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# The evolution of large cavities and disc eccentricity in circumbinary discs

Abstract : We study the mutual evolution of the orbital properties of high-mass ratio, circular, co-planar binaries and their surrounding discs, using 3D Smoothed Particle Hydrodynamics simulations. We investigate the evolution of binary and disc eccentricity, cavity structure, and the formation of orbiting azimuthal overdense features in the disc. Even with circular initial conditions, all discs with mass ratios q > 0.05 develop eccentricity. We find that disc eccentricity grows abruptly after a relatively long time-scale (∼400-700 binary orbits), and is associated with a very small increase in the binary eccentricity. When disc eccentricity grows, the cavity semimajor axis reaches values $a_{\rm cav}\approx 3.5\, a_{\rm bin}$ . We also find that the disc eccentricity correlates linearly with the cavity size. Viscosity and orbit crossing appear to be responsible for halting the disc eccentricity growth - eccentricity at the cavity edge in the range ecav ∼ 0.05-0.35. Our analysis shows that the current theoretical framework cannot fully explain the origin of these evolutionary features when the binary is almost circular (ebin ≲ 0.01); we speculate about alternative explanations. As previously observed, we find that the disc develops an azimuthal overdense feature in Keplerian motion at the edge of the cavity. A low-contrast overdensity still co-moves with the flow after 2000 binary orbits; such an overdensity can in principle cause significant dust trapping, with important consequences for protoplanetary disc observations.
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https://hal-insu.archives-ouvertes.fr/insu-03711450
Contributor : Nathalie POTHIER Connect in order to contact the contributor
Submitted on : Friday, July 1, 2022 - 1:49:38 PM
Last modification on : Saturday, July 2, 2022 - 3:40:12 AM

### Citation

Enrico Ragusa, Richard Alexander, Josh Calcino, Kieran Hirsh, Daniel J. Price. The evolution of large cavities and disc eccentricity in circumbinary discs. Monthly Notices of the Royal Astronomical Society, 2020, 499, pp.3362-3380. ⟨10.1093/mnras/staa2954⟩. ⟨insu-03711450⟩

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1 Benchmarks

Three single-part linear elastic benchmark problems

A benchmark of three single-part mechanical analyses solved with CAEplex using the free and open source tools Gmsh and Fino as the back-ends against a the commercial closed-source code. As the license renders this software privative, only results can be compared as output by a black-box. Even more, experts say that this “reference”software cannot export point-wise numerical data so only maximum scalar values can be compared instead of the full dependence on $$x$$, $$y$$ and $$z$$ of the unknowns. The comparison was made trying to keep the same number of nodes in the two codes similar while meshing the 3D CAD models with second-order tetrahedral elements.

Parallelepiped whose Young’s modulus is a function of the temperature

A benchmark thermal-mechanical problem that has an analytical solution was solved using fino. Even though the original problem is defined over a simple parallelepiped, this reports uses a fully unstructured grid to validate the code using arbitrary jacobians and not just structured hexahedra. The way fino handles dependence of parameters on space, in this case through intermediate distributions such as the temperature, is illustrated by the simplicity of the input files used to solve both the thermal and the mechanical problems. The comparison between the numerical and the analytical solutions are within the order of $$10^{-6}$$ even for coarse grids.

Thermo-elastic expansion of finite cylinders

The problem of the thermal expansion of a cylinder under a know temperature distribution (given as an algebraic expression of $$x$$, $$y$$ and $$z$$) is solved with the free and open source tool fino using a finite-element formulation and the results are compared to those reported by Veeder in 1967 using a power series approach. The original problem was posed to study deformations and stresses on the uranium pellets located inside nuclear reactor fuel elements under heavy temperature gradients, which continues to be an interesting problem even nowadays. Not only does this report solve again the problem and compare the results with the original ones, but it also illustrates how the fino back-end works and what its distinctive features are.

2 Technical reports

Evaluation of partial derivatives of pointwise node-based data over unstructured grids

This brief note introduces a proposal to obtain the nodal values $$P_j$$ of a partial derivative of a pointwise node-centered scalar field such as displacements $$u_j$$ or temperatures $$T_j$$ obtained with the finite-element method. We require that the nodal values of the derivative $$P_j$$ interpolated with the shape functions $$h_j(\vec{x})$$ give the same internal product against each of the shape functions $$h_j(\vec{x})$$ than the derivative of the original field (say $$\partial h/\partial x$$) computed by using the derivative of the shape functions, say $$\partial h_j/\partial x$$. This requirement involves the mass matrix $$M$$, which we diagonalize to reduce the computational effort needed to obtain the nodal values $$P_j$$. We show a snippet of computational code as implemented in Seamplex’ Fino.

Description of the computational tool wasora

Wasora is a free computational tool designed to aid cognizant experts to analyze complex systems by solving mathematical problems by means of a high-level plain-text input file containing a syntactically-sweetened description of definitions and instructions. Some of its main features are:

• evaluation of algebraic expressions
• one and multi-dimensional function interpolation
• scalar, vector and matrix operations
• numerical integration, differentiation and root finding of functions
• possibility to solve iterative and/or time-dependent problems
• adaptive integration of systems of differential-algebraic equations
• I/O from files and shared-memory objects (with optional synchronization using semaphores)
• execution of arbitrary code provided as shared object files
• parametric runs using quasi-random sequence numbers to efficiently sweep a sub-space of parameter space
• solution of systems of non-linear algebraic equations
• non-linear fit of scattered data to one or multidimensional functions
• non-linear multidimensional optimization
• management of unstructured grids
• complex extensions by means of plugins

Besides solving general math problems usually associated with engineering analysis, the code is designed in such a way that particular (and potentialy complex) computations may be implemented as plugins (such as computations based on the finite element method or dedicated neutronic codes) that run over the framework, taking advantage of all the common background wasora provides. This technical document introduces the code and describes its main features by walking through the rationale behind its design and the types of problems that are suitable to be tackled with wasora.

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# Tag Info

0

If $R^n\simeq R^{n+1}$ then $R^{n+1}\simeq R^{n+2}$ (why?), so $R^n\simeq R^{n+2}$ and so on.

1

In $K \otimes_R C$ all tensors are elementary, i.e., they all have the form $x \otimes c$. That comes from using a common denominator for any finite list of elements in $K$ (as ratios of elements of $R$). The kernel of the $R$-linear mapping $C \rightarrow C \otimes_R K$ given by $c \mapsto 1 \otimes c$ is precisely the torsion submodule $C_{\rm tor}$, so ...

0

Hint on well-definedness. define $\Phi: C \times R \to C \otimes_R K$ by $c \times r \mapsto c \otimes r$ , then by the universal property of tensor $\Phi$ induces a $R$-hom $C \otimes_{R} R \to C \otimes_R K$ with $c \otimes r \mapsto c \otimes r$, of course, it is well-defined.

5

Hint: $R=\mathbb{Z}$, $C=\mathbb{Z}/2\mathbb{Z}$. There is no problem in $\phi$ being well defined; there's some problem in showing it's injective. ;-) The map is well defined, because such is the map $\tilde\phi\colon C\to C\otimes_RK$ defined by $$x\mapsto x\otimes1$$ and composing with the natural isomorphism $C\otimes_RR\to C$ gives your $\phi$.

0

Consider a domain $R$ with a nontrivial maximal ideal $I$. All nonzero modules of $R/I$ are injective. It looks like you are expecting that $M$ being $R/I$-injective implies $M$ is $R$-divisible. But that isn't true, considering you can take $M$ to be any nondivisible $R$ module and then look at $R/I$ for any maximal ideal to get an injective $R/I$ module.

0

Try to show any proper subgroups of $\Bbb Z(p^\infty)$ is finite (by showing the contrapositive).

0

So given $M$ an injective torsion-free module, we have that $\varphi_r$ is an isomorphism. Note that we can view $M$ as a $K(R)$ module (the field of fractions) in the following manner: Define $(a/b)\cdot m$ to be $\varphi^{-1}_b(am)$, as $\varphi_b$ is an isomorphism this is well defined. Hence, we have that $M$ is a $K(R)$-module, that is, a ...

2

The answer is that such matrices are always similar over $\def\Z{\Bbb Z}\Z$ (conjugate in $GL_2(\Bbb Z)$). The question is deeper however than it might look at first, and as far as I can see any solution requires some somewhat subtle arithmetic considerations. A few things are easy: $A,B$ always have determinant$~1$ (from the constant coefficients of the ...

2

By the snake lemma there is an exact sequence $$0\rightarrow\mathrm{Ker} \ \alpha\rightarrow 0 \rightarrow \mathrm{Ker} \ \gamma \rightarrow \mathrm{Coker} \ \alpha\rightarrow 0\rightarrow\mathrm{Coker} \ \gamma\rightarrow 0,$$ whence $\mathrm{Ker} \ \alpha=0$ and $\mathrm{Coker} \ \gamma=0$; also $\mathrm{Ker} \ \gamma\cong\mathrm{Coker} \ \alpha$ so that ...

0

If $U : \mathsf{Mod}(R) \to \mathsf{Set}$ is the forgetful functor, then there is a natural inclusion map $\mathrm{End}(1_{\mathsf{Mod}(R)}) \hookrightarrow \mathrm{End}(U)$. But $U$ is represented by $R$, hence $\mathsf{End}(U) \cong \mathsf{End}(R) \cong Z(R)$ by Yoneda. The map $Z(R) \cong \mathrm{End}(U)$ is given by $c \mapsto (m \mapsto cm)$ (by the ...

2

Take a set of basis $e_1,\ldots,e_n$ of $V$ and let $x=\sum_ {i}x^ie_i,~y=\sum_ {j}y^je_j$, then $$x\otimes y=\sum_ {i,j}(x^ie_i)\otimes(y^je_j)=\sum_ {i,j}x^iy^je_i\otimes e_j$$ $$y\otimes x=\sum_ {i,j}y^jx^ie_j\otimes e_i$$ The symmetry implies $$x^iy^j=x^jy^i$$ That is, $$\frac{x^i}{y^i}=\frac{x^j}{y^j}=a$$ for some constant $a$.

3

So for such $x,y\neq 0$, you want to show $x\otimes y=y\otimes x$ in $V\otimes_F V$ if $x=ay$ for some $a\in F$. Then $x\otimes y=ay\otimes y=y\otimes ay=y\otimes x$. Since this seems by far to be the easier half of the problem, I am beginning to wonder if you meant to ask about the other direction. Suppose $x,y$ are linearly independent. As such, this ...

3

The straightforward answer for "what is a basis for $\mathbb{C}/\mathbb{Q}$" is that we don't know. The sneaky answer is that we do know there is one, because any maximal linearly independent set is a basis, and exists by Zorn's Lemma. This shows that $\mathbb{C}$ is a free $\mathbb{Q}$-module, since that concept is literally equivalent to the existence of ...

-3

Off course given set generates C(Q) but is not L.I. as you may notice 2^(1/2) and all its rational multiples are there. In fact you can't write precisely what will be the basis. The basis can be written iff vector space is finite dimensional.

5

Note first that for any ring $R$ and any $R$-module $M$, if the cardinality $\lvert M\rvert$ is infinite and greater than $\lvert R \rvert$, then any generating set of $M$ (in particular, any basis) must have cardinality equal to $\lvert M\rvert$ (this is simple combinatorics). In particular, any basis of $\bf C$ over $\bf Q$ must have cardinality of the ...

0

If $F(x^2)$ is not $F(x)$ then $[F(x) : F(x^2)] = 2$, and $[F(x) : F] = [F(x) : F(x^2)] [F(x^2) : F]$ is even.

2

Any basis for $\Bbb C$ as a $\Bbb Q$ vector space must be infinite, since any finite dimensional $\Bbb Q$ vector space is countable, but $\Bbb C$ is not. Constructing such a basis requires the axiom of choice.

2

Consider the ring $R=F_2[Z]/(Z^2)=M$ where $F_2$ is the field of two elements. This is a self-injective ring, so $M$ is an injective $R$-module. But now consider $x=1$ and $r=Z$, where I abuse notation for the images of $1$ and $Z$ in this ring. Saying that there exists $y\in R$ such that $yZ=1$ implies that $Z$ is a unit, but it is clearly not since it is ...

1

Edit: I misread the question. If you only want to determine whether some particular pair of matrices $(A,B)$ are similar to each other, you should solve the two equations $PA=BP$ and $|\det P|=1$. You may find Dario Alpern's online solver useful. (My original answer.) $x^2+x+1$ is the characteristic polynomial of $A=\pmatrix{a&b\\ c&d}$ if and only ...

3

Suppose $\eta$ is a natural tranformation. Its value at the left $R$-module $R$ is a map $\eta_R:R\to R$ of left $R$-modules. You can easily check that $c=\eta_R(1)$, the image of $1\in R$, belongs to the center $Z(R)$ of $R$. If now $M$ is any $R$-module, and $m\in M$, there is a unique map $f:R\to M$ such that $f(1)=m$. Using naturality of $\eta$, we see ...

1

Well, if you know the characteristic polynomial, you know the trace ($-1$) and the determinant ($1$), so you should be able to write the general matrix using two (integer) unknowns [for example, if the $1, 1$ entry is $a$ then the $2, 2$ entry is $-1-a$). Now, you have your two possible matrices $M_1$ and $M_2.$ Being similar, means that there exists a ...

4

Nope. Consider two $4 \times 4$ matrices: $$\begin{pmatrix}0 & & & \\ & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}$$ $$\begin{pmatrix}0 & 1 & & \\ 0 & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}$$ (where I've written the matrices in Jordan block form; ...

1

I think the author mean that we define an addition on $M\oplus N$, but not on $M\times N$. Moreover we consider $M\times N$ as a generating set for formal sums $(a,b)+(c,d)$. As to second question I think this expression is not nice and explains nothing.

0

I'm assuming you mean $M$ to be a submodule of $V$. By definition, $KM$ is the set of $K$-linear combinations of vectors in $M$. This is a $K$-subspace of $V$. In fact, since $M$ itself is finitely generated over $R$, the set $KM$ is equal to the set of $K$-linear combinations of a set of generators $x_1,\ldots,x_n\in M$ for $M$ as an $R$-module. In general, ...

0

Hint: What you're talking about are the induced isomorphisms of a non-degenerate (not necessarily perfect) pairing. The definition of $P'$ is, however, a bit different. Here is the right one: $$P': A \to B^*, a \mapsto P(a,-)$$ So the image of an $a \in A$ is really a dual map $B \to R$. Are those modules finitely generated over $R$?

1

The element $xab^{-1}$ is an element in the localization $S^{-1}M$ where $S = R \setminus \{0\}$. There is a natural isomorphism $M \otimes_R S^{-1}R \simeq S^{-1}M$ defined by $m \otimes \frac{a}{b} \mapsto \frac{ma}{b}$. Here $S^{-1}R$ is the ring $K$. In general writing $x_1 \otimes x_2 = x_1x_2$ doesn't make sense unless you are invoking some type of ...

2

The notation $R \cap \mathfrak m$ is sometimes used to denote the inverse image of $\mathfrak m$ in $R$. It is a prime ideal $\mathfrak p$ of $A$. There is a canonical morphism $R_{\mathfrak p} \to A_{\mathfrak m}$, and the $A_{\mathfrak m}$-module $M_{\mathfrak m}$ acquires an $R_{\mathfrak p}$-module structure via this map.

9

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...

1

First, let's settle the issue of the empty tensor product. Consider the identity map $id: W\to W$ for an arbitrary $R$-module $W$. This is an $R$-balanced map out of the collection $\{W\}\cup \emptyset$ and so descends to a map $W\otimes E\to W$. But we know that $(W,id)$ satisfies the universal property of the tensor product of $W$, so $W\cong W\otimes E$ ...

3

$E^{\otimes 0}$ should be the identity of $\otimes$, i.e. the base ring $R$.

5

For ease of notation, let $A = \operatorname{im}\varphi_1 \subset N$, $T = \operatorname{im}\varphi_2 \subset S$, and $Q = T/\operatorname{im} (\varphi_2\circ\varphi_1) = T/\varphi_2(A)$. Consider the map $\psi = \pi \circ \varphi_2 \colon N \to Q$, where $\pi \colon T \to Q$ is the canonical projection. The kernel of $\psi$ is $$\ker \psi = \psi^{-1}(0) = ... 0 The proof does not exclude the possibility that the restriction of V to N is irreducible. Then U is necessarily equal to V. For example V could be 1-dimensional. 2 We are given N and that will give us the prime factors p and q as:$$N = 91 = p \times q = 7 \times 13$$We need the Euler Totient Function of the modulus, hence we get:$$\varphi(N) = \varphi(91) = (p-1)(q-1) = 6 \times 12 = 72$$Now, we choose an encryption exponent 1 \lt e \lt \varphi(N) = 72. We were told to pick an an e \lt 6, so lets ... 3 As you remarked, injective modules are divisible, that is, rM=M for all r\in R, r\ne 0. The key step is to show that every non-zero homomorphism f:M\to R is surjective. Let x\in M such that f(x)\ne 0. Set r=f(x). Since rM=M there exists y\in M such that ry=x. Then rf(y)=r, so f(y)=1, and this is enough. Now use that M is ... 0 \mathbb{Z}G is a free module over \mathbb{Z}, with G as basis. Try proving that F is (isomorphic to) a free \mathbb{Z}-module over X\times G. 4 As P is f. g. we have an exact sequence 0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0, Q denoting the kernel of the map A^n\rightarrow P. As P is projective, this exact sequence splits, A^n\cong Q\oplus P. The exact sequence A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0 shows P to be f. p. (where Q\oplus P\rightarrow ... 0 F is sums of elements of X with coefficients in \mathbb{Z}G, which is the same as sums of \{gx:g\in G, x\in X\} with coefficients in \mathbb{Z}. F is not a free \mathbb{Z}-module over X (over \{1_G\cdot x:x\in X\}) because X does not generate F as a \mathbb{Z}-module. 1 Do you understand that \Bbb Z G\cong \Bbb Z^{|G|}\not\cong\Bbb Z as \Bbb Z-modules when |G|>1? 0 As Martin suggested, a nice reference is Eisenbud - Commutative Algebra with a view towards algebraic geometry The definition of Fitting ideals that can be found there is indeed in term of exterior algebra, and precisely is the following (it consists of 2 parts) Def: Let \varphi:F\to G be a map of free modules over a ring R. We define the ... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \Bbb Z-module) with an action of the ring R, defined by R \to Aut(M), but the main problem is, that not every image of \mu is an automorphism. So, in fact, you could have \varphi \in End(M) \setminus Aut(M) as an image of ... 0 Using formulae used here,$$\frac{386}{97}=3+\frac{95}{97}=3+\frac1{\frac{97}{95}}=3+\frac1{1+\frac2{95}}=3+\frac1{1+\frac1{\frac{95}2}}=3+\frac1{1+\frac1{47+\frac12}}$$So, the previous convergent of \displaystyle \frac{386}{97} is \displaystyle3+\frac1{1+\frac1{47}}=3+\frac{47}{48}=\frac{191}{48}$$\implies386\cdot48-191\cdot97=1\implies ...

0

There is likely a typo in the question. Using the Euclidean algorithm, we have that: \begin{align*} 386 &= 3(97) + 95 \\ 97 &= 1(95) + 2 \\ 95 &= 47(2) + 1 \end{align*} Working backwards, we see that: \begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\ &= 48(386) - 191(97)\\ ...

1

Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \cdot p - 386\cdot q = 1$$ Then $97^{-1} \equiv p \mod 386$.

1

Hint. If $n=p^k$, $k\ge 1$, then $0=p^k\Bbb Z/p^k\Bbb Z<p^{k-1}\Bbb Z/p^k\Bbb Z<\cdots<\Bbb Z/p^k\Bbb Z$ is a series of composition of length $k$. Can you generalize this to the case $n=p_1^{k_1}\cdots p_r^{k_r}$, $k_i\ge 1$ and $r\ge 2$? Edit. Let's try the next step: $n=p^kq^l$. Then $$0<p^{k-1}q^l\Bbb Z/p^kq^l\Bbb Z<\cdots<pq^l\Bbb ... 5 The answer is yes. Hint: You can easily see that considering the kernel, image resp. of an R-homomorphism. Conversely, if assumptions ii) or iii) hold for such R-homomorphisms, then M has to be simple, where you can consider the possible submodules of M to check. 0 Let's denote by \times the “external” direct sum just to avoid confusion. Saying that$$ N_1+N_2+\dots+N_n=N_1\oplus N_2\oplus\dots\oplus N_n $$means that the obvious homomorphism$$ N_1\times N_2\times\dots\times N_n\to N_1+N_2+\dots+N_n $$is injective. Hint for the second part: you can define homomorphisms$$ M_i\to ...

1

Claim: $N = f(K) + f(L)$. Certainly, $N \supseteq f(K) + f(L)$. For the reverse containment, let $n \in N$. As $f$ is surjective, there exists $m \in M$ such that $f(m) = n$. Since $M = K + L$, there exist $k \in K$ and $\ell \in L$ such that $m = k + \ell$. Now, $f(k) + f(\ell) = n$, demonstrating that $N \subseteq f(K) + f(L)$. Claim: $N = f(K) ... 3 It's obvious (from surjectivity) that$f(K)+f(L)=N$. Now take$x\in f(K)\cap f(L)$. Then$x=f(y)=f(z)$with$y\in K$and$z\in L$. Since$y-z\in \ker f$we get$y\in L$, so$y\in K\cap L$and therefore$y\in\ker f$. Thus we get$x=f(y)=0$. 8 I'm assuming that$\mathbb Z_n$means$\mathbb Z/n\mathbb Z$. I'm also assuming that "subproduct" means "subgroup of the product". If this is the case then yes, the subgroup generated by$(1, 1, \ldots) \in \prod_{n > 1}\mathbb Z_n$is isomorphic to$\mathbb Z$. To see that this is the case note that we can always define a homomorphism out of$\mathbb ...

1

The second part follows at once from split short sequences' theorem: look at $$0\longrightarrow N\stackrel g\longrightarrow M\stackrel\pi\longrightarrow M/g(N)\longrightarrow 0$$ Prove the above is an exact sequence (i.e., $\;g\;$ is one-to-one, $\;\pi\;$ is onto and $\;\text{Im}\,g=\ker\pi\;$ , with $\;\pi\;$ the canonical projection). Well now, the ...

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# Basket Tranches value and risk

#### gautamFRM

##### New Member
Hi David,

When we find correlation in CDS, I understand following :
high correlation makes senior tranches (i.e., high n to default for basket CDS) more expensive (all other things equal) and junior tranches (e.g., 1st to default) less expensive;
low correlation makes senior tranches less expensive and junior tranches more expensive. You can see in my post that the binomial distribution can be used to illustrate this.

Also, In terms of payoffs of basket CDs,
More Risky will have more payoff and more price/Value

------

So my confusion is:

in case of higher correlation, senior tranche will have more value or less value ?
A: Senior is more risk ... so more payoff/more price,
But in the following question: you said "senior tranche decreases in value due to higher correlation"

I think i am missing something major here.

Should I think in terms of risk or value ?

Thanks,
Abhi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi Abhi,

I think it can get confusing. Can we assume a two-bond portfolio, where each bond has PD = 10%. Consider a 2nd-to-default basket CDS.
• If default correlation = 0, 2TD triggers with probability = 10%*10% = 1.0%.
• If default correlation increases to 100%, probability of 2TD trigger increases to 10.0%; i.e., as correlation increases, the probability that both bonds default increases (here, the senior nth-to-default tranche triggers)
• So the common (robust) dynamic is: higher default correlation increases (decreases) the risk [i.e., probability of defaults that reference] of the senior (junior) tranche
GARP sometimes prefers to refer to value of the tranche rather than the spread (cost of protection) as in the errata question here that you linked to, specifically:
The fund’s chief economist predicts that the default probability will decrease significantly and that the default correlation will increase. Based on this prediction, which of the following is a good strategy to pursue? Correct answer: d. Sell the senior tranche and buy the equity tranche.

Explanation: The decrease in probability of default would increase the value of the equity tranche. Also, a default of the equity tranche would increase the probability of default of the senior tranche, due to increased correlation, reducing its value. Thus, it is better to go long the equity tranche and short the senior tranche.

One challenge (for me, at least) is due to the different between the tranche (which is analogous to a bond) and protection on the tranche (which is a CDS, just a basket CDS). Consider the analogous bond (~ tranche) and CDS (~ senior basket CDS on tranche):
• Start with bond (tranche) at value of $X due to 3% spread (over riskfree) such that CDS spread =3%. Now assume risk of bond increases: • If bond gets riskier, 3% spread increases to (eg) 5% --> price (value) of bond decreases. But, hedged by increase in CDS spread to 5% and increase in price of CDS. I hope that helps, thanks, #### maheshs ##### Member Hi David The screen grab is from the credit risk notes pg 57 (Malz chapter 9) I don't fully understand the statement "But when default rates are relatively high, an increase in correlation can materially increase the IRR of the equity tranche but also increase the losses to the senior bond tranche" I understand the bit about losses to the senior tranche but the not the part that refers to the increase in IRR of equity tranche. I find it contradictory to the question raised earlier in the thread that talks about increase in value of equity tranche when the default probability reduces - perhaps, i am mixing up things. If you get a chance, can you help clarify this please? Thanks Mahesh #### ShaktiRathore ##### Well-Known Member Subscriber You are confused between default rates and correlation. And high value does not imply high IRR. I would think and understand the above results as follows: 1)a At high correlation b/w the tranches and high default probability: High default rate of probability implies high risky equity tranche, because as default rate of senior tranches increases implies risk of losses to equity tranche increase due to higher correlation so that there are losses to senior tranche but at the same time equity tranche IRR increases, as risk increases return increases same logic implies that as risk of equity tranche increase IRR increases. b) at high correlation and low default probability b/w tranches: this implies that senior tranches becomes less risky due to low default probability which implies due to high correlation the riskiness of the equity tranche also decreases so that the equity tranche value increases you can think of it as that future same cash flows of the equity tranche are discounted now at low discount rate due to reduction in risk so that the value of equity tranche increases. This does not imply high IRR but infact if i am not wrong a low IRR. thanks #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber Hi maheshs, I agree with ShaktiRathore: I don't see the conflict unless we confuse default rate (PD) with default correlation. Ceteris paribus, a higher default probability will correspond to a lower value (or higher spread) for all tranches (equity, junior, and senior). This is illustrated in Malz page 326, see below, by continuously lower values as we move to the right (the X-asix is default probability). Keep in mind that he graphed equity value, but junior/senior as losses, so their higher plots correspond with lower values. Unlike default probability (which is directionally the same for all tranches), default correlation equivocates for the junior (mezzanine; some authors will say mezzanine is "complicated") and is DIFFERENT for the Equity and Senior tranches. This is also illustrated by Malz (below), who is also (I think) consistent with the above thread. I think we can still comprehend it, by analogy, with the same 2-bond portfolio above. That is, a portfolio contains two bonds, each with PD = 10%. Consider two extreme cases: • Zero correlation: A first-to-default (the imperfect analog to equity tranche) has trigger probability = 1-(1-10%)^2 = 19%. A second-to-default (analog to senior tranche) has trigger probability = 10%^2 = 1.0%. • Now spike up to perfect correlation of 1.0 (an "increase in default correlation"). Notice the opposite direction of the two tranches: the "equity" tranche has trigger probability = 10%, which is a DECREASE from 19%, but the "senior" tranche trigger probability of 10% is an INCREASE from 1%! In this way, with increasing correlation, the value of equity increases (PD decreases) while the value of senior decreases (PD increases) The simple dynamics of the 2-bond example appear to be consistent with Malz below: • With respect to equity value, for any given default probability (x value), an increase in correlation (rho) implies an increase in value (compare three lines) • With respect to senior value, for any given default probability (x value), an increase in correlation (rho) implies an increase in LOSSES, which corresponds to a decrease in value Malz page 326: #### k.simpson ##### New Member I have a variant on the OP's question. Suppose that you are of the view that correlation is higher than the market thinks it is. There is a CDS basket available on 100 names (equal value), divided into senior and subordinated baskets (sub detaches at 20). What would be the trade most consistent with your belief? Should you: A) Short the senior and go long the subordinated basket? B) Long the senior and go short the subordinated basket? My reasoning (assuming I'm confident my view will play out) was to go long the senior (if the market assigns less chance of the senior tranche being impaired, then this protection is undervalued by my view. It is cheap so buy it), and short the sub (again, the market views the subordinated tranche as more risky than I see it, due to my higher correlation belief, so I will sell them protection). The answer was A, the opposite. Am I missing something fundamental? Is it somehow that buying the basket is actually selling protection, and that's why the answers are opposite? Any clarity would be greatly appreciated! TIA #### k.simpson ##### New Member The question also mentions that this is a basket of 100 single name CDS swaps. As far as I understand it, then, that's essentially a standard basket that is then tranched into a senior/sub combo. Any suggestions on this and where I might have gone wrong? ShaktiRathore, your help was valuable last time Thanks if you can offer it again. I'm thinking that my confusion may not be from not understanding the implications of rising/falling correlations, but rather the trading aspect. #### Johnson ##### New Member Hi David, I understood the effect on the values of the tranches with increasing correlations and default probability. Malz has in the same chapter page 331 provided the impact on the credit var with changing correlations and default prob. I am not able to understand it. How is it impacted? Thanks #### Unusualskill ##### Member Hi Abhi, I think it can get confusing. Can we assume a two-bond portfolio, where each bond has PD = 10%. Consider a 2nd-to-default basket CDS. • If default correlation = 0, 2TD triggers with probability = 10%*10% = 1.0%. • If default correlation increases to 100%, probability of 2TD trigger increases to 10.0%; i.e., as correlation increases, the probability that both bonds default increases (here, the senior nth-to-default tranche triggers) • So the common (robust) dynamic is: higher default correlation increases (decreases) the risk [i.e., probability of defaults that reference] of the senior (junior) tranche GARP sometimes prefers to refer to value of the tranche rather than the spread (cost of protection) as in the errata question here that you linked to, specifically: One challenge (for me, at least) is due to the different between the tranche (which is analogous to a bond) and protection on the tranche (which is a CDS, just a basket CDS). Consider the analogous bond (~ tranche) and CDS (~ senior basket CDS on tranche): • Start with bond (tranche) at value of$X due to 3% spread (over riskfree) such that CDS spread =3%. Now assume risk of bond increases:
• If bond gets riskier, 3% spread increases to (eg) 5% --> price (value) of bond decreases. But, hedged by increase in CDS spread to 5% and increase in price of CDS.
I hope that helps, thanks,

@David Harper CFA FRM

Came across this post while studying. May I ask how would you explain this dynamic: Higher default correlation decreases the risk of the junior tranche.

For my understanding, "higher default correlation increases the risk of the senior tranche", I explain using this idea: When there is a higher possibility of defaulting together for junior and senior tranche. Thus, when junior tranche defaults, senior tranche is also more likely default, resulting in investors in the senior tranche suffer loss, thus is risker (higher spread) for senior tranche.

However, I cannot think of a way to explain the idea in similar manner for this "Higher default correlation decreases the risk of the junior tranche"? Thank you!


text:
## Physics: Principles with Applications (7th Edition)

The ratio of the masses is $\frac{2}{1}$
Let the masses be $m_1$ and $m_2$. By the law of conservation of momentum: $m_1v_1 = m_2v_2$ This can be written as: $v_1 = \frac{m_2v_2}{m_1}$ Let's assume that $m_1$ has twice the kinetic energy. Therefore, $\frac{1}{2}m_1v_1^2 = 2\times \frac{1}{2}m_2v_2^2$ $m_1(\frac{m_2v_2}{m_1})^2 = 2\times m_2v_2^2$ $\frac{m_2^2}{m_1} = 2m_2$ $\frac{m_2}{m_1} = 2$ The ratio of the masses is $\frac{2}{1}$, where the larger fragment with half as much kinetic energy has double the mass of the smaller fragment.


text:
# Prospective Design Analysis

library(PRDA)

Given the hypothetical population effect size and the required power level, the function prospective() performs a prospective design analysis. Prospective design analysis allows to define the sample size needed to obtain a required level of power computing also the associated inferential risks. Function arguments are:

retrospective(effect_size, power, ratio_n = 1,
test_method = c("pearson", "two_sample", "welch",
"paired", "one_sample"),
alternative = c("two_sided","less","greater"),
sig_level = .05, ratio_sd = 1, B = 1e4,
tl = -Inf, tu = Inf, B_effect = 1e3,
sample_range = c(2, 1000), tol = .01,
display_message = TRUE)

Complete arguments description is provided in the function documentation ?prospective. In the following sections, instead, different examples are presented. For further details about design analysis see Altoè et al. (2020) and Bertoldo, Zandonella Callegher, and Altoè (2020).

## Prospective Design Analysis for Correlation

To conduct a prospective design analysis considering a correlation between two variables, we need to specify test_method = "pearson" (default option). Note that, only Pearson’s correlation is available, while the Kendall’s $$\tau$$ and Spearman’s $$\rho$$ are not implemented.

### Example 1: Pearson’s correlation

Consider a study that will evaluate the correlation between two variables. Knowing from the literature that we expect an effect size of $$\rho = .25$$, which is the required sample size to obtain a power of 60%? We can use the function prospective() setting the argument test_method = "pearson".

set.seed(2020) # set seed to make results reproducible

prospective(effect_size = .25, power = .60, test_method = "pearson",
display_message = TRUE)
#> Evaluate n = 501
#> Estimated power is 1
#>
#> Evaluate n = 251
#> Estimated power is 0.98
#>
#> Evaluate n = 126
#> Estimated power is 0.81
#>
#> Evaluate n = 64
#> Estimated power is 0.53
#>
#> Evaluate n = 95
#> Estimated power is 0.69
#>
#> Evaluate n = 80
#> Estimated power is 0.61
#>
#> Evaluate n = 72
#> Estimated power is 0.57
#>
#> Evaluate n = 76
#> Estimated power is 0.59
#>
#>  Design Analysis
#>
#> Hypothesized effect:  rho = 0.25
#>
#> Study characteristics:
#>    test_method   sample_n1   sample_n2   alternative   sig_level   df
#>    pearson       76          NULL        two_sided     0.05        74
#>
#> Inferential risks:
#>    power   typeM   typeS
#>    0.593   1.277   0
#>
#> Critical value(s): rho  =  ± 0.226

The default option display_message = TRUE prints the different steps to find the required sample size and the progress bar. Note that, however, the progress bar is available only when effect_size is defined as a function. In the output, we have the summary information about the hypothesized population effect, the study characteristics, and the inferential risks. To obtain a power of around 60% the required sample size is $$n = 76$$, the associated Type M error is almost 1.30 and the Type S error is approximately 0. Finally, the critical values (i.e., the minimum absolute effect size value that would result significant) are $$\rho = \pm.226$$. Note that correlation tests were conducted considering "two_sided" alternative hypothesis and a significance level of .05 (the default settings).

## Prospective Design Analysis for Means Comparison

To conduct a retrospective design analysis considering means comparisons, we need to specify the appropriate t-test (i.e., One-sample, Paired, Two-sample, or Welch’s t-test) using the argument test_method. Arguments specifications for the different t-tests are presented in the following Table.

Test test_method Other required arguments
One-sample t-test one_sample ratio_n = NULL
Paired t-test paired ratio_n = 1
Two-sample t-test two_sample ratio_n
Welch’s t-test welch ratio_n and ratio_sd

### Example 2: Paired t-test

Imagine that we are planning a study where the same group is measured twice (e.g., pre- and post-test). Knowing from the literature that we expect an effect size of $$d = .35$$, which is the required sample size to obtain a power of 80%? We can use the function prospective() specifying the corresponding arguments. We use the option test_method = one_sample for paired t-test and set ratio_n = 1.

prospective(effect_size = .35, power = .8, test_method = "paired",
ratio_n = 1, display_message = FALSE)
#>
#>  Design Analysis
#>
#> Hypothesized effect:  cohen_d = 0.35
#>
#> Study characteristics:
#>    test_method   sample_n1   sample_n2   alternative   sig_level   df
#>    paired        66          66          two_sided     0.05        65
#>
#> Inferential risks:
#>    power   typeM   typeS
#>    0.798   1.124   0
#>
#> Critical value(s): cohen_d  =  ± 0.246

To obtain a power of 80%, the required sample size is $$n=66$$, the associated Type M error is around 1.10 and the Type S error is approximately 0. Finally, the critical values (i.e., the minimum absolute effect size value that would result significant) are $$d = \pm 0.246$$.

### Example 3: Two-sample t-test

Imagine now the case where two groups (e.g., treatment and control group) will be compared. However, we know in advance that the sample size in the two groups will be different (e.g., the number of participants in the treatment group could be limited due to strict selecting criteria). We can define the ratio between the sample size in the first group and in the second group using the ratio_n argument. Again, we hypothesize an effect size of $$d = .35$$, but this time we specify a one-sided alternative hypothesis and a significance level of .10. We can do that using respectively the arguments alternative and sig_level.

prospective(effect_size = .35, power = .80, ratio_n = .5,
test_method = "two_sample", alternative = "great", sig_level = .10,
display_message = FALSE)
#>
#>  Design Analysis
#>
#> Hypothesized effect:  cohen_d = 0.35
#>
#> Study characteristics:
#>    test_method   sample_n1   sample_n2   alternative   sig_level   df
#>    two_sample    55          110         greater       0.1         163
#>
#> Inferential risks:
#>    power   typeM   typeS
#>    0.802   1.167   0
#>
#> Critical value(s): cohen_d  =  0.213

The option test_method = "two_sample" is used to consider a two-sample t-test. To obtain a power of 80%, we would need at least 55 participants in the first group and 110 participants in the second group. The associated Type M error is almost 1.20 and the Type S error is approximately 0. Finally, the critical value is $$d = .213$$.

### Example 4: Welch’s t-test

Consider again the previous example, but this time we do not assume homogeneity of variance between the two groups. We suppose, instead, that the ratio between the standard deviation of the first group and of the second group is 1.5. In this case the appropriate test is the Welch’s t-test. We set the option test_method = "welch" and specify the argument ratio_sd.

prospective(effect_size = .35, power = .80, ratio_n = .5, test_method = "welch",
ratio_sd = 1.5, alternative = "great", sig_level = .10,
display_message = FALSE)
#>
#>  Design Analysis
#>
#> Hypothesized effect:  cohen_d = 0.35
#>
#> Study characteristics:
#>    test_method   sample_n1   sample_n2   alternative   sig_level   df
#>    welch         63          126         greater       0.1         90.403
#>
#> Inferential risks:
#>    power   typeM   typeS
#>    0.799   1.17    0
#>
#> Critical value(s): cohen_d  =  0.212

Now, to obtain a power of 80%, we would need at least 63 participants in the first group and 126 participants in the second group. The associated Type M error is almost 1.20 and the Type S error is approximately 0. Finally, the critical value is $$d = .212$$. Results are really close to the previous ones.

## Population effect size distribution

Defining the hypothetical population effect size as a single value could be limiting. Instead, researchers may prefer to use a probability distribution representing their uncertainty regarding the hypothetical population effect. Note that this could be interpreted as a prior distribution of the population effect in a Bayesian framework.

To define the hypothetical population effect size (effect_size) according to a probability distribution, it is necessary to specify a function that allows sampling values from a given distribution. The function has to be defined as function(n) my_function(n, ...), with only one single argument n representing the number of samples (e.g., function(n) rnorm(n, mean = 0, sd = 1)). See vignette("retrospective") for further details.

### Example 5: Effect size distribution

Consider the same scenario as in the correlation example (Example 1). This time we define the hypothesized effect size according to a normal distribution with mean .30 and standard deviation .10. Moreover, to avoid unreasonable values we truncate the distribution between .15 and .45.

prospective(effect_size = function(n) rnorm(n, .3, .1), power = .60,
test_method = "pearson", tl = .15, tu = .45, B_effect = 500,
B = 500, display_message = FALSE)
#> Truncation could require long computational time
#>
#>  Design Analysis
#>
#> Hypothesized effect:  rho ~ rnorm(n, 0.3, 0.1) [tl =  0.15 ; tu = 0.45 ]
#>    n_effect   Min.   1st Qu.   Median   Mean   3rd Qu.   Max.
#>    500        0.15   0.241     0.302    0.3    0.356     0.448
#>
#> Study characteristics:
#>    test_method   sample_n1   sample_n2   alternative   sig_level   df
#>    pearson       52          NULL        two_sided     0.05        50
#>
#> Inferential risks:
#>         Min.    1st Qu.   Median   Mean       3rd Qu.   Max.
#> power   0.176   0.412     0.6060   0.582236   0.75      0.936
#> typeM   1.019   1.143     1.2675   1.361914   1.50      2.338
#> typeS   0.000   0.000     0.0000   0.000464   0.00      0.033
#>
#> Critical value(s): rho  =  ± 0.273

Note that we adjusted B_effect and B to find a good trade-off between computational times and results accuracy. Differently from previous outputs, we have now a summary for the sampled effects distribution and for the inferential risks.

## Graphical representation

Currently there are no personalized plot functions in {PRDA}. However, it is easy to access all the results and use them to create the plots according to your needs.

The function prospective() returns a list with class "design_analysis" that contains:

• design_analysis - a character string indicating the type of design analysis (prospective or retrospective).
• call_arguments - a list with all the arguments passed to the function.
• effect_info - a list with all the information regarding the considered hypothetical population effect size. In particular, in effect_samples we find the vector with the sampled effects (or unique value in the case of a single value).
• test_info - a list with all the information regarding the test performed.
• prospective_res - a data frame with the results of the design analysis (i.e., power, typeM, and typeS).

Output complete description is provided in the function help page ?prospective.

da_fit <- prospective(effect_size = function(n) rnorm(n, .3, .1), power = .60,
test_method = "pearson", tl = .15, tu = .45,
B_effect = 500, B = 500, display_message = FALSE)
#> Truncation could require long computational time

str(da_fit, max.level = 1)
#> List of 5
#>  $design_analysis: chr "prospective" #>$ call_arguments :List of 16
#>  $effect_info :List of 6 #>$ test_info      :List of 7
#>  \$ prospective_res:'data.frame': 500 obs. of  3 variables:
#>  - attr(*, "class")= chr [1:2] "design_analysis" "list"

Similarly to the examples provided in vignette("retrospective"), results can be used to create the plots according to your needs. See vignette("retrospective") for further details.

### References

Altoè, Gianmarco, Giulia Bertoldo, Claudio Zandonella Callegher, Enrico Toffalini, Antonio Calcagnì, Livio Finos, and Massimiliano Pastore. 2020. “Enhancing Statistical Inference in Psychological Research via Prospective and Retrospective Design Analysis.” Frontiers in Psychology 10. https://doi.org/10.3389/fpsyg.2019.02893.

Bertoldo, Giulia, Claudio Zandonella Callegher, and Gianmarco Altoè. 2020. “Designing Studies and Evaluating Research Results: Type M and Type S Errors for Pearson Correlation Coefficient.” Preprint. PsyArXiv. https://doi.org/10.31234/osf.io/q9f86.

Gelman, Andrew, and John Carlin. 2014. “Beyond Power Calculations: Assessing Type S (Sign) and Type M (Magnitude) Errors.” Perspectives on Psychological Science 9 (6): 641–51. https://doi.org/10.1177/1745691614551642.


text:
# Probability and expectancy problem [closed]

if we choose a size $$25$$ subset of the set: $$\{1,2,....100\}$$ what is the expectancy of the number of sequential pairs in the subset? expectancy still confuses me, can anybody help?

## closed as off-topic by Namaste, user21820, RRL, Lord_Farin, DidDec 19 '18 at 18:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, user21820, RRL, Lord_Farin, Did
If this question can be reworded to fit the rules in the help center, please edit the question.

• – StubbornAtom Dec 10 '18 at 18:56

Let $$X_i$$ be an indicator random variable that $$i$$ and $$i+1$$ ($$i\in \{1,2,...,99\}$$) are in a choosen subset.

Then $$P(X_i = 1) = {{98\choose 23}\over {100\choose 25}}= {6\over 99}$$ Since $$X= X_1+...+X_{99}$$ we have $$E(X) = E(X_1)+...+E(X_{99}) = 99{6\over 99} = 6$$

• Is that better @copper.hat? – Aqua Dec 5 '18 at 18:44
• Yes thanks. ${}{}$ – copper.hat Dec 5 '18 at 18:49

Let's assign a label $$1,2 \dots 25$$ (arbitrary order) to each of the selected numbers. Let $$A_{i,j}$$ with $$1\le i be $$1$$ if elements with labels $$i,j$$ have sequential values, $$0$$ otherwise.

Then $$X = \sum A_{i,j}$$ counts the number of sequential pairs, which is what we want, and

$$E(X)=\sum A_{i,j}$$

But $$E(X)=\sum E[A_{i,j}]= n_p P(A_{i,j} = 1) = n_p \frac{99}{\binom{100}{2}}$$

Where $$n_p = \binom{25}{2}=300$$ is the number of pairs. Then

$$E(X) = \frac{300 \times 99} { \binom{100}{2}}=6$$


text:
# Description

the same way it replies to
Will this resolve the issue?

@scatenag thinks the problem is that we have not a link to an image (Eg. image.svg)
This causes latex to fail.

In search of a solution for a fails with latex compiler:

# Related Objects

### Event Timeline

moranegg triaged this task as Normal priority.Dec 14 2019, 12:54 AM

Maybe "This might be resolved with T2146 ?" - T2149 is this one :)...
Anyway .. No, even if T2146 is resolved, this is unrelated.

Actually, the problem is related on how to include badges in latex.
I mean: in the swapguide we edit a .md and it will become readable via github we interface (here everithing is ok).
But .. we generate alse a latex file where the image

`[![SWH](https://archive.softwareheritage.org/badge/origin/https://github.com/Unipisa/TAUmus)](https://archive.softwareheritage.org/browse/origin/https://github.com/Unipisa/TAUmus/)`

becomes

`\href{https://archive.softwareheritage.org/browse/origin/https://github.com/Unipisa/TAUmus/}{\includegraphics{https://archive.softwareheritage.org/badge/origin/https://github.com/Unipisa/TAUmus}}`

The problem is that the latex compiler doesnt find andy "file" at the url even if is there..
for this reason I suggested to serve also a badge.svg file to help those situations...

but .. again it is a latex&badges problem, not a swh problem .. so you I can mark this as resolved ..

scatenag closed this task as Resolved.Dec 16 2019, 10:10 AM


text:
# What is the momentum of an object of mass $m$, moving with a velocity $v$? $(a).\ (mv)^2$$(b).\ mv^2$$(c).\ \frac{1}{2}mv^2$$(d).\ mv$

#### Complete Python Prime Pack

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Option $(d)$ is correct.

Momentum is known as the product of the mass and velocity of a moving object.

If a body of mass $m$ is moving with velocity $v$,

then the momentum of the moving body, $P=mv$

Momentum is a vector quantity.

The SI unit of momentum is kilogram-meter per second and is written as kg-m/s.

$P=m$

Updated on 10-Oct-2022 13:22:32


text:
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Please, prove that if we have two variables (equal sample size) $X$ and $Y$ and the variance in $X$ is greater than in $Y$, then the sum of squared differences (i.e., squared euclidean distances) ...
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### How to use G Power 3 to calculate statistical power in mixed design ANOVA with unequal group sample sizes

In G power 3, ANOVA repeated measures within-between interaction: Only the total sample size is reported assuming equal sample size for the two groups. My questions are: How would it work if the ...
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text:
My Math Forum (http://mymathforum.com/math-forums.php)
-   Elementary Math (http://mymathforum.com/elementary-math/)

 lazio December 1st, 2017 09:10 AM

If the rent for a house is £895 per month and is being rented by 2 people. One is accepted to pay £20 more per week, what is going to be monthly payment for each of those people?

Is it going to be £407.50 for one and £487.50 for the other OR £367.50 and £527.50 for the other?

I'm pretty sure it's the first statement.

The second has been calculated by my mate who is arguing that £895/2 = 447.50 and £447.50 - £80 = £367.50 and 447.50 + £50 = £527.50.

My logic is £895 - 80 = £815 £815 / 2 = £407.50 £407.50 + £80 = £487.50 for one and the rest for the other.

 idontknow December 1st, 2017 09:33 AM

First $\displaystyle X+Y=895$ and $\displaystyle X=Y+20\cdot 4=Y+80$
$\displaystyle 2Y+80=895$
$\displaystyle Y=407.5$
The first statement is true

 lazio December 1st, 2017 09:41 AM

That's exactly what I mean. How can I in the simplest way explain this? My mate doesn't understand algebra, I assume. He doesn't understand the fact that you shouldn't divide the whole amount and -80 from one amount and +80 from the other but -40 and +40 so the gap could equal £80.

 skipjack December 1st, 2017 10:26 AM

Both methods can be made to work, but I would first question the assumption that you both make that a month contains 4 weeks.

 JeffM1 December 1st, 2017 02:38 PM

If you work from the premise that a month has 4 weeks, there is a simple way to explain it.

An even split would be $\dfrac{895}{2} = 447.5.$

No difficulty there.

Now compute $447.5 + 40 = 487.5.$ Straight forward, no algebra.

And compute $447.5 - 40 = 407.5.$

So if you pay 487.5 and your mate pays 407.5 that sums to

$487.5 + 407.5 = 895.0.$ Simple arithmetic.

And what is the difference in the rent paid?

$487.5 - 407.5 = 80 = 4 \times 20.$ Simple arithmetic.

If you ignore how you got the answer and just explain why the answer pays the rent and has you paying 80£ more, all you need is very basic arithmetic.

So this is the correct answer if what your agreement was REALLY involved a differential of 80 quid a month. But it is not correct if your agreement REALLY involved a differential of 20 quid a week because the answer calculated assumes that there are only 48 weeks in the year when in fact there are just about exactly 52. If the latter is the case, then here is what is called for.

Your monthly payment $= 490.83.$

Your mate's monthly payment $= 895 - 490.83 = 404.17.$

Now you are paying together $490.83 + 404.17 = 895.$

And you are paying $490.83 - 404.17 = 86.66$ more each month.

That makes sense because there are a bit more than 4 weeks in the average month. Now let's how it works out over a year.

Annually, you pay $12 \times 490.83 = 5889.96$

Annually, your mate pays $12 \times 404.17= 4850.04.$

Notice the sum totals $10740 = 12 \times 895.$

In total, you pay $5889.06 - 4850.04 = 1039.02$ more in a year.

And $52 \times 20 = 1040.$

At the end of the year you owe him a quid.

 Country Boy January 30th, 2018 12:44 PM

Quote:
 Originally Posted by lazio (Post 585103) If the rent for a house is £895 per month and is being rented by 2 people. One is accepted to pay £20 more per week, what is going to be monthly payment for each of those people? Is it going to be £407.50 for one and £487.50 for the other OR £367.50 and £527.50 for the other? I'm pretty sure it's the first statement. The second has been calculated by my mate who is arguing that £895/2 = 447.50 and £447.50 - £80 = £367.50 and 447.50 + £50 = £527.50.
I presume that last should be 447.50+ £80= £527.50.
But 527.50- 367.50= £160, not £80. To get one that is £80 more than the other you need to add and subtract £80/2= £40 to and from £447.50.

Then you would get £447.50- £40= £407.50 and £447.50+ £40= £487.50 as in the first method.

Quote:
 My logic is £895 - 80 = £815 £815 / 2 = £407.50 £407.50 + £80 = £487.50 for one and the rest for the other.


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# Flow Rate and Orifice Diameter

I have a electrical/electronics background and have limited knowledge of Fluid Mechanics. So, i will try to be as clear as possible.

I am currently working with brake fluid in an ABS system and to effectively construct my alternative control strategy,i have following questions:

1> Flow-rate: I just need to know how to measure the flow-rate across an orifice? I know of the formula

$$Q = CA\sqrt{\frac{2\Delta P}{d}}$$

$C$ - discharge coefficient

$\Delta P$ - Change in pressure

$d$ - density of the fluid

$A$ - Area of the orifice

Now, this formula should pretty much give me the answer, but my real question is; if i want to calculate simply the flow-rate across a circular orifice of different diameters. So, $Q1$ for $D1$,$Q2$ for $D2$ ... $Qn$ for $Dn$. I am simply interested in finding the flow rate without the consideration of pressure change or velocity etc.

Just to clarify, FLOW RATE through a PIPE of varying diameters.

I will be grateful for any and every suggestion!

• Well, flow rate through a pipe of varying diameters is constant. it is given by the cross-sectional area at the point you are measuring multiplied by the area-average velocity at this cross-section. I did not understand what it is that you are asking. the pressure change and velocity aid the measurement of the flow rate, via the Bernoulli principle. – aditya kp Oct 10 '13 at 21:33
• @adityakp for the given ABS, i want to define the value of Q as a function of X and ΔP where X is the diameter of the orifice. So, if i know ΔP and X(its controlled by a solenoid) i get a certain value of Q. – sheetansh Oct 11 '13 at 12:29
• @MikeDunlavey : Thanks mike for your answer. isnt C(discharge coefficient) a constant for a given system? – sheetansh Oct 11 '13 at 12:38
• @MikeDunlavey : thanks for your comment, which pretty much answered my question! i am able to relate orifice diameter and Q. As for discharge coefficient, as the values of same are rather small in my system, i will take it into account as mathematical factor in my Matlab simulations(if any queries arise, will ask them again) but for the sake of of this question i have my answer! thanks again! Please, put up the same comment as the answer so the question can be closed! – sheetansh Oct 14 '13 at 6:25

OK, area $A$ is just $\pi D^2/4$. The real question is: What is $C$? It depends on the shape of the orifice (and Reynolds number). There are some quick-and-dirty approximations here. It depends on the orifice geometry, like whether its edges are rounded or sharp.


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# Color Palettes in HCL Space

October 12, 2012
By

(This article was first published on Trestle Technology, LLC » R, and kindly contributed to R-bloggers)

This is a quick follow-up to my previous post about Color Palettes in RGB Space. Achim Zeileis had commented that, perhaps, it would be more informative to evaluate the color palettes in HCL (polar LUV) space, as that spectrum more accurately describes how humans perceive color. Perhaps more clear trends would emerge in HCL space, or color palettes would more closely hug their principal component.

For the uninitiated, a good introduction to HCL color spaces is available at this site, or from Mr. Zeileis' own paper here.

We'll start by loading in some code written previously (or slightly modified to support HCL). We can plot out an image of the different color palettes we're evaluating once again.

source("../Correlation.R")
par(mfrow = c(6, 3), mar = c(2, 1, 1, 1))
for (i in 1:18) {
palette <- getSequentialLuv(i, allSequential)
image(mat, col = hex(palette), axes = FALSE, main = i)
}

## HCL-Space

The fundamental question for this analysis was how these color palettes move through HCL space, as opposed to RGB space, which was considered last time.

An interactive visualization of palette #2 in 3-Dimensional HCL space

### R2 Values

As we did in the previous analysis, we can use the principal component of each palette, to compute the R2 value to quantify how well the data aligns to this component.

Palette #2 with the Principal Component.

We'll recycle some of the old functions, and create a new one that calculates the R2 values in HCL space.

#' Compute the proportion of variance accounted for by the given number of components
#'
#' @author Jeffrey D. Allen \email{Jeffrey.Allen@@UTSouthwestern.edu}
propVar <- function(pca, components=1){
#proportion of variance explained by the first component
pv <- pca$sdev^2/sum(pca$sdev^2)[1]
return(sum(pv[1:components]))
}

#' Calculate the R-squared values of all 18 color palettes
#'
#' @author Jeffrey D. Allen \email{Jeffrey.Allen@@UTSouthwestern.edu}
calcR2Sequential <- function(){
library(rgl)
R2 <- list()
for (i in which(sequential[,2] == 9)){
palette <- RGBToLuv(sequential[i:(i+8), 7:9])
pca <-plotLuvCols(palette[,"L"],palette[,"C"], palette[,"H"], pca=1)
R2[[length(R2)+1]] <- propVar(pca,1)
}
return(R2)
}

calcR2RGBSequential <- function() {
library(rgl)
R2 <- list()
for (i in which(sequential[, 2] == 9)) {
palette <- sequential[i:(i + 8), 7:9]
pca <- plotCols(palette$R, palette$G, palette$B, pca = 1) cat(i, ": ", propVar(pca, 1), "\n") R2[[length(R2) + 1]] <- propVar(pca, 1) } return(R2) }  ### Comparison of R2 Values Let's compare the R2 values computed from RGB space to those computed in HCL space # R2 Values in HCL Space r2 <- calcR2Sequential() r2 <- unlist(r2) names(r2) <- 1:18 # R2 Values in RGB Space rgbr2 <- calcR2RGBSequential()  ## 34 : 0.981 ## 76 : 0.9097 ## 118 : 0.9541 ## 160 : 0.9847 ## 202 : 0.9552 ## 244 : 0.9663 ## 286 : 0.9674 ## 328 : 0.9273 ## 370 : 0.9344 ## 412 : 0.9593 ## 454 : 0.9049 ## 496 : 0.9007 ## 538 : 0.9954 ## 580 : 0.9752 ## 622 : 0.9846 ## 664 : 0.9292 ## 706 : 0.9311 ## 748 : 1  rgbr2 <- unlist(rgbr2) names(rgbr2) <- 1:18 # plot the comparison plot(rgbr2 ~ r2, ylab = "RGB R2 Values", xlab = "HCL R2 Values", main = "RGB vs. HCL R2 Values")  pv <- anova(lm(rgbr2 ~ r2))$"Pr(>F)"[1]

As seen clearly in the plot, these two variables are not correlated (p-value of 0.5232), so there's definitely a difference in how these palettes move through HCL vs. RGB space.

## Color Ratings

One hypothesis of this analysis was that because HCL space corresponds more directly to our perception of color, perhaps a smoother or more linear path through HCL space would have greater consequence on the visual appeal of the color palette than it would in RGB space. To test this, we can repeat the same analysis as we had before to see if a small R2 value is significantly correlated with the visual appeal of a color palette.

colorPreference <- read.csv("../turk/output/Batch_790445_batch_results.csv",
header = TRUE, stringsAsFactors = FALSE)
colorPreference <- colorPreference[, 28:29]
colorPreference[, 1] <- substr(colorPreference[, 1], 44, 100)
colorPreference[, 1] <- substr(colorPreference[, 1], 0, nchar(colorPreference[,
1]) - 4)
colnames(colorPreference) <- c("palette", "rating")

prefList <- split(colorPreference[, 2], colorPreference[, 1])
prefList <- prefList[order(as.integer(names(prefList)))]

### R2 Values

We can calculate the R2 values for each palette as previously discussed and compare to see if it's associated with the aesthetic appeal of a palette.

r2

##      1      2      3      4      5      6      7      8      9     10
## 0.9614 0.9794 0.9799 0.9388 0.9708 0.8902 0.9651 0.9579 0.9725 0.9842
##     11     12     13     14     15     16     17     18
## 0.9325 0.9129 0.9603 0.9358 0.9568 0.9611 0.9785 1.0000

plot(avgs ~ r2, main = "Linearity of Color Palette vs. Aesthetic Appeal", xlab = "R-squared Value",
ylab = "Average Aesthetic Rating")
abline(lm(avgs ~ r2), col = 4)

pv <- anova(lm(avgs ~ r2))$"Pr(>F)"[1]  Oddly, this result contradicts the previous result; it seems that adhering to a linear path through HCL space is actually inversely related with the aesthetic appeal on these palettes. We should note, of course, that the p-value for this correlation is not significant, as it stands (p-value = 0.6598). Regardless, the negative trend seems fairly obvious on inspection and, if you were to exclude the two "outlier" palettes on the left side which have stark breaks in HCL space on either end of the spectrum, you get a significant (0.01) negative correlation. Indeed, these same two palettes were the same outliers in the previous plot comparing RGB to HCL R2 values. Obviously, as it stands, the conclusion of the analysis is that there is no correlation in these palettes between linearity in HCL space and aesthetic appeal, but it is curious that the hinted correlation is actually negative. ### Dimensional Spread Another point of interest is the "spread" or coverage of a palette across one particular axis. For instance, it may be interesting to compare a color palette which varies only in luminance to one which varies only in chroma to see if one type of progression is more aesthetically appealing. We can summarize such a phenomenon by calculating the distance between each color in a palette along a given axis (luminosity, for instance), which we'll label "ΔLuminosity" the . Extracting the median of these points may provide some insight into the movement of a particular palette along an axis. plotDimensionalScoring <- function(palettes, scoring, ...) { ranges <- data.frame(L = numeric(0), C = numeric(0), H = numeric(0)) for (i in 1:length(palettes)) { colPal <- RGBToLuv(palettes[[i]]) # get the differences between each color on each axis colPal <- diff(colPal) colPal <- abs(apply(colPal, 2, median)) ranges[i, ] <- colPal } plot3d(0, 0, 0, xlab = "L", ylab = "C", zlab = "H", type = "n", xlim = c(-0.5, max(ranges[, 1] + 1)), ylim = c(-0.5, max(ranges[, 2] + 1)), zlim = c(-0.5, max(ranges[, 3] + 1)), ...) cols <- heat_hcl(n = 101) for (i in 1:nrow(ranges)) { plot3d(ranges[i, 1], ranges[i, 2], ranges[i, 3], type = "p", add = TRUE, pch = 19, size = 10, col = cols[101 - (round(scoring[i], 2) * 100)], ...) } ranges } normAvgs <- avgs - min(avgs) normAvgs <- normAvgs/max(normAvgs) ranges <- plotDimensionalScoring(allSequential, normAvgs)  Each palette now has one median Δ value for each axis. We can plot these in 3D space to see where the palettes fall. We'll go ahead and color-code these based on their average visual appeal to see if we can observe any trends or "hotspots" in which palettes are consistently rated as visually appealing based only on their median movement along an axis. You must enable Javascript to view this page properly. A plot of all color palettes' median change in HCL space. We're descending into fairly non-scientific analysis here, but I noticed a pattern when viewing these points along the chroma and luminosity axes. It seems like there's something of a pattern resulting in consistently positioned high-ranked palettes. I interpolated a heatmap based on these ratings on the C and L axes. library(akima) resolution = 64 cSeq <- seq(min(ranges[, 2]), max(ranges[, 2]), length.out = resolution) lSeq <- seq(min(ranges[, 1]), max(ranges[, 1]), length.out = resolution) a <- interp(x = ranges[, 1], y = ranges[, 2], z = normAvgs, xo = lSeq, yo = cSeq, duplicate = "mean") filled.contour(a, color.palette = heat_hcl, xlab = "Luminance", ylab = "Chroma", main = "Visual Appeal Across Chroma & Luminance")  maxL <- lSeq[which.max(apply(a$z, 2, max, na.rm = TRUE))]

## Warning: no non-missing arguments to max; returning -Inf

maxC <- cSeq[which.max(apply(a$z, 1, max, na.rm = TRUE))]  ## Warning: no non-missing arguments to max; returning -Inf  You can see the peak of visual appeal at luminance = 6.374 and chroma = 8.332. A few examples of palettes using this "optimized" chroma and luminance spacing... par(mfrow = c(6, 3), mar = c(2, 1, 1, 1)) for (i in 1:18) { barplot(rep(1, 9), col = sequential_hcl(9, h = (20 * i), c. = c(100, 33.6), l = c(40, 84.59), power = 1)) }  ### Individual Axis Analysis There is one palette that seems to be a bit of an outlier in most regards. Of all the palettes, the greyscale palette has the lightest L value, and the lowest C and H values. If you calculate the Z-scores on each color axis, this palette has consistently higher scores; below is a plot of the average Z-score across the 3 axes. barplot(apply(abs(scale(ranges)), 1, mean), col = c(rep("#BBBBBB", 17), "#CC7777"), xlab = "Palette #", ylab = "Average Z-score Across {H, C, L} Axes")  Because of this, I decided to exclude it from all further calculations of significance. Where appropriate, I'll still plot it graphically but distinguish it from the other palettes. Rather than relying on a 3D plot, we can examine each axis individually to see if there's a significant association between a palette's median variation on that access and its visual appeal. (The greyscale plot will be plotted with a triangle symbol in these plots, and was excluded to p-value and regression calculations). filRanges <- ranges[1:17, ] filNormAvgs <- normAvgs[1:17] plot(normAvgs ~ ranges[, "H"], pch = c(rep(1, 17), 2), main = "Hue vs. Visual Appeal", xlab = expression("Median " * Delta * Hue), ylab = "Average Aesthetic Rating") abline(lm(filNormAvgs ~ filRanges[, "H"]), col = 2) pvH <- anova(lm(filNormAvgs ~ filRanges[, "H"]))$"Pr(>F)"[1]
text(3, 0.1, paste("p-value =", round(pvH, 3)))

plot(normAvgs ~ ranges[, "C"], pch = c(rep(1, 17), 2), main = "Chroma vs. Visual Appeal",
xlab = expression("Median " * Delta * Chroma), ylab = "Average Aesthetic Rating")
abline(lm(filNormAvgs ~ filRanges[, "C"]), col = 2)
pvC <- anova(lm(filNormAvgs ~ filRanges[, "C"]))$"Pr(>F)"[1] text(3, 0.1, paste("p-value =", round(pvC, 3)))  plot(normAvgs ~ ranges[, "L"], pch = c(rep(1, 17), 2), main = "Luminance vs. Visual Appeal", xlab = expression("Median " * Delta * Luminance), ylab = "Average Aesthetic Rating") abline(lm(filNormAvgs ~ filRanges[, "L"]), col = 2) pvL <- anova(lm(filNormAvgs ~ filRanges[, "L"]))$"Pr(>F)"[1]
text(5, 0.1, paste("p-value =", round(pvL, 3)))

As shown on the plots, there is no observable correlation with regards to movement through hue with visual appeal; there is a significant negative correlation between median variation in chroma and the visual appeal; and there seems to be a negative (though non-significant) correlation between median variation in luminance and visual appeal.

## Conclusion

### Comparison to Colorspace Palettes

It was mentioned that colorspace generates their palettes in HCL space, so the trends should more obviously emerge for such palettes. As it turns out, the palettes in HCL not necessarily linear, even in HCL space. Many palettes are fairly linear on at least one axis, but often have at least one color at either end which is completely non-linear, causing the R2 values to often be lower than they were in colorbrewer palettes. For instance:

csp <- as(hex2RGB(sequential_hcl(n = 9)), "polarLUV")@coords
pca <- plotLuvCols(csp[, "L"], csp[, "C"], csp[, "H"], 1)

A plot of the "sequential_hcl" color palette from the colorspace package.

After inspecting the code, the hue value in this color palette never changes. So it seems that we're encountering a substantial rounding error when you get the the dark/light edges of the color palette.

### Summary

I'm still a bit perplexed by the results I observer, but I suppose this analysis hints at a few things.

First of all, I realized that it very much matters which color space you're working in when designing color palettes. As the lack of a correlation between the R2 values from RGB space and HCL space show, the movement of different palettes through these spaces are drastically different and don't seem to be correlated.

Second, it looks like there is some potential to optimize the spacing of colors in a given palette by finding these "hotspots" in the Luminance vs. Chroma plot. On the other hand, movement along the hue axis for a given palette doesn't seem to have much of an effect on the visual appeal.

Finally, it seems that the rounding error on hex codes and 255-value RGB values becomes significant for especially dark or bright ends of a palette. This may put into question the effectiveness of such analysis in 3D space. Of course, the hue value represents a complete circle, so a hue of 359 is very close to a hue of 0, which wouldn't be captured in a naive 3D analysis. Even accounting for this problem, there are other nuances of this color space which can't be captured in 3D space, and may jeopardize the effectiveness of PCA in such a space.

### Future Work

As mentioned previously, it will be interesting to analyze not just the visual appeal of a color palette, but the effectiveness of a palette at communicating information. Hopefully the next post will be able to capture some of that information.

As always, comments are welcome! I'm well outside the scope of my training and expertise, so I'd be happy to hear any critiques or concerns about methodology.

### Acknowledgements

R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...


text:
# How to Use thePower Rule of Exponents

## Summary

The formalized definition of the power rule of exponents is...

$$(x^{a})^{b} = x^{ab}$$

Basically, it tells us that if we have a power raised to another exponent, we can multiply the two exponents to simplify the expression. You can also use this rule in reverse.

The power rule of exponents is also known as the "power property of exponents" or the "powers of powers" rule.

## How to Use the Power Rule of Exponents

1. Identify the exponent that is being raised to another exponent.
2. Multiply the exponents.
3. Write the simplified power with the result of Step 2 as the exponent of the original base.

## Examples

$$(2^{3})^{4}=2^{12}$$

$$(7^{5})^{3}=7^{15}$$

$$(y^{9})^{2}=y^{18}$$

## Why Does It Work?

To understand why the power rule of exponents works, let's calculate $$(2^{3})^{4}$$ with order of operations instead of using the power rule.

$$(2^{3})^{4}$$

Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (3) tells me that I need to multiply 2 by itself 3 times.

$$(2 \times 2 \times 2)^{4}$$

The exponent outside the parentheses (4) tells me that I need to multiply all the stuff inside the parentheses by itself 4 times.

$$(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times2 \times 2) \times (2 \times2 \times 2)$$

So, when $$(2^{3})^{4}$$ is fully expanded, we have twelve 2's multiplied by each other.

We can rewrite that as $$2^{12}$$.

So, that is why...

$$(2^{3})^{4}= 2^{12}$$

The reason you multiply the exponents is because the outside exponent multiplies the effect of the inside exponent. This is also true when variables are involved.

$$(y^{9})^{2}$$

Following the order of operations, I'll simplify the stuff inside parentheses first. The exponent (9) tells me that I need to multiply y by itself 9 times.

$$(yyyyyyyyy)^{2}$$

The exponent outside the parentheses (2) tells me that I need to multiply the stuff inside the parentheses by itself 2 times.

$$(yyyyyyyyy)(yyyyyyyyy)$$

There are 18 y's multiplied together, which is $$y^{18}$$.

So, that is why...

$$(y^{9})^{2}= y^{18}$$

## Reverse Power Rule of Exponents

Sometimes, it can be helpful to use the reverse power rule of exponents to write a power with one exponent as a power of powers.

$$x^{ab}=(x^{a})^{b}$$

For example, if you were trying to factor perfect squares or perfect cubes, you may want to write $$x^{30}$$ like this...

$$x^{30}=(x^{15})^{2}$$

OR

$$x^{30}=(x^{10})^{3}$$

Or if you were trying to rewrite rational exponents as roots, you may want to write $$x^{\frac{3}{2}}$$ like this...

$$x^{\frac{3}{2}}=$$$$(x^{\frac{1}{2}})^{3}$$

OR

$$x^{\frac{3}{2}}=$$$$(x^{3})^{\frac{1}{2}}$$

This would be helpful because an exponent of $$\frac{1}{2}$$ is the same thing as a square root

### What if I have multiple exponents?

#### Show/Hide

If you have multiple exponents (as powers within powers), you should multiply all of the exponents.

$$[(3^{4})^{7}]^{2}=3^{56}$$

$$[(x^{2})^{5}]^{6}=x^{60}$$

$$[(y^{3})^{7}]^{4}=y^{84}$$

I'll simplify an example using order of operations, so you can see WHY we multiply all of the exponents...

$$[(8^{2})^{4}]^{3}$$

I'll start by simplifying the stuff inside the inner-most set of parentheses. The exponent (2) tells me to multiply 8 by itself two times.

$$[(8\times8)^{4}]^{3}$$

The next exponent (4) tells me to multiply the stuff inside the inner parentheses by itself 4 times.

$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]^{3}$$

The last exponent tells me to multiply the stuff inside the outermost parentheses by itself 3 times.

$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times$$

$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]\times$$

$$[(8\times8)\times(8\times8)\times(8\times8)\times(8\times8)]$$

There are twenty-four 8's multiplied by each other, so…

$$[(8^{2})^{4}]^{3}=8^{24}$$

### What if I have multiple numbers in the base?

#### Show/Hide

If you have multiple numbers inside the parentheses, you can use the distributive property of exponents to distribute the outer exponent to each factor within the parentheses.

If any of the bases don't have an exponent written, remember that they have an "invisible" exponent of 1

$$(a^{2}b^{5})^{3}=a^{6}b^{15}$$

$$(2x^{7})^{3}=8x^{21}$$

$$(x^{2}y^{3}z^{4})^{5}=x^{10}y^{15}z^{20}$$

I'll simplify an example with order of operations, so you can see WHY we distribute the exponent to each factor inside the parentheses...

$$(x^{3}y^{4}z)^{2}$$

I'll simplify the inside of the parentheses first. The exponent (3) tells me that I need to multiply x by itself 3 times. The exponent (4) tells me that I need to multiply y by itself 4 times. The z has an "invisible" exponent of 1, so it is multiplied once.

$$(xxxyyyyz)^{2}$$

The outer exponent tells me that I need to multiply the stuff inside parentheses by itself 2 times.

$$(xxxyyyyz)(xxxyyyyz)$$

There are 6 x’s, 8 y’s, and 2 z's multiplied together, so…

$$(x^{3}y^{4}z)^{2}=x^{6}y^{8}z^{2}$$

### What if the base is a fraction?

#### Show/Hide

If you have a fraction inside the parentheses, you can use the division version of the distributive property of exponents to distribute the outside exponent to the numerator and denominator of the fraction.

$$(\frac{7^2}{5^3})^4 = \frac{7^{8}}{5^{12}}$$

$$(\frac{a^6}{9^4})^2 = \frac{a^{12}}{9^{8}}$$

$$(\frac{x^{12}}{y^3})^5 = \frac{x^{60}}{y^{15}}$$

In this example, you can see WHY we distribute the exponent to the numerator and denominator...

$$(\frac{x^2}{y^5})^4$$

I'll simplify the inside of the parentheses first. The exponent (2) tells me that I need to multiply x by itself 2 times. The exponent (5) tells me that I need to multiply y by itself 5 times.

$$(\frac{xx}{yyyyy})^{4}$$

The next exponent tells me I need to multiply the stuff inside the parentheses by itself 4 times.

$$(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})(\frac{xx}{yyyyy})$$

There are 8 x’s in the numerator and 20 y’s in the denominator, so...

$$(\frac{x^2}{y^5})^4 = \frac{x^{8}}{y^{20}}$$

### What if the exponent is negative?

#### Show/Hide

If one (or more) of the exponents is negative, multiply the exponents together using the rules for multiplying negative numbers

$$(2^{-4})^{3}= 2^{-12}$$

$$(x^{4})^{-7}= x^{-28}$$

$$(y^{-3})^{-6}= y^{18}$$

In this example, you can see WHY we can multiply the exponents as negative numbers...

$$(5^{-3})^{2}$$

I'll start by simplifying the inside of the parentheses first. The (-3), tells me that I need to multiply 5 by itself 3 times in the denominator of a fraction.

$$(\frac{1}{5\times5\times5})^{2}$$

The next exponent (2) tells me that I need to multiply the stuff inside the parentheses by itself two times.

$$(\frac{1}{5\times5\times5})(\frac{1}{5\times5\times5})$$

There are six 5's multiplied in the denominator of a fraction, so...

$$(5^{-3})^{2}= 5^{-6}$$

## Free Online Practice Problems

Khan Academy - Powers of Powers

Khan Academy - Powers of Products & Quotients (Structured Practice)

Khan Academy - Powers of Products & Quotients (Integer Exponents)

Khan Academy - Powers of Products & Quotients

## Free Printable Worksheets

MathWorksheets4Kids - Power Rule of Exponents

Math Aids - Products to a Power Worksheet Generator

Math Aids - Quotients to a Power Worksheet Generator

To use the worksheet generator from Math Aids, select the options you want for your worksheet and then click the "Create It" button.


text:
Math Help - help for formula and proof of product of series

1. help for formula and proof of product of series

I can't figure out the formula, and without it I can't get much of a proof...

Hi, notice $1-\frac{1}{i^2} = \frac{(i-1)(i+1)}{i^2}$, distribute the product and use the factorial notation.


text:
# Math Help - A little concept of group actions.

1. ## A little concept of group actions.

I know that given a group G and a set S, a group action of G on the S is a function F:S ---> S satisfying some properties.
But what really confuses me arises from a question:
Suppose G is a group and A is an abelian normal subgroup of G, how does G/A operates on A by conjugation?
First, "operates" means "acts"?
If so, then given an element gA of G/A, is the action like... gA $.$a=(gA)a(gA)-1=(gA)a(g-1A)=gAaAg-1=gAg-1=A ?
See, this is where it confuses me.
Group actions by conjugation should take its element gA instead of the basic element ga, but what it follows is that I got a set A as the result of the action instead of an element of A.

I did it by its basic definition, but ended up with a set? Can anyone tell me what's wrong with my equation?

2. ## Re: A little concept of group actions.

Hi Rita,

I think your work was heading in the right direction. I will try to formalize what you were saying below.

Let $\cdot$ be the group multiplication. G/A acts on A in the following way: Let $gA\in G/A.$ Then the action $\star: G/A \times A\rightarrow A$ is defined by $gA\star a=g\cdot a\cdot g^{-1}.$ Note that the definition of $\star$ is legitimate, because $A$ is normal in $G$ so $g\cdot a\cdot g^{-1}\in A.$ The tough work is showing that $\star$ is well-defined. That $\star$ is well-defined follows from the fact that $A$ is abelian.

Does this help clear things up? Let me know if anything is unclear. Good luck!

3. ## Re: A little concept of group actions.

here is what we need to show:

that if gA = g'A, we get the same conjugate of a if we conjugate by g or g'.

recall that two cosets gA and g'A are equal if and only if g'-1g is in A.

so g'-1g = a', for some a' in A. therefore:

g = g'a'. now pick an arbitrary element a in A. then:

gag-1 = (g'a')a(g'a')-1 = g'(a'aa'-1)g'-1.

but A is abelian, so a'aa'-1 = aa'a'-1 = ae = a. hence:

gag-1 = g'(a'aa'-1)g'-1 = g'ag'-1, as desired.

EDIT:

here, our set S is A = {e,a1,a2,.....} (the subscripts aren't "really fair" A might be uncountable, so we might need a larger index set than the natural numbers).

given an element of a group G/A we need to come up with a bijective mapping A-->A "induced" by an element of G/A.

the mapping in this case is a--->gag-1, so (gA).a = gag-1.

we need A normal in G to be sure that gag-1 is in A (or else we don't have a mapping from A to A). we know this mapping is bijective, because it's an element of Inn(G), and inner automorphisms are bijective (even when restricted to a subset A of G).

we need A to be abelian, because that is what guarantees that it doesn't matter which "g" we pick, any g in the coset gA will gives the same conjugate of A.

let's pick a non-abelian group G, with a normal abelian subgroup A, and see how one of these things might work:

we'll pick G = S3, and A = {e,(1 2 3), (1 3 2)} = A3. the elements of G/A are {A, (1 2)A} (G/A has order 2).

so we are going to define (gA).a = gag-1.

for example, A.a = a for any element a of {e, (1 2 3), (1 3 2)} (because A = eA). it's pretty clear that since A is abelian, conjugating an element of A with another element of A just gives back our original element.

the interesting part is what happens when we let (1 2)A act on A:

we can conjugate by any element of (1 2)A = {(1 2), (2 3), (1 3)}. let's conjugate by (1 2) and see what happens:

(1 2)e(1 2)-1 = (1 2)e(1 2) = e.
(1 2)(1 2 3)(1 2) = (1 2)(1 3) = (1 3 2)
(1 2)(1 3 2)(1 2) = (1 2)(2 3) = (1 2 3)

note this is the same map we get if we conjugate by (2 3) instead:

(2 3)e(2 3) = e
(2 3)(1 2 3)(2 3) = (2 3)(1 2) = (1 3 2)
(2 3)(1 3 2)(2 3) = (2 3)(1 3) = (1 2 3)

if we write A = {e,a,a2} (where a = (1 2 3)) then:

A (the coset) acts on A (the set) as the identity map:

e-->e
a-->a
a2-->a2

(1 2)A (the coset) acts on A (the set) as the mapping:

e-->e
a-->a2
a2-->a

in other words, any element of A3 induces the identity map of A3, and any 2-cycle induces the inversion map on A3. (which switches a and a-1).

be careful! we can't ALWAYS let a quotient group act on a subgroup. if we have a quotient group G/N, we can get a mapping from N-->N by letting G act on N by conjugation (since N is normal), but there is no guarantee that we can get an action of G/N on N (interestingly enough we CAN get an action of a subgroup H of G on N by conjugation, though), unless N is abelian, too.

4. ## Re: A little concept of group actions.

Oh! So the key point is how the conjugation forms. It maks sense a lot, thank you soooo much.


text:
## Sea Rocks

For Decorations and areal map layout tools.

Moderator: Plugin Moderators

KINGTUT10101
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### Sea Rocks

Some rocks for your oceans. These will spawn naturally in the water and can be found in the decoration tab. There are only three different types at the moment, but I might make more in the future of I get requests to do so. These are the improved version of ocean rocks.

NisekoKilla
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Glory to Arstotzka.
it would be good if rocks can only spawn near coasts because it would be really strange that rock is in the middle of the ocean.
Psst!
I have some good animes over there!
Press me if you're brave enouch
Don't worry this is just url to the MyAnimeList :d...
My coub Click me

KINGTUT10101
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That's what I tried to do, but it didn't work like I wanted it to. I set the height to -175, but it didn't work.

Lobby
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Nice 8)

You might add some more contrast and shadows so the rocks look more plastic:

Also you might load all frames from a single image by providing the width w for each frame, height h for each frame and the amount of frames count (they will be loaded from left to right from image bmp):

Code: Select all

[
{
"type":"decoration",
"height":1,
"width":1,
"needs land":false,
"needs water":true,
"author":"Kingtut 101 and Theo (Water Texture)",
"title":"Sea Rocks",
"id":"oceanrocks.kt101",
"price":1000,
"frames":
[
{"bmp":"searocks.png","w":32,"h":15,"count":3,"handle y":3}
],
"spawn":
{
"p":0.1,
"height":-175,
},
"draw ground":true
}
]
Also I recommend to use a positive value for radius as behavior for negative ones is undefined.

Last edited by Lobby on Wed May 31, 2017 15:29, edited 2 times in total.

danekbel
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@Lobby
I tried to use this code, but...http://www.theotown.com/forum/userpix/412_s70531161412_1.jpg
But json checker said that this code is valid
http://www.theotown.com/forum/userpix/412_s70531161403_1.jpg
some text.

Lobby
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It's a problem with special characters which are a result of using tabs. I replaced them with spaces, so it should work now.

The problem is the so called non-breaking space character, which seems to be used by code boxes automatically. To solve the issue I'll adjust the plugin loader, I hope that will fix some issues in the future

KINGTUT10101
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Thanks for the help with the json.

KINGTUT10101
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Two questions to Lobby:

Can I use those textures?

What is radius? Is it how far apart they spawn from each other?

Lobby
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I tried to make it clear:

Your rockets can only spawn on tiles whose height is in between height - radius and height + radius for radius being a positive number. They will spawn more likely on tiles whose height is near height but at max with probability p.

KINGTUT10101
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So will having a high radius make them more or less common?
(Sorry it's still a bit confusing for me)

Lobby
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More common

KINGTUT10101
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Thank you

danekbel
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I used Lobby's code and I can't find them in decoration tab. Is there any wat to place them by yourself?
some text.

Lobby
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They should be listed in decorations

### Who is online

Users browsing this forum: No registered users and 2 guests


text:
## Intermediate Algebra: Connecting Concepts through Application

1. Exponential equation 2. $x = 7$
1. Exponential equation For example: $7^{x} + 3 = 0$ is an exponential equation. (The variable is the exponent.) $x^{7} + 3 =0$ is a power equation. (The variable is raised to the power of a number.) 2. Solve $5(\frac{1}{2})^{x} - \frac{2}{32} = 3(\frac{1}{2})^{x} - \frac {3}{64}$ $5(\frac{1}{2})^{x} - 3(\frac{1}{2})^{x} = \frac{2}{32} - \frac{3}{64}$ $(5-3)((\frac{1}{2})^{x}) = \frac{1}{64}$ $2((\frac{1}{2})^{x}) = \frac{1}{64}$ $(\frac{1}{2})^{x} = \frac{1}{128}$ $2^{-x} = 128^{-x}$ $2^{-x} = 2^{-7}$ $-x = -7$ $x = 7$ Check: $5(\frac{1}{2})^{7} - \frac{2}{32} \overset{?}{=} 3(\frac{1}{2})^{7} - \frac {3}{64}$ $5(\frac{1}{128}) - \frac{2}{32} \overset{?}{=} 3(\frac{1}{128}) - \frac {3}{64}$ $(\frac{5}{128}) - \frac{2}{32} \overset{?}{=} (\frac{3}{128}) - \frac {3}{64}$ $-\frac{3}{128} = -\frac{3}{128}$


text:
# Knot with genus $1$ and trivial Alexander polynomial?

I would like to know whether there exists a knot $K$ with genus $g(K)=1$ and trivial Alexander polynomial $\Delta(K) \doteq 1$.

A linked question could be: does there exist a Whitehead double with genus $1$?

Thanks to all!

-
Both the questions have affirmative answers. Moreover, it is true that the Whitehead double of every non-trivial knot is a knot with genus $1$.
In fact, if you consider the pattern $K'$ contained in the solid torus $D^2 \times S^1$, it is quite easy to construct a Seifert surface $S$ for $K'$ with genus $1$ and contained in the solid torus. Such a surface is built with two rings and one of them is twisted.
Added (Jim Conant): Here are a couple of pictures of the genus one Seifert surface for $Wh(K)$.


text:
# 2014 June(72C) Math Question 1

Choice C

To calculate the mean of n numbers, take the sum of the n numbers and divide it by n. $$\frac{370+310+380+340+310}{5}=342$$

# 2014 June(72C) Math Question 2

Choice H

A discount is a percentage that is subtracted from a number.

For example, a 10% discount of $30 is$3 (10% converted to a decimal is .10 and .10 x 30 is 3).

$12 x 25% =$3

# 2014 June(72C) Math Question 3

Choice A

$$450 = c\cdot 10^3 \\$$ $$\Rightarrow x=0.45$$

# 2014 June(72C) Math Question 4

Choice H

% increase = Increase ÷ Original Number × 100

$12 × 106% =$12.72

# 2014 June(72C) Math Question 5

Choice E

In a Geometric Sequence each term is found by multiplying the previous term by a constant.

The constant for the sequence in question is -3. Therefore the sequence is:

1, -3, 9, -27, 81, -243, 729, ...

# 2014 June(72C) Math Question 6

Choice K

A square root of a number a is a number y such that y^2 = a; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is a. $$\sqrt{a} = 36 \\ ⇒a = 36^2 = 1296$$

Choice C

# 2014 June(72C) Math Question 55

Choice E

The square root must be …

# 2014 June(72C) Math Question 56

Choice F

It can be observed that …

# 2014 June(72C) Math Question 57

Choice E

The absolute value of a …

# 2014 June(72C) Math Question 58

Choice J

The smallest angle must be …

# 2014 June(72C) Math Question 59

Choice D

To reduce a fraction to …

# 2014 June(72C) Math Question 60

Choice J

As we can see from …