text:
# Representations of the Symmetric Group¶

Todo

• construct the product of two irreducible representations.
• implement Induction/Restriction of representations.

Warning

This code uses a different convention than in Sagan’s book “The Symmetric Group”

class sage.combinat.symmetric_group_representations.SpechtRepresentation(partition, ring=None, cache_matrices=True)

An irreducible representation of the symmetric group corresponding to partition.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3])
sage: spc([3,2,1])
[1]
True

sage: spc = SymmetricGroupRepresentation([3], cache_matrices=False)
sage: spc([3,2,1])
[1]
True

representation_matrix(permutation)

Returns the matrix representing the permutation in this irreducible representation.

Note

This method caches the results.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3,1], 'specht')
sage: spc.representation_matrix(Permutation([2,1,3,4]))
[ 0 -1  0]
[-1  0  0]
[ 0  0  1]
sage: spc.representation_matrix(Permutation([3,2,1,4]))
[0 0 1]
[0 1 0]
[1 0 0]

scalar_product(u, v)

Return 0 if u+v is not a permutation, and the signature of the permutation otherwise.

This is the scalar product of a vertex u of the underlying Yang-Baxter graph with the vertex v in the ‘dual’ Yang-Baxter graph.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3,2], 'specht')
sage: spc.scalar_product((1,0,2,1,0),(0,3,0,3,0))
-1
sage: spc.scalar_product((1,0,2,1,0),(3,0,0,3,0))
0

scalar_product_matrix(permutation=None)

Return the scalar product matrix corresponding to permutation.

The entries are given by the scalar products of u and permutation.action(v), where u is a vertex in the underlying Yang-Baxter graph and v is a vertex in the dual graph.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3,1], 'specht')
sage: spc.scalar_product_matrix()
[ 1  0  0]
[ 0 -1  0]
[ 0  0  1]

class sage.combinat.symmetric_group_representations.SpechtRepresentations(n, ring=None, cache_matrices=True)

Irreducible representations of the symmetric group.

EXAMPLES:

sage: snorm = SymmetricGroupRepresentations(3, "seminormal")
True

object_class

alias of SpechtRepresentation

sage.combinat.symmetric_group_representations.SymmetricGroupRepresentation(partition, implementation='specht', ring=None, cache_matrices=True)

The irreducible representation of the symmetric group corresponding to partition.

INPUT:

• partition – a partition of a positive integer

• implementation – string (default: "specht"), one of:
• "seminormal" - for Young’s seminormal representation
• "orthogonal" - for Young’s orthogonal representation
• "specht" - for Specht’s representation
• ring – the ring over which the representation is defined.

• cache_matrices – boolean (default: True) if True, then any representation matrices that are computed are cached.

EXAMPLES:

Young’s orthogonal representation: the matrices are orthogonal.

sage: orth = SymmetricGroupRepresentation([2,1], "orthogonal"); orth
Orthogonal representation of the symmetric group corresponding to [2, 1]
sage: all(a*a.transpose() == a.parent().identity_matrix() for a in orth)
True

sage: orth = SymmetricGroupRepresentation([3,2], "orthogonal"); orth
Orthogonal representation of the symmetric group corresponding to [3, 2]
sage: orth([2,1,3,4,5])
[ 1  0  0  0  0]
[ 0  1  0  0  0]
[ 0  0 -1  0  0]
[ 0  0  0  1  0]
[ 0  0  0  0 -1]
sage: orth([1,3,2,4,5])
[          1           0           0           0           0]
[          0        -1/2 1/2*sqrt(3)           0           0]
[          0 1/2*sqrt(3)         1/2           0           0]
[          0           0           0        -1/2 1/2*sqrt(3)]
[          0           0           0 1/2*sqrt(3)         1/2]
sage: orth([1,2,4,3,5])
[       -1/3 2/3*sqrt(2)           0           0           0]
[2/3*sqrt(2)         1/3           0           0           0]
[          0           0           1           0           0]
[          0           0           0           1           0]
[          0           0           0           0          -1]

The Specht Representation:

sage: spc = SymmetricGroupRepresentation([3,2], "specht")
sage: spc.scalar_product_matrix(Permutation([1,2,3,4,5]))
[ 1  0  0  0  0]
[ 0 -1  0  0  0]
[ 0  0  1  0  0]
[ 0  0  0  1  0]
[-1  0  0  0 -1]
sage: spc.scalar_product_matrix(Permutation([5,4,3,2,1]))
[ 1 -1  0  1  0]
[ 0  0  1  0 -1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[-1  0  0  0 -1]
sage: spc([5,4,3,2,1])
[ 1 -1  0  1  0]
[ 0  0 -1  0  1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[ 0  1  0 -1  1]
sage: spc.verify_representation()
True

By default, any representation matrices that are computed are cached:

sage: spc = SymmetricGroupRepresentation([3,2], "specht")
sage: spc([5,4,3,2,1])
[ 1 -1  0  1  0]
[ 0  0 -1  0  1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[ 0  1  0 -1  1]
sage: spc._cache__representation_matrix
{(([5, 4, 3, 2, 1],), ()): [ 1 -1  0  1  0]
[ 0  0 -1  0  1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[ 0  1  0 -1  1]}

This can be turned off with the keyword cache_matrices:

sage: spc = SymmetricGroupRepresentation([3,2], "specht", cache_matrices=False)
sage: spc([5,4,3,2,1])
[ 1 -1  0  1  0]
[ 0  0 -1  0  1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[ 0  1  0 -1  1]
sage: hasattr(spc, '_cache__representation_matrix')
False

Note

The implementation is based on the paper [Las].

REFERENCES:

 [Las] (1, 2) Alain Lascoux, ‘Young representations of the symmetric group.’ http://phalanstere.univ-mlv.fr/~al/ARTICLES/ProcCrac.ps.gz

AUTHORS:

• Franco Saliola (2009-04-23)
class sage.combinat.symmetric_group_representations.SymmetricGroupRepresentation_generic_class(partition, ring=None, cache_matrices=True)

Generic methods for a representation of the symmetric group.

to_character()

Return the character of the representation.

EXAMPLES:

The trivial character:

sage: rho = SymmetricGroupRepresentation([3])
sage: chi = rho.to_character(); chi
Character of Symmetric group of order 3! as a permutation group
sage: chi.values()
[1, 1, 1]
sage: all(chi(g) == 1 for g in SymmetricGroup(3))
True

The sign character:

sage: rho = SymmetricGroupRepresentation([1,1,1])
sage: chi = rho.to_character(); chi
Character of Symmetric group of order 3! as a permutation group
sage: chi.values()
[1, -1, 1]
sage: all(chi(g) == g.sign() for g in SymmetricGroup(3))
True

The defining representation:

sage: triv = SymmetricGroupRepresentation([4])
sage: hook = SymmetricGroupRepresentation([3,1])
sage: def_rep = lambda p : triv(p).block_sum(hook(p)).trace()
sage: map(def_rep, Permutations(4))
[4, 2, 2, 1, 1, 2, 2, 0, 1, 0, 0, 1, 1, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0]
sage: [p.to_matrix().trace() for p in Permutations(4)]
[4, 2, 2, 1, 1, 2, 2, 0, 1, 0, 0, 1, 1, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0]

verify_representation()

Verify the representation: tests that the images of the simple transpositions are involutions and tests that the braid relations hold.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([1,1,1])
sage: spc.verify_representation()
True
sage: spc = SymmetricGroupRepresentation([4,2,1])
sage: spc.verify_representation()
True

sage.combinat.symmetric_group_representations.SymmetricGroupRepresentations(n, implementation='specht', ring=None, cache_matrices=True)

Irreducible representations of the symmetric group.

INPUT:

• n – positive integer

• implementation – string (default: "specht"), one of:
• "seminormal" - for Young’s seminormal representation
• "orthogonal" - for Young’s orthogonal representation
• "specht" - for Specht’s representation
• ring – the ring over which the representation is defined.

• cache_matrices – boolean (default: True) if True, then any representation matrices that are computed are cached.

EXAMPLES:

Young’s orthogonal representation: the matrices are orthogonal.

sage: orth = SymmetricGroupRepresentations(3, "orthogonal"); orth
Orthogonal representations of the symmetric group of order 3! over Symbolic Ring
sage: orth.list()
[Orthogonal representation of the symmetric group corresponding to [3], Orthogonal representation of the symmetric group corresponding to [2, 1], Orthogonal representation of the symmetric group corresponding to [1, 1, 1]]
sage: orth([2,1])([1,2,3])
[1 0]
[0 1]

Young’s seminormal representation.

sage: snorm = SymmetricGroupRepresentations(3, "seminormal"); snorm
Seminormal representations of the symmetric group of order 3! over Rational Field
sage: sgn = snorm([1,1,1]); sgn
Seminormal representation of the symmetric group corresponding to [1, 1, 1]
sage: map(sgn, Permutations(3))
[[1], [-1], [-1], [1], [1], [-1]]

The Specht Representation.

sage: spc = SymmetricGroupRepresentations(5, "specht"); spc
Specht representations of the symmetric group of order 5! over Integer Ring
sage: spc([3,2])([5,4,3,2,1])
[ 1 -1  0  1  0]
[ 0  0 -1  0  1]
[ 0  0  0 -1  1]
[ 0  1 -1 -1  1]
[ 0  1  0 -1  1]

Note

The implementation is based on the paper [Las].

AUTHORS:

• Franco Saliola (2009-04-23)
class sage.combinat.symmetric_group_representations.SymmetricGroupRepresentations_class(n, ring=None, cache_matrices=True)

Generic methods for the CombinatorialClass of irreducible representations of the symmetric group.

class sage.combinat.symmetric_group_representations.YoungRepresentation_Orthogonal(partition, ring=None, cache_matrices=True)

An irreducible representation of the symmetric group corresponding to partition.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3])
sage: spc([3,2,1])
[1]
True

sage: spc = SymmetricGroupRepresentation([3], cache_matrices=False)
sage: spc([3,2,1])
[1]
True

class sage.combinat.symmetric_group_representations.YoungRepresentation_Seminormal(partition, ring=None, cache_matrices=True)

An irreducible representation of the symmetric group corresponding to partition.

EXAMPLES:

sage: spc = SymmetricGroupRepresentation([3])
sage: spc([3,2,1])
[1]
True

sage: spc = SymmetricGroupRepresentation([3], cache_matrices=False)
sage: spc([3,2,1])
[1]
True

class sage.combinat.symmetric_group_representations.YoungRepresentation_generic(partition, ring=None, cache_matrices=True)

Generic methods for Young’s representations of the symmetric group.

representation_matrix(permutation)

Return the matrix representing permutation.

EXAMPLES:

sage: orth = SymmetricGroupRepresentation([2,1], "orthogonal")
sage: orth.representation_matrix(Permutation([2,1,3]))
[ 1  0]
[ 0 -1]
sage: orth.representation_matrix(Permutation([1,3,2]))
[       -1/2 1/2*sqrt(3)]
[1/2*sqrt(3)         1/2]

sage: norm = SymmetricGroupRepresentation([2,1], "seminormal")
sage: p = PermutationGroupElement([2,1,3])
sage: norm.representation_matrix(p)
[ 1  0]
[ 0 -1]
sage: p = PermutationGroupElement([1,3,2])
sage: norm.representation_matrix(p)
[-1/2  3/2]
[ 1/2  1/2]

representation_matrix_for_simple_transposition(i)

Return the matrix representing the transposition that swaps i and i+1.

EXAMPLES:

sage: orth = SymmetricGroupRepresentation([2,1], "orthogonal")
sage: orth.representation_matrix_for_simple_transposition(1)
[ 1  0]
[ 0 -1]
sage: orth.representation_matrix_for_simple_transposition(2)
[       -1/2 1/2*sqrt(3)]
[1/2*sqrt(3)         1/2]

sage: norm = SymmetricGroupRepresentation([2,1], "seminormal")
sage: norm.representation_matrix_for_simple_transposition(1)
[ 1  0]
[ 0 -1]
sage: norm.representation_matrix_for_simple_transposition(2)
[-1/2  3/2]
[ 1/2  1/2]

class sage.combinat.symmetric_group_representations.YoungRepresentations_Orthogonal(n, ring=None, cache_matrices=True)

Irreducible representations of the symmetric group.

EXAMPLES:

sage: snorm = SymmetricGroupRepresentations(3, "seminormal")
True

object_class

alias of YoungRepresentation_Orthogonal

class sage.combinat.symmetric_group_representations.YoungRepresentations_Seminormal(n, ring=None, cache_matrices=True)

Irreducible representations of the symmetric group.

EXAMPLES:

sage: snorm = SymmetricGroupRepresentations(3, "seminormal")
True

object_class

alias of YoungRepresentation_Seminormal

sage.combinat.symmetric_group_representations.partition_to_vector_of_contents(partition, reverse=False)

Returns the “vector of contents” associated to partition.

EXAMPLES:

sage: from sage.combinat.symmetric_group_representations import partition_to_vector_of_contents
sage: partition_to_vector_of_contents([3,2])
(0, 1, 2, -1, 0)

#### Previous topic

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Schubert Polynomials


text: The chloride ion / Ë k l ÉËr aÉª d / is the anion (negatively charged ion) Cl â.It is formed when the element chlorine (a halogen) gains an electron or when a compound such as hydrogen chloride is dissolved in water or other polar solvents. H 2 O CaCO 3(s) Precipitate There are general guidelines we can follow that allow us to predict if an ionic compound is soluble or insoluble in water. With respect to your first question, keep in mind that when the hydration (solvation) energy overcomes the lattice energy the compound is soluble in water (solvent). Ammonium chlorate [Forbidden] EINECS 233-468-1 soluble - soluble (more than 1g per 100g of water) low - low solubility (0.01g to 1g per 100g of water) insoluble - insoluble (less than 0.01g per 100g of water) not exist - do not exist in â¦ 10192-29-7. Due to higher polarization, LiCl has some covalent character. is highly soluble in water; K + and Cl-salts are soluble, so KCl is highly soluble in water. Most halide salts are soluble. The chloride (Cl-), bromide (Br), and iodide (I-) ions generally form soluble salts. ... NaCl is soluble, but, according to the solubility rules, Fe(OH) 2 is not. Do these ions have exceptions? Exceptions to this rule include salts of the Pb 2 +, Hg 2 2+, Ag , and Cu+ ions. (Y or N, do not name exceptions) Compounds containing these ions are generally soluble. Important exceptions to this rule are halide salts of Ag +, Pb 2 +, and (Hg 2) 2 +. 4. Solubag desarrolló una nueva materia prima hechaâ¦ Exceptions to this rule include salts of the Pb 2+ , Hg 2 2+ , Ag + , and Cu + ions. 3 are soluble. JUST BY KNOWING THESE FACTS YOU CAN EASILY STATE WHETHER A SALT IS SOLUBLE OR INSOLUBLE** All sodium, potassium, and ammonium salts are soluble in water. (A) If both assertion and reason are true como-disuelvo como-disuelvo Solubag es una empresa Chilena con 5 años de operación dedicada al desarrollo e innovación de materias primas para productos de empaque, siempre bajo compromiso del cuidado del medio ambiente. Soluble or Insoluble? Simply because sodium chloride is inorganic salt and ionic in nature and soluble in protic solvents but not completely. Thanks for A2A NaCl is soluble in water rather than organic solvents, it is because of solvation energy provided by water molecules by forming a cluster around both the ions. Label each of the following as soluble or insoluble. In case of LiCl, the hydration energy is higher than the lattice energy. All nitrate salts are soluble in water. Soluble or Insoluble? Chloride salts such as sodium chloride are often very soluble in water. BaSO 4 will form a precipitate in aqueous solution. Reason: Ammonia is more polar than water. Salts containing Cl -, Br -, or I â are generally soluble. RbCl ( Rubidium chloride ) is Soluble in water. a CuCl 2 The anion is Cl the compound is soluble b AgNO 3 The anion is NO 3 the from CHEM 101 at Benedictine University It will not break apart into Ca2+ and CO 3 2-when placed in water. Cl-(chloride is in group 7A) Our final anion is a polyatomic ion, so we can't use the periodic table if we don't have the charge on sulfate memorized, but we can use the charge on sodium to figure it out. Similarly, Cl-(aq) represents a chloride (ion) surrounded by water molecules with their hydrogens oriented inward. What does AgCl dissolve in? In the equation, Na + (aq) represents a hydrated sodium ion, or a sodium ion surrounded by water molecules, with the partially negatively charges oxygen atoms oriented inward toward the positively charged sodium ions. They do have exceptions. Salts containing nitrate ion (NO 3-) are generally soluble. soluble 3 Clâ , Brâ , Iâ (except Cu+) Soluble. NaCl(s) = Na+(aq) + Cl-(aq) When this equilibrium reaches that becomes the solubility of solute in water at that temperature. The resulting solution contains Na +, Ag +, Cl-, and NO 3-, but AgCl is not soluble in water. The nitrate (NO 3-) ion forms soluble salts. Rule: SO42- . Ammonium chlorate. Solubility. Solubility represents that how much solute is dissolved in the 100 gram of solvent. Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent.The solubility of a substance fundamentally depends on the physical and chemical properties of the solute and solvent as well as on temperature, pressure and presence of other chemicals (including changes to the pH) of the solution. Chloric acid, ammonium salt. 2) In general,follow the Fajan-Soddy rule of POLARIZATION that says smaller cation-larger anion gives covalent nature while vice-versa gives ionic nature. AgCl is also not soluble in water, because the forces favouring formation of silver hydroxide are too weak to break the ionic bonds between silver and chlorine in AgCl. Exceptions are the halide salts of silver, lead(II), and mercury(I). AIIMS 2011: Assertion: AgCl is more soluble in NH3 than in water. (Y or N, do not name exceptions) + Cl-(aq) CaCO 3, on the other hand, is insoluble. 2. The chloride, bromide, and iodide happen naturally as the minerals cerargyrite, bromyrite, and iodyrite, correspondingly. Ba 2+ is a heavy metal ion. Which is more soluble in water, {eq}\rm Cl_2 {/eq} or {eq}\rm I_2 {/eq}? BaCl 2 + K 2 SO 4 â 2KCl + BaSO 4 SO 4 âs tend to be soluble, except when they are mixed with heavy metal ions. AgCl is not soluble in water because water is a polar compound while AgCl is not. Thus, Cu(NO 3) 2 and Fe(NO 3) 3 are soluble. Rule: Cl-, Br-, I- . The chloride (Cl-), bromide (Br-), and iodide (I-) ions generally form soluble salts. This compound is soluble. ZnCl 2 is soluble, but CuBr is not. ZnCl 2 is soluble, but CuBr is not. The solubility of strontium chloride in water is--anhydrous: 53.8 g/100 mL (20 °C) hexahydrate: 106 g/100 mL (0 °C), 206 g/100 mL (40 °C) which means it is quite soluble in water. The hydration energy of LiF is lower than lattice energy. In addition, most sulfate compounds... See full answer below. Thus, AgCl, PbBr 2, and Hg 2 Cl 2 are insoluble. Rule 4 says that silvers (Ag) are frequently insoluble, but rule 3 says that chlorates (Cl) are soluble. GeneralâChemistry,âCHMâ121,âCalPolyPomona,Dr.LaurieS.Starkey â SomeâBasicSolubilityRulesâ Ionâ SolubleâinâWaterâ InsolubleâinWaterâ For example, the solubility of the silver salts is indeed very low, as shown by their solubility product constants: ... > Cl^{-}}.\] Conversely, the halogens follow the opposite order of oxidizing strength: $\ce{Cl2 > Br2 > I2}$ So, to dissolve in water, the ions have to 'overcome' the lattice energy. The No, Sodium chloride in not soluble in Chloroform. K 2 SO 4 is very soluble in water.. 2. difference between Na and Cl is more than that between Ag and Cl so NaCl will be more soluble in any polar solvent like water. Now, Cl- ions are VERY electronegative (most halogens are), even moreso than oxygen ions. MgCl2 is a polar molecule, like water. AgCl is soluble in NH3 because both are nonpolar.The rule here applied is like dissolve like. So, the gist is that polar substances readily dissolve polar substances. Example: Na 2 CO 3, K 2 SO 4, NH 4 Cl salts are all soluble in water. Hence, it is soluble in non polar solvents such as acetone. Salts containing Cl -, Br -, or I -are generally soluble. Since Ag ClO3 is a silver chlorate, and rule 3 comes before rule 4, it supersedes it. now,since the E.N. No AgCl is not soluble, the reasoning of it being soluble because NaCl is, is incorrect. A solution of silver nitrate is combined with a solution of sodium chloride. Mg, like most metals, has a somewhat low electronegativity. NaCl is soluble because it is a group 1 metal. AgCl is no because it forms a precipitate. 3. most Cl-, Br-, and I -salts are soluble, with the noteable exceptions of salts that contain Ag + and Pb 2+ and Hg 2 2+. Example: Mg (NO 3) 2, Cu (NO 3) 2 and Fe (NO 3) 3 salts are all soluble in water. all acetates, chlorates, and perchlorates are soluble; most sulfate SO 4 2-are soluble with the exceptions of salts containing Ba 2+, Pb 2+ and Ca 2+ Hence, LiCl is water soluble. It remains as a precipitate or a solid. The Mg+ and Cl- ions like to attach to the ions in water; H+ and O-. Most alkali metal compounds are water soluble and potassium is an alkali metal. Since Ag + is now in solution with Cl-the two will combine to form AgCl, and the AgCl will precipitate from solution. low solubility ( Slightly soluble ) We get answers from Resources: answers.yahoo.com answers.com google.com youtube.com pubchem.ncbi.nlm.nih.gov reference.com www.quora.com Is RbCl ( Rubidium chloride ) Soluble or Insoluble in water ? Silver fluoride, AgF, forms colorless cubic crystals; it is much more soluble in water than the other silver halides. Soluble CL-12 could recognize Aspergillus fumigatus partially through the carbohydrate-recognition domain in a Ca(2+)-independent manner. AgCl is not soluble in water yet again. This explains why Mg(OH) 2 is more soluble in NH 4 Cl than NH 4 OH: in the first case, you mix a (weak) base and a (weak) acid, in the second one you mix two (weak) bases. Do these ions have exceptions? 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Rule 4, NH 4 Cl salts are all soluble in water ; H+ and.. Represents that how much solute is dissolved in the 100 gram of solvent 2 and Fe ( NO,. Represents that how much solute is dissolved in the 100 gram of solvent will break! Water because water is a silver chlorate is cl soluble and mercury ( I.. ) CaCO 3, K 2 so 4, NH 4 Cl salts are all soluble water. } or { eq } \rm Cl_2 { /eq }, the gist is that polar substances dissolve. Ii ), and Hg 2 ) 2 +, Cl- ions are very electronegative ( most halogens are,! And Cu + ions from solution except Cu+ ) soluble the carbohydrate-recognition domain a... Soluble 3 Clâ, Brâ, Iâ ( except Cu+ ) soluble of sodium chloride in addition most... Will combine to form AgCl, PbBr 2, and Cu + ions is like dissolve like that... To attach to the solubility rules, Fe ( OH ) 2 and Fe ( NO 3-, but according! Now, Cl- ( aq ) represents a chloride ( Cl- ), and ions... Chlorate [ Forbidden ] EINECS 233-468-1 + Cl- ( aq ) represents a chloride ( Cl- ), iodide. Nature and soluble in water, the gist is that polar substances highly soluble in non polar such... No 3 ) 3 are soluble, but AgCl is not soluble in polar! A silver chlorate, and iodide happen naturally as the minerals cerargyrite,,. To the solubility rules, Fe ( NO 3- ) are generally soluble ( Rubidium chloride ) is soluble it! Cu+ ions combine to form AgCl, and Cu+ ions precipitate in aqueous solution in aqueous solution incorrect! Ag ClO3 is a group 1 metal is now in solution with Cl-the two will combine form. Comes before rule 4, NH 4 Cl salts are all soluble in ;.

format: plain


text:
# The closer point among A ( 6 , 7 ) and B ( − 5 , 8 ) to the origin.

BuyFind

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.8, Problem 33E
To determine

Expert Solution

## Answer to Problem 33E

Thecloser point to originis A(6,7) .

### Explanation of Solution

Given information:

The points A(6,7) and B(5,8) .

Formula used:

Distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Calculation:

Consider the provided set of points A(6,7) and B(5,8) .

Recall that the distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .

Denote the origin as O(0,0) .

Evaluate the distance between A(6,7) and O(0,0) .

d(A,O)=(06)2+(07)2=36+49=85

Next evaluate the distance between B(5,8) and O(0,0) .

d(B,O)=(0(5))2+(08)2=25+64=89

Observe that d(B,O)>d(A,O) .

That is the distance between the point A(6,7) and the origin is less than the distance the point B(5,8) and the origin.

Therefore, the point A(6,7) is near to origin.

Thus, the closer point to origin is A(6,7) .

### Have a homework question?

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text:
# Professor Peter Hydon

Professor of Mathematics
Director of Division of Computing, Engineering, and Mathematical Sciences

Peter had a broad mathematical education: BSc in Mathematics at Bath, followed by a master’s degree (Part III of the Mathematics Tripos) and PhD at the University of Cambridge. After three years’ postdoctoral work at the University of Leeds he spent two years in a research post at the University of Northumbria before joining the University of Surrey’s Department of Mathematics in 1996 as a Lecturer. He was promoted to Reader in 2000 and Professor in 2006. He held a wide variety of roles and responsibilities, and was Head of Department for the two years before he joined Kent in 2015 as Head of SMSAS.After five years in that role, in January 2020, Peter stepped into the role of Division of Computing, Engineering, and Mathematical Sciences.Member of Senate and various University and Faculty committees, Peter comments "I also have many opportunities to represent SMSAS nationally and more widely".

## Research interests

Most of Peter's research aims to develop methods for solving, simplifying or approximating a given system of equations, by exploiting the system’s algebraic and geometric structures.Multisymplectic systems, with applications to geometric mechanicsGeometric integration (developing numerical methods that preserve structural features of the system being approximated)Automorphisms of Lie algebras He has written two graduate textbooks, both published by Cambridge University Press, which introduce newcomers to these research areas. His website has details of these and other publications.

## Supervision

Gianluca Frasca Caccia - Finite Difference Conservation Laws

## Publications

### Article

• Frasca-Caccia, G. and Hydon, P. (2019). Locally conservative finite difference schemes for the Modified KdV equation. Journal of Computational Dynamics [Online] 6:307-323. Available at: http://dx.doi.org/10.3934/jcd.2019015.
Finite diffrence schemes that preserve two conservation laws of a given partial differential equation can be found directly by a recently-developed symbolic approach. Until now, this has been used only for equations with quadratic nonlinearity. In principle, a simplified version of the direct approach also works for equations with polynomial nonlinearity of higher degree. For the Modified Korteweg-de Vries equation, whose nonlinear term is cubic, this approach yields several new families of second-order accurate schemes that preserve mass and either energy or momentum. Two of these families contain Average Vector Field schemes of the type developed by Quispel and co-workers. Numerical tests show that each family includes schemes that are highly accurate compared to other mass-preserving methods that can be found in the literature.
• Mansfield, E., Rojo-Echeburua, A., Hydon, P. and Peng, L. (2019). Moving Frames and Noether’s Finite Difference Conservation Laws I. Transactions of Mathematics and its Applications [Online] 3. Available at: https://doi.org/10.1093/imatrm/tnz004.
We consider the calculation of Euler--Lagrange systems of ordinary difference equations, including the difference Noether's Theorem, in the light of the recently-developed calculus of difference invariants and discrete moving frames. We introduce the difference moving frame, a natural discrete moving frame that is adapted to difference equations by prolongation conditions.

For any Lagrangian that is invariant under a Lie group action on the space of dependent variables, we show that the Euler--Lagrange equations can be calculated directly in terms of the invariants of the group action. Furthermore, Noether's conservation laws can be written in terms of a difference moving frame and the invariants. We show that this form of the laws can significantly ease the problem of solving the Euler--Lagrange equations, and
we also show how to use a difference frame to integrate Lie group invariant difference equations. In this Part I, we illustrate the theory by applications to Lagrangians invariant under various solvable Lie groups. The theory is also generalized to deal with variational symmetries that do not leave the Lagrangian invariant.

Apart from the study of systems that are inherently discrete, one significant application is to obtain geometric (variational) integrators that have finite difference approximations of the continuous conservation laws embedded \textit{a priori}. This is achieved by taking an invariant finite difference Lagrangian in which the discrete invariants have the correct continuum limit to their smooth counterparts.
We show the calculations for a discretization of the Lagrangian for Euler's elastica, and compare our discrete solution to that of its smooth continuum limit.
• Frasca-Caccia, G. and Hydon, P. (2018). Simple bespoke preservation of two conservation laws. IMA Journal of Numerical Analysis [Online]. Available at: https://doi.org/10.1093/imanum/dry087.
Conservation laws are among the most fundamental geometric properties of a partial differential equation
(PDE), but few known finite difference methods preserve more than one conservation law. All conservation
laws belong to the kernel of the Euler operator, an observation that was first used recently to
construct approximations symbolically that preserve two conservation laws of a given PDE. However,
the complexity of the symbolic computations has limited the effectiveness of this approach. The current
paper introduces some key simplifications that make the symbolic-numeric approach feasible. To
illustrate the simplified approach, we derive bespoke finite difference schemes that preserve two discrete
conservation laws for the Korteweg-de Vries (KdV) equation and for a nonlinear heat equation. Numerical
tests show that these schemes are robust and highly accurate compared to others in the literature.
• Grant, T. and Hydon, P. (2013). Characteristics of Conservation Laws for Difference Equations. Foundations of Computational Mathematics [Online] 13:667-692. Available at: http://dx.doi.org/10.1007/s10208-013-9151-2.
Each conservation law of a given partial differential equation is determined (up to equivalence) by a function known as the characteristic. This function is used to find conservation laws, to prove equivalence between conservation laws, and to prove the converse of Noether's Theorem. Transferring these results to difference equations is nontrivial, largely because difference operators are not derivations and do not obey the chain rule for derivatives. We show how these problems may be resolved and illustrate various uses of the characteristic. In particular, we establish the converse of Noether's Theorem for difference equations, we show (without taking a continuum limit) that the conservation laws in the infinite family generated by Rasin and Schiff are distinct, and we obtain all five-point conservation laws for the potential Lotka-Volterra equation.
• Fisher, D., Gray, R. and Hydon, P. (2013). Automorphisms of real Lie algebras of dimension five or less. Journal of Physics A: Mathematical and Theoretical [Online] 46:1-18. Available at: http://iopscience.iop.org/article/10.1088/1751-8113/46/22/225204.
The Lie algebra version of the Krull-Schmidt Theorem is formulated and proved. This leads to a method for constructing the automorphisms of a direct sum of Lie algebras from the automorphisms of its indecomposable components. For finite-dimensional Lie algebras, there is a well-known algorithm for finding such components, so the theorem considerably simplifies the problem of classifying the automorphism groups. We illustrate this by classifying the automorphisms of all indecomposable real Lie algebras of dimension five or less. Our results are presented very concisely, in tabular form.
• Hydon, P. and Mansfield, E. (2011). Extensions of Noether’s Second Theorem: from continuous to discrete systems. Proceedings of the Royal Society A- Mathematical Physical and Engineering Sciences [Online] 467:3206-3221. Available at: http://dx.doi.org/10.1098/rspa.2011.0158.
A simple local proof of Noether's Second Theorem is given. This proof immediately leads to a generalization of the theorem, yielding conservation laws and/or explicit relationships between the Euler–Lagrange equations of any variational problem whose symmetries depend on a set of free or partly constrained functions. Our approach extends further to deal with finite-difference systems. The results are easy to apply; several well-known continuous and discrete systems are used as illustrations.
• Mansfield, E. and Hydon, P. (2001). On a variational complex for difference equations. Contemporary Mathematics 285:195-205.

### Book section

• Mansfield, E. and Hydon, P. (2001). Towards approximations which preserve integrals. In: Mourrain, B. ed. Proceedings of the 2001 International Symposium on Symbolic and Algebraic Computation. New York, USA: ACM, pp. 217-222. Available at: http://dx.doi.org/10.1145/384101.384131.
We investigate the algorithmic approximation of ordinary differential equations having a known conservation law, with finite difference schemes which inherit a discrete version of the conservation law. We use the method of moving frames on a multispace due to Olver. We assume that the system of ODEs to be studied has a variational principle and that the conservation law arises from a variational symmetry via Noether's theorem.

### Book

• Hydon, P. (2014). Difference Equations by Differential Equation Methods. [Online]. Cambridge University Press: Cambridge University Press. Available at: http://www.cambridge.org/gb/academic/subjects/mathematics/differential-and-integral-equations-dynamical-systems-and-co/difference-equations-differential-equation-methods?format=HB&isbn=9780521878524.
Most well-known solution techniques for differential equations exploit symmetry in some form. Systematic methods have been developed for finding and using symmetries, first integrals and conservation laws of a given differential equation. Here the author explains how to extend these powerful methods to difference equations, greatly increasing the range of solvable problems. Beginning with an introduction to elementary solution methods, the book gives readers a clear explanation of exact techniques for ordinary and partial difference equations. The informal presentation is suitable for anyone who is familiar with standard differential equation methods. No prior knowledge of difference equations or symmetry is assumed. The author uses worked examples to help readers grasp new concepts easily. There are 120 exercises of varying difficulty and suggestions for further reading. The book goes to the cutting edge of research; its many new ideas and methods make it a valuable reference for researchers in the field.
Last updated


text:
# 2.1.3.6 SetEventHandlerForObject

## Description

It sets up an event handler for an automation object.

## Syntax

int SetEventHandlerForObject( Object * pObj, LPCSTR lpcszEventName, Function * pOCFunc )

## Parameters

pObj
[input]pointer to an existing COM object reference which must have been initialized.
lpcszEventName
[input]event name (it can retrieved from the object library for the given object)
pOCFunc
[input] pointer to a Function object which has been intialized with an existing Origin C function.

## Return

0 if the event handler was set up successfully, 2 if an events COM interface cannot be retrieved from the object, 3 if the event name was wrong.

## Examples

EX1

// This is the event handler which catches the event. The prototype must agree
// with the event's declaration, which can be found out by inspecting
// the object library for each particular object.
void    MyEventHandler()
{
printf("Event hanled\r\n");
}
void    SetEventHandlerForObject_ex1()
{
// Declare a function object (the function MyEventHandler must
// have been defined before)
Function        fnEventHandler(MyEventHandler);

Object obj = CreateObject("SomeObject.SomeObject");

int        iRet;
// This sets up the event handler. It tells the object obj: Call the function
// MyEventHandler when event "SomeEvent" happens.
iRet = SetEventHandlerForObject(&obj, "SomeEvent", &fnEventHandler);
if (0 == iRet)
out_str("Event handler set up successfully.");
}


text:
# How to find the smallest flabby sheaf containing a given sheaf ?

None of the spaces C^k (\mathbb{R}^n), with 0 \leq k \leq \infty, is a flabby sheaf. However, they are respectively contained in the smallest flabby sheaves C^k_{nd} (\mathbb{R}^n) of functions f : \mathbb{R}^n \longrightarrow \mathbb{R} for which there exist \Gamma \subset \mathbb{R}^n, \Gamma closed and nowhere dense, such that f restricted to \mathbb{R}^n \setminus \Gamma is C^k-smooth.

Is any similar way known to construct flabby sheaves which contain given sheaves of functions defined on some topological space X ?

-
Since you already wrote latexy formulas, you cound wrap them in $ and have them rendered! :) – Mariano Suárez-Alvarez Jul 28 '10 at 15:27 add comment ## 1 Answer There is the "flabbification" or "flasquification" functor used in the Godement resolution. Namely, given a sheaf$\mathcal{F}$on$X$, let$\mathcal{F}_x$be the stalk at a point$x$. Then we define the sheaf$\Phi(\mathcal{F})$to have sections $$\Gamma(U, \Phi(\mathcal{F}))=\prod_{x\in U} \mathcal{F}_x$$ with the obvious restriction maps. (This is the same as endowing the etale space of$\mathcal{F}$with the trivial topology and considering the sheaf of sections.) There is a natural injection$\mathcal{F}\to \Phi(\mathcal{F})$sending a section of$\mathcal{F}$to its stalks. Note that$\Phi$is functorial; indeed, it is the right adjoint to the natural inclusion of the full subcategory (Flasque sheaves on$X$)$\hookrightarrow$(Sheaves on$X$). See for example this Wikipedia article or Godement's book on sheaf theory. Edit: One can also mimic your construction of$C^k_{nd}$as follows. Let$\mathcal{F}_{nd}$have global sections given by$\bigcup \Gamma(X-\partial U, \mathcal{F})/\sim$where the union is taken over all open sets$U$, and we say$(f, X-\partial U)\sim (f', X-\partial U')$if$f=f'$when restricted to$X-(\partial U\cup \partial U')$. Then local sections will be restrictions of these global sections. This will always be flabby, but will not necessarily have a morphism$\mathcal{F}\to \mathcal{F}_{nd}$unless$\mathcal{F}$has enough sections (for example, if$\mathcal{F}$is fine, as in your example). - Thank you for the answer. It seems however that the Godement flabbification is not the one which goes from$C^k$to$C^k_{nd}$. And then, what is the difference ? More precisely, given a topological space$X$and a sheaf$F$of functions from open subsets$U \subseteq X$to some set$E$, can one similar to the above, and not to Godement, construct some smallest flabby sheaf$F_{nd}$of fucntions from open subsets$U \subseteq X$to$E$, so that$F_{nd}$extends$F$? – ron l winger Jul 28 '10 at 14:52 I've added another construction to address your concerns. I'm not convinced it's useful though--partly because I'm not sure what you mean by "smallest." If you can come up with some functorial property you're looking for, that would help. – Daniel Litt Jul 28 '10 at 15:12 What I mean by "smallest" seems to happen with$C^k_{nd}$when compared with$C^k$. Indeed, it seems that one cannot squeeze a flabby sheaf between the two of them, unless it is$C^k_{nd}$itslef. But then, perhaps, instead of the "smallest", one should rather talk about a "minimal" one ? – Elemer E Rosinger Jul 29 '10 at 14:07 Sorry, I sent the previous comment from the office of Prof. Rosinger. ron l winger – Elemer E Rosinger Jul 29 '10 at 14:09 Ah -- you claim that$C^k_{nd}$has no flabby subsheaf containing$C^k$. Can you prove this? I think it's false. For example, one can require that the$f|_\Gamma$(where$\Gamma$is your closed nowhere dense set) equals zero; this gives a smaller flabby sheaf containing$C^k\$, unless I'm mistaken. –  Daniel Litt Jul 29 '10 at 14:17
show 2 more comments


text:
# Chemical shift

In nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is the resonant frequency of a nucleus relative to a standard in a magnetic field. Often the position and number of chemical shifts are diagnostic of the structure of a molecule.[1][2][3] Chemical shifts are also used to describe signals in other forms of spectroscopy such as photoemission spectroscopy.

Some atomic nuclei possess a magnetic moment (nuclear spin), which gives rise to different energy levels and resonance frequencies in a magnetic field. The total magnetic field experienced by a nucleus includes local magnetic fields induced by currents of electrons in the molecular orbitals (note that electrons have a magnetic moment themselves). The electron distribution of the same type of nucleus (e.g. 1H, 13C, 15N) usually varies according to the local geometry (binding partners, bond lengths, angles between bonds, and so on), and with it the local magnetic field at each nucleus. This is reflected in the spin energy levels (and resonance frequencies). The variations of nuclear magnetic resonance frequencies of the same kind of nucleus, due to variations in the electron distribution, is called the chemical shift. The size of the chemical shift is given with respect to a reference frequency or reference sample (see also chemical shift referencing), usually a molecule with a barely distorted electron distribution.

## Operating frequencyEdit

The operating (or Larmor) frequency ω0 of a magnet is calculated from the Larmor equation[4]

${\displaystyle \omega _{0}=\gamma B_{0}\,,}$

where B0 is the actual strength of the magnet in units like teslas or gauss, and γ is the gyromagnetic ratio of the nucleus being tested which is in turn calculated from its magnetic moment μ and spin number I with the nuclear magneton μN and the Planck constant h:[citation needed]

${\displaystyle \gamma ={\frac {\mu \,\mu _{\mathrm {N} }}{hI}}\,.}$

Thus, the proton operating frequency for a 1 T magnet is calculated as:

${\displaystyle \omega _{0}=\gamma B_{0}={\frac {2.79\times 5.05\times 10^{-27}\,{\rm {J/T}}}{6.62\times 10^{-34}\,{\rm {Js}}\times {\tfrac {1}{2}}}}\times 1\,{\rm {T}}=42.5\,{\rm {MHz}}\,.}$

## Chemical shift referencingEdit

Chemical shift δ is usually expressed in parts per million (ppm) by frequency, because it is calculated from:[5]

${\displaystyle \delta ={\frac {\nu _{\mathrm {sample} }-\nu _{\mathrm {ref} }}{\nu _{\mathrm {ref} }}}\,,}$

where νsample is the absolute resonance frequency of the sample and νref is the absolute resonance frequency of a standard reference compound, measured in the same applied magnetic field B0. Since the numerator is usually expressed in hertz, and the denominator in megahertz, δ is expressed in ppm.

The detected frequencies (in Hz) for 1H, 13C, and 29Si nuclei are usually referenced against TMS (tetramethylsilane) or DSS, which by the definition above have a chemical shift of zero if chosen as the reference. Other standard materials are used for setting the chemical shift for other nuclei.

Thus, an NMR signal observed at a frequency 300 Hz higher than the signal from TMS, where the TMS resonance frequency is 300 MHz, has a chemical shift of:

${\displaystyle {\frac {300\,{\rm {Hz}}}{300\times 10^{6}\,{\rm {Hz}}}}=1\times 10^{-6}=1\,{\rm {ppm\,.}}}$

Although the absolute resonance frequency depends on the applied magnetic field, the chemical shift is independent of external magnetic field strength. On the other hand, the resolution of NMR will increase with applied magnetic field.

## The induced magnetic fieldEdit

The electrons around a nucleus will circulate in a magnetic field and create a secondary induced magnetic field. This field opposes the applied field as stipulated by Lenz's law and atoms with higher induced fields (i.e., higher electron density) are therefore called shielded, relative to those with lower electron density. The chemical milieu of an atom can influence its electron density through the polar effect. Electron-donating alkyl groups, for example, lead to increased shielding while electron-withdrawing substituents such as nitro groups lead to deshielding of the nucleus. Not only substituents cause local induced fields. Bonding electrons can also lead to shielding and deshielding effects. A striking example of this are the pi bonds in benzene. Circular current through the hyperconjugated system causes a shielding effect at the molecule's center and a deshielding effect at its edges. Trends in chemical shift are explained based on the degree of shielding or deshielding.

Nuclei are found to resonate in a wide range to the left (or more rare to the right) of the internal standard. When a signal is found with a higher chemical shift:

• the applied effective magnetic field is lower, if the resonance frequency is fixed (as in old traditional CW spectrometers)
• the frequency is higher, when the applied magnetic field is static (normal case in FT spectrometers)
• the nucleus is more deshielded
• the signal or shift is downfield or at low field or paramagnetic

Conversely a lower chemical shift is called a diamagnetic shift, and is upfield and more shielded.

## Diamagnetic shieldingEdit

In real molecules protons are surrounded by a cloud of charge due to adjacent bonds and atoms. In an applied magnetic field (B0) electrons circulate and produce an induced field (Bi) which opposes the applied field. The effective field at the nucleus will be B = B0Bi. The nucleus is said to be experiencing a diamagnetic shielding.

## Factors causing chemical shiftsEdit

Important factors influencing chemical shift are electron density, electronegativity of neighboring groups and anisotropic induced magnetic field effects.

Electron density shields a nucleus from the external field. For example, in proton NMR the electron-poor tropylium ion has its protons downfield at 9.17 ppm, those of the electron-rich cyclooctatetraenyl anion move upfield to 6.75 ppm and its dianion even more upfield to 5.56 ppm.

A nucleus in the vicinity of an electronegative atom experiences reduced electron density and the nucleus is therefore deshielded. In proton NMR of methyl halides (CH3X) the chemical shift of the methyl protons increase in the order I < Br < Cl < F from 2.16 ppm to 4.26 ppm reflecting this trend. In carbon NMR the chemical shift of the carbon nuclei increase in the same order from around −10 ppm to 70 ppm. Also when the electronegative atom is removed further away the effect diminishes until it can be observed no longer.

Anisotropic induced magnetic field effects are the result of a local induced magnetic field experienced by a nucleus resulting from circulating electrons that can either be paramagnetic when it is parallel to the applied field or diamagnetic when it is opposed to it. It is observed in alkenes where the double bond is oriented perpendicular to the external field with pi electrons likewise circulating at right angles. The induced magnetic field lines are parallel to the external field at the location of the alkene protons which therefore shift downfield to a 4.5 ppm to 7.5 ppm range. The three-dimensional space where a diamagnetic shift is called the shielding zone with a cone-like shape aligned with the external field.

Induced magnetic field of alkenes in external magnetic fields, field lines in grey.

The protons in aromatic compounds are shifted downfield even further with a signal for benzene at 7.73 ppm as a consequence of a diamagnetic ring current.

Alkyne protons by contrast resonate at high field in a 2–3 ppm range. For alkynes the most effective orientation is the external field in parallel with electrons circulation around the triple bond. In this way the acetylenic protons are located in the cone-shaped shielding zone hence the upfield shift.

Induced magnetic field of alkynes in external magnetic fields, field lines in grey.

## Magnetic properties of most common nucleiEdit

1H and 13C are not the only nuclei susceptible to NMR experiments. A number of different nuclei can also be detected, although the use of such techniques is generally rare due to small relative sensitivities in NMR experiments (compared to 1H) of the nuclei in question, the other factor for rare use being their slender representation in nature and organic compounds.

Magnetic properties of common nuclei[6]
Isotope Occurrence
in nature

(%)
Spin number I Magnetic moment μ
(μN)
(e × 10−24 cm2)
Operating frequency at 7 T
(MHz)
Relative sensitivity
1H 099.984 1/2 2.79628 0 300.13 1
2H 000.016 1 0.85739 0.0028 046.07 0.0964
10B 018.8 3 1.8005 0.074 032.25 0.0199
11B 081.2 3/2 2.6880 0.026 096.29 0.165
12C 098.9 0 0 0 000 0
13C 001.1 1/2 0.70220 0 075.47 0.0159
14N 099.64 1 0.40358 0.071 021.68 0.00101
15N 000.37 1/2 −0.28304 0 030.41 0.00104
16O 099.76 0 0 0 000 0
17O 000.0317 5/2 −1.8930 −0.0040 040.69 0.0291
19F 100 1/2 2.6273 0 282.40 0.834
28Si 092.28 0 0 0 000 0
29Si 004.70 1/2 −0.5548 0 059.63 0.0785
31P 100 1/2 1.1205 0 121.49 0.0664
35Cl 075.4 3/2 0.92091 −0.079 029.41 0.0047
37Cl 024.6 3/2 0.68330 −0.062 024.48 0.0027

1H, 13C, 15N, 19F and 31P are the five nuclei that have the greatest importance in NMR experiments:

• 1H because of high sensitivity and vast occurrence in organic compounds
• 13C because of being the key component of all organic compounds despite occurring at a low abundance (1.1%) compared to the major isotope of carbon 12C, which has a spin of 0 and therefore is NMR-inactive.
• 15N because of being a key component of important biomolecules such as proteins and DNA
• 19F because of high relative sensitivity
• 31P because of frequent occurrence in organic compounds and moderate relative sensitivity

## Other chemical shiftsEdit

The related Knight shift (first reported in 1949) is observed with pure metals. The NMR chemical shift in its present-day meaning first appeared in journals in 1950. Chemical shifts with a different meaning appear in X-ray photoelectron spectroscopy as the shift in atomic core-level energy due to a specific chemical environment. The term is also used in Mössbauer spectroscopy, where similarly to NMR it refers to a shift in peak position due to the local chemical bonding environment. As is the case for NMR the chemical shift reflects the electron density at the atomic nucleus.[7]


text:
# filters.ferry¶

The ferry filter is used to stash intermediate variables as part of processing data. For example, a common scenario is to keep both the original value and the reprojected X and Y variables in a scenario that uses the filters.reprojection filter. In the normal case, the X and Y data would be overwritten with the new longitude and latitude values as part of the reprojection. The ferry filter will allow you to keep this around for later use.

## Example¶

In this scenario, we are doing what is described above – stashing the pre-projection X and Y values into the StatePlaneX and StatePlaneY dimensions. Future processing, can then operate on these data.

{
"pipeline":[
"uncompressed.las",
{
"spatialreference":"EPSG:2993",
"filename":"../las/1.2-with-color.las"
},
{
"type":"filters.ferry",
"dimensions":"X = StatePlaneX, Y=StatePlaneY"
},
{
"type":"filters.reprojection",
"out_srs":"EPSG:4326+4326"
},
{
"type":"writers.las",
"scale_x":"0.0000001",
"scale_y":"0.0000001",
"filename":"colorized.las"
}
]
}

## Options¶

dimensions
A list of dimensions whose values should be copied to the specified dimensions. The format of the option is <from>=<to>, <from>=<to>,... Spaces are ignored. ‘from’ dimensions must exist and have been created by a reader or filter. ‘to’ dimensions will be created if necessary.


text:
# Thread: Fixed points in a permutation, mean and variance

1. ## Fixed points in a permutation, mean and variance

What are the mean and variance of the number of fixed elements, that is, elements left in the same position, of a randomly selected permutation of n elements? (Hint: Let X denote the number of fixed points of a random permutation, Write X=X1+X2+...+Xn wherre Xi=1 if the permuation fixes the ith element and Xi=0 otherwise.)

A google search and the solution in the book I'm using says the variance is 1 but neither provides a proof. I got (n-1)/n when I did it. Does someone know how to get 1? Thanks.

2. Originally Posted by discprobques
What are the mean and variance of the number of fixed elements, that is, elements left in the same position, of a randomly selected permutation of n elements? (Hint: Let X denote the number of fixed points of a random permutation, Write X=X1+X2+...+Xn wherre Xi=1 if the permuation fixes the ith element and Xi=0 otherwise.)
Tell us what you found to be the mean and how it is found.


text:
## Useless

√♥ = ?

cos ♥ = ?

D ♥ = ?

I ♥ = ?

F{♥} = (1/√(2π)) ∫(-∞,∞)♥ e−iωt dt = ?

My normal approach is useless here.

## 線性回歸

### 第 1 題

In a study of five industrial areas, a researcher obtained these data relating the average number of units of a certain pollutant in the air and the incidence (per 100,000 people) of a certain disease:

 Units of pollutant Incidence of disease 3 4 5 8 10 48 52 58 70 96

Find the equation of the least-squares line y = Ax + B (to two decimal points places.)

Given that A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$ and B = \frac {\sum y - A \sum x} n$B = \frac {\sum y - A \sum x} n$.

x y x2 xy
3 48 9 144
4 52 16 208
5 58 25 290
8 70 64 560
10 96100 960
303242142162

A = \frac {5 \cdot 2162 - 30 \cdot 324} {5 \cdot 214 - 30^2} = \frac{109}{17} \approx 6.411764705882353

B = \frac {324 - \left( 109/17 \right) \cdot 30} {5} = \frac{2238}{85} \approx 26.32941176470588

## 導函數

### 第 2 題，A 卷

Use the definition of the derivative \displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ to find \displaystyle \frac d{dx} \sin x $\displaystyle \frac d{dx} \sin x$.

\frac d{dx} \sin x = \lim_{h \to 0} \frac {\sin \left( x + h \right) - \sin x} h

\sin \left( x + h \right) = \sin x \cos h + \cos x \sin h

\begin{align*}
\frac d{dx} \sin x &= \lim_{h \to 0} \frac {\sin x \cos h + \cos x \sin h - \sin x} h \\
&= \left( \sin x \right) \left( \lim_{h \to 0} \frac {\cos h - 1} h \right) + \left( \cos x \right) \lim_{h \to 0} \frac {\sin h} h
\end{align*}

\frac d{dx} \sin x = \cos x

## 切線與法線

### 第 3 題，A 卷

Find the equation of the tangent line to the curve y3xy2 + xy = 14 3 at (1, −2).

### 第 3 題，B 卷

… the normal line ….

y3xy2 + xy = 3

3y2y′ − (2xyy′ + y2) + (xy′ + y) = 0

(3y2 − 2xy + x) y′ = y2y

 y' = \frac{y^2 - y}{3y^2 - 2xy + x}
 y' \left( 1, -2 \right) = \frac6{17}

 y + 2 = \frac6{17} \left( x - 1 \right)

 y + 2 = \frac{-17}6 \left( x - 1 \right)

## 導數

### 第 4 題

Given \displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x} $\displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x}$, find f′(2)

f' \left( x \right) = \frac {-2x \left( x-1 \right)^3 - 3x^2 \left( x-1 \right)^2} {x+1} + \frac {x^2 \left( x-1 \right)^3} {\left( x+1 \right)^2}

f' \left( 2 \right) = \frac{-44}9

### 第 6 題，B 卷

Given f(x) = (x2 + 1)sin x, find f′(π/2)

f(x) = exp(sin x ln(x2 + 1))

\frac d{dx} \sin x \ln \left( x^2 + 1 \right) = \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1}

f' \left( x \right) = \left( x^2 + 1 \right)^{\sin x} \left( \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1} \right)

f' \left( \frac \pi 2 \right) = \left( \frac {\pi^2} 4 + 1 \right) \left( \frac \pi {\pi^2/4 + 1} \right) = \pi

## 微分

### 第 7 題

In the late 1830s, French physiologist Jean Poiseuille discovered the formula we use today to predict how much the radius of a particular clogged artery decreases the normal volume of flow. His formula,

V = kr4

says that volume of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant times the fourth power of the tube’s radius r. How dose does a 10% decrease in r affect V?

#### 官方解法

\frac{dV}{dr} = 4kr^3

Δr = -0.1r

\Delta V \approx \frac{dV}{dr} \Delta r

ΔV ≈ -0.4kr4 = -0.4V

V 減少 40%

#### 直接解法

V0 = kr04

r = 0.9r0

V = k (0.9r0)4 = 0.94 V0

\frac {V - V_0} {V_0} = 0.9^4 - 1 = -0.3439

## 線性近似

### 第 8 題

Use the differentials to approximate the quantity \sqrt{4.6}$\sqrt{4.6}$ to four decimal points places.

f(x) ≈ f(a) + f′(a) (xa)

\sqrt x \approx \sqrt a + \frac {x - a} {2 \sqrt a}

x = 4 為起點，

\begin{align*}
\sqrt{4.6} &\approx 2 + \frac {4.6 - 4} {2 \cdot 2} = 2.15 \\
\sqrt{4.6} &\approx 2.15 + \frac {4.6 - 2.15^2} {2 \cdot 2.15} = \frac{3689}{1720} \approx 2.144767441860465 \\
\sqrt{4.6} &\approx \frac{3689}{1720} + \frac {4.6 - \left( 3689/1720 \right)^2} {2 \cdot \left( 3689/1720 \right)} \approx 2.144761058962219
\end{align*}

\sqrt{4.6} \approx 2.1448

## 應用題

### 第 9 題

Water boils at 212℉ at sea level and 200℉ at an elevation of 6000 ft. Assume that the boiling point B varies linearly with altitude α. Find the function B = f(α) that describes the dependence. Comment on whether a linear function gives a realistic model.

f(α) = mα + k

f(0) = k = 212

f(6000) = 6000m + 212 = 200

m 為 -0.002，所以

f(α) = -0.002α + 212


text:
# DEHUFF

Time Limit: 3000MS

Memory Limit: 65536K

## Description

A certain data compression technique involves creating a table of variable-length binary codes where one or more binary digits are used to represent a single letter of an alphabet. Usually, letters that occur most frequently in words generated using this alphabet will have shorter binary codes than those used less frequently. For example, in an alphabet consisting of the letters A through Z, in general the letter E appears in more words than the letter Q; therefore it would be expected that E would have a shorter binary code than does Q. Given a sample string using at least one of each letter in an alphabet, along with the entire binary encoding of that sample string, you should be able to generate at least one binary code table for each letter in the alphabet. For example, consider the sample string: "CAB" which contains each letter of the alphabet {A,B,C}. If the binary encoding of "CAB" is "01011" then the (only) binary code table is: C = 0 A = 10 B = 11 The binary codes for each character are prefix codes in that no code in the set can be the initial binary string for any other code (so A = 01, B = 011 would not be allowed). For this problem, you will write a program that determines binary code tables for sample strings and their binary encodings. If there is a single binary code table solution, then you will print it out (sorted). If more than one binary code table can be generated from the given set of data, you will print "MULTIPLE TABLES". Note: For a given alphabet, the entire code space will be used; that is, there will be no unused codes.

## Input

Input consists of a single line consisting of an integer value N representing the number of datasets that follow. Each dataset consists of two lines. The first line is the sample string that contains at least one of each letter (or space) in the alphabet. The second line is the binary encoding of the sample string. Note: Sample strings may only contain upper case letters and spaces.

## Output

For each dataset, print a line that identifies the dataset in the format: "DATASET #n" where n is the dataset number (1 through N). If more than one binary code table can be generated to represent the alphabet, print "MULTIPLE TABLES" on a new line and move onto the next dataset. If only one binary code table can be generated, for each character in the alphabet, display the letter, a space, and equal sign (=), a space and the binary code for that letter. Display the table in order sorted by the ASCII value of each character of the alphabet.

## Sample Input

3
CAB
01011
HELLO WORLD
111011011110111101111100111111111111011111101111010
ABCDEFGHI
010110111011110111110111111011111110001011111111

## Sample Output

DATASET #1
A = 10
B = 11
C = 0
DATASET #2
= 0
D = 10
E = 110
H = 1110
L = 11110
O = 111110
R = 1111110
W = 1111111
DATASET #3
MULTIPLE TABLES

## Source

Greater New York 2002


text:
Project 2: Amazons

## Introduction

The game of Amazons is a simple but rather interesting board game, usually for two players. It was invented in 1988 by Walter Zamkauskas of Argentina, and originally called El Juego de las Amazonas (now a trademark of Ediciones de Mente). The board is a 10 by 10 chessboard. Each player gets four "amazons" (represented as chess queens), white for the player who moves first, and black for the opponent. Initially, the board is set up as in the diagram on the left of Figure 1.

Figure 1. On the left: an Amazon board showing the standard numbering of squares, and the initial placement of the pieces. On the right: the board that results after the two moves d1-d7(g7) and d10-c9(h4).

Each move consists of two parts: first, an amazon of the appropriate color makes a chess queen move—any non-zero number of squares in a straight line horizontally, vertically, or diagonally with the restriction that the piece may not move onto or through an occupied square. Pieces are not captured in this game; the board keeps getting fuller. Next, the piece that moved "throws a spear" from her final position. Spears move exactly like pieces, and are subject to the same restrictions. The spear sticks permanently in the (previously empty) square in which it lands. That square is counted as being occupied for the rest of the game (so no piece or spear may move through it). For example, the right diagram in Figure 1 shows the result of two moves. From the initial position, white moved from d1 to d7 and from there threw a spear to g7. Then black moved from d10 to c9, and from there threw a spear to h4.

A player who has no legal move loses. For example, after black moves a7-a6(a7) in Figure 2, he will have used up all his moves, and white (who still has plenty of room) will win after his next move. Draws are impossible.

Figure 2. A lost position for black. After a7-a6(a7), black has no move

## Notation

A square is denoted by a column letter followed by a row number (as in e4). Columns are enumerated from left to right with letters a through j. Rows are enumerated from the bottom to the top with numbers 1 through 10. An entire move then consists of the starting and ending position of the piece that is moved, followed, in parentheses, by the position to which the spear is thrown. Thus, d10-c9(h4) means "Move from d10 to c9 and then throw a spear to h4."

## Commands

When running from the command line, the program will accept the following commands, which may be preceded by whitespace.

• new: End any game in progress, clear the board to its initial position, and set the current player to white.
• A move, either in the format described in Notation or (for convenience) with blanks replacing punctuation, as in g1 c5 e7.
• seed N: If the AIs are using random numbers for move selection, this command seeds their random-number generator with the integer N. Given the same seed and the same opposing moves, an AI should always make the same moves. This feature makes games reproducible.
• auto C: Make the C player an automated player. Here, C is "black" or "white", case-insensitive.
• manual C: Make the C player a human player (entering moves as manual commands).
• dump: Print the current state of the board in exactly the following format:

===
- - - - - - B - - -
- - B - - - - - - -
- - - - - - - - - -
B - - W - - S - - B
- - - - - - - - - -
- - - - - - - - - -
W - - - - - - S - W
- - - - - - - - - -
- - - - - - - - - -
- - - - - - W - - -
===

Here, B denotes a black queen, W a white queen, and S a spear. You must not use the === lines for any other output).

• quit: Exit the program.

Feel free any other commands you think might be nice.

## Output

When an AI plays, it should print out each move that it makes using exactly the format

 * a1-c3(c6)

(with asterisk shown). Do not print these lines out for a manual player's moves.

When one side wins, the program should print out one of

 * White wins.
* Black wins.

(also with periods) as appropriate. Do not use the * character in any other output you produce.

You may prompt a manual player for input using the form

...>

where "..." may be any text. The grading scripts will discard any text from the beginning of a line up to a > character.

Your job is to write a program to play Amazons. To run it in text mode, use the command

java -ea amazons.Main

to enter commands from the terminal or use

java -ea amazons.Main INPUT

to feed it commands from file INPUT.

The AI in your program should be capable of finding a win that is within 10 moves. The branching factor for Amazons is quite high at the beginning of a game, but rapidly declines thereafter. Therefore we suggest that you choose a maximum search depth for a move that depends on how full the board is. Experiment a bit to see what works. The autograder will allow 3 minutes for a fully automated game.

The GUI is an optional (extra credit) part of this project. We will actually do automatic testing only on the commands

java -ea amazons.Main

and

java -ea amazons.Main INPUT

## Staff Program

The staff-amazons program on the instructional machines runs our solution to the project. This version has additional bells and whistles that you are not required to duplicate. It is not the standard for this project, just as example of a solution. In particular, your GUI (if you do it) need not look anything like ours. The autograder will use this program in some of its tests to check your program's output.

## Testing

We have only provided a token UnitTest file; you can add additional unit test files and list them in UnitTest so that they all get run by

java -ea amazons.UnitTest

(which is what make check does).

The integration test program, test-amazons, is not part of your skeleton. We will be providing it slightly later. To run test-amazons, you'll use

python3 testing/test-amazons TESTFILE-1.in

to run TESTFILE-1.in through your program and

python3 testing/test-amazons TESTFILE-1.in TESTFILE-2.in

to run two programs simultaneously so that each one sends all of its AI's moves (such as "* c3f6(i6)" as described previously) to the other program. (Replace "TESTFILE" with the actual name of your test file.)

Each .in and input file should start with a Unix-style command for running a program, such as

 java -ea amazons.Main

(You will probably use just this command; the autograder will sometimes use this line to run the staff solution against your program.) The rest of the .in file is fed to this program as the standard input, except for lines that start with "*" in the first column, which are special instructions to the testing script.

• The command *time MOVE GAME puts a time limit of MOVE seconds on each move in a game and GAME seconds for one side's moves in an entire game (i.e., an entire sequence of moves controlled by one of the move/win commands below).
• The command *move means "wait for the program to output an AI move, and then continue with the script." When used with the two-argument form of test-amazons, it also sends this move as input to the other program.
• The command *move/win is intended for use when both players are AIs, and means "wait for the program to output a complete sequence of AI moves, followed by "* ... wins." It does not print either the moves or the "win" message.
• The command *move/win+ is the same as *move/win, but also prints the "* ... wins." message.
• The command *win+ waits for a "* ... wins." message from the program and prints it.
• All lines that don't start with * are sent to the program being tested.

A few other commands apply only to the two-argument form of test-amazons. They are intended to allow two programs to play each other.

• The command *remote move/win means "Wait for an AI move from the other program, give it to this program, then execute a *move command. Repeat until one side sends a win message. Do not print the moves or win message.
• The command *remote move/win+ is the same as *remote move/win, but prints the "win" message.

The idea with these two commands is that one of the two scripts will, at a certain point, contain the commands

*move
*remote move/win

and the other will contain

*remote move/win

so that the first sends a move from its AI to the other program, which then waits for a response from its AI to send back, and so forth.

For the remote commands, both programs should generate "wins" messages, and test‑amazons will check that they are the same.

The test-amazons script throws out any other output from either program except for properly formatted board dumps, as are supposed to be produced by the dump command described previously. You can see all the output by running it with

 python3 testing/test-amazons --verbose TESTFILE-1.in

or

 python3 testing/test-amazons --verbose TESTFILE-1.in TESTFILE-2.in

which will show all the commands sent to each program and all their output.

The test-amazons program will report an error if a program hangs or times out, or if it exits abnormally (with an exception or an exit code other than 0). Finally, if there is a file TESTFILE-1.std or TESTFILE-2.std, test-amazons will check it against the output from the program for TESTFILE-1.in (likewise for TESTFILE-2.std against the output for TESTFILE-2.in).

## Extra Credit

First get your program working, and then, if you feel the urge, try the extra-credit GUI (Graphical User Interface). If you do, the option

java -ea amazons.Main --display

should create the GUI. If you don't implement the GUI, this option should cause your program to exit with a non-zero code via System.exit(2). Your GUI does not have to look at all like ours.

The board is an obvious place to start. We have provided suggestions for methods that you can use if you want, but you are not required to do so. We have structured the skeleton so that the different kinds of player (ordinary human at the keyboard using text commands, AI, or human using a GUI) are represented as different subtypes of a type Player, an example of how using OOP can cut down on pervasive conditional tests for types of player.


text:
# Definition:Half-Plane/Open/Right

## Definition

Let $\PP$ be a Cartesian plane.

Let $\LL$ be the $y$-axis.

The open right half-plane $\HH_{\text {OR} }$ is the area of $\PP$ on the right of $\LL$.

That is, where $x > 0$:

$\HH_{\text {OR} } := \set {\tuple {x, y}: x \in \R_{>0} }$


text:
### Step 1

Create a new document that is 11" wide by 8.5" tall. The color mode should be set to CMYK to ensure that colors are printable. Set raster effects to 300ppi for the highest quality print results.

Next, set up your layers. I like to start with a layer named after the character (in this case, "Suko"), with these sub layers:

• g – for character-specific guides (optional)
• d – for top details and line work
• sh – for shading and shine details that will be filled paths
• c – for color and shading with Live Paint

Feel free to use longer names that make more sense to you. You should also create a template layer at this point for your foundation drawing. I dim mine to 60% and set to unprintable in case I forget to delete it.

Center the template image on the artboard and lock your template layer. My pencil sketch is available here. I used a reference for the pose, but the clothing and character design are original. Suko is a character from my comic series, Green Corner.

The swatches for this file, if you'd like to use them, are available here. They are labeled and organized, so it should be easy to tell what each swatch is for. Note that the "main blue" swatch is a spot color so that I can use tints of it.

### Step 2

With the "d" layer active, start drawing with one of the calligraphic brushes with a black stroke set to 0.05 pt. Using a small stroke weight allows for a more delicate line. The brushes are angled to give subtle line variation. If you need a copy of the brushes, you can download them here.

It's easiest to start by drawing simpler lines first and leave the finer details such as facial features until later. I use the Pen Tool for drawing most lines. Use your pencil sketch as a guide, but don't be afraid to make corrections if you feel a line will look better if adjusted. Switch between the two brushes as necessary, if you find a line to be too thin or thick in a certain place.

A note about using these brushes: If you get the variation you like from a brush, but the weight needs adjusting, then adjust the stroke weight slightly for that path. Try 0.03 pt for finer lines, 0.07 pt for thicker lines. If a path has an unusual thickness in a certain area (like around the fingers and pointy ears), cut the path on a point with the Scissors Tool (C). Cutting on a point helps maintain the integrity of the path but can give you more control over the line quality.

### Step 3

Sometimes it's easiest to draw a continuous line and then cut out the section you don't need to be visible. The guide through the leg in the image below represents a continuous path that was cut, with the cut section converted to a guide (Command + 5). I move all character guides to the "g" layer as they are created. Make sure you're back on the "d" layer before you continue drawing.

### Step 4

For the eyes, I used the Ellipse Tool to create the iris, pupil, and shine. Remember to cut off and delete the excess of the iris. Alternatively, you can draw the sides of the iris with the Pen Tool. Depending on the view of the face, the Ellipse Tool may not be useful for the iris.

For symmetrical and nearly identical features, such as the eyes and eyebrows, draw one and copy. If you need to flip the copy, use the Reflect Tool (O) and hold down Option while flipping to copy. Then adjust with the selection tools as needed by moving, rotating, and reshaping. The image below shows the process of drawing one eye and eyebrow to creating two of each.

### Step 5

Your line work should be mostly done at this point and look something like the image below. Continue adding some more details, like the rose on its own layer above (using the same sub layer structure), but don't draw the thorns yet. Also, add the belt buckle by using the Rounded Rectangle Tool and the Subtract From Shape Area button in the pathfinder palette.

### Step 6

Now is a good time to make sure that the facial features are correct. Unlock and show everything on all layers and sub layers. Rotate everything 90 degrees clockwise with the Selection Tool (V), while holding down the Shift key.

Adjust the facial features as needed, but ignore how the brush styles look altered. Use outline view (Command + Y) to see the paths without styles if that helps you while adjusting. I changed Suko's expression at this point to a slight smile. When you're happy with the changes, rotate everything back to the original position.

### Step 7

When the line work is done, select and copy all of the paths on your "d" layer. Just click on the circle on the right side of the layer to select all paths on that sub layer and press Command + C. Then lock all the layers you are not using and activate the "c" layer. Make sure that Paste Remembers Layers is off.

Paste the copied line work on the "c" layer and delete the belt buckle. Compound paths can't be added to a Live Paint group. Then, select everything on the "c" layer and convert it to a Live Paint group by pressing Command + Option + X. You may receive a warning about brush styles being discarded. This is why our line work is on a separate layer.

### Step 8

Once you have your Live Paint group, hide your other layers. Then, use the Direct Selection Tool (A) to select any superfluous paths that you won't need for Live Paint and delete them. Once you've deleted all the extras and top details, it should look like the image below.

### Step 9

Select your Live Paint group and set the fill and stroke to none. Press the Forward Slash key then X then the Forward Slash key to do this quickly. Make the "d" layer visible again, but make sure only the "c" layer is unlocked. Now you can use the Live Paint bucket (K) to fill in your flat colors. Fillable areas are highlighted with a red outline, like in the image below.

### Step 10

If a path doesn't seem to fill correctly, pull points with the Direct Selection Tool (A), or extend paths with the Pen Tool (P), as shown in the image below. Outline view is very helpful for this, so switch back and forth as you need to by pressing Command + Y.

You may find it easiest to have the Live Paint group deselected and to view in normal view when filling areas. Sometimes the fill color won't stick in outline view or if any paths are selected.

### Step 11

After a while, your flat color should look something like the image below. The belt buckle is on the "d" layer above and filled with a gradient. Use the Gradient Tool to adjust the angle of the gradient. Don't forget to change the fill color for the eyebrows and paint the irises.

### Step 12

Use the Pen Tool to draw in lines for the whites of the eyes. The fastest way to do this is to click on an end point of the lower lash line then continue the path up to the upper lash line. Deselect by pressing Command + Shift + A and do the other side. Make sure to do both eyes.

When you go back to normal view, you may want to adjust the positioning of the points for the eye whites, so that they don't awkwardly extend past the lash lines.

### Step 13

Now we can move on to shading. Lock your "Live Paint" group and choose a bright color that you are not coloring with. I usually use pink. Make sure you have no brush selected and set your stroke to 0.25 pt. You need to see the lines, but you don't want them too thick as you need to make sure the shaping is correct.

Draw outlines for the areas where you will add your shading. Some areas are easier to create shading outlines for by copying existing paths, ignore those for now. If you shaded your sketch well, use outline view to use your sketch as a guide. When you're done outlining, use the Magic Wand Tool (Y) to select all the paths.

### Step 14

With your shading paths selected, press the Forward Slash key to change the stroke to none then send them to the back of the layer by pressing Command + Shift + Left Bracket key. Unlock the "Live Paint" group and drag your selected paths into it.

The easiest way to drag them is to select the square on the right side of the layer (see the topmost red square in the image below) and drag that into the Live Paint group. When done correctly, you will see only your "Live Paint" group in the layer, but your paths will still be selected.

### Step 15

Deselect the paths and fill with your shading colors. Make adjustments as needed if there are gaps between paths that prevent you from filling. When the bulk of the shading is completed for Live Paint, move to your "sh" layer to start adding extra details.

I like to add a shaded accent to the eyes (alternatively, the accent can be lighter than the iris), and add small shading paths for the lips and the nose. You can also add shiny details to the skin and clothing. Use shiny accents on the skin minimally, you usually don't want you character to look like they have greasy skin.

### Step 16

To make the line work more subtle, you can go back to your "d" layer and start cutting your paths (but don't delete anything) any time a line crosses an area of one color to another. then, select the lines for an area and change the stroke color to be a little darker than your shading color. Leave the pupils black. You can use a dark brown for the eyelashes. Note that a really dark brown may print as a rich black.

For the hair shine, use a white fill set to 60% transparency on the "sh" layer. Having the hair shine below the line work allows the lines to remain dominant while you add dimension.

### Step 17

Suko is wearing jeans, so let's fix the details on his pants to add some thread. Select the paths you want to change and discard the brush style. Then change the color to a dark yellow or light, warm brown. Set the stroke weight to 0.5pt with a 2pt dash and 1pt gap. You can also draw any additional paths you want to look like thread.

### Step 18

The rose is drawn and colored the same way as the rest of the art. However, you will need to copy some of the paths from Suko's "d" layer to overlap the edges where his fingers touch the rose. Cut off and delete any excess.

For the thorns, I draw little chevrons that overlap the stem. The thorns will use the same base color and stroke as the stem. The image below shows the thorns overlapping the stem in outline view and the initial shading. You can add some shine details to the rose as well. Use a pale pink for the shine on the rose petals, and use the skin base color on the leaves set to 60% transparency.

### Step 19

To create a shadow for the character, there are two methods. Either you can use the "live paint" group to create a drop shadow (set to multiply and 40% transparency), or you can create a shadow path. If you want to be able to rotate or modify the shape of the shadow, you need to use a path for the shadow.

Since I want to rotate and modify the shadow shape, I copied the outer paths from the "d" layer and pasted them onto a new layer below all other layers, the "shadow" layer. Convert these paths into a Live Paint group and fill with a single color (I used a 40% gray).

Make sure you leave the appropriate blank holes that coincide with the art above. When you're done filling, expand the "Live Paint" group by clicking the Expand button in the top toolbar or selecting Object > Live Paint > Expand from the menus. Then click on Merge in the pathfinder palette to remove any overlapping paths and simplify the shape. You will now have a compound path. If there are any stray points in the middle of your compound shape, select them with the Direct Select Tool (A) and delete.

### Step 20

Soften the edges of the shadow by selecting Effect > Stylize > Feather from the menus. Set the feather radius to 10 px. Adjust the positioning of the main shadow path and scale to 99% horizontal, 95% vertical (Object > Transform > Scale).

Rotate the path slightly. Use the Ellipse Tool to create a shadow for the rose. Set the feather radius to 8 px, and don't forget to move the path a little below the rose.

When you're done, it should look something like the image below. Remember to set your shadows to multiply if you will be adding a background. Depending on your background color, you may also want to adjust the fill color of the shadows later.

### Step 21

Now's the time to double-check that we haven't missed any details.

We're missing the shading cast by Suko's hairline, so we'll add that now. Lock your "Live Paint" group. Unlock the "d" layer and select only the paths you need to copy for one area (e.g. the hairline), copy them and re-lock the layer.

Then, paste in front on your "c" layer and offset the paths by moving them with the Selection Tool (V). You may want to discard the brush style and change to a pink stroke again. Cut off any excess of the paths you just moved. When you have deleted all the excess, select the paths with the Magic Wand Tool (Y) and drag to the "Live Paint" group as before. Now you can fill in the shading for that area. Fix any gaps as needed.

### Conclusion

And here's the final image with a simple, gradient background. You can use a more detailed background if you prefer. I hope you enjoyed this tutorial and learned something new!

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New answers tagged comets

2

A bit late for an answer, but maybe still interesting: National Geographic called comet McNaught the biggest comet measured: In the case of comet McNaught, Ulysses found that the plasma tail slowed the solar wind inside the bow wave to half its normal speed. The solar wind usually travels at about 435 miles (700 kilometers) a second at that distance from ...

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Van-der-Waals forces and gravitation are the main answer. The main property keeping together bodies smaller than a few $100 {\rm km}$ are their constituents surface forces between the grains (van-der-Waals) and chemical/ molecular bonding within the grains themselves. It's the same mechanic which makes the dust and grease in your flat stick to surfaces and ...

13

Yes, comets spin although measuring it can be tricky due to the coma and outgassing from the nucleus. It's easiest to measure the rotation period when the comet is inactive near aphelion although this is more difficult as the comet is fainter, necessitating a large amount of large (>4m) telescope time which is difficult to obtain. Searching the JPL Small ...

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# Can I use Table without the appropriate bounds?

I´d like to use the Table command without defining n, so that I can have a list of n arguments.

It is possible for the Sum command, but I guess not for the Table command, or is it? Something like

n = 5; eSO = Table[e[i], {i, 1, n}]; ebar[1] = EBAU - Sum[ebar[i], {i, 2, n}];
a[1] = A - Sum[a[i], {i, 2, n}]; b[1] = B - Sum[b[i], {i, 2, n}];

Cost = Table[1/2/a[i]*(ebar[i] - e[i])^2 + b[i]*(Sum[e[i], {i, n}])^2 /2, {i, 1, n}];

FOC = Table[D[Sum[Cost[[i]], {i, n}] == 0, e[i]], {i, 1, n}];

solsSO = eSO /. Solve[FOC, eSO];

solsSO = Flatten[solsSO];

ESO = Simplify[Sum[solsSO[[i]], {i, n}]]

but a bit more complicated in the end.

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If you want that your m is not fixed but variable you could try something like: wi=Table[Table[Sin[i],{i,1,m}],{m,1,10}] –  partial81 Jun 25 '12 at 13:27
I'm not sure I understand you correctly. Are you looking for this: Table[w[i], {i, {1, 2, 4, 3, 7, 2, 11}}], i.e. calling Table with a list of predefined index values? –  István Zachar Jun 25 '12 at 13:30
Id like a list with dimension n, so id get. {w[1],w[2],...,w[n-1],w[n]} –  Max M Jun 25 '12 at 13:51
You might like to have a look at this question, which demonstrates how to represent indeterminate-length (or infinite) lists in Mathematica. –  Oleksandr R. Jun 25 '12 at 15:46
I can't think of a way of doing this atm, but you might consider changing the title of your post to attract more attention. As I see it, the problem boils down to solving a system of equations with a indeterminate number of variables. –  sebhofer Jun 27 '12 at 7:48

It's not possible because it makes as much sense as wanting to have list of n elements with n being undefined. The most similar thing you can do is to have a symbolic representation of that, that evaluates to what you want when n gets a numeric value. For that, you can either define your own

symbolicTable[exp_, it:{_, __?NumericQ| _List}]:=Table[exp, it]

If you then use symbolicTable just like Table, it will only evaluate when the iterator bounds are numeric and remain unevaluated when they are not.

ooor, just turn off the warning message you get when you try to use Table with a non-numeric argument

Off[Table::iterb]

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Thx for you reply, but does this really help me with my problem? Maybe, I dont really understand your proposal. I still wont be able to have a list with n arguments then, will I? –  Max M Jun 25 '12 at 13:23
@Max: we might be more helpful if you tell us your actual problem, that is, the problem that requires this construction you speak of. –  Ｊ. Ｍ. Jun 25 '12 at 13:25
Its a rather simple optimization in economics. n would be the number of countries for which I have some cost function over which i minimize. Therefore I d like to have a list for the n countries where each argument represents one country, then do the differentiation and solve it. (still haveing trouble using minimize or sth similar). I can understand if that doesnt work since the program probably needs appropiate bounds for the list and the differentiation. if wi = Table[w[i], {i, 1, m}] is already not possible then the rest wont be either, i guess. –  Max M Jun 25 '12 at 13:38
@MaxM So do you want to do an analytical or a numerical calculation? –  sebhofer Jun 25 '12 at 14:03
@MaxM, I don't think Table can be used like that for analytical problems. You probably have to adapt your question giving more specifics of the problem at hand, to get real help –  Rojo Jun 25 '12 at 14:13


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# Linear Algebra Question (Kronecker Delta?)

1. Sep 19, 2013

### FeDeX_LaTeX

The problem statement, all variables and given/known data

For fixed m ≥ 1, let $\epsilon(i,j)$ denote the m x m matrix $\epsilon(i,j)_{rs} = \delta_{ir}\delta_{js}$, where i,j may denote any integers in the range 1 ≤ i,j ≤ m.

(a) When m = 4, write out all $\epsilon(i,j)$ explicitly and label them correctly.

The attempt at a solution

My course hasn't started yet and all I know about the Kronecker delta/Levi-Civita epsilon notation is what little I've seen in videos.

They haven't stated the parameters of r and s, but I guess that doesn't matter since $\delta_{i,j} = 1$ if and only if i = j, and 0 otherwise?

All I can think of at the moment is this (can anyone tell me if this is correct?):

$\epsilon(1,1)_{rs} = \delta_{1r}\delta_{1s} = \delta_{11}\delta_{11}$
$\epsilon(1,2)_{rs} = \delta_{1r}\delta_{2s} = \delta_{11}\delta_{22}$
$\epsilon(1,3)_{rs} = \delta_{1r}\delta_{3s} = \delta_{11}\delta_{33}$
(...) etc. making 16 values in total. Is this the kind of thing they're looking for?

I still can't quite understand what I'm doing -- how does $\delta_{11}\delta_{22}$ represent an entry in a 4 x 4 matrix, for instance? Does that mean that both the entries in row 1 column 1 AND row 2 column 2 have value 1?

(I apologise for deleting the 'relevant equations' section -- I literally don't know which ones are relevant.)

2. Sep 19, 2013

### tiny-tim

Hi FeDeX_LaTeX!
No, they want you write out eg $\epsilon(1,1)$ as a matrix:

$\epsilon(1,1)_{11}\ \ \epsilon(1,1)_{12}$
$\epsilon(1,1)_{21}\ \ \epsilon(1,1)_{22}$

3. Sep 19, 2013

### FeDeX_LaTeX

Thanks for the reply -- I'm confused... how did you know $\epsilon(1,1)$ should be a 2 x 2 matrix?

4. Sep 19, 2013

### Ray Vickson

He does not say that $\epsilon(1,1)$ is a 2 x 2 matrix; he just shows you some of what you need to compute. The question actually say m = 4, so you need a 4 x 4 matrix.

5. Sep 19, 2013

### FeDeX_LaTeX

So I'm completing a 4 x 4 matrix consisting of only $\epsilon(1,1)$?

6. Sep 19, 2013

### Office_Shredder

Staff Emeritus
$$\epsilon(1,1)$$ IS a 4x4 matrix. The r,s entry of it is given by $\delta_{1r} \delta_{s1}$.

In particular you will notice that many of the entries of this matrix are zero. Can you identify what values of r and s make $\delta_{1r} \delta_{s1}$ non-zero?

7. Sep 19, 2013

### Ray Vickson

It reads to me like it wants you to write out all 16 of the 4 x 4 matrices, although I don't see what the point of doing that could possibly be---unless, maybe, there is a pattern that it is important to exploit later in some context.

8. Sep 19, 2013

### FeDeX_LaTeX

Sorry, I wasn't aware it represented a matrix -- I thought it just represented some number permutations. So $\epsilon(1,1)$ is just going to give me a 4 x 4 matrix with zero for every entry apart from the top-left, right?

r = s = 1 makes it non-zero?

9. Sep 19, 2013

### Office_Shredder

Staff Emeritus
That's correct. The other ones can be solved similarly