text:
## Solving the Ising Model using a Mixed Integer Linear Program Solver (Gurobi)

I came across an interesting thing, that finding the minimizer of the Ising model is encodable as a mixed integer linear program.

The Ising model is a simple model of a magnet. A lattice of spins that can either be up or down. They want to align with an external magnetic field, but also with their neighbors (or anti align, depending on the sign of the interaction). At low temperatures they can spontaneously align into a permanent magnet. At high temperatures, they are randomized. It is a really great model that contains the essence of many physics topics.

Linear Programs minimize linear functions subject to linear equality and inequality constraints. It just so happens this is a very solvable problem (polynomial time).

MILP also allow you to add the constraint that variables take on integer values. This takes you into NP territory. Through fiendish tricks, you can encode very difficult problems. MILP solvers use LP solvers as subroutines, giving them clues where to search, letting them step early if the LP solver returns integer solutions, or for bounding branches of the search tree.

How this all works is very interesting (and very, very roughly explained), but barely matters practically since other people have made fiendishly impressive implementations of this that I can’t compete with. So far as I can tell, Gurobi is one of the best available implementations (Hans Mittelman has some VERY useful benchmarks here http://plato.asu.edu/bench.html), and they have a gimped trial license available (2000 variable limit. Bummer.). Shout out to CLP and CBC, the Coin-Or Open Source versions of this that still work pretty damn well.

Interesting Connection: Quantum Annealing (like the D-Wave machine) is largely based around mapping discrete optimization problems to an Ising model. We are traveling that road in the opposite direction.

So how do we encode the Ising model?

Each spin is a binary variable $s_i \in {0,1}$

We also introduce a variable for every edge. which we will constrain to actually be the product of the spins. $e_{ij} \in {0,1}$. This is the big trick.

We can compute the And/Multiplication (they coincide for 0/1 binary variables) of the spins using a couple linear constraints. I think this does work for the 4 cases of the two spins.

$e_{ij} \ge s_i +s_j - 1$

$e_{ij} \le s_j$

$e_{ij} \le s_i$

The xor is usually what we care about for the Ising model, we want aligned vs unaligned spins to have different energy. It will have value 1 if they are aligned and 0 if they are anti-aligned. This is a linear function of the spins and the And.

$s_i \oplus s_j = s_i + s_j - 2 e_{ij}$

Then the standard Hamiltonian is

$H=\sum B_i s_i + \sum J_{ij} (s_i + s_j - 2 e_{ij})$

Well, modulo some constant offset. You may prefer making spins $\pm 1$, but that leads to basically the same Hamiltonian.

The Gurobi python package actually let’s us directly ask for AND constraints, which means I don’t actually have to code much of this.

We are allowed to use spatially varying external field B and coupling parameter J. The Hamiltonian is indeed linear in the variables as promised.

After already figuring this out, I found this chapter where they basically do what I’ve done here (and more probably). There is nothing new under the sun. The spatially varying fields B and J are very natural in the field of spin glasses.

https://onlinelibrary.wiley.com/doi/10.1002/3527603794.ch4

For a while I thought this is all we could do, find the lowest energy solution, but there’s more! Gurobi is one of the few solvers that support iteration over the lowest optimal solutions, which means we can start to talk about a low energy expansion. https://www.gurobi.com/documentation/8.0/refman/poolsearchmode.html#parameter:PoolSearchMode

Here we’ve got the basic functionality. Getting 10,000 takes about a minute. This is somewhat discouraging when I can see that we haven’t even got to very interesting ones yet, just single spin and double spin excitations. But I’ve got some ideas on how to fix that. Next time baby-cakes.

(A hint: recursion with memoization leads to some brother of a cluster expansion.)

from gurobipy import *
import matplotlib.pyplot as plt
import numpy as np

# Create a new model
m = Model("mip1")

m.Params.PoolSearchMode = 2
m.Params.PoolSolutions = 10000

# Create variables
N = 10

xor = {}
B = np.ones((N,N)) #np.random.randn(N,N)
J = 1. #antialigned
H = 0.
for i in range(N-1):
for j in range(N-1):
#for d in range(2)
xor[i,j,0] = spins[i,j] + spins[i+1,j] - 2*links[i,j,0]
xor[i,j,1] = spins[i,j] + spins[i,j+1] - 2*links[i,j,1]
H += J*xor[i,j,0] + J*xor[i,j,1]

for i in range(N):
for j in range(N):
H += B[i,j]*spins[i,j]

#B[i,j] = 1.
#quicksum([2*x, 3*y+1, 4*z*z])

#LinExpr([1.]*N*N,spins)

# Set objective
m.setObjective(H, GRB.MINIMIZE)

# Add constraint: x + 2 y + 3 z <= 4
#m.addConstr(x + 2 * y + 3 * z <= 4, "c0")

# Add constraint: x + y >= 1
#m.addConstr(x + y >= 1, "c1")

m.optimize()

#for v in m.getVars():
#    print(v.varName, v.x)

print('Obj:', m.objVal)
print('Solcount:', m.SolCount)
for i in range(m.SolCount):
m.Params.SolutionNumber = i #set solution numbers
print("sol val:", m.Xn)
print("sol energy:", m.PoolObjVal)
print(spins[0,0].Xn)

ising = np.zeros((N,N))
for i in range(N):
for j in range(N):
ising[i,j] = spins[i,j].Xn
plt.matshow(ising)

plt.show()

Here’s the ground antiferromagnetic state. Cute.

## OSQP and Sparsegrad: Fast Model Predictive Control in Python for an inverted pendulum

OSQP is a quadratic programming solver that seems to be designed for control problems. It has the ability to warm start, which should make it faster.

I heard about it in this Julia talk

They have some really cool technology over there in Julia land.

The problem is setup as a sequential quadratic program. I have a quadratic cost to try to bring the pendulum to an upright position.

The equations of motion are roughly

$\ddot{\theta}I=-mgL\sin(\theta)+mfL\cos(\theta)$

$\ddot{x}=f$

$I=\frac{1}{3}mL^2$

We don’t include the back reaction of the pole on the cart. It is basically irrelevant for our systems and overly complicated for no reason. The interesting bit is the nonlinear dynamics and influence of the cart acceleration.

I write down obeying the equations of motion as a linear constraint between timesteps. I use sparsegrad to get a sparse Jacobian of the equations of motion. The upper and lower (l and u) bounds are set equal to make an equality constraint.

Our setup has a finite track about a meter long. Our motors basically are velocity control and they can go about +-1m/s. Both of these can be encapsulated as linear constraints on the position and velocity variables. $l \le Ax \le u$

$\phi(x)=0$

$\phi(x) \approx \phi(x_0) + \partial \phi(x_0) (x - x_0)$

$A= \partial \phi(x_0)$

$l=u=\partial \phi(x_0) x_0 - \phi_0(x_0)$

Similarly for finding the quadratic cost function in terms of the setpoint $x_s$. $\frac{1}{2}x^T P x + q^Tx= \frac{1}{2}(x-x_s)P(x-x_s)$ Expanding this out we get

$q = - Px_s$

I run for multiple iterations to hopefully converge to a solution (it does). The initial linearization might be quite poor, but it gets better.

The OSQP program seems to be able to run in under 1ms. Nice! Initial tests of our system seem to be running at about 100Hz.

Could probably find a way to make those A matrices faster than constructing them entirely anew every time. We’ll see if we need that.

I very inelegantly smash all the variables into x. OSQP and scipy sparse don’t support multiIndex objects well, best as I can figure. Vectors should be single dimensioned and matrices 2 dimensioned.

Still to be seen if it works on hardware. I was already having infeasibility problems. Adjusting the regularization on P seemed to help.

import sparsegrad.forward as forward
import numpy as np
import osqp
import scipy.sparse as sparse
import matplotlib.pyplot as plt
import time

#def f(x):
#	return x**2

N = 1000
NVars  = 5
T = 5.0
dt = T/N
dtinv = 1./dt
#Px = sparse.eye(N)
#sparse.csc_matrix((N, N))
# The three deifferent weigthing matrices for x, v, and external force
reg = sparse.eye(N)*0.01
# sparse.diags(np.arange(N)/N)
P = sparse.block_diag([reg,reg ,sparse.eye(N), 1*reg,1*reg])
#P[N,N]=10
THETA = 2
q = np.zeros((NVars, N))
q[THETA,:] = np.pi
#q[N,0] = -2 * 0.5 * 10
q = q.flatten()
q= -P@q
#u = np.arr

x = np.random.randn(N,NVars).flatten()
#x = np.zeros((N,NVars)).flatten()
#v = np.zeros(N)
#f = np.zeros(N)

g = 9.8
L = 0.5
gL = g * L
m = 1.0 # doesn't matter
I = L**2 / 3
Iinv = 1.0/I
print(Iinv)

def constraint(var, x0, v0, th0, thd0):
#x[0] -= 1
#print(x[0])
x = var[:N]
v = var[N:2*N]
theta = var[2*N:3*N]
a = var[4*N:5*N]
xavg, vavg, thetavg, thdotavg = map(lambda z: (z[0:-1]+z[1:])/2, dynvars)
dx, dv, dthet, dthdot = map(lambda z: (z[1:]-z[0:-1])*dtinv, dynvars)
vres = dv - a[1:]
xres = dx - vavg
torque = -gL*np.sin(thetavg) + a[1:]*L*np.cos(thetavg)
thetdotres = dthdot - torque*Iinv
thetres = dthet - thdotavg

return x[0:1]-x0, v[0:1]-v0, theta[0:1]-th0, thetadot[0:1]-thd0, xres,vres, thetdotres, thetres
#return x[0:5] - 0.5

def getAlu(x, x0, v0, th0, thd0):
gt = np.zeros((2,N))
gt[0,:] = 0.1 # x is greaer than 0
gt[1,:] = -1 #veclotu is gt -1m/s
gt = gt.flatten()
lt = np.zeros((2,N))
lt[0,:] = 0.8
lt[1,:] = 1 # velocity less than 1m/s
lt = lt.flatten()

z = sparse.bsr_matrix((N, N))
ineqA = sparse.bmat([[sparse.eye(N),z,z,z,z],[z,sparse.eye(N),z,z,z]]) #.tocsc()
print(ineqA.shape)
#print(ineqA)
cons = constraint(forward.seed_sparse_gradient(x), x0, v0, th0, thd0)
A = sparse.vstack(map(lambda z: z.dvalue, cons)) #  y.dvalue.tocsc()
print(A.shape)
totval = np.concatenate(tuple(map(lambda z: z.value, cons)))
temp = A@x - totval

A = sparse.vstack((A,ineqA)).tocsc()

#print(tuple(map(lambda z: z.value, cons)))
print(temp.shape)
print(lt.shape)
print(gt.shape)
u = np.concatenate((temp, lt))
l = np.concatenate((temp, gt))
return A, l, u

#A = y.dvalue.tocsc()
#print(y.dvalue)
#sparse.hstack( , format="csc")
A, l, u = getAlu(x, 0.5, 0, 0, 0)
m = osqp.OSQP()
m.setup(P=P, q=q, A=A, l=l, u=u) #  **settings
results = m.solve()
print(results.x)

#plt.plot(results.x[3*N:4*N])
#plt.plot(results.x[4*N:5*N])
start = time.time()
iters = 100
for i in range(iters):
A, l, u = getAlu(results.x, 0.5, 0, 0, 0)
print(A.shape)
#print(len(A))
m.update(Ax=A.data, l=l, u=u)
results = m.solve()
end = time.time()
print("Solve in :", iters / (end-start) ,"Hz")

plt.plot(results.x[:N], label="x")
plt.plot(results.x[N:2*N], label="v")
plt.plot(results.x[2*N:3*N], label="theta")
plt.plot(results.x[4*N:5*N], label="force")
plt.legend()
plt.show()
#m.update(q=q_new, l=l_new, u=u_new)

## Approximating Compiling to Categories using Type-level Haskell: Take 2

The previous episode is here .

Summary: I’m trying to use typelevel programming in Haskell to achieve some of the aims of Conal Elliott’s compiling to categories GHC plugin. The types of highly polymorphic tuple functions are enough to specify the implementation. We aren’t going to be able to piggy-back off of GHC optimizations (a huge downside), but we can reify lambdas into other categories and avoid the scariness of plugins.

The current implementation github source is here

JESUS CHRIST.

Of course, Oleg already did it. This is a file where he builds the implementation of a polymorphic function from the type signature. Instead of tuples, he is focusing on higher order functions with deep nesting of (->).

The trick that I was missing is in the IsFunction typeclass at the very end, which is only achievable as a Incoherent instances.

I would never have had the courage to use an Incoherent instance if I hadn’t seen a higher authority do it first. It has turned out in my typelevel journey that many instances that I’ve been tempted to make overlapping or incoherent don’t actually have to be, especially with the availability of closed type families. I think you really do need Incoherent in this application because type variables stay polymorphic forever.

To the best of my knowledge, if you need to differentiate between a tuple type (a,b) and an uninstantiated polymorphic value a’ like we do when deconstructing the input type of our lambda, you need to use an Incoherent instance. Since a’ could hypothetically eventually be unified to some (a,b) we should not be able to do different things for the two cases without stepping outside the usual laws of typeclass resolution.

New features of the implementation:

• The new implementation does not require the V annotation of the input like previous version by using the previously mentioned. This is super cool because now I can throw the stock Prelude.fst into toCcc.
• I changed how the categorical implementation is built, such that it short circuits with an ‘id’ if a large structure is needed from the input, rather than deconstructing all the way to every piece of the input. Lots of other optimizations would be nice (vital?), but it is a start.
• I also implemented a FreeCat GADT so that we can see the implementation in ghci.
• I switched from using Data.Proxy to the type annotations extensions, which is a huge readability win.
• I added a binary application operator binApp, which is useful for encapsulating categorical literals as infix operators into your lambda expressions.
• General cleanup, renaming, and refactoring into library files.

A couple typelevel tricks and comments:

You often have to make helper typeclasses, so don’t be afraid to. If you want something like an if-then-else in your recursion, it is very likely you need a form of the typeclass that has slots for ‘True or ‘False to help you pick the instance.

If possible, you often want the form

(a~Something) => MyClass 'True a

rather than

Myclass 'True Something

The type inference tends to be better.

Here are some usage examples of toCcc.

example6 = toCcc (\x -> x) 'a'

example7 = toCcc @FreeCat (\(x, y) -> x)

example8 = toCcc @FreeCat (\(x, y) -> y)
example8andahalf = toCcc' (Proxy :: Proxy FreeCat) (\(x, y) -> y)
example9 = toCcc @FreeCat (\(x, y) -> (y,x))
example10 = toCcc @FreeCat (\(( x, z), y) -> (y,x))
swappo = toCcc @FreeCat $\((x, z), y) -> (x,(z,y)) example11 = toCcc @(->)$ \(x,y) -> binApp addC x y
example12 = toCcc @(->) $\(x,y) -> App negateC x -- infix synonyms (+++) = binApp addC (***) = binApp mulC example13 = toCcc @(->)$ \(x,(y,z)) -> x +++ (y *** z)

My partially implemented version of some of Conal’s category typeclasses. Should I switch over to using the constrained-categories package?

{-# LANGUAGE GADTs, StandaloneDeriving, NoImplicitPrelude  #-}

module Cat where
import Control.Category
import Prelude hiding ((.))

class Category k => Monoidal k where
par :: k a c -> k b d -> k (a,b) (c,d)

class Monoidal k => Cartesian k where
fst :: k (a,b) a
snd :: k (a,b) b
dup :: k a (a,a)

fan f g = (par f g) . dup

data FreeCat a b where
Comp :: FreeCat b c -> FreeCat a b -> FreeCat a c
Id :: FreeCat a a
Fst :: FreeCat (a,b) a
Snd :: FreeCat (a,b) b
Dup :: FreeCat a (a,a)
Par :: FreeCat a b -> FreeCat c d -> FreeCat (a,c) (b,d)

deriving instance Show (FreeCat a b)

instance Category FreeCat where
(.) = Comp
id = Id

instance Monoidal FreeCat where
par = Par

instance Cartesian FreeCat where
fst = Fst
snd = Snd
dup = Dup

instance Monoidal (->) where
par f g = \(x,y) -> (f x, g y)

instance Cartesian (->) where
fst (x,y) = x
snd (x,y) = y
dup x = (x,x)

class NumCat k where
mulC :: Num a => k (a,a) a
negateC :: Num a => k a a
addC :: Num a => k (a,a) a

instance NumCat (->) where
mulC = uncurry (*)
negateC = negate

The actual implementation of toCcc

{-# LANGUAGE DataKinds,
AllowAmbiguousTypes,
TypeFamilies,
TypeOperators,
MultiParamTypeClasses,
FunctionalDependencies,
PolyKinds,
FlexibleInstances,
UndecidableInstances,
TypeApplications,
NoImplicitPrelude,
ScopedTypeVariables #-}
module CCC (
CCC
, App (App)
, binApp
, toCcc'
, toCcc ) where

import Cat
import Data.Proxy
import Data.Type.Equality (type (==))
import Prelude (Bool(..))
import Control.Category

-- The unfortunate Incoherent instances I need to force polymorphic values
class IsTup a b | a -> b
instance {-# INCOHERENT #-} (c ~ 'True) => IsTup (a,b) c
instance {-# INCOHERENT #-} (b ~ 'False) => IsTup a b

data Leaf n = Leaf

data Z
data S a

data App f a = App f a

f  x = App f x
binApp f a b = App f (a,b)

class Cartesian k => CCC k a a' b' | a a' -> b' where
toCcc :: a -> k a' b'
instance (Tag a,
Build k a b a' b',
Cartesian k)
=> CCC k (a->b) a' b' where
toCcc f = build @k @a @b @a' @b' res where  -- build (Proxy :: Proxy labels) (Proxy :: Proxy b) res where
res = f val

toCcc' :: CCC k f a' b' => Proxy k -> f -> k a' b'
toCcc' _ f = toCcc f

class Tag a where
val :: a

instance (IsTup a flag, Tag' a Z flag n) => Tag a where
val = val' @a @Z @flag

class Tag' a n (flag :: Bool) n'| a n flag -> n' where
val' :: a

instance (IsTup a flaga,
IsTup b flagb,
Tag' a n flaga n'',
Tag' b n'' flagb n') => Tag' (a,b) n 'True n'  where
val' = (val' @a @n @flaga, val' @b @n'' @flagb)

instance (a ~ Leaf n) => Tag' a n 'False (S n)  where
val' = Leaf

type family Or a b where
Or 'True b = 'True
Or a 'True = 'True
Or 'False 'False = 'False

class In a b flag | a b -> flag where
instance (
((a,b) == c) ~ isc,
flag' ~ Or flaga flagb,
flag ~ Or flag' isc,
In a c flaga,
In b c flagb) => In (a,b) c flag
instance ((Leaf n == c) ~ flag) => In (Leaf n) c flag

class Build k input key a' b' | input key a' -> b' where
build :: Cartesian k => key -> k a' b'

instance (
iseq ~ ((a,b) == key),
In a key isinleft,
In b key isinright,
Cond k iseq isinleft isinright (a,b) key a' b'
) => Build k (a,b) key a' b' where
build key = cond @k @iseq @isinleft @isinright @(a,b) @key @a' key

instance (Leaf n ~ b, a' ~ b') => Build k (Leaf n) b a' b' where
build _ = id

class Cond k iseq isinleft isinright input key a b | iseq isinleft isinright input key a -> b where
cond :: Cartesian k => key -> k a b
-- Find the key is in the input
instance (a ~ b) => Cond k 'True x x input key a b where
cond _ = id
instance (Build k a key a' c', (a',b') ~ ab) => Cond k 'False 'True x (a,b) key ab c' where -- get those input types inferred baby!
cond key = (build @k @a @key @a' key) . fst
instance (Build k b key b' c', (a',b') ~ ab) => Cond k 'False 'False 'True (a,b) key ab c' where
cond key = (build @k @b @key @b' key) . snd

-- Otherwise destruct on the key
instance (Build k input key1 a' c',
Build k input key2 a' d') => Cond k 'False 'False 'False input (key1,key2) a' (c',d') where
cond (key1,key2) = fan (build @k @input @key1 @a' key1) (build @k @input @key2 @a' key2)

instance (Build k input key a' b',
f ~ k b' c')
=> Cond k 'False 'False 'False input (App f key) a' c' where
cond (App f key) = f . (build @k @input @key @a' key)

-- Could I replace almost everything with App? A very powerful construct
-- This is a of some relation to defunctionalization like in HList
-- Maybe I should build a typelevel FreeCat and then do compilation passes on it

{-
type family (StripLeaf a) where
StripLeaf (a,b) = (StripLeaf a, StripLeaf b)
StripLeaf (Leaf n a) = a
-}


text:
# Finding derivative of Z^(7/3)

Tags:
1. Sep 18, 2016

### Pouyan

Consider the principal branch of the function

f(z)= z7/3

Find f'(-i) and write it in the form a+bi

My attemp is :

I know zc = exp(c logz)
and the derivative of that is : (c/z) * exp(c Logz)
That is in this case (7/3)*(i) *exp((7/3)*Log-i) = f'(-i)
I know that Log(-i) = Log(1) + i(-pi/2)= -i pi/2
and exp((7/3)(-i pi/2)) = (cos(7pi/6)-i*sin(7pi/6))
and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

But I see the answer is : -(7/6)(1+isqrt(3))!
What is wrong with my algorithm ?

Last edited: Sep 18, 2016
2. Sep 18, 2016

### ehild

Include (i) into the exponent.

3. Sep 18, 2016

### Pouyan

-i is in the exponent

4. Sep 18, 2016

### Staff: Mentor

Is there some reason you can't use the derivative formula $\frac d {dz} z^k = k z^{k - 1}$?
Then $f'(-i) = \frac 7 3 (-i)^{4/3} = \frac 7 3 (e^{-i \pi/2})^{4/3}$
Simplify this last expression to get to the form in your textbook.

5. Sep 18, 2016

### Ray Vickson

Easier: if $f(z) = z^n$ then $f'(z) = n z^{n-1}$. Apply that to $n = 7/3$ and $z = -i$.

Anyway,
$$\exp((7/3)(-i \pi/2)) = \cos(7\pi/6)-i \sin(7\pi/6) =- \frac{\sqrt{3}}{2} + \frac{1}{2} i$$.

6. Sep 18, 2016

### Staff: Mentor

That's what I said.

7. Sep 18, 2016

### ehild

i is also a factor:
Write the multiplicative i in exponential form and collect the exponents.

8. Sep 18, 2016

### Pouyan

Thank you so much for your help. I actually mixed up this problem with the formula

Zm/n = exp((m/n)Log|z|)exp(i(m/n)(Arg z + 2kpi))
but now I see I could solve it easier

9. Sep 18, 2016

### Ray Vickson

For some reason, your message did not appear on my screen until after I pressed the "enter" key. That sort of thing happens to me quite often, and I have no idea why.

In fact, I did not see your message until I had logged off and later, logged on again!

format: plain


text:
forex software

Create and Test Forex Strategies

forex software

Help Contents

Tools

eas-guide:preset-indicators

Preset Indicators

Overview

The Preset Indicators are used by the Generator when composes new strategies.

When the Generator's option “The strategies use Preset Indicators” is enabled, each newly generated strategy starts with the preset indicators.

If you use the Reactor with Optimization on, it first composes new strategies with the corresponding Preset Indicators, but it is possible the Optimizer to change their numeric parameters. This happens when the “Optimize Preset Indicators” option is enabled.


text:
# System size dependence of associated yields in hadron-triggered jets

Abstract : We present results on the system size dependence of high transverse momentum di-hadron correlations at $\sqrt{s_{NN}}$ = 200 GeV as measured by STAR at RHIC. Measurements in d+Au, Cu+Cu and Au+Au collisions reveal similar jet-like correlation yields at small angular separation ($\Delta\phi\sim0$, $\Delta\eta\sim0$) for all systems and centralities. Previous measurements have shown that the away-side yield is suppressed in heavy-ion collisions. We present measurements of the away-side suppression as a function of transverse momentum and centrality in Cu+Cu and Au+Au collisions. The suppression is found to be similar in Cu+Cu and Au+Au collisions at a similar number of participants. The results are compared to theoretical calculations based on the parton quenching model and the modified fragmentation model. The observed differences between data and theory indicate that the correlated yields presented here will provide important constraints on medium density profile and energy loss model parameters.
Keywords :
Type de document :
Article dans une revue
Physics Letters B, Elsevier, 2010, 683, pp.123-128. <10.1016/j.physletb.2009.12.020>

http://hal.in2p3.fr/in2p3-00375987
Contributeur : Dominique Girod <>
Soumis le : jeudi 16 avril 2009 - 14:42:57
Dernière modification le : mercredi 27 juillet 2016 - 14:48:48

### Citation

B.I. Abelev, M. M. Aggarwal, Z. Ahammed, B. D. Anderson, D. Arkhipkin, et al.. System size dependence of associated yields in hadron-triggered jets. Physics Letters B, Elsevier, 2010, 683, pp.123-128. <10.1016/j.physletb.2009.12.020>. <in2p3-00375987>

### Métriques

Consultations de la notice


text:
# Stabilizer code

{{#if:|

}}

{{#if:April 2007||}} {{#if:|

}}

{{#if:||[[Category:Articles needing expert attention{{#if:| since {{{date}}}}}|Stabilizer code]]}} {{#if:|

}}

An $[n,k]$ stabilizer code is defined to be the vector space $V_S$ stabilized by a subgroup $S$ of $G_n$ such that $-I\notin S$ and $S$ has $n-k$ independent and commuting generators, $S=\langle g_1,\ldots,g_{n-k}\rangle$. These conditions are are required so that the vector space stabilized is non-trivial. Denote this code $C(S)$.

## References

{{#if:|

}}

{{#switch:||Talk=[[Category:Articles lacking sources {{#if:June 2007|from June 2007|}}]]{{#if:June 2007|{{#ifexist:Category:Articles lacking sources from June 2007||}}|}}|}}


text:
Partial wave analysis - incoming/outgoing?

1. Apr 13, 2013

JoePhysicsNut

In the chapter on partial wave analysis in Griffiths's Introduction to Quantum Mechanics, he considers a spherically symmetric potential and says that for large r, the radial part of Schrodinger's equation becomes,

$\frac{d^{2}u}{dr^{2}}≈-k^{2}u$

with a general solution of

$u(r)=C\exp{ikr}+D\exp{-ikr}$.

He then says the first term represents the outgoing wave and the second term is the incoming wave. Why is that the case? These two differ by $\pi$ in the complex plane, but I don't see how that enables one to make the "incoming"/"outgoing" distinction.

2. Apr 13, 2013

fzero

Act on the solution with the radial momentum operator $\hat{p}_r = -i\hbar \partial_r$ and compare the signs of the two terms.

3. Apr 13, 2013

JoePhysicsNut

Thanks! The operator yields the momentum for the first term and (-1)*momentum for the second. A negative magnitude for momentum does not make sense, so therefore it is to be evaluated for times t<0 making it the incoming wave. Is that the argument?

4. Apr 13, 2013

fzero

We're computing the momentum, not just the magnitude. If $k$ is real, then the magnitude of momentum is always positive. We're also not discussing time dependence here. The argument is simply that, if $k>0$, then one solution carries momentum in the $+r$ direction, while the other carries it in the $-r$ direction. If we did add in the time dependence, we could see this more explicitly, but it isn't necessary.

5. Apr 15, 2013

Meir Achuz

A factor $e^{-i\omega t}$ is understood. This makes one form outgoing and the other ingoing.


text:
# scipy.special.roots_chebyu¶

scipy.special.roots_chebyu(n, mu=False)[source]

Gauss-Chebyshev (second kind) quadrature.

Computes the sample points and weights for Gauss-Chebyshev quadrature. The sample points are the roots of the n-th degree Chebyshev polynomial of the second kind, $$U_n(x)$$. These sample points and weights correctly integrate polynomials of degree $$2n - 1$$ or less over the interval $$[-1, 1]$$ with weight function $$f(x) = \sqrt{1 - x^2}$$.

Parameters: n : int quadrature order mu : bool, optional If True, return the sum of the weights, optional. x : ndarray Sample points w : ndarray Weights mu : float Sum of the weights

#### Previous topic

scipy.special.roots_chebyt

#### Next topic

scipy.special.roots_chebyc


text:
# Pang Yan Han's blog

## Note on Think Stats: Given the PMF of a random variable X, compute the variance of X

23 Jul 2015, by Pang Yan Han

Disclaimer: Opinions expressed on this blog are solely my own and do not express the views or opinions of my employer(s), past or present.

This week, I started using some free time to read Think Stats by Allen B. Downey. So far it’s been a pretty good book. I was reading the start of Chapter 2.4, where the author says to download the Pmf.py file from http://thinkstats.com/Pmf.py . Curious of what’s inside, I read through the code and encountered this method in the Pmf class:

def Var(self, mu=None):
"""Computes the variance of a PMF.

Args:
mu: the point around which the variance is computed;
if omitted, computes the mean

Returns:
float variance
"""
if mu is None:
mu = self.Mean()

var = 0.0
for x, p = self.d.iteritems():
var += p * (x - mu)**2
return var

Without going into details of the Pmf.Mean method and what the Pmf.d dictionary stores, I hope you trust me and the author that Pmf.Mean computes the mean of the PMF of a random variable $X$ correctly using the following formula:

$$\mathbb{E} [X] = \sum_{i=1}^{n} p_i X_i$$

and that Pmf.d is a dictionary which maps values the random variable can take on, to the probabilities of appearance.

What kind of surprised me about the Pmf.Var code above was this part:

var = 0.0
for x, p = self.d.iteritems():
var += p * (x - mu)**2
return var

which translates to:

$$Var(X) = \sum_{i=1}^{n} p_i * (X_i - \mu)^{2}$$

Ok, I gotta admit, my Probability chops are kind of rusty. To cut myself some slack, I did not recall doing exercises to compute variance from a pmf, nor did I recall seeing this formula in the lecture notes and textbooks I’ve used for several introductory Probability and Statistics classes. Or maybe my memory is just plain bad. (See Afterthoughts section). So I was doubtful and the one thought going on in my mind is: Is that really how you compute the variance of a random variable $X$ given its pmf? I did a little googling on computing variance and one of the top few results was this familiar formula: $Var ( X ) = \mathbb{E} [ X^2 ] - ( \mathbb{E} [ X ] )^{2}$

which looks quite different from $\sum_{i=1}^{n} p_i * (X_i - \mu)^2$. Could it be that they’re actually equivalent? Let’s prove it.

We want to show that:

$$Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = \sum_{i=1}^{n} p_i * (X_i - \mu)^2$$

It’s actually easier to start from $\sum_{i=1}^{n} p_i * (X_i - \mu)^2$, so let’s start from that.

Proof:

$$\sum_{i=1}^{n} p_i * (X_i - \mu)^2 \\ = \sum_{i=1}^{n} p_i * ({X_i}^2 - 2 X_i \mu + {\mu}^2) \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 - 2 p_i X_i \mu + p_i {\mu}^2] \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 ] - 2 \mu \sum_{i=1}^{n} [ p_i X_i ] + {\mu}^2 \sum_{i=1}^{n} p_i \\ = \mathbb{E}[X^2] - 2 \mu \mathbb{E}[X] + {\mu}^2 \cdot 1 \\ = \mathbb{E}[X^2] - 2 \mu \cdot \mu + {\mu}^2 \\ = \mathbb{E}[X^2] - {\mu}^2 \\ = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$

A proof starting from $\mathbb{E}[X^2] - (\mathbb{E}[X])^2$ requires a little bit of trickery, namely the use of $a + (-a) = 0$, but it’s not too complicated, so let’s do it.

Proof:

$$\mathbb{E}[X^2] - (\mathbb{E}[X])^2 \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 ] - \mathbb{E}[X] \cdot \mathbb{E}[X] \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 ] - \mathbb{E}[X] \cdot \sum_{i=1}^{n} [ p_i X_i ] \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 ] - \mu \sum_{i=1}^{n} [ p_i X_i ] \\ = \sum_{i=1}^{n} [ p_i {X_i}^2 - \mu p_i X_i ] \\ = \sum_{i=1}^{n} [ p_i X_i (X_i - \mu) ] \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu + \mu) \cdot (X_i - \mu) ] \\ = \sum_{i=1}^{n} [ p_i ({X_i}^{2} - X_i \mu - X_i \mu + {\mu}^{2} + X_i \mu - {\mu}^{2}) \\ = \sum_{i=1}^{n} [ p_i (({X_i}^{2} - X_i \mu - X_i \mu + {\mu}^{2}) + (X_i \mu - {\mu}^{2})) ] \\ = \sum_{i=1}^{n} [ p_i (({X_i}^{2} - 2 X_i \mu + {\mu}^{2}) + (X_i \mu - {\mu}^{2})) ] \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 + p_i (X_i \mu - {\mu}^{2}) ] \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 ] + \sum_{i=1}^{n} [ p_i (X_i \mu - {\mu}^{2}) ] \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 ] + \mu \sum_{i=1}^{n} [ p_i X_i ] - \sum_{i=1}^{n} [ p_i {\mu}^{2} ] \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 ] + \mu \cdot \mu - {\mu}^2 \sum_{i=1}^{n} p_i \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 ] + {\mu}^2 - {\mu}^2 \cdot 1 \\ = \sum_{i=1}^{n} [ p_i (X_i - \mu)^2 ]$$

So there we have it, $Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = \sum_{i=1}^{n} p_i (X_i - \mu)^2$. More importantly, $\sum_{i=1}^{n} p_i (X_i - \mu)^2$ is indeed the correct way to compute the variance of a random variable $X$ given its pmf.

## Afterthoughts

Halfway through this post I concluded that it’s really my poor memory. Granted, I typically compute variance using $\mathbb{E}[X^2] - ([X])^2 \$ instead of $\sum_{i=1}^{n} p_i * (X_i - \mu)^2$, as do most students in an introductory Probability class I guess. Or you just use a computer program to do it and not worry about what goes on in the background. Ah wells. More reason to brush up on the fundamentals of Probability. It’s a good exercise though =)


text:
# How do you write the equation of the line with slope -4 and y - intercept (0,3)?

Here slope $m = - 4$
$y - {y}_{1} = m \times \left(x - {x}_{1}\right)$
$\implies y - 3 = - 4 \times \left(x - 0\right)$
$\implies y + 4 x - 3 = 0$


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2019, 18:45

# Problem Solving (PS)

 new topic Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  4   5   6   7   8   9   10   11  ...  528    Next Search for:
Filter
Topics Author Replies   Views Last post

Forum Experts

 EMPOWERgmat PS Forum Expert - Ask Me Anything about PS and Math   Go to page: 1, 2 Tags: Source: EMPOWERgmat
EMPOWERgmatRichC

26

4821

26 Oct 2019, 19:33

 Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math   Go to page: 1 ... 8, 9, 10 Tags: Source: Veritas Prep

190

9154

02 Dec 2019, 21:57

Announcements

522
ER
 100 Hardest and easiest questions for PS as of November 2019
Bunuel

12

104376

29 Jul 2019, 07:22

1937
ER
 GMAT PS Question Directory by Topic & Difficulty
bb

0

755645

22 Feb 2012, 11:27

## GMAT Club Member Recommended Prep Products:

Topics

5
 * represents a certain function for all real numbers. If (1*)*=6
FilipaFontao

1

630

08 Sep 2018, 05:07

41
ER
 \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?
faiint

10

7304

22 Aug 2019, 04:09

8
ER
 &, #, and each represents a different digit, and & multiplied by #
Bunuel

8

1595

27 Dec 2016, 16:42

ER
 is defined as (2x+y)/(x-2y). What is the value of <1/2, 2/3 >-
MathRevolution

3

217

13 Sep 2019, 09:09

ER
 is the largest integer less than or equal to x and [x] is the smal
MathRevolution

5

852

19 Jul 2018, 05:27

16
ER
 "A" and "B" run around a circular track starting from the same point s

8

3616

29 Oct 2018, 04:14

12
ER
 "Ice-Cold" Ice-cream factory produces only tricolor ice-cream products
reto

6

3444

19 Jun 2019, 10:55

18
ER
 "Randy's" bar serves customers every weeknight from 5 pm to
WoundedTiger

11

4318

24 Sep 2018, 09:59

2
ER
 "Tires R' us" have 4 times more tires for sale than any regular tire
Bunuel

6

277

13 Sep 2019, 16:41

9
ER
 "You were a quarter of my age when I was twice of your present age"
0akshay0

5

2230

15 Jan 2018, 08:34

9
ER
 #p is defined as 2p-20 for any number p. What is p, if #(#(#p))=4?
reto

6

2290

13 Feb 2019, 03:39

153
ER
 † and ¥ represent nonzero digits, and (†¥)² - (¥†)²   Go to page: 1, 2 Tags: Difficulty: 700-Level,  Number Properties,  Source: Veritas Prep
WoundedTiger

25

17417

05 Feb 2019, 20:48

21
 ∆ABC is a right triangle in XY-plane such that AB is parallel to the
disenapati

3

786

24 Jan 2019, 10:44

2
ER
 ∆ABC is a right-angled isosceles triangle, and ∠B is the right angle
Bunuel

6

557

13 Oct 2018, 01:53

11
ER
 | (|a| + 4)*(|b| - 3) | = 16. How many pairs of integers (a,
chondro48

2

576

07 Aug 2019, 03:58

25
ER
 | 1 - | 2 - | 3 - | 4 – π | | | |
MathRevolution

4

823

18 Nov 2019, 07:34

4
ER
 |-4| (|-20|-|5|)= ?
Abakup

8

7278

03 Jul 2019, 00:48

13
ER
 |−4|(|−20| − |5|) =
Bunuel

7

3788

13 Sep 2019, 23:09

16
ER
 |√3-2|+|3-√2|+|5+√3|+|1-√2|=?
MathRevolution

5

940

02 Oct 2018, 05:52

11
ER
 |1 - √2| + |1 - √3| - |√2 + √3| = ?
MathRevolution

10

1505

06 Feb 2019, 11:57

ER
 |2-3|-|2-5||=?
MathRevolution

3

845

09 Oct 2018, 10:40

19
 |a+b|/|a|+|b| + |b+c|/|b|+|c| +|c+a|/|c|+|a|=M
nick1816

10

827

28 Sep 2019, 05:02

51
ER
 |a|=|b|, which of the following must be true :
himanshuhpr

14

9618

19 Sep 2019, 03:32

15
ER
 |x + 1| < |x^2 + 2x + 2|
gmatbull

16

3224

01 Sep 2016, 17:15

40
ER
 |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and
EgmatQuantExpert

17

3089

01 Nov 2019, 03:13

26
ER
 |x + 3| = |y -4|, where x and y are non-zero integers. If |x| < 5 and
gmatbusters

10

1293

07 Jul 2019, 18:03

82
ER
 |x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| +
papillon86

17

14469

26 Aug 2019, 00:39

306
ER
 |x+3| - |4-x| = |8+x|. How many solutions does the equation   Go to page: 1, 2, 3, 4 Tags: Difficulty: 700-Level,  Absolute Values/Modulus
guerrero25

60

33332

22 Jul 2019, 11:32

13
ER
 −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most acc
Bunuel

5

835

04 Sep 2019, 22:24

ER
 −2^4 − (x^2 − 1)^2 =
Bunuel

2

242

18 Apr 2019, 12:28

 −2( 3 − x [(5 + y − 2)/x] − 7 + 2 ⋅3^2) =

1

373

08 Jul 2019, 09:42

37
ER
 √[2√63 + 2/(8+3√7)] =
rahulms

19

10434

20 Aug 2019, 05:12

10
ER
 √{(4.8)(10^9)} is closest to which of the following? A. 7,000
MathRevolution

7

2978

12 Nov 2018, 00:22

2
ER
 √2209?
AbdulMalikVT

4

208

02 Sep 2019, 19:32

7
ER
 √2450 =
Bunuel

6

1443

12 Nov 2018, 18:29

7
ER
 √3 is what percent of √12?
MathRevolution

8

998

29 Jul 2018, 01:52

6
ER
 √3+√2/√3-√2 is approximately between which two numbers?
MathRevolution

7

1237

20 Jan 2017, 09:14

5
ER
 √5 percent of 5√5 =
Bunuel

4

8660

15 Aug 2019, 12:28

7
ER
 √576800 is between
Bunuel

4

1602

04 Jul 2018, 01:02

ER
 √75+2√3=? A. 3√3 B. 5√3 C. 7√3 D. 8√3
MathRevolution

3

791

18 Jun 2017, 18:01

1
ER
 √m4=64 and m>0, what is the value of cubic root of m?
MathRevolution

7

825

14 Jul 2017, 10:36

ER
 √r + 2/√r = 4. What is the value of r + 4/r?
MathRevolution

2

285

28 Mar 2019, 01:41

11
ER
 ∛(328 - 800 - 48) is
GMATPrepNow

7

2064

18 Aug 2019, 06:53

ER
 ♠n denotes the number obtained when n is rounded to the nearest tenth.
Bunuel

3

2444

29 Aug 2016, 12:50

2
 $5783 is divided among A, B and C in such a way that if$28 ,$37,$18
LUCIFER1703

2

302

28 Jun 2019, 11:44

1
 $1200 is invested at a given interest rate for two years. WoundedTiger 5 2737 19 Apr 2018, 12:23 5 ER $2,000 is deposited in a savings account that pays 6% annual interest
Bunuel

4

2253

10 Sep 2018, 12:07

1
ER
 $2000 is deposited into a savings account that earns interest at the Bunuel 3 681 05 Feb 2018, 10:27 4 ER $3000 is distributed among A, B and C such that A gets 2 3 rd of w
cfc198

4

569

02 Jul 2019, 05:27

7
ER
 $4800 becomes$6000 in 4 years at a certain rate of compound interest.
mangamma

1

434

07 Mar 2019, 07:42

 new topic Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous    1  ...  4   5   6   7   8   9   10   11  ...  528    Next Search for:


text:
Found 2 result(s)

### 27.09.2018 (Thursday)

#### AmplitudesandhiddensymmetriesinsupersymmetricChern-Simonsmattertheories

Regular Seminar Karthik au:Inbasekar'>Karthik Inbasekar (Tel Aviv U.)

 at: 14:00 QMWroom G O Jones 610 abstract: Chern-Simons theories coupled to fundamental matter have a wide variety of applications ranging from Quantum Hall effect to Quantum gravity via AdS/CFT. These theories enjoy a strong weak duality that has been tested to a very good accuracy via large N computations, such as thermal partition functions, and S matrices. Supersymmetric Chern-Simons theories are equally interesting since they have a self duality and relate to non-supersymmetric theories via RG flows. In the N=2,3 supersymmetric Chern-Simons matter theories, the four point amplitude computed to all orders in the 't Hooft coupling is not renormalised! It is a unique situation in a quantum field theory that the scattering amplitude doesn't receive loop corrections. This indicates the presence of powerful symmetry structures within the theory. This also suggests that higher point amplitudes may be easier to compute using four point amplitudes as building blocks. These higher point amplitudes not only serve as a testing tool for duality but also a probe into the symmetry structure of the theory. As a first step towards this goal, we begin by computing arbitrary n point tree level amplitudes in the N=2 theory via BCFW recursion relations. We then show that the four point tree level amplitude enjoys a dual superconformal symmetry. Since the all loop four point amplitude is tree level exact, it follows that the dual superconformal symmetry is exact to all loops. This is in contrast to highly supersymmetric examples such as N=4 SYM and N=6 ABJM, where the dual superconformal symmetry is in general anomalous. Furthermore, we show that the superconformal and dual superconformal symmetries generate an infinite dimensional Yangian symmetry for the four point amplitude. If these symmetries persist to higher point amplitudes, this suggests that the N=2 superconformal Chern-Simons matter theory may be integrable.

### 26.09.2018 (Wednesday)

#### AmplitudesandhiddensymmetriesinsupersymmetricChern-Simonsmattertheories

Regular Seminar Karthik au:Inbasekar'>Karthik Inbasekar (Tel Aviv University)

 at: 14:00 ICroom H503 abstract: Chern-Simons theories coupled to fundamental matter have a wide variety of applications ranging from Quantum Hall effect to Quantum gravity via AdS/CFT. These theories enjoy a strong weak duality that has been tested to a very good accuracy via large N computations, such as thermal partition functions, and S matrices. Supersymmetric Chern-Simons theories are equally interesting since they have a self duality and relate to non-supersymmetric theories via RG flows. In the N=2,3 supersymmetric Chern-Simons matter theories, the four point amplitude computed to all orders in the 't Hooft coupling is not renormalised! It is a unique situation in a quantum field theory that the scattering amplitude doesn't receive loop corrections. This indicates the presence of powerful symmetry structures within the theory. This also suggests that higher point amplitudes may be easier to compute using four point amplitudes as building blocks. These higher point amplitudes not only serve as a testing tool for duality but also a probe into the symmetry structure of the theory. As a first step towards this goal, we begin by computing arbitrary n point tree level amplitudes in the N=2 theory via BCFW recursion relations. We then show that the four point tree level amplitude enjoys a dual superconformal symmetry. Since the all loop four point amplitude is tree level exact, it follows that the dual superconformal symmetry is exact to all loops. This is in contrast to highly supersymmetric examples such as N=4 SYM and N=6 ABJM, where the dual superconformal symmetry is in general anomalous. Furthermore, we show that the superconformal and dual superconformal symmetries generate an infinite dimensional Yangian symmetry for the four point amplitude. If these symmetries persist to higher point amplitudes, this suggests that the N=2 superconformal Chern-Simons matter theory may be integrable.


text:
Abstract

This article presents a theoretical and numerical study of an innovative joint using cold-formed steel sections. The motivation for the study of this connection is the ease of manufacturing and assembly that it provides. The profiles are made of cold-formed lipped channel sections, which are welded to form closed built-up sections on the columns and open built-up lipped sections to the beams. The beams use endplates connected by bolts (threaded bars) to the columns. The study evaluates the connection’s initial stiffness of 19 models, where the following parameters were varied: the thickness of the profiles and endplates, the height of the column sections and the diameter of the bolts. A theoretical and a numerical study were developed: the numerical study was performed using finite elements through the commercial software ANSYS, whereas the theoretical study was made based on the component method, prescribed by Eurocode 3, that does not include the design of the connection analyzed herein. Thus, aiming to enable the design of joints composed of cold-formed lipped channel sections, the analysis results were compared and an adjustment coefficient, proportional to the slenderness of the column’s plates, was proposed. The coefficient was introduced to the stiffness component that represents the column web in compression in the mechanical model. The ratio between the coefficients’ numerical and theoretical values presented a maximum variation of 11%, which was considered satisfactory.

Keywords:
cold-formed; numerical analysis; semi-rigid joints

1. Introduction

The beam-column connections of steel profiles are commonly considered as rigid or pinned. According to Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010., pinned joints should be capable of transmitting the internal forces, without developing significant moments, which might adversely affect the members or the structure as a whole. On the other hand, joints classified as rigid may be assumed to have sufficient rotational stiffness to justify analysis based on full continuity. These simplifications are used in designs due to the difficulty of analyzing and scaling the real behavior of the joints. However, it is known that some of the connections do not have this idealized behavior and have partial transmission capacity of the bending moment and rotation. These joints are called semi-rigid and its use should be an economic solution for structural designs.

The first study analyzing rotational stiffness was developed by Wilson and Moore (1917)WILSON W. M., MOORE, H. F. Tests to determine the rigidity of riveted joints in steel structures. University of Illinois, 1917. (Bulletin n.104).. Since then, numerous works addressed the subject, contributing to the development of normative prescriptions, either using the similarity of the beam-to-column endplate joints with the behavior of T-stub connections, or based on finite elements analysis. Krishnamurthy (1978)KRISHNAMURTHY, N. A fresh look at bolted end-plate behaviour and design. AISC Engineering Journal, v.15, 2nd Quarter, p. 39-49, April, 1978. recommended a design method for endplate connections, and Ribeiro et al. (1998)RIBEIRO, L. F. L., GONÇALVES, R. M., CASTIGLIONI, C. A. Beam-to-column end plate connections - an experimental analysis. Journal of Constructional Steel Research, Surrey. London, v. 46, p. 264-266, 1998. performed several experimental tests and analyzed the influence of the endplate thickness and the bolt diameter in these joints, using Krishnamurthy’s recommendations. Lima et al. (2004)LIMA, L. R. O. DE, VELLASCO, P. C. G. da S. , SILVA, L. A. P. S. da, ANDRADE, S. A. L. de. Experimental evaluation of extended endplate beam-to-column joints subjected to bending and axial force. Engineering Structures, Estados Unidos, v. 46, n.7, p. 1-15, 2004. and Silva et al. (2004)SILVA, L. A. P. S. da, LIMA, L. R. O. DE , DE LIMA, L.R.O. , VELLASCO, P. C. G. da S. , ANDRADE, S. A. L. de (.) Behaviour of flush end-plate beam-to-column joints under bending and axial force. Steel and Composite Structures. Coréia, v. 4, n.2, p. 77-94, 2004. studied the influence of axial forces on beam-to-column joints, showing that the presence of axial force on the beam significantly modified the joint response.

Garifullin et al. (2017)GARIFULLIN, M., PAJUNEN, S., MELA, K., HEINISUO, M., HAVULA, J. Initial in-plane rotational stiffness of welded RHST joints with axial force in main member. Journal of Constructional Steel Research, v. 139, p. 353-362, 2017. evaluated the existing calculation approach for the initial rotational stiffness of welded rectangular hollow sections T joints. Based on experimental results, the authors proposed an improved equation for the component ‘main member flange in bending’. Zhao et al. (2017)ZHAO, X., DAI, L., WANG, T., SIVAKUMARAN, K. S., CHEN, Y. A theoretical model for the rotational stiffness of storage rack beam-to-upright connections. Journal of Constructional Steel Research, v. 133, p. 269-281, 2017. presented a mechanical model based on the component method to predict the initial rotational stiffness of boltless connections in cold-formed steel storage racks. The comparisons between the numerical and experimental results showed that the mechanical model can properly evaluate the initial rotational stiffness of the studied connections. Furthermore, Bučmys et al. (2018)BUČCMYS, Ž., DANIŪNAS, A., JASPART, J., DEMONCEAU, J. A component method for cold-formed steel beam-to-column bolted gusset plate joints. Thin-Walled Structures, v. 123, p. 520-527, 2018. studied cold-formed steel bolted gusset plate connections based on the component method approach. A three-spring mechanical model and a technique to calculate joint stiffness were presented. Based on the experimental and numerical results, the authors showed that the stiffness coefficients that are listed in Eurocode 3 part 1-8 could be applied to the studied joints.

Within this context, this article analyzes the initial stiffness of an innovative joint composed by cold-formed steel sections, using the component method proposed in Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010.. This analytical method evaluates the stiffness of some types of predefined connections that allow presenting the known moment-rotation behavior. In this work, the equations proposed by the Eurocode method received an adjustment coefficient proportional to the slenderness of the column’s plates to assimilate the difference in geometric characteristics of this new joint. The results of this new formulation were compared with finite element numerical models, developed in software the ANSYS 12.0 (2012)ANSYS Version 12.0. Theory Reference, ANSYS, Inc., 2012.. The results were considered satisfactory and will be presented in the next sections.

2. Numerical analysis

2.1 Geometric properties

The profiles used in the numerical and theoretical analyses are made of cold-formed lipped channel sections, which are welded to form a closed built-up section on the columns and open built-up lipped section to the beams, which have welded endplates. These plates are connected to the column through six bolts (threaded bars) and nuts, as shown in Figure 1.

Figure 1
Beam-column connection.

The study of the connection was performed through the variation of the geometry of the components. The models considered are presented in Table 1. The beam (tw), column (tc) and endplate (tp) thicknesses were varied, along with the height of the tubular columns (hc) and bolt diameters (D). These variations were performed because these parameters directly contribute to the rigidity of the components present in the connection.

Table 1
Connection components geometry.

2.2 Properties of materials

The numerical modeling used materials with the following mechanical properties: modulus of elasticity (E) equals to 205 GPa, Poisson coefficient (u) equals to 0.3, yield strength equals to 250 MPa, equivalent to steel grade A36, for all elements. As in Maggi (2005)MAGGI, Y. I., GONÇALVES, R. M., LEON, R. T., RIBEIRO, L. F. L. Parametric analysis of steel bolted endplate connections using finite element modeling. Journal of Constructional Steel Research, v. 61, n.5, p. 689-708, 2005., the tangent modulus (Et) used in this study is 10% of the longitudinal modulus of elasticity (E). The material’s nonlinearity was considered with a bilinear stress-strain diagram, which divides the elastic and plastic behavior into two straight-line segments.

2.3 Modeling method

The numerical models were performed through ANSYS (2012)ANSYS Version 12.0. Theory Reference, ANSYS, Inc., 2012., a commercial software that uses the Ansys Parametric Design Language (APDL). This language uses text files where commands are entered to set the following: parameters that define the geometry, material types and elements, mesh, loads, and boundary conditions.

The shell element "SHELL181" was used on the beams and columns. In plates and bolts, the most suitable was the volumetric element "SOLID95". Finally, to represent the geometric discontinuities between the components, the elements used were "TARGE170", "CONTA175" and "CONTA174", always used in pairs in the connection. The contact elements are used between:

• Bolt and endplate;

• Endplate and column flange;

• Column stiffeners;

• Bolt and column.

The characteristics of the mentioned elements are: “SHELL181” - a plan element with four nodes with six degrees of freedom and translations in the x, y and z axes; “SOLID95” - a volumetric element with twenty nodes with translations in the x, y and z axes. The "CONTA174" is used when the base elements ("SOLID95") are volumetric and "CONTA175" are used with surface elements ("SHELL181"), both used with the target element “TARGE170”.

3. Theoretical analysis

3.1 Component method

The theoretical analysis was performed according to the component method, proposed by Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010.. The component method is based on the plastic distribution of tensile forces in the bolt-rows. This means that the force on any row is determined not only by its distance to the line of rotation of the connection, but also by the geometry of each bolt-row. The coefficients are associated in series in the lines and later associations are performed in parallel between them.

3.2 Connection coeffificients

The coefficients used in the connection are presented herein and the calculation procedure proposed by the method of the components are presented in Figure 2. The component method for the joint analyzed can be applied considering three coefficients: the column web in compression (k2), the endplate in bending (k5) and bolts in tension (k10), as shown in the full mechanical model presented.

Figure 2
Rotational stiffness calculation procedure.(a) Beam-to- column joint, (b) Mechanical model.

The effective stiffness of the associated “springs” in series is calculated for each bolt-row by Equation 1:

(1) $k eff , n = 1 ∑ 1 k i , n$

Where ki,n represents the value of the i-th coefficient in the n-th bolt-row. The next step is to calculate the equivalent stiffness, keq, of the bolt-row, now associating in parallel, by Equation 2:

(2) $k eq = Σ k eff , n · h n z eq$

Where hn is the distance between the bolt-row and the compression center; keff,n is the equivalent stiffness of the associated line in series and zeq is given by the Equation 3:

(3) $z eq = Σ k eff , n · h n 2 Σ k eff , n · h n$

Finally, the rotational stiffness is determined by Equation 4:

(4) $S j = E · z 2 μ 1 k 1 + 1 k 2 + 1 k eq$

Where E is the modulus of elasticity of the steel; k1 and k2 are the stiffness coefficients of components 1 and 2 respectively; z is the lever arm and µ is the ratio of rigidities Sj, ini/ Sj obtained through Equation 5:

(5) $if M j , Sd ≤ 2 3 · M j , Rd = > μ = 1 if 2 3 · M j , Rd ≤ M j , Sd ≤ M j , Rd = > μ = 1 . 5 . M j , Sd M j , Rd 2 . 7$

Although the method presents 15 components, only 3 interfere in the initial stiffness of the connection object of this work: k2 - Column web in compression; k5 - Endplate in bending and k10 - Bolts in tension.

3.2.1 Coefficient k2 - Column web in compression

The stiffness coefficient for the non-stiffened column web in compression is given by Equation 6:

(6) $k 2 = 0 . 7 . b eff , c , wc . t c h c$

Where hc is the column height, and beff,c,wc is given by the expression beff,c,wc = tw+ 2√2 ap+ 5(tc+s)+ sp; where: tc is the column web thickness; ; tw is the beam web thickness; ap is the weld throat between the flange of the beam and the endplate; s is the weld throat in welding profiles; and sp is the length obtained by diffusion at 45º on the endplate.

3.2.2 Coefficient k5 - Endplate in bending

The stiffness coefficient for the non-stiffened endplate in bending component, submitted to bending is similar to the previous coefficient and is given by the Equation 7:

(7) $k 5 = 0 . 9 · l eff · t p 3 m 3$

Where leff and m are geometric coefficients and tp is the endplate thickness.

3.2.3 Coefficient k10 - Bolts in tension

The stiffness coefficient relative to the bolts in tension is given by the Equation 8:

(8) $k 10 = 1 . 6 · A s L b$

Where Lb is the bolt elongation length (Equation 9) and As cross-section area of the bolt.

(9) $L b = h c + 2 t p$

4. Results

4.1 Numerical results

Considering the 19 models described in Table 1, four comparisons are made in Figure 3, that shows the connections’ Moment-Rotation behavior. In Figure 3 (a), the endplate thickness is varied in the 4 models compared; in Figure 3 (b), the column and beam thicknesses are both varied in the 7 models analyzed, while in Figure 3 (c) the diameter of the bolts and in Figure 3 (d) the column height are the parameter that are altered.

Figure 3
Variation of the initial stiffness of the models.

The deformed shapes of the joint are presented in Table 2, that shows the deformation of the 3 components that interfere in the initial stiffness of the connection; column 1 shows the initial deformation at a stage where the bending moment was 160 N.m, and column two and three where the moments were 720 N.m and 2000 N.m, respectively.

Table 2
Displacement vector sum of the connection's components (mm).

4.2 Coefficient adjustment proposition

Since Equation 6 does not represent the k2 component, due to different tubular profile buckling modes of an open built-up lipped section, it was necessary to add a coefficient “Q”, obtained based on the numerical analysis data. This coefficient, given by Equation 10, is proportional to the slender profile, which correlates the profiles presented by the component method used by Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010., with formed lipped channel sections used in the present study.

(10) $Q = 1 25 · h c t c$

Thus, the coefficient k2 is replaced by Equation 11, considering two column’s webs (2tc).

(11) $k 2 = 1 25 · h c t c · 0 . 7 . b eff , c , wc . 2 t c h c$

4.3 Evaluation of the initial stiffness of the connection

The multiplication of the coefficient “Q” in the formulation of k2, and the modifications presented, provided an equation very well estimated in the numerical analysis. Figure 4 shows the representative curve of the theoretical and numerical results of LVC01 model, containing the curves with and without the consideration of the coefficient “Q” and the limits of rigidity proposed by Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010..

Figure 4
Initial stiffness evaluation for the theoretical and numerical LVC01 model.

Finally, Table 3 shows the initial stiffness value for the theoretical and numerical results, the last column presents the ratio between the theoretical (considering “Q”) and numerical model.

Table 3
Ratio between theoretical and numerical results.

5. Discussion

From the data in Figure 3, it is possible to notice that the endplate thickness is an important factor responsible for the connections’ stiffness behavior, mainly because it can be subjected to the occurrence of bending, as shown in Table 2, that causes displacements in the beam. Likewise, the raise of the column and beam thicknesses increases connection rigidity, as shown in Figure 3 (b), due to higher values for the moment of inertia that prevent greater bending effects in the column webs.

The diameter of the bolts was another determinant factor for the analysis, considering that in two of the bolts lines, they were subjected to tension. In Figure 3 (b), it is observed that there was no direct relationship between the increase of the bolt diameter and the connections’ rigidity; in fact, the cases with higher diameter values presented a lower rigidity behavior, which can be explained by the fact that the higher diameter values provided less bending in the bolts located in the lower region of the connection, making the column web receive all of the load coming from the lower flange of the beam.

Even though the column height variation did not represent high values of dispersion in the Moment-Rotation curves, as shown in Figure 3 (d), it is noticed that the raise of the height, that represents a higher value of slenderness (height/thickness), generates a lower value of rigidity, leading to the effect seen in the column web shown in Table 2.

To enable the design of the connection analyzed herein, from the numerical analysis data, a coefficient “Q” was proposed, and its incorporation in the component method provided close results to the numerical values, as shown in Figure 4 and Table 3.

6. Conclusions

The general component method is used to calculate the stiffness coefficients of specific components and is limited by some criteria. This article presented a connection between cold-formed sections, and through the addition of some related coefficients, it was possible to use the mechanical model, used by Eurocode 3 (2010)EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010., in a connection outside the established limits.

The adjustment coefficient “Q”, proportional to the slenderness of the column’s plates, was introduced to the coefficient k2 (the stiffness component of the mechanical model for the non-stiffened column web in compression) and provided the ratio between numerical and theoretical coefficients with a maximum variation of 11%, which was considered satisfactory.

Even though the study herein presented is limited to specific geometries of cold-formed lipped channel sections, the results showed that the design of this type of connection with the consideration of its initial stiffness can be allowed, using the component method with the adjustment coefficient proposed.

Acknowledgments

The authors would like to thank UFOP, CAPES and CNPq for the financial support received for the development of this research.

References

• ANSYS Version 12.0. Theory Reference, ANSYS, Inc., 2012.
• BUČCMYS, Ž., DANIŪNAS, A., JASPART, J., DEMONCEAU, J. A component method for cold-formed steel beam-to-column bolted gusset plate joints. Thin-Walled Structures, v. 123, p. 520-527, 2018.
• EUROCODE 3: Design of steel structures. Part 1-8: Design of joints, 2010.
• GARIFULLIN, M., PAJUNEN, S., MELA, K., HEINISUO, M., HAVULA, J. Initial in-plane rotational stiffness of welded RHST joints with axial force in main member. Journal of Constructional Steel Research, v. 139, p. 353-362, 2017.
• LIMA, L. R. O. DE, VELLASCO, P. C. G. da S. , SILVA, L. A. P. S. da, ANDRADE, S. A. L. de. Experimental evaluation of extended endplate beam-to-column joints subjected to bending and axial force. Engineering Structures, Estados Unidos, v. 46, n.7, p. 1-15, 2004.
• KRISHNAMURTHY, N. A fresh look at bolted end-plate behaviour and design. AISC Engineering Journal, v.15, 2nd Quarter, p. 39-49, April, 1978.
• MAGGI, Y. I., GONÇALVES, R. M., LEON, R. T., RIBEIRO, L. F. L. Parametric analysis of steel bolted endplate connections using finite element modeling. Journal of Constructional Steel Research, v. 61, n.5, p. 689-708, 2005.
• RIBEIRO, L. F. L., GONÇALVES, R. M., CASTIGLIONI, C. A. Beam-to-column end plate connections - an experimental analysis. Journal of Constructional Steel Research, Surrey London, v. 46, p. 264-266, 1998.
• SILVA, L. A. P. S. da, LIMA, L. R. O. DE , DE LIMA, L.R.O. , VELLASCO, P. C. G. da S. , ANDRADE, S. A. L. de (.) Behaviour of flush end-plate beam-to-column joints under bending and axial force. Steel and Composite Structures Coréia, v. 4, n.2, p. 77-94, 2004.
• WILSON W. M., MOORE, H. F. Tests to determine the rigidity of riveted joints in steel structures University of Illinois, 1917. (Bulletin n.104).
• ZHAO, X., DAI, L., WANG, T., SIVAKUMARAN, K. S., CHEN, Y. A theoretical model for the rotational stiffness of storage rack beam-to-upright connections. Journal of Constructional Steel Research, v. 133, p. 269-281, 2017.

Publication Dates

• Publication in this collection
Oct-Dec 2018


text:
# Computing the PDF of the sum of N moves of an empirical PDF for USDJPY 1-minute moves

[Cross posted.]

Per-minute tick data for USDJPY is available here. Suppose we download this file to usdjpy.txt and then save it into a Numpy array in Python 3 as follows:

import numpy as np
data=[x.split(',') for x in data][1:]
jpy=np.array([float(close) for (ticker,yy,time,open,high,low,close,vol) in data])

The per-minute returns in USDJPY, expressed in basis points, will be:

rjpy=10000.0*np.diff(jpy)/jpy[0:-1]

Define a histogram function and empirical PDF function as follows:

def histc(X,bins):
map_to_bins = np.digitize(X,bins)
r = np.zeros(bins.shape)
for i in map_to_bins:
r[i-1] += 1
return [r,map_to_bins]

def epdf(S,numIntervals=100):
minS=np.min(S)
maxS=np.max(S)
intervalWidth=(maxS-minS)/numIntervals
x=np.arange(minS,maxS+intervalWidth/2.,intervalWidth)
[ncount,ii]=histc(S,x)
if ncount[1]>len(S)/2:
medS=np.median(S)
minS=0.8*medS
maxS=1.2*medS
intervalWidth=(maxS-minS)/numIntervals
x=np.arange(minS,maxS+intervalWidth/2.,intervalWidth)
[ncount,ii]=histc(S,x)
relativefreq=ncount/sum(ncount)
return (x,relativefreq)

The empirical PDF of USDJPY 1-minute pip returns is then:

(x,rf)=epdf(djpy,numIntervals=1000)

which if we plot it

from bokeh.plotting import figure, show
from bokeh.io import output_notebook
output_notebook()
simple=[(x[i],int(100*rf[i])) for i in range(rf.shape[0]) if int(100*rf[i]) > 0]
X=np.array([x for x,y in simple])
Y=np.array([y for x,y in simple])
p=figure(plot_width=600,plot_height=200,tools="pan,wheel_zoom,box_zoom,reset,resize")
p.line(X,Y)
show(p)

Looks like this:

Now suppose I want to know what could happen in an hour (60 one-minute moves). Following the answer to this question, I could convolve the EPDF above 60 times and I should get the right answer. I think this would look something like this:

def step_pdf(pdf1,pdf2):
pdf=np.convolve(pdf1,pdf2)
pdf=(pdf[0:-1:2]+pdf[1::2])/2
pdf=np.append(np.array([0]),pdf)
pdf=pdf/pdf.sum()
return pdf

from functools import reduce
pdf60=reduce(step_pdf,[rf for i in range(60)])

If I then plot the new pdf60 on top of the old pdf

p=figure(plot_width=600,plot_height=200,tools="pan,wheel_zoom,box_zoom,reset,resize")
p.line(x,rf,color='red')
p.line(x,pdf60,color='blue')
show(p)

I see (call this “Convolution PDF60”):

The blue line is my 60-minute PDF from the above 60-fold convolution. It is smoother, which I expect, but it is still roughly in the same range as the original 1-minute PDF, which I do not expect. So now I will try a more constructive way of generating the 60-minute PDF: I will construct as many 60-minute samples randomly as I have 1-minute samples, by summing randomly selected vectors of size 60 from my original population of 1-minute moves. Then I will compute the empirical PDF of the result. I completely trust this construction, so I will use it as a benchmark against my original construction. So:

n=djpy.shape[0]
draws=np.random.randint(0,n,size=(n,60))
djpy60=np.array([djpy[draws[i]].sum() for i in range(n)])
(x,pdf60)=epdf(djpy60,numIntervals=1000)

Now if I plot pdf60:

p=figure(plot_width=600,plot_height=200,tools="pan,wheel_zoom,box_zoom,reset,resize")
p.line(x,rf,color='red')
p.line(x,pdf60,color='blue')
show(p)

I see a much wider distribution of 60-minute moves, which corresponds much more strongly to my intuition (call this “Monte Carlo PDF60”):

Question: Why aren’t my Convolution PDF60 and my Monte Carlo PDF60 in agreement?

# What proves that a random process with zero diffusion is not a martingale?

[Cross-posted.]

Consider the process $dX_t=W_t dt+0 dW_t$, alternatively $X_t=\int_0^t W_s ds$. $W_t$ is Brownian motion. I read a proof that $X_t$ is a martingale that simply states “Because the diffusion of $dX_t$ is 0, $X_t$ is not a martingale.”

By definition, a stochastic process $X_t$ adapted to a filtration $\{F_t\}$ is a martingale iff $E(|X_t|) <\infty, t \geq 0$ and $E(X_t|{\cal F}_s)=X_s, 0\leq s

Question: What exactly about either of these conditions establishes that if a random process has 0 diffusion, it is not a martingale?

I am asking because I see the 0-diffusion condition used often for this purpose, but in the above example, of a process which is still random even though it has a zero diffusion, I don’t get it.

# Write expectation of brownian motion conditional on filtration as an integral?

[Cross-post.]

Let $W_t$ be a Brownian motion, so $W_t=z_t \sqrt{t}$ where $z_t \in N(0,1)$ and the pdf of $z$ is
$f(z)=\frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}}$. So

$E(W_t)=\int_{-\infty}^{\infty} W_t f(z) dz =\int_{-\infty}^{\infty} z \sqrt{t} \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} dz =\int_{0}^{\infty} (z+(-z)) \sqrt{t} \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} dz=0$

Now suppose ${\cal F}_t$ is the natural filtration for $W_t$. By construction of Brownian motion, we are given that $E(W_t|{\cal F}_s)=W_s, 0\leq s\leq t$.

Question: How do I write $E(W_t|{\cal F}_s)$ as a Riemann integral expression similar to the Riemann integral expression of $E(W_t)$ given above?

Note: I have done extensive Google search on this, without finding any responsive exposition. If this question is beside the point, please explain why. If it’s on point, please answer with the Riemann integral expression.

# Empirical PDF from Empirical CDF

(cross post)

Suppose I do an experiment $N$ times and get a vector $X$ of results. Let $C_X(y)$ be the empirical cumulative distribution function of $X$. Suppose $X$ is sorted so that $x_1 \leq x_2 \cdots \leq x_N$. Approximately,
$C_X(y)=0\textrm{ if }y \leq x \textrm{ for all }x \in X$
$C_X(y)=1\textrm{ if }y > x \textrm{ for all }x \in X$
$C_X(y)=\frac{i+\frac{y-x_i}{x_{i+1}-x_i}}{N} \textrm{ if } x_i \leq y \leq x_{i+1}$

Question: What is the most efficient way to compute the corresponding empirical PDF of $X$? Just interpolate through the histogram?

# Two-step empirical CDF from one-step empirical CDF

(cross-post)

Suppose I have a random variable $X_i$ which changes by $X_{i+1}=X_i+\delta_i$ from one timestep to the next. Suppose I do an experiment where I observe $N$ values $d_1,d_2,\ldots,d_N$ of $\delta_0$ and make an experimental CDF of $\delta_0$, by sorting $d$ so that $d_1 \leq d_2 \cdots \leq d_N$, and then approximating the CDF of $y$ as $\frac{i}{N}$ where $d_i \leq y \leq d_{i+1}$.

Question: What is the most efficient way to compute the empirical CDF of two steps of $X$, assuming that the process for going from $X_i$ to $X_{i+1}$ follows the same empirical distribution? The brute force way that occurs to me is to create the set $E={d_i+d_j: 1\leq i\leq N, 1\leq j\leq N}$, sorting $E=e_1,\ldots,e_{N^2}$ and then approximating the two-step CDF of $y$ as $\frac{i}{N^2}$ where $e_i \leq y \leq e_{i+1}$.


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnlilne Submission ㆍMy Manuscript - For Reviewers - For Editors
 On (disk, annulus) pairs of Heegaard splittings that intersect in one point Bull. Korean Math. Soc. 2009 Vol. 46, No. 1, 99-105 Printed January 1, 2009 Jung Hoon Lee Korea Institute for Advanced Study Abstract : Let $M=H_1\cup_S H_2$ be a Heegaard splitting of a $3$-manifold $M$, $D$ be an essential disk in $H_1$ and $A$ be an essential annulus in $H_2$. Suppose $D$ and $A$ intersect in one point. First, we show that a Heegaard splitting admitting such a $(D,A)$ pair satisfies the disjoint curve property, yet there are infinitely many examples of strongly irreducible Heegaard splittings with such $(D,A)$ pairs. In the second half, we obtain another Heegaard splitting $M=H'_1\cup_{S'} H'_2$ by removing the neighborhood of $A$ from $H_2$ and attaching it to $H_1$, and show that $M=H'_1\cup_{S'} H'_2$ also has a $(D,A)$ pair with $|D\cap A|=1$. Keywords : Heegaard splitting, essential annulus, disjoint curve property MSC numbers : Primary 57N10, 57M50 Downloads: Full-text PDF


text:
Kriss S.

# The cost for 7 dance lessons is $82.The cost for 11 lessons is$122. Write a linear equation to find the ntotal cost C for L lessons.

Then use the equation to find the cost of 4 lessons

## 1 Expert Answer

By:

Jason L. answered • 11/29/16

Tutor
4.8 (6)

Graduate Student Who Loves to Do Math

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.


text:
# Definition:Disjoint Union (Set Theory)/Also known as

## Disjoint Union: Also known as

A disjoint union in the context of set theory is also called a discriminated union.

In Georg Cantor's original words:

We denote the uniting of many aggregates $M, N, P, \ldots$, which have no common elements, into a single aggregate by
$\tuple {M, N, P, \ldots}$.
The elements in this aggregate are, therefore, the elements of $M$, of $N$, of $P$, $\ldots$, taken together.


text:
# Gyroscopic coupling

1. Sep 28, 2009

### eichfeld

I am not really solving an exercise for homework so hopefully this general forum is OK for my question.

For my thesis, what is attitude control of a flying object with a reaction wheel, I am learning the basics from this NASA Technical Report:

http://www.hot.ee/ronn/design_glob_anal_spacecraft_att_control.pdf [Broken] (I mirrored it because currently NASA website publishing this is down)

Things are quite clear to me until eq (22) in PDF page 20, where the 3rd term is "gyroscopic coupling". I am unfamiliar with this phenomenon and can not grasp it from the equation. Could someone please explain me the meaning of this term? I do not see how can change of angular velocity depend only in the current velocity. Is it somehow the same thing as precession?

Last edited by a moderator: May 4, 2017
2. Sep 29, 2009

### Cleonis

I did some googling, and I did get the impression is that the expression 'gyroscopic coupling' is used for the case of experiencing consequences of gyroscopic precession.

If so then the most extreme example of gyroscopic coupling was the flight behavior of aircrafts with a single rotatory engine. With a rotatory engine the cylinders are arranged radially, and the entire motorblock is rotating. That was good for cooling the cilinders, but it was a huge rotating mass in a relatively light aircraft. If the pilot tried to pitch up or down, the aircraft wouldn't pitch but yaw, and vice versa. To fly those aircrafts the pilot's had to anticipate the gyroscopic precession and compensate for it in advance.

Cleonis

3. Sep 29, 2009

### rcgldr

For a helicopter, the pilots control inputs are moved about 90 degrees out of phase, to compensate for gyroscopic precession. A roll torque results in a pitch response and vice versa. Yaw isn't an an issue since it shares an axis with the main rotor.

4. Sep 30, 2009

### twowheelsbg

Hi, guys, sorry for the side orientated question,
but i am also confused exploring the gyroscope theory.
Could someone point out a web link with easy to understand explanations
of gyro basics, in particular forced precession equations and gyroscopic moment.

regards

5. Oct 6, 2009

### eichfeld

Ok. I seem to understand what the term is. So in this case the Wa is velocity of precession, h is the moment of impuls percessing and the product without inertia term is the torque of precession?

Usually in textbooks they explain precession by applying torque to L and getting W from it, but it should also vice-versa, right?

6. Oct 6, 2009

### Cleonis

What is your question? What you wrote is not clear enough.

Preferably, if your question is about a mathematical formula, enter the formula in LaTeX markup.
Check out the https://www.physicsforums.com/misc/howtolatex.pdf" [Broken]

Cleonis

Last edited by a moderator: May 4, 2017
7. Oct 6, 2009

### D H

Staff Emeritus
This term arises because angular momentum is expressed in the body frame. This is of course a rotating reference frame, so the rotational equations of motion need to reflect this fact. Other names for this term include "inertial torque" (c.f. "inertial force", the fictitious forces due to frame acceleration/rotation) and "Euler torque" (c.f. the terms in Euler's equations).

An easy way to see how this term arises is to use the identity that relates the time derivative of a vector quantity q as ascertained from the perspective of an inertial frame versus a rotating frame:

$$\left(\frac {d\mathbf q}{dt}\right)_{\text{inertial}} = \left(\frac {d\mathbf q}{dt}\right)_{\text{rotating}} +\,\, \boldsymbol{\omega}\times \mathbf q$$

This is true for all vector quantities, not just position. So, apply it to angular momentum.

$$\left(\frac {d\mathbf L}{dt}\right)_{\text{rotating}} = \left(\frac {d\mathbf L}{dt}\right)_{\text{inertial}} -\,\, \boldsymbol{\omega}\times \mathbf L$$

The first term on the left hand side is simply the external torque. The second term is gyroscopic / inertial / Euler torque.


text:
Home » Use System Variables windowStartTime and windowEndTime in a Synapse Pipeline

# Use System Variables windowStartTime and windowEndTime in a Synapse Pipeline

In this article, you’ll learn how you can get metadata from your pipeline run inside your pipeline’s notebooks (or other activities). The documentation on this subject is relatively limited, so I hope it will prevent a lot of wasted time.

In Airflow, you have macros, to refer to the metadata of a pipeline run. In Kubeflow, you have variables on various levels. In Azure Synapse (and in Data Factory), you have the windowEndTime and windowStartTime system variables of the Tumbling Window trigger. To get them from the pipeline into a Synapse notebook, you should take the following steps.

## Set pipeline default parameters

First, specify the default values for both parameters, on the level of the pipeline. You can choose the names of both parameters, but keep them short and consistent as they will be referred to in a later step.

## Set trigger run parameters

Next, create the trigger which will start the pipeline.

• Click Trigger > New/Edit
• Click the dropdown > Click ‘+ New’
• Select Tumbling Window and adequate recurrence
• When you submit, the next window will appear. Add the proper variables.
@trigger().outputs.windowStartTime
@trigger().outputs.windowEndTime

## Set activity parameters

Finally, on the level of the (notebook) activity, you can set the base parameters. On the left hand side, you’ll set the names of your Python variables. On the right hand side, you can load them with the parameters from the pipeline level.

## Use the parameters inside a notebook

Inside the notebook, you should create a cell where you add all the variables in which you want to store the window start and window end time. You should also specify a default value for when the parameters aren’t loaded (I set them to None).

window_start_time = None
window_end_time = None

Finally, you should set the cell as a parameter cell. You can do this by clicking the three dots on the left hand side of the cell.

⚠If you simply want to extract a date from your newly created variables, you should take the first 10 characters of these strings.

window_start_time = window_start_time[0:10]
window_end_time = window_end_time[0:10]

Great success!

### Say thanks, ask questions or give feedback

Technologies get updated, syntax changes and honestly… I make mistakes too. If something is incorrect, incomplete or doesn’t work, let me know in the comments below and help thousands of visitors.


text:
# Under what conditions is $(A+B)(A-B) = (A^2 - B^2)$ for two $n \times n$ matrices $A$ and $B$?

so my approach to this problem was

to view $A$ as a matrix of the form $\begin{bmatrix}a_1 & a_2 & a_3 & \dots & a_n\end{bmatrix}$ and $B$ as $\begin{bmatrix}b_1 & b_2 & b_3 & \dots & b_n\end{bmatrix}$

define variable $C = (A+B)$ and $D = (A-B)$

$C = \begin{bmatrix}c_1 & c_2 & c_3 & \dots & c_n\end{bmatrix}$ $D = \begin{bmatrix}d_1 & d_2 & d_3 & \dots & d_n\end{bmatrix}$

$CD = A^2 - B^2$

$Cd_1 + Cd_2 +\dots + Cd_n = (a_1A - b_1B) + (a_2A + b_2B) + (a_3B + b_3B) + \dots + (a_nB + b_nB)$

substitute back for $d$ and $C$

$(a_1-b_1)(A+B) + (a_2 - b_2)(A+B) + \dots = (a_1A - b_1B) + (a_2A + b_2B) + (a_3B + b_3B)+ \dots$

$(a_1A + A_1B -B_1A - b_1B) + (a_2A + A_2B -B_2A - b_2B) + \dots = (a_1A - b_1B) + (a_2A + b_2B) + \dots$

at this point I'm stuck, because I feel like I'm on the wrong track because my answer does not match correctly with the solution. How does one solve this problem?

You can make it simpler. $$(A+B)(A-B)=A^2+BA-AB-B^2$$ When is this equal to $A^2-B^2$?
$(A+B)(A-B)=A^2+BA-AB-B^2$ its equal to $A^2-B^2$ when $AB=BA$ i mean A,B must commute


text:
ARTICLES
443 Article(s)
1, 2, 3, 4, 5, 6
Document(s)
Title
ABSTRACT: Bovine spongiform encephalopathy (BSE) created a global European crisis in the 1980s and 90s, with very serious health and economic implications. Classical BSE now appears to be under control, to a great extent as a result of a global resea...
We report on new measurements of the electric Generalized Polarizability (GP) of the proton $\alpha_E$ in a kinematic region where a puzzling dependence on momentum transfer has been observed, and we have found that $\alpha_E = (5.3 \pm 0.6_{stat} \p... We report on new measurements of the electric Generalized Polarizability (GP) of the proton$\alpha_E$in a kinematic region where a puzzling dependence on momentum transfer has been observed, and we have found that$\alpha_E = (5.3 \pm 0.6_{stat} \p...


text:
# What is a QC chemist?

## What is a QC chemist?

A Quality Control (QC) chemist utilizes chemistry lab skills to test and measure materials, in a manufacturing or pharmaceutical field.

## What does a QC chemist do?

What is a Quality Control Chemist. The quality control chemists are the ones who ensure that everything is done properly. To do this, they prepare, measure, and test samples, lab materials, and products according to industry-specific standard procedures

## How do you become a quality control chemist?

Requirements and Qualifications

• Bachelor’s degree in chemistry, biology, or a related field.
• 1+ years of experience in a manufacturing lab.
• Knowledge of analytical testing techniques, such as FTIR and UV-VIS.
• Able to write extensive reports on lab results.
• Must pass a background check and drug test.

## What is the work of QC Chemist?

Job DescriptionA Quality Control Chemist prepares and tests samples from all phases of the manufacturing process to ensure that the product quality meets the standards, prepares documents that report test results and is responsible for preserving workplace safety while handling hazardous materials.

## How do you become a QC Chemist?

To become a Quality Assurance Chemist, you must have a bachelor’s degree in chemistry or a closely related field like biology or biochemistry. However, it would be best if you develop the necessary laboratory skills for fulfilling the required duties.

## What chemist makes most money?

Top Jobs for Chemistry Majors

• Chemistry Teacher. Average Base Pay: $53,000. • Chemical Engineer. Average Base Pay:$77,600.
• Doctor. Average Base Pay: $200,000. • Forensic Scientist. Average Base Pay:$56,000.

## Why are chemists paid so little?

Unfortunately, there is a much greater need to graduate students then there is for industry jobs and thus by supply and demand, the wage of chemist is comparatively low to something like a doctor where there is actually a constant shortage of doctors.

## What is the work of QC chemist?

Job DescriptionA Quality Control Chemist prepares and tests samples from all phases of the manufacturing process to ensure that the product quality meets the standards, prepares documents that report test results and is responsible for preserving workplace safety while handling hazardous materials.

## Is QC a good career?

Job DescriptionA Quality Control Chemist prepares and tests samples from all phases of the manufacturing process to ensure that the product quality meets the standards, prepares documents that report test results and is responsible for preserving workplace safety while handling hazardous materials.

## Do chemists get paid well?

Job TitleSalaryCipla QC Chemist salaries – 3 salaries reportedu20b911,771/moFresher Than Fresh QC Chemist salaries – 2 salaries reportedu20b925,893/moUltraTech Cement QC Chemist salaries – 2 salaries reportedu20b935,735/moAlkem Laboratories QC Chemist salaries – 2 salaries reportedu20b919,848/mo16 more rows


text:
# UPSSSC FOREST GUARD REASONING QUIZ

Attempt now to get your rank among 99 students!

## Question 1:

If "FOREST" is coded as "GPSFTU" then how "BUTTON" will be coded?

## Question 2:

In a code language, 'PROVEN' is written as 'PSQOFW'. How will 'MONGER' be written in that language?

## Question 3:

An accurate clock shows 8 a.m. Through how many degree will the hour hand rotate when clock shows 2 p.m.

## Question 4:

An accurate clock shows 6 O’clock in the morning. Through how many degree will the hour hand rotate when the clock shows 4 O’clock in the evening?

## Question 5:

What day of the week was 5 February 2008?

## Question 6:

If the 4th day of a month is Sunday, which of the following would be the 6th day before the 30th day of that month?

## Question 7:

If $P \% Q$ means $P$ is the father of $Q$, $P!Q$ means $P$ is the sister $Q$ and $P{ }^{*} Q$ means $P$ is the daughter of $Q$, then which of the following shows that $\mathrm{P}$ is the husband of S's sister?

## Question 8:

K said to L, "You are my husband's son-in-law's daughter." How is L related to K?

## Question 9:

Ram is taller than Geetha. Vimala is taller than Ram but shorter than Meenu. Bindu is taller than Geetha but shorter than Ram. Who among the given people is the shortest?

## Question 10:

Naresh is 19th from the left end and Naresh is 29th from the Right end in the row. How many students are in the Row?


text:
# Can you simplify this term?

$$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$

• @EricS. Yes all are scalars and you may assume $r>0$. – emcor Dec 16 '15 at 9:56
• If you rewrite both enumerator and denominator into a single fraction, i.e. of the form $X=\frac{\frac{\alpha}{\beta}}{\frac{\gamma}{\delta}}$, you can rewrite that into $X=\frac{\alpha}{\beta}\frac{\delta}{\gamma}$ and a lot of terms will drop – Eric S. Dec 16 '15 at 9:58
• @EricS. It would be very nice if you could do that... ; ) – emcor Dec 16 '15 at 10:00
• Just be patient. Take lots of paper and multiply top and bottom by common terms. first by $(1+r)^T$. That will cause $\frac {1-c}{(1+4)^{T+1}}$ to reduce to $\frac{1-c}{1+r}$. Then by r. Eventually the whole thing should become manageable. Takes patience though. – fleablood Jan 7 '16 at 17:44

$$\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}= \frac{r^2}{r^2}\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}= \frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}=\\ \frac{(1+r)^{T+1}}{(1+r)^{T+1}}\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{r{c}(1+r)^{T+1}+r^2({1-c}){(1+r)}-r^2(1+r)^{T+1}}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{r\left({c}-r\right)(1+r)^{T+1}+r^2({1-c}){(1+r)}}= \frac{{c}(1+r)^{T+1}+r^2({1-c})}{\left(({c}-r)(1+r)^{T}+r({1-c})\right){(1+r)r}}$$ $$%=\frac{c-r^2 (c-1) (r+1)^{-T-1}}{r(c -r)-r^2(c-1) (r+1)^{-T}}$$

would this help?

• Thank you. Can you please also show the steps how to get there? – emcor Dec 16 '15 at 10:03

$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^{T}}-1}$
$====================$
$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c-r}{r}+\frac{1-c}{(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r^2(1+r)^{T+1}}}{\frac{(c-r)(1+r)^{T}+(1-c)r}{r(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)}}{\frac{(c-r)(1+r)^{T}+(1-c)r}1}$
$====================$
$X=\frac{(c(1+r)^{T+1}+(1-c)r^2)1}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$
$====================$
$X=\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$


text:
1
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :
A
$${1 \over {10}}$$
B
$${1 \over {17}}$$
C
$${1 \over {11}}$$
D
$${1 \over {12}}$$
2
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
Four persons can hit a target correctly with probabilities $${1 \over 2}$$, $${1 \over 3}$$, $${1 \over 4}$$ and $${1 \over 8}$$ respectively. if all hit at the target independently, then the probability that the target would be hit, is
A
$${{25} \over {32}}$$
B
$${{25} \over {192}}$$
C
$${{1} \over {192}}$$
D
$${{7} \over {32}}$$
3
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:
A
2
B
3
C
4
D
5
4
JEE Main 2019 (Online) 8th April Morning Slot
+4
-1
Let A and B be two non-null events such that A $$\subset$$ B . Then, which of the following statements is always correct ?
A
P(A|B) = 1
B
P(A|B) = P(B) – P(A)
C
P(A|B) $$\le$$ P(A)
D
P(A|B) $$\ge$$ P(A)
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination


text:
# hcl bond diagram

zhuravlova.me 9 out of 10 based on 1000 ratings. 800 user reviews.

Hydrogen chloride Hydrogen chloride is a diatomic molecule, consisting of a hydrogen atom H and a chlorine atom Cl connected by a polar covalent bond. The chlorine atom is much more electronegative than the hydrogen atom, which makes this bond polar. Hydrochloric acid | HCl PubChem Hydrochloric acid | HCl or ClH | CID 313 structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities ... Drawing the Lewis Structure for HCl (Hydrochloric Acid) Drawing the Lewis Structure for HCl (Hydrochloric Acid) Another straight forward Lewis structure. You have a total of 8 valence electrons available to fill the octets of Chlorine and Hydrogen in the HCl Lewis structure. Chemistry Molecular Structure (28 of 45) Sigma(s p) Bond Example In this video I will explain the sigma(s p) bond of hydrochloric acid, HCl. Skip navigation Sign in. Search. ... Molecular Structure (21 of 45) Bonding Theory Basics Hydrogen ... GCSE CHEMISTRY Covalent Bonding in a Hydrogen Chloride ... Hydrogen can only form 1 bond. and make a hydrogen chloride molecule (HCl). This is a picture of a hydrogen chloride molecule. and the chlorine atom can count 8 electrons in its outer shell. react further with other chlorine or hydrogen atoms. chlorine and hydrogen atoms form a single covalent bond. What Is the Lewis Dot Structure for HCL? | Reference The Lewis dot structure for HCl has hydrogen's symbol, H, on the left side, chlorine's symbol, Cl, on the right side, two dots between the two and an additional six dots around Cl. Hydrogen has a single valence electron, while chlorine has seven valence electrons. Hydrochloric acid | HCl | ChemSpider Structure, properties, spectra, suppliers and links for: Hydrochloric acid, 7647 01 0. Spectra and Molecular Structure – HCl & DCl odinity Spectra and Molecular Structure – HCl & DCl. To describe the rotation of an object, the classical mechanics form would describe the energy as where I is the moment of inertia, ω is the angular velocity, and it is then put into terms of L, the angular momentum, μ, the reduced mass, and r, the bond length. What is the bond order of HCl? | Yahoo Answers What is the bond order of HCl? use the drawing of energy diagram for HCl to predict the bond order. HCl has 8 valance electrons and the bond order is 3, but it's WRONG so how do i find the bond order. How to Draw Covalent Bonding Molecules HCl 1:25 Chlorine is in group 7 so it has seven electrons in the outer shell. HCl has single bond so it has two electrons in the overlap. Cl2 2:23 Acetaminophen | C8H9NO2 PubChem Acetaminophen. Acetaminophen is a widely used nonprescription analgesic and antipyretic medication for mild to moderate pain and fever. Harmless at low doses, acetaminophen has direct hepatotoxic potential when taken as an overdose and can cause acute liver injury and death from acute liver failure. Bond Energies Chemistry LibreTexts For example, the bonds of two water molecules are broken to form hydrogen and oxygen. $2H_2O \rightarrow 2H_2 O_2$ Energy is always required to break a bond, which is known as bond energy. While the concept may seem simple, bond energy serves a very important purpose in describing the structure and characteristics of a molecule. Chlorine dioxide Chlorine dioxide is a chemical compound with the formula ClO 2.This yellowish green gas crystallizes as bright orange crystals at −59 °C. As one of several oxides of chlorine, it is a potent and useful oxidizing agent used in water treatment and in bleaching.. The compound has been fraudulently marketed as a cure for a wide range of diseases, including childhood autism. 10.5: Molecular Orbital Theory Chemistry LibreTexts Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $$\PageIndex{14}$$ that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. HCN Lewis Structure, Molecular Geometry, Shape, and Polarity Understanding of HCL can be beneficial to you if you are a chemistry freak. In this article, you can get a detailed understanding of HCN Lewis structure and Molecular geometry. You must be aware that to understand the molecular geometry of HCL; one must know HCL Lewis structure. The bond dissociation energy of HCl is________when ... The bond dissociation energy of HCl is_____when..? ... The bond is contracted due to the unique high electronegativity of F. In order to achieve this overlap the F(a) 2px e⁻ pair interacts with the the F(b) 2px e⁻ pair causing repulsion that weakens the σ bond; the same is true for the 2py e⁻ pair AOs. The situation is a trade off. Dipole moment (video) | Chemical bonds | Khan Academy So we're more concerned with analyzing a dipole moment in terms of the molecular structure, so let's go ahead and look at the dot structure for HCl. So if I look at this covalent bond between the hydrogen and the chlorine, I know that that covalent bond consists of two electrons. MO Diagrams for Diatomic Molecules UCI Department of ... Summary MO Theory • LCAO MO Theory is a simple method for predicting the approximate electronic structure of molecules. • Atomic orbitals must have the proper symmetry and energy to interact and form molecular orbitals. • Photoelectron spectroscopy provides useful information on the energies of atomic orbitals. • Next we’ll see that symmetry will help us treat larger molecules in Wolfram|Alpha Widgets: "Lewis structure" Free Chemistry ... Get the free "Lewis structure" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Chemistry widgets in Wolfram|Alpha. Hydrogen Chloride: Formula, Structure & Properties Video ... HCl is a strong acid, and dissociates completely in water. It is formed by covalent bonds between hydrogen and chloride ions. HCl has many commercial uses, including use in the production of steel ... Lewis Dot Structure for HCl | Chemical Bonding | Success ... Chemical Bonding: Lewis Dot Structure for HCl (4 of 6) The Lewis Structure for HCl (hydrochloric acid) is one of the easier dot structures to draw. When you draw the structure remember that Hydrogen (H) only needs two valence electrons to have a full outer shell. VALENCE BOND THEORY (VBT) | HYBRIDIZATION | SP | SP2 | SP3 ... Introduction to valence bond theory, VBT and hybridization concept in detail. Types hybridization i.e., sp, sp2, sp3, sp3d, sp3d2, sp3d3 hybridization ... of two chlorine atoms overlap along the inter nuclear axis and thus by forming a σ p p bond. 3) HCl molecule: * In the ground state, the electronic configuration of hydrogen atom is 1s 1. BOND ENTHALPY (BOND ENERGY) chemguide Bond dissociation enthalpy and mean bond enthalpy. Simple diatomic molecules. A diatomic molecule is one that only contains two atoms. They could be the same (for example, Cl 2) or different (for example, HCl). The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms everything being in the gas ... Molecular orbital diagram The bond order of diatomic nitrogen is three, and it is a diamagnetic molecule. The bond order for dinitrogen (1σ g 2 1σ u 2 2σ g 2 2σ u 2 1π u 4 3σ g 2) is three because two electrons are now also added in the 3σ MO. The MO diagram correlates with the experimental photoelectron spectrum for nitrogen. 7.5 Strengths of Ionic and Covalent Bonds – Chemistry A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound. Chemistry Chapter 11 Flashcards | Quizlet On the phase diagram shown to the right, segment _____ corresponds to the conditions of temperature and pressure under which the solid and the gas of the substance are in equilibrium. A) AB B) AC Structure and bonding Revision 1 Higher Chemistry ... Structure and bonding Elements are held together in different ways and the properties of chemical compounds are determined by the bonding between atoms and the attractive intermolecular forces ... Vibration Rotation Spectrum of HCl Bond Length of HCl A bond length for the HCl molecule can be calculated from the HCl spectrum by assuming that it is a rigid rotor and solving the Schrodinger equation for that rotor. For a free diatomic molecule the Hamiltonian can be anticipated from the classical rotational kinetic energy Types of chemical bonding, Ionic, covalent and metallic ... In this topic three types of chemical bonding are are going to be covered. They are ionic bonding, covalent bonding and, metallic bonding. You will learn how to draw dot cross diagrams for molecules and ionic compounds, also their electronic configurations. 3 Ways to Calculate Bond Order in Chemistry wikiHow How to Calculate Bond Order in Chemistry. On the atomic level, bond order is the number of bonded electron pairs between two atoms. In diatomic nitrogen (N≡N), for instance, the bond order is 3 because there are 3 chemical bonds linking t... Structure and bonding | Organic chemistry | Science | Khan ... Let's review the basics of chemical bonds including dot structures, hybridization, bond line structures, electronegativity, and polarity. We will also discuss how bonding and intermolecular forces relate to physical properties such as boiling point.