text: For example, a lithium atom (Z=3, A=7 amu) contains three protons (found from Z), three electrons (as the number of protons is equal to the number of electrons in an atom), and four neutrons (7 – 3 = 4). For C, the isotopic mass is exactly 12, since the atomic mass unit is defined as 1/12 of the mass of C. For other isotopes, the isotopic mass is usually within 0.1 u of the mass number. CC BY-SA 3.0. http://en.wikibooks.org/wiki/A-level_Chemistry/OCR/Atoms,_Bonds_and_Groups/Atoms_and_Reactions/Atoms The appellee’s brief (the response brief) must have a red cover. Read this article to learn how to fix this problem. What is Mass number The mass number is the total number of neutrons and protons in the nucleus of the atom. The mass number is defined as the total number of protons and neutrons in an atom. For example, a sample of chlorine contains 75.8% chlorine-35 and 24.2% chlorine-37, giving an average atomic mass of 35.5 atomic mass units. If you don't want to convert the whole column, you can select one or more cells instead. 10 mol of carbon dioxide has a mass of 440 g. Definitions Element showing its atomic number and mass number. Isotopes of the same element will have the same atomic number but different mass numbers. Mass number = atomic number + number of neutrons Consider Table below which shows data from the first six elements of the periodic table. The mass number gives an estimate of the isotopic mass measured in atomic mass units (u). The atomic mass, on the other hand, is the number of both the protons and neutrons present in the nucleus of the element.Electrons do not usually weigh much, so the atomic mass is … Measured numbers. Sponsors & Contributors. P.) 1, 16, 19, and 20, M.A.C. An atom with eight protons is an oxygen atom, for example. Example: What is the molar mass of ethanol (C 2 H 5 OH)?. For example, a lithium atom (Z=3, A=7 amu) contains three protons (found from Z), three electrons (as the number of protons is equal to the number of electrons in an atom), and four neutrons (7 – 3 = 4). Select a column. So, these numbers that we have here, just as a review, these are atomic mass. For example, Chlorine-37 and Argon-37 have the same mass number. That 92 is the atomic number, which is the number of protons. The mass of an atom consists of the mass of the nucleus plus that of the electrons, so the atomic mass unit is not exactly the same as the mass of the proton or neutron. It is denoted by Z. e.g: i.e., atomic number = 20 When using the full name of an element, not its symbol, mass number come after the name of an element with a hyphen in between, for example, sodium-23, carbon-12, iron-56. The baryonic number, which is equivalent to the atomic mass number, has to remain constant for a reaction. Mass number of an element is more related with the mass of it. Subtract the number of protons from the atomic weight. Explains the calculation of mass number. For example, nitrogen has 7 protons and 7 neutrons in its nucleus, giving it a mass number of 14. Select a column with this problem. The total number of protons and neutrons in the nucleus of an atom. Relative atomic mass (rounded to the nearest whole number): H = 1, C = 12, O = 16 We have moved all content for this concept to for better organization. Try the given examples, or type in your own The mass number is represented as the left superscript to the symbol of an element. It is denoted by Z. e.g: i.e., atomic number = 20 The mass number is commonly cited in distinguishing among the isotopes of an element, all of which have the same atomic number (number of protons) and are represented by the same literal symbol; for example, the two best known isotopes of uranium (those with mass numbers 235 and 238) are designated uranium-235 (symbolized 235 U) and uranium-238 (238 U). 3717 Cl has a mass number of 37. We welcome your feedback, comments and questions about this site or page. Wikibooks The number of protons determines an element’s atomic number and is used to distinguish one element from another. For example, an atom of carbon has 6 protons and 6 neutrons. The mass number gives an estimate of the isotopic mass measured in atomic mass units (u). Together, the number of protons and the number of neutrons determine an element’s mass number. How many neutrons are there in one atom of ? So let me move down here so we can look at the definition for the mass number. The American Heritage® Student Science Dictionary, Second Edition. Sentence Examples. The number of electrons can also be different in atoms of the same element, thus producing ions (charged atoms). ... For example, if you push two objects under the same conditions with the same amount of force, the object with the lower mass will accelerate faster. Its nucleus contains 17 protons and 20 neutrons. mass number The total number of protons and neutrons in the nucleus of an atom. To find the number of neutrons, use the periodic table of elements again. For example, the atomic mass of chlorine (Cl) is 35.45 amu because chlorine is composed of several isotopes, some (the majority) with an atomic mass of 35 amu (17 protons and 18 neutrons) and some with an atomic mass of 37 amu (17 protons and 20 neutrons). K.E = ½ mv 2. Answer:-(i) Atomic Number : The number of protons present in the nucleus of an atom is called atomic number. 1. problem and check your answer with the step-by-step explanations. 2 AlCl 3 + 3 H 2 O(ℓ) → Al 2 O 3 + 6 HCl(g). CC BY-SA 3.0. http://en.wiktionary.org/wiki/atomic_number When measuring the mass of an atom, we actually measure the mass of the nucleus. Lithium, for example, has three protons and four neutrons, giving it a mass number of 7. In this example it's cell E23. 1. In the periodic table, atoms are arranged according to their atomic number. For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3  ionic states. CC BY-SA 3.0. http://cnx.org/content/m44390/latest/?collection=col11448/latest Mass number of an element is more related with the mass of it. Knowing the mass number and the atomic number of an atom allows you to determine the number of neutrons present in that atom by subtraction. Since an element’s isotopes have slightly different mass numbers, the atomic mass is calculated by obtaining the mean of the mass numbers for its isotopes. Mass Number (A) = Number of Protons + Number of Neutrons The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. Mass convert scattered numbers stored as text to numbers in Excel. Solution: Step 1: Look up the relative atomic masses of the atoms from the periodic table. c = speed of light (3×108 m/s) The kinetic energy mass formula is given as. For example, Cl (17 protons and 18 neutrons) has a mass number of 35 and an isotopic mass of 34.96885. Example 1: The hydrogen atom doesn’t have a neutron but it has 1 proton, so the mass number of hydrogen is 1. In the periodic table, atoms are arranged according to their atomic number. How many neutrons are there in one atom of ? $\text{Number of neutrons} = \text{mass number} - \text{atomic number}$ The mass number of carbon-13 is 13. Cesium 137, a radioactive cesium isotope with a mass number of 137 and a half-life of 33 years, is used for gamma irradiation of certain foods and for radiation therapy. Give any two uses of isotopes. For example, all hydrogen atoms, and only hydrogen atoms, contain one proton and have an atomic number of 1. Embedded content, if any, are copyrights of their respective owners. Example 2 of Calculations Involving Atomic Number and Mass Number Example 1: What is the symbol, atomic number, mass number, and number of protons, electrons, and neutrons of Sulfur? Calculate the mass if the weight of a body is 80 N. Solution: Given, weight of the body = 80 N. The mass of the body is expressed by. Solution: Step 1: Look up the relative atomic masses of the atoms from the periodic table. Consider an example of sodium, which has a mass number of 23. The small contribution of mass from electrons is disregarded in calculating the mass number. Q.8:- Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. The number of protons and neutrons combined to give us the mass number of an atom. Mass number and isotopic mass. Rest your cursor here. problem solver below to practice various math topics. m = w / g. m = 8 0 / 9.8. m = 8.163 kg. Formula for Mass: Definition & Examples 4:59 Halogens on the Periodic Table: Properties, Reactivity & Uses 4:19 Hydroxide Ion: Definition & Formula 3:38 If the numbers which stored as text are scattered in different places in your worksheet, you can try the following method. Subtract the atomic number from the atomic mass. CC BY-SA 3.0. http://www.boundless.com//biology/definition/atomic-mass–2 Isotopes are simply specifying the number of neutrons and protons (together called nucleons) in the atom. First, an example of a nuclear symbol: 6 14 C. Make sure to remember that the lower number is the atomic number and the upper number is the mass number. It is very important to note that while the average atomic mass of magnesium is 24.31 amu, and the average atomic mass of carbon is 12.01 amu, this is not the mass of any individual atom. atomic massThe average mass of an atom, taking into account all its naturally occurring isotopes. Example 10. Isotopes of elements are distinguished by their mass number; for example, carbon-12 and carbon-14 have mass numbers of 12 and 14 respectively. That is because the mass of an electron is negligible when compared to a proton or a neutron. For example, all hydrogen atoms, and only hydrogen atoms, contain one proton and have an atomic number of 1. mass number The total number of protons and neutrons in the nucleus of an atom. Mass is an example of a measured number. Main Difference – Atomic Number vs Mass Number. The number of neutrons can vary to produce isotopes, which are atoms of the same element that have different numbers of neutrons. Knowing the mass number and the atomic number of an atom allows you to determine the number of neutrons present in that atom by subtraction. Find the number of neutrons. Please submit your feedback or enquiries via our Feedback page. For example, for silicon-28 (28Si), the 28 is rounded from the mass, which is 27.976927 amu. R. App. The number of neutrons in an atom, however, can vary somewhat and remain the same element; these varieties are called isotopes. When a number is given following an element name, this is its isotope, which basically states the mass number. Thus, its number of protons is 11. Wiktionary : an integer that approximates the mass of an isotope and that designates the number of nucleons in the nucleus the symbol for carbon of mass number 14 is 14 C or C 14 — compare atomic mass More from Merriam-Webster on mass number Given an atomic number (Z) and mass number (A), you can find the number of protons, neutrons, and electrons in a neutral atom. Medical Definition of mass number : an integer that approximates the mass of an isotope and that designates the number of nucleons in the nucleus the symbol for carbon of mass number 14 is 14C or C14 — compare atomic mass For example, the atomic number of sodium is 11. CC BY 3.0. http://en.wiktionary.org/wiki/atomic_number, http://www.boundless.com//biology/definition/atomic-mass–2, http://en.wikibooks.org/wiki/A-level_Chemistry/OCR/Atoms,_Bonds_and_Groups/Atoms_and_Reactions/Atoms, http://cnx.org/content/m44390/latest/?collection=col11448/latest, https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/, Determine the relationship between the mass number of an atom, its atomic number, its atomic mass, and its number of subatomic particles. The appellant's principal appellate brief must have a blue cover. All carbon atoms, and only carbon atoms, contain six protons and have an atomic number of 6. Mass is a combination of the total number of atoms, the density of the atoms, and the type of atoms in an object. The helium atom has a mass number of 4 and atomic number of 2. Mass is a combination of the total number of atoms, the density of the atoms, and the type of atoms in an object. Example 2: Oxygen atom has 8 protons and 8 neutrons. Example of Mass Production . Q.8:- Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. When numbers are actually stored in cells as text, problems with calculations or sorting can occur. So, Carbon-12, which has an atomic mass number of 12, has 6 neutrons (12 nucleons - 6 protons = 6 neutrons). For Example, Hydrogen-3 Corresponds To A Hydrogen Isotope With A Mass Number Of 3 ? tarting on Tuesday, November 13, 2018, the RMV is transitioning its new license/ID card issuance format from a nine-digit alphanumeric code consisting of one letter (“S”) and eight numbers to one consisting of two letters (“SA”) and seven number. The atomic number and the mass number of an element is usually written in the following way. Example of Mass Production . Example 1. For 12 C, the isotopic mass is exactly 12, since the atomic mass unit is defined as 1/12 of the mass of 12 C. For other isotopes, the isotopic mass is usually within 0.1 u of the mass number. Remember that the atomic number is the same as the number of protons, which you have already identified. 3. Massachusetts Rules of Appellate Procedure (Mass. Example: What is the molar mass of ethanol (C 2 H 5 OH)?. Scientists determine the atomic mass by calculating the mean of the mass numbers for its naturally-occurring isotopes. These are atomic mass. For our boron example, 11 (atomic mass) – 5 (atomic number) = 6 neutrons Atomic number is used to define the type of element a material or substance is.It is the number of protons present in an element’s nucleus. The number of protons determines an element’s atomic number (Z) and distinguishes one element from another. However, it is not giving the exact mass … For example, if you were to pick up an individual carbon atom (pretend that's possible!) E = energy. The mass number is the combined number of protons and neutrons in a nucleus, so it's protons and neutrons, and it's symbolized by A. Neutral atoms of each element contain an equal number of protons and electrons. Where, m = mass, v = velocity. Give any two uses of isotopes. The American Heritage® Student Science Dictionary, Second Edition. The weight of an object, such as 8.887 grams. Question: Isotopes Are Often Named By Placing The Mass Number (the Sum Of The Number Of Portons And The Number Of Neutrons) After The Name Of The Element. To find the number of neutrons, you will need to subtract the atomic number from the atomic mass. The number of neutrons is variable, resulting in isotopes, which are different forms of the same atom that vary only in the number of neutrons they possess. When we see the atomic masses of elements in the periodic table, these are actually the averages of … An element’s mass number (A) is the sum of the number of protons and the number of neutrons. When mass is reported as 0.5237 g, as shown on this scale, it is more precise than a mass reported as 0.5 g. Examples of measured numbers: The diameter of a coin, such as 10.2 mm. The mass of an atom consists of the mass of the nucleus plus that of the electrons, so the atomic mass unit is not exactly the same as the mass of the proton or neutron. Helium has an atomic number of 2 since it has two protons in its nucleus. For example, carbon’s atomic number (Z) is 6 because it has 6 protons. Henry Ford, founder of the Ford Motor Company, developed the assembly line technique of mass production. mass numberThe sum of the number of protons and the number of neutrons in an atom. It is represented using the letter ‘A.’ As both protons and neutrons are present in the nucleus of an atom, they are together called nucleons. Boundless vets and curates high-quality, openly licensed content from around the Internet. This particular resource used the following sources: http://www.boundless.com/ The mass of atoms is measured in terms of the atomic mass unit, which is defined to be 1/12 of the mass of an atom of carbon-12, or 1.660538921 × 10 −24 gram. This time, find the element's atomic weight, and round it to the nearest whole number. Protons and neutrons both weigh about one atomic mass unit or amu. As atoms often undergo alpha decay that eliminates two protons and two neutrons from a radioactive nucleus, the mass number of elements can change accordingly. Solution. mass number. The kinetic energy mass formula is given as. Try the free Mathway calculator and Atomic number is used to define the type of element a material or substance is.It is the number of protons present in an element’s nucleus. Type a number “ 1 ” into a blank cell, and then copy it with pressing Ctrl + C keys. Atomic Number vs Mass Number Atoms are characterized by their atomic numbers and mass numbers. Carbon-13, which has an atomic mass number of 13, has 7 neutrons (13 nucleons - 6 protons = 7 neutrons). Here's another nuclear symbol: 92 238 U. How many moles of HCl will be produced when 249 g of AlCl 3 are reacted according to this chemical equation?. The number of protons in an atomic nucleus determines which element the atom is. ... and inside the parentheses, type a cell reference that contains text stored as numbers. All carbon atoms, and only carbon atoms, contain six protons and have an atomic number of 6. The atomic mass, on the other hand, is the number of both the protons and neutrons present in the nucleus of the element.Electrons do not usually weigh much, so the atomic mass is … The number is superscripted to the left of the symbol of sodium, which is Na (see the below figure). The mass number gives you the total number of protons and neutrons in the nucleus of that element. Atomic Number vs Mass Number Atoms are characterized by their atomic numbers and mass numbers. Helium has an atomic number of 2 since it has two protons in its nucleus. then it would either be 12.00 amu or 13.000335 amu. Given an atomic number (Z) and mass number (A), you can find the number of protons, neutrons, and electrons in a neutral atom. Definitions Element showing its atomic number and mass number. Here's a brief tutorial about nuclear symbols. For example, the mass number of argon atoms and calcium atoms can both be 40. Boundless Learning Provide Feedback. Lithium has an atomic number of 3 since it has three protons in its nucleus. The mass of atoms is measured in terms of the atomic mass unit, which is defined to be 1/12 of the mass of an atom of carbon-12, or 1.660538921 × 10 −24 gram. Atomic Mass and Mass Number Example Hydrogen has three natural isotopes: 1 H, 2 H, and 3 H. Each isotope has a different mass number. We will do this in two steps: convert the mass of AlCl 3 to moles and then use the balanced chemical equation to find the number of moles of HCl formed. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. (5 votes) m = (E / c2) Where, m = mass. Answer:-(i) Atomic Number : The number of protons present in the nucleus of an atom is called atomic number. Copyright © … Example 2. For example, the atomic mass of chlorine (Cl) is 35.45 amu because chlorine is composed of several isotopes, some (the majority) with an atomic mass of 35 amu (17 protons and 18 neutrons) and some with an atomic mass of 37 amu (17 protons and 20 neutrons). 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### Author Topic: Halloween Challenge 2  (Read 197 times)

#### Victor Ivrii

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##### Halloween Challenge 2
« on: October 21, 2016, 11:57:10 AM »

#### Roro Sihui Yap

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##### Re: Halloween Challenge 2
« Reply #1 on: October 21, 2016, 05:47:44 PM »
10a)
$$u(x,t) = f(x+t) + g(x-t)$$

From $u|_{t=0} = 0$, $\ \ f(x) + g(x) = 0$ $\rightarrow f(x)= -g(x)$
From $u|_{x=0} = 0$, $\ \ f(t) + g(-t) = 0 \ \$
Using $f(x)= -g(x), \$we get $\ -g(t) + g(-t) =0$ $\rightarrow g(-t)= g(t)$

From $u|_{x=a} = 0$, $\ \ f(a+t) + g(a -t) = 0$ $\rightarrow g(t-a) = g(t+a)$ $\rightarrow g(x) = g(x +2a)\label{A}$
From $u|_{t=b} = 0$, $\ \ f(x+b) + g(x-b) = 0$ $\rightarrow g(x-b) = g(x+b)$ $\rightarrow g(x) = g(x+2b)\label{B}$

$g$ is a periodic function, suppose g(x) is not constant function, let the period of $g$ be $p$, the smallest number such that $g(x) = g(x+np)$ where $n$ is any integer and $p \ne 0$,

Then, $2a = k_1p$ and $2b = k_2p$ where $k_1,k_2$ are integers
$$\frac{2a}{k_1} = \frac{2b}{k_2} \implies \frac{a}{b} = \frac{k_1}{k_2}$$

Contradiction, we assumed a and b are not commensurable. Thus $g(x)$ must be a constant function, suppose $g(x) = k$, then $f(x) = -k$ where k is any number

$$u(x,t) = f(x+t) + g(x-t) = k - k = 0$$
The only solution is the trivial solution
« Last Edit: October 21, 2016, 05:51:09 PM by Roro Sihui Yap »

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Let $P(x)$ and $Q(x)$ be arbitrary predicates. Which of the following statements is always TRUE?

1. $\left(\left(\forall x \left(P\left(x\right) \vee Q\left(x\right)\right)\right)\right) \implies \left(\left(\forall x P\left(x\right)\right) \vee \left(\forall xQ\left(x\right)\right)\right)$
2. $\left(\forall x \left(P\left(x\right) \implies Q\left(x\right)\right)\right) \implies \left(\left(\forall x P\left(x\right)\right) \implies \left(\forall xQ\left(x\right)\right)\right)$
3. $\left(\forall x\left(P\left(x\right) \right) \implies \forall x \left( Q\left(x\right)\right)\right) \implies \left(\forall x \left( P\left(x\right) \implies Q\left(x\right)\right)\right)$
4. $\left(\forall x \left( P\left(x\right)\right) \Leftrightarrow \left(\forall x \left( Q\left(x\right)\right)\right) \right) \implies \left(\forall x \left (P\left(x\right) \Leftrightarrow Q\left(x\right)\right)\right)$

This type question is also solved to convert Digital logic
how to solve these questions?

Procedure to be followed: For each option assume LHS to be TRUE and try to make RHS be False by selecting some values which makes LHS true in every condition.
(one can also start from RHS assuming it to be false and trying to make LHS to be true)

A. $\mathbf{[ (\forall x (P(x)} ∨ \mathbf{Q(x))) ]} ⟹ [ (∀ \mathbf{x P(x))} ∨ \mathbf{(∀xQ(x))} ]$

Let us assume the domain of $x$ such that, for the first half of the values $P(x)$ is True & $Q(x)$ is False while for the other half, $Q(x)$ is True and $P(x)$ is False.

LHS: Since for LHS to be true either $P(x)$ or $Q(x)$ should be true. Hence our assumption of the domain will make LHS be TRUE.

RHS: for $(∀\mathbf{x P(x))}$ or $(∀ \mathbf{x Q(x))}$ to be true, atleast one of the $P(x)$ or $Q(x)$ must be true for all values of $x$, which is not possible as per assumption we have made.Thus, RHS becomes FALSE.

Thus, $T → F$ makes statement A False.

C. $[ ∀\mathbf{x(P(x))} ⟹ ∀\mathbf{x(Q(x))} ] ⟹ [ ∀\mathbf{x(P(x)⟹Q(x))} ]$

Let us assume some values of $P(x)$ and $Q(x)$ as follows :

$$\begin{array}{l|l|l} \text{x} & \text{P(x)} & \text{ Q(x)} \\ \hline \text{x_1} & \text{F} & \text{T} \\ \text{x_2} & \text{T} &\text{F} \end{array}$$
LHS: for assumed domain $∀ \mathbf{ x(P(x)) }$ will becomes False and  $∀\mathbf{x(Q(x))}$ will also be false. Since $F → F$ is True, LHS becomes TRUE.

RHS: for $x_1$ $\mathbf{[(P(x)⟹Q(x))}$ is True and for $x_2$ $\mathbf{(P(x)⟹Q(x))}$ is False. As a whole $∀\mathbf{x(P(x)⟹Q(x))}$ becomes False, thus RHS becomes FALSE.

Thus, $T→F$ makes statement C as False.

D. $[ ∀\mathbf{x(P(x))} ⇔ \mathbf{(∀x(Q(x))) ]} ⟹ [ ∀\mathbf{x(P(x) ⇔ Q(x)) ]}$

if we assume same domain as in above option C,then observations are as following :

LHS:  $∀\mathbf{x(P(x))}$ becomes false as $x_1$ is false. Also $(∀\mathbf{x(Q(x)))}$ becomes false as $x_2$ is false.Thus $F ↔ F$ implies LHS is TRUE.

RHS: $\mathbf{(P(x) ⇔ Q(x))}$ will be false for both $x_1$ and $x_2$. Hence $∀\mathbf{x(P(x) ⇔ Q(x))}$ becomes False which makes RHS to be FALSE.

Thus $T→F$ makes statement D as False.

B. $[ ∀\mathbf{x(P(x) ⟹ Q(x)) ] ⟹ [ (∀xP(x)) ⟹ (∀xQ(x)) ]}$

As we are assuming LHS to be TRUE then we'll not make any selection in which $P(x)$ is True and $Q(x)$ is false as it will make our assumption false.Thus values can be like this:

$$\begin{array}{l|l|l}\text{x} & \text{P(x)} & \text{ Q(x)} \\\hline \text{x_1} & \text{T} & \text{T} \\ \text{x_2} & \text{F} &\text{T} \\ \text{x_3} & \text{F} &\text{F} \end{array}$$
LHS: $\mathbf{(P(x) ⟹ Q(x))}$ becomes TRUE for each value and thus $∀\mathbf{x(P(x) ⟹ Q(x))}$ become TRUE.

RHS: $\mathbf{(∀xP(x)) \text{ and } (∀xQ(x))}$ both becomes false for assumed values which implies $F→F$ and thus makes RHS to be TRUE.

$T→T$ makes statement B to be TRUE. Thus Answer is B.

by

Thanks for the video.

Can someone please provide a algebraic proof for option B?

When I try to derive it, I get,

$\forall x [ P(x) \rightarrow Q(x) ] \ \rightarrow \ [\exists x P(x) \rightarrow \forall x Q(x)]$

$\exists$(x) means true for atleast one x.

$\forall$(x) means true for each and every x.
This is what I was looking for. This can be applied to any problem and can be solved in a couple of minutes. Thanks a lot man!!

Sir @

typo in option C right side

F → T is T

Let P: Student is a girl.

and Q: Student is smart.

Option B says: IF for all student x if x is a girl then the student is smart THEN if the whole class comprises of girls then the whole class comprises of smart students.

check my comment above.
@bikram @manu00x can you explain why C & D are not true?
@arjun sir your approach to solve these type of question by taking a set of values and try to false is really awesome ,i always follow that approach and feel calm

For these kind of quetion use method.

Put RHS is false then try to make LHS is true . if this happen then statemnt is false otherwise true.

Or

Put LHS is True then try to make RHS is False . if this happen then statemnt is false otherwise true.

How? will you brief it?

Can u please brief on how to arrive at the solution using above method

@anirudh,
Can u pls explain that method using all the options ?   Because there is no resouces n=or nothing i knew which give a general easy approch to these kind of question :(

If you can do it would be a great help
yes prashant sir that is very nice approach Arjun sir and @bhuv explain that approach very clearly


text:
com.opengamma.strata.market.curve

## Class CurveGroupDefinitionBuilder

• java.lang.Object
• com.opengamma.strata.market.curve.CurveGroupDefinitionBuilder

• public final class CurveGroupDefinitionBuilder
extends Object
A mutable builder for creating instances of CurveGroupDefinition.
• ### Method Summary

All Methods
Modifier and Type Method and Description
CurveGroupDefinitionBuilder addCurve(CurveDefinition curveDefinition, Currency currency, RateIndex index, RateIndex... otherIndices)
Adds the definition of a curve to the curve group definition which is used to provide discount rates and forward rates.
CurveGroupDefinitionBuilder addCurve(CurveName curveName, Currency currency, RateIndex index, RateIndex... otherIndices)
Adds a curve to the curve group definition which is used to provide discount rates and forward rates.
CurveGroupDefinitionBuilder addDiscountCurve(CurveDefinition curveDefinition, Currency currency, Currency... otherCurrencies)
Adds the definition of a discount curve to the curve group definition.
CurveGroupDefinitionBuilder addDiscountCurve(CurveName curveName, Currency currency, Currency... otherCurrencies)
Adds the definition of a discount curve to the curve group definition.
CurveGroupDefinitionBuilder addForwardCurve(CurveDefinition curveDefinition, Index index, Index... otherIndices)
Adds the definition of a forward curve to the curve group definition.
CurveGroupDefinitionBuilder addForwardCurve(CurveName curveName, Index index, Index... otherIndices)
Adds the definition of a forward curve to the curve group definition.
CurveGroupDefinitionBuilder addSeasonality(CurveName curveName, SeasonalityDefinition seasonalityDefinition)
Adds a seasonality to the curve group definition.
CurveGroupDefinition build()
Builds the definition of the curve group from the data in this object.
CurveGroupDefinitionBuilder computeJacobian(boolean computeJacobian)
Sets the 'compute Jacobian' flag of the curve group definition.
CurveGroupDefinitionBuilder computePvSensitivityToMarketQuote(boolean computePvSensitivityToMarketQuote)
Sets the 'compute PV sensitivity to market quote' flag of the curve group definition.
CurveGroupDefinitionBuilder name(CurveGroupName name)
Sets the name of the curve group definition.
• ### Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Method Detail

• #### name

public CurveGroupDefinitionBuilder name(CurveGroupName name)
Sets the name of the curve group definition.
Parameters:
name - the name of the curve group, not empty
Returns:
this builder
• #### computeJacobian

public CurveGroupDefinitionBuilder computeJacobian(boolean computeJacobian)
Sets the 'compute Jacobian' flag of the curve group definition.
Parameters:
computeJacobian - the flag indicating if the Jacobian matrices should be computed and stored in metadata or not
Returns:
this builder
• #### computePvSensitivityToMarketQuote

public CurveGroupDefinitionBuilder computePvSensitivityToMarketQuote(boolean computePvSensitivityToMarketQuote)
Sets the 'compute PV sensitivity to market quote' flag of the curve group definition.

If set, the Jacobian matrices will also be calculated, even if not requested.

Parameters:
computePvSensitivityToMarketQuote - the flag indicating if present value sensitivity to market quotes should be computed and stored in metadata or not
Returns:
this builder

public CurveGroupDefinitionBuilder addDiscountCurve(CurveDefinition curveDefinition,
Currency currency,
Currency... otherCurrencies)
Adds the definition of a discount curve to the curve group definition.
Parameters:
curveDefinition - the discount curve configuration
otherCurrencies - additional currencies for which the curve can provide discount factors
currency - the currency for which the curve provides discount rates
Returns:
this builder

public CurveGroupDefinitionBuilder addDiscountCurve(CurveName curveName,
Currency currency,
Currency... otherCurrencies)
Adds the definition of a discount curve to the curve group definition.

A curve added with this method cannot be calibrated by the market data system as it does not include a curve definition. It is intended to be used with curves which are supplied by the user.

Parameters:
curveName - the name of the curve
otherCurrencies - additional currencies for which the curve can provide discount factors
currency - the currency for which the curve provides discount rates
Returns:
this builder

public CurveGroupDefinitionBuilder addForwardCurve(CurveDefinition curveDefinition,
Index index,
Index... otherIndices)
Adds the definition of a forward curve to the curve group definition.
Parameters:
curveDefinition - the definition of the forward curve
index - the index for which the curve provides forward rates
otherIndices - the additional indices for which the curve provides forward rates
Returns:
this builder

public CurveGroupDefinitionBuilder addForwardCurve(CurveName curveName,
Index index,
Index... otherIndices)
Adds the definition of a forward curve to the curve group definition.

A curve added with this method cannot be calibrated by the market data system as it does not include a curve definition. It is intended to be used with curves which are supplied by the user.

Parameters:
curveName - the name of the curve
index - the index for which the curve provides forward rates
otherIndices - the additional indices for which the curve provides forward rates
Returns:
this builder

public CurveGroupDefinitionBuilder addCurve(CurveDefinition curveDefinition,
Currency currency,
RateIndex index,
RateIndex... otherIndices)
Adds the definition of a curve to the curve group definition which is used to provide discount rates and forward rates.
Parameters:
curveDefinition - the definition of the forward curve
currency - the currency for which the curve provides discount rates
index - the index for which the curve provides forward rates
otherIndices - the additional indices for which the curve provides forward rates
Returns:
this builder

public CurveGroupDefinitionBuilder addCurve(CurveName curveName,
Currency currency,
RateIndex index,
RateIndex... otherIndices)
Adds a curve to the curve group definition which is used to provide discount rates and forward rates.

A curve added with this method cannot be calibrated by the market data system as it does not include a curve definition. It is intended to be used with curves which are supplied by the user.

Parameters:
curveName - the name of the curve
currency - the currency for which the curve provides discount rates
index - the index for which the curve provides forward rates
otherIndices - the additional indices for which the curve provides forward rates
Returns:
this builder

public CurveGroupDefinitionBuilder addSeasonality(CurveName curveName,
SeasonalityDefinition seasonalityDefinition)
Adds a seasonality to the curve group definition.
Parameters:
curveName - the name of the curve
seasonalityDefinition - the seasonality associated to the curve
Returns:
this builder
• #### build

public CurveGroupDefinition build()
Builds the definition of the curve group from the data in this object.
Returns:
the definition of the curve group built from the data in this object


text:
# "Prove the following equation is an identity"

#### materialartist09

I've always struggled with these kinds of problems, since there's so many kinds of ways to do it. Often, whenever I try to solve and prove identities, I end up nowhere. :/ So some tips and advice would be awesome.

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#### greg1313

Forum Staff
Using $\sin2x = 2\sin x\cos x$:

$$\sin^2 x=4\sin^2 \frac x2 \cos^2 \frac x2 \implies \sin^2 \frac x2 = \frac{\sin^2 x}{4\cos^2 \frac x2}$$

Using $\cos \frac x2 = \pm\sqrt{\frac{1 + \cos x}{2}}$ and a Pythagorean identity:

$$\frac{\sin^2 x}{4\cos^2 \frac x2} = \frac{1 - \cos^2 x}{2 + 2\cos x}$$

Divide top and bottom of $\frac{1 - \cos^2 x}{2 + 2\cos x}$ by $\sin^2 x$:

$$\sin^2 \frac x2 = \frac{\sin^2 x}{4\cos^2 \frac x2} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$$

#### materialartist09

$$\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}$$
I'm very confused, how did sin^2x equal that?

The only identity I see that matches it is a power-reducing formula which gives me 1-cos x over 2.

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#### romsek

Math Team
$$\displaystyle \sin^2\! x=4\sin^2\! \frac x2 \cos^2\! \frac x2 \implies \sin^2\! \frac x2 = \frac{\sin^2\! x}{4\cos^2\! \frac x2}$$
I'm very confused, how did sin^2x equal that?

The only identity I see that matches it is a power-reducing formula which gives me 1-cos x over 2.
$\sin^2(x) = \sin^2\left(2 \frac x 2\right) =$

$\left(\sin\left(2 \frac x 2\right)\right)^2 =$

$\left(2 \sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\right)^2 =$

$4 \sin^2\left(\frac x 2\right)\cos^2\left(\frac x 2\right)$

Last edited by a moderator:
1 person

#### Shariq Faraz

after writing the R.H.S. in terms of sin(x) and cos(x)

using the formula cos(2x) = 1 - 2(sinx)^2

#### Attachments

• 104.5 KB Views: 7
1 person

#### greg1313

Forum Staff
$$\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$$

1 person

#### studiot

I've always struggled with these kinds of problems, since there's so many kinds of ways to do it. Often whenever i try to solve and prove identities I end up nowhere :/ So some tips and advice would be awesome.
The thing to understand is the difference between an identity and an equality .

For an equation in x, like you have

An identity has the same value or result, or is true
for all x or for every x.

An equality has the same value or result, or is true
for some x but not for every x.

So to prove that an equation is an identity you aim is to prove that it holds good for every value of x.

So

$$\displaystyle {x^2} - 4 = 0$$

is true for some values of x and is an equation

but

$$\displaystyle {\left( {x - 4} \right)^2} = {x^2} - 8x + 16$$

is true for all values of x so is an identity

Does this help?

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#### skipjack

Forum Staff
In both of greg1313's solutions, there's a sign error in the final step.

Note that these solutions are easier to find by starting with the rather complicated right-hand side of the original equation and seeking to simplify it.

The right-hand side of the original equation has removable discontinuities where $\cos x$ = -1, so the original equation isn't, strictly speaking, an identity.

1 person

#### materialartist09

$$\displaystyle \sin^2 \frac x2 = \frac{1 - \cos x}{2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{2 + 2\cos x} = \frac{\csc^2 x - \cot^2 x}{2\csc^2 x - 2\csc x\cot x}$$
okay i understand, thank you so much!

I do have 1 more question though, looking at my sheet, I can't really figure out how 1-cos^2x over 2+2cosx equals the right side of the equation.

#### romsek

Math Team
okay i understand, thank you so much!

I do have 1 more question though, looking at my sheet, I can't really figure out how 1-cos^2x over 2+2cosx equals the right side of the equation.
divide everything by $\sin^2(x)$


text:
# Find: a² + b² + c²

#### Albert

##### Well-known member
$\dfrac {1234567891011121314151617} {7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=?$

(use of computer is not allowed!)

#### Opalg

##### MHB Oldtimer
Staff member
Re: find :a^2+b^2+c^2=?

$\dfrac {1234567891011121314151617} {7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=?$

(use of computer is not allowed!)
Reminds me of this thread. First step is to check that numerator and denominator have the same number of digits (25). Then my little pocket calculator (assuming that I'm allowed to use it) gives the approximations as $$\frac{1234}{7161}\approx 0.1723,$$ $$\frac{12345678}{71615141} \approx 0.17238.$$ It looks as though the first three digits are $1,\ 7,\ 2$, with $1^2+7^2+2^2 = 54.$ I have not tried to prove this carefully as in that previous thread.

#### Albert

##### Well-known member
Re: find :a^2+b^2+c^2=?

$\dfrac {1234567891011121314151617} {7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=?$

(use of computer is not allowed!)
Let: $\dfrac{1234}{7162}<A=\dfrac {1234567891011121314151617} {7161514131211101987654321}=0.abc----<\dfrac{1235}{7160}$

$0.1722<A<0.1725$

$\therefore a=1,b=7,c=2$

and $a^2+b^2+c^2=54$


text:
# Embedding initialization and layer initialization

i was making a nn with just categorical features , hence used nn.Embedding , after which i applied linear layer!
and i found out that the output distribution does not nave 1 as standard deviation, seems to me because embedding are initalized with normal(0,1) distribution and layer with uniform distribution!
hence if the std is not 1 with increasing depth of network std of output must tend towards 0 , hece vanishing gradients!!
so should i change the type of initialization or it will work fine , with no problems??
becuz to me seems to be a bit odd!!!

i tried a deep network of 8 layers and found that output std is 0.09 and same is the gradients std , hence it seems to me a vanishing gradients!


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# compute the Kähler moduli of an elliptic curve

Say given elliptic curve $\{ (x,y) | y^2 = (x^2-1)(x^2-k^2) \}$, what is the right form of the K$\ddot{a}$hler form and how to compute the K$\ddot{a}$hler moduli of this elliptic curve? Thank you.

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Put it into the more usual form of $y^2=f(x)$ with $f$ cubic (see Cassels book on elliptic curves for how to do this; Cassels gives lots of transformations for getting general genus 1 curves into this form, including the type you have here) and then it's just $dy/x$ (with the new coordinates). – Kevin Buzzard Sep 14 '10 at 6:38
Kevin, you have described an element of $H^0(\Omega^1)$. Isn't the Kahler form a 2-form, giving the hyperplane class in $H^2$ (so it should be a (1,1) form on the elliptic curve, not a (1,0) form). – Emerton Sep 14 '10 at 7:01
@Dan: $H^{1,1}$ is 1 dimensional. You just have to write down the Kähler form. Then the Kähler moduli is just a 1 dimensional space given by scalings of the Kähler form. – Kevin H. Lin Sep 14 '10 at 8:25
@Emerton: sounds like I misunderstood the question. I'd delete my comment were it not for the fact that it would make your comment look meaningless :-) Yes, I described a global holomorphic 1-form. I thought these were called Kaehler 1-forms by some people and assumed this was what the questioner was asking about. – Kevin Buzzard Sep 14 '10 at 9:51
Can't we make use of Kevin's comment? Take his (1,0) form \alpha and set \omega = i * \alpha \wedge \overline{\alpha}. Then \omega is a real (1,1)-form, and Kahler if it is non-degenarate (and positive, but take -\omega if it's negative). – Gunnar Þór Magnússon Sep 14 '10 at 11:20

I see what you mean, but I am still confused with the Fubini-Study like Kahler form here. We can choose a volume form (Kahler form) up to scaling, so the volume (Kahler moduli) will be a multiple of some area A. On the other side, the complex structure is normalised as $\tau$ ( say our lattice is $Z+Z\tau$). By the mirror map, $\tau \arrow \rho= b + iA$, my question is how can I fix $b$ and $A$? – Dan Sep 14 '10 at 21:55


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Measurement of excitation spectra in the {}^{12}{\rm C}(p,d) reactionnear the \eta^{\prime} emission threshold

# Measurement of excitation spectra in the ${}^{12}$C$(p,d)$ reaction near the $η'$ emission threshold

## Abstract

Excitation spectra of were measured in the reaction near the emission threshold. A proton beam extracted from the synchrotron SIS-18 at GSI with an incident energy of 2.5 GeV impinged on a carbon target. The momenta of deuterons emitted at were precisely measured with the fragment separator FRS operated as a spectrometer. In contrast to theoretical predictions on the possible existence of deeply bound mesic states in carbon nuclei, no distinct structures were observed associated with the formation of bound states. The spectra were analyzed to set stringent constraints on the formation cross section and on the hitherto barely-known -nucleus interaction.

###### pacs:
13.60.Le, 14.40.Be, 25.40.Ve, 21.85.+d

Present address: ]GSI Helmholtzzentrum für Schwerionenforschung GmbH, Planckstraße 1, 64291 Darmstadt, Germany

-PRiME/Super-FRS Collaboration

The mass of the meson is exceptionally large (958 MeV/) compared with other mesons in the same pseudo-scalar multiplet. This large mass is known as the “U(1) problem” raised by Weinberg Weinberg1975 (). Since the origin of the exceptionally large mass is attributed to the effect of the anomaly in QCD (Quantum ChromoDynamics) under the presence of spontaneous breaking of chiral symmetry, it is natural to expect a weakening of such an anomaly effect in a nuclear medium, where chiral symmetry may partially be restored Hayano2010 (); Leupold2010 (). This suppression of the anomaly effect would lead to a large mass reduction in a nuclear medium Jido2012 (). Recent model calculations have predicted a 4–15% mass reduction at normal nuclear density Costa2005 (); Nagahiro2006 (); Bass2006 (); Sakai2013 (). Thus, the observation of such a modification would provide novel insights into strongly-interacting many-body systems and the QCD vacuum.

A mesic atom or nucleus, which is a bound state of a meson and a nucleus, will offer a testing ground for the investigation of in-medium meson properties which can be different from those in the QCD vacuum due to partial restoration of chiral symmetry Hayano2010 (); Leupold2010 (). For example, deeply-bound pionic atoms in nuclei, in which a modification of the isovector part of the s-wave pion-nucleus potential manifests itself, are very well-established systems Yamazaki2012 (); Suzuki:2002ae (). Distinct peak structures corresponding to pionic states with different configurations, especially pions in the atomic state, can be identified in the missing-mass spectrum of the (,) reaction Gilg1999ua (). In contrast to invariant-mass spectroscopy of hadrons decaying into multi-particle final states in a nuclear medium, missing-mass spectroscopy is an alternative approach providing experimental access to in-medium meson properties.

In addition a quantum many-body system, such as an atomic nucleus, exhibits various phenomena which cannot be observed in a two-body system. The strong interaction is indeed “strong” in that a typical binding energy of a particle relative to its rest energy is large, in contrast to the electro-magnetic interaction which binds an electron and an atomic nucleus. Hence, the sub-threshold behavior of two-body interactions is of importance in building up a many-body system. This may be also the case for the -nucleon interaction.

The mass reduction of the meson leads to an attractive interaction between an meson and a nucleus. Hence, an experimental observation of -nucleus bound states may provide a direct clue to deduce the mass at normal nuclear density. It should be noted that a clear distinction of neighboring levels in a spectrum requires a rather narrow absorption width. Numerical results of an -nucleus optical potential based on a chiral unitary model show that the depth of the imaginary part of the potential (half absorption width) is much smaller than that of the real part (mass reduction), regardless of the strength of the two-body interaction Nagahiro2012 ().

Experimental information on low-energy -nucleus interaction is very limited. A recent measurement of photoproduction and the emission from nuclei at CBELSA/TAPS was utilized to evaluate the -nucleus optical potential for the first time Nanova2013 (). The absorption width, for the average momentum of , was deduced to be 15–25 at normal nuclear density, from a measurement of the transparency ratio of mesons propagating in various nuclear targets Nanova2012 (). Furthermore, the excitation function and the momentum distribution of mesons, produced and emitted from , were compared with a model calculation with the real part of the potential as a free parameter leading to a potential depth of  Nanova2013 (). In contrast, a small scattering length with a real part consistent with zero was obtained from the excitation function of the reaction very close to the threshold Czerwinski2014 ().

As can be inferred from the -nucleon and -nucleus interactions, which have been investigated in a variety of reactions  Machner2015 (); Haider2015 (), it is far from straightforward to extrapolate the on-shell scattering amplitude to the subthreshold region, pointing to the importance of meson-nucleus bound systems, serving as a unique probe of subthreshold behavior.

A direct measurement of the in-medium mass reduction was proposed by spectroscopy of bound states in a nucleus by measuring an excitation spectrum of in the reaction near the emission threshold Itahashi2012 (). Theoretical calculations indicated the possible existence of such bound states. One based on Nambu–Jona-Lasinio model predicted a sufficiently attractive potential of  MeV Costa2005 (); Nagahiro2006 () to accommodate in a carbon nucleus provided that the absorption is not too strong. Considerations on the momentum transfer and the differential cross section of the elementary reaction led to a preferable incident proton energy of 2.5 GeV.

As discussed in Ref. Itahashi2012 (), excitation energy spectra were predicted for different assumed -nucleus interactions expressed by attractive real and absorptive imaginary potential depths MeV at the nuclear center. The predicted spectra demonstrated the experimental sensitivity for the bound states and the advantage of an unbiased spectral analysis in an inclusive measurement that surpasses the disadvantages of a small signal-to-noise ratio mainly arising from large cross sections of multiple pion production in the reaction. Thus, we aimed at a region where attraction is relatively strong and absorption is weak by achieving extremely good statistics of relative errors and a moderate resolution of  MeV over the spectral region of interest.

The excitation spectra were measured near the emission threshold using the reaction at zero degrees. We employed a proton beam with the energy of  MeV extracted from the synchrotron SIS-18 at GSI, Darmstadt, impinging on a carbon target of natural isotopic composition with the thickness of g/cm. The beam intensity was /s and the spill length and the cycle were 4 s and 7 s, respectively. The intensity was directly measured with an uncertainty of % by inserting a detector SEETRAM Jurado2002 () in the beam near the target. The typical horizontal beam spot size was 1 mm.

The emitted deuterons had a momentum of  MeV/ at the production threshold after energy loss in the target and were momentum-analyzed by the FRS Geissel:1991zn () used as a spectrometer with a specially developed ion optical setting. The central focal plane (F2) was momentum-achromatic and the final focal plane (F4) was dispersive with a designed momentum resolving power of . The F4 dispersion was measured to be  cm/%. The deuteron tracks were measured by two sets of MWDCs (multi-wire drift chambers) separately installed at a distance of about one meter near F4 as depicted in Fig. 1.

Four sets of plastic scintillation counters installed at F2 and F4 identified the particles by measuring the time-of-flight (TOF) for signal deuterons ( 150 ns) and for background protons ( 132 ns). Count rates at F4 during the spill extraction were kHz for protons and kHz for deuterons. 99.5% of the protons were rejected in the TOF based triggers while % of deuterons were selected. The DAQ live rate ranged between 30–40% and the recorded trigger rate varied around 1 kHz during the spill extraction. After further selection based on the waveform analyses of PMT signals of the scintillation counters rejecting multiple hits, we achieved deuteron overkill in the analysis with a very small proton contamination fraction of the deuteron identified events.

During the measurement, we accumulated data to cover a wider energy region between and MeV around the emission threshold by scaling all FRS magnets with seven factors between 0.980 and 1.020. An excitation energy range of  MeV was covered in one setting by the central acceptance region, where one could safely rely on the momentum acceptance of the spectrometer. The momentum acceptance was determined by a code MOCADI Iwasa:1997bm () simulating the particle tracks based on the geometries of the magnets, the ion optical transport, the apertures of the beam pipes, effects of the materials, and the detector performances. The uncertainties in the acceptance estimation were taken into account in the subsequent analyses.

The central momenta of the seven settings were calibrated by using an elastic D reaction on a deuterated polyethylene (CD) target with the thickness of g/cm using a  MeV proton beam. Nearly mono-energetic deuterons with the momentum of  MeV/ were emitted forward within the solid angle of the FRS (horizontal: mrad and vertical: mrad). We also evaluated the ion optical aberrations from the measured correlations between positions and angles and adopted the corrections. The calibration data were taken every eight hours during the total 70.5 hours of production measurement and confirmed the stability of the spectrometer system.

The momentum spectra for seven central-momentum settings were combined after the acceptance and optical aberration corrections by fitting the spectra in the overlap regions of neighboring settings. Note here that this procedure decreased the degrees of freedom of the data, which turned out to cause minor influences as we discuss below.

Figure 2 (top) shows the measured excitation spectrum of the reaction near the emission threshold. The excitation energy relative to the threshold  MeV is shown by the bottom axis and the deuteron momentum by the top axis. The ordinate is the double differential cross section of the reaction at zero degrees which has an error of % in the absolute scale mainly as a result of the uncertainties in the incident beam intensities and in the solid angle. The systematic error associated with the excitation energy is deduced to be 1.7 MeV, mainly owing to the uncertainties of the beam energies.

No distinct narrow structure has been observed in the excitation spectrum in spite of the extremely good statistical sensitivity at a level of better than 1%. The measured cross section steadily increases from 4.9 to 5.7 b/(sr MeV) within the measured excitation energy region from to MeV and agrees within an order of magnitude with the simulated cross sections of the quasi-free processes  Itahashi2012 (), where denotes a nucleon in a carbon nucleus. The measured spectrum has been fitted over the whole region by a third-order polynomial displayed as solid gray curve where and the number of degrees of freedom are 119 and 121, respectively. For positive excitation energies, the fit may include increasing contributions from quasi-free production, estimated to reach the order of some 10 nb/(sr MeV) at MeV Nagahiro2013 (). The fitting residues are shown in the bottom panel with envelopes of two standard deviations. No significant structure is observed. Note that the jumps in the envelope reflect edges between two neighboring spectrometer settings.

The inset displays the measured momentum distribution of elastically scattered deuterons in a calibration measurement D with the CD target after subtracting the carbon contribution, demonstrating the symmetrical response of the momentum measurement. The energy resolution during the production runs has been estimated by fitting the spectrum with a Gaussian to yield  MeV over the whole energy region after considering the energy losses in the targets and the momentum spreads of the incident beams.

Since no distinct structure has been observed in the spectrum, we have deduced upper limits of the formation cross section of mesic nuclei as a function of the excitation energy and the width (FWHM). We have assumed that the spectrum has two components, a smooth continuous part that is expressed by a third-order polynomial and a formation cross section of -mesic nuclei given by a Voigt function, i.e. a Lorentzian with the width folded by a Gaussian with the width of . The spectrum has been fitted by the sum of the two components within a region of 35 MeV around the Lorentzian center with the height of the Voigt function and four coefficients of the polynomial as free parameters. We have evaluated upper limits at the 95% confidence level assuming the probability density functions to be Gaussian, integrating them in the physical non-negative regions, and finding points where 95% of the probability-density-function areas have been covered by the regions beneath.

Repeating the above procedure by changing the Lorentzian positions and widths, we have obtained upper limits for the region and  MeV as presented by the solid curves in Fig. 3. The limits in the differential cross section / are indicated by the vertical axis on the left and those in the Lorentzian heights /( ) by the three scales on the right side for each . Typical systematic errors are indicated by the vertical bars, which arise from the uncertainties in the beam intensity, the beam energy, the spectrometer acceptance, and the spectral resolution, and by moving the fitting boundaries by 5 MeV. We have set upper limits particularly rigid near the emission threshold: 0.1–0.2 b/sr for MeV, 0.2–0.4 b/sr for MeV, and 0.3–0.6 b/sr for MeV. The above analysis has been checked by a nearly identical procedure on the separate spectra for the seven spectrometer settings. Simultaneous fitting on each spectrum has yielded results shown by the dashed curves in Fig. 3, which are consistent with those in the analysis of the combined spectrum.

The resulting upper limits near the threshold of the peak height nb/(sr MeV) exclude the existence of narrow peak structures with the height as large as 40 nb/(sr MeV) expected for potential parameters  MeV Nagahiro2013 (), which was predicted by a theoretical calculation based on the NJL model Nagahiro2006 () and the absorption width suggested by the measured transparency ratio Nanova2012 ().

For further discussion of the constraints set on the -nucleus interaction, we have compared the combined spectrum with theoretical spectra described in Ref. Nagahiro2013 () in a space of potential parameters . For each potential-parameter combination of  MeV  MeV and  MeV  MeV, the spectrum has been fitted in an energy region between and MeV by a sum of a third-order polynomial and a theoretical spectrum scaled by a scale-parameter and folded by a Gaussian with the spectral resolution . Wider ( MeV) and narrower ( MeV) fit-regions have also been tested. In a similar way as above, upper limits of the scale-parameter have been evaluated at the 95% C.L. and are displayed on a potential-parameter plane in Fig. 4 after linear-interpolation between the calculated points. The dashed curves show in a band accounting for the estimated systematic errors.

Looking closely at Fig. 4, one finds that more stringent constraints (expressed by smaller ) are set for larger and smaller . Potential-parameter sets giving smaller than 1 are excluded by the 95% C.L. within the present analysis. Note here that theoretical calculations are subject to an uncertainty1 of a factor of 2 originating in the estimate of the cross section of the elementary process of 30 b/sr Itahashi2012 (), which has not yet been determined experimentally, and the experimental determination is of particular importance. Thus, the contour, for instance, corresponds to a C.L. 95% upper limit for the case that the absolute theoretical cross section was over-estimated by a factor of .

In conclusion, we have conducted a high accuracy measurement of the excitation spectrum of the reaction near the emission threshold. We accomplished targeted statistical significance, spectral resolution, and overall accuracy in the measurement. No distinct structures are observed in the spectrum. Thus, a strongly attractive potential of MeV predicted by the NJL model Nagahiro2006 () is rejected for a relatively shallow imaginary potential of MeV Nanova2012 (). The present experiment has only limited sensitivity for relatively weak attraction implied by the scattering length Czerwinski2014 () and suggested by the photoproduction experiments Nanova2013 (). In the near future, we will extend the experimental sensitivity by constructing a detector system to efficiently select events originating from the formation of -mesic nuclei by tagging the decay particles in an experiment at GSI/FAIR. We also consider possibilities of using other reaction channels such as .

The authors acknowledge the GSI staffs for their support and staffs of Institut für Kernphysik, Forschungszentrum Jülich for a test at COSY. We also thank Dr. M. Tabata for a specially developed aerogel. Y.K.T. acknowledges financial support from Grant-in-Aid for JSPS Fellows (No. 258155) and H.F. from Kyoto University Young Scholars Overseas Visit Program. This work is partly supported by a MEXT Grants-in-Aid for Scientific Research on Innovative Areas (Nos. JP24105705 and JP24105712), JSPS Grants-in-Aid for Scientific Research (S) (No. JP23224008) and for Young Scientists (A) (No. JP25707018), by the National Natural Science Foundation of China (No. 11235002) and by the Bundesministerium für Bildung und Forschung.

### Footnotes

1. From Fig. 3 in Ref. Grishina2000 (), one can read the boundaries in the total cross section of for the c.m. energy region of interest. The resultant range is in agreement with a calculation by K. Nakayama Nakayama2012 ().

### References

1. S. Weinberg, Phys. Rev. D 11, 3583 (1975).
2. R. S. Hayano and T. Hatsuda, Rev. Mod. Phys. 82, 2949 (2010).
3. S. Leupold, V. Metag, and U. Mosel, Int. J. Mod. Phys. E 19, 147 (2010).
4. D. Jido, H. Nagahiro, and S. Hirenzaki, Phys. Rev. C 85, 032201(R) (2012).
5. P. Costa, M. C. Ruivo, C. A. de Sousa, and Yu. L. Kalinovsky, Phys. Rev. D 71, 116002 (2005).
6. H. Nagahiro, M. Takizawa, and S. Hirenzaki, Phys. Rev. C 74, 045203 (2006).
7. S. D. Bass and A. W. Thomas, Phys. Lett. B634, 368 (2006).
8. S. Sakai and D. Jido, Phys. Rev. C 88, 064906 (2013).
9. K. Suzuki et al., Phys. Rev. Lett. 92, 072302 (2004).
10. T. Yamazaki, S. Hirenzaki, R. S. Hayano, and H. Toki, Phys. Rep. 514, 1 (2012).
11. H. Gilg et al., Phys. Rev. C 62, 025201 (2000) and K. Itahashi et al., ibid., 025202.
12. H. Nagahiro, S. Hirenzaki, E. Oset, and A. Ramos, Phys. Lett. B709, 87 (2012).
13. M. Nanova et al., Phys. Lett. B727, 417 (2013).
14. M. Nanova et al., Phys. Lett. B710, 600 (2012).
15. E. Czerwiński et al., Phys. Rev. Lett. 113, 062004 (2014).
16. H. Machner, J. Phys. G 42, 043001 (2015).
17. Q. Haider and L.-C. Liu, Int. J. Mod. Phys. E 24, 1530009 (2015).
18. K. Itahashi et al., Prog. Theor. Phys. 128, 601 (2012).
19. B. Jurado, K.-H. Schmidt, and K.-H. Behr, Nucl. Instrum. Meth. A483, 603 (2002).
20. H. Geissel et al., Nucl. Instrum. Meth. B70, 286 (1992).
21. N. Iwasa et al., Nucl. Instrum. Meth. B126, 284 (1997).
22. H. Nagahiro et al., Phys. Rev. C 87, 045201 (2013).
23. V. Yu. Grishina et al., Phys. Lett. B475, 9 (2000).
24. K. Nakayama, private communication (2012).
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# Revision history [back]

This expression has no analytical solution, therefore you need to solve it numerically. For that, you can use find_root:

sol = find_root(0.08*x-0.0064 == log(x), 0,2); sol

which will give you,

1.0835990137881888

This expression has no analytical solution, therefore you need to solve it numerically. For that, you can use find_root: in a given range $0\lt x \lt 2$:

sol = find_root(0.08*x-0.0064 == log(x), 0,2); sol

which will give you,

1.0835990137881888


text:
# Question re interpreting MarkScheme - Moments (Edexcel A-level)

rsk
Attached is a Q from an AS mechanics paper. I'm perplexed about part (c) in which we are told that the tensions in the two cables are equal.
My physics brain says take moments about the CoM of the beam so eliminating the mass of the beam and making the moments of the two cables equal and opposite, which reduces the problem to a very simple one, and gives the correct answer.

Markscheme warns about 'correct' answers which ignore the mass of the beam and offers 4 different options for taking moments, none of them around the CoM of the beam. Can anyone tell me why this method is invalid?

#### Attachments

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Homework Helper
Gold Member
How is the warning exactly phrased? It may be an indirect warning not to ignore the mass of the beam. If the system is in equilibrium, the sum of the moments will be zero with reference to any point. No matter where you choose your reference, you will end up with two equations and two unknowns. Another possible reason for the warning is that if you take the reference point at the CoM of the rod, you will answer part (b) before part (a) and they don't want you to do that.

Staff Emeritus
Homework Helper
Your method is fine. The simplest explanation would be that whoever wrote that warning had a brain fart.

I just noticed on the requirements for the moments equation that "all terms required." Perhaps your method is acceptable but the solution requires you to explicitly include the term ##100 g(0)##. That, however, would seem inconsistent with the provided expressions as they don't include a term for the moment due to the tension when the lever arm is 0.

rsk
Homework Helper
Gold Member
Your method is fine. The simplest explanation would be that whoever wrote that warning had a brain fart.

I just noticed on the requirements for the moments equation that "all terms required." Perhaps your method is acceptable but the solution requires you to explicitly include the term ##100 g(0)##. That, however, would seem inconsistent with the provided expressions as they don't include a term for the moment due to the tension when the lever arm is 0.
I tried to be charitable but, after digesting your comments, I'll go with "brain fart."

rsk and vela
rsk
Thanks both.
I suppoe my method doesn't include those lines above where the vertical forces are balanced - but that's completely unnecessary when we don't need to know T.
I don't like to think that students would be penalised for this - for spotting the most efficient solucion - in an exam.

@kuruman - the warning is just what you see there on the bottom of the markscheme "watch out for "correct" answers which ignore the 100g" - so in other words it's a warning for the markers and not for the students.

Homework Helper
Gold Member
Let's hope that this scheme provides just guidelines in an attempt to achieve uniformity and that the markers would recognize a correct solution when they see one. Am I correct in assuming that the takers of this test see just their scores but not the marked papers?

rsk
rsk
Schools do have the right to request the papers (or copies) if they want to appeal grades but it's not the norm and there's a fee involved.

I also hope that a decent marker would accept this (or raise it up the chain of command) but the worry is always over who is marking - there are often too few experienced teachers willing to give up their holidays for this and the worry that it's someone else just applying the MS to the letter (that's why they are often so detailed).

I'll try not to worry about it. I must be in a good place at the moment if I can find time to worry about hypotheticals which will probably never happen.

rsk
An update. So concerned was I about good students possibly losing marks that I got in touch with the exam board.

They reply that the CoM method is indeed the best, but that some students don't see the best method so other methods have to be included in the MS.

I still think that the CoM method is so simple that the 6 marks are in no way justified and that whoever wrote the paper didn't spot that solution but I have at least been reassured that, at the marking stage, students would not have been penalised.

Homework Helper
Gold Member
There has to be a level of trust in the markers and in the belief that if a student obtains the correct numerical answer, one has to examine the solution carefully to determine if it is correct outside the marking criteria. By the same token, students need to document their solutions carefully. Here is an example of a simplified solution to part (c) that eliminates quite a bit of algebra and does not look like any of the suggested solutions.

1. The ropes are symmetrically disposed about the CoM of the rod and have equal tensions.
2. This means that if the ropes are removed and replaced by a pivot at the CoM of the rod, the assembly will remain at equilibrium.
3. The 60g gymnast is 2 m from the pivot whilst the 48g gymnast is at (x - 4) m from the pivot.
Thus, the moment balance equation is $$60g~\text{[N]}\times 2 ~\text{[m]}=48g~\text{[N]} \times (x-4~\text{[m]})\implies x=6.5~\text{m}.$$It is the marker's responsibility to override the various suggested solutions and give full marks for this one with perhaps a bonus for thinking outside the box. However, it behooves the student to provide statements 1 and 2 to justify the equation in 3.

rsk
Yes, and I can't see students giving statements 1 and 2 as you write them.
However I will ensure that from now I encourage to show this kind of thinking, albeit briefly
ie take moments about CoM: eliminate 100g force as moment =0; moments of Ts are equal and opposite.

In fact, I will keep this question to use as an example with them of why explicitly stating the steps in the reasoning is so important.

Last edited:
kuruman


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## Algebra 2 (1st Edition)

$$\frac{6x+18}{\left(x-3\right)\left(x+6\right)}$$
Ensuring that we have like denominators and then attempting to simplify, we find: $$\frac{4\left(x+6\right)}{\left(x-3\right)\left(x+6\right)}+\frac{2\left(x-3\right)}{\left(x+6\right)\left(x-3\right)}\\ \frac{4\left(x+6\right)+2\left(x-3\right)}{\left(x-3\right)\left(x+6\right)}\\ \frac{6x+18}{\left(x-3\right)\left(x+6\right)}$$