text:
# Thread: Proofs! Externally Tangent Circles

1. ## Proofs! Externally Tangent Circles

2 circles – one of radius 5, the other of radius 8 – intersect at exactly one point and the center of each circle lies outside the other circle. A line is externally tangent to both circles. Find the distance between the two points of tangency. Draw a figure and label it. Justify your answer completely with postulates and theorems. A two-column proof is suggested.

THANKS SO MUCH FOR HELPING ME!!

2. More generally:

Consider two circles with radii $r,R$ then the value of line externally tangent to both circles is given by $2\sqrt{r\cdot R}$

3. thanks but can you be more specific and include proofs using postulates and theorems?

4. Originally Posted by cbbplanet
thanks but can you be more specific and include proofs using postulates and theorems?
Hello,

I've attached a diagram of the situation. (The light grey circles are necessary to construct the tangent points)

To calculate the distance between the tangent points you are dealing with a right triangle, which I've coloured grey.

The length of the hypotenuse is (R + r)
One leg (coloured violet) has the length (R-r)

Now use Pythagorean theorem:

$d^2 + (R-r)^2 = (R+r)^2$ Expand the brackets:

$d^2 + R^2 - 2rR + r^2 = R^2 + 2rR + r^2$ . Subtract on both sides r² and R²:

$d^2 = 4rR~\implies~d = 2\sqrt{r \cdot R}$

5. I NEED AN EMERGENCY ASSISTANCE!!! the proofs are due tommorrow.

i have figured everything out but can someone tell me how to prove the grey shade triangle is a right triangle? THANKS!!!

6. Originally Posted by cbbplanet
I NEED AN EMERGENCY ASSISTANCE!!! the proofs are due tommorrow.

i have figured everything out but can someone tell me how to prove the grey shade triangle is a right triangle? THANKS!!!
One of its angle is the angles between a tangent and a radius to a circle and hence right.

RonL


text:
FANDOM

154 Pages

The V-Cube 8 is an 8×8×8 version of Rubik's Cube. Unlike the original puzzle (but like the 4×4×4 and 6×6×6 cubes), it has no fixed facets: the center facets (36 per face) are free to move to different positions. The design was covered by Panagiotis Verdes' patent from 2007 but Verdes Innovations SA did not produce it for sale until 2014. Other 8×8×8 cubes are produced by the Chinese companies ShengShou and YuXin.

Methods for solving the 3×3×3 cube work for the edges and corners of the 8×8×8 cube, as long as one has correctly identified the relative positions of the colors — since the center facets can no longer be used for identification.

Mechanics Edit

The puzzle consists of 296 pieces ("cubies") on the surface. There are also 84 movable pieces entirely hidden within the interior of the cube, as well as six fixed pieces attached to the central "spider" frame. The V-Cube 9 uses essentially the same mechanism, except that on the latter these hidden pieces (corresponding to the center rows) are made visible.

There are 216 center pieces which show one color each, 72 edge pieces which show two colors each, and eight corner pieces which show three colors. Each piece (or sextet of edge pieces) shows a unique color combination, but not all combinations are present (for example, there is no edge piece with both red and orange sides, since red and orange are on opposite sides of the solved Cube). The location of these cubes relative to one another can be altered by twisting the layers of the Cube 90°, 180° or 270°, but the location of the colored sides relative to one another in the completed state of the puzzle cannot be altered: it is fixed by the distribution of color combinations on edge and corner pieces.

Currently, the 8×8×8 is produced with white plastic as a base, with red opposite orange, blue opposite green, and yellow opposite black. The V-Cube version has one center piece marked the letter V. Versions with black plastic and a white face, with the other colors remaining the same, are also available.

The V-Cube has rounded sides like the V-Cube 7, whereas the Chinese versions have flat faces. On all versions the outermost layers are thicker than the middle ones. Without this alteration there would be no way to connect the corner pieces to the rest of the mechanism.

PermutationsEdit

There are 8 corners, 72 edges and 216 centers.

Any permutation of the corners is possible, including odd permutations. Seven of the corners can be independently rotated, and the orientation of the eighth depends on the other seven, giving 8!×37 combinations.

There are 216 centers, consisting of nine sets of 24 pieces each. Within each set there are four centers of each color. Centers from one set cannot be exchanged with those from another set. Each set can be arranged in 24! different ways. Assuming that the four centers of each color in each set are indistinguishable, the number of permutations is reduced to 24!/(246) arrangements. The reducing factor comes about because there are 24 ways to arrange the four pieces of a given color. This is raised to the sixth power because there are six colors. The total number of center permutations is the permutations of a single set raised to the ninth power, 24!9/(2454).

There are 72 edges, consisting of 24 inner, 24 intermediate and 24 outer edges. These cannot be flipped (because the internal shape of the pieces is asymmetrical), nor can an edge from one set exchange places with an edge from another set. The six edges in each matching sextet are distinguishable, since corresponding edges are mirror images of each other. Any permutation of the edges in each set is possible, including odd permutations, giving 24! arrangements for each set or 24!3 total, regardless of the position or orientation any other pieces.

Assuming the cube does not have a fixed orientation in space, and that the permutations resulting from rotating the cube without twisting it are considered identical, the number of permutations is reduced by a factor of 24. This is because the 24 possible positions and orientations of the first corner are equivalent because of the lack of fixed centers. This factor does not appear when calculating the permutations of N×N×N cubes where N is odd, since those puzzles have fixed centers which identify the cube's spatial orientation.

This gives a total number of permutations of

$\frac{8! \times 3^7 \times 24!^{12}}{24^{55}} \approx 3.52 \times 10^{217}$

The entire number is 35 173 780 923 109 452 777 509 592 367 006 557 398 539 936 328 978 098 352 427 605 879 843 998 663 990 903 628 634 874 024 098 344 287 402 504 043 608 416 113 016 679 717 941 937 308 041 012 307 368 528 117 622 006 727 311 360 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000.

The V-Cube has one center piece marked with a V, which distinguishes it from the other three in its set. This increases the number of patterns by a factor of four to 1.41×10218, although any of the four possible positions for this piece could be regarded as correct.

SolutionsEdit

There are a number of methods that can be used to solve a V-Cube 8. One method is to first group the center pieces of common colors together, then to match up edges that show the same two colors. Once this is done, turning only the outer layers of the cube allows it to be solved like a 3×3×3 cube. However, certain positions that cannot be solved on a standard 3×3×3 cube may be reached. For instance, a single sextet of edges may be inverted, or the cube may appear to have an odd permutation (that is, two pieces must be swapped, which is not possible on the 3×3×3 cube). These situations are known as parity errors, and require special algorithms to be solved.

Another similar approach to solving this cube is to first pair the edges, and then the centers. This, too, is vulnerable to the parity errors described above.

Other methods solve the cube by solving a cross and the centers, but not solving any of the edges and corners not needed for the cross, then the other edges would be placed similar to the 3x3 Fridrich method.

Some methods are designed to avoid the parity errors described above. For instance, solving the corners and edges first and the centers last would avoid such parity errors. Once the rest of the cube is solved, any permutation of the center pieces can be solved. Note that it is possible to apparently exchange a pair of face centers by cycling 3 face centers, two of which are visually identical.

RecordsEdit

The World Cube Association does not keep records for this puzzle as of 2018. Mattia Furlan claims the unofficial world record with a time of 4:36.3 minutes.

SourcesEdit

1. http://en.wikipedia.org/wiki/V-Cube_8

format: plain


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# Liouville's theorem (Hamiltonian)

A separate article is about Liouville's theorem in complex analysis: see Liouville's theorem (complex analysis).

In mathematical physics, Liouville's theorem, named after the French mathematician Joseph Liouville, is a key theorem in classical statistical and Hamiltonian mechanics. It asserts that the phase-space distribution function is constant along the trajectories of the system - that is that the density of system points in the vicinity of a given system point travelling through phase-space is constant with time..

Liouville's theorem is also important in the study of symplectic topology, where it is formulated rather differently.

 Contents

## Liouville equation

The Liouville equation describes the time evolution of phase space distribution function (while density is the correct term from mathematics, physicists generally call it a distribution). Consider a dynamical system with coordinates [itex]q_i[itex] and conjugate momenta [itex]p_i[itex], where [itex]i=1,\dots,d[itex]. Then the phase space distribution [itex]\rho(p,q)[itex] determines the probability [itex]\rho(p,q)\,d^dq\,d^dp[itex] that a particle will be found in the infinitesimal phase space volume [itex]d^dq\,d^dp[itex]. The Liouville equation governs the evolution of [itex]\rho(p,q;t)[itex] in time [itex]t[itex]:

[itex]\frac{d\rho}{dt}=\frac{\partial\rho}{\partial t}+\sum_{i=1}^d\left(\frac{\partial\rho}{\partial q_i}\dot{q}_i+\frac{\partial\rho}{\partial p_i}\dot{p}_i\right)=0.[itex]

Time derivatives are denoted by dots, and are evaluated according to Hamilton's equations for the system. This equation demonstrates the conservation of density in phase space (which was Gibbs's name for the theorem). Liouville's theorem states that

The distribution function is constant along any trajectory in phase space.

A simple proof of the theorem is to observe that the evolution of [itex]\rho[itex] is defined by the continuity equation:

[itex]\frac{\partial\rho}{\partial t}+\sum_{i=1}^d\left(\frac{\partial(\rho\dot{q}_i)}{\partial q_i}+\frac{\partial(\rho\dot{p}_i)}{\partial p_i}\right)=0.[itex]

and notice that the difference between this and Liouville's equation are the terms

[itex]\rho\sum_{i=1}^d\left(\frac{\partial\dot{q}_i}{\partial q_i}+\frac{\partial\dot{p}_i}{\partial p_i}\right)=\rho\sum_{i=1}^d\left(\frac{\partial H}{\partial q_i\,\partial p_i}-\frac{\partial H}{\partial p_i \partial q_i}\right)=0,[itex]

where [itex]H[itex] is the Hamiltonian, and Hamilton's equations have been used. That is, viewing the motion through phase space as a 'fluid flow' of system points, the theorem that the convective derivative of the density [itex]d \rho/dt[itex] is zero follows from the equation of continuity by noting that the 'velocity field' [itex](\dot p , \dot q)[itex] in phase space has zero divergence (which follows from Hamilton's relations).

Another illustration is to consider the trajectory of a cloud of points through phase space. It is straightforward to show that as the cloud stretches in one coordinate – [itex]p_i[itex] say – it shrinks in the corresponding [itex]q_i [itex] direction so that the product [itex]\Delta p_i \Delta q_i [itex] remains constant.

## Physical interpretation

The expected total number of particles is the integral over phase space of the distribution:

[itex]N=\int d^dq\,d^dp\,\rho(p,q)[itex]

(A normalizing factor is conventionally included in the phase space measure, but has been omitted here.) In the simple case of a particle moving in Euclidean space under a force field [itex]\mathbf{F}[itex] with coordinates [itex]\mathbf{x}[itex] and momenta [itex]\mathbf{p}[itex], Liouville's theorem can be written

[itex]\frac{\partial\rho}{\partial t}+\mathbf{v}\cdot\nabla_\mathbf{x}\rho+\frac{\mathbf{F}}{m}\cdot\nabla_\mathbf{p}\rho=0,[itex]

where [itex]\mathbf{v}=\dot{\mathbf{x}}[itex] is the velocity. In astrophysics this is called the Vlasov equation, and is used to describe the evolution of a large number of collisionless particles moving in a gravitational potential.

In classical statistical mechanics, the number of particles [itex]N[itex] is very large, (typically of order Avogadro's number, for a laboratory-scale system). Setting [itex]\partial\rho/\partial t=0[itex] gives an equation for the stationary states of the system and can be used to find the density of microstates accessible in a given statistical ensemble. The stationary states equation is satisfied by [itex]\rho[itex] equal to any function of the Hamiltonian [itex]H[itex]: in particular, it is satisfied by the Maxwell-Boltzmann distribution [itex]\rho\propto e^{-H/kT}[itex], where [itex]T[itex] is the temperature and [itex]k[itex] the Boltzmann constant.

See also canonical ensemble and microcanonical ensemble

## Other formulations

The theorem is often restated in terms of the Poisson bracket:

[itex]\frac{\partial\rho}{\partial t}=-\{\,\rho,H\,\}[itex]

or the Liouville operator or Liouvillian,

[itex]\hat{\mathbf{L}}=\sum_{i=1}^{d}\left[\frac{\partial H}{\partial p_{i}}\frac{\partial}{\partial q_{i}}-\frac{\partial H}{\partial q_{i}}\frac{\partial }{\partial p_{i}}\right],[itex]

as

[itex]\frac{\partial \rho }{\partial t}+{\hat{L}}\rho =0.[itex]

Another way to formulate Liouville's theorem is to say that a phase-space volume [itex]\Gamma[itex] is conserved under time translation. If

[itex]\int_\Gamma d^dq\,d^dp = C,[itex]

and [itex]\Gamma(t)[itex] is the set of points in phase-space which the points of [itex]\Gamma[itex] can evolve into at time [itex]t[itex], then

[itex]\int_{\Gamma(t)} d^dq\,d^dp = C,[itex]

for all times [itex]t[itex]. That is, phase space volumes are conserved. Since time-evolution in Hamiltonian mechanics is a canonical transformation this can be proved by showing that all canonical transformations have unit Jacobian.

In terms of a symplectic geometry the theorem states that the 2-form S, formed from the wedge product of [itex]\Delta p_i[itex] and [itex]\Delta q_i[itex] has a Lie derivative for its Hamiltonian evolution (given by the Poisson bracket with respect to the vector field {H, } which vanishes).

## See also

• Canonical quantization yields a quantum-mechanical version of this theorem. This procedure, often used to devise quantum analogues of classical systems, involves describing a classical system using Hamiltonian mechanics. Classical variables are then re-interpreted as quantum operators, while Poisson brackets are replaced by commutators. In this case, the resulting equation is
[itex]\frac{\partial}{\partial t}\rho=-\frac{i}{\hbar}[\rho,H][itex]
where ρ is the density matrix.
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Generic Puiseux series

# Generic Puiseux series

AbstractAlgebra.jl allows the creation of Puiseux series over any computable commutative ring $R$.

Puiseux series are power series of the form $a_jx^{j/m} + a_{j+1}x^{(j+1)/m} + \cdots + a_{k-1}x^{(k-1)/m} + O(x^{k/m})$ for some integer $m > 0$ where $i \geq 0$, $a_i \in R$ and the relative precision $k - j$ is at most equal to some specified precision $n$.

The generic Puiseux series module is implemented in src/generic/PuiseuxSeries.jl.

As well as implementing the Series Ring interface, the Puiseux series module in AbstractAlgebra.jl implements the generic algorithms described below.

All of the generic functionality is part of the Generic submodule of AbstractAlgebra.jl. This is exported by default so that it is not necessary to qualify function names.

## Types and parent objects

The types of generic polynomials implemented by AbstractAlgebra.jl are Generic.PuiseuxSeriesRingElem{T} and Generic.PuiseuxSeriesFieldElem{T}.

Both series element types belong to the union type Generic.PuiseuxSeriesElem.

Puiseux series elements belong directly to either AbstractAlgebra.RingElem or AbstractAlgebra.FieldElem since it is more useful to be able to distinguish whether they belong to a ring or field than it is to distinguish that they are Puiseux series.

The parent types for Puiseux series, Generic.PuiseuxSeriesRing{T} and Generic.PuiseuxSeriesField{T} respectively, belong to AbstractAlgebra.Ring and AbstractAlgebra.Field respectively.

The default precision, string representation of the variable and base ring $R$ of a generic Puiseux series are stored in its parent object.

## Puisuex series ring constructors

In order to construct Puiseux series in AbstractAlgebra.jl, one must first construct the ring itself. This is accomplished with any of the following constructors.

PuiseuxSeriesRing(R::AbstractAlgebra.Ring, prec_max::Int, s::AbstractString; cached::Bool = true)
PuiseuxSeriesRing(R::AbstractAlgebra.Field, prec_max::Int, s::AbstractString; cached::Bool = true)
PuiseuxSeriesField(R::AbstractAlgebra.Field, prec_max::Int, s::AbstractString; cached::Bool = true)

Given a base ring R, a maximum relative precision and a string s specifying how the generator (variable) should be printed, return a typle S, x representing the Puiseux series ring and its generator.

By default, S will depend only on S, x and the maximum precision and will be cached. Setting the optional argument cached to false will prevent this.

Here are some examples of constructing various kinds of Puiseux series rings and coercing various elements into those rings.

Examples

julia> R, x = PuiseuxSeriesRing(ZZ, 10, "x")
(Puiseux series ring in x over Integers, x+O(x^11))

julia> S, y = PuiseuxSeriesField(QQ, 10, "y")
(Puiseux series field in y over Rationals, y+O(y^11))

julia> f = R()
0+O(x^10)

julia> g = S(123)
123//1+O(y^10)

julia> h = R(BigInt(1234))
1234+O(x^10)

julia> k = S(y + 1)
1//1+y+O(y^10)

## Big-oh notation

Series elements can be given a precision using the big-oh notation. This is provided by a function of the following form, (or something equivalent for Laurent series):

O(x::SeriesElem)

Examples

julia> R, x = PuiseuxSeriesRing(ZZ, 10, "x")
(Puiseux series ring in x over Integers, x+O(x^11))

julia> f = 1 + 2x + O(x^5)
1+2*x+O(x^5)

julia> g = 2x^(1//3) + 7x^(2//3) + O(x^(7//3))
2*x^1//3+7*x^2//3+O(x^7//3)

What is happening here in practice is that O(x^n) is creating the series 0 + O(x^n) and the rules for addition of series dictate that if this is added to a series of greater precision, then the lower of the two precisions must be used.

Of course it may be that the precision of the series that O(x^n) is added to is already lower than n, in which case adding O(x^n) has no effect. This is the case if the default precision is too low, since x on its own has the default precision.

## Puiseux series implementation

Puiseux series have their maximum relative precision capped at some value prec_max. This refers to the internal Laurent series used to store the Puiseux series, i.e. the series without denominators in the exponents.

The Puiseux series type stores such a Laurent series and a scale or denominator for the exponents. For example, $f(x) = 1 + x^{1/3} + 2x^{2/3} + O(x^{7/3})$ would be stored as a Laurent series $1 + x + 2x^2 + O(x^7)$ and a scale of $3$..

The maximum precision is also used as a default (Laurent) precision in the case of coercing coefficients into the ring and for any computation where the result could mathematically be given to infinite precision.

In all models we say that two Puiseux series are equal if they agree up to the minimum absolute precision of the two power series.

Thus, for example, $x^5 + O(x^{10}) == 0 + O(x^5)$, since the minimum absolute precision is $5$.

Sometimes it is necessary to compare Puiseux series not just for arithmetic equality, as above, but to see if they have precisely the same precision and terms. For this purpose we introduce the isequal function.

For example, if $f = x^2 + O(x^7)$ and $g = x^2 + O(x^8)$ and $h = 0 + O(x^2)$ then $f == g$, $f == h$ and $g == h$, but isequal(f, g), isequal(f, h) and isequal(g, h) would all return false. However, if $k = x^2 + O(x^7)$ then isequal(f, k) would return true.

There are a number of technicalities that must be observed when working with Puiseux series. As these are the same as for the other series rings in AbstractAlgebra.jl, we refer the reader to the documentation of series rings for information about these issues.

## Basic ring functionality

All Puiseux series provide the functionality described in the Ring and Series Ring interfaces with the exception of the pol_length and polcoeff functions.

Examples

julia> S, x = PuiseuxSeriesRing(ZZ, 10, "x")
(Puiseux series ring in x over Integers, x+O(x^11))

julia> f = 1 + 3x + x^3 + O(x^10)
1+3*x+x^3+O(x^10)

julia> g = 1 + 2x^(1//3) + x^(2//3) + O(x^(7//3))
1+2*x^1//3+x^2//3+O(x^7//3)

julia> h = zero(S)
0+O(x^10)

julia> k = one(S)
1+O(x^10)

julia> isone(k)
true

julia> iszero(f)
false

julia> U = base_ring(S)
Integers

julia> v = var(S)
:x

julia> T = parent(x + 1)
Puiseux series ring in x over Integers

julia> g == deepcopy(g)
true

julia> t = divexact(2g, 2)
1+2*x^1//3+x^2//3+O(x^7//3)

julia> p = precision(f)
10//1

## Puiseux series functionality provided by AbstractAlgebra.jl

The functionality below is automatically provided by AbstractAlgebra.jl for any Puiseux series.

Of course, modules are encouraged to provide specific implementations of the functions described here, that override the generic implementation.

### Basic functionality

coeff(a::Generic.PuiseuxSeriesElem, n::Int)
coeff(a::Generic.PuiseuxSeriesElem, n::Rational{Int})

Return the coefficient of the term of exponent $n$ of the given power series. If $n$ exceeds the current precision of the power series or does not correspond to a nonzero term of the Puiseux series, the function returns a zero coefficient.

modulus(a::Generic.PuiseuxSeriesElem{T}) where {T <: ResElem}

Return the modulus of the coefficients of the given Puiseux series.

isgen(a::Generic.PuiseuxSeriesElem)

Return true if the given Puiseux series is arithmetically equal to the generator of its Puiseux series ring to its current precision, otherwise return false.

isunit(a::Generic.PuiseuxSeriesElem)

Return true if the given Puiseux series is arithmetically equal to a unit, i.e. is invertible, otherwise return false.

Examples

julia> R, t = PuiseuxSeriesRing(QQ, 10, "t")
(Puiseux series field in t over Rationals, t+O(t^11))

julia> S, x = PuiseuxSeriesRing(R, 30, "x")
(Puiseux series field in x over Puiseux series field in t over Rationals, x+O(x^31))

julia> a = O(x^4)
0//1+O(t^10)+O(x^4)

julia> b = (t + 3)*x + (t^2 + 1)*x^2 + O(x^4)
(3//1+t+O(t^10))*x+(1//1+t^2+O(t^10))*x^2+O(x^4)

julia> k = isgen(gen(R))
true

julia> m = isunit(-1 + x^(1//3) + 2x^2)
true

julia> n = valuation(a)
4//1

julia> p = valuation(b)
1//1

julia> c = coeff(b, 2)
1//1+t^2+O(t^10)

### Division

inv(M::Generic.MatrixElem{T}) where {T <: RingElement}

Given a non-singular $n\times n$ matrix over a ring, return an $n\times n$ matrix $X$ such that $MX = I_n$, where $I_n$ is the $n\times n$ identity matrix. If $M$ is not invertible over the base ring an exception is raised.

Examples

julia> R, x = PuiseuxSeriesRing(QQ, 30, "x")
(Puiseux series field in x over Rationals, x+O(x^31))

julia> a = 1 + x + 2x^2 + O(x^5)
1//1+x+2//1*x^2+O(x^5)

julia> b = R(-1)
-1//1+O(x^30)

julia> c = inv(a)
1//1-x-x^2+3//1*x^3-x^4+O(x^5)

julia> d = inv(b)
-1//1+O(x^30)

### Special functions

exp(a::AbstractAlgebra.RelSeriesElem)

Return the exponential of the power series $a$.

exp(a::Generic.LaurentSeriesElem)

Return the exponential of the power series $a$.

exp(a::Generic.PuiseuxSeriesElem{T}) where T <: RingElement

Return the exponential of the given Puiseux series $a$.

sqrt(a::Generic.PuiseuxSeriesElem{T}) where T <: RingElement

Return the square root of the given Puiseux series $a$.

Examples

julia> R, t = PolynomialRing(QQ, "t")
(Univariate Polynomial Ring in t over Rationals, t)

julia> S, x = PuiseuxSeriesRing(R, 30, "x")
(Puiseux series ring in x over Univariate Polynomial Ring in t over Rationals, x+O(x^31))

julia> T, z = PuiseuxSeriesRing(QQ, 30, "z")
(Puiseux series field in z over Rationals, z+O(z^31))

julia> a = 1 + z + 3z^2 + O(z^5)
1//1+z+3//1*z^2+O(z^5)

julia> b = z + 2z^2 + 5z^3 + O(z^5)
z+2//1*z^2+5//1*z^3+O(z^5)

julia> c = exp(x + O(x^40))
1//1+x+1//2*x^2+1//6*x^3+1//24*x^4+1//120*x^5+1//720*x^6+1//5040*x^7+1//40320*x^8+1//362880*x^9+1//3628800*x^10+1//39916800*x^11+1//479001600*x^12+1//6227020800*x^13+1//87178291200*x^14+1//1307674368000*x^15+1//20922789888000*x^16+1//355687428096000*x^17+1//6402373705728000*x^18+1//121645100408832000*x^19+1//2432902008176640000*x^20+1//51090942171709440000*x^21+1//1124000727777607680000*x^22+1//25852016738884976640000*x^23+1//620448401733239439360000*x^24+1//15511210043330985984000000*x^25+1//403291461126605635584000000*x^26+1//10888869450418352160768000000*x^27+1//304888344611713860501504000000*x^28+1//8841761993739701954543616000000*x^29+1//265252859812191058636308480000000*x^30+O(x^31)

julia> d = divexact(x, exp(x + O(x^40)) - 1)
1//1-1//2*x+1//12*x^2-1//720*x^4+1//30240*x^6-1//1209600*x^8+1//47900160*x^10-691//1307674368000*x^12+1//74724249600*x^14-3617//10670622842880000*x^16+43867//5109094217170944000*x^18-174611//802857662698291200000*x^20+77683//14101100039391805440000*x^22-236364091//1693824136731743669452800000*x^24+657931//186134520519971831808000000*x^26-3392780147//37893265687455865519472640000000*x^28+O(x^29)

julia> f = exp(b)
1//1+z+5//2*z^2+43//6*z^3+193//24*z^4+O(z^5)

julia> h = sqrt(a)
1//1+1//2*z+11//8*z^2-11//16*z^3-77//128*z^4+O(z^5)


text:
# Math Help - Trig Derivative Problem.

1. ## Trig Derivative Problem.

Hello everyone, any assistance would be greatly appreciated.

Let $f(x)$ $=$ $12xsinxcosx$ . Find $f'(\frac{-3\pi}{2})$.

I attempt to differentiate the function using the product rule:
I use the product rule for $sinx cosx$
$sinx(-sinx)+cosx(cosx)=-sin^2x+cos^2x$

And then $12x(-sin^2x+cos^2x)$
$=12x(-1)+(-sin^2x+cos^2x)(12)$
$=-12x-12sin^2x+12cos^2x$

Using the derivative of $f(x)$, I try to evaluate $f'(\frac{-3\pi}{2})$.
$-12(\frac{-3\pi}{2})-12sin^2(\frac{-3\pi}{2})+12cos^2(\frac{-3\pi}{2})$

I end up getting $18\pi-12$

However, the book answer is just $18\pi$.
Can someone tell me where I went wrong? Thanks in advance.

2. ## Re: Trig Derivative Problem.

You're best off if you use the identity \displaystyle \begin{align*} \sin{(2x)} \equiv 2\sin{(x)}\cos{(x)} \end{align*}. Then \displaystyle \begin{align*} f(x) = 12x\sin{(x)}\cos{(x)} = 6x\sin{(2x)} \end{align*}. Then by the product rule we have

\displaystyle \begin{align*} f(x) &= 6x\sin{(2x)} \\ f'(x) &= 6\left[ 2x\cos{(2x)} + \sin{(2x)} \right] \\ f'\left( -\frac{3\pi}{2} \right) &= 6\left[ 2\left( -\frac{3\pi}{2} \right) \cos{( -3\pi )} + \sin{(-3\pi)} \right] \\ &= 6\left[ -3\pi \left( -1 \right) + 0 \right] \\ &= 18\pi \end{align*}

3. ## Re: Trig Derivative Problem.

Thank you very much for the quick response Prove It.


text:
# Discrete math. Finding a perfect square.

The problem is: Find all natural numbers $n$ for which $2^n + 1$ is a perfect square?

I am having a bit of trouble finding a generic way of finding these numbers. Of course the first obvious solution is $n = 3.$ For which we have $8 + 1 = 3^2.$ Anyone has any smart ideas?

## 2 Answers

$$2^n+1=m^2\implies 2^n=(m-1)(m+1).$$ This implies that $(m-1)$ and $(m+1)$ are both powers of $2$. In particular, $m-1$ divides $m+1$ so that $m-1$ divides $(m+1)-(m-1)=2$. This means that $m-1$ is either $2$ or $1$. Only $m-1=2$ works, so $m=3$ and $n=3$ constitute the only solution.

• Glorious leader has done it again! He is truly all-knowing! – PyRulez Apr 13 '15 at 23:58

Method $$\#1:$$

If $$2^n+1=m^2\iff2^n=(m+1)(m-1)$$

We can easily test for $$n\le2$$

For $$n>2,$$ clearly $$m$$ is odd

and we have $$2^{n-2}=\frac{m-1}2\cdot\frac{m+1}2$$

But as $$\displaystyle\frac{m+1}2-\frac{m-1}2=1,\left(\frac{m-1}2,\frac{m+1}2\right)=1,$$ at least one of them is odd

But each divides $$2^{n-2},$$ the odd must be $$\pm1$$

If $$\displaystyle\frac{m-1}2=1, m=3$$ which is a legitimate solution

If $$\displaystyle\frac{m+1}2=-1, m=-3$$ which is also a valid solution

But, $$\displaystyle\frac{m-1}2=-1$$ and $$\displaystyle\frac{m+1}2=1$$ makes $$2^n=0$$

Method $$\#2:$$

If $$n\ge1,2^n+1$$ is odd, for the existence of a square we need $$2^n+1=(2a+1)^2\iff2^n=4a(a+1),4$$ must divide $$2^n\implies n\ge2$$

So, we have $$2^{n-2}=a(a+1)$$ and again, $$(a+1,a)=(a+1-a,a)=1$$

As exactly, one of $$a,a+1$$ is odd

Now any odd divisor of $$2^{n-2}$$ must be $$\pm1$$

Case $$\#1:$$ If $$a$$ is odd

Case $$\#2:$$ If $$a+1$$ is odd


text:
# Not fooled by randomness

September 10, 2012
By

(This article was first published on Portfolio Probe » R language, and kindly contributed to R-bloggers)

The paper is “Not Fooled by Randomness: Using Random Portfolios to Analyze Investment Funds” by Roberto Stein.  Here is an explanation of the idea of random portfolios.

## Favorite sentence

The real question here is whether we’re actually measuring skill, or these are still measures of performance, so influenced by extraneous factors that the existence of the funds manager’s skill cannot be ascertained.

Skill and performance are often conflated, but they are separate.  Performance depends not only on skill but on the constraints that are faced.

## Real data

The exciting part of the paper for me is the analysis of a group of U.S. equity mutual funds using monthly data from 2005 through 2010.

Figure 1 shows some information on the funds in the study.

Figure 1: Annual turnover and number of names in the funds that are studied.

An executive summary is that there was reasonably good evidence that at least one (Jensen) of the 20 exhibited skill, but much less evidence of that once expenses are included.  There were also two that showed negative skill over the time period, that is, seemed to actively destroy value for the investor.

Figure 2 shows the returns for portfolios that have the constraints:

• assets from the constituents of the S&P 500
• no more than 100 names in the portfolio
• long-only
• maximum weight is 5% (at the start of 2006, there will be drift)

The returns are from the start of 2006 through 2011.

Figure 2: Distribution of random portfolio returns (2006-2011): simple returns (blue), log returns (gold). The shape for the simple returns is very similar to Figure 1 in the Stein paper — they are both for 6 year periods, but they are a year off.  This plot has 2011 instead of 2005.

We expect the simple returns to have a positive skew.  Our prior for the log returns is that they might be symmetric.

## Multi-fund models

The paper starts with a discussion of papers that use a bootstrap methodology on a universe of funds.  In particular, the papers:

• Kosowski, R., Timmermann, A., Wermers, R., White, H., 2006, “Can Mutual Fund Stars Really Pick Stocks? New Evidence from a Bootstrap Analysis”, Journal of Finance, Vol. LXI, No. 6, 2551-2595. (SSRN version)
• Cuthbertson, K., Nitzsche, D., O’Sullivan, N., 2008, “UK mutual fund performance: skill or luck?”, Journal of Empirical Finance 15, 613-634. (SSRN version)
• Fama, E., K. French, 2010, “Luck versus skill in the cross-section of mutual fund returns”, Journal of Finance, Vol. LXV-5, 1915-1947. (SSRN version)

The techniques of these papers have several problems:

• They demand having data on a big universe of funds.
• They depend on a model of fund returns.
• They implicitly assume that all funds face similar constraints.
• Even if the above were not problems, you don’t get specific information about any particular fund.

In contrast, evaluating skill with random portfolios is much simpler and more informative.

## Constrained random portfolio analysis

One methodology for assessing fund management skill with random portfolios is:

1. Do random trades through the evaluation period starting with the positions held by the fund at the start of the period.  The trades obey the fund constraints and mimic the turnover of the fund (and cash flow, if known).
2. Repeat step 1 multiple times.
3. Compare the period return of the fund to the distribution of returns of the random portfolios.

The portfolios created in the first two steps are pure luck given the decisions that the fund actually faced.  We infer skill if the real fund does not look like it fits in the “luck” distribution.  More color on this methodology can be found at this performance measurement page.

The Stein paper introduces an alternative way of using random portfolios to distinguish skill from luck.  A limitation of the technique just described is that it only uses the return for the whole period.  It throws away all the information about how that was achieved.

The paper describes a means of using the whole series of returns within the period.  Steps 1 and 2 remain the same.  Step 3 is more complicated.

## Conclusion

The final sentence of the paper is one that I agree with:

In all, [random portfolio analysis] is an important addition to the finance analysis toolkit.

## Appendix R

#### read in data from a table in pdf

The data to get Figure 1 was taken out of the paper by (in Windows) copying the body of Table I to the clipboard and then doing the R command:

notFoolFunds <- matrix(readClipboard(), nrow=20,
byrow=TRUE)

This is a chararcter matrix but we are really interested in the numbers.  So make a new matrix with the two columns we care about and modify the result to be numeric:

notFoolFundsNum <- notFoolFunds[, 4:5]
mode(notFoolFundsNum) <- "numeric"

Figure 1 is sort of:

plot(notFoolFundsNum)

#### random portfolio returns

The random portfolios for Figure 2 were generated (10,000 of them) by:

require(PortfolioProbe)
rp100w05 <- random.portfolio(1e4, prices=sp5.close[1,],
gross=1e6, long.only=TRUE, port.size=100,
max.weight=.05)

Then the returns are computed:

sret.rp100w05 <- valuation(rp100w05,
prices=sp5.close[c language="(1,1511),"][/c], returns="simple")
lret.rp100w05 <- valuation(rp100w05,
prices=sp5.close[c language="(1,1511),"][/c], returns="log")

Of course, we could have just transformed the simple returns into log returns or vice versa.

A plot of the density of the simple returns is:

plot(density(sret.rp100w05 * 100))

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text:
# Limiting skysphere's y movement

## Recommended Posts

Hi, Iv got a skysphere that is centred around the camera using a shader (multiplying the verts positions by the view, then the proj matrix and outputting xyww). Anyway, in the scene there is also a sea plane. As the camera moves up and down the skysphere follows the camera, however because of this the clouds on the skysphere appear to move up and down whilst the sea plane remains in the same place (as it should). Can anyone come up with an idea for limiting the cameras movement in the Y direction so this is not so apparent? Cheers.

##### Share on other sites
Uh, I don't suppose you can set the sky sphere's scale *really* big and just have it follow the X and Z movements on the player, could you? i.e.
void CSkyObject::UpdatePosition(){     skysphere.pos.x = player.pos.x;     skysphere.pos.z = player.pos.z;}

Is there a reason why you haven't tried that already?

## Create an account

Register a new account

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text:
# [texhax] nested macro and counter generation [better]

Uwe Lueck uwe.lueck at web.de
Sat May 10 00:19:52 CEST 2008

... my former list may not have helped you, sorry. But isn't it simply

\newcommand{\CREATE}[1]{%
\newcounter{#1}\expandafter \newcommand \csname #1\endcsname}

? Then use \CREATE like \newcommand with the only difference that you omit
the backslash starting the desired command name, or like \newenvironment
omitting the final argument.

But it is not entirely clear to me what you mean about the counters.
Do you mean you may need more than one counter per one new command? Then perhaps

\makeatletter
\newcommand{\CREATE}[2][\empty\empty]{% effect: \def\@fortmp{\empty}%
\@for\CREATEid:=#1\do{\newcounter{#2\CREATEid}}%
\expandafter \newcommand \csname #2\endcsname}
\makeatother

(tested). The \@for is a LaTeX internal.
The mandatory argument would be the command name (without backslash),
again use \CREATE similarly to \newcommand as above. If you want more
than one counter, use the optional argument for a comma-separated list.
E.g., \CREATE[a,b,c]{foo}[1]{\typeout{#1!!}} will obtain counter names
fooa, foob, fooc.

The last \CREATEid could be preceded with a, e.g., -', to obtain counter names,
e.g., foo-a, foo-b, foo-c:

\makeatletter
\newcommand{\CREATE}[2][\empty\empty]{%
\@for\CREATEid:=#1\do{%
\newcounter{#2\expandafter \ifx\CREATEid\empty \else -\CREATEid \fi}}%
\expandafter \newcommand \csname #2\endcsname}
\makeatother

but \thefoo-a would break. Then you might define something like

\newcommand{\The}[1]{\csname the#1\endcsname}

to type \The{foo-a} instead of \thefoo-a. (Tested without-option and with [a,b,c]. )

And it is not clear wether you want that the counters are reset by others.

HTH -- Uwe.
____________________________________________________________
Susan Dittmar <SDittmar at eureca.de> schrieb am 08.05.2008 13:42:03:

for a LaTeX document I would like to define a macro CREATE, which
- takes some arguments itself
- creates and accesses a counter with a name of the first argument
concatenated with some identifying letters (I need more than one such
counter)
- creates a macro with a name equal to the first argument, where other
arguments of the CREATE macro define specifics of this macro, and
which itself takes some arguments
`


text:
IntervalSurgeon

Introduction

IntervalSurgeon presents functions for intersecting, overlapping, piling and annotating integer-bounded intervals. Underlying algorithms are written in C++ for efficiency where appropriate (with the help of the Rcpp package). A typical use case would be for manipulating genomic intervals.

For the purposes of this package, intervals are represented as two-column integer matrices where the inclusive start points are in the first column and the exclusive end points are in the second column.

library(IntervalSurgeon)

starts <- 3*1:10
ends <- 5*1:10

intervals <- cbind(starts, ends)

The lengths of the intervals are therefore: intervals[,2]-intervals[,1].

A key concept in IntervalSurgeon is breaking ranges which contain intervals into 'sections' delimited by the unique start/end points in the set. The sections for a set of intervals x is therefore a two-column matrix representing a set of intervals which partitions the range of x by the sorted start and end points. The sorted start and end points can be obtained using the breaks function (so named to reflect start/end points of intervals frequently being referred to as 'breakpoints' in genomics), which is equivalent to: sort(unique(as.integer(intervals)))). The sections can be computed from the sorted start and end points using the sections function.

One can use the depth and pile functions respectively to obtain the depth of intervals over their 'sections' (within which the depth is constant), and the row indices of original intervals which cover each section.

The flatten function returns a non-touching, non-overlapping set of intervals (as a matrix) which overlap at least one of a given set.

sectioned <- sections(breaks(intervals))
flattened <- flatten(intervals)
depths <- depth(intervals)
piles <- pile(intervals)

Detached intervals

IntervalSurgeon provides functions which are optimised for dealing with detached (i.e. non-overlapping and non-touching) intervals which are sorted and non-empty. Here, we generate two such sets of intervals:

x_starts <- 10*1:10
x <- cbind(x_starts, x_starts + 5)

y_starts <- 20*1:5 + 1
y <- cbind(y_starts, y_starts + 7)

We can determine that they are indeed detached, sorted and non-empty using the detached_sorted_nonempty function.

detached_sorted_nonempty(x)
## [1] TRUE

The IntervalSurgeon functions for operating on such sets of detached, sorted, non-empty intervals are analogues of the set operation functions in the base package, namely: intersect, union and setdiff. Here, the function names are in plural (i.e. with extra s's on the end).

For a given set of sorted, non-overlapping intervals, some of the start points might be the same as the end points of adjacent intervals. These intervals are therefore 'touching' and can be 'stitched' together using the stitch function. If there were overlaps, the flatten function can be used to generate a set of sorted disjoint intervals. Note that flatten will also stitch touching intervals, although the stitch function is faster (albeit requiring intervals to be sorted).

Finding overlaps between sets of intervals

Information about overlaps between sets of intervals can be obtained by 'joining' the sets. This is analogous to an SQL inner-join of two tables, and can be performed on sets of intervals using the join function. This function returns a matrix containing all overlaps of intervals from each set. Each row in the returned matrix corresponds to a specific overlap of intervals with one from each of the sets passed to the function. The nth element in a row contains the row index of the interval in the nth set of intervals passed to the function. Depending on the value of the output argument, there may two additional columns giving the start and end coordinates of the overlap (the default: output="intervals", no extra columns (output="indices") or one additional column giving the row index of the 'section' of the complete set of intervals (output="sections", see ?sections).

x <- cbind(3*1:8, 5*1:8)
y <- cbind(4*1:4, 7*1:4)
join_xy <- join(x, y)

head(join_xy)
##      [,1] [,2] [,3] [,4]
## [1,]    1    1    4    5
## [2,]    2    1    6    7
## [3,]    2    2    8   10
## [4,]    3    2    9   14
## [5,]    3    3   12   15
## [6,]    4    2   12   14

One common task would be to tag intervals with overlapping intervals, perhaps from a different set. For example, this might be useful for tagging a set of genomic intervals with the names of genes which they overlap. The annotate function is provided for this exact purpose.

x <- cbind(0, c(a=10, b=20, c=30))
y <- cbind(c(A=0, B=10, C=20), c(5, 15, 25))

annotate(x, y)
## $a ## [1] "A" ## ##$b
## [1] "A" "B"
##
## $c ## [1] "A" "B" "C" Use with genomic intervals Genomic intervals are often represented in R as data.frames with columns corresponding to chromosome name, start position and end position. Obviously intervals do not intersect if they are on different chromosomes, so in order to manipulate such intervals with IntervalSurgeon, genome-wide operations must be performed chromosome-at-a-time. Using split to create a list of chromosome specific data.frames, or looping over the names of chromosomes and subsetting the original table, before cbinding/as.matrixing the start and end columns then makes the data accessible to the IntervalSurgeon functions. regions_annotated_with_genes <- lapply( c(1:22, "X", "Y"), function(chromosome) { regions_on_chr <- as.matrix(regions[regions$chr == chromosome,c("start", "end")])
genes_on_chr <- as.matrix(genes[genes\$chr == chromosome,c("start","end")])
annotate(regions_on_chr, genes_on_chr)
}
)


text:
# Properties

 Label 1800.2.a.e Level $1800$ Weight $2$ Character orbit 1800.a Self dual yes Analytic conductor $14.373$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$1800 = 2^{3} \cdot 3^{2} \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1800.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$14.3730723638$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 600) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

 $$f(q)$$ $$=$$ $$q - 3q^{7} + O(q^{10})$$ $$q - 3q^{7} - 2q^{11} + 3q^{13} + 6q^{17} - 7q^{19} + 6q^{23} + 2q^{29} - 5q^{31} - 10q^{37} - 12q^{41} - 3q^{43} - 10q^{47} + 2q^{49} + 6q^{59} - 13q^{61} - 7q^{67} + 4q^{71} + 6q^{73} + 6q^{77} - 8q^{79} - 6q^{83} - 16q^{89} - 9q^{91} + 7q^{97} + O(q^{100})$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 0
0 0 0 0 0 −3.00000 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$2$$ $$-1$$
$$3$$ $$-1$$
$$5$$ $$1$$

## Inner twists

This newform does not admit any (nontrivial) inner twists.

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1800.2.a.e 1
3.b odd 2 1 600.2.a.b 1
4.b odd 2 1 3600.2.a.bl 1
5.b even 2 1 1800.2.a.t 1
5.c odd 4 2 1800.2.f.e 2
12.b even 2 1 1200.2.a.q 1
15.d odd 2 1 600.2.a.i yes 1
15.e even 4 2 600.2.f.d 2
20.d odd 2 1 3600.2.a.i 1
20.e even 4 2 3600.2.f.o 2
24.f even 2 1 4800.2.a.bd 1
24.h odd 2 1 4800.2.a.bp 1
60.h even 2 1 1200.2.a.b 1
60.l odd 4 2 1200.2.f.c 2
120.i odd 2 1 4800.2.a.bc 1
120.m even 2 1 4800.2.a.bs 1
120.q odd 4 2 4800.2.f.z 2
120.w even 4 2 4800.2.f.k 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
600.2.a.b 1 3.b odd 2 1
600.2.a.i yes 1 15.d odd 2 1
600.2.f.d 2 15.e even 4 2
1200.2.a.b 1 60.h even 2 1
1200.2.a.q 1 12.b even 2 1
1200.2.f.c 2 60.l odd 4 2
1800.2.a.e 1 1.a even 1 1 trivial
1800.2.a.t 1 5.b even 2 1
1800.2.f.e 2 5.c odd 4 2
3600.2.a.i 1 20.d odd 2 1
3600.2.a.bl 1 4.b odd 2 1
3600.2.f.o 2 20.e even 4 2
4800.2.a.bc 1 120.i odd 2 1
4800.2.a.bd 1 24.f even 2 1
4800.2.a.bp 1 24.h odd 2 1
4800.2.a.bs 1 120.m even 2 1
4800.2.f.k 2 120.w even 4 2
4800.2.f.z 2 120.q odd 4 2

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(1800))$$:

 $$T_{7} + 3$$ $$T_{11} + 2$$ $$T_{13} - 3$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T$$
$3$ $$T$$
$5$ $$T$$
$7$ $$3 + T$$
$11$ $$2 + T$$
$13$ $$-3 + T$$
$17$ $$-6 + T$$
$19$ $$7 + T$$
$23$ $$-6 + T$$
$29$ $$-2 + T$$
$31$ $$5 + T$$
$37$ $$10 + T$$
$41$ $$12 + T$$
$43$ $$3 + T$$
$47$ $$10 + T$$
$53$ $$T$$
$59$ $$-6 + T$$
$61$ $$13 + T$$
$67$ $$7 + T$$
$71$ $$-4 + T$$
$73$ $$-6 + T$$
$79$ $$8 + T$$
$83$ $$6 + T$$
$89$ $$16 + T$$
$97$ $$-7 + T$$


text:
# Product of two Lebesgue integrable functions

In general, I know that it is not necessarily the case that the product of two Lebesgue integrable functions $f,g$ will be Lebesgue integrable. But I was reading in a textbook that if at least one of these functions is bounded, then their product will be Lebesgue integrable. How can we prove this statement? I'd appreciate some input on this, thanks.

-
This is the easiest case of Hölder's inequality and it would have been as quick to write what Didier wrote as to fetch this link. Still, it's worthwhile to read this page. –  t.b. Aug 6 '11 at 18:08

If $|f|\le M$ almost everywhere, then $|fg|\le M|g|$ almost everywhere hence $\displaystyle\int|fg|\le M\int|g|$ is finite because $g$ is integrable.


text:
3,528 views

The asymptotic upper bound solution of the recurrence relation given by $T(n) = 2T \left( \frac{n}{2} \right) +\frac{n}{\lg \: n}$ is

1. $O(n^2)$
2. $O(n \:\lg \: n )$
3. $O(n \:\lg \:\lg \: n)$
4. $O(\lg \:\lg \: n)$

@Rupendra What is the problem with applying master theorem?

Option C) by master's theorem. A = 2, B = 2 and K =1 and P = -1.

As a = bk and p = -1

answer is  O (nlog22 log log n) = O(n log log n)

master's theorem : 2)-b case answer is  O (nlog22 log log n) = O(n log log n)

This is wrong explanation:

master theorm can't be applied here...

$$T(n)=2T\left(\frac{n}{2}\right)+\frac{n}{\lg n}$$

For Master theorem, $a = 2, b = 2, n^{\log _b a} = n$. $f(n) = \frac{n}{\lg n}$. We cannot say $f(n) = O(n^{\log_b a - \epsilon})$, as the difference between $n$ and $\frac{n}{\lg n}$ is not polynomial. So, we cannot apply Master theorem. So, trying substitution. Since, we have a lg term, we can try all powers of 2.

$T(1) = 1$, assuming.

$T(2) = 2T(1) + \frac{2}{\lg 2} = 4$

$T(4) = 8 + 2 = 10$

$T(8) = 20 + \frac{8}{3} = 22.66$

$T(16) = 45.3 + 4 = 49.3$

$T(32) = 98.6 + 6.4 = 105$

$T(64) = 210 + 10.6 = 220.6$

Not able to reach a conclusion. We can see that the recurrence is between case 1 and case 2 of Master theorem. So, it is $\Omega \left(n\right)$ and $O\left(n \lg n\right)$. So, lets solve the recurrence directly.

$T(n)=2T\left(\frac{n}{2}\right)+\frac{n}{\lg n}$

$= 2^2 T\left(\frac{n}{4}\right)+\frac{n}{\lg n - 1} + \frac{n}{\lg n}$

$\dots$

$= 2^{\lg n} T(1) + \frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{\lg n}$

$= n + n \left( \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{\lg n}\right)$

$=n + n \left(\lg \lg n + \gamma \right)\\= \Theta\left(n \lg \lg n\right)$

The sum of $\lg n$ terms in HP approximated to $\lg \lg n +\gamma$ where $\gamma$ is Euler Mascheroni constant.

by

More conceptual method..

Harmonic Series Summation causes little trouble.

Should I byheart selected answer master's theorem to get nice rank.

Becz, there come certain concepts that become really tough to grasp like

in TOC - Decidability table u gave..

not a necessity to by heart that. I'll not do that. I'll solve by this method only- and as you told Harmonic series is the only issue, but it is easier to by heart that :)

And for decidability also, I don't recommend studying that table- always know the reason and derive the answer- table I gave just for quick verification. Mostly GATE questions come from basic areas..
sir we can do it directly by extended masters theorem right?

Thank you for your answer.I also solve this question using this equation.Got this question from Cormen but some where I saw we can't solve this question using Master theorem (now I  can't remember the source.)so I google it.Stack overflow also saying can't solve it.(you can find the link above.I can't understand the explanation).By applying this equation I got wrong answer here https://gateoverflow.in//3354/gate2008-it_44.
Comparing https://gateoverflow.in//3354/gate2008-it_44 and aT(n/b) + sqrt n..
a = sqrt 2 , b = 2, f(n) = root (n)
n^ (loga base b) = n^1/2 = root n
T(n) = O( sqrt(n)*logn)
According to asymptotic notation options A,B,C are correct..

But they asked exact solution (no option is in asymptotic notation) so checking initial value may eliminate some options.
T(1) = 1 and only option A satisfy this .

Here Masters theorem not gives wrong result, but from given option we have to refine exact solution because no option is in asymptotic form..
Thank you laser :) I think option 4 is not even asymptotically correct.
I will not be online for 1 week.If there is any mistake please comment.I will check it after 1 week

Option C)

MASTER THEOREM

A = 2, B = 2 and K =1 and P = -1.

As a = bk and p = -1

answer is  O (nlog22 log log n) = O(n log log n)

CASE 2b here

### 1 comment

yes answer is c well my aprroach is much easier .

i forgot to return the value which i let just

1
1,798 views


text:
gnuastro-commits
[Top][All Lists]

## [gnuastro-commits] master 24757bf: Description of new true clump critera

 From: Mohammad Akhlaghi Subject: [gnuastro-commits] master 24757bf: Description of new true clump critera corrected in book Date: Thu, 17 May 2018 21:14:51 -0400 (EDT)

branch: master

Description of new true clump critera corrected in book

In the last few commits, a small modification was made to the criteria to
choose true clumps: the noise is now the actual noise taken under the river
pixel used (with maximum value). It is not calculated from the clump's
flux-weighted central position any more. This is more realistic (especially
now that we are only subtracting two pixel values).
---
NEWS              | 13 ++++++++-----
doc/gnuastro.texi | 21 ++++++++++++---------
2 files changed, 20 insertions(+), 14 deletions(-)

diff --git a/NEWS b/NEWS
index b609f4f..4dc4f6c 100644
--- a/NEWS
+++ b/NEWS
@@ -156,11 +156,14 @@ GNU Astronomy Utilities NEWS                          -*-
outline -*-
--detquant     ==> --snquant
- By default the output detection map is a binary image (values of 0 or 1).
- With no output name, the output has a _detected.fits' suffix.
-    - [Now in Segment]: For finding true clumps, the difference in the peak
-      of the clump and the highest valued river pixel, divided by the noise
-      standard deviation are used, not the total signal-to-noise ratio. In
-      initial tests, this algorithm was much more promising in detecting
+
+  Segment:
+    - [Previously in NoiseChisel]: For finding true clumps, the difference
+      in the peak of the clump and the highest valued river pixel, divided
+      by the noise standard deviation are used. Until now, the total
+      signal-to-noise ratio was used as a criteria. In initial tests, this
+      algorithm was much more promising in detecting clumps over strong

Table:
--column: multiple columns (comma separated) can be used in one
diff --git a/doc/gnuastro.texi b/doc/gnuastro.texi
index 35ffae1..6bc688f 100644
--- a/doc/gnuastro.texi
+++ b/doc/gnuastro.texi
@@ -15587,20 +15587,23 @@ detect the diffuse and extended emission, but in
segmentation, you want to
detect sharp peaks.

@item
-The criteria to select true from false clumps is the peak signal-to-noise
-ratio. This value is calculated from a clump's peak value (@mymath{C_c})
-and the highest valued river pixel around that clump (@mymath{R_c}). Both
-are calculated on the convolved image (signified by the @mymath{c}
-subscript). To avoid absolute differences, it is then divided by the input
-(not convolved) Sky standard deviation under that clump (@mymath{\sigma})
-as shown below.
+The criteria to select true from false clumps is the peak significance. It
+is defined to be the difference between the clump's peak value
+(@mymath{C_c}) and the highest valued river pixel around that clump
+(@mymath{R_c}). Both are calculated on the convolved image (signified by
+the @mymath{c} subscript). To avoid absolute values (differing from dataset
+to dataset), @mymath{C_c-R_c} is then divided by the Sky standard deviation
+under the river pixel used (@mymath{\sigma_r}) as shown below:

+
+When @option{--minima} is given, the nominator becomes

The input Sky standard deviation dataset (@option{--std}) is assumed to be
for the unconvolved image. Therefore a constant factor (related to the
convolution kernel) is necessary to convert this into an absolute peak
-signal-to-noise address@hidden get an estimate of the standard deviation
address@hidden get an estimate of the standard deviation
correction factor between the input and convolved images, you can take the
following steps: 1) Mask (set to NaN) all detections on the convolved image
with the @code{where} operator or @ref{Arithmetic}. 2) Calculate the

`


text:
# How to motivate inner product spaces

What is the most motivating way to introduce general inner product spaces? I am looking for examples which have a real impact. For Euclidean spaces we relate the dot product to the angle between the vectors which most people find tangible. How can we extend this idea to the inner product of general vectors spaces such as the set of matrices, polynomials, functions?

-
I'd say Fourier series. Decomposing general functions into sines and cosines exactly like we decompose an ordinary vector into its Cartesian components. (BTW, I think this should be a CW) –  Giuseppe Negro Sep 2 '12 at 11:43
Fourier series sound like a really good motivation. You could also be more general, and think of functions on an abstract measure space $X$. Then the inner product $\left< f,g \right> = \int_X f \cdot g$ feels like a natural way to give $\mathcal{C}(X)$ more structure, and $\mathrm{L}^2(X)$ can then be viewed as a natural completion of $\mathcal{C}(X)$. –  Feanor Sep 2 '12 at 12:18

One of the uses for inner products is to give an identification between the space and its dual. If the space is finite-dimensional then this is an isomorphism, otherwise it gives an injection $E \rightarrow E^*$. Indeed suppose you are given a non-degenerate bilinear form $(,)$. Then for $x \in E$ set $x^* = (y \mapsto (x,y)) \in E^*$. This provides an injective linear mapping.


text:
# Positive semigroups and abstract Lyapunov equations

Positive semigroups and abstract Lyapunov equations We consider abstract equations of the form $$\mathcal {A}x=-z$$ A x = - z on a locally convex space, where $$\mathcal {A}$$ A generates a positive semigroup and $$z$$ z is a positive element. This is an abstract version of the operator Lyapunov equation $$A^*P+PA=-Q$$ A ∗ P + P A = - Q from control theory. It is proved that under suitable assumptions existence of a positive solution implies that $$-\mathcal {A}$$ - A has a positive inverse, and the generated semigroup is asymptotically stable. We do not require that $$z$$ z is an order unit, or that the space contains any order units. As an application, we generalize Wonham’s theorem on the operator Lyapunov equations with detectable right hand sides to reflexive Banach spaces. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Positivity Springer Journals

# Positive semigroups and abstract Lyapunov equations

, Volume 19 (1) – Mar 4, 2014
21 pages

1

/lp/springer_journal/positive-semigroups-and-abstract-lyapunov-equations-DJf0qNAnFM
Publisher
Springer Basel
Subject
Mathematics; Fourier Analysis; Operator Theory; Potential Theory; Calculus of Variations and Optimal Control; Optimization; Econometrics
ISSN
1385-1292
eISSN
1572-9281
D.O.I.
10.1007/s11117-014-0279-3
Publisher site
See Article on Publisher Site

### Abstract

We consider abstract equations of the form $$\mathcal {A}x=-z$$ A x = - z on a locally convex space, where $$\mathcal {A}$$ A generates a positive semigroup and $$z$$ z is a positive element. This is an abstract version of the operator Lyapunov equation $$A^*P+PA=-Q$$ A ∗ P + P A = - Q from control theory. It is proved that under suitable assumptions existence of a positive solution implies that $$-\mathcal {A}$$ - A has a positive inverse, and the generated semigroup is asymptotically stable. We do not require that $$z$$ z is an order unit, or that the space contains any order units. As an application, we generalize Wonham’s theorem on the operator Lyapunov equations with detectable right hand sides to reflexive Banach spaces.

### Journal

PositivitySpringer Journals

Published: Mar 4, 2014

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