text:
• Today, 05:18
Oh yes, that's a typo... I meant $k$... (Nerd)
3 replies | 61 view(s)
• Today, 03:50
You mean the derivative in respect to $t$ ? (Thinking)
25 replies | 423 view(s)
• Today, 00:02
You wound up with $x^2\sin(x)$ on the RHS before integrating, but it should be $x\sin(x)$. :)
1 replies | 26 view(s)
• Yesterday, 19:23
When you multiply through by $\mu(x)$, you should have: \frac{d}{dx}(\sin(x)y)=2 And then integrate: \sin(x)y=2x+c_1 ...
2 replies | 33 view(s)
• June 19th, 2018, 19:21
Towards the end, when you divide through by $x$, you want: y(x)=\frac{e^x}{x}+\frac{c}{x} You mistakenly divided the constant by $e^x$.
1 replies | 39 view(s)
• June 19th, 2018, 15:31
Yes, but you used it on the original ODE, not the one in standard linear form. :)
4 replies | 87 view(s)
• June 18th, 2018, 07:29
Wilmer replied to a thread Present Value Maths in Other Topics
Book is correct. Are you using a calculator? You're probably entering the data wrongly...we can't tell... Try it this way: u = (1.02/1.10)^10 v =...
2 replies | 63 view(s)
• June 18th, 2018, 06:19
Thanks steenis ... No worries at all ... Thanks for all your help ... Peter
10 replies | 214 view(s)
• June 18th, 2018, 06:04
Thanks Steenis ... That proof seems really clear ... Will work through it again shortly... Peter
4 replies | 80 view(s)
• June 18th, 2018, 05:18
Sorry Steenis ... I don't understand you ... Can you give me a hint as to what is wrong ...? Peter
10 replies | 214 view(s)
• June 18th, 2018, 04:51
Thanks steenis ... most helpful ... Can see that the short exact sequence $0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)}... 10 replies | 214 view(s) • June 18th, 2018, 01:22 ======================================================================== Since I could not see any specific errors, I have completed the proof... 4 replies | 80 view(s) • June 17th, 2018, 14:46 I have thought of the following algorithm: We put an antenna$k$meters east of the westernmost house. We continue to the east, by placing an... 3 replies | 61 view(s) • June 17th, 2018, 12:39 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations I can't imagine trying to use the internet on a telephone. I'm sorry scrolling on a telephone is such a chore...they should fix that. 8 replies | 113 view(s) • June 17th, 2018, 11:00 Hello!!! (Wave) We consider a long country road with$n$houses placed along of it (we think of the road as a big line segment). We want to put... 3 replies | 61 view(s) • June 17th, 2018, 05:04 Peter started a thread Deveno ... in Chat Room Deveno is much missed ... especially by those who frequent the Linear and Abstract Algebra Forum ... Deveno's pedagogical abilities were as... 0 replies | 53 view(s) • June 17th, 2018, 04:40 I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ... I am currently studying Chapter 10: Introduction to Module Theory ...... 4 replies | 80 view(s) • June 17th, 2018, 00:59 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations Are you using Tapatalk by any chance? I have code in place to let me know when posts have been edited, so I don't miss added content (it's better to... 8 replies | 113 view(s) • June 17th, 2018, 00:52 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations When you multiply by$\mu(x)$, you get: \sec(x)y'+\tan(x)\sex(x)y=\sec(x)\sin(2x) This can be written as: ... 8 replies | 113 view(s) • June 16th, 2018, 23:10 Thanks Steenis ... You have shown that$R^{(n)} / N \cong M$where$N = \text{ Ker } f$... ... ... ... ... (1) ... and we have by... 10 replies | 214 view(s) • June 16th, 2018, 19:51 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations -\ln(\cos(x))=\ln\left((\cos(x))^{-1}\right)=\ln(\sec(x)) 8 replies | 113 view(s) • June 16th, 2018, 19:38 MarkFL replied to a thread [SOLVED] 2.2.3 de with tan x in Differential Equations \mu(x)=\exp\left(\int \tan(x)\,dx\right)=e^{\Large\ln(\sec(x))}=\sec(x) 8 replies | 113 view(s) • June 16th, 2018, 09:40 https://percentagecalculator.net/ "percent" means "per hundred" Simple example: 200 increases to 218: that's "9 per hundred", right?... 4 replies | 111 view(s) • June 16th, 2018, 00:29 300+300\cdot\frac{1666}{100}=300(1+16.66)=300\cdot17.66=5298 Here, we have taken 300, and added 1666% of 300 to it. However if we multiply 300 by... 4 replies | 111 view(s) • June 16th, 2018, 00:03 Thanks steenis ... but not sure if I follow .. ... but will try ... as follows ... We have an epimorphism$f:R^{(n)} \longrightarrow M$... 10 replies | 214 view(s) More Activity ### 73 Visitor Messages 1. Yes, you are correct. The first post in a thread may be edited within 2 hours, and all other posts within 24 hours. However, if you run over the 24 hour limit, I will lift the limit for you temporarily since you are adding such great content. 2. Hey Balarka, according the the settings in the ACP it is 5 minutes. 3. Hey Balarka, Daniel is upgrading the forum software to a completely new platform. It should be back up soon. 4. I agree that fun shouldn't introduce wrong concepts. PS: Yes, I'll add one when I start the next post. 5. Ok, then you are looking at rigor but as I said these tutorials will be for fun. I want a high school student to understand it. It would be great to have your comments when I start posting bout that, though. I know I can learn from you as I am haven't read that much about these concepts. 6. What is your definition of Regularization ? 7. We can extend the zeta function to analytic function in the whole complex plane except at 1. By definition we have $$\zeta(s) =\sum \frac{1}{n^s}$$ So zeta on the left is analytic for all$s \neq 1$but the sum on the right does only converge for$Re(s) >1\$.
So the analytic continuation of zeta has enabled us to give values to the divergent sum on the right.
8. Hey Balarka , the idea of regularization is giving a finite value for divergent series or integrals. Consider the following
$$\zeta(-2)= 0$$
$$1+4+9+\cdots = 0$$
9. Okay, thanks. I think the former one is just what I wanted.
10. Hey Balarka, I do not know of a closed for that sum. But there are some identities related to this problem:
\begin{align*} \sum_{k=1}^p \frac{H_k^{(2)}}{k}+\sum_{k=1}^p \frac{H_k}{k^2} &= H_p^{(3)}+H_p^{(2)}H_p \\ \sum_{k=1}^p \frac{H_k^{(2)}}{k}+\sum_{k=1}^p \frac{(H_k)^2}{k} &= \frac{(H_p)^3+3H_p H_p^{(2)}+2H_p^{(3)}}{3} \end{align*}
To prove these, we can use induction.
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#### Basic Information

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I know a thing or two about topology. Find number theory interesting but don't know much about it.
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Some of my notes on number theoretic topics are on a few primality tests and on Hardy & Littlewood's result (incomplete). The other articles are on quintics : about a brief description of Kiepert algorithm and a short introduction on quintic-solving algorithms, respectively. I have also written up an introductory thread on Riemann Hypothesis

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text:
# Distributions Provided by the **mniw** Package In mniw: The Matrix-Normal Inverse-Wishart Distribution

## Wishart Distribution

The Wishart distribution on a random positive-definite matrix $\XX_{q\times q}$ is is denoted $\XX \sim \wish(\PPs, \nu)$, and defined as $\XX = (\LL \ZZ)(\LL \ZZ)'$, where:

• $\PPs_{q\times q} = \LL\LL'$ is the positive-definite matrix scale parameter,
• $\nu > q$ is the shape parameter,
• $\ZZ_{q\times q}$ is a random lower-triangular matrix with elements

[ Z_{ij} \begin{cases} \iid \N(0,1) & i < j \ \ind \chi^2_{(\nu-i+1)} & i = j \ = 0 & i > j. \end{cases} ]

The log-density of the Wishart distribution is [ \log p(\XX \mid \PPs, \nu) = -\tfrac{1}{2} \left[\mathrm{tr}(\PPs^{-1} \XX) + (q+1-\nu)\log |\XX| + \nu \log |\PPs| + \nu q \log(2) + 2 \log \Gamma_q(\tfrac \nu 2)\right], ] where $\Gamma_n(x)$ is the multivariate Gamma function defined as [ \Gamma_n(x) = \pi^{n(n-1)/4} \prod_{j=1}^n \Gamma\big(x + \tfrac 1 2 (1-j)\big). ]

### Inverse-Wishart Distribution

The Inverse-Wishart distribution $\XX \sim \iwish(\PPs, \nu)$ is defined as $\XX^{-1} \sim \wish(\PPs^{-1}, \nu)$. Its log-density is given by [ \log p(\XX \mid \PPs, \nu) = -\tfrac 1 2 \left[\mathrm{tr}(\PPs \XX^{-1}) + (\nu+q+1) \log |\XX| - \nu \log |\PPs| + \nu q \log(2) + 2 \log \Gamma_q(\tfrac \nu 2)\right]. ]

### Properties

If $\XX_{q\times q} \sim \wish(\PPs,\nu)$, the for a nonzero vector $\aa \in \mathbb R^q$ we have [ \frac{\aa'\XX\aa}{\aa'\PPs\aa} \sim \chi^2_{(\nu)}. ]

## Matrix-Normal Distribution

The Matrix-Normal distribution on a random matrix $\XX_{p \times q}$ is denoted $\XX \sim \MN(\LLa, \SSi_R, \SSi_C)$, and defined as $\XX = \LL\ZZ \UU + \LLa$, where:

• $\LLa_{p \times q}$ is the mean matrix parameter,
• ${\SSi_R}_{p \times p} = \LL\LL'$ is the row-variance matrix parameter,
• ${\SSi_C}_{q \times q} = \UU'\UU$ is the column-variance matrix parameter,
• $\ZZ_{q\times q}$ is a random matrix with $Z_{ij} \iid \N(0,1)$.

The log-density of the Matrix-Normal distribution is [ \log p(\XX \mid \LLa, \SSi_R, \SSi_C) = -\tfrac 1 2 \left[\mathrm{tr}\big(\SSi_C^{-1}(\XX-\LLa)'\SSi_R^{-1}(\XX-\LLa)\big) + \nu q \log(2\pi) + \nu \log |\SSi_C| + q \log |\SSi_R|\right]. ]

### Properties

If $\XX_{p \times q} \sim \MN(\LLa, \SSi_R, \SSi_C)$, then for nonzero vectors $\aa \in \mathbb R^p$ and $\bb \in \mathbb R^q$ we have [ \aa' \XX \bb \sim \N(\aa' \LLa \bb, \aa'\SSi_R\aa \cdot \bb'\SSi_C\bb). ]

## Matrix-Normal Inverse-Wishart Distribution

The Matrix-Normal Inverse-Wishart Distribution on a random matrix $\XX_{p \times q}$ and random positive-definite matrix $\VV_{q\times q}$ is denoted $(\XX,\VV) \sim \mniw(\LLa, \SSi, \PPs, \nu)$, and defined as [ \begin{aligned} \XX \mid \VV & \sim \MN(\LLa, \SSi, \VV) \ \VV & \sim \iwish(\PPs, \nu). \end{aligned} ]

### Properties

The MNIX distribution is conjugate prior for the multivariable response regression model [ \YY_{n \times q} \sim \MN(\XX_{n\times p} \bbe_{p \times q}, \VV, \SSi). ] That is, if $(\bbe, \SSi) \sim \mniw(\LLa, \OOm^{-1}, \PPs, \nu)$, then [ \bbe, \SSi \mid \YY \sim \mniw(\hat \LLa, \hat \OOm^{-1}, \hat \PPs, \hat \nu), ] where [ \begin{aligned} \hat \OOm & = \XX'\VV^{-1}\XX + \OOm & \hat \PPs & = \PPs + \YY'\VV^{-1}\YY + \LLa'\OOm\LLa - \hat \LLa'\hat \OOm \hat \LLa \ \hat \LLa & = \hat \OOm^{-1}(\XX'\VV^{-1}\YY + \OOm\LLa) & \hat \nu & = \nu + n. \end{aligned} ]

## Matrix-t Distribution

The Matrix-$t$ distribution on a random matrix $\XX_{p \times q}$ is denoted $\XX \sim \MT(\LLa, \SSi_R, \SSi_C, \nu)$, and defined as the marginal distribution of $\XX$ for $(\XX, \VV) \sim \mniw(\LLa, \SSi_R, \SSi_C, \nu)$. Its log-density is given by [ \begin{aligned} \log p(\XX \mid \LLa, \SSi_R, \SSi_C, \nu) & = -\tfrac 1 2 \Big[(\nu+p+q-1)\log | I + \SSi_R^{-1}(\XX-\LLa)\SSi_C^{-1}(\XX-\LLa)'| \ & \phantom{= -\tfrac 1 2 \Big[} + q \log |\SSi_R| + p \log |\SSi_C| \ & \phantom{= -\tfrac 1 2 \Big[} + pq \log(\pi) - \log \Gamma_q(\tfrac{\nu+p+q-1}{2}) + \log \Gamma_q(\tfrac{\nu+q-1}{2})\Big]. \end{aligned} ]

### Properties

If $\XX_{p\times q} \sim \MT(\LLa, \SSi_R, \SSi_C, \nu)$, then for nonzero vectors $\aa \in \mathbb R^p$ and $\bb \in \mathbb R^q$ we have [ \frac{\aa'\XX\bb - \mu}{\sigma} \sim t_{(\nu -q + 1)}, ] where [ \mu = \aa'\LLa\bb, \qquad \sigma^2 = \frac{\aa'\SSi_R\aa \cdot \bb'\SSi_C\bb}{\nu - q + 1}. ]

## Random-Effects Normal Distribution

Consider the multivariate normal distribution on $q$-dimensional vectors $\xx$ and $\mmu$ given by [ \begin{aligned} \xx \mid \mmu & \sim \N(\mmu, \VV) \ \mmu & \sim \N(\lla, \SSi). \end{aligned} ] The random-effects normal distribution is defined as the posterior distribution $\mmu \sim p(\mmu \mid \xx)$, which is given by [ \mmu \mid \xx \sim \N\big(\GG(\xx-\lla) + \lla, \GG\VV\big), \qquad \GG = \SSi(\VV + \SSi)^{-1}. ] The notation for this distribution is $\mmu \sim \re(\xx, \VV, \lla, \SSi)$.

## Hierarchical Normal-Normal Model

The hierarchical Normal-Normal model is defined as [ \begin{aligned} \yy_i \mid \mmu_i, \bbe, \SSi & \ind \N(\mmu_i, \VV_i) \ \mmu_i \mid \bbe, \SSi & \iid \N(\xx_i'\bbe, \SSi) \ (\bbe, \SSi) & \sim \mniw(\LLa, \Omega^{-1}, \PPs, \nu), \end{aligned} ] where:

• ${\yy_i}_{q\times 1}$ is the response vector for subject $i$,
• ${\mmu_i}_{q\times 1}$ is the random effect for subject $i$,
• ${\VV_i}_{q\times q}$ is the error variance for subject $i$,
• ${\xx_i}_{p\times 1}$ is the covariate vector for subject $i$,
• $\bbe_{p \times q}$ is the random-effects coefficient matrix,
• $\SSi_{q \times q}$ is the random-effects error variance.

Let $\YY_{n\times q} = (\rv \yy n)$, $\XX_{n\times p} = (\rv \xx n)$, and $\TTh_{n \times q} = (\rv \mmu n)$. If interest lies in the posterior distribution $p(\TTh, \bbe, \SSi \mid \YY, \XX)$, then a Gibbs sampler can be used to cycle through the following conditional distributions: [ \begin{aligned} \mmu_i \mid \bbe, \SSi, \YY, \XX & \ind \re(\yy_i, \VV_i, \xx_i'\bbe, \SSi) \ \bbe, \SSi \mid \TTh, \YY, \XX & \sim \mniw(\hat \LLa, \hat \OOm^{-1}, \hat \PPs, \hat \nu), \end{aligned} ] where $\hat \LLa$, $\hat \OOm$, $\hat \PPs$, and $\hat \nu$ are obtained from the MNIW conjugate posterior formula with $\YY \gets \TTh$.

## Try the mniw package in your browser

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text:
College Physics (7th Edition)

(a) There is complete transfer of energy for $m_1=m_2$.
Kinetic energy is given by $K=\frac{p^2}{2m}$. Initial kinetic energy for this elastic collision, $K_i=\frac{p^2}{2m_1}$ For a $m_1$ hitting a stationary $m_2$ and complete transfer of energy, Final kinetic energy is $K_f=\frac{p^2}{2m_2}$, since momentum is conserved. From the second condition for elastic collision - \begin{align*} K_i&=K_f\\ or,\, \frac{p^2}{2m_1}&=\frac{p^2}{2m_2}\\ or,\, m_1 &=m_2 \end{align*}


text:
# Grouping sequential values in R

I have a list of numbers I need to group by similarity (differences being 1 between each). For example, in a list of [198, 202, 207, 218, 219, 220], 190 would be put into a list, 202 would be put into another list, 207, into a list, and 218 219 220 into another list.

-
Do you just want to put all numbers that are 1 unit apart into the same cluster? Are you just looking for code / how that can be done in R? –  gung Jul 10 at 20:45

Although some clustering algorithms will do this, there's a simpler way: sort the input, create a binary vector indicating whether each neighboring pair is dissimilar or not, prefix that with a 1 for the initial value of the input, and compute its cumulative sum. This creates a new group identifier each time a dissimilar pair is seen.

x <- c(198, 202, 207, 218, 219, 220, 220)
y <- sort(x)
g <- cumsum(c(1, abs(y[-length(y)] - y[-1]) > 1))
by(y, g, identity)

The last line isn't really needed in R: g already holds the essential information. I stuck it there only because a list was requested. Here's what g looks like in the example:

> rbind(g, y)
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
g    1    2    3    4    4    4    4
y  198  202  207  218  219  220  220

The groups are 1 --> (198), 2 --> (202), 3 --> (207), 4 --> (218, 219, 220, 220).

If you don't want duplicates to appear, clean the input first using unique.

On my system this code groups an input of four million integers in one second. Most of that is spent in the sorting step.

The same algorithm is readily applied on other platforms such as Stata, Matlab, Mathematica, Python, etc.

-

This can be accomplished with hierarchical clustering using the hclust function in R. Here's some example code. The cutree function allows you to specify the distance (height) at which the tree should be cut and returns the cluster for each data point.

numbers_to_cluster <- c(0:10, 20:30, 40:50)
distances_between_numbers <- dist(x = numbers_to_cluster, method = "euclidean")
hierarchical_clustering <- hclust(d = distances_between_numbers, method = "single")
clusters <- cutree(tree = hierarchical_clustering, h = 1)

-


text:
### Reduction Modulo $2^{448}-2^{224}-1$

Kaushik Nath and Palash Sarkar

##### Abstract

An elliptic curve known as Curve448 defined over the finite field $\mathbb{F}_p$, where $p=2^{448}-2^{224}-1$, has been proposed as part of the Transport Layer Security (TLS) protocol, version 1.3. Elements of $\mathbb{F}_p$ can be represented using 7 limbs where each limb is a 64-bit quantity. This paper describes efficient algorithms for reduction modulo $p$ that are required for performing field arithmetic in $\mathbb{F}_p$ using 7-limb representation. A key feature of our work is that we provide the relevant proofs of correctness of the algorithms. We also report efficient 64-bit assembly implementations for key generation and shared secret computation phases of the Diffie-Hellman key agreement protocol on Curve448. Timings results on the Haswell and Skylake processors demonstrate that the new 64-bit implementations for computing the shared secret are faster than the previously best known 64-bit implementations.

Note: Minor revision

Available format(s)
Category
Public-key cryptography
Publication info
Preprint. MINOR revision.
Keywords
Curve448Goldilocks primemodulo reductionelliptic curve cryptographyDiffie-Hellman key agreement.
Contact author(s)
kaushikn_r @ isical ac in
palash @ isical ac in
History
2022-01-06: last of 3 revisions
See all versions
Short URL
https://ia.cr/2019/1304

CC BY

BibTeX

@misc{cryptoeprint:2019/1304,
author = {Kaushik Nath and Palash Sarkar},
title = {Reduction Modulo $2^{448}-2^{224}-1$},
howpublished = {Cryptology ePrint Archive, Paper 2019/1304},
year = {2019},
note = {\url{https://eprint.iacr.org/2019/1304}},
url = {https://eprint.iacr.org/2019/1304}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Tag Info

21

None of the interesting equations in physics can be derived from simpler principles, because if they could they wouldn't give any new information. That is, those simpler principles would already fully describe the system. Any new equation, whether it's the Navier-Stokes equations, Einstein's equations, the Schrodinger equation, or whatever, must be ...

18

They are derivable from classical mechanics using either the continuum or molecular points of view. Starting with a continuum view, one applies conservation of mass, momentum, and energy to a control volume and the result is the Navier Stokes equations. The Navier Stokes equations, in the usual form, apply to Newtonian fluids, that is fluids whose stress ...

17

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? ...

12

Frank White's Viscous Fluid Flow book contains a good list of these "exact" solutions. I am not sure if it is complete though. I've provided links to a few of the solutions. Steady flow between a fixed and moving plate Axially moving concentric cylinders Flow between rotating concentric cylinders Hagan-Poiseuille flow Combined Couette-Poiseuille flow ...

12

I once asked Putterman after a similar colloquium what he meant by this statement, and his answer was "long time tails". Long time tails are fractional powers that appear in the long time behavior of correlation functions, see, for example, here and here. These fractional powers are seen in molecular dynamics (they are more difficult to see experimentally), ...

11

Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...

11

There is a simple general argument for why you get small-scale motion from large-scale motion in any nonlinear nonintegrable continuous mechanical system, whether it is fluids, or electromagnetic waves interacting with charged plasmas, or surface waves on water, or anything nonlinear at all. This argument must break down for those special cases where the ...

10

Actually, there are two different viscosity coefficients. You can see this from the stress tensor $$\sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k}$$ which has the two ...

10

This is difficult to answer, because the answer depends on whether the answer is positive or negative. Although most people, including myself, expect a positive answer, a negative answer is actually possible, unlike say, for the question of P!=NP, or the well-definedness/mass-gap of gauge theory where we are 100% sure that we know the answer already (in the ...

10

Turbulence is indeed an unsolved problem both in physics and mathematics. Whether it is the "greatest" might be argued but for lack of good metrics probably for a long time. Why it is an unsolved problem from a mathematical point of view read Terry Tao (Fields medal) here : ...

7

The equation is correct--- the (laminar) flow at small Reynolds number is given by making the flow be along the pipe, and substituting into the Navier Stokes equations, which reduces to your thing. The one issue is the sign--- $\Delta P$ is negative if you mean the flow is going to be in the positive z direction. I will absorb the constants and consider the ...

5

Strictly speaking, turbulence doesn't exist in two dimensions. The energy cascade required for turbulence to develop (transfer energy from large scales to small scales) is due to the (incompressible for illustration) vorticity equation: $\frac{D\vec{\omega}}{Dt} = \left(\vec{\omega}\cdot\nabla\right)\vec{v} + \nu\nabla^2\vec{\omega}$ specifically the ...

5

I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items: Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ... Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum ...

5

Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...

5

For a single-component fluid, the conservation of mass follows $$\left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right)$$ In terms of a ...

4

It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ...

4

My answer will focus just on the mathematical parts pertaining to partial differential equations. Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables. For example the heat equation (where $\boldsymbol{x}$ is the position vector in 1, 2 or 3D, doesn't matter) $$\partial_t ... 4 Solutions of the form$$ cos(x_i)e^{-x_j}$$are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with i,j\in\{1,2,3\}). x_j is then the wall normal direction. This is actually quite well ... 4 I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second. It's easier to explain if we temporarily imagine \phi represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of \phi due to the ... 4 I don't think that such a computation of a theoretical limit of accuracy is possible. There are several sources of uncertainty in weather models: initial and boundary data, parameterizations, numerical instability, rounding and approximation errors of the numerical scheme employed to solve the Navier-Stokes equations for the atmosphere. The term ... 3 I am not sure how useful this "back of the envelope" calculation of reliability of Numerical Weather Prediction is going to be. Several of the assumptions in the question are not correct, and there are other factors to consider. Here are some correcting points: The Weather is 3 dimensional and resides on the surface of the planet up to a height of at ... 3 Let me here just derive the equation (6.11) that follows the sentence, you mention. The Navier-Stokes equation (6.6a) reads$$\partial_t v_i + v_j\partial_j v_i = -\partial_i p + f_i + \nu~\partial_j\partial_j v_i.$$The incompressibility condition (6.6b) reads$$\partial_jv_j=0. $$Hence we have in the unprimed and the primed point that$$\partial_t ...

3

I'm far from expert and my advice is basically to read Terry Tao's piece linked in John Rennie's comment. (Read Terry Tao is always good advice.) But I would like to make a couple of points. Singularities are not impossible or unphysical Mainly, I wanted to clear up the idea that, if we find smooth solutions that became singular in finite time that would ...

3

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...

3

You almost give the answer in your question: With increasing flow rates, the inertial forces become larger than the viscous forces We can state it more precisely: The inertial forces scale quadratically with the flow speed $U$: $\rho \mathbf u\nabla \mathbf u\sim \rho U^2/L$, while the viscous forces scale linearly: $\mu \nabla^2\mathbf u \sim \mu ... 3 For Newtonian fluids (such as water and air), the viscous stress tensor,$T_{ij}$, is proportional to the rate of deformation tensor,$D_{ij}$: $$D_{ij} = \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)$$ $$T_{ij} = \lambda\Delta\delta_{ij} + 2\mu D_{ij}$$ where$\Delta \equiv D_{11} + D_{22} + D_{33}$. The ... 3 Most importantly, the Navier-Stokes equations are based on a continuum assumption. This means that you should be able to view the fluid as having properties like density and velocity at infinitely small points. If you look at e.g. liquid flows in nanochannels or gas flows in microchannels you could be in a regime where this assumption breaks down. As far as ... 3 IANAFD but I'll stick my neck out and say this: resolving the Clay problem one way or another won't cause people doing CFD to lose any more sleep than they already do. First of all, Jean Leray proved the existence of weak solutions to Navier Stokes in$R^3$way back in the 1930s, and that is pretty much what matters for the task of getting numerical ... 3 There is a normalized form, though it's properly called the dimensionless Euler equations. The way to do it is define: scale time$t_0$scale density$\rho_0$scale length$L_0\$ and then derive the scales from these: $$v_0 = \frac{L_0}{t_0},\quad p_0=\rho_0v_0^2$$ NB: it is possible to use other combinations, but I find that these are often the ...

3

There are known solutions to the Navier-Stokes equations. A simple example would be laminar shear-driven flow between two moving plates. Just as in the case of Einstein's equations, the known solutions regard simple situations with particular boundary conditions; a general solution that covers all possible cases is not known in either case. One should not ...

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


text:
# Is there a policy about font size?

I notice that recent posts are using \large in LaTeX when it is not really needed or similarly display math in the title of the post. The impression is that it is just pure visual competition for attention grabs. Is there a policy about font size? I fear that if there is none, soon posts we be typed using \Huge...

There is general agreement that titles should not include displaystyle math so as to conserve vertical space on the front page (see here). I don't think it is a stretch to expand this to include other forms deliberate enlarging of titles, or otherwise making titles conspicuous in lists of questions (e.g., using colours as in "$\color{red}{\text{NEED HELP NOW!}}$"). I would feel free to edit these elements out of question titles.

• I don't think it is a stretch...: Was that intentional?! ;-) – cardinal Dec 8 '13 at 15:48

The query Displayed math in title returns 48 posts. Some of them may have been brought into shape after the database dump -- although it's quite fresh right now, made only about an hour ago. Some are pretty old. In any case, the query may help editors to keep this trend under control as far as displays are concerned.

LaTeX abuse in the body of a post is not as easy to identify with SEDE query.

• I just check many of these. I only found one or two that were actually problematic. The rest just had two expressions in with single dollar signs on each end put next to each other, so there was a double dollar sign in the middle, which is what the search picked up on. – Potato Dec 13 '13 at 0:33

In the body of a post, this may well be caused by a user not knowing how to adjust a web browser correctly, or a user using a somewhat broken browser. In any case, inappropriate scaling of large amounts of mathematics seems to be a problem that should be edited away whenever possible.

• For the final answer, it should be $0$ k. – Felix Marin Dec 13 '13 at 6:26
• @FelixMarin, I don't understand what you're saying at all. If it is a joke, I don't get it. – dfeuer Dec 13 '13 at 7:10
• It's not a joke. Sometimes, after a long answer it's fine to use color and a little bit font enlargement to emphasize the 'answer main result'. Just the 'final one', not all the answer. – Felix Marin Dec 13 '13 at 19:36
• @FelixMarin, "$0$ k" is not normally used to mean "OK", hence my confusion. – dfeuer Dec 13 '13 at 19:43
• Historically (in some version en.wikipedia.org/wiki/List_of_proposed_etymologies_of_OK ) OK was 0 k. as it means 'zero killed' – Felix Marin Dec 13 '13 at 19:56


text:
# How should we handle home modification questions?

We have received some questions about home modifications. The answer typically boils down to "It is probably more complicated that you imagine. Hire an engineer."

Each question is similar, but unique.

How should we handle these questions?

2. Redirect them all to one typical question (i.e. a target question for "close as duplicate")?
3. Others?

Some sample questions:

• Additional options - We can update the "on-topic" page in the help center to explicitly state those questions are off-topic. We can also create a custom off-topic close reason. Custom close reasons are generally reserved for high volume issues though. – user16 Dec 5 '15 at 15:48
• My comment is "I think if this question in the form "What are the necessary structural engineering parameters that need to be considering when removing the supporting post?" IMHO This type of question is valuable for this site" This type of question can help poster get educated on engineering perspective. – Mahendra Gunawardena Dec 6 '15 at 2:07
• This meta discussion is certainly useful in discussing this from a liability side, as this falls under the same liability issues as anything else. – Trevor Archibald Dec 6 '15 at 4:44

At a minimum, I think we need to close off the questions that are asking about structural modifications. Personally, I'm neutral towards whether we close them after updating the on-topic help page, as a custom close reason, or as duplicates of a canonical "You need to hire an Engineer" type question. For that matter, I wouldn't object to closing them purely as "too broad" either.

If I had to pick one, I would lean towards updating the on-topic page to indicate that questions requiring a licensed Engineer to answer are simply too broad for the scope of the site and therefore off-topic for the site.

Regarding answering the questions, even if it's to the effect of "You really, really, really need to hire a licensed Engineer." I'd discourage answering those types of questions as the answers don't provide lasting value for the site.

By way of example, let's take the answer from one of the example questions you listed. It's a well written answer, and currently has 8 up votes. But the crux of the answer boils down to this:

NO! GOD NO! WHAT ARE YOU DOING!

I don't think we want answers like that to become commonplace on the site.

A cleaner way of handling that would have been to leave a comment stating "There are far too many variables at work for a meaningful answer to be provided. Please consult with a licensed Engineer in your area to have this evaluated." And if you feel compelled to comment upon the sanity of the project at the same time, then, well, that's what comments are sometimes used for.

It's probably worth pointing out that questions that are closed are likely going to be deleted. I think our community is better off focusing on answering questions that are more likely to be kept around and add value to our site.

It's probably worth noting that 2nd and 3rd examples provided are a little different than the first. The roof structure and large aquarium questions were more about evaluating what was already there as opposed to modifying the integrity of the structure.

And that also brings up a related comment from Mahendra:

I think if this question in the form "What are the necessary structural engineering parameters that need to be considering when removing the supporting post?" IMHO This type of question is valuable for this site

We need to walk a careful path here. On the one hand, the suggestion represents an Atwood Transform as mentioned in the article Q&A is hard. But we need to make sure that the transformed question remains reasonably scoped. With the remove support posts question, I'm not convinced that an Atwood Transform would work on that question. I think the resulting question would remain too broad and not a good fit for the site.

• I'm not totally sold on the dupe-in-lieu-of-off-topic approach but I think it's worth a try. Let's see if we can make this an exceptionally high-quality canonical question to use for closing questions that we can't (or shouldn't) answer from behind a keyboard. (That's not all of the potential questions we're talking about here but it's a good chunk - maybe a majority.) – Air Dec 8 '15 at 17:58
• n.b. the [home-improvement] tag is at this point an uncomfortable experiment that may ultimately need to die. Let's see how it goes. – Air Dec 8 '15 at 18:01
• Adding onto the comment, if OP seeking engineering solutions such as "I would like to remove the supporting post, what are the alternate methods to distribute the load" or "How can evaluate the force acting on these posts? or "What do I need to consider if I need reinforce the supporting beam If I remove the supporting post?" These type of questions will help OP be educate when seeking a consultant engineer. – Mahendra Gunawardena Dec 9 '15 at 17:38
• @air (and others) I would imagine that this topic/concern must show up on diy.SE frequently. Should we research how they handle it over there? – hazzey Dec 9 '15 at 21:23
• @hazzey I did a cursory search of their Meta site but didn't find any sort of fully-formed policy. I also looked at existing questions involving the modification of existing structural elements (e.g., A, B, C) and saw that the M.O. over there seems to be simply to answer as normal with caveats as necessary. I think some migrations will be possible and others we'll simply need to turn down in one way or another. – Air Dec 9 '15 at 22:51

I have to comment on the approach taken by Air - I think it is wildly successful and brings out a unique approach to this site.

Having been on for a while, I've seen quite a few questions shut down for "not enough information" or "you need to hire a professional". While this is a valid response to questions, I believe from what I've seen that most people who ask us questions simply don't really understand the information that is needed.

From that perspective, I like the Atwood Transform proposed - especially if we can be proactive like Air and make several good canonical questions. I'd like to add a few more. With enough views, these will pop up in searches and likely help not only with traffic, but help a poster from asking a question only to get shut down - instead they will see that "yeah I shouldn't even ask this now - I need to follow the advice of this Q&A and seek help / get more information".

That way, instead of shutting down a question with "you need more information", we can shut them down with "you need more information, and goto this question to find out what information you need to ask this question". To put it more succinctly, let's be the opposite of this image and help them understand:

• engineering.stackexchange.com/questions/6493/… - This is actually happening to me, so I figured I'd throw it in as well. Since "knocking out a wall" isn't the only kind of do-it-yourself engineering in the modern home. – Mark Dec 10 '15 at 17:02
• It's too early to say whether the approach of writing a canonical question and then closing its permutations as duplicates rather than as off-topic will be successful or not. I'm a bit nonplussed by the answers on the trial balloon question so far. – Air Dec 14 '15 at 16:28


text:
# NCERT solutions for Exercise 9.3 Class 8 Maths Chapter 9 Algebraic Expressions And Identities

1.Carry out the multiplication of the expressions in each of the following pairs:
(i) $$4p, q + r$$

(ii) $$ab, a – b$$

(iii) $$a + b, 7a^2b^2$$

(iv) $$a^2 – 9, 4a$$

(v) $$pq + qr + rp, 0$$

Multiplication of the expressions are as follows:
(i) $$4p \times (q + r) = (4p \times q) + (4p \times r) = 4pq + 4pr$$

(ii) $$ab, a – b = ab \times (a – b) = (ab \times a) – (ab \times b) = a^2b – ab^2$$

(iii) $$(a + b) \times 7a^2b^2 = (a \times 7a^2b^2) + (b \times 7a^2b^2) = 7a^3b^2 + 7a^2b^3$$

(iv) $$(a2 – 9) \times 4a = (a^2 \times 4a) – (9 \times 4a) = 4a^3 – 36a$$

(v) $$(pq + qr + rp) \times 0 = 0$$

2. Complete the table.

Products of the given expressions in the table are done as follows:
.(i) $$a \times (b + c + d) = (a \times b) + (a \times c) + (a \times d) = ab + ac + ad$$

(ii) $$(x + y – 5) (5xy) = (x \times 5xy) + (y \times 5xy) – (5 \times 5xy) = 5x^2y + 5xy^2 – 25xy$$

(iii) $$p \times (6p^2 – 7p + 5) = (p \times 6p^2) – (p \times 7p) + (p \times 5) = 6p^3 – 7p^2 + 5p$$

(iv) $$4p^2q^2 \times (p^2 – q^2) = 4p^2q^2 \times p^2 – 4p^2q^2 \times q^2 = 4p^4q^2 – 4p^2q^4$$

(v) $$(a + b + c) \times (abc) = (a \times abc) + (b \times abc) + (c \times abc) = a^2bc + ab^2c + abc^2$$

Therefore, the completed table is as follows:

3.Find the products
(i)$$(a^2)\times(2a^{22})\times(4a^{26})$$

(ii)$$(\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)$$

(iii)$$(-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)$$

(iv)$$x\times x^2\times x^3\times x^4$$

(i)$$(a^2)\times(2a^{22})\times(4a^{26})$$

$$=(1\times 2\times 4)\times a^{2+22+26}=8a^{50}$$

(ii)$$(\frac{2}3xy)\times(-\frac{9}{10}x^2y^2)$$

$$=(\frac{2}3\times (-\frac{9}{10}))\times x^{(1+2})y^{(1+2)}$$

(iii)$$(-\frac{10}3pq^3)\times(\frac{6}{5}p^3q)$$

$$=(-\frac{10}3\times\frac{6}{5})\times p^{(1+3)}q^{(1+3)}$$

$$=-4p^4q^4$$

(iv)$$x\times x^2\times x^3\times x^4$$

$$= x^{(1+2+3+4)}=x^{10$$

4.(a) Simplify: $$3x(4x – 5) + 3$$ and find its values for (i) x = 3 (ii) x = $$\frac { 1 }{ 2 }$$.
(b) Simplify: $$a(a^2 + a + 1) + 5$$ and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1

(a) Given the expression: $$3x(4x – 5) + 3 = 4x \times 3x – 5 \times 3x + 3 = 12x^2 – 15x + 3$$
(i) So for x = 3, we have

$$12 \times (3)2 – 15 \times 3 + 3 = 12 \times 9 – 45 + 3 = 108 – 42 = 66$$

(ii)For x=$$\frac{1}2$$, we have:

$$\Rightarrow 12(\frac{1}2)^2 - 15(\frac{1}2)+3$$

$$\Rightarrow 12\times(\frac{1}4) - \frac{15}2+3$$

$$\Rightarrow 3 - \frac{15}2+3$$

$$\Rightarrow \frac{6-15+6}2=\frac{12-15}2=-\frac{3}2$$

(b) We have $$a(a^2 + a + 1) + 5$$

$$= (a^2 \times a) + (a \times a) + (1 \times a) + 5$$

$$= a^3 + a^2 + a + 5$$

(i) For a = 0, we have:

$$= (0)^3 + (0)^2 + (0) + 5 = 5$$

(ii) For a = 1, we have:

$$= (1)^3 + (1)^2 + (1) + 5 = 1 + 1 + 1 + 5 = 8$$

(iii) For a = -1, we have:

$$= (-1)^3 + (-1)^2 + (-1) + 5 = -1 + 1 – 1 + 5 = 4$$

5.(a) $$Add: p(p – q), q(q – r) and r(r – p)$$
(b) $$Add: 2x(z – x – y) and 2y(z – y – x)$$

(c) $$Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)$$

(d) $$Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)$$

(a) $$p(p – q) + q(q – r) + r(r – p)$$

=$$(p \times p) – (p \times q) + (q \times q) – (q \times r) + (r \times r) – (r \times p)$$[Opening the brackets]

= $$p^2 – pq + q^2 – qr + r^2 – rp$$

=$$p^2 + q^2 + r^2 – pq – qr – rp$$

(b) $$2x(z – x – y) + 2y(z – y – x)$$

= $$(2x \times z) – (2x \times x) – (2x \times y) + (2y \times z) – (2y \times y) – (2y \times x)$$[Opening the brackets]

= $$2xz – 2x^2 – 2xy + 2yz – 2y^2 – 2xy$$

$$= -2x^2 – 2y^2 + 2xz + 2yz – 4xy$$

$$= -2x^2 – 2y^2 – 4xy + 2yz + 2xz$$

(c) $$4l(10n – 3m + 2l) – 3l(l – 4m + 5n)$$

$$= (4l \times 10n) – (4l \times 3m) + (4l \times 2l) – (3l \times l) – (3l \times -4m) – (3l ×\times 5n)$$[Opening the brackets]

$$= 40ln – 12lm + 8l^2 – 3l^2 + 12lm – 15ln$$

$$= (40ln – 15ln) + (-12lm + 12lm) + (8l^2 – 3l^2)$$

$$= 25ln + 0 + 5l^2$$

$$= 25ln + 5l^2$$

$$= 5l^2 + 25ln$$

(d)$$[4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]$$

$$= (-4ac + 4bc + 4c^2) – (3a^2 + 3ab + 3ac – 2ab + 2b^2 – 2bc)$$[Opening the brackets]

$$= -4ac + 4bc + 4c^2 – 3a^2 – 3ab – 3ac + 2ab – 2b^2 + 2bc$$

$$= -3a^2 – 2b^2 + 4c^2 – ab + 6bc – 7ac$$


text:
# Step Function vs Simple Function

Unfortunately, my book (Royden-Fitzpatrick's Real Analysis) offers no definition of a step function but merely compares its role in Riemann integration to the role played by linear combinations of characteristic functions in Lebesgue integration. From what I gather, these are simple function in which he defines as follows:

A real-valued function $\varphi$ is defined on a measurable set $E$ is simple provided it is measurable and takes only a finite number of values. This means that $\varphi = \sum_{i=1}^n c_k \chi_{E_k}$ where $E_k = \{x \in E \mid \varphi(x) = c_k\}$

However, here and here the step function is defined in the very same way, the only difference being that the $E_k$ are taken to be intervals. But I have read elsewhere that the set of step functions and simple functions do not coincide. What am I misunderstanding? The reason I ask is because the following problem I am working could be rendered trivial depending on how one interprets these terms (at least it seems it could be):

Let $I$ be a closed, bounded interval and $E$ a measurable subset of $I$. Let $\epsilon > 0$. Show that there is a step function $h$ on $I$ and a measurable subset $F$ of $I$ for which $h = \chi_E$ on $F$ and $m(I-F) < \epsilon$.

I suppose my question comes down to: how am I to understand how the author is using the term "step function" so that the problem makes sense but isn't triviaL?

• I don't think you are misunderstanding anything. A step function is a special case of a simple function, in which the sets $E_k$ are intervals. For a simple function, the $E_k$ can be any measurable sets. So, every step function is a simple function, but not vice versa.
– user169852
Oct 20 '17 at 15:32
• @Bungo Oh. Well that's reassuring. It is customary to take the sums as finite? Oct 20 '17 at 15:54
• Yes, the sums are finite for both step functions and simple functions. These types of functions are often used to approximate other functions via limiting arguments. It would be a pain if we had to incorporate an additional limiting argument because the step/simple functions themselves were defined with an infinite sum.
– user169852
Oct 20 '17 at 16:00
• For this particular question, it is trivial to make a simple function $h$ with the desired property (you could just pick $F=I$). But with the restriction to step functions it really becomes an approximation problem (when $E$ is more complex than just a finite union of intervals). Apr 15 at 17:57

Royden and Fitzpatrick define step function later in their text on page 69. As you mentioned, a step function is a special case of a simple function in that the $$E_k$$ are contiguous intervals.


text:
+0

# Find rational fraction

0
407
4

Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

Guest Dec 2, 2015

#2
+18956
+40

Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

$$\begin{array}{rcl} 8\cdot \arctan{ ( \frac{1}{10} ) } - \arctan{ ( \frac{a}{b} ) } &=& \frac{\pi}{4}\\ \end{array}$$

$$\text{Formula: } \quad \boxed{~ \tan{(2x)} = \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } ~}$$

I.  We set:    $$x = \arctan{ ( \frac{1}{10} ) } \quad \text{ and } \quad \tan{(x)} = \frac{1}{10}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 2 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} }= \dfrac{2\cdot \frac{1}{10} } { 1- \frac{1}{10}\cdot \frac{1}{10} } = \frac{20}{99}\\ \qquad 2 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{20}{99} ) } \end{array} }$$

II. We set:   $$x = 2\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{20}{99} ) }\quad \text{ and } \quad \tan{(x)} = \frac{20}{99}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 4 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{20}{99} } { 1- \frac{20}{99}\cdot \frac{20}{99} } = \frac{3960}{9401}\\ \qquad 4 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{3960}{9401} ) } \end{array} }$$

III. We set:   $$x = 4\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{3960}{9401} ) }\quad \text{ and } \quad \tan{(x)} = \frac{3960}{9401}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 8 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{3960}{9401} } { 1- \frac{3960}{9401}\cdot \frac{3960}{9401} } = \frac{74455920}{72697201}\\ \qquad 8 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{74455920}{72697201} ) } \end{array} }$$

IV.

$$\text{Formula: } \quad \boxed{~ \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } ~}$$

We set:   $$x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1$$

$$\small{ \begin{array}{rcl} \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } &=& \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } = 1\\\\ \qquad \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } &=& 1\\\\ \qquad \frac{74455920}{72697201}+\frac{a}{b} &=& 1- \frac{74455920}{72697201}\cdot \frac{a}{b}\\\\ \qquad \frac{a}{b}+ \frac{74455920}{72697201}\cdot \frac{a}{b} &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( 1 + \frac{74455920}{72697201} \right) &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( \frac{72697201+74455920}{72697201} \right) &=& \frac{72697201-74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot ( 72697201+74455920 ) &=& 72697201-74455920\\\\ \qquad \frac{a}{b} &=& \dfrac{ 72697201-74455920 } { 72697201+74455920 }\\\\ \qquad \frac{a}{b} &=& \dfrac{ -1758719 } { 147153121 }\\\\ \hline \\ \tan{(x+y)} = 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }+\arctan{ ( \frac{a}{b} ) })} &=& 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) })} &=& 1 \\ 8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) } &=& \arctan{(1)} \\\\ \mathbf{ 8\arctan{ ( \dfrac{1}{10} ) }-\arctan{ ( \dfrac{ 1758719 } { 147153121 } ) } } & \mathbf{=} & \mathbf{\dfrac{\pi} {4}} \end{array} }$$

heureka  Dec 2, 2015
Sort:

#2
+18956
+40

Pi/4=8arctan(1/10) - arctan(a/b). Find "a" and "b", so that the equality is true. Thank you very much for any help.

$$\begin{array}{rcl} 8\cdot \arctan{ ( \frac{1}{10} ) } - \arctan{ ( \frac{a}{b} ) } &=& \frac{\pi}{4}\\ \end{array}$$

$$\text{Formula: } \quad \boxed{~ \tan{(2x)} = \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } ~}$$

I.  We set:    $$x = \arctan{ ( \frac{1}{10} ) } \quad \text{ and } \quad \tan{(x)} = \frac{1}{10}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 2 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} }= \dfrac{2\cdot \frac{1}{10} } { 1- \frac{1}{10}\cdot \frac{1}{10} } = \frac{20}{99}\\ \qquad 2 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{20}{99} ) } \end{array} }$$

II. We set:   $$x = 2\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{20}{99} ) }\quad \text{ and } \quad \tan{(x)} = \frac{20}{99}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 4 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{20}{99} } { 1- \frac{20}{99}\cdot \frac{20}{99} } = \frac{3960}{9401}\\ \qquad 4 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{3960}{9401} ) } \end{array} }$$

III. We set:   $$x = 4\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{3960}{9401} ) }\quad \text{ and } \quad \tan{(x)} = \frac{3960}{9401}$$

$$\small{ \begin{array}{rcl} \tan{(2x)} = \tan{( 8 \arctan{ ( \frac{1}{10} ) } )} &=& \frac{ 2\tan{(x)} }{ 1- \tan{(x)}\cdot \tan{(x)} } =\dfrac{2\cdot \frac{3960}{9401} } { 1- \frac{3960}{9401}\cdot \frac{3960}{9401} } = \frac{74455920}{72697201}\\ \qquad 8 \arctan{ ( \frac{1}{10} ) } &=& \arctan{ ( \frac{74455920}{72697201} ) } \end{array} }$$

IV.

$$\text{Formula: } \quad \boxed{~ \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } ~}$$

We set:   $$x = 8\arctan{ ( \frac{1}{10} ) } = \arctan{ ( \frac{74455920}{72697201} ) } \quad \text{ and } \quad \tan{(x)} = \frac{74455920}{72697201}\\ y = \arctan{ ( \frac{a}{b} ) } \quad \text{ and } \quad \tan{(y)} = \frac{a}{b}\\ \tan{(x+y)} = 1$$

$$\small{ \begin{array}{rcl} \tan{(x+y)} = \frac{ \tan{(x)}+\tan{(y)} }{ 1- \tan{(x)}\cdot \tan{(y)} } &=& \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } = 1\\\\ \qquad \dfrac{ \frac{74455920}{72697201}+\frac{a}{b} }{ 1- \frac{74455920}{72697201}\cdot \frac{a}{b} } &=& 1\\\\ \qquad \frac{74455920}{72697201}+\frac{a}{b} &=& 1- \frac{74455920}{72697201}\cdot \frac{a}{b}\\\\ \qquad \frac{a}{b}+ \frac{74455920}{72697201}\cdot \frac{a}{b} &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( 1 + \frac{74455920}{72697201} \right) &=& 1-\frac{74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot \left( \frac{72697201+74455920}{72697201} \right) &=& \frac{72697201-74455920}{72697201}\\\\ \qquad \frac{a}{b}\cdot ( 72697201+74455920 ) &=& 72697201-74455920\\\\ \qquad \frac{a}{b} &=& \dfrac{ 72697201-74455920 } { 72697201+74455920 }\\\\ \qquad \frac{a}{b} &=& \dfrac{ -1758719 } { 147153121 }\\\\ \hline \\ \tan{(x+y)} = 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }+\arctan{ ( \frac{a}{b} ) })} &=& 1 \\ \tan{(8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) })} &=& 1 \\ 8\arctan{ ( \frac{1}{10} ) }-\arctan{ ( \frac{ 1758719 } { 147153121 } ) } &=& \arctan{(1)} \\\\ \mathbf{ 8\arctan{ ( \dfrac{1}{10} ) }-\arctan{ ( \dfrac{ 1758719 } { 147153121 } ) } } & \mathbf{=} & \mathbf{\dfrac{\pi} {4}} \end{array} }$$

heureka  Dec 2, 2015
#3
+91786
+5

Thanks Heureka, that is a fabulous answer. :)

Melody  Dec 2, 2015
#4
+10

Brilliant work, heureka!. Thank you very much for  the outstanding technical work and the final answer. I noticed that both "a" and "b" are primes! Wasn't expecting that at all. A brief question: couldn't this be solved by " continued fraction" method? That is: Pi/4= 8arctan(1/10) - arctan(1)=0.011951......etc. Then we take the tangent of this, which comes to=0.011951625545203353.......which is the answer of your two numbers: 1,758,719 / 147,153,121=0.011951625545203353........etc.

P.S. I plugged this into "WolframAlpha" and sure enough it gives this continued fraction:

[0; 83, 1, 2, 27, 1, 2, 1, 1, 1, 2, 3, 4, 6, 1, 6], When you work this backwards, it gives the fraction you found!:1758719/147153121=0.01195162554520335317930497716049121377452810.....

THANKS A LOT.

Guest Dec 2, 2015

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text:
# Use the appropriate test to decide if they converge

Use the appropriate test to decide whether the following serie converges or not:

$$\sum \limit_{n=1} ^{\infty} \frac{3n^2 - 2n +1}{2n^2 + 5}$$

Related Calculus and Beyond Homework Help News on Phys.org
LeonhardEuler
Gold Member
Try the n-th term test for divergence.

LeonhardEuler said:
Try the n-th term test for divergence.

b) $$\sum \limit_{n=1} ^{\infty} \frac{3n^2 - 2n + 1}{2n^4 + 5}$$

c) $$\sum \limit_{n=1} ^{\infty} \frac{n^3 4^n}{3(n!)}$$

d) $$\sum \limit_{n=1} ^{\infty} \frac{2 + 3 sin~n}{5n^2 + 2}$$

LeonhardEuler
Gold Member
When you look at b, you should see that for large n it behaves like $\frac{1}{2n^2}$. What test will let you use that fact?

c) is pretty straght foward. When you see everything in with powers and factorials you should think of the comparison test or the root test.

d)Similar to b). The sin basically has no effect because the series that this looks like converges absolutely.

LeonhardEuler said:
When you look at b, you should see that for large n it behaves like $\frac{1}{2n^2}$. What test will let you use that fact?

c) is pretty straght foward. When you see everything in with powers and factorials you should think of the comparison test or the root test.

d)Similar to b). The sin basically has no effect because the series that this looks like converges absolutely.
For a) can I just say this series does not converge as a_n does not tend to zero as n --> infinity but to 2/3 instead.

I have never heard of this n-th term test. Could someone quickly do it for a) please, just so I can understand it. Much appreciated :-)

LeonhardEuler
Gold Member
The n-th term test can be used exactly like you said. If the terms in the sequence do not approach zero then the series can not converge. That is all you need to say.


text:
MEAN() returns an iid model applied to the formula's response variable.

MEAN(formula, ...)

Arguments

formula Model specification. Not used.

Value

A model specification.

Details

The model does not use any specials, and so everything on the formula's right-hand-side will be ignored.

Specials

window

The window special is used to specify a rolling window for the mean.

window(size = NULL)

 size The size (number of observations) for the rolling window. If NULL (default), a rolling window will not be used.

library(tsibbledata)
#> 1  <MEAN>


text:
# Preface¶

## Aims and scope¶

This book addresses an important class of mathematical problems (the Riemann problem) for first-order hyperbolic partial differential equations (PDEs). Such PDEs arise when modeling wave propagation in applications such as fluid dynamics, traffic flow, elasticity, and electromagnetics. The solution of the Riemann problem captures essential information about such models, and is the key ingredient in modern numerical methods for their solution.

The book covers the fundamental ideas related to classical Riemann solutions, including their special structure and the types of waves that arise, as well as the ideas behind fast approximate solvers for the Riemann problem. The emphasis is on the general ideas, but each chapter delves into a particular application.

It is assumed that the reader has some knowledge of hyperbolic problems in advance. This book is intended to supplement rather than replace standard textbooks on the theory and numerical solution of hyperbolic PDEs. It should also be useful as a tool for students or researchers seeking to better understand the nature of Riemann solutions to a particular hyperbolic system.

The notation generally follows that of (LeVeque 2002), which covers most of the topics included here and many others. We have kept citations to a minimum in these notebooks, but there are a wealth of other books and papers available in the literature. In particular, some other books that have substantial discussion of Riemann problems and related topics include (Lax 1972), (LeVeque 1990), (Godlewski and Raviart 1996), (Kröner 1997), (Trangenstein 2009), (Toro 2013), and (Hesthaven 2018).

## Book formats¶

This book is available in three formats:

• The traditional print medium, in the form of a physical book.

• As a collection of Jupyter notebooks with executable code and interactive animations, available from the Github repository https://github.com/clawpack/riemann_book. These are not meant to be read on Github, but copied to a local machine or cloud service and run there; see below.

• As a set of webpages that are rendered from the Jupyter notebooks. These webpages contain some animations, but the interactive widgets from the notebooks do not work and it is not possible to modify and re-run cells.

For the full experience, run the notebooks!

### The print version¶

Most of the figures in this book are generated by Python programs. Most of the code for those figures is in files in the Github repository linked above, but some of it is included inline in the text at the point where each figure appears. The reader may choose to skip the code sections, especially on a first reading.

### The Jupyter notebook version¶

In the electronic version of this book, each chapter is a Jupyter notebook. The Jupyter Notebook is a free, open-source web application that allows you to create and share documents that contain live code, equations, visualizations and explanatory text. The notebooks that form this text have some additional benefits not present in the print version:

• Some of the code in the notebooks generates animations, showing how solutions vary in time or with respect to a parameter.
• Other code in the notebooks generates interactive "widgets" that allow the reader to vary multiple parameters or plot different quantities, in order to gain deeper insight and understanding.
• The reader can also modify the code itself, perhaps to vary parameters that we (the authors) didn't think of. We encourage readers to use this to satisfy their own curiosity and enhance their learning.

Although some Python code is included directly in the notebooks, for more clarity much of the code has been "hidden" in Python modules that are imported near the top of each module. We also provide links in each notebook (and each chapter of the printed book) to the online location of these modules in case you wish to dive deeper into the code. Most of this code is in the exact_solvers subdirectory of the Github repository. Other code used primarily for producing plots can be found in the utils subdirectory.

### How to access and run the Jupyter notebook version¶

In order to run the code in the Jupyter notebook version of the book, you will need to have certain tools and libraries installed on a computer, either locally or in the cloud. There are multiple convenient ways to do this.

• On your own computer: follow the installation instructions
• With Docker: follow these instructions to use the Dockerfile that is included in the Github repository.
• With Binder: The cloud service binder allows you to start up a notebook server running notebooks from any Git repository (assuming it specifies the dependencies properly). You can try it out for these notebooks at this link. This should start up a notebook server on a Jupyterhub that lets you execute all the notebooks in your browser with no local installation of the notebooks or dependencies required.

## Overview¶

This book consists of several parts, each divided into several chapters (as independent Jupyter notebooks). The printed book contains only the first two parts, which cover the fundamentals of exact and approximate solutions of Riemann problems; the remaining parts consist of online notebooks that delve into more advanced topics. In a graduate course on hyperbolic conservation laws, the first two parts could be incorporated into the curriculum, while the later chapters might provide suggestions for student projects.

Part I covers the classical theory of the Riemann problem and its solution, introducing the reader to essential concepts including characteristics; similarity solutions; shock waves and the Rankine-Hugoniot jump conditions; rarefactions and Riemann invariants; genuinely nonlinear and linearly degenerate waves; and entropy conditions. Building up from scalar hyperbolic PDEs and linear systems, it culminates with discussions of the full Riemann solution for the shallow water and Euler systems.

Part II introduces approximate Riemann solvers, which are an essential tool in numerical algorithms for hyperbolic PDEs. This part discusses the most important approaches to such solvers: from simple Lax-Friedrichs and HLL solvers to linearizations, including Roe solvers. Potential issues with these methods (such as the need for entropy fixes and lack of positivity) are addressed. Part II concludes with comparisons of canonical approximate solvers for the shallow water and Euler systems.

The Index.html notebook contains links to the chapters of Parts I and II.

The remaining parts are still a work in progress at the time of this writing, and are not included in the print or PDF version of the book. They can be found in the Github repository and additional topics may be added at a later date.

## Writing a book with Jupyter notebooks¶

Although the primary goal of this book is to present the mathematics of Riemann solvers in a novel and educational manner, an additional goal from the beginning has been to explore the use of Jupyter notebooks for this sort of publishing venture, and to help popularize Jupyter notebooks within the applied mathematics community. We believe that notebooks are a great tool for both teaching and research, and an exciting new format for publishing. It has been an interesting challenge to figure out ways to present material that works well in the notebooks and that also translates into html files and a physical book, and we hope that our experiences not only help convey the subject matter but also can serve as a template for other publishing projects or as an impetus to develop better approaches.

All of the notebooks and associtated Python code are of course available in the github repository, and in addition we have written some wiki pages on various other tools we have found useful in developing this book. In particular, bookbook has been used in converting notebooks to latex/pdf files, and tools for working with notebooks such as nbdime and nbstripout have been invaluable for helping to manage modifications to notebooks in the context of git. Without git and github this project would have never come together, since the three authors have been employed on three different continents for most this project, and many discussions about how to do things, what to include, how to edit notebooks, etc. have taken place via pull requests on github. As of this writing there are 183 closed pull requests on this repository, which provide a history of many of our discussions and decisions.

Numerous other books and lecture notes have been written that make extensive use of Jupyter notebooks, and they are also becoming increasingly popular as a supplement to research publications. See the Gallery of interesting Jupyter notebooks for other examples and inspiration. We hope that more applied mathematicians will explore this platform for future publications.

## Acknowledgments¶

We are indebted to the people who created the tools we have used to develop these notebooks, and in particular to the developers of Jupyter. Fernando Perez and several others helped us out at many points during the development of this book. Jake Vanderplas originally developed the JSAnimation package for us to use in Clawpack, and helped us with a number of other notebook issues.

The initial idea for this book arose during discussions at the Clawpack Developers Workshop held at the University of Utah, just before the 2015 SIAM Conference on Computational Science and Engineering. Many members of the Clawpack community have helped support this project, in particular Kyle Mandli, an early adopter who has provided valuable feedback, as did Lisa Davis.

We are delighted that SIAM will be publishing the printed version of these notebooks, and are appreciative of their willingness to support this experiment, and of their patience as we have worked out various kinks in the process. We are also very pleased that they have allowed the notebook repository to remain freely available on github, with licensing that allows you to not only use the notebooks but to adapt them for your own non-commercial purposes. Although running the notebooks is the best way to experience this material, we hope you will also buy the book and support SIAM -- or better yet, write your own book and publish with them.


text:
# Could logistic regression be used to detect large errors in least squares regression?

I have the following linear model:

$w^*=\text{arg min}_w\sum_{i=1}^N (Y_i-\sum_{j=1}^M X_{i,j}*w_j)^2$

Let be $T \in N^*$ and $e_i=|Y_i-\sum_{j=1}^M X_{i,j}*w_j|$.

It's possible using logistic regression to predict which errors will be less than $T$ ( $e_i<T$) and greater or equal with $T$ ($e_i \ge T$)?

I will give more in formations for a better understanding:

$N$ represent the number of observations. My data has the following property: the histogram of errors using multiple linear regression has a Laplace distribution. My data come from digital images represented on 8 bits. The $Y_i$ are current pixels and $X_ij$ are neighborhoods pixels. I want to predict which pixels produce errors less than $T$. I want to know what R functions can I use to make a test? T is nit very large, it has the values between 1 and 15 in general.

-
 What do you mean by "$N^*$"? What kinds of values can $Y_i$ have? To what data, precisely, do you propose applying logistic regression? – whuber♦ May 29 '12 at 17:29

It would be possible in the case where the original model was heteroskedastic and the heteroskedasticity was related to the covariates. For example,

$y_i \sim \text{N}(x_i^T\beta, \sigma^2x_{i,1}^2)$

where the variance of the $i^{th}$ observation is proportional to the square of the first covariate.

One can imagine, in non-Normal regression situations, similar structures that don't require heteroskedasticity per se. However, the default assumptions of regression models in general involve the errors being independent of the regressors, and the variance of the errors being constant.

On the other hand, if doing, for example, Poisson regression, you're out of the linear model world, but since the variance of the "error" is proportional to the mean, it follows that it is related to the covariates, and such a logistic regression would work - although it would convey no information not already conveyed by the results of the Poisson regression, which fully specify the conditional distributions of the $y_i | x_i$. In the generalized linear / additive model framework, where the likelihood is fully specified, the only way you'll be able to add information to the initial regression by using the logistic regression you suggest is if the initial regression has misspecified (usually by ignoring) the structure of the residuals, e.g., ignored the heteroskedasticity in the linear model presented above.

Nonetheless, your suggestion might reveal something about the structure of the residuals in an exploratory analysis. I suspect, though, that effectively discretizing the residuals by $< T$ or $\ge T$ would usually decrease the information content of them more than it would help clarify the analysis - unless it was an outlier analysis, perhaps.

-
 +1 for heteroskedasticity, which I should have thought of myself. – gung May 29 '12 at 18:31

You are trying to use logistic regression to find a possible structure to your residuals. Residuals should be unstructured. If logistic regression picks up something the model is misspecified. Great.

Important: logistic regression looks for a very specific structure. Your method works only if residuals are broken in a very specific way.

For this sort of detection use non-linear regressions. Kernel or Additive regression on the squared/log residuals are far more useful.
Even better: they will allow you to model the whole variance. This answers your question of where the large errors really are.

-
 What means that "Residuals should be unstructured"? – Ion Caciula May 29 '12 at 19:22 +1, I think what he means is that standard regression assumes the residuals are independent & identically distributed as a Gaussian w/ constant variance. If this is true, then by definition the logistic regression won't be able to find a relationship b/t $P(e_i\geq T)$ & your covariates. This is what I was trying to get at, although CarrKnight has given a clearer & more comprehensive version. – gung May 29 '12 at 20:36 Exactly! Thank you Gung. – CarrKnight May 30 '12 at 19:49

I think this would only work if the logistic regression model had access to relevant covariates that are missing from the OLS model. In such a situation (i.e., model misspecification), there could be regions in which the observed response values diverge strongly from the predicted values and the logistic regression model would have the requisite information to detect them.

-


text:
NULL
Section
All sections
Countries | Regions
Countries | Regions
Article Types
Article Types
Year
Volume
Issue
Pages
IMR Press / RCM / Volume 23 / Issue 7 / DOI: 10.31083/j.rcm2307246
Open Access Review
Cardiac Magnetic Resonance and Ventricular Arrhythmia Risk Assessment in Chronic Ischemic Cardiomyopathy: An Unmet Need?
Show Less
1 Arrhytmia Section, Cardiology Department, Miguel Servet University Hospital, 50009 Zaragoza, Spain
*Correspondence: beatrizjg86@gmail.com (Beatriz Jáuregui)
Rev. Cardiovasc. Med. 2022, 23(7), 246; https://doi.org/10.31083/j.rcm2307246
Submitted: 1 April 2022 | Revised: 21 May 2022 | Accepted: 27 May 2022 | Published: 28 June 2022
(This article belongs to the Special Issue Role of Cardiovascular Magnetic Resonance in Cardiology)
This is an open access article under the CC BY 4.0 license.
Abstract

Ischemic cardiomyopathy (ICM) constitutes a major public health issue, directly involved in the prevalence and incidence of heart failure, ventricular arrhythmias (VA) and sudden cardiac death (SCD). Severe impairment of left ventricular ejection fraction (LVEF) is considered a high-risk marker for SCD, conditioning the criteria that determine an implantable cardiac defibrillator (ICD) placement in primary prevention according to current clinical guidelines. However, its sensitivity and specificity values for the prediction of SCD in ICM may not be highest. Myocardial characterization using cardiac magnetic resonance with late gadolinium enhancement (CMR-LGE) sequences has made it possible to answer clinically relevant questions that are currently not assessable with LVEF alone. There is growing scientific evidence in favor of the relationship between fibrosis evaluated with CMR and the appearance of VA/SCD in patients with ICM. This evidence should make us contemplate a more realistic clinical value of LVEF in our daily clinical decision-making.

Keywords
cardiac magnetic resonance
ischemic cardiomyopathy
left ventricular ejection fraction
myocardial infarction
arrhythmogenic substrate
ventricular tachycardia
sudden cardiac death
1. Introduction

Ischemic cardiomyopathy (ICM) and its consequences are estimated to be the cause of at least 80% of sudden cardiac deaths (SCDs) (around 240,000–280,000 per year) in Western countries [1]. From autopsy data, around 30% of all the ICM-related SCDs lack acute coronary findings, but are instead linked to the presence of an old ischemic myocardial scar [2]. The biological processes of myocardial healing, left ventricular (LV) remodeling and scar changes involve the development of a potential arrhythmogenic substrate capable of causing reentrant ventricular arrhythmias (VA), the main cause of arrhythmic SCD in chronic ICM [3].

Ventricular fibrillation (VF) constitutes the final VA leading to arrhythmic SCD. Polymorphic ventricular tachycardias (VTs) are usually related to acute ischemia and can quickly degenerate into VF [4]. Monomorphic VTs (MVTs) are due to anatomical/functional reentry related to the presence of myocardial scar/fibrosis, and may degenerate into VF owing to primary rhythm disorganization, hemodynamic instability and secondary global ischemia. In patients with chronic ICM, VAs causing SCD are mostly reentrant MVTs [5] that require the presence of a histologic substrate for reentry. However, several external triggers may also be required for their initiation, such as neurohormonal alterations, mechanical factors (e.g., myocardial stretching in conditions of volume/pressure overload), and ionic alterations [6, 7].

Severe LV systolic dysfunction is considered a high-risk marker for SCD [8]. In current clinical practice guidelines [9, 10], the presence of a left ventricular ejection fraction (LVEF) $<$35% in chronic ICM is the criterion that determines an implantable cardiac defibrillator (ICD) placement in primary prevention of SCD [11, 12]. However, only 20–30% of the patients included in randomized clinical trials on primary prevention receive appropriate ICD shocks after 4 years of follow-up, indicating limited positive predictive value of LVEF as a risk marker for SCD [11, 12]. In addition, approximately 65% of patients with SCD have normal or only mildly depressed LV systolic function (i.e., LVEF between 35% and 50%) [13, 14, 15]. Therefore, severe LV dysfunction alone is a not sufficiently specific marker for SCD, although it could be useful when used with other predictors or as part of a multivariable risk score including various clinical parameters, such as age, New York Heart Association (NYHA) functional class, renal function, QRS width, coexistence of atrial fibrillation, VT inducibility during electrophysiological studies (EPS), myocardial hypertrophy, heart rate variability, etc. [16, 17, 18, 19]. However, such scoring algorithms have not yet been validated in large prospective series. Besides, these markers are not very specific when it comes to discriminating the risk of SCD versus the risk of non-sudden death [16, 17, 20, 21].

2. Conventional Arrhythmia Risk Assessment in Ischemic Cardiomyopathy
2.1 Left Ventricular Ejection Fraction

Mortality and SCD risk increase when LVEF decreases below 50% [22, 23], and particularly when it becomes $<$40% [24]. On the other hand, due to the improvement in reperfusion therapies, there are relatively few patients with very low LVEF after an acute coronary event [25, 26, 27]. However, there is no evidence of a cause-effect relationship between an abnormal LVEF and the occurrence of SCD. In fact, although the absolute risk of SCD is higher in patients with lower LVEF, the total number of SCDs is higher in patients with higher LVEF [28]. In addition, current LVEF values used to define high-risk populations (typically $<$35%) have poor sensitivity, failing to identify up to 50% of patients at risk for SCD [29].

Current evidence for the choice of LVEF as the main criterion for ICD implantation in primary prevention in patients with ICM is based on the classical trials MADIT-I and II [11, 30], CABG-PATCH [31], MUSTT [32], DINAMIT [33], and SCD-HeFT [12]. Although in the latest clinical practice guidelines the presence of LVEF $<$35% remains the criterion that currently determines implantation [11, 12], it has not been shown to be a parameter either sensitive or specific for the prediction of SCD in ICM [34, 35, 36, 37, 38, 39]. In the recent PRESERVE EF study [40] it was found that 9 of 575 patients (1.6%) with ICM and LVEF $>$40% (mean 50.8%) had appropriate ICD therapies due to sustained VA after a mean clinical follow-up of 32 months. These patients, who did not have a standard indication for ICD implantation in primary prevention, had been indicated for ICD because they had some non-invasive additional risk factor (see below) and were inducible during EPS.

2.2 Other Parameters

Traditionally, other clinical variables apart from LVEF have been tested in order to improve the risk stratification of VA/SCD in patients with ICM. Some of these variables have been early detection of potential triggers (premature ventricular complexes and non-sustained VTs), presence of late potentials on the signal-averaged ECG, T wave alternans, inducibility during EPS, analysis of autonomic functional markers, or the use of multivariable clinical models. The available evidence has confirmed that none of them was sufficiently reliable for arrhythmic risk stratification [19, 41, 42, 43, 44, 45, 46, 47, 48, 49]. Therefore, their use cannot be recommended for making relevant clinical decisions.

3. Current Status and Role of Cardiac Magnetic Resonance for Arrhythmia Risk Assessment in Ischemic Cardiomyopathy
3.1 Role of Cardiac Magnetic Resonance to Calculate Left Ventricular Ejection Fraction

As previously mentioned, LVEF is a suboptimal risk marker for VA and SCD, and yet it constitutes the main criterion on which current clinical practice guidelines base the indication for ICD implantation in primary prevention of SCD [11, 12]. LVEF has been classically obtained using echocardiography, which has shown to generally overestimate LVEF by 3–7% compared to cardiac magnetic resonance (CMR), a more reproducible technique [50]. These differences might justify a better risk stratification using CMR [51, 52], despite no published ICD trial has included the use of CMR-derived LVEF. On the other hand, there is evidence [53] suggesting that the use of a weighted CMR score after a ST-segment–elevation myocardial infarction, including CMR-LVEF and other variables (myocardial salvage index, microvascular obstruction, and myocardial hemorrhage), was independently associated with major adverse cardiovascular events, and may provide incremental prognostic stratification as compared with GRACE score and echo-LVEF [53].

3.2 Role of Cardiac Magnetic Resonance to Identify the Arrhythmogenic Substrate

Far beyond LVEF, other clinical predictors for arrhythmia events may be found; for example, in a recent meta-analysis including 17 studies with LVEF values ranging from 25 to 59% [54], the presence of a coronary chronic total occlusion (CTO), particularly when affecting the infarct-related artery, was associated with a 1.68-fold increase in the occurrence of VT/VF or appropriate ICD therapy. Aiming to non-invasively characterize the post-infarction myocardial injury, cardiac magnetic resonance with late gadolinium enhancement (CMR-LGE) has made it possible to answer clinically relevant questions that are currently not assessable with LVEF alone or with the use of other non-invasive imaging modalities. In fact, the diagnostic and prognostic role of CMR in ICM has already been broadly recognized in more recent articles, guidelines and consensus documents [55, 56, 57, 58]. Numerous studies have shown a link between the presence of macroscopic fibrosis evaluated with CMR and the appearance of VA in patients with ICM. In a recent meta-analysis by Disertori et al. [59] including 2850 patients from 19 different studies, patients without LGE had an annual arrhythmia/SCD event rate of 1.7%, compared with an event rate of 8.6% per year in patients with LGE. Moreover, 100% of those with events had macroscopic fibrosis on CMR-LGE [59, 60]. This has been supported in modern, large prospective series of patients with ICM and non-ischemic etiologies [61].

The ability of CMR-LGE to detect myocardial fibrosis has been supported by histological correlation studies [62, 63]. The presence of slow and heterogeneous electrical conduction associated with fibrosis may favor reentry, increasing vulnerability to VT/VF onset [6, 64, 65, 66]. Areas with different levels of fibrosis coexist in the gray zone, heterogeneous tissue, or border zone (BZ), resulting in simultaneous regions of viable and non-viable myocardium, which can be identified with CMR-LGE [64]. In addition, heterogeneous tissue channels (HTC), or border zone channels (BZC), have been correlated with functional, slow conduction channels identifiable by endocardial voltage maps during electroanatomical mapping (EAM) [67, 68, 69, 70], as well as with the critical isthmuses of VT reentry circuits [71].

Despite the evidence that size and heterogeneity of the myocardial scar could be predictors of VA and mortality, it remains unknown whether CMR, beyond the use of LVEF, could facilitate the clinical decision-making in relation to arrhythmia risk stratification. In this sense, a previous prospective study by Klem et al. [72], based on a cohort of ischemic patients with a wide range of LVEF values, analyzed the risk of arrhythmic events: Patients with LVEF $>$30% and fibrosis $>$5% of total myocardial mass had a higher VA risk than those patients with LVEF $>$30% and small scars ($<$5%), but similar to that of patients with LVEF $<$30%. Similarly, those patients with LVEF $<$30% and fibrosis $<$5% had a risk of arrhythmic events similar to that of patients with LVEF $>$30%. More recently [73], the presence of myocardial fibrosis and the mass of heterogeneous tissue (a.k.a. BZ), both analyzed with CMR-LGE, were evaluated in relation to the occurrence of SCD and the composite event of SCD or VA. Of 947 patients analyzed retrospectively [73], there were 29 cases of SCD (2.96%) and 80 cases of SCD/VA (8.17%) after a median follow-up of 5.82 years. The visual presence of myocardial fibrosis or heterogeneous tissue on CMR was strongly associated with the appearance of SCD and the composite event. In contrast, the associations between LVEF $<$35% and the development of SCD, or the composite SCD/VA, were much weaker in a competing risk analysis [73]. In addition, all patients having events (SCD/VA) showed evident fibrosis on CMR. Furthermore, the cut-off point of LVEF $<$35% was very weakly associated with both the occurrence of SCD or the composite SCD/VA [73]. Additionally, from all the scar variables that can be assessed with CMR, the amount (mass) of BZ and, particularly, the amount of BZ mass being part of tissue corridors (BZC) appear to be the most powerful variables associated with the development of reentrant VA [74, 75].

Including CMR-derived variables, the currently ongoing European PROFID project will aim to develop a clinical prediction model for SCD in ICM in order to improve the selection of ICD candidates. Two randomized trials will validate the usefulness of this prediction model according to the LVEF (PROFID-Preserved, NCT04540289; and PROFID-Reduced, NCT04540354). The preliminary results presented in EHRA Congress 2022 indicate a potentially relevant role of CMR-LGE; particularly regarding the extent of scar and BZ. Therefore, un updated model including these variables is to be expected.

Apart from evaluating macroscopic fibrosis using LGE, CMR is also capable of assessing diffuse interstitial myocardial fibrosis, which is related to adverse remodeling in patients with ICM [76, 77]. This is usually performed using T1-mapping techniques. Diffuse fibrosis has been shown to be an independent predictor of VA [78]. However, it should be remarked that T1-mapping is limited to just a single slice of myocardium, assuming that the diffuse fibrosis affects uniformly the remote myocardium. Therefore, more studies would be required to explore the possibilities of this method.

3.3 Role of Cardiac Magnetic Resonance during Ventricular Tachycardia Substrate Ablation Procedures

In addition to the aforementioned, it has recently been shown that the QRS morphology of potential inducible VTs and the location of the responsible circuit in each case can be accurately predicted using computer simulation models trained with CMR-LGE images from infarcted pigs [79]. This may suggest that our current post-processing methods for BZ analysis could eventually be complemented with even more sophisticated tools (computer models, machine learning, etc.) to identify reentrant circuits through CMR findings. Other studies have strongly supported the usefulness of CMR in identifying and characterizing the arrhythmogenic substrate in secondary prevention. Integrated 3D reconstructions of the scar obtained from CMR images are capable of accurately depicting not only the location of the arrhythmogenic substrate, but also its structure and the presence of BZC, which constitute the therapeutic target during substrate ablation procedures [80, 81, 82]. These studies have demonstrated improved clinical outcomes and better procedural efficiency when taking advantage of the CMR-derived data on the arrhythmogenic substrate.

4. Conclusions

There is growing scientific evidence in favor of the relationship between CMR-LGE findings and the appearance of VA/SCD, particularly in ICM but also in non-ischemic cardiomyopathies. This evidence should make us contemplate a more realistic clinical value of LVEF in our daily clinical decision-making. Nevertheless, additional studies are still required. Standardization of image acquisition methods and measurements, as well as analysis of the diagnostic performance of each evaluable scar-derived parameter are issues to be resolved. Furthermore, the dynamic nature of scar remodeling makes it difficult to set an appropriate time point for CMR acquisition [83]. Finally, there is an urgent need for reliable post-processing tools allowing reproducible analysis of raw CMR images. Undoubtedly, the key for diagnosis and prognosis in our patients lies within them.

Author Contributions

BJ conceived the topic and wrote the first version of the manuscript. NC, TO, CL-P and AA discussed the contents and significantly contributed to the final version of the manuscript.

Ethics Approval and Consent to Participate

Not applicable.

Acknowledgment

Not applicable.

Funding

This research received no external funding.

Conflict of Interest

The authors declare no conflict of interest.

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Publisher’s Note: IMR Press stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Rank Dependent Weighted Average Utility Models for Decision Making under Ignorance or Objective Ambiguity

Nicolas Gravel
• Function : Author
Thierry Marchant
• Function : Author

Abstract

The paper provides an axiomatic characterization of a family of rank dependent weighted average utility criteria applicable to decisions under ignorance or objective ambiguity. A decision under ignorance is described by the finite set of its final consequences while a decision under objective ambiguity is described by a finite set of probability distributions over a set of final consequences. The criteria characterized are those that assign to every element in a set a weight that depends upon the rank of this element if it was available for sure (or non-ambiguously) and that compare sets on the basis of their weighted utility for some utility function. A specific subfamily of these criteria that requires the weights to be proportional to each other is also characterized.

Dates and versions

hal-03817362 , version 1 (17-10-2022)

Identifiers

• HAL Id : hal-03817362 , version 1

Cite

Nicolas Gravel, Thierry Marchant. Rank Dependent Weighted Average Utility Models for Decision Making under Ignorance or Objective Ambiguity. 2022. ⟨hal-03817362⟩

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