the paradoxes). Rather, the original problem was a dilemma for non-classical logicians of all stripes.

Now, to the detail: “To define the relevant disjunction, we must talk about the property of “entailing a contradiction if conjoined with p”. But to do this we are treating “entails” like it was a connective in the language” — I think that’s wrong. (In my original paper, by the way, I wasn’t defining negation, but rather, the intuitionistic conditional, and coding up Curry’s paradox, but the difference between what I said and Hartry’s re-telling is not material for this discussion.) It’s true that we need to talk of such things as what propositions entail what propositions, but everyone doing logic needs to do that. The point of the selection is to select which propositions to disjoin, and the argument only applies to those whose logic supplies the resources to allow us to make that selection. Given a proposition p, does the class of all propositions inconsistent with p have a disjunction? In some logics, the answer is ‘yes’, in others, the answer is ‘no’.

My main target, of course, is someone like Priest, who uses logics in which propositions are sets of (normal and non-normal) worlds, where disjunction of propositions is the union of those sets. Inconsistency (in the salient sense here) is having an empty intersection with, the relevant classes of propositions just does have a union, and this union is itself a set of worlds — proposition. The algebra of propositions in these models supports the reasoning, whether or not we have a connective in the object language that does the job… But now to say that even though the algebra of propositions is closed under that operation, such a thing _cannot_ be expressed in the object language in pain of paradox? This is exactly the kind of failure of semantic closure that people like Priest have argued against, forever.

That’s the dialectic of the argument as I originally construed it, anyway.

2. Hi Greg,

Thanks a lot for the feedback – that was very helpful!

“I agree that Field has a way out, as he describes, but I think it comes at a significant cost. “

Maybe I was missing something with Fields response then, because it seemed to me consistent with what he said that there are no indeterminate cases of entailment (at least, he didn’t give a positive reason to think there would be, just a reason to think there might be.)

“My initial statement of the problem wasn’t directed at Field, of course (I’d been working on this before Hartry was working on non-classical approaches to the paradoxes). Rather, the original problem was a dilemma for non-classical logicians of all stripes.”

Sorry to make it all about Field! Although, I do think that this is a good test case (and obviously, Field thinks it is highly relevant to his project!)

“It’s true that we need to talk of such things as what propositions entail what propositions, but everyone doing logic needs to do that. The point of the selection is to select which propositions to disjoin, and the argument only applies to those whose logic supplies the resources to allow us to make that selection. Given a proposition p, does the class of all propositions inconsistent with p have a disjunction?”

Right. I wasn’t so much objecting to the existence of such a disjunction, but more to our ability, as finite beings, to specify it. The way Field cashes out validity in terms of conditional assertion, rather than the assertion of a conditional, won’t allow us to simply say “that disjunction, whose disjuncts are just those things which when conjoined with p entail a contradiction” (if we cash out argument validity instead as the validity of a conditional, we can say things like this, but this kind of validity won’t do for the argument.) This is inability to specify the disjunction is not entirely to do with it being infinite, but rather to do with the way we need to specify the disjuncts. Adding an “entailment” *connective* generates Curry like paradoxes of it’s own – I take this distinction between conditional assertion and assertion of a conditional to be a nice philosophical motivation for not adding such a connective into the language.

“My main target, of course, is someone like Priest, who uses logics in which propositions are sets of (normal and non-normal) worlds, where disjunction of propositions is the union of those sets. Inconsistency (in the salient sense here) is having an empty intersection with, the relevant classes of propositions just does have a union, and this union is itself a set of worlds — proposition. The algebra of propositions in these models supports the reasoning, whether or not we have a connective in the object language that does the job… But now to say that even though the algebra of propositions is closed under that operation, such a thing _cannot_ be expressed in the object language in pain of paradox? This is exactly the kind of failure of semantic closure that people like Priest have argued against, forever.”

I see. This argument seems closer to the argument you labelled “The Algebra of Propositions” in your paper, rather than the infinitary disjunction argument which could be carried out purely in the object language (no?) The latter argument seems stronger to me, since it applies also to people who deny the existence of a (classical) intended model (such as Field).

3. Here are a couple of things I’m puzzled about the argument in the “Infinitary Disjunction” section – maybe you could help me.

Consider the version of the argument for strong negation rather conditional (call strong negation, Negation, for short.) I take that the existence of such disjunctions guarantees the following:

“For each Negation free formula, $\phi$, there is a Negation free infinitary disjunction equivalent to the Negation of $\phi$.”

What I’d like to know is (a) whether there is a Negation free equivalent of any formula (including formulae with embedded Negations) (b) whether you can code the function taking the Gödel number of a sentence to the Gödel number of it’s Negation, (c) whether the fact above should allow the introduction of a Negation connective if (a) obtains (and if it doesnt?)

4. Somewhat off topic, but since you brought it up… I grant that it makes little sense for “entails” to embed on the conception that “A entails B” is really an assertion about the conditional assertion of B upon A… but what about:

(E) That the king of France is bald entails that France has a king entails that something entails that France has a king.

We could either conclude, with Field’s conception of entailment, that (E) is nonsense. But it doesn’t strike me as nonsense. So I’m inclined to think that “entails” does embed.

5. Hi Colin. Thanks for the comment!

Yes that is worrying. I take it your choice of example was supposed to prevent me from interpreting these as valid assertions of conditionals (rather than valid conditional assertions)?

That said – the problem of Curry’s paradox for an embeddable entailment conditional seems to me quite general. It seems like you have a burden of proof there, at least of defending against Curry style arguments, if you are to claim entailment is embeddable.

6. […] finally, a smattering of logic: Andrew Bacon discusses Restall’s Paradox, Kenny Easwaran endorses a particular kind of probabilistic proof, and I consider a slight […]

7. For what it’s worth, the credit for interpreting “entails” or the turnstile as conditional assertion is really due to Dana Scott in a series of papers from the early 70s.

8. Thanks! Do you have any references?

9. “On Engendering an Illusion of Understanding” http://www.jstor.org/discover/10.2307/2024952?uid=3739832&uid=2&uid=4&uid=3739256&sid=21102000500171

“Completeness and Axiomatizability in many-valued logic”


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Contents

Contents

Idea

A Mealy machine is a particular type of finite state automaton.

A Mealy machine with input alphabet $A$ and output alphabet, $B$ is just a deterministic finite automaton?,which produces a letter of the output alphabet, $B$ at every transition step. More formally

Definition

A finite Mealy machine, $\mathbf{A}$, consists of

• $Q$, a finite set of states;

• $A$, an input alphabet;

• $B$, an output alphabet;

• $q_0$, an initial state;

• $\delta: Q\times A\to Q$, a transition function;

and

• $\lambda:Q\times A \to B$, which associates an output symbol with each transition.

Again as for Moore machines, there is an output response function $\omega_\mathbf{A}:A^*\to B^*$ defined s follows:

if $x= x_1 \ldots x_n\in A^*$ and the states that $\mathbf{A}$ passes through when processing this string are $q_0, q_1, \ldots, q_n$, so diagrammatically

$q_0\xrightarrow{x_1} q_1\xrightarrow{x_2}\ldots \xrightarrow{x_n} q_n,$

define $\omega_\mathbf{A}(x)= \lambda(q_0,x_1)\ldots \lambda(q_{n-1},x_n)$.

Mealy machines and Moore machines have essentially the same descriptive power.

References

Last revised on July 18, 2019 at 09:57:09. See the history of this page for a list of all contributions to it.


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Authors:
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Award ID(s):
Publication Date:
NSF-PAR ID:
10135902
Journal Name:
34th IEEE International Parallel & Distributed Processing Symposium
4. Abstract Given two datasets (or tables) A and B and a search distance $$\epsilon$$ ϵ , the distance similarity join, denoted as $$A \ltimes _\epsilon B$$ A ⋉ ϵ B , finds the pairs of points ( $$p_a$$ p a , $$p_b$$ p b ), where $$p_a \in A$$ p a ∈ A and $$p_b \in B$$ p b ∈ B , and such that the distance between $$p_a$$ p a and $$p_b$$ p b is $$\le \epsilon$$ ≤ ϵ . If $$A = B$$ A = B , then the similarity join is equivalent to a similarity self-join, denoted as $$A \bowtie _\epsilon A$$ A ⋈ ϵ A . We propose in this paper Heterogeneous Epsilon Grid Joins ( HEGJoin ), a heterogeneous CPU-GPU distance similarity join algorithm. Efficiently partitioning the work between the CPU and the GPU is a challenge. Indeed, the work partitioning strategy needs to consider the different characteristics and computational throughput of the processors (CPU and GPU), as well as the data-dependent nature of the similarity join that accounts in the overall execution time (e.g., the number of queries, their distribution, the dimensionality, etc.). In addition to HEGJoin , we design in this paper amore »


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Q&A

# 2 bit subtractor using full subtractor

+2
−1

Can we make a n-bit subtractor using full subtractors or if we want do subtract one binary from another we must use full adders and use the complement of the 2nd number to do the subtraction?

I have only found examples which we use the complement of the second number and using full adders if it is possible to create a n-bit substractor using only full subtractors and without any adders or complements i would really appreciate it if he/she posted the circuit because i am stuck.

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#### 1 comment thread

+3
−0

We don't just give direct answers to homework (or homework-like) questions here, but can help you work thru the problem on your own.

You might consider trying to synthesize a subtractor from scratch, just like you would an adder. The fundamental building block of an adder is the half-adder. It takes in two bits to add, and produces the sum and carry bits. Since there are only 4 possible combinations of inputs, a truth table nicely describes its operation exhaustively:

  In1   In2   Sum   Car
---   ---   ---   ---
0     0     0     0
0     1     1     0
1     0     1     0
1     1     0     1

That should be enough to create a circuit of basic logic gates to implement the half-adder. You then chain together two of these per bit to make the full adder. This process is well documented in many place out there, very likely in your textbook too.

Now consider doing the same thing for subtraction. You have two input bits, and two output bits. In this case the output bits are the difference and the borrow.

When you get done with this, it might be instructive to compare that result to constructing a subtractor by complementing one of the input numbers, adding it to the other input number, and adding 1 to the result. Note that adding 1 to the result is the same as if the carry input to the first bit were 1.

I know how to make a full subtractor i just dont know how to connect multiple of them in order to make subtraction of 2 binary numbers with more than 1 bit

That makes little sense. If you really had a full subtractor, then you'd know what the input and output signals were, and then it would be obvious how to chain them to subtract wide numbers.

A full subtractor has three inputs and two outputs. The inputs are the two data bits to subtract, and the borrow input. The outputs are the resulting difference bit and the borrow output. You really should be able to figure out how to chain such blocks to subtract multi-bit numbers.

I cant understand which pins to connect with which pins.

Pins of what exactly? What exactly do you think the pins of this thing are?

Again, we are not here to do your homework for you. You have to put in some effort and be trying to learn. Show a block diagram of this thing, its input and output signals, and explain what you think each of these signals do. You may realize how to hook it up in the process of trying to explain it. Or, if there is a misconception, we'll know better how to help you understand it.

This is getting away from the original question. Instead of adding this to your existing question (and thereby causing confusion), it is probably better to show the diagram, input and output signals, and ask about connecting up the block in a separate question.

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#### 1 comment thread

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# News

## Analytical theory of nonlinear metamaterials: design on demand

Monday, 11 October 2010
Poutrina et al. developed an analytical approach allowing to relate a particular geometry of a metamaterial unit cell incorporating nonlinear elements to the effective nonlinear susceptibilities characterizing the resulting effective medium. The approach allows the prediction, analysis, and possible enhancement of the nonlinear response of a metacrystal, exploiting and proving a long-predicted concept of the increased nonlinear response in metamaterial structures. The theory shows an excellent...

## Quantitative characterization of nonlinear metamaterials

Wednesday, 6 October 2010
In a paper published in Applied Physics Letters, Larouche et al. present the first quantitative characterization of nonlinear metamaterials. Using an approach based on transfer matrices previously developed in the CMIP, they determine the quadratic nonlinear magnetic susceptibility of a slab of varactor-loaded split ring resonators to be 3.5 m/V. This gigantic value confirms qualitative predictions found in the literature and supports the validity of an analytical model developed by Poutrina et...

## Acoustic cloaking transformations from attainable material properties

Tuesday, 5 October 2010
The acoustic "sisters" of electromagnetic invisibility cloaks are long-sought devices that could defeat sonars. The existence of two types of acoustic waves in elastic media, however, makes acoustic cloaking very hard to achieve. The paper by Yaroslav Urzhumov and co-workers, published recently in the New Journal of Physics (July 12, 2010), is the first report on the feasibility of building an acoustically undetectable elastic cloak. Acoustic cloaking transformations from attainable material...

## Plasmonic Multi-Mode Interference Couplers

Wednesday, 21 October 2009
The MMI coupler is a key component in optical integrated circuits. Its main function is to combine or split optical signals, enabling the development of powerful multiplexing devices and integrated sensors. In a paper published in Optics Express (September 28, 2009 issue), Yu-Ju Tsai and co-workers investigate metallic MMI couplers that support long-range surface plasmons instead of conventional electromagnetic waves. Although the use of metals introduces material losses in the system, the...

## Electromagnetic source transformations using superellipse equations

Tuesday, 12 May 2009
Transformation optics can be used to design media with unique properties for the manipulation of electromagnetic waves. The transformation optical approach can be used not only to control the path of light, but also to change the way that currents and charges (or sources) behave when they interact with light. The interaction of electromagnetic waves is fundamental in the design of antennas.  In the latest issue of Applied Physics Letters, Jeffery Allen and colleagues demonstrate several...

## Duke to lead program on Transformation Optics sponsored by the Army Research Office

Tuesday, 12 May 2009

## Broadband non-resonant metamaterial used for ground plane cloaking

Friday, 16 January 2009
There has been a large amount of interest generated by recent work on invisibility cloaking. While the concept of invisibility has many corollaries in popular fiction which inspire the imagination, from the Harry Potter stories to Predator, researchers are taking only the first steps- both theoretical and experimental- towards these sci-fi goals. The general approach being used, which combines transformation optics with metamaterials operating in the microwave region of the electromagnetic...

## The transformation optical approach to antenna design

Monday, 23 June 2008
Transformation optics is a new tool for the design of electromagnetic media. Using transformation optics, researchers can visualize the interaction of certain complex materials with light as a warping of space. The physical basis underlying this connection, however, extends also to sources of light—electric currents, for example. In a recent paper published in Optics Express, Nathan Kundtz and colleagues illustrate the application of transformation optics to electric currents. In one example,...


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# Branching processes, regularity of

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2010 Mathematics Subject Classification: Primary: 60J80 [MSN][ZBL]

A property of branching processes ensuring that the number of particles at any moment of time is finite. The problem of the regularity of a branching process is usually reduced to the problem of uniqueness of the solution of some differential or integral equation. For instance, in a continuous-time branching process the differential equation

$$\frac{\partial F (t; s) }{\partial t } = \ f (F (t; s))$$

with the initial condition $F(0; s) = s$ has a unique solution if and only if, for any $\epsilon > 0$, the integral

$$\int\limits _ {1 - \epsilon } ^ { 1 } { \frac{dx}{f (x) } }$$

is divergent. In the branching Bellman–Harris process the generating function $F(t; s)$ of the number of particles is the solution of the non-linear integral equation

$$\tag{* } F (t; s) = \ \int\limits _ { 0 } ^ { t } h (F (t - u; s)) dG (u) + s (1 - G (t)),$$

where $G(t)$ is the distribution function of the lifetimes of particles and $h(t)$ is the generating function of the number of daughter particles ( "direct descendants" ) of a single particle. If, for given $t _ {0} , c _ {1} , c _ {2} > 0$ and an integer $n \geq 1$, the inequalities

$$c _ {1} t ^ {n} \leq G (t) \leq c _ {2} t ^ {n}$$

are valid for all $0 \leq t \leq t _ {0}$, the solution of equation (*) is unique if and only if the equation

$$\frac{d ^ {n} \phi }{dt ^ {n} } = \ h ( \phi ) - 1$$

with initial conditions

$$\phi (0) = 1,\ \ \phi ^ {(r)} (0) = 0,\ r = 1 \dots n - 1,$$

has a unique solution

$$0 \leq \phi (t) \leq 1.$$

For a branching process described by equation (*) to be regular, it is necessary and sufficient for the integral

$$\int\limits _ { 0 } ^ \epsilon { \frac{dx}{x ^ {1-1/n } (1-h(1-x)) ^ {1/n} } }$$

to diverge for any $\epsilon > 0$.

#### References

 [S] B.A. Sewastjanow, "Verzweigungsprozesse" , Akad. Wissenschaft. DDR (1974) (Translated from Russian) MR0408018 Zbl 0291.60039

#### Comments

Additional references can be found in the article Branching process.

How to Cite This Entry:
Branching processes, regularity of. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Branching_processes,_regularity_of&oldid=46156
This article was adapted from an original article by B.A. Sevast'yanov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article


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# Closed Subset of Real Numbers with Lower Bound contains Infimum

## Theorem

Consider the real number line as a metric space under the usual metric.

Let $A \subseteq \R$ such that $A$ is closed in $\R$ and $A \neq \varnothing$.

Let $A$ be bounded below.

Then $A$ contains its infimum.

## Proof

$\map \inf A \in \map \cl A$
$A = \map \cl A$

Therefore $\map \inf A \in A$.

$\blacksquare$


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# Tangent lines to circles

In Euclidean plane geometry, a tangent line to a circle is a line that touches the circle at exactly one point, never entering the circle's interior. Tangent lines to circles form the subject of several theorems, and play an important role in many geometrical constructions and proofs. Since the tangent line to a circle at a point P is perpendicular to the radius to that point, theorems involving tangent lines often involve radial lines and orthogonal circles.

## Tangent lines to one circle

A tangent line t to a circle C intersects the circle at a single point T. For comparison, secant lines intersect a circle at two points, whereas another line may not intersect a circle at all. This property of tangent lines is preserved under many geometrical transformations, such as scalings, rotation, translations, inversions, and map projections. In technical language, these transformations do not change the incidence structure of the tangent line and circle, even though the line and circle may be deformed.

The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference. Conversely, the perpendicular to a radius through the same endpoint is a tangent line. The resulting geometrical figure of circle and tangent line has a reflection symmetry about the axis of the radius.

By the power-of-a-point theorem, the product of lengths PM·PN for any ray PMN equals to the square of PT, the length of the tangent line segment (red).

No tangent line can be drawn through a point within a circle, since any such line must be a secant line. However, two tangent lines can be drawn to a circle from a point P outside of the circle. The geometrical figure of a circle and both tangent lines likewise has a reflection symmetry about the radial axis joining P to the center point O of the circle. Thus the lengths of the segments from P to the two tangent points are equal. By the secant-tangent theorem, the square of this tangent length equals the power of the point P in the circle C. This power equals the product of distances from P to any two intersection points of the circle with a secant line passing through P.

The angle θ between a chord and a tangent is half the arc belonging to the chord.

The tangent line t and the tangent point T have a conjugate relationship to one another, which has been generalized into the idea of pole points and polar lines. The same reciprocal relation exists between a point P outside the circle and the secant line joining its two points of tangency.

If a point P is exterior to a circle with center O, and if the tangent lines from P touch the circle at points T and S, then ∠TPS and ∠TOS are supplementary (sum to 180°).

If a chord TM is drawn from the tangency point T of exterior point P and ∠PTM ≤ 90° then ∠PTM = (1/2)∠TOM.

### Compass and straightedge constructions

It is relatively straightforward to construct a line t tangent to a circle at a point T on the circumference of the circle:

• A line a is drawn from O, the center of the circle, through the radial point T;
• The line t is the perpendicular line to a.
Construction of a tangent to a given circle (black) from a given exterior point (P).

Thales' theorem may be used to construct the tangent lines to a point P external to the circle C:

• A circle is drawn centered on the midpoint of the line segment OP, having diameter OP, where O is again the center of the circle C.
• The intersection points T1 and T2 of the circle C and the new circle are the tangent points for lines passing through P, by the following argument.

The line segments OT1 and OT2 are radii of the circle C; since both are inscribed in a semicircle, they are perpendicular to the line segments PT1 and PT2, respectively. But only a tangent line is perpendicular to the radial line. Hence, the two lines from P and passing through T1 and T2 are tangent to the circle C.

Another method to construct the tangent lines to a point P external to the circle using only a straightedge:

• Draw any three different lines through the given point P that intersect the circle twice.
• Let ${\displaystyle A_{1},A_{2},B_{1},B_{2},C_{1},C_{2}}$ be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to P.
• Let D be the point where the lines ${\displaystyle A_{1}B_{2}}$ and ${\displaystyle A_{2}B_{1}}$ intersect,
• Similarly E for the lines ${\displaystyle B_{1}C_{2}}$ and ${\displaystyle B_{2}C_{1}}$.
• Draw a line through D and E.
• This line meets the circle at two points, F and G.
• The tangents are the lines PF and PG. [1]

### Tangential polygons

A tangential polygon is a polygon each of whose sides is tangent to a particular circle, called its incircle. Every triangle is a tangential polygon, as is every regular polygon of any number of sides; in addition, for every number of polygon sides there are an infinite number of non-congruent tangential polygons.

#### Tangent quadrilateral theorem and inscribed circles

A tangential quadrilateral ABCD is a closed figure of four straight sides that are tangent to a given circle C. Equivalently, the circle C is inscribed in the quadrilateral ABCD. By the Pitot theorem, the sums of opposite sides of any such quadrilateral are equal, i.e.,

${\displaystyle {\overline {AB}}+{\overline {CD}}={\overline {BC}}+{\overline {DA}}.}$

This conclusion follows from the equality of the tangent segments from the four vertices of the quadrilateral. Let the tangent points be denoted as P (on segment AB), Q (on segment BC), R (on segment CD) and S (on segment DA). The symmetric tangent segments about each point of ABCD are equal, e.g., BP=BQ=b, CQ=CR=c, DR=DS=d, and AS=AP=a. But each side of the quadrilateral is composed of two such tangent segments

${\displaystyle {\overline {AB}}+{\overline {CD}}=(a+b)+(c+d)={\overline {BC}}+{\overline {DA}}=(b+c)+(d+a)}$

proving the theorem.

The converse is also true: a circle can be inscribed into every quadrilateral in which the lengths of opposite sides sum to the same value.[2]

This theorem and its converse have various uses. For example, they show immediately that no rectangle can have an inscribed circle unless it is a square, and that every rhombus has an inscribed circle, whereas a general parallelogram does not.

## Tangent lines to two circles

The external (above) and internal (below) homothetic center S of the two circles.

For two circles, there are generally four distinct lines that are tangent to both (bitangent) – if the two circles are outside each other – but in degenerate cases there may be any number between zero and four bitangent lines; these are addressed below. For two of these, the external tangent lines, the circles fall on the same side of the line; for the two others, the internal tangent lines, the circles fall on opposite sides of the line. The external tangent lines intersect in the external homothetic center, whereas the internal tangent lines intersect at the internal homothetic center. Both the external and internal homothetic centers lie on the line of centers (the line connecting the centers of the two circles), closer to the center of the smaller circle: the internal center is in the segment between the two circles, while the external center is not between the points, but rather outside, on the side of the center of the smaller circle. If the two circles have equal radius, there are still four bitangents, but the external tangent lines are parallel and there is no external center in the affine plane; in the projective plane, the external homothetic center lies at the point at infinity corresponding to the slope of these lines.[3]

### Outer tangent

Finding outer tangent. Two circles' outer tangents.

The red line joining the points ${\displaystyle (x_{3},y_{3})}$ and ${\displaystyle (x_{4},y_{4})}$ is the outer tangent between the two circles. Given points ${\displaystyle (x_{1},y_{1})}$, ${\displaystyle (x_{2},y_{2})}$ the points ${\displaystyle (x_{3},y_{3})}$, ${\displaystyle (x_{4},y_{4})}$ can easily be calculated with help of the angle ${\displaystyle \alpha }$:

{\displaystyle {\begin{aligned}x_{3}&=x_{1}+r\cos({\tfrac {\pi }{2}}-\alpha )\\y_{3}&=y_{1}+r\sin({\tfrac {\pi }{2}}-\alpha )\\x_{4}&=x_{2}+R\cos({\tfrac {\pi }{2}}-\alpha )\\y_{4}&=y_{2}+R\sin({\tfrac {\pi }{2}}-\alpha )\\\end{aligned}}}

Here R and r notate the radii of the two circles and the angle ${\displaystyle \alpha }$ can be computed using basic trigonometry. You have ${\displaystyle \alpha =\gamma -\beta }$ with ${\displaystyle \gamma =-\arctan \left({\tfrac {y_{2}-y_{1}}{x_{2}-x_{1}}}\right)}$ and ${\displaystyle \beta =\arcsin \left({\tfrac {R-r}{\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}\right)}$. [4][not in citation given (See discussion.)]

### Inner tangent

Inner tangent. The external tangent lines pass through the internal homothetic center.

An inner tangent is a tangent that intersects the segment joining two circles' centers. Note that the inner tangent will not be defined for cases when the two circles overlap.

### Construction

The bitangent lines can be constructed either by constructing the homothetic centers, as described at that article, and then constructing the tangent lines through the homothetic center that is tangent to one circle, by one of the methods described above. The resulting line will then be tangent to the other circle as well. Alternatively, the tangent lines and tangent points can be constructed more directly, as detailed below. Note that in degenerate cases these constructions break down; to simplify exposition this is not discussed in this section, but a form of the construction can work in limit cases (e.g., two circles tangent at one point).

#### Synthetic geometry

Let O1 and O2 be the centers of the two circles, C1 and C2 and let r1 and r2 be their radii, with r1 > r2; in other words, circle C1 is defined as the larger of the two circles. Two different methods may be used to construct the external and internal tangent lines.

External tangents
Construction of the outer tangent

A new circle C3 of radius r1 − r2 is drawn centered on O1. Using the method above, two lines are drawn from O2 that are tangent to this new circle. These lines are parallel to the desired tangent lines, because the situation corresponds to shrinking both circles C1 and C2 by a constant amount, r2, which shrinks C2 to a point. Two radial lines may be drawn from the center O1 through the tangent points on C3; these intersect C1 at the desired tangent points. The desired external tangent lines are the lines perpendicular to these radial lines at those tangent points, which may be constructed as described above.

Internal tangents
Construction of the inner tangent

A new circle C3 of radius r1 + r2 is drawn centered on O1. Using the method above, two lines are drawn from O2 that are tangent to this new circle. These lines are parallel to the desired tangent lines, because the situation corresponds to shrinking C2 to a point while expanding C1 by a constant amount, r2. Two radial lines may be drawn from the center O1 through the tangent points on C3; these intersect C1 at the desired tangent points. The desired internal tangent lines are the lines perpendicular to these radial lines at those tangent points, which may be constructed as described above.

#### Analytic geometry

Let the circles have centres c1 = (x1,y1) and c2 = (x2,y2) with radius r1 and r2 respectively. Expressing a line by the equation ${\displaystyle ax+by+c=0,}$ with the normalization a2 + b2 = 1, then a bitangent line satisfies:

ax1 + by1 + c = r1 and
ax2 + by2 + c = r2.

Solving for ${\displaystyle (a,b,c)}$ by subtracting the first from the second yields

aΔx + bΔy = Δr

where Δx = x2 − x1, Δy = y2 − y1 and Δr = r2 − r1.

If ${\displaystyle d={\sqrt {(\Delta x)^{2}+(\Delta y)^{2}}}}$ is the distance from c1 to c2 we can normalize by X = Δx/d, Y = Δy/d and R = Δr/d to simplify equations, yielding the equations aX + bY = R and a2 + b2 = 1, solve these to get two solutions (k = ±1) for the two external tangent lines:

a = RX − kY√(1 − R2)
b = RY + kX√(1 − R2)
c = r1 − (ax1 + by1)

Geometrically this corresponds to computing the angle formed by the tangent lines and the line of centers, and then using that to rotate the equation for the line of centers to yield an equation for the tangent line. The angle is computed by computing the trigonometric functions of a right triangle whose vertices are the (external) homothetic center, a center of a circle, and a tangent point; the hypotenuse lies on the tangent line, the radius is opposite the angle, and the adjacent side lies on the line of centers.

(XY) is the unit vector pointing from c1 to c2, while R is ${\displaystyle \cos \theta }$ where ${\displaystyle \theta }$ is the angle between the line of centers and a tangent line. ${\displaystyle \sin \theta }$ is then ${\displaystyle \pm {\sqrt {1-R^{2}}}}$ (depending on the sign of ${\displaystyle \theta }$, equivalently the direction of rotation), and the above equations are rotation of (XY) by ${\displaystyle \pm \theta ,}$ using the rotation matrix:

${\displaystyle {\begin{pmatrix}R&\mp {\sqrt {1-R^{2}}}\\\pm {\sqrt {1-R^{2}}}&R\end{pmatrix}}}$
k = 1 is the tangent line to the right of the circles looking from c1 to c2.
k = −1 is the tangent line to the right of the circles looking from c2 to c1.

The above assumes each circle has positive radius. If r1 is positive and r2 negative then c1 will lie to the left of each line and c2 to the right, and the two tangent lines will cross. In this way all four solutions are obtained. Switching signs of both radii switches k = 1 and k = −1.

#### Vectors

Finding outer tangent. Circle tangents.

In general the points of tangency t1 and t2 for the four lines tangent to two circles with centers v1 and v2 and radii r1 and r2 are given by solving the simultaneous equations:

{\displaystyle {\begin{aligned}(t_{2}-v_{2})\cdot (t_{2}-t_{1})&=0\\(t_{1}-v_{1})\cdot (t_{2}-t_{1})&=0\\(t_{1}-v_{1})\cdot (t_{1}-v_{1})&=r_{1}^{2}\\(t_{2}-v_{2})\cdot (t_{2}-v_{2})&=r_{2}^{2}\\\end{aligned}}}

These equations express that the tangent line, which is parallel to ${\displaystyle t_{2}-t_{1},}$ is perpendicular to the radii, and that the tangent points lie on their respective circles.

These are four quadratic equations in two two-dimensional vector variables, and in general position will have four pairs of solutions.

### Degenerate cases

Two distinct circles may have between zero and four bitangent lines, depending on configuration; these can be classified in terms of the distance between the centers and the radii. If counted with multiplicity (counting a common tangent twice) there are zero, two, or four bitangent lines. Bitangent lines can also be generalized to circles with negative or zero radius. The degenerate cases and the multiplicities can also be understood in terms of limits of other configurations – e.g., a limit of two circles that almost touch, and moving one so that they touch, or a circle with small radius shrinking to a circle of zero radius.

• If the circles are outside each other (${\displaystyle d>r_{1}+r_{2}}$), which is general position, there are four bitangents.
• If they touch externally at one point (${\displaystyle d=r_{1}+r_{2}}$) – have one point of external tangency – then they have two external bitangents and one internal bitangent, namely the common tangent line. This common tangent line has multiplicity two, as it separates the circles (one on the left, one on the right) for either orientation (direction).
• If the circles intersect in two points (${\displaystyle |r_{1}-r_{2}|), then they have no internal bitangents and two external bitangents (they cannot be separated, because they intersect, hence no internal bitangents).
• If the circles touch internally at one point (${\displaystyle d=|r_{1}-r_{2}|}$) – have one point of internal tangency – then they have no internal bitangents and one external bitangent, namely the common tangent line, which has multiplicity two, as above.
• If one circle is completely inside the other (${\displaystyle d<|r_{1}-r_{2}|}$) then they have no bitangents, as a tangent line to the outer circle does not intersect the inner circle, or conversely a tangent line to the inner circle is a secant line to the outer circle.

Finally, if the two circles are identical, any tangent to the circle is a common tangent and hence (external) bitangent, so there is a circle's worth of bitangents.

Further, the notion of bitangent lines can be extended to circles with negative radius (the same locus of points, ${\displaystyle x^{2}+y^{2}=(-r)^{2},}$ but considered "inside out"), in which case if the radii have opposite sign (one circle has negative radius and the other has positive radius) the external and internal homothetic centers and external and internal bitangents are switched, while if the radii have the same sign (both positive radii or both negative radii) "external" and "internal" have the same usual sense (switching one sign switches them, so switching both switches them back).

Bitangent lines can also be defined when one or both of the circles has radius zero. In this case the circle with radius zero is a double point, and thus any line passing through it intersects the point with multiplicity two, hence is "tangent". If one circle has radius zero, a bitangent line is simply a line tangent to the circle and passing through the point, and is counted with multiplicity two. If both circles have radius zero, then the bitangent line is the line they define, and is counted with multiplicity four.

Note that in these degenerate cases the external and internal homothetic center do generally still exist (the external center is at infinity if the radii are equal), except if the circles coincide, in which case the external center is not defined, or if both circles have radius zero, in which case the internal center is not defined.

### Applications

#### Belt problem

The internal and external tangent lines are useful in solving the belt problem, which is to calculate the length of a belt or rope needed to fit snugly over two pulleys. If the belt is considered to be a mathematical line of negligible thickness, and if both pulleys are assumed to lie in exactly the same plane, the problem devolves to summing the lengths of the relevant tangent line segments with the lengths of circular arcs subtended by the belt. If the belt is wrapped about the wheels so as to cross, the interior tangent line segments are relevant. Conversely, if the belt is wrapped exteriorly around the pulleys, the exterior tangent line segments are relevant; this case is sometimes called the pulley problem.

## Tangent lines to three circles: Monge's theorem

For three circles denoted by C1, C2, and C3, there are three pairs of circles (C1C2, C2C3, and C1C3). Since each pair of circles has two homothetic centers, there are six homothetic centers altogether. Gaspard Monge showed in the early 19th century that these six points lie on four lines, each line having three collinear points.

## Problem of Apollonius

Animation showing the inversive transformation of an Apollonius problem. The blue and red circles swell to tangency, and are inverted in the grey circle, producing two straight lines. The yellow solutions are found by sliding a circle between them until it touches the transformed green circle from within or without.

Many special cases of Apollonius's problem involve finding a circle that is tangent to one or more lines. The simplest of these is to construct circles that are tangent to three given lines (the LLL problem). To solve this problem, the center of any such circle must lie on an angle bisector of any pair of the lines; there are two angle-bisecting lines for every intersection of two lines. The intersections of these angle bisectors give the centers of solution circles. There are four such circles in general, the inscribed circle of the triangle formed by the intersection of the three lines, and the three exscribed circles.

A general Apollonius problem can be transformed into the simpler problem of circle tangent to one circle and two parallel lines (itself a special case of the LLC special case). To accomplish this, it suffices to scale two of the three given circles until they just touch, i.e., are tangent. An inversion in their tangent point with respect to a circle of appropriate radius transforms the two touching given circles into two parallel lines, and the third given circle into another circle. Thus, the solutions may be found by sliding a circle of constant radius between two parallel lines until it contacts the transformed third circle. Re-inversion produces the corresponding solutions to the original problem.

## Generalizations

The concept of a tangent line and tangent point can be generalized to a pole point Q and its corresponding polar line q. The points P and Q are inverses of each other with respect to the circle.

The concept of a tangent line to one or more circles can be generalized in several ways. First, the conjugate relationship between tangent points and tangent lines can be generalized to pole points and polar lines, in which the pole points may be anywhere, not only on the circumference of the circle. Second, the union of two circles is a special (reducible) case of a quartic plane curve, and the external and internal tangent lines are the bitangents to this quartic curve. A generic quartic curve has 28 bitangents.

A third generalization considers tangent circles, rather than tangent lines; a tangent line can be considered as a tangent circle of infinite radius. In particular, the external tangent lines to two circles are limiting cases of a family of circles which are internally or externally tangent to both circles, while the internal tangent lines are limiting cases of a family of circles which are internally tangent to one and externally tangent to the other of the two circles.[5]

In Möbius or inversive geometry, lines are viewed as circles through a point "at infinity" and for any line and any circle, there is a Möbius transformation which maps one to the other. In Möbius geometry, tangency between a line and a circle becomes a special case of tangency between two circles. This equivalence is extended further in Lie sphere geometry.

## References

1. ^
2. ^
3. ^ Paul Kunkel. "Tangent circles". Whistleralley.com. Retrieved 2008-09-29.
4. ^ Libeskind, Shlomo (2007), Euclidean and Transformational Geometry: A Deductive Inquiry, pp. 110&ndash, 112 (online copy, p. 110, at Google Books)
5. ^ Kunkel, Paul (2007), "The tangency problem of Apollonius: three looks" (PDF), BSHM Bulletin: Journal of the British Society for the History of Mathematics, 22 (1): 34–46, doi:10.1080/17498430601148911


text:
The nodal period (or draconic period) of a satellite is the time interval between successive passages of the satellite through either of its orbital nodes,[1][2] typically the ascending node. This type of orbital period applies to artificial satellites, like those that monitor weather on Earth, and natural satellites like the Moon.

It is distinct from the sidereal period, which measures the period with respect to reference stars seemingly fixed onto a spherical background, since the location of a satellite's nodes precess over time.[3] For example, the nodal period of the Moon is 27.2122 days[4] (one draconic month), while its sidereal period is 27.3217 days[5] (one sidereal month).

## Near-Earth satellites

The oblate figure of the Earth has important effects of the orbits of near-Earth satellites.[6] An expression for the nodal period (Tn) of a near circular orbit, such that the eccentricity (ε) is almost but not equal to zero, is the following:[7]

${\displaystyle T_{n}={\frac {2\pi a^{\frac {3}{2}}}{\mu ^{\frac {1}{2}}}}\left(1-{\frac {3J_{2}\left(4-5\sin ^{2}i\right)}{4\left({\frac {a}{R}}\right)^{2}{\sqrt {1-\varepsilon ^{2}}}\left(1+\varepsilon \cos \omega \right)^{2}}}-{\frac {3J_{2}\left(1-\varepsilon \cos \omega \right)^{3}}{2\left({\frac {a}{R}}\right)^{2}\left(1-\varepsilon ^{2}\right)^{3}}}\right)}$

where ${\displaystyle a}$ is the semi-major axis, ${\displaystyle \mu }$ is the gravitational constant, ${\displaystyle J_{2}}$ is a perturbation factor due to the oblateness of the earth, ${\displaystyle i}$ is the inclination, ${\displaystyle R}$ is the radius of the earth and ${\displaystyle \omega }$ is the argument of the perigee.


text:
# At what distance from a 27 mW point source of electromagnetic waves is the electric field...

## Question:

At what distance from a {eq}27 \ mW {/eq} point source of electromagnetic waves is the electric field amplitude {eq}0.010 \ V / m {/eq}?

## Energy Density of Electromagnetic waves

The energy density of Electromagnetic wave is defined as the energy stored by the electric field component and magnetic field component of Electromagnetic wave.

Mathematically, {eq}U_{total}=\ U_{E}+\ U_{B}\\ U_{total}=\ \dfrac{\epsilon_{0} E_{0}^{2}}{2}+\ \dfrac{B_{0}^{2}}{2\mu_{0}} {/eq}


text:
# Logarithmic coefficients and a coefficient conjecture for univalent functions

Logarithmic coefficients and a coefficient conjecture for univalent functions Let $${{\mathcal {U}}}(\lambda )$$ U ( λ ) denote the family of analytic functions f(z), $$f(0)=0=f'(0)-1$$ f ( 0 ) = 0 = f ′ ( 0 ) - 1 , in the unit disk $${\mathbb {D}}$$ D , which satisfy the condition $$\big |\big (z/f(z)\big )^{2}f'(z)-1\big |<\lambda$$ | ( z / f ( z ) ) 2 f ′ ( z ) - 1 | < λ for some $$0<\lambda \le 1$$ 0 < λ ≤ 1 . The logarithmic coefficients $$\gamma _n$$ γ n of f are defined by the formula $$\log (f(z)/z)=2\sum _{n=1}^\infty \gamma _nz^n$$ log ( f ( z ) / z ) = 2 ∑ n = 1 ∞ γ n z n . In a recent paper, the present authors proposed a conjecture that if $$f\in {{\mathcal {U}}}(\lambda )$$ f ∈ U ( λ ) for some $$0<\lambda \le 1$$ 0 < λ ≤ 1 , then $$|a_n|\le \sum _{k=0}^{n-1}\lambda ^k$$ | a n | ≤ ∑ k = 0 n - 1 λ k for $$n\ge 2$$ n ≥ 2 and provided a new proof for the case $$n=2$$ n = 2 . One of the aims of this article is to present a proof of this conjecture for $$n=3, 4$$ n = 3 , 4 and an elegant proof of the inequality for $$n=2$$ n = 2 , with equality for $$f(z)=z/[(1+z)(1+\lambda z)]$$ f ( z ) = z / [ ( 1 + z ) ( 1 + λ z ) ] . In addition, the authors prove the following sharp inequality for $$f\in {{\mathcal {U}}}(\lambda )$$ f ∈ U ( λ ) : \begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2} \le \frac{1}{4}\left( \frac{\pi ^{2}}{6}+2\mathrm{Li\,}_{2}(\lambda )+\mathrm{Li\,}_{2}(\lambda ^{2})\right) , \end{aligned} ∑ n = 1 ∞ | γ n | 2 ≤ 1 4 π 2 6 + 2 Li 2 ( λ ) + Li 2 ( λ 2 ) , where $$\mathrm{Li}_2$$ Li 2 denotes the dilogarithm function. Furthermore, the authors prove two such new inequalities satisfied by the corresponding logarithmic coefficients of some other subfamilies of $${\mathcal {S}}$$ S . http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Monatshefte f�r Mathematik Springer Journals

# Logarithmic coefficients and a coefficient conjecture for univalent functions

, Volume 185 (3) – Feb 3, 2017
13 pages

/lp/springer_journal/logarithmic-coefficients-and-a-coefficient-conjecture-for-univalent-SA13S0uw0c
Publisher
Springer Journals
Subject
Mathematics; Mathematics, general
ISSN
0026-9255
eISSN
1436-5081
D.O.I.
10.1007/s00605-017-1024-3
Publisher site
See Article on Publisher Site

### Abstract

Let $${{\mathcal {U}}}(\lambda )$$ U ( λ ) denote the family of analytic functions f(z), $$f(0)=0=f'(0)-1$$ f ( 0 ) = 0 = f ′ ( 0 ) - 1 , in the unit disk $${\mathbb {D}}$$ D , which satisfy the condition $$\big |\big (z/f(z)\big )^{2}f'(z)-1\big |<\lambda$$ | ( z / f ( z ) ) 2 f ′ ( z ) - 1 | < λ for some $$0<\lambda \le 1$$ 0 < λ ≤ 1 . The logarithmic coefficients $$\gamma _n$$ γ n of f are defined by the formula $$\log (f(z)/z)=2\sum _{n=1}^\infty \gamma _nz^n$$ log ( f ( z ) / z ) = 2 ∑ n = 1 ∞ γ n z n . In a recent paper, the present authors proposed a conjecture that if $$f\in {{\mathcal {U}}}(\lambda )$$ f ∈ U ( λ ) for some $$0<\lambda \le 1$$ 0 < λ ≤ 1 , then $$|a_n|\le \sum _{k=0}^{n-1}\lambda ^k$$ | a n | ≤ ∑ k = 0 n - 1 λ k for $$n\ge 2$$ n ≥ 2 and provided a new proof for the case $$n=2$$ n = 2 . One of the aims of this article is to present a proof of this conjecture for $$n=3, 4$$ n = 3 , 4 and an elegant proof of the inequality for $$n=2$$ n = 2 , with equality for $$f(z)=z/[(1+z)(1+\lambda z)]$$ f ( z ) = z / [ ( 1 + z ) ( 1 + λ z ) ] . In addition, the authors prove the following sharp inequality for $$f\in {{\mathcal {U}}}(\lambda )$$ f ∈ U ( λ ) : \begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2} \le \frac{1}{4}\left( \frac{\pi ^{2}}{6}+2\mathrm{Li\,}_{2}(\lambda )+\mathrm{Li\,}_{2}(\lambda ^{2})\right) , \end{aligned} ∑ n = 1 ∞ | γ n | 2 ≤ 1 4 π 2 6 + 2 Li 2 ( λ ) + Li 2 ( λ 2 ) , where $$\mathrm{Li}_2$$ Li 2 denotes the dilogarithm function. Furthermore, the authors prove two such new inequalities satisfied by the corresponding logarithmic coefficients of some other subfamilies of $${\mathcal {S}}$$ S .

### Journal

Monatshefte f�r MathematikSpringer Journals

Published: Feb 3, 2017

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text:
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
At the end of the day, we will want to be looking at this true-in-$$\mathfrak{N}$$ $$\Pi$$-formula $$\sigma$$ that we have constructed and be able to say to it, "There is no deduction of you." The construction of the formula $$\sigma$$ will involve rather detailed analysis of the collections of deductions, so it will be convenient, if initially messy, to have a way to translate deductions into natural numbers. Thus for example, rather than saying "This long sequence of formulas is a deduction of the formula $$0 = 1$$", we could say "The number 42 is a code for a deduction of $$0 = 1$$."
The other advantage of having this coding is that it will allow us to code up statements about numbers that code up statements. For example, it might be the case that the number 24601 is a code for the statement "The number 42 is a code for a deduction of $$0 = 1$$." Or even (and this is the key idea) 24601 might be the code for the statement "There is no number that is a code for a deduction of the formula for which I am the code." The rather messy details of this will be covered in Chapter 6.
This all hinges on the facts that it is easy to code statements as numbers and decode numbers to see what statements they encode. Also, it is easy to check whether a potential deduction is, in fact, a deduction. Recall that a deduction is nothing more than a finite sequence of formulas, each one of which is either an axiom or follows from previous formulas in the sequence via a rule of inference. Thus, once we decide on a way to code formulas and to code sequences of formulas, it will not be a problem to examine a number and decide if that number codes up a deduction or not. Thus our path forward will be to fix our coding scheme, prove that the coding is nice, use the coding scheme in order to construct the formula $$\sigma$$, and then prove that $$\sigma$$ is both true and not provable. This route to the Incompleteness Theorem is followed in Chapters 5 and 6.


text:
# Makes a navbar for a website with object oriented php

I do this as sort of hobby and want to no if I can improve.
This is a simplified version of my code that normal works with bootstrap classes and id's.
At the bottom there is the input of the classes.
I have already a working version connected with SQL Database.
but want to take it a step further with object oriented php.

The next step for me is to create the content page underneath the navbar.
I work with a .htaccess file that redirects everything to index.php, and then use the URI to show the right content.

The main class

<?php

class Main
{
public $URL; public$PageName;
public function __construct($PageName) {$this->PageName = $PageName;$this->URL = $_SERVER['REQUEST_URI']; } public static function header() { echo '<!DOCTYPE html>   ... '; } public static function navstart() { echo '
'; } public static function navend() { echo '
'; } } the navitem class is for making the items in the nav bar
 class navItem extends Main { public function __construct($PageName)
{
parent::__construct($PageName);$ActiveUrl = ($this->URL == "/".$this->PageName) ? " active " : " ";
echo     '

PageName.'">'.$this->PageName.'
  '; } } The nav dropdown class to make the dropdown in the navbar class navDropdown extends Main { public function __construct($PageName, $Item) { parent::__construct($PageName);
$shortURL = substr($this->URL, 0, strpos($this->URL, "_"));$ActiveUrl = ($shortURL == "/".$this->PageName) ? " active " : " ";
echo '

'.$this->PageName.'

';
foreach ($Item as$Item) {
$ActiveUrl = ($this->URL == "/".$this->PageName.'_'.$Item) ? " active " : " ";
echo '
PageName.'_'.$Item.'" >'.$Item.'';
}
echo '

  ';
}
}

the variables to make the page

main::navstart();
$test = new navDropDown("test", array('test1' , 'test2', 'test3' , 'test4'));$home = new navItem("home");
$about = new navItem("about");$pages = new navDropDown("pages",array("one", "two", "three" ));
\$contact = new navItem("contact");
main::navend();
• I think you need to make some edits, especially in the markup. Currently the following is in a code block section and it shouldn't be: the navitem class is for making the items in the nav bar. – pacmaninbw Apr 22 at 22:00

I think you have some reading/research to do as I'm not sure you understand OOP...

## What is Main?

• Main is setup as a class, used as a namespace in instantiation, then extended(why?).
• Generally in the context it seems to be intended to be used App or <AppName> are better options

## When to extend

• Why do all elements extend Main?
• It appears to be solely for it's constructor which it never uses itself. (Extra Baggage in Main)
• Main also only contains functions the other elements never need or use (Extra Baggage in the Sub-Objects)
• Extending a class makes more sense if the classes extended from it share functionality

## What even is an Object? ##

• You seem to be using your objects as functions disguised as things.
• For example: when you create a navItem:
• You create the object
• It's controller prints itself to the screen
• You capture it in a variable never to be used again. Why?

## Who gets to be an Object?

• navItem is an object navBar Isn't
• navDropdown is an object navDropdownItem isn't
• Thanks I Will do soms reading and try again – Banaan Apr 22 at 21:15


text:
# Revision history [back]

The solutions are correct but they are not fully simplified.

sage: f= x^2 - (1 + I)*x + 6 + 3*I
sage: solve(f==0,x)
[x == -1/2*sqrt(-10*I - 24) + 1/2*I + 1/2, x == 1/2*sqrt(-10*I - 24) + 1/2*I + 1/2]

For each solution, you can simplify its right-hand-side as follows:

sage: [s.rhs().full_simplify() for s in solve(f==0,x)]
[3*I, -2*I + 1]


text:
# What are kernel initializers and what is their significance?

I was looking at code and found this:

model.add(Dense(13, input_dim=13, kernel_initializer='normal', activation='relu'))

I was keen to know about kernel_initializer but wasn't able to understand it's significance?

The neural network needs to start with some weights and then iteratively update them to better values. The term kernel_initializer is a fancy term for which statistical distribution or function to use for initialising the weights. In case of statistical distribution, the library will generate numbers from that statistical distribution and use as starting weights.

For example in the above code, normal distribution will be used to initialise weights. You can use other functions (constants like 1s or 0s) and distributions (uniform) too. All possible options are documented here.

Additional explanation: The term kernel is a carryover from other classical methods like SVM. The idea is to transform data in a given input space to another space where the transformation is achieved using kernel functions. We can think of neural network layers as non-linear maps doing these transformations, so the term kernels is used.

• Is there anything to consider, regarding the values these initializers produce? E.g. there are some that produce only positive values and some are centered around zero. Would you choose an initializer that produces values in the range that will be expected for the layer? Oct 29, 2021 at 10:24

In the activation function, the initialization is important to train the neural network. I pasted the suggestion about kernel from DeeplearingAI webpage.

Initializing kernel weights with inappropriate values will lead to divergence or a slow-down in the training of your neural network. The illustrated the exploding/vanishing gradient problem with simple symmetrical weight matrices, the observation generalizes to any initialization values that are too small or too large.

Consider the case where every kernel weight is initialized slightly larger than the identity matrix like: $$\begin{bmatrix}1.5 & 0 \\ 0 & 1.5\end{bmatrix}$$     When activation is used in backward propagation, this leads to the exploding gradient problem. That is, the gradients of the cost with the respect to the parameters are too big. This leads the cost to oscillate around its minimum value.
Similarly, consider the case where every kernel weight is initialized slightly smaller than the identity matrix.$$\begin{bmatrix}0.5 & 0 \\ 0 & 0.5\end{bmatrix}$$


text:
# Calculus Limit

Limit is one of the important topics of calculus. When a function cannot be defined at some point, in such cases we can look at what the function is approaching to as it gets closer to that point. This is known as limit. There are times when the value of the function is same as the function limit. And at times they vary too.

$lim_{x -> a} h (x) = H$

This implies that as h (x) approaches to H as x gets closer to ‘a’.

This implies that there exist e > 0 such that h (x) lies in the interval (H – e, H + e) that is |h (x) - H| < e with existence of d > 0 such that x lies in either (a – d, a) or (a, a + d) that is 0 < |x – a| < d.

Consider the function $\frac{(5 x – 5 x^2)}{(x – 1)}$. This function is not defined at x = 1. So at this particular point we can take limit of this function and see as to what it approaches to when x -> 1.

That is $lim_(x -> 1)$ $\frac{(5 x – 5 x^2)}{(x – 1) }$

= $lim_(x -> 1)$ $\frac{5 x (1 – x)}{(x – 1) }$

= $lim_{x -> 1}$ (-5 x )
= - 5
It is not necessary that we use limits for only difficult functions. They can be used in simple functions too. Also at times we use limits sideways like a+ or a- that is right side of the point or left side of the point. This happens mostly in case of discontinuous functions. But the concept of limits solved a great term $\frac{1}{x}$ when x approaches to zero. With the help of limits we can say that as x gets closer to zero, $\frac{1}{x}$ approaches to zero.

## Properties

Few important properties on limits are as follow:

1) $Lim_(x -> b)$ [m h (x)] = m $lim_(x -> b)$ h (x)

2) $Lim_(x -> b)$ [g (x) $\pm$ h (x)] = $lim_(x -> b)$ g (x) $\pm$ lim_(x -> b) h (x)

3) $Lim_(x -> b)$ [f (x) . h (x)] = $lim_(x -> b)$ f (x) . $lim_(x -> b)$ h (x)

4) $Lim_(x -> b)$ [$\frac{f (x)}{h (x)}$] = $\frac{lim_{x -> b}f (x)}{lim_{x -> b} h (x)}$

only if lim_(x -> b) h (x) $\neq$ 0

5) $Lim_(x -> b)$ $[h (x)]^n$ = $[lim_(x -> b) h (x)]^n$ ;(where n $\epsilon$ R)

6) $Lim_(x -> b) m = m$ ; where m is a constant

7) $Lim_(x -> b) x^n = b^n$

## Examples

Couple of examples on limits are given below:

Example 1: Find $lim_(x -> 1)$ [$\frac{(x^3 – 1)}{(x – 1)}$]

Solution: Here when x = 1, then the function cannot be defined as 1 – 1= 0 which cannot be possible according to the definition of rational expressions. Thus we use limits to find the value of the given function as x approaches to 1.

$lim_(x -> 1)$ [$\frac{(x^3 – 1)}{(x – 1)}$]

= $lim_(x -> 1)$ [$\frac{(x – 1) (x^2 + x + 1)}{(x – 1)}$]

= $lim_(x -> 1)$ ($x^2$ + x + 1)

= 1 + 1 + 1

= 3

Example 2: Evaluate $lim_{y \rightarrow 9}$ $\frac{3-\sqrt{y}}{9-y}$

Solution: Put y= 9 in $\frac{3-\sqrt{y}}{9-y}$, we get

$\frac{3-3}{9-9}$ = $\frac{0}{0}$ form

Reduced expression to avoid 0/0 form

Divide and multiply fraction by 3 + $\sqrt{y}$

$\frac{3-\sqrt{y}}{9-y}$ $\times$ $\frac{3+\sqrt{y}}{3+\sqrt{y}}$

= $\frac{(3^2-(\sqrt{y})^2}{(9-y)(3+\sqrt{y})}$

= $\frac{9-y}{(9-y)(3+\sqrt{y})}$

= $\frac{1}{(3+\sqrt{y})}$

By substituting y = 9 in above, we have

= $\frac{1}{(3+\sqrt{9})}$

= $\frac{1}{6}$

=> $lim_{y \rightarrow 9}$ $\frac{3-\sqrt{y}}{9-y}$ = $\frac{1}{6}$