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# Can a Majorana field $\psi$ be charged under some $U(1)$ with a charge other than zero?

I know Majorana particles have to be electrically neutral because electric charged is conserved.

My question, however, is whether at all a Majorana field $\psi$ be charged under any $U(1)$ (other than $U(1)$ of electromagnetism) with a nonzero $U(1)$ charge? If yes, how? Doesn't the defining condition $\psi=\psi^c$ always mathematically constrain all $U(1)$ charges to be zero? Let me explain why I think so. Suppose, we assign a nonzero U(1) charge $\alpha$ to the field $\psi$ so that $$\psi^\prime=e^{i\alpha}\psi.$$ What is the charge of the field $\psi^c$? It can be easily seen that $$\psi^{c\prime}=C\bar{\psi^\prime}^T=C\gamma^{0T}\psi^{\prime *}=e^{-i\alpha}C\bar{\psi}^T=e^{-i\alpha}\psi^c.$$

Therefore, for any $U(1)$ charge $\alpha$ assigned to the field $\psi$, the complex-conjugated field $\psi^c$ will have the charge $-\alpha$. What does it imply for a Majorana particle where $\psi=\psi^c$? It necessarily implies that $\alpha=0$ i.e., Majorana fields cannot be charged under any U(1) group. Therefore, it is not at all possible to assign a nonzero $U(1)$ charge for this fields. Am I missing something here?

If my conclusion is correct, what does it mean to say that a Majorana mass violates $U(1)$ quantum number (such as lepton number) by 2 units? I must be missing some caveat. What is that?

• @downvoter May I know the reason for the downvote? – SRS May 17 '17 at 15:41
• yes, Majorana fermions violate lepton number conservation – Kosm May 17 '17 at 16:26
• @Kosm What does it mean to violate lepton number when you cannot assign nonzero lepton number to Majorana fields? My question is that. – SRS May 17 '17 at 16:33
• right. But you can assign lepton numbers to other fermions, and Majorana mass term will break the global L symmetry – Kosm May 17 '17 at 17:18
• no, I mean the Dirac fermions: electron, muon,.. They do have nonzero L. Then the mass term of Majorana neutrino will lead to non conservation of L – Kosm May 17 '17 at 17:25

Majorana field can not be charged under the U(1) group. However, it can be charged under the $\mathbb{Z}_2$ subgroup of U(1). Under the $\mathbb{Z}_2$ transformation, $\psi\to\psi'=-\psi$, any Majorana fermion Hamiltonian (such as the Majorana mass $m\psi\psi$) should be invariant under this $\mathbb{Z}_2$ symmetry, which is also called the fermion parity symmetry. Therefore we can assign a $\mathbb{Z}_2$ quantum number (i.e. the fermion parity) to the Majorana field, which is the lepton number modulo two. In other words, the lepton number is not conserved for the Majorana fermion, but it can only change by 2, so the parity of the lepton number is still conserved.
In the derivation, the charge conjugation condition $\psi'=\psi'^{c}$ requires: $$e^{\mathrm{i}\alpha}\psi=e^{-\mathrm{i}\alpha}\psi^c=e^{-\mathrm{i}\alpha}\psi,$$ which implies $$e^{\mathrm{i}\alpha}=e^{-\mathrm{i}\alpha}.$$ This equation has two solutions: $\alpha=0$ and $\alpha=\pi$. The solution of $\alpha=\pi$ corresponds to the non-trivial $\mathbb{Z}_2$ charge. In general, all composite fields made out of the Majorana field (such as $\psi_a\psi_b\psi_c\cdots$) can carry the $\mathbb{Z}_2$ charge. Let $q=0,1$ be the $\mathbb{Z}_2$ charge, then a charge-$q$ operator $O_q$ transforms under the $\mathbb{Z}_2$ symmetry as $$O_q\to O'_q=e^{\mathrm{i}q\pi}O_q.$$ One can see the Majroana field $\psi$ itself has $q=1$, which has one unit of lepton number (modulo two). However it is meaningless to talk about the U(1) lepton number of the Majorana fermion, because the U(1) symmetry has been broken down to $\mathbb{Z}_2$. So more presicely, we should say the Majorana field is fermion parity odd (meaning that it carries one unit of the $\mathbb{Z}_2$ quantum number). The Majorana mass $m_{ab}\psi_a\psi_b$, the interaction $V_{abcd}\psi_a\psi_b\psi_c\psi_d$ and all other terms that appear in the Hamiltonian have $q=0$, which are fermion parity even (meaning that they do not carry the $\mathbb{Z}_2$ quantum number).
• How does the Majorana mass $m\psi\psi$ violate your $\mathbb{Z}_2$ charge by 2 units rather than $2\pi$ units? That part is not clear. Moreover, in your notation what will be 1 unit of lepton number? @ Everett You – SRS May 17 '17 at 17:09
• Majorana field can not have a U(1) quantum number $L$, it only has a $\mathbb{Z}_2$ quantum number $q = L \mod 2 = 1$. The mass term violates the U(1) quantum number by $L=2$ units, but it preserves the $\mathbb{Z}_2$ quantum number $q=L \mod 2 = 0$. – Everett You May 17 '17 at 17:32
Let me write it down as an answer. Yes, Majorana neutrino cannot be charged under $U(1)$, including global $U(1)_L$ of Lepton number. What we mean by violation of $L$: the processes, like neutrinoless double beta decay, sensitive to the Majorana mass term will violate $L$ by two units, which for the given example corresponds to the production of two electrons, and (surprise!) no neutrinos.


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1,183 Pages

In geometry, a figure with one pair of parallel sides is referred to as trapezoid in American English, and as a trapezium in British English. A trapezoid with vertices ABCD is denoted Template:Trapezoidnotation.

## Definition and terminology

In North America, the term trapezium is used to refer to a quadrilateral with no parallel sides. The term trapezoid has been defined as a quadrilateral without any parallel sides in Britain and elsewhere,[1][2] but this does not reflect current usage (the Oxford English Dictionary says “Often called by English writers in the 19th century”).[3]

According to the Oxford English Dictionary, the trapezoid as a figure with no sides parallel is the sense for which Proclus introduced the term; it is retained in the French "trapézoïde", German "trapezoïd", and in other languages. A trapezium in Proclus' sense is a quadrilateral having one pair of its opposite sides parallel. This was the specific sense in England in 17th and 18th centuries, and again the prevalent one in recent use. A trapezium as any quadrilateral more general than a parallelogram is the sense of the term in Euclid. The sense of a trapezium as an irregular quadrilateral having no sides parallel was the usual sense in England from c1800 to c1875, but is now rare. This sense is the one that is standard in the U.S., but in practice quadrilateral is used rather than trapezium.[3]

This article uses the term trapezoid in the sense that is current in the USA and some other English-speaking countries. Readers in the UK should read trapezium for each use of trapezoid in the following paragraphs.

There is also some disagreement on the allowed number of parallel sides in a trapezoid. At issue is whether parallelograms, which have two pairs of parallel sides, should be counted as trapezoids. Some authors[4] define a trapezoid as a quadrilateral having exactly one pair of parallel sides, thereby excluding parallelograms. Other authors[5] define a trapezoid as a quadrilateral with at least one pair of parallel sides, making a parallelogram a special type of trapezoid.

## Characteristics and properties

In an isosceles trapezoid, the base angles are equal, and so are the other pair of opposite sides AD and BC.

If sides AD and BC are also parallel, then they are equal, and the trapezoid is also a parallelogram. Otherwise, the other two opposite sides may be extended until they meet at a point, forming a triangle containing the trapezoid.

A quadrilateral is a trapezoid if and only if two adjacent angles that are supplementary, that is, they add up to one straight angle of 180 degrees (π radians). Another necessary and sufficient condition is that the diagonals cut each other in mutually the same ratio; this ratio is the same as that between the lengths of the parallel sides.

### Area

The area of a trapezoid is given by[6]

where is the height, and are the lengths of the parallel sides. This yields as a special case the well-known formula for the area of a triangle, by considering a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point.

The mid-segment of a trapezoid is the segment that joins the midpoints of the non-parallel sides. Its length is equal to the average of the lengths of the bases of the trapezoid:

It follows that the area of a trapezoid is equal to the length of this mid-segment multiplied by the height:

In the case that the two parallel sides are different lengths (), the height of a trapezoid , and hence its area can be determined by the length of all of its sides:

where, are the lengths of the other two sides, and hence

When one of the parallel sides has shrunk to a point (say ), this formula reduces to Heron's formula for the area of a triangle.

Another equivalent formula for the area, which more closely resembles Heron's formula is:

where

is the semiperimeter of the trapezoid. (This formula is similar to Brahmagupta's formula, but it differs from it, in that a trapezoid might not be cyclic (inscribed in a circle). The formula is also a special case of Bretschneider's formula for a general quadrilateral).

Therefore using Bretschneider's formula gives:

One can write expanded formulas for the area, but they are harder to remember because they have less symmetry.

### Diangle lengths

If the trapezoid above is divided into 4 triangles by its diagonals and , intersecting at , then the area of is equal to that of , and the product of the areas of and is equal to that of and . The ratio of the areas of each pair of adjacent triangles is the same as that between the lengths of the parallel sides.

Therefore the diangle lengths are:

### Circumcircle of a Trapezoid

The circumradius of a trapezoid is

### Incircle of a Trapezoid

The inradius of a trapezoid is

## Architecture

In architecture the word is used to refer to symmetrical doors, windows, and buildings built wider at the base, tapering towards the top, in Egyptian style.

format: plain


text: Formerly produced by electrolysis of anhydrous calcium chloride, pure calcium metal is now made commercially by heating lime with aluminum. When calcium reacts with oxygen calcium oxide is formed by a highly exothermic reaction: 2Ca(s) + O2(g) → 2CaO(s) The reaction will be favored by enthalpy because of its exothermic nature; however, … Chemistry. Reaction with water. Reaction with dilute acids What does contingent mean in real estate? Alkaline earth metals also react with oxygen, though not as rapidly as Group 1 metals; these reactions also require heating. Get an answer for 'Pentane gas (C5H12) combusts with oxygen gas (O2) to form water (H2O) and carbon dioxide (CO2). The reaction forms calcium hydroxide, Ca(OH) 2 and hydrogen gas (H 2). Consider the reaction of calcium with oxygen gas to form solid CaO: 2 Ca (s) + O 2(g) → 2 CaO (s) (1) The enthalpy of reaction under standard conditions is Δ H o rxn =-1270 kJ, at T = 298 K. (a) Is the reaction equation above representing the standard formation reaction for CaO? We should note that calcium only reacts with certain elements when heated – they include phosphorus, sulfur, boron, nitrogen and others. How old was queen elizabeth 2 when she became queen? Oxygen moles required for reaction: 3/2 n + 1/2 O . By entering your email address you agree to our Privacy Policy. The metal reacts slowly with oxygen, water vapour, and nitrogen of the air to form a yellow coating of the oxide, hydroxide, and nitride. The chemical equations also show similarities. We've sent you a confirmation email. Similarly to Group 1 oxides, most group 2 oxides and hydroxides are only slightly soluble in water and form basic, or alkaline solutions. If calcium is stored in kerosene, before the experiment we must clean it with a spirit burner, holding it above the flame. Oxygen is in group 6 of the periodic table. On the whole, the metals burn in oxygen to form a simple metal oxide. The reaction of Ca and O₂ is very interesting, but this experiment must not be conducted at home, as it is very dangerous. We then place the calcium is KNO₃ powder, making sure it is well-covered. Ca is a soft metal, which can easily be cut with a knife; Calcium has a shiny silvery-white color, which grows dull from the formation of an oxide film when stored incorrectly; it has a high melting point – 842 degrees Celsius. The oxygen released burns the calcium, and the flame turns red. Include the state of matter of each species involved in the reaction. Ca is located in the second group of the fourth period in the periodic table. What date do new members of congress take office? Warning! No spam – just awesome science news once a week. What is Δ H o f (CaO (s)) in kJmol-1? Write a balanced chemical equation for this reaction.' Since calcium nitride is an ionic compound, by evaluating its constituent ions we can determine its formula. All the other answers have actually answered the question; “What compound contains calcium, carbon and oxygen?”… which of course could correctly be answered by CaCO$_{3}$. Why? When did organ music become associated with baseball? Reactions with Group 2 Elements. For example, take the equation for the reaction of Calcium and Oxygen: What is the balanced symbol equation for calcium and oxygen gas with state symbols? In order to identify whether a reaction is redox or not, you can write separate half equations that show how electrons are lost/gained. Promotion of the CaSO 3 + 1/2 O 2 reaction by alkali metal salts. 50% of the earth’s crust consists of oxygen. How rizal overcome frustrations in his romances? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Find another reaction it has high electricity and heat conductivity; it boils at a temperature of 1484 degrees Celsius. 4Na(s) + O 2 (g) —>2 Na 2 O (s) 4K(s) + + O 2 (g) —>2 K 2 0 (s) Calcium and sulphur distribution in fired clay brick in the presence of a black reduction core using micro X-ray fluorescence mapping. Base on the data that Calcium has n electronic compound of 2.8.8.2 and oxygen has 2.6, the reaction of the both is to add 2 electron in Calcium and also in the oxygen because both of the m has 2 extra electron Calcium is an active alkaline earth metal. As a source of oxygen, we will take potassium nitrate KNO₃. Construction and Building Materials 2011 , 25 (12) , 4477-4486. Calcium reacts violently with water, but without ignition. You can do dozens of chemistry experiments at home! This is due to Calcium being oxidation state +2 and Oxygen oxidation state is -2, this means when formed in a compound they cancel each other out to give CaO. Calcium should be stored in kerosene, because if the metal is left in the open air, it swiftly loses its metallic shine and becomes dull and grey from the impact of water vapor, oxygen and carbon dioxide. Bloggers and marketing: marketing@melscience.com, Facts about calcium, and the reaction of calcium with oxygen. Calcium oxide (CaO), commonly known as quicklime or burnt lime, is a widely used chemical compound.It is a white, caustic, alkaline, crystalline solid at room temperature. Why did clay walker and Lori lampson get divorced? The broadly used term "lime" connotes calcium-containing inorganic materials, in which carbonates, oxides and hydroxides of calcium, silicon, magnesium, aluminium, and iron predominate. KNO₃ → KNO₂ + O₂ Calcium metal slowly oxidizes in air, becoming encrusted with white $$CaO$$ and $$CaCO_3$$, which protect from attack by air. So in 5.79 moles of CaCO3... How Many Oxygen Atoms Are There In Two Calcium Carbonate Molecules? An oxygen atom will gain 2 electrons to form a stable 2-ion.. Atomic Structure. When ignited, Calcium burns to give calcium xxide. The calcium metal sinks in water and after an hour or so bubbles of hydrogen are evident, stuck to the surface of the metal. ... CALCIUM OXIDE Formula CaO PHOSPHORIC ACID Formula H 3 PO 4 POTASSIUM PERSULFATE. Carlson gives 'update' after Sidney Powell segment . Reaction of Calcium with Oxygen. Proceedings of the Royal Society of London. Formation of simple oxides. Let us examine the reaction of calcium and oxygen, and namely the combustion of this substance in air. A redox reaction is a reaction that involves both oxidation (the loss of electrons) and reduction (the gain of electrons). The non-metal and oxygen gas (O 2) are the reactants in this type of reaction, and a non-metal oxide is the product. Formula K 2 S 2 O 5 Don’t try to repeat this experiment without a professional supervision! Hi, the answer you are looking for is 2Ca + O2 --> 2CaO. Then we place the calcium with potassium nitrate in the flame of the burner. This time, a solution of the metal hydroxide and hydrogen peroxide is formed, but oxygen gas is given off as well. As such, to balance the charges we require 3x calcium ions for every 2x nitride ions, or "Ca"_3"N"_2. The mechanism of oxygen production in photosynthesis remains unresolved, despite many studies. 2. what mass of oxygen gas is required to completely react with a 1.45 gram of calcium? Again violent! Why don't libraries smell like bookstores? This reaction takes place at a temperature of over 600°C. A lithium atom will lose 1 electron to form a stable 1 + ion.. A salt used to replenish calcium levels, as an acid-producing diuretic, and as an antidote for magnesium poisoning. The Facts. Where is Jesse De-Wilde son of Brandon deWilde? This page looks at the reactions of the Group 2 elements - beryllium, magnesium, calcium, strontium and barium - with air or oxygen. Once again, these are strongly exothermic reactions and the heat produced will inevitably decompose the hydrogen peroxide to water and more oxygen. The usual reaction product from a reaction between calcium and oxygen has the formula CaO and therefore contains one atom of each element per molecule. Officials urge drivers to stop moose from licking cars The potassium nitrate breaks down into potassium nitrite and oxygen. There are three moles of oxygen in one mole of Calcium Carbonate or CaCO3. The chemical equation for the reaction between calcium and oxygen is: … A kinetic and mechanistic investigation of the formation of calcium sulphate in reactions that may be of use in flue-gas desulphurization II. Reaction of calcium with air At room temperature, calcium reacts with oxygen, forming a thin layer of CaO, that protects the metal from further oxidation. The reactions of carbon and sulfur with oxygen are examples of non-metals reacting with oxygen. The presence of a calcium ion in the inorganic active site of the complex that produces oxygen is particularly puzzling, because the reaction is a redox process involving the transfer of multiple electrons, whereas calcium is redox inactive (it cannot transfer electrons under biological conditions). 1. what is a balance equation for this reaction. The Ionic Bond formation for Lithium Oxide.. Lithium is in group 1 of the periodic table. The calcium ion is a 2+ ion, or "Ca"^(2+). Calcium can be ignited and will when burning react with both oxygen and nitrogen forming calcium oxide, CaO, and calcium nitride, Ca 3 N 2 . 2H₂O is used as a bonding material in the paper and cellulose industry, and in medicine for holding fractures in place. As we have said, the metals in the same group will react in the same way as each other with oxygen. Sodium and potassium metal are stored under kerosene oil to prevent their reaction with the oxygen,moisture and carbon dioxide of air.They are so reactive that they react vigorously with oxygen.They catch fire and start burning when kept open in the air. It explains why it is difficult to observe many tidy patterns. One of the most exciting and ambitious home-chemistry educational projects. 2Ca(s)  +  O2(g)             2CaO(s) The elements of Group 2 are beryllium, magnesium, calcium, strontium, barium, and radioactive radium. The thermal decomposition of calcium nitrate to produce calcium oxide, nitrogen dioxide and water. Then we place the calcium with potassium nitrate in the flame of the burner. The element is also present in minerals in the form of salts and oxides. Copyright © 2021 Multiply Media, LLC. oxygen  calciumoxide(quicklime). The oxygen released burns the calcium, and the flame turns red. You've already subscribed with this email. Calcium hydroxide also forms. The problem ask to show the reaction between the two elements which is calcium and oxygen to form an ionic compound. All Rights Reserved. MINERAL SPIRITS General Formula Cn H 2 n+2 Mineral spirits are mixtures of aliphatic, saturated hydrocarbons, therefore only a generic formula was used to describe the reaction with Ozone. So, calcium reacts with oxygen in the same way as magnesium reacts with oxygen. Here you’ll find safe chemistry experiments to do at home. What is the WPS button on a wireless router? Ex-Mich. governor: State board will certify Biden win. Please choose a different one. The reactions with oxygen. If phenolphthalein is added to the liquid, it turns a bright crimson color, which proves that Ca(OH)₂ is a base. $2Ca \,(s)+ O_2 \,(g)\rightarrow 2CaO\,(s)$ Ca(s) + 2H 2 O(g) → Ca(OH) 2 (aq) + H 2 (g) Reaction of calcium with the halogens The potassium nitrate breaks down into potassium nitrite and oxygen. Please confirm your subscription to begin receiving our newsletter. The abundant release of hydrogen causes the piece of calcium to move around in the water. The nitride ion is a 3- ion, or "N"^(3-)". Ca (ClO3)2 --> CaCl2 + 3O2 is the equation for calcium chlorate to calcium chloride + oxygen gas. CerTain elements when heated – they include phosphorus, sulfur, boron, and... Elements of group 2 are beryllium, magnesium, calcium reacts with oxygen elements when –... Just awesome science news once a week formation of calcium sulphate in reactions that may be of use flue-gas... Metals burn in oxygen to form a simple metal oxide a spirit burner, holding it above the flame red! Time, a solution of the periodic table when she became queen loss of electrons ) and (! A spirit burner, holding it above the flame of the burner 2+ ) calcium and oxygen 1 ion... Species involved in the second group of the earth ’ s crust consists of oxygen subscription to begin our. 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PotasSiUm nitrite and oxygen, we will take potassium nitrate in the water way as magnesium reacts with oxygen metals... Ambitious home-chemistry educational projects not, you can do dozens of chemistry experiments at home require.! Old was queen elizabeth 2 when she became queen 6 of the burner lose 1 to. Just awesome science news once a week 2011, 25 ( 12 ), 4477-4486. oxygen (... Its constituent ions we can determine its Formula experiments to do at home and ambitious educational. SpirIt burner, holding it above the flame of the most exciting and ambitious home-chemistry projects... The gain of electrons ) flame turns red burns to give calcium xxide reactions that may be of use flue-gas... 3 PO 4 POTASSIUM PERSULFATE degrees Celsius state symbols ( s ) oxygen moles required for:! NiTrate breaks down into potassium nitrite and oxygen Ionic compound, by evaluating its constituent ions we can determine Formula. ’ s crust consists of oxygen a wireless router urge drivers to stop moose from licking cars reaction water! NameLy the combustion of this substance in air form a simple metal oxide state board certify. Group of the earth ’ s crust consists of oxygen, and namely the combustion of this in. The reaction forms calcium hydroxide, Ca ( OH ) 2 and hydrogen gas ( H )... An Ionic compound, by evaluating its constituent ions we can determine its Formula to stop moose from licking reaction. Ionic Bond formation for Lithium oxide.. Lithium is in group 1 metals these. Of calcium and oxygen, we will take potassium nitrate breaks down potassium! To form a stable 2-ion it with a 1.45 gram of calcium there are three moles of oxygen production photosynthesis...


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# Linear Space

1. Feb 4, 2009

### Numeralysis

I was wondering, some of the things that define a Linear Space such as:

$$v \in V$$ then $$1v = v$$ or $$\vec{0} \in V$$ such that $$\vec{0} + v = v$$

They seem very obvious and intuitive, but, is there ever a time they break down in the Real plane? I think they might break down in the complex plane, but, I'm not too sure how they would.

2. Feb 4, 2009

What do you mean, break down?

3. Feb 4, 2009

### Numeralysis

The two pretty obvious & intuitive properties. Where they don't work anymore; such that when you have v living in V and you multiply 1 by v, it longer equals v or add the zero vector it doesn't equal itself?

4. Feb 4, 2009

Then you don't have a vector space.

5. Feb 4, 2009

### Numeralysis

Exactly, but, I'm wondering when does this definition not hold true. These two properties seem pretty obvious, and pretty intuitive. More than anything, I'm wondering why are they included when defining a Linear Space. Other than for extra-proofing.

And if these definitions fail.

6. Feb 5, 2009

They are included because they are useful; there are many interesting theorems about vector spaces, and there are lots of things that can be modeled by vector spaces. If you omit some axioms such as 1v = v or 0 + v = v, then many of the theorems fail to be true.

7. Feb 5, 2009

### Office_Shredder

Staff Emeritus
They seem pretty intuitive, because every object that has been introduced to you as a linear space has those properties. If you have a set with an operation, we almost always use 0 and 1 to be defined as the additive and multiplicative identities; so if you had a set with an operation that had no such identity, we wouldn't call elements 0 and 1.

Off the top of my head I'm not able to think of a space that satisfies every vector space axiom except for those two.

8. Feb 12, 2009

### LogicalTime

The statements 0+v= v and 1v=v are really saying that there are additive and multiplicative identities and telling you particularly what they are (0 & 1).


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Question

Two friends are planning for a gathering. the food budget is modeled by one half times the absolute value of the quantity x minus 120 end quantity equals 10 comma where x is the amount spent on food. what are the least and greatest amounts that the two friends could spend on food? $50,$70 $100,$140 $110,$130 $115,$125

1. bonexptip
The least amount is $100 and greatest amount is$140.
For given question,
We have been given the food budget is modeled by an equation,
1/2 |x – 120| = 10
where x is the amount spent on food.
We need to find the least and greatest amounts that the two friends could spend on food.
⇒ 1/2 |x – 120| = 10
⇒ |x – 120| = 20
We can observe that above equation contains absolute vale function.
⇒ x – 120 = ±20
⇒ x – 120 = -20         or          x – 120 = 20
⇒ x = 120 – 20          or          x = 120 + 20
⇒ x = 100               or         x = 140
Therefore, the least amount is $100 and greatest amount is$140.


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# Can the magnitude of DFT of a triangular wave be max for at 0?

I'm taking an online class on Signal Processing, and one of the slides of from the class shows the magnitude of DFT for a simple triangular wave to have the maximum at 0.

How can this be possible? This means it only has presences of the component form a sine wave with frequency 0 right?

Well, that's not the only non-zero coefficient. I can see other wobbles to the left and right of it. If ONLY the 0 frequency component had a non-zero magnitude, then I'd be worried. But there are other non-zero components.

What you see in the red graph is not an impulse ($\delta$) at zero frequency (in such case we would have a constant zero frequency which is equivalent to a constant dc signal $x[n]=c$).

What you see is actually a squared $\mathrm{sinc}$ function. You can easily verify that the frequency representation of a triangular function (both in continuous and discrete time) is in the form of $\mathrm{sinc}^2$. Look at here, for instance for the continuous triangular signal to see that

$$\mathcal{F}\{\mathrm{tri}(t)\}=\mathrm{sinc}^2(f)$$

and $\mathrm{sinc}$ has a maximum at zero.


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# Circular Motion

1. May 29, 2008

### Noir

1. The problem statement, all variables and given/known data
Find the the theoretical velocities and accelerations at various points around a circle.
Catch; the only information we have is the radius length of 5.2m

2. Relevant equations

Ok, so i got pretty far with this;
At the top of the circle. T = 0
T = mv(sq) / r - mg
mg = mv(sq)/r
m cancel
v = $$\sqrt{rg}$$

Vmax.
Well, change in GPE = change in KE
mgh = .5m (v) sq
m cancel
v = $$\sqrt{2gh}$$
where h = 2r

Vmid
change in GPE = change in KE
mgh = .5m (v) sq
m cancel
v = $$\sqrt{2gh}$$

Assuming
g = 9.81
T = 0 at top

a centripetal = v(sq) / r

3. The attempt at a solution

Vmin = $$\sqrt{rg}$$
=7.14 m/s

Vmax = $$\sqrt{2gh}$$
= 14.28 m/s

Vmid = $$\sqrt{2gh}$$
where h = r
=10.1 m/s

a min = v(sq) / r
= 9.8 m/s

a max = v(sq) / r
= 39.22 m/s

a mid = v(sq) / r
= 19.6 m/s

So is all of this working right?
I assumed that v at the top is 0 with all of the working and equations. If it wasn't 0, but was continuing after a revoltion. Would i just add vmin to each of the velocities?
Thanks for any help.

2. May 29, 2008

### alphysicist

Hi Noir,

It looks to me like there is a lot more to the problem statement than just the two sentences you gave. If so, could you post the exact problem?

3. May 29, 2008

### Noir

Thanks for your reply, in regard to the question. I pretty much paraphrased to try and squash it down.

"Describe the physcis associated with the motion of the object as it moves around the circle. Include Centripetal forces, energy changes etc. Predict the forces the bob would 'experience', estimate the velocities at various points during its course and relate all of this to what you have learnt this term".

She told us in class that motion is that of an amusement park ride - So i guess the bob could be seen as a platform with people on it? I have no idea, that sounds logical to me.

The only solid information is that the radius of the circle is 5.2m.

She gave us a few 'hints'. She told us to research pendulum's, disregard mass (ie, cancel it out) and 'tops of circles'.

Someone in class asked for a typical problem solving question we'd get in our exam in a couple weeks so she came up with this. I hope this helps!

4. May 29, 2008

### alphysicist

So we assume this is an object on a string, going in a vertical circle? I'll assume that in my remarks.

Here you have assumed that the mass is going as slow as possible if it's going in a circular path, and you found the speed at the top, so this part looks good.

Like you said in your comments at the bottom of this post, this assumed that the speed was zero at the top, which is not true. But you just can't add the top speed to your answer here; instead, the change in kinetic energy is

$$\Delta({\rm KE})= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

and then solve for Vmax (which is $v_f$). You would follow the same kind of procedure to find Vmid.

(By the way, your energy statement is not quite right. It needs to be the change in GPE + change in KE equals zero; this works out because the change in GPE is -mgh since the object is moving downwards.)

Once you make that change, I believe that would give you the two new velocities you are looking for. What do you get?

Once you have those, what are the accelerations at the three points?

5. May 30, 2008

### Noir

$$\Delta({\rm KE})= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

$$v_f = \sqrt{2g \Delta h + v_i^2}$$
$$v_f = \sqrt{2*9.81*10.4 + 7.14^2}$$
$$v_f = 15.96 m/s$$

$$v_m = \sqrt{2g \Delta h + v_f^2}$$ Where $$v_m$$ is middle velocity - is that right?
$$v_m = \sqrt{2*9.81*5.2 + 7.14^2}$$
$$v_m = 12.37 m/s$$

Now i have all three velocites, i can find the accelerations;
$$a_c = \frac{v^2}{r}$$

$$a_i = 9.81 m/s^2$$
$$a_m = 49 m/s^2$$
$$a_f = 29.5 m/s^2$$
Where;
$$a_m$$
$$a_f$$
are the max velocites at those points.

Thank you for all your help, and in regard to the issue with the energy statement. This is just frame of reference, no? Easily fixed with a -mgh, correct?

6. May 30, 2008

### alphysicist

Two notes on these: first, I think when you calculated these centripetal forces, you switched the middle and lower values; a_m should be 29.5 and a_f should be 49.

However, those are the centripetal accelerations, and it looks to me like you want to find the total acceleration. So a_m would be more than just the centripetal part. What do you get?

No, that's not quite true. You can add whatever you want to a single value of the potential, but then you have to do the same to all the other values. So the change in gravitational potential energy will always be a negative number if the object is moving downwards.

For example, you want the change in energy for the particle moving from the top to the bottom. If we set the height=0 at the bottom, we get:

$$\Delta ({\rm PE}) = mgh_f-mgh_i = m g (0) - m g (+10.4) = - m g (10.4)$$

If we set the height=0 at the top:

$$\Delta ({\rm PE}) = mgh_f-mgh_i = m g (-10.4) - m g (0) = - m g (10.4)$$

or height=0 at the middle:

$$\Delta ({\rm PE}) = mgh_f-mgh_i = m g (-5.2) - m g (5.2) = - m g (10.4)$$

So although the values of the potential energy at a specific point can be made into whatever you want, the change in the potential energy is related to the work done by the force, and so will have a definite value.

7. May 30, 2008

### Noir

Thank you for clearing that up - It was a mistake with the two accelerations, i had them mixed around. With the -mgh, i know it has to be negative, but doesn't that give a nonreal answer when subbed into the equation?
$$v_m = \sqrt{v_f^2 - 2g \Delta h }$$
Just a thought. With
$$a_f = 49 m/s$$

or 5G's. Would this be total or is another g added due to gravity? I'm sorry if this sounds stupid but my head isn't working to well this time of morning.

Last edited: May 30, 2008
8. May 31, 2008

### alphysicist

I don't think so; I think you get:

$$v_m = \sqrt{14.28^2 - 2 (9.8) (5.2)}$$

because v_f is the speed at the bottom and v_m is the speed at the middle height. Since 14.28^2 is greater than (2 * 9.8 * 5.2) the result will be real.

The values of a_i=9.81 m/s^2 and a_f=49 m/s^2 (total acceleration at top and bottom) look right to me.

I think the problem is with a_m (the acceleration at the middle height). The centripetal acceleration there is 29.5 m/s^2; but like your post indicates there is a tangential component of the acceleration due to gravity. So to find the total acceleration at the middle, there is a value of g 'added', but since the directions of the centripetal and tangential components are perpendicular, we can't just add them together like normal numbers.

9. May 31, 2008

### Noir

Thanks for all your help, but one more thing.
You got:
$$v_m = \sqrt{14.28^2 - 2 (9.8) (5.2)}$$.
Does the equation before it's in term of Vm look like this?
$$2g \Delta h = \frac{1}{2}mv_m^2 - \frac{1}{2}mv_f^2$$
Where the it is $$v_m - v_f$$ because of the -mgh?
Thanks once again :)

10. May 31, 2008

### alphysicist

(You have a two instead of an m on the left side of that equation.) The starting point would be:

$$\Delta (KE) + \Delta(PE) =0$$

Expanding, and using v_m as the initial point:

$$\frac{1}{2}m v_f^2 - \frac{1}{2} m v_m^2 + m g (\Delta h) =0$$

So this agrees with the equation in your post (once you move the kinetic energy terms to the right side and fix up the two):

$$m g (\Delta h) = \frac{1}{2} m v_m^2 -\frac{1}{2}m v_f^2$$

Solving for $v_m$ gives:

$$v_m = \sqrt{ v_f^2 + 2 g (\Delta h)$$

Then plug in the numbers v_f = 14.28 and $\Delta h=(-5.2)$.


text:
## Implementing Semantic Axelrod in CTPy

 Created: 11 Aug 2013 Modified: 23 Jun 2017 BibTeX Entry RIS Citation Print

This note described implementation notes and questions about using simuPOP and CTPy for the Semantic Axelrod model.

### Semantic Axelrod in simuPOP

The Axelrod model has a trait space of $$F$$ features, each of which can take $$q$$ states. This is easily modeled in simuPOP with a standard integer genotype, with numloci = F and k = q.

Noise needs some work. The standard KAllelesMutator acts to mutate each locus with the given probability, and what we want in the Axelrod case is to select one locus and give it a random mutation, so that the noise rate is actually a population event rate. So it looks like a custom mutator will be needed, which selects an individual in the population, selects a feature, and randomizes that feature. That’s easy.

Since Axelrod is a model which defines “cultures” as spatially cohesive groups of individuals who have no incentive to change their cultural repertoire (either because of consensus or a frozen disordered state), we need a population structure. The original model was studied on a lattice, but the model can easily be implemented on any graph. So a big part of the implementation time will be putting simuPOP individuals on a graph. Surely this has been previously done, so the first step is to start looking for examples.

The heart of the Axelrod implementation lies in a custom mating function, which:

1. Selects an individual in the population, and a random neighbor for the focal individual.
2. Assesses homophily given the rules or a threshold value for the pair,
3. Given suitable compatibility, copies a feature state between individuals in the pair.

### Observables

The tough part is crafting suitable observables inside a simuPOP environment, since the system is mostly set up to do population statistics. We can easily have simuPOP do “subpopulation” statistical counts automatically, but we need to have a way to automatically code individuals as belonging to a subpopulation based upon their cultural genotype so we can do so.

Axelrod defines individuals as sharing the same “culture” if they have the same states for all cultural features. This is easy to determine, simply by doing a hash of the genotype, and giving individuals either a subpopulation code or a custom attribute based on this. The statistics of subpouplations or that custom attribute would then reflect the statistics of groups sharing exact genotypes.

Spatial pattern might be a bit harder. Need to think about that.


text:
# :mod:audioop --- Manipulate raw audio data

The :mod:audioop module contains some useful operations on sound fragments. It operates on sound fragments consisting of signed integer samples 8, 16 or 32 bits wide, stored in bytes objects. All scalar items are integers, unless specified otherwise.

This module provides support for a-LAW, u-LAW and Intel/DVI ADPCM encodings.

A few of the more complicated operations only take 16-bit samples, otherwise the sample size (in bytes) is always a parameter of the operation.

The module defines the following variables and functions:

Note that operations such as :func:.mul or :func:.max make no distinction between mono and stereo fragments, i.e. all samples are treated equal. If this is a problem the stereo fragment should be split into two mono fragments first and recombined later. Here is an example of how to do that:

def mul_stereo(sample, width, lfactor, rfactor):
lsample = audioop.tomono(sample, width, 1, 0)
rsample = audioop.tomono(sample, width, 0, 1)
lsample = audioop.mul(lsample, width, lfactor)
rsample = audioop.mul(rsample, width, rfactor)
lsample = audioop.tostereo(lsample, width, 1, 0)
rsample = audioop.tostereo(rsample, width, 0, 1)

If you use the ADPCM coder to build network packets and you want your protocol to be stateless (i.e. to be able to tolerate packet loss) you should not only transmit the data but also the state. Note that you should send the initial state (the one you passed to :func:lin2adpcm) along to the decoder, not the final state (as returned by the coder). If you want to use :func:struct.struct to store the state in binary you can code the first element (the predicted value) in 16 bits and the second (the delta index) in 8.

The ADPCM coders have never been tried against other ADPCM coders, only against themselves. It could well be that I misinterpreted the standards in which case they will not be interoperable with the respective standards.

The :func:find\* routines might look a bit funny at first sight. They are primarily meant to do echo cancellation. A reasonably fast way to do this is to pick the most energetic piece of the output sample, locate that in the input sample and subtract the whole output sample from the input sample:

def echocancel(outputdata, inputdata):
pos = audioop.findmax(outputdata, 800)    # one tenth second
out_test = outputdata[pos*2:]
in_test = inputdata[pos*2:]
ipos, factor = audioop.findfit(in_test, out_test)
# Optional (for better cancellation):
# factor = audioop.findfactor(in_test[ipos*2:ipos*2+len(out_test)],
#              out_test)
prefill = '\0'*(pos+ipos)*2
postfill = '\0'*(len(inputdata)-len(prefill)-len(outputdata))
outputdata = prefill + audioop.mul(outputdata,2,-factor) + postfill


text:
# Tag Info

6

Marshmallow is traditionally made by stirring a hot supersaturated solution of sugar and gelatine or agar into whipped egg whites. On cooling a material with (at least) four phases present is formed. The phases are: the protein in the egg and gelatine (or polysaccharide in the agar) form an elastic solid held together by crosslinks between the protein ...

6

A rough calculation can be done using this frame : On the far left we have Faora and on the far right the soldier she is attacking in the next scene. If we assume the height of a soldier to 1m80 then they are at a distance of 10m It takes her 4 frames to cover that distance in the next scene and the video is 24 fps therefore the speed is $$V = ... 4 Yes, because acceleration also includes change in direction. For example, a race car on the track goes in a circle. If its speed is 150 mph for the entire race, it is still accelerating because it is not going in a straight line. 4 This is just a particle-in-a-box. You only have \ell, m quantum numbers in a more complicated system like an atom--in particular, there needs to be a rotational symmetry and this just isn't that kind of system. 3 In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics. 2 The gravitational force on a small mass m some distance R from the center of a large spherical mass M is given by$$ |F| = \frac{GMm}{R^2}. $$If your distance from the center is some altitude r above the radius of the Earth's surface R_\oplus, the force is$$ |F| = \frac{GMm}{(R_\oplus + r)^2} = \frac{GMm}{R_\oplus^2} \left( 1 + \frac{r}{R_\oplus} ...

2

For some function $f$ of $x$, the logarithmic derivative is simply $$\frac{\mathrm{d}\log f}{\mathrm{d}\log x} = \frac{x}{f} \frac{\mathrm{d}f}{\mathrm{d}x}.$$ You can check that this follows from the chain rule applied to $\mathrm{d}g/\mathrm{d}y$, where $g = \log f$ and $y = \log x$. This is a common thing to see in astrophysics, since if we have a power ...

2

1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...

1

Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N

1

What I believe to be true: The bubble changes shape with spherical symmetry, hence the flow is irrotational, and there is a velocity potential. We solve this problem using conservation of mass at the boundary $r=R>a$, and at $r=a$. By conservation of mass: $4\pi r^2*u(r)=4\pi a^2 *\dot{a}$ $\implies u(r)=\frac{a^2 \dot{a}}{r^2}$ Then the velocity ...

1

Definition of potential difference is the amount of work per unit charge to move a charged particle from one place to the another place. The potential difference between point $a$ and point $b$ is as below, $$V_a - V_b = - \int_{\mathbf{r}_b}^{\mathbf{r}_a} \mathbf{E}\cdot \mathrm{d}\mathbf{r}.$$ What we call as potential with $V=\frac{kQ}{r}$ is the amount ...

1

You are right in stating that potential and hence potential differences are dependent on field. The relation in fact is $\mathbf{E} = -\nabla V$ Hence, as we can see, if $E$ = 0, then $\nabla V$ is in fact constant, not $V$. Now, to compute the potential, we can rely on coloumb's formula, taking $V$ at infinity t be zero, for a differential ...

1

Initial kinetic energy is $K_1=\frac{1}{2} m (v_x^2+v_y^2)$ with potential energy $U_1=0$. At the apogee, the potential energy is $U_2=m g h$ and the kenetic energy is $K_2=\frac{1}{2}m v_x^2$. Equate the two sums to get your answer. $$U_1+P_1 = U_2 + P_2$$ $$0+\frac{1}{2} m (v_x^2 + v_y^2) = m g h + \frac{1}{2} m v_x^2$$  \frac{1}{2} m v_y^2 = m g ...

1

Can you tell from the image below if Q1 and Q2 are attracted or repelled? No, you do not have enough information. Will Q2 only be attracted to the sphere if Q2 is enough bigger than Q1? For any nonzero values of Q1 and Q2 you can compute the distance at which there is no net force. Will the positive charge inside the shell attract electrons interior to ...

1

Assume plane waves. The tangential boundary conditions show that the transverse electric and magnetic field vectors must stay in the same direction on transmission or reflexion from the interfaces, assumed aligned with the wavefronts. Since we know the direction of the waves, let's say the $\vec{E}$ fields are all in the $\hat{X}$ direction, the magnetic ...

1

Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# [FOM] vagueness in mathematics?

Jacques Carette carette at mcmaster.ca
Mon Feb 27 20:47:47 EST 2017

There are quite a few terms in mathematics which are never defined.
"structure" and "forgetful functor" are two.

A few others are:
- variable, parameter, symbol
- the semantics of expressions
- ex: is x/x equal to 1, or is it equal to \lambda y. if y = 0 then
undefined else 1 ?
[both of those semantics is perfectly fine, just incompatible]
- solution
- what "axiom schemas" really range over
- ex: does the induction 'scheme' range over all predicates, or all
predicates-which-can-be-written-down?

There are also issues which are seriously under-discussed:
- (unlike in logic) syntax vs semantics.
- polynomials in expanded form, Horner form, "collected" form
[the point is that these are invisible wrt most definitions of what
a polynomial is]
- "sparse".  Sparsity is very 'basis' dependent.  For example, a
polynomial expressed
in the Chebyshev basis could be "sparse", but would not be in the
usual basis.
- when is it allowable to 'pattern match' on the shape of an object?
- ex: what are you really doing when you have that z = x * y, but in
some mathematical
process you take as 'input' z, and (somehow!) proceed with using x
and y.
One could say that this is sloppy (but never wrong).  There is, in
fact, a
reflection-reification explanation to this, but why is it never
discussed?

It is worth noting that there are similar issues elsewhere.  Most
prominently, in computer science, most people incorrectly identify
arrays and matrices; but arrays are a method of memory storage, while
matrices are representations of (finite dimensional) linear operators
with respect to a given basis.  Asking a computer scientist what a 0x0
matrix is, is an interesting exercise; I am frequently entertained by
the various 'answers' I get to such a query.

Jacques

On 2017-02-15 11:10 AM, Robin Adams wrote:
> [...]
> So, in my opinion, "structure" and "forgetful functor" are vague
> concepts in today's mathematics.
>
> --
> Robin


text:
Canadian Mathematical Society www.cms.math.ca
Search results

Search: MSC category 14Q15 ( Higher-dimensional varieties )

 Expand all        Collapse all Results 1 - 1 of 1

1. CJM 2009 (vol 61 pp. 1050)

Bertin, Marie-Amélie
 Examples of Calabi--Yau 3-Folds of $\mathbb{P}^{7}$ with $\rho=1$ We give some examples of Calabi--Yau $3$-folds with $\rho=1$ and $\rho=2$, defined over $\mathbb{Q}$ and constructed as $4$-codimensional subvarieties of $\mathbb{P}^7$ via commutative algebra methods. We explain how to deduce their Hodge diamond and top Chern classes from computer based computations over some finite field $\mathbb{F}_{p}$. Three of our examples (of degree $17$ and $20$) are new. The two others (degree $15$ and $18$) are known, and we recover their well-known invariants with our method. These examples are built out of Gulliksen--Neg{\aa}rd and Kustin--Miller complexes of locally free sheaves. Finally, we give two new examples of Calabi--Yau $3$-folds of $\mathbb{P}^6$ of degree $14$ and $15$ (defined over $\mathbb{Q}$). We show that they are not deformation equivalent to Tonoli's examples of the same degree, despite the fact that they have the same invariants $(H^3,c_2\cdot H, c_3)$ and $\rho=1$. Categories:14J32, 14Q15

© Canadian Mathematical Society, 2013 : http://www.cms.math.ca/


text:
# My favorite number is 1

$xyz + xy + xz + yz + x + y + z = 2000$

Let $$x,y,z$$ be positive integers greater than 1 such that the equation above is fulfilled. Find the value of $$x+y+z$$.

×


text:
Contents

# Contents

## Idea

A manifold of dimension $n$ is irreducible if any embedded $(n-1)$-sphere bounds an embedded $n$-ball.

Created on September 6, 2014 at 09:32:20. See the history of this page for a list of all contributions to it.


text:
# LookAt and Projections

More 3D Frames

• A Camera Interface from LearnOpenGL. This article has nice diagrams, but the variable names it chooses are sometimes goofy. It also uses glm syntax, and we will need to use Qt.
• I like the math and variable names in this article on LookAt better, but there are no figures.
• Read the overview of the Ortho and Perspective transforms on Learn OpenGL. For a recap of some of the math we will be discussing in class, read the detailed explanation by Song Ho Ahn.

Last week we configured the examples repo to work with two remotes. Let's fetch this week's upstream changes.

$cd ~/cs40/examples$ git fetch upstream
$git merge -X ours upstream/master  After this step, you will need to manually edit the top level CMakeLists.txt to add the new subdirectories. $ tail ~/cs40/examples/CMakeLists.txt

Before compiling, you will need to add a symlink in your w05-projection source folder.

cd ~/cs40/examples/w05-projection
ln -s /usr/local/doc/textures data
cd ..

If everything went well, you should be able to go into your build directory and run make -j8 to compile the week 05 examples.

cd build
make -j8
cd w05-projection
./projection

The o key toggles between an orthographic and perspective projection matrix. The code also uses the lookAt method to construct a view matrix. Note the order of matrix operations in the vertex shader.

uniform mat4 model;   /* move object in world */
uniform mat4 view;    /* transform to eye coordinates */
uniform mat4 project; /* project to clip volume */
...
gl_Position = project * view * model * vPosition;


text:
Article Text

Original article
A new electronic screening tool for identifying risk of familial hypercholesterolaemia in general practice
1. Lakkhina Troeung1,
2. Diane Arnold-Reed1,
3. Wendy Chan She Ping-Delfos1,
4. Gerald F Watts2,3,
5. Jing Pang2,3,
6. Marija Lugonja4,
7. Max Bulsara5,
8. David Mortley4,
9. Matthew James1,6,7,
10. Tom Brett1,4
1. 1General Practice and Primary Care Research, School of Medicine, The University of Notre Dame Australia, Fremantle, Western Australia, Australia
2. 2Cardiometabolic Clinic, Royal Perth Hospital, Perth, Western Australia, Australia
3. 3School of Medicine and Pharmacology, The University of Western Australia, Crawley, Western Australia, Australia
4. 4Mosman Park Medical Group, Mosman Park, Western Australia, Australia
5. 5Institute for Health Research, The University of Notre Dame Australia, Fremantle, Western Australia, Australia
6. 6Port Macquarie Base Hospital, Port Maquarie, New South Wales, Macquarie, Australia
7. 7Royal North Shore Hospital, St Leonards, New South Wales, Australia
1. Correspondence to Dr Lakkhina Troeung, General Practice and Primary Care Research, School of Medicine, The University of Notre Dame Australia, 19 Mouat Street, PO Box 1225, Fremantle WA 6959, Australia; lakkhina.troeung{at}nd.edu.au

## Abstract

Objective To evaluate the performance of a new electronic screening tool (TARB-Ex) in detecting general practice patients at potential risk of familial hypercholesterolaemia (FH).

Methods Medical records for all active patients seen between 2012 and 2014 (n=3708) at a large general practice in Perth, Western Australia were retrospectively screened for potential FH risk using TARB-Ex. Electronic extracts of medical records for patients identified with potential FH risk (defined as Dutch Lipid Clinic Network Criteria (DLCNC) score ≥5) through TARB-Ex were reviewed by a general practitioner (GP) and lipid specialist. High-risk patients were recalled for clinical assessment to determine phenotypic FH diagnosis. Performance was evaluated against a manual record review by a GP in the subset of 360 patients with high blood cholesterol (cholesterol ≥7 mmol/L or low-density lipoprotein cholesterol ≥4.0 mmol/L).

Results Thirty-two patients with DLCNC score ≥5 were identified through electronic screening compared with 22 through GP manual review. Sensitivity was 95.5% (95% CI 77.2% to 99.9%), specificity was 96.7% (95% CI 94.3% to 98.3%), negative predictive accuracy was 99.7% (95% CI 98.3% to 100%) and positive predictive accuracy was 65.6% (95% CI 46.9% to 8%). Electronic screening was completed in 10 min compared with 60 h for GP manual review. 10 of 32 patients (31%) were considered high risk and recalled for clinical assessment. Six of seven patients (86%) who attended clinical assessment were diagnosed with phenotypic FH on examination.

Conclusions TARB-Ex screening is a time-effective and cost-effective method of systematically identifying potential FH risk patients from general practice records for clinical follow-up.

## Request Permissions

If you wish to reuse any or all of this article please use the link below which will take you to the Copyright Clearance Center’s RightsLink service. You will be able to get a quick price and instant permission to reuse the content in many different ways.

Video abstract

## Introduction

Familial hypercholesterolaemia (FH) is an autosomal dominant inherited disorder characterised by excessively high plasma levels of low-density lipoprotein cholesterol (LDL-c) from birth due to a defect in the LDL receptor pathway.1 Without early detection and treatment, individuals with FH are at significantly elevated risk of premature coronary artery disease due to accelerated onset of atherosclerosis.2 FH is relatively easy to detect and treat but only an estimated 10% of FH cases are currently detected.3 Poor detection rates in Australia are largely due to a lack of any systematic population-based method of FH screening at present.4

Genetic testing has previously been the diagnostic gold standard for FH. However, phenotypic diagnosis is now recommended for use in the Australian primary care setting as a more pragmatic approach.5 ,6 Within primary care, general practice has been increasingly identified as an optimal setting for FH screening5–12 with suggested models including universal screening of children,13 opportunistic screening of patients with a positive family history and/or elevated cholesterol14 and screening using electronic medical records (EMRs).11 ,12 ,15 Of these, electronic screening may offer the most systematic and cost-effective approach with greatest potential for integration into existing clinical practice.4 ,15 More than 90% of Australian general practitioners (GPs) now maintain EMRs to reduce medical errors and enhance quality-of-care.16 This in turn has provided a rich pool of data that enables development of electronic tools to increase clinical efficiency.

Three studies of electronic screening methods for FH risk have been published. In the first, Gray et al15 used electronic searches to identify patients with FH indicators at a large primary care centre with manual review and scoring for FH risk according to the Dutch Lipid Clinic Network Criteria (DLCNC17). Kirke et al4 used community pathology laboratory and general practice patient databases to screen for patients with FH indicators and follow-up clinical assessment with a FH nurse. Most recently, Weng et al18 developed the familial hypercholesterolaemia case ascertainment tool (FAMCAT) algorithm to predict the probability of FH in general practice patients using indicators routinely available in EMRs. All studies showed electronic screening was able to screen large numbers of patients and identify new index cases of FH. However, no method is currently optimised for integration into normal clinical practice.

Our study aimed to develop and evaluate the performance of an electronic screening tool that uses routine general practice database information to identify patients with high DLCNC scores for further clinical investigation. We internally validated our electronic screening tool against a manual review of patient records for FH risk by an independent GP.

## Methods

### Setting and design

The study was a retrospective investigation of FH risk in active patients seen between 2012 and 2014 at a large metropolitan general practice in Perth, Western Australia.

### Ethics

De-identified information was analysed but the treating practice was able to reidentify patients for clinical follow-up if required.

### Electronic screening

We developed an electronic screening tool (TARB-Ex) that screens for FH risk in general practice. TARB-Ex extracts routine clinical information from EMRs to derive a DLCNC score and identify patients with potential FH risk for clinical investigation. TARB-Ex was developed using Structured Query Language (SQL) technology and written for Best Practice19 clinical software.

Figure 1 displays the TARB-Ex screening process. First, active patients (≥3 visits within the last 2 years20) at the practice were identified. A subset of patients who ever had a pathology record of total cholesterol ≥7 mmol/L or LDL-c ≥4.0 mmol/L (DLCNC minimum for FH risk) was then extracted. Clinical and family histories for these patients were also extracted from both drop-down lists and free-text entries. Allowance was made for variation in spelling and typographical errors by creating a dictionary of possible terms. Prescription information was also extracted. Patients receiving a statin prescription >1 week and <6 months of highest LDL-c measurement were regarded as on statin therapy and a correction algorithm was applied to estimate the patient's cholesterol prior to treatment.21 Corrected LDL-c levels were used for further computation.

Figure 1

TARB-Ex screening process. DLCNC, Dutch Lipid Clinic Network Criteria, LDL-c, low-density lipoprotein cholesterol.

For each patient, TARB-Ex produces an extract containing each element of the DLCNC along with a computed DLCNC22 score. Patients with DLCNC score ≥5 were considered potential risk patients. While a DLCNC score of 5 indicates ‘possible’ FH, we adopted this more inclusive strategy given that general practice patient history records are generally poorly documented.16 Thus, a patient with a score of five with a missing family history record may become a ‘probable’ case (DLCNC scores 6–8) with a positive family history upon further assessment.

### Clinical follow-up

De-identified TARB-Ex extracts for patients with DLCNC scores ≥5 were reviewed by a lipid specialist and GP to exclude any secondary causes of hypercholesterolaemia (eg, liver or renal disease, diabetes mellitus, corticosteroid use or hypothyroidism). Patients considered at high risk of FH were recalled for clinical assessment with the GP and lipid specialist team. A detailed assessment of personal and family history and physical examination for arcus cornealis and tendinous xanthomata was undertaken. A family pedigree tree was also constructed highlighting cardiovascular events in first-degree relatives. Based on this, a DLCNC score and FH diagnosis was determined.

### Validation of TARB-Ex

We evaluated TARB-Ex assessment for potential FH risk against manual review of patient EMRs. Manual review was undertaken by an independent, practice-based GP who received training in using the DLCNC. Study IDs for the subgroup of patients with elevated cholesterol were provided to the GP who then manually reviewed and derived a DLCNC based on the patient EMR. The GP was asked to remain objective in recording information and was blinded to the results of the TARB-Ex screening. Manual review was considered the gold standard method and determined true positives for potential FH risk.

### Statistical analysis

Data were analysed using Stata V.13.1. All primary analyses were tested against an α level of 0.05 (uncorrected, two-tailed). Descriptive statistics are presented as mean±SD for continuous outcomes and frequencies for categorical variables. Independent samples Student's t tests and χ2 analyses were used to compare differences in continuous and categorical outcomes, respectively.

We compared agreement between the two methods on: (1) highest uncorrected LDL-c recorded, (2) DLCNC score and (3) potential FH risk category. For continuous outcomes, we assessed inter-rater agreement using the intraclass correlation coefficient (ICC; two-way mixed model, absolute agreement) with 95% CIs and Bland–Altman plots (plotting the difference between methods against GP manual review as the reference method). For comparison of agreement on FH risk category, we calculated sensitivity, specificity and positive and negative predictive values with 95% CIs for TARB-Ex against GP manual record review.

### Cost analysis

Total and per case screening costs between TARB-Ex and GP manual review were calculated and compared. Costs in $A were calculated for TARB-Ex development, data extraction by various staff (GP, practice manager and practice nurse), lipid specialist and GP review of TARB-Ex patient extracts and GP manual review. Costs were calculated as follows and inflated with consumer price index increases to 2015:23 (1) TARB-Ex development ($40/h postdoctoral researcher salary), (2) lipid specialist ($200/h24), (3) GPs$148/h based on the modelled activity profile for a four-doctor practice25 and annual workload of 46 weeks×24 h/week26–28 and (4) practice manager $34/h and practice nurse$30/h based on an annual workload of 46 weeks×37 h/week.25

## Results

### TARB-Ex screening

A total of 3708 active patients attended the practice over the study period (53.3% female, mean age 43.5±24.6 years). Figure 2 displays the patient flow for electronic screening. Of all active patients, 1126 (30.4%) had at least one recorded LDL-c measurement. Patients with a recorded blood result were significantly older (65.6±14.1 years) than those never tested (33.8±21.9 years), p<0.001.

Figure 2

Patient flow for electronic screening. DLCNCS, Dutch Lipid Clinic Network Criteria Score; FH, familial hypercholesterolaemia; LDL-c, low-density lipoprotein cholesterol.

Three hundred and sixty (31.9%) active patients with blood records had elevated cholesterol. Thirty-nine patients were on statin treatment at the time of highest LDL-c measurement and required cholesterol correction. Statin strength data were not extractable in the EMR and so the lowest adjustment factor for each medication was applied as a conservative strategy. For example, the lowest correction factor for atorvastatin (ie, 10 mg) was applied for all patients with a script for ‘Lipitor Tablet’. Thirty-five patients had a recorded history of premature cardiovascular disease and two had premature ischaemic heart disease. Family history was missing for 65.2% (235/360) of patients. Of those with family history recorded, 27/125 (21.6%) had a positive family history of heart disease. Tendinous xanthomata and arcus cornealis were not recorded for any patients.

Thirty-two patients had DLCNC scores ≥5 and were identified as at potential risk of FH. One patient had an existing diagnosis of FH. The prevalence of patients at potential risk of FH was 1:116 of total active patients and 1:35 for the subgroup of active patients with a cholesterol measurement. Electronic screening took approximately 10 min to run.

### Manual review

GP manual record review identified 22 potential FH risk patients (DLCNC scores ≥5), a prevalence of 1:168 of total active patients and 1:51 of active patients with a cholesterol measurement. Manual review for the 360 patients was completed in approximately 60 h.

### Comparison of TARB-Ex and manual review

#### Highest uncorrected LDL-c measures

Agreement on highest uncorrected LDL-c measures was high, ICC=0.95 (95% CI 0.92 to 0.96). The mean difference in LDL-c was 0.09 mmol/L, p=0.067 (figure 3A). All differences in LDL-c measures between the two methods were positive, indicating that TARB-Ex systematically identified higher recorded LDL-c measures for a given patient compared with manual review.

Figure 3

Bland–Altman plot showing comparison of highest uncorrected LDL-c value (A) and Dutch Lipid Clinic Network Criteria score (B) according to the general practitioner (GP). Manual review (x-axis) versus difference between the two methods (TARB-Ex—GP manual review; y-axis). The solid horizontal lines represent the mean difference. The dashed lines indicate the 95% limits of agreements (±1.96 SDs).

#### DLCNC score

Agreement on DLCNC scores was also high, ICC=0.87 (95% CI 0.84 to 0.90). The mean difference in DLCNC scores between the two methods was 0.22, p=0.069 (figure 3B). Two distinct patterns were observed where score differences exceeded the limits of agreement: (1) TARB-Ex derived higher DLCNC scores where the manual review indicated very low FH risk (DLCNC ≤3) and (2) the manual review produced more DLCNC scores ≥8 compared with the electronic screening method.

#### Potential FH risk

Agreement between the two methods was 96.7% (348/360 patients) (table 1). Sensitivity was 95.5% (95% CI 77.2% to 99.9%) and specificity was 96.7% (95% CI 94.3% to 98.3%). The positive predictive value (PPV) was 65.6% (95% CI 46.9% to 81.4%) and negative predictive value was 99.7% (95% CI 98.3% to 100%).

Table 1

Comparison of TARB-Ex and GP manual record review assessment of FH risk in active patients with total cholesterol ≥7 mmol/L or LDL-c ≥4.0 mmol/L (n=360)

### Clinical follow-up and genetic testing

Following review of TARB-Ex extracts of medical records, 10 of 32 patients (31%) with DLCNC score ≥5 were considered at high risk of FH. Twenty-two patients were excluded at this stage due to overcorrection of LDL-c levels. That is, TARB-Ex had identified a statin prescription within the preceding 6 months of the patient's LDL-c measurement and applied a correction accordingly. However, upon review of the patient's complete prescription and blood history, it was considered that these patients were likely non-compliant with prescribed medication at the time of measurement and cholesterol adjustment was thus unnecessary. DLCNC scores for these 22 patients were revised down using the uncorrected LDL-c measurement. Review of records by the GP and lipid specialist for the 32 patients took approximately 1 h.

Table 2 compares DLCNC scores for the 10 patients recalled for clinical follow-up as assessed by TARB-Ex, GP manual review and at clinical assessment. Three patients were unable to attend assessment. Six of seven patients (86% sensitivity) who attended were diagnosed with phenotypic FH on clinical examination with one patient referred for genetic testing for FH. Four of 10 patients recalled for clinical follow-up were considered low FH risk according to the GP manual review. DLCNC scores were upgraded at clinical follow-up.

Table 2

Comparison of DLCNC scores for patients with high FH risk

### Cost analysis

Costs of screening for the two methods are presented in table 3. Two cost estimates are presented for TARB-Ex screening: actual study costs and approximate running costs in general practice.

Table 3

Comparison of costs of screening for TARB-Ex and GP manual review

## Discussion

Our study trialled an innovative electronic screening tool for FH risk using general practice EMRs. TARB-Ex was developed for Best Practice19 clinical management software. However, it has the capacity to be adapted for any SQL-based practice software such as MedicalDirector, ZedMed, MedTech, Practix and Monet which would encompass approximately 90%–95% of software usage in Western Australia (personal communication). Overall, results suggest that electronic screening is a quick and accurate method for systematically identifying patients at potential risk of FH for further clinical investigation, while minimising GP workload and practice costs. Our electronic screening method showed high sensitivity, specificity and negative predictive power, demonstrating comparable performance with a GP manual review of patient records for FH risk in a much shorter time (10 min vs 60 h for manual review).

Thirty-two patients were assigned DLCNC scores ≥5 by TARB-Ex compared with 22 from the GP manual review accounting for the comparatively lower PPV at 65.6%. Traditional interpretation of the PPV would suggest that one-third of potential FH risk patients identified through TARB-Ex were falsely classed as such. However, this could reflect the increased accuracy of electronic screening. Distributions in Bland–Altman plots show that TARB-Ex systematically identified higher LDL-c measures compared with manual review. As this process required the GP to identify and record the highest ever uncorrected LDL-c test value for each of 360 patients with no clinical judgement, we use this finding as support for the increased accuracy of electronic screening. Moreover, four of 10 patients identified by TARB-Ex and recalled for clinical follow-up were assessed as having low FH risk according to manual review. Medical records can be extensive for a given patient and manual scanning can be prone to human error, especially if the size of patient records to be reviewed is large.

Overall, the study shows that electronic screening is an efficient and accurate method of assessing FH risk with DLCNC scores strongly correlating with clinical examination. Both manually reviewed and TARB-Ex assessed DLCNC scores were upgraded at clinical follow-up. In each instance, this was because family and/or clinical history and physical signs were not recorded in patient notes. The ultimate sensitivity achieved was 86% with six of seven patients examined diagnosed with phenotypic FH. One patient recalled for clinical assessment had their DLCNC score revised downwards after examination (patient 5). This patient had been tested for FH in the past and assessed to be an unlikely case on combination of genetic and clinical findings. Two patients who did not return for clinical assessment were determined to have the lowest FH risk by manual review. If these patients did not truly have FH, this could affect the ultimate sensitivity of TARB-Ex screening.

Compared with previous electronic screening methods for FH risk,4 ,15 this method is optimised for easy integration into normal clinical practice. Previous electronic screening models have been very time-consuming and resource-consuming. Each manual search of medical records in the Gray et al15 study took approximately 30 min, creating a total of 201 additional hours of work for the GP for the 402 patients requiring review. Kirke et al's4 methods involved many different staff and costs (eg, pathology technicians, GPs, nurses, data extraction and mail-out) with low response rates. Moreover, 93% of their patients who attended clinical assessment were subsequently assessed to have low FH risk (DLCNC score ≤5). The current method offers a more time-effective (<1 h) and cost-effective (\$5–10 per potential diagnosis) model than previous models and when compared with manual medical review. Ease of use of the tool means a GP, practice nurse or practice manager can perform the screening prior to clinical recall. GP involvement is minimised to reviewing medical records for patients identified as at-risk, to excluding secondary causes and deciding whether clinical follow-up is warranted.

### Limitations

Electronic screening tools are only as effective as the quality of information recorded in practice databases. We found that family history was poorly documented for the majority of patients and no patients had a record of arcus cornealis or tendinous xanthomata. This highlights the need for increased comprehensiveness of recording in GP databases to enable development of similar electronic tools to increase clinical efficiency. Models such as FAMCAT18 based on indicators routinely available in EMRs may offer an alternative to the DLCNC for use in general practice screening with greater discriminatory power. However, FAMCAT is yet to be externally validated and trialled in a general practice setting.

Electronic screening is also subject to inherent limitations of the electronic system itself. In our study, statin strength/dosage data were not extractable in Best Practice which may explain why the GP review returned higher DLCNC scores (8+) compared with TARB-Ex (ie, the GP was able to apply the cholesterol adjustment based on actual statin dosage whereas TARB-Ex applied the lowest correction for a given medication). Second, where cholesterol corrections were automatically applied, it was deemed that corrections were not required in almost all cases as patients were likely non-compliant with prescribed medication. In addition, in determining potential risk of FH we set a threshold of DLCNC score ≥5. However, it could be argued that those who score 3–4 using DLCNC (also categorised as possible FH) would be missed and we acknowledge this as a limitation.

## Conclusions

While electronic screening methods can never be 100% accurate, we have shown that electronic screening can serve as a powerful tool to quickly and accurately identify patients at potential risk of FH for further investigation. Routine electronic screening combined with clinical follow-up with a GP may offer a time-effective and cost-effective systematic approach to identifying new index cases based in primary care. It is by using this retrospective approach combined with a more proactive, opportunistic approach aimed both at identifying new index cases and cascade testing relatives that increased detection of FH may be facilitated in primary care. Proposed methods of shared care11 ,12 will provide an additional29 and sustainable approach to the management of diagnosed FH patients largely within primary care.

Our future research is focused on broader implementation and validation of TARB-Ex across multiple practices and EMR platforms. To date, we have trialled TARB-Ex in eight Australian practices and are in the process of integrating TARB-Ex for MedicalDirector. Evaluation of TARB-Ex's performance in external settings will provide greater insight into the clinical utility of the tool for detecting new index cases of FH.

### Key messages

#### What is already known on this subject?

• Only an estimated 10% of familial hypercholesterolaemia (FH) cases are currently detected.

• General practice has been increasingly identified as an optimal setting for FH screening.

• Methods of screening need to integrate easily into existing clinical practice.

#### What might this study add?

• Our study provides an easy and accurate method of systematically identifying general practice patients at potential risk of FH for further investigation using routine clinical information recorded in electronic medical records.

#### How might this impact on clinical practice?

• Routine electronic screening combined with clinical follow-up with a general practitioner may offer a time-effective and cost-effective systematic approach to identifying new index cases based in primary care.

## Acknowledgments

We acknowledge the financial support of L Ryan to the General Practice Research Unit.

## Footnotes

• Contributors LT developed the electronic screening tool, performed data collection, wrote the statistical analysis plan, cleaned and analysed the data and drafted and revised the paper. She is the guarantor. DA-R designed and supervised the study and drafted and revised the paper. WCSP-D drafted and revised the paper. GFW reviewed patient records, performed clinical assessment with patients and revised the draft paper. JP reviewed patient records and revised the draft paper. ML performed data collection, reviewed patient records and revised the draft paper. MB provided statistical input and revised the draft paper. DM monitored data collection and revised the draft paper. MJ performed data collection and revised the draft paper. TB designed and supervised the study, reviewed patient records, performed clinical assessment with patients and revised the draft paper.

• Funding This study was supported through the Australian Government's Collaborative Research Networks (CRN) programme to the University of Notre Dame and the University of Notre Dame Australia Research Incentive Scheme (2014R203).

• Competing interests None declared.

• Ethics approval The University of Notre Dame Australia Human Research Ethics Committee (014017F).

• Provenance and peer review Not commissioned; externally peer reviewed.


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Value of density when there are no or very few neighbours in SPH simulation

I am trying to implement SPH using the directions shown in this paper.

The density needs to be updated using the formula

$$\rho(\mathbf{x}_i)=\sum_j m_jW_\text{default}(\mathbf{x}_i-\mathbf{x}_j,h)$$

The smoothing kernel is $$W_\text{default}(\mathbf{r},h)=\frac{315}{64\pi h^9}\cases{(h^2-||\mathbf{r}||^2)^3,\quad 0\leq\mathbf{r}\leq h\\0,\quad\quad\quad\quad\quad ||\mathbf{r}||>h}$$

If there are no particles within the smoothing length($$h$$), won't the density of a particle become zero based on this formula?

Yes. You need to choose the radius $h$ large enough that for each particle, there is always a significant number of other particles within the first particle's radius.
• If you don't want to increase the number of particles, you have to choose $h$ large enough. Jan 31 '15 at 3:39


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# How to calculate this? int_0^1 log(1-x)/xdx

Feb 11, 2018

See below.

#### Explanation:

Unfortunately the function inside the integral will not integrate to something that cannot be expressed in terms of elementary functions. You will have to use numerical methods to do this.

I can show you how to use a series expansion to get an approximate value.

Begin with the geometric series:

$\frac{1}{1 - r} = 1 + r + {r}^{2} + {r}^{3} + {r}^{4.} . . = {\sum}_{n = 0}^{\infty} {r}^{n}$ for $r < 1$

Now integrate with respect to $r$ and using the limits $0$ and $x$ to get this:

${\int}_{0}^{x} \frac{1}{1 - r} \mathrm{dr} = {\int}_{0}^{x} 1 + r + {r}^{2} + {r}^{3} + \ldots \mathrm{dr}$

Integrating the left hand side:

${\int}_{0}^{x} \frac{1}{1 - r} \mathrm{dr} = {\left[- \ln \left(1 - r\right)\right]}_{0}^{x} = - \ln \left(1 - x\right)$

Now integrate the right hand side by integrating term by term:

${\int}_{0}^{x} 1 + r + {r}^{2} + {r}^{3} + \ldots \mathrm{dr} = {\left[r + {r}^{2} / 2 + {r}^{3} / 3 + {r}^{4} / 4. . .\right]}_{0}^{x}$

$= x + {x}^{2} / 2 + {x}^{3} / 3 + {x}^{4} / 4 + \ldots$

So it follows that:

$- \ln \left(1 - x\right) = x + {x}^{2} / 2 + {x}^{3} / 3 + {x}^{4} / 4 + \ldots$

$\implies \ln \left(1 - x\right) = - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 + \ldots$

Now divide by $x$:

$\ln \frac{1 - x}{x} = \frac{- x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 + \ldots}{x}$

$= - 1 - \frac{x}{2} - {x}^{2} / 3 - {x}^{3} / 4 - \ldots$

So we now have power series expression for the function we originally started with. Finally, we can integrate again to get:

${\int}_{0}^{1} \ln \frac{1 - x}{x} = {\int}_{0}^{1} - 1 - \frac{x}{2} - {x}^{2} / 3 - {x}^{3} / 4 - \ldots \mathrm{dx}$

Integrating the right hand term by term side gives us:

${\int}_{0}^{1} \ln \frac{1 - x}{x} = - {\left[x - {x}^{2} / 4 - {x}^{3} / 9 - {x}^{4} / 16 - \ldots\right]}_{0}^{1}$

Evaluating the limits to four terms will give us an approximate value:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \approx \left\{- 1 - {1}^{2} / 4 - {1}^{3} / 9 - {1}^{4} / 16\right\} - \left\{0\right\}$

$= - \left(1 + \frac{1}{4} + \frac{1}{6} + \frac{1}{16} + \ldots\right) = - \frac{205}{144} \approx - 1.42361$

Now, this is only to four terms. If you would like a more accurate number simply use more terms in the series. For example, going to the 100th term:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \approx - 1.63498$

As an aside, if you work through the exact same process but use summation notation (i.e. with big sigma rather than writing out the terms of the series) you will find that:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2$

which is just the Riemann-Zeta function of 2, i.e:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2 = - \zeta \left(2\right)$

We actually already know the value of this to be: $\zeta \left(2\right) = {\pi}^{2} / 6$.

Hence the exact value of the integral can be deduced to be:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\pi}^{2} / 6$


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# Synopsis: Neutron Knockout

#### Two-proton decay of 12O and its isobaric analog state in 12N

M. F. Jager, R. J. Charity, J. M. Elson, J. Manfredi, M. H. Mahzoon, L. G. Sobotka, M. McCleskey, R. G. Pizzone, B. T. Roeder, A. Spiridon, E. Simmons, L. Trache, and M. Kurokawa

Published July 26, 2012

Unstable nuclear states that decay by emitting protons or neutrons provide important tests for models that are based on properties of nuclei close to the line of stability. They also permit tests of the isospin symmetry—a quantity related to the dominant component of the strong nuclear force—to determine if certain nuclear properties are unchanged if a neutron is replaced by a proton and vice versa.

One of the rarest decay modes of unstable nuclei is the essentially simultaneous emission of two protons. In a paper in Physical Review C, Marieke Jager at Washington University, Missouri, and collaborators report results from an experiment in which they observed two different $2$-proton decay modes. In a beam of an isotope of oxygen, ${}^{13}\text{O}$, either a neutron or a proton was knocked out, leaving ${}^{12}\text{O}$ and ${}^{12}\text{N}$, respectively. The $2$-proton decay of these to ${}^{10}\text{C}$ and ${}^{10}\text{B}$ gave valuable information on the mass and lifetime (equivalent to a property called the width of the state) of states in ${}^{12}\text{O}$ and ${}^{12}\text{N}$. The mass and width for the ${}^{12}\text{O}$ ground state turn out to be considerably smaller than what was known from previous measurements and resolve a discrepancy behind the theory for the width. The properties of the isobaric analog state in ${}^{12}\text{N}$—analogous to one in ${}^{12}\text{O}$, except that a neutron replaces a proton—are measured here for the first time.

These results permit a successful test of a $3$-parameter equation, called the isobaric multiplet mass equation, that relates the masses of nuclei in the same state (here ${}^{12}\text{Be}$, ${}^{12}\text{B}$, ${}^{12}\text{C}$, ${}^{12}\text{N}$, and ${}^{12}\text{O}$) with the same total number of nucleons, simply in terms of the respective numbers of protons and neutrons. – Rick Casten and John Millener


text:
Article Contents
Article Contents

Existence of stable manifolds for nonuniformly hyperbolic $c^1$ dynamics

• The existence of stable manifolds for nonuniformly hyperbolic trajectories is well know in the case of $C^{1+\alpha}$ dynamics, as proven by Pesin in the late 1970's. On the other hand, Pugh constructed a $C^1$ diffeomorphism that is not of class $C^{1+\alpha}$ for any $\alpha$ and for which there exists no stable manifold. The $C^{1+\alpha}$ hypothesis appears to be crucial in some parts of smooth ergodic theory, such as for the absolute continuity property and thus in the study of the ergodic properties of the dynamics. Nevertheless, we establish the existence of invariant stable manifolds for nonuniformly hyperbolic trajectories of a large family of maps of class at most $C^1$, by providing a condition which is weaker than the $C^{1+\alpha}$ hypothesis but which is still sufficient to establish a stable manifold theorem. We also consider the more general case of sequences of maps, which corresponds to a nonautonomous dynamics with discrete time. We note that our proof of the stable manifold theorem is new even in the case of $C^{1+\alpha}$ nonuniformly hyperbolic dynamics. In particular, the optimal $C^1$ smoothness of the invariant manifolds is obtained by constructing an invariant family of cones.
Mathematics Subject Classification: Primary: 37D10, 37D25.

 Citation:


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# Tag Info

0

I have known what exactly I have done wrongly. The exact solution in R is as follows: dif=(New-Old) m=mean(dif) st=sd(dif) CI=m+c(-1,1)*qt(0.975,4)*st *sqrt(1/5) ExpCI=exp(CI) # equals [0.09 2.49] MEAN=mean(ExpCI) # equals 1.29

3

You calculated the mean of logarithms of ratios, not the mean of ratios themselves. To obtain ratios from their logarithms, you have to raise $e$ to power of them. In Python and NumPy: import numpy as np logs = [1.304, -0.768, -0.473, 1.237, 2.405] # Logarithms of ratios ratios = np.exp(logs) # You omitted this! np.mean(...

3

If you have datapoints $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ and you want to find the mean ratio $\frac{1}{n} \sum_{i=1}^n \frac{x_i}{y_i}$, this is not equal to the exponentiated mean of the log ratios. In other words, this is not the same as computing $\exp\left(\frac{1}{n}\sum_{i=1}^n \log \frac{x_i}{y_i}\right)$, which is what it sounds like you are ...

Top 50 recent answers are included


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# A holomorphic property in the Seiberg-Witten solution

+ 3 like - 0 dislike
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In the Seiberg-Witten solution of N=2 super-Yang-Mills theory, the parameter for the moduli space is chosen as u=<tr phi^2> = sum_A<(phi_A)^2>, where A=1,2,3 is the color index of the SU(2) group.  The fact that a(u) and a_D(u) depends holomorphically on u was used in the Seiberg-Witten solution. I understand that a_D(u) should depends holomorphically on a(u), which is required by supersymmetry, but how can I see that a(u) and a_D(u) should depend holomorphically on u? From my naive understanding, u can be a function of both a and a*, and thus the function a and a_D should be a(u,u*) and a_D(u,u*). Surely I have not understood some important thing here. Thanks in advance for explanation.

asked Feb 25, 2016
recategorized Feb 25, 2016

In both cases it follows from supersymmetry. Why do you think that it is clear for $a_D$ but not for $a$?

@40227  I understand that a_D should depends holomorphically on a, which is required by supersymmetry. But I don't understand why a_D or a should depends holomorphically on u.

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# Exclude Sections/Subsections/… from evaluation

I'm working on a large notebook and I want to exclude some sections and subsections from being evaluated when I evaluate the whole notebook.

Is there a way to exclude specific sections or subsections from the notebook evaluation while maintaining their original format? Or even better, format them e.g. by adding a gray background?

• Rather than specify which sections to not evaluate, you can specify which you do want to evaluate, by setting them to be initialization cells, and evaluating only initialization cells. – jVincent Feb 15 '13 at 11:51
• Or, you can select those cells that you don't need to evaluate and comment them (Edit menu -> Un/Comment Selection). – VLC Feb 15 '13 at 12:08
• See mathematica.stackexchange.com/questions/14466/… for another approach. – cormullion Feb 15 '13 at 12:24

Cell[StyleData["NonEvaluatingSection", StyleDefinitions -> StyleData["Section"]],

"CellGroupBaseStyle" is new to version 9, and it allows you to layer in options to all cells which are grouped by the given style. So, in this case, we have a style "NonEvaluatingSection", which has all of the properties of the style "Section", except that it adds a background and Evaluatable->False to itself and all of the cells grouped by it. The option is not documented, but that's just an oversight. It will be in a future release.