text:
# 6.5.4. Duplicate record fields¶

DuplicateRecordFields
Implies: DisambiguateRecordFields 8.0.1

Allow definition of record types with identically-named fields.

Going beyond DisambiguateRecordFields (see Record field disambiguation), the DuplicateRecordFields extension allows multiple datatypes to be declared using the same field names in a single module. For example, it allows this:

module M where
data S = MkS { x :: Int }
data T = MkT { x :: Bool }

Uses of fields that are always unambiguous because they mention the constructor, including construction and pattern-matching, may freely use duplicated field names. For example, the following are permitted (just as with DisambiguateRecordFields):

s = MkS { x = 3 }

f (MkT { x = b }) = b

Field names used as selector functions or in record updates must be unambiguous, either because there is only one such field in scope, or because a type signature is supplied, as described in the following sections.

While DuplicateRecordFields permits multiple fields with the same name in a single module, it does not permit a field and a normal value binding to have the same name. For that, use NoFieldSelectors.

## 6.5.4.1. Selector functions¶

Fields may be used as selector functions only if they are unambiguous, so this is still not allowed if both S(x) and T(x) are in scope:

bad r = x r

Warning: the type-based disambiguation rules described in the remainder of this section are being removed (see the proposal DuplicateRecordFields without ambiguous field access). The -Wambiguous-fields option will warn about code that relies on these rules. In a future GHC release, such code will produce ambiguity errors.

An ambiguous selector may be disambiguated by the type being “pushed down” to the occurrence of the selector (see Type inference for more details on what “pushed down” means). For example, the following are permitted:

ok1 = x :: S -> Int

ok2 :: S -> Int
ok2 = x

ok3 = k x -- assuming we already have k :: (S -> Int) -> _

In addition, the datatype that is meant may be given as a type signature on the argument to the selector:

ok4 s = x (s :: S)

However, we do not infer the type of the argument to determine the datatype, or have any way of deferring the choice to the constraint solver. Thus the following is ambiguous:

bad :: S -> Int

Even though a field label is duplicated in its defining module, it may be possible to use the selector unambiguously elsewhere. For example, another module could import S(x) but not T(x), and then use x unambiguously.

In a record update such as e { x = 1 }, if there are multiple x fields in scope, then the type of the context must fix which record datatype is intended, or a type annotation must be supplied. Consider the following definitions:

data S = MkS { foo :: Int }
data T = MkT { foo :: Int, bar :: Int }
data U = MkU { bar :: Int, baz :: Int }

Without DuplicateRecordFields, an update mentioning foo will always be ambiguous if all these definitions were in scope. When the extension is enabled, and there is exactly one type that has all the fields being updated, that type will be used. For example:

f x = x { foo = 3, bar = 2 }

Here f must be updating T because neither S nor U have both fields.

If there are multiple types with all the fields, type information may be used to disambiguate which record type is meant. Warning: the following rules are being removed (see the proposal DuplicateRecordFields without ambiguous field access). The -Wambiguous-fields option will warn about code that relies on these rules. In a future GHC release, such code will produce ambiguity errors.

• Use the type being pushed in to the record update, as in the following:

g1 :: T -> T
g1 x = x { foo = 3 }

g2 x = x { foo = 3 } :: T

g3 = k (x { foo = 3 }) -- assuming we already have k :: T -> _

• Use an explicit type signature on the record expression, as in:

h x = (x :: T) { foo = 3 }

The type of the expression being updated will not be inferred, and no constraint-solving will be performed, so the following will be rejected as ambiguous:

let x :: T
x = blah
in x { foo = 3 }

\x -> [x { foo = 3 },  blah :: T ]

\ (x :: T) -> x { foo = 3 }

## 6.5.4.3. Import and export of record fields¶

When DuplicateRecordFields is enabled, an ambiguous field must be exported as part of its datatype, rather than at the top level. For example, the following is legal:

module M (S(x), T(..)) where
data S = MkS { x :: Int }
data T = MkT { x :: Bool }

However, this would not be permitted, because x is ambiguous:

module M (x) where ...

Similar restrictions apply on import.


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# Divisors in a Euclidean Domain

Let $R$ be a Euclidean Domain. If $(a,b)=1$, and $a$ divides $bc$, I would like a hint on how to prove that $a$ divides $c$.

-

$R$ is a UFD since it is an Euclidean Domain. Now, it follows easily your assertion.

-
Be careful, using the above could amount to a circular proof. –  Bill Dubuque Jan 20 '12 at 3:49

HINT $\rm\quad a\ |\ ac,bc\ \Rightarrow \ a\ |\ (ac,bc) = (a,b)\ c = c$

Or, in Bezout form $\rm\ a\ |\ (ax+by)c = c$

-

Emiliocba's answer is correct: the desired conclusion holds in any UFD domain (and even a bit more generally, e.g. in a GCD-domain).

However, the fact that a Euclidean domain is a UFD is really the concatenation of two facts:

(i) A Euclidean domain is a PID: indeed, every ideal is generated by any element of minimal norm.
(ii) A PID is a UFD.

The first fact is quicker to prove: what I said above is the entire idea of the proof, and fleshing it out takes maybe two more lines.

Thus it may be helpful to give a proof of the desired conclusion in an arbitrary PID. The method here is familiar from elementary number theory:

since $a$ and $b$ are relatively prime in the PID $R$, there exist elements $x,y \in R$ such that $ax+by = 1$. Moreover, to say that $a | bc$ is to say that there exists $d \in R$ with $ad = bc$. Thus

$c = acx + bcy = acx + ady = a(cx+dy)$,

so $a | c$.

-

Try thinking about ideals and use the fact that if $a|bc$ then $(bc)\subset (a)$. And any ideal is contained in a prime one.

-

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# Faà di Bruno's Formula/Proof 2

## Theorem

Let $D_x^k u$ denote the $k$th derivative of a function $u$ with respect to $x$.

Then:

$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$

## Proof

For convenience, let:

$w_j := D_u^j w$
$u_k := D_x^k u$

Then:

 $\ds D_x \left({w_j}\right)$ $=$ $\ds w_{j + 1} u_1$ Derivative of Composite Function $\ds D_x \left({u_k}\right)$ $=$ $\ds u_{k + 1}$ Definition of Higher Derivative

Thus:

 $\ds D_x^1 w$ $=$ $\ds w_1 u_1$ Derivative of Composite Function $\ds D_x^2 w$ $=$ $\ds \left({w_2 u_1 u_1 + w_1 u_2}\right)$ Derivative of Composite Function: Second Derivative $\ds D_x^3 w$ $=$ $\ds \left({\left({w_3 u_1 u_1 u_1 + w_2 u_2 u_1 + w_2 u_1 u_2}\right) + \left({w_2 u_1 u_2 + w_1 u_3}\right)}\right)$ Derivative of Composite Function: Third Derivative or directly

Analogously, let a corresponding tableau be set up of set partitions as follows:

 $\ds \mathcal D^1$ $=$ $\ds \set 1$ $\ds \mathcal D^2$ $=$ $\ds \set {\set 2 \mid \set 1} + \set {2, 1}$ $\ds \mathcal D^3$ $=$ $\ds \set {\set 3 \mid \set 2 \mid \set 1} + \set {\set {3, 2} \mid \set 1} + \set {\set 2 \mid \set {3, 1} } + \set {\set 3 \mid \set {2, 1} } + \set {3, 2, 1}$

Thus inspired, let $a_1 a_2 \cdots a_j$ denote a partition of the set $\set {1, 2, \ldots, n-1}$.

Let:

 $\ds \mathcal D a_1 a_2 \cdots a_j$ $:=$ $\ds \set n a_1 a_2 \cdots a_j$ $\ds$  $\, \ds + \,$ $\ds \left({a_1 \cup \set n}\right) a_2 \cdots a_j$ $\ds$  $\, \ds + \,$ $\ds a_1 \left({a_2 \cup \set n}\right) \cdots a_j$ $\ds$  $\, \ds + \,$ $\ds \cdots$ $\ds$  $\, \ds + \,$ $\ds a_1 a_2 \cdots \left({a_j \cup \set n}\right)$

This rule is isomorphic to:

 $\ds D_x \left({w_j u_{r_1} u_{r_2} \ldots u_{r_j} }\right)$ $=$ $\ds w_{j + 1} u_1 u_{r_1} u_{r_2} \ldots u_{r_j}$ $\ds$  $\, \ds + \,$ $\ds w_j u_1 u_{r_1 + 1} u_{r_2} \ldots u_{r_j}$ $\ds$  $\, \ds + \,$ $\ds w_j u_1 u_{r_1} u_{r_2 + 1} \ldots u_{r_j}$ $\ds$  $\, \ds + \,$ $\ds \cdots$ $\ds$  $\, \ds + \,$ $\ds w_j u_1 u_{r_1} u_{r_2} \ldots u_{r_j + 1}$

where the term $w_j u_{r_1} u_{r_2} \ldots u_{r_j}$ corresponds to a partition $a_1 a_2 \cdots a_j$.

Thus there exists a bijection from $\mathcal D^n$ to $D_x^n w$.

Thus, collecting like terms in $D_x^n w$, we obtain a sum of terms:

$c \left({k_1, k_2, \ldots}\right) w_j u_i^{k_1} u_2^{k_2} \ldots$

where:

$j = k_1 + k_2 + \cdots$

and:

$n = k_1 + 2 k_2 + \cdots$

and where:

$c \left({k_1, k_2, \ldots}\right)$ is the number of partitions of $\set {1, 2, \ldots, n}$ into $j$ subsets where there are $k_t$ subsets with $t$ elements.

Consider an array of $k_t$ boxes of capacity $t$.

The number of ways to put $n$ different elements into these boxes is the multinomial coefficient:

$\dbinom n {1, 1, \ldots, 1, 2, 2, \ldots, 2, 3, 3, \ldots, 3, 4, \ldots} = \dfrac {n!} {\left({1!}\right)^{k_1} \left({2!}\right)^{k_2} \left({3!}\right)^{k_3} \cdots}$

It remains to divide by $k_1! \, k_2! \, k_3! \ldots$ corresponding to the number of ways each group of $k_t$ can be permuted

Hence:

$c \left({k_1, k_2, \ldots}\right) = \dfrac {n!} {k_1! \left({1!}\right)^{k_1} \, k_2! \left({2!}\right)^{k_2} k_3! \left({3!}\right)^{k_3} \cdots}$

and the result follows.

$\blacksquare$

## Source of Name

This entry was named for Francesco Faà di Bruno.


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# Prove that if $f$ is separable and irreducible polynomial then the Galois group of $f$ is transitive

1. Prove that if $f$ is separable and irreducible polynomial then the Galois group of $f$ is transitive.
2. Prove also that even though the Galois group of $f$ is transitive not every permutation of the roots occur.

It is not difficult to give an example for the second statement e.g. Galois group of $x^4-2$ over $\mathbb{Q}$, but I don't know how to prove the first statement, any help would be great.

• Transitivity of a group is meaningful only if you specify the set the group is acting on (and the action). Ok, in this context you are clearly asking about its action on the zeros of $f$. Hint: If $S$ is any subset of the roots forming an orbit under the action of the Galois group, can you show that the coefficients of $p(x)=\prod_{s\in S}(x-s)$ is stable under the Galois group and hence, by the Galois correspondence its coefficients are ____. Furthermore, $p(x)\mid f(x)$, so _________ – Jyrki Lahtonen Aug 14 '15 at 16:33
• So the orbit is the set of the roots of $f$ so the action is transitive. Thanks. – delueze Aug 14 '15 at 16:40
• You are welcome. You are also welcome to flesh the argument to an answer yourself. That way you get more feedback, and we can check that no lingering holes remain in your reasoning. – Jyrki Lahtonen Aug 14 '15 at 16:42
• @JyrkiLahtonen do we really need to invoke the Galois correspondence to conclude ? Can't it be more explicit ? – Weier Mar 20 at 10:02


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You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version.

# 7.10: Single Variable Division Equations

Difficulty Level: At Grade Created by: CK-12
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Progress
Practice Single Variable Division Equations
Progress
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Have you ever been to a theater with limited seating? Well, Marc and Kara had this happen with their grandparents.

Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it too.

If this was the division, how many people did the manager divide up to start with?

Write a division equation and solve it to complete this dilemma.

This Concept is all about single variable division equations. You will know how to do this by the end of the Concept.

### Guidance

Sometimes, you will see equations that have division in them. Remember, that we can use a fraction bar to show division.

To solve an equation in which a variable is divided by a number, we can use the inverse of division––multiplication. We can multiply both sides of the equation by that number to solve it.

We must multiply both sides of the equation by that number because of the Multiplication Property of Equality , which states:

if $a=b$ and $c \neq 0$ , then $a \times c=b \times c$ .

So, if you multiply one side of an equation by a nonzero number, $c$ , you must multiply the other side of the equation by that same number, $c$ , to keep the values on both sides equal.

Now let's apply this information.

$k \div (-4)=12$ .

In the equation, $k$ is divided by -4. So, we can multiply both sides of the equation by -4 to solve for $k$ . You will need to use what you know about multiplying integers to help you solve this problem. It may help to rewrite $k \div (-4)$ as $\frac{k}{-4}$ .

$k \div (-4) &= 12\\\frac{k}{-4} &= 12\\\frac{k}{-4} \times (-4) &= 12 \times (-4)\\\frac{k}{-4} \times \frac{-4}{1} &= -48\\\frac{k}{\cancel{-4}} \times \frac{\cancel{-4}}{1} &= -48\\\frac{k}{1} &= -48\\k &= -48$

The –4's will cancel each other out when they are divided. Then we multiply.

The value of $k$ is –48.

Remember the rules for multiplying integers will apply when working with these equations!! Think back and use them as you work.

$\frac{n}{1.5}=10$

In the equation, $n$ is divided by 1.5. So, we can multiply both sides of the equation by 1.5 to solve for $n$ .

$\frac{n}{1.5} &= 10\\\frac{n}{1.5} \times 1.5 &= 10 \times 1.5\\\frac{n}{1.5} \times \frac{1.5}{1} &= 15\\\frac{n}{\cancel{1.5}} \times \frac{\cancel{1.5}}{1} &= 15\\\frac{n}{1} &= 15\\n &= 15$

The value of $n$ is 15.

When an equation has division in it, you can use the Multiplication Property of Equality to solve it, and then the property has been useful. Always remember to think about the inverse operations and associate them with the different properties. This will help you to keep it all straight and not get mixed up.

Now it is time for you to practice solving a few of these equations.

Solve each equation for the missing variable.

#### Example A

$\frac{x}{-2}=5$

Solution: $x = -10$

#### Example B

$\frac{y}{5}=6$

Solution: $y = 30$

#### Example C

$\frac{b}{-4}=-3$

Solution: $b = 12$

Here is the original problem once again.

Marc and Kara went to see a play with their grandparents. When they arrived at the theater, the manager divided their group and several other people into six smaller groups. Each of these groups was lead to a section of the theater where there were empty seats. Each group had six people in it too.

If this was the division, how many people did the manager divide up to start with?

Write a division equation and solve it to complete this dilemma.

First, we can write an equation.

Some number of people divided by six is six.

$\frac{x}{6} = 6$

Now we can solve it by multiplying.

$6 \times 6 = 36$

$x = 36$

### Vocabulary

Here are the vocabulary words in this Concept.

Isolate the Variable
this means that we want to work to get the variable alone on one side of the equals.
Inverse Operation
Opposite operation
Division Property of Equality
states that we can solve an equation by multiplying the same value to both sides of the equation.
Multiplication Property of Equality
states that we can solve an equation by dividing both sides of the equation by the same value.

### Guided Practice

Here is one for you to try on your own.

Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid for his share was $4.25. a. Write an equation to represent $c$ , the total cost, in dollars, of the bill for lunch. b. Determine the total cost of the bill. Answer Consider part $a$ first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Because the friends split the bill evenly, write a division equation to represent the problem. $& \underline{Three \ friends} \ \underline{evenly \ split} \ the \ \underline{total \ cost} \ldots amount \ each \ friend \ paid \ \underline{was} \ \underline{\4.25} \ldots\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\& \qquad \ c \qquad \qquad \qquad \ \div \qquad \qquad \quad 3 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ = \quad 4.25$ This equation, $c \div 3=4.25$ , represents $c$ , the total cost of the lunch bill. Next, consider part $b$ . Solve the equation to find the total cost, in dollars, of the lunch bill. $c \div 3 &= 4.25\\\frac{c}{3} &= 4.25\\\frac{c}{3} \times 3 &= 4.25 \times 3\\\frac{c}{\cancel{3}} \times \frac{\cancel{3}}{1} &= 12.75\\c &= 12.75$ The total cost of the lunch bill was$12.75.

### Video Review

Here is a video for review.

### Practice

Directions: Solve each single-variable division equation for the missing value.

1. $\frac{x}{5}=2$

2. $\frac{y}{7}=3$

3. $\frac{b}{9}=-4$

4. $\frac{b}{8}=-10$

5. $\frac{b}{8}=-10$

6. $\frac{x}{-3}=-10$

7. $\frac{y}{18}=-20$

8. $\frac{a}{-9}=-9$

9. $\frac{x}{11}=-12$

10. $\frac{x}{3}=-3$

11. $\frac{x}{5}=-8$

12. $\frac{x}{1.3}=3$

13. $\frac{x}{2.4}=4$

14. $\frac{x}{6}=1.2$

15. $\frac{y}{1.5}=3$

### Vocabulary Language: English

Division Property of Equality

Division Property of Equality

The division property of equality states that two equal values remain equal if both are divided by the same number. For example: If $2x = 8$, then $1x = 4$.
Inverse Operation

Inverse Operation

Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
Isolate the variable

Isolate the variable

To isolate the variable means to manipulate an equation so that the indicated variable is by itself on one side of the equals sign.
Multiplication Property of Equality

Multiplication Property of Equality

The multiplication property of equality states that if the same constant is multiplied to both sides of the equation, the equality holds true.

Dec 21, 2012

Mar 22, 2015


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# Changeset 1363 for docs/HPCA2012/01-intro.tex

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Aug 24, 2011, 8:08:45 AM (8 years ago)
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Intro, related work section

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 r1349 In the end, as summarized by Figure~\ref{perf-energy} our Parabix-based XML parser improves the performance by %?$\times$ performance %by ?$\times$ and energy efficiency %by ?$\times$ several-fold compared to widely-used software parsers and approaching the performance of to widely-used software parsers, approaching the %?$cycles/input-byte$ performance of ASIC XML Overall we make the following contributions in this paper. 1) We introduce the Parabix architecture, tool chain and run-time 1) We outline the Parabix architecture, tool chain and run-time environment and describe how it may be used to produce efficient XML parser implementations on a variety of commodity processors. parsing. 2) We compare our Parabix XML parsers against conventional parsers 2) We compare Parabix XML parsers against conventional parsers and assess the improvement in overall performance and energy efficiency on each platform.   We are the first to compare and contrast SSE/AVX extensions across Section~\ref{section:multithread} then looks at the multithreading of the Parabix XML parser using pipeline parallelism. Section~\ref{section:related} discusses related work, while Section~\ref{section:conclusion} concludes the paper.


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# Thread: Relation between a series and its alternating version.

1. ## Relation between a series and its alternating version.

Hello! Here is something that I've been wondering for quite a while. Consider the following two infinite series:

$S = \sum_{k\ge0}\frac{1}{(ak+b)(ck+d)}, ~~ T = \sum_{k\ge0}\frac{(-1)^k}{(ak+b)(ck+d)}$

Assuming that all issues of convergence are sorted, what's the relation (if any) between the values of S and T?

2. I don't know if I have correctly understood your question. In general, if $u_k\geq 0$ for all k and $\sum_{k\geq 0}u_k,\;\sum_{k\geq 0}(-1)^ku_k$ are both convergent (with sums S and T respectively) then, it is clear that $\sum_{k\geq 0}u_{2k}$ is convergent (denote by R its sum). Then, according to the algebra of series : S+T=R .

3. Originally Posted by FernandoRevilla
I don't know if I have correctly understood your question. In general, if $u_k\geq 0$ for all k and $\sum_{k\geq 0}u_k,\;\sum_{k\geq 0}(-1)^ku_k$ are both convergent (with sums S and T respectively) then, it is clear that $\sum_{k\geq 0}u_{2k}$ is convergent (denote by R its sum). Then, according to the algebra of series : S+T=R .
I think you have understood my question correctly, thank you. What about $\sum_{k\ge 0}(-1)^ku_{2k}$? I think it's convergent, but I'm not sure how it relates S and T. Basically, what I wanted to see if there is a way to find S and T by the just finding one of them.

4. Originally Posted by TheCoffeeMachine
Basically, what I wanted to see if there is a way to find S and T by the just finding one of them.

Consider the series:

$(i)\quad 0+\dfrac{1}{1^2}+0+\dfrac{1}{2^2}+0+\dfrac{1}{3^2} +\ldots$

$(ii)\quad \dfrac{1}{1^2}+0+\dfrac{1}{2^2}+0+\dfrac{1}{3^2}+0 +\ldots$

In both cases $S=\pi^2/6$ . However $T=-\pi^2/6$ in the first case and $T=\pi^2/6$ in the second one.


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# American Institute of Mathematical Sciences

September  2020, 10(3): 345-354. doi: 10.3934/naco.2020006

## Sliding mode control for uncertain T-S fuzzy systems with input and state delays

 School of science, Shenyang University of Technology, Shenyang, Liaoning 110870, China

* Corresponding author: Ruxia Zhang

Received  May 2019 Revised  August 2019 Published  February 2020

Fund Project: The first author is supported by is supported by National Nature Science Foundation under grant NO.61673099 and Provincial Education Department Key Project (LZGD2017039)

In this paper, the problem of sliding mode control (SMC) for uncertain T-S (Tagaki-Sugeno) fuzzy systems with input and state delays is investigated, in which the nonlinear uncertain terms are unknown, and also unmatched. For the T-S fuzzy model of the controlled object, a method based on sliding mode compensator is designed, and the system is controlled by sliding mode. Based on solving linear matrix inequalities (LMI), we obtain the design method of sliding mode and controller. The sufficient conditions for the asymptotical stability of the sliding mode dynamics are given by using LMI technique and the Lyapunov stability theory, and it has been shown that the state trajectories can be driven onto the sliding surface in a finite time. Finally, a numerical example is provided to illustrate the effectiveness of the proposed theories.

Citation: Yuan Li, Ruxia Zhang, Yi Zhang, Bo Yang. Sliding mode control for uncertain T-S fuzzy systems with input and state delays. Numerical Algebra, Control and Optimization, 2020, 10 (3) : 345-354. doi: 10.3934/naco.2020006
##### References:
 [1] Abdennebi, Nizar and Mansour, A new sliding function for discrete predictive sliding mode control of time delay systems, International Journal of Automation and Computing, 10 (2013), 288-295. [2] P. Balasubramaniam and T. Senthilkumar, Delay - Dependent robust stabilization and H$\infty$ control for uncertain stochastic T-S fuzzy systems with discrete interval and distributed time-varying delays, International Journal of Automation & Computing, 10 (2013), 18-31. [3] T. S. Chiang, C. S. Chiu and P. Liu, Adaptive TS-FNN control for a class of uncertain multi-time-delay systems: The exponentially stable sliding mode- based approach, International Journal of Adaptive Control & Signal Processing, 23 (2010), 378-399. [4] Q. Gao, H. Huang and G. Feng, Robust H$\infty$ stabilization of uncertain T-S fuzzy systems via dynamic integral sliding mode control, Ifac Proceedings Volumes, 46 (2013), 485-490. [5] S. Guo, F. Zhu and L. Xu, Unknown input observer design for Takagi-Sugeno fuzzy stochastic system, International Journal of Control Automation & Systems, 13 (2015), 1003-1009. [6] C. Han, G. Zhang and L. Wu, Sliding mode control of T-S fuzzy descriptor systems with time-delay, Journal of the Franklin Institute, 349 (2012), 1430-1444.  doi: 10.1016/j.jfranklin.2011.07.001. [7] M. Kchaou, H. Gassara and A. E. Hajjaji, Dissipativity-based integral sliding mode control for a class of Takagi-Sugeno fuzzy singular systems with time-varying delay, Control Theory & Applications Iet, 8 (2014), 2045-2054.  doi: 10.1049/iet-cta.2014.0101. [8] Y. M. Li and Y. Y. Li, Fuzzy control for nonlinear uncertain T-S fuzzy systems with time-varying delays, Applied Mechanics and Materials, 6 (2013), 341-342. [9] R. Li and Q. Zhang, Robust H$_\infty$ sliding mode observer design for a class of Takagi-Sugeno fuzzy descriptor systems with time-varying delay, Applied Mathematics & Computation, 337 (2018), 158-178.  doi: 10.1016/j.amc.2018.05.008. [10] Z. Liu and C. Gao, A new result on robust H$\infty$ control for uncertain time-delay singular systems via sliding mode control, Complexity, 21 (2016), 165-177.  doi: 10.1002/cplx.21793. [11] R. M. Nagarale and B. M. Patre, Exponential function based fuzzy sliding mode control of uncertain nonlinear systems, International Journal of Dynamics & Control, 4 (2016), 67-75.  doi: 10.1007/s40435-014-0117-2. [12] T. Niknam and M. H. Khooban, Fuzzy sliding mode control scheme for a class of non-linear uncertain chaotic systems, Iet Science Measurement & Technology, 7 (2013), 249-255. [13] L. Ren, S. Xie and Z. G. Miao, Fuzzy robust sliding mode control of a class of uncertain systems, Journal of Central South University, 23 (2016), 2296-2304. [14] D. B. Salem and W. Saad, Integral sliding mode control for systems with time-varying input and state delays, International Conference on Engineering & Mis, (2017), 978–982. [15] A. Si-Ammour, S. Djennoune and M. Bettayeb, A sliding mode control for linear fractional systems with input and state delays, Communications in Nonlinear Science & Numerical Simulation, 14 (2009), 2310-2318.  doi: 10.1016/j.cnsns.2008.05.011. [16] H. Wang, B. Zhou and R. Lu, New stability and stabilization criteria for a class of fuzzy singular systems with time-varying delay, Journal of the Franklin Institute, 351 (2014), 3766-3781.  doi: 10.1016/j.jfranklin.2013.02.030. [17] Y. Wang, Y. Xia and H. Li, A new integral sliding mode design method for nonlinear stochastic systems, Automatica, 90 (2018), 304-309.  doi: 10.1016/j.automatica.2017.11.029. [18] J. Wu, Robust stabilization for uncertain T-S fuzzy singular system, International Journal of Machine Learning & Cybernetics, 7 (2016), 699-706. [19] Y. Xia, H. Yang, M. Fu and P. Shi, Sliding mode control for linear systems with time-varying input and state delays, Circuits Systems, and Signal Processing, 30 (2011), 629-641.  doi: 10.1007/s00034-010-9237-x. [20] L. Xiao, H. Su and J. Chu, Sliding mode prediction tracking control design for uncertain systems, Asian Journal of Control, 9 (2010), 317-325.  doi: 10.1111/j.1934-6093.2007.tb00417.x. [21] X. G. Yan, S. Spurgeon and Y. Korlov, Output feedback control synthesis for non-linear time- delay systems using a sliding-mode observer, IMA Journal of Mathematical Control and Information, 31 (2014), 501-518.  doi: 10.1093/imamci/dnt028. [22] S. Y. Yoon and Z. Lin, Robust output regulation of linear time-delay systems: A state predictor approach, International Journal of Robust & Nonlinear Control, 26 (2016), 1686-1740.  doi: 10.1002/rnc.3374. [23] J. Yu and Z. Yi, Stability analysis and fuzzy control for uncertain delayed T-S nonlinear systems, International Journal of Fuzzy Systems, 18 (2016), 1-8.  doi: 10.1007/s40815-016-0203-z. [24] Y. Zhang, Robust stability and H$\infty$ control of discrete-time uncertain impulsive systems with time-varying delay, Circuits Systems & Signal Processing, 35 (2016), 3882-3912. [25] Y. Zhao, J. Shen and D. Chen, New stability criterion for discrete-time genetic regulatory networks with time-varying delays and stochastic disturbances, Mathematical Problems in Engineering, 2016 (2016), 1-13.  doi: 10.1155/2016/7634680.

show all references

##### References:
 [1] Abdennebi, Nizar and Mansour, A new sliding function for discrete predictive sliding mode control of time delay systems, International Journal of Automation and Computing, 10 (2013), 288-295. [2] P. Balasubramaniam and T. Senthilkumar, Delay - Dependent robust stabilization and H$\infty$ control for uncertain stochastic T-S fuzzy systems with discrete interval and distributed time-varying delays, International Journal of Automation & Computing, 10 (2013), 18-31. [3] T. S. Chiang, C. S. Chiu and P. Liu, Adaptive TS-FNN control for a class of uncertain multi-time-delay systems: The exponentially stable sliding mode- based approach, International Journal of Adaptive Control & Signal Processing, 23 (2010), 378-399. [4] Q. Gao, H. Huang and G. Feng, Robust H$\infty$ stabilization of uncertain T-S fuzzy systems via dynamic integral sliding mode control, Ifac Proceedings Volumes, 46 (2013), 485-490. [5] S. Guo, F. Zhu and L. Xu, Unknown input observer design for Takagi-Sugeno fuzzy stochastic system, International Journal of Control Automation & Systems, 13 (2015), 1003-1009. [6] C. Han, G. Zhang and L. Wu, Sliding mode control of T-S fuzzy descriptor systems with time-delay, Journal of the Franklin Institute, 349 (2012), 1430-1444.  doi: 10.1016/j.jfranklin.2011.07.001. [7] M. Kchaou, H. Gassara and A. E. Hajjaji, Dissipativity-based integral sliding mode control for a class of Takagi-Sugeno fuzzy singular systems with time-varying delay, Control Theory & Applications Iet, 8 (2014), 2045-2054.  doi: 10.1049/iet-cta.2014.0101. [8] Y. M. Li and Y. Y. Li, Fuzzy control for nonlinear uncertain T-S fuzzy systems with time-varying delays, Applied Mechanics and Materials, 6 (2013), 341-342. [9] R. Li and Q. Zhang, Robust H$_\infty$ sliding mode observer design for a class of Takagi-Sugeno fuzzy descriptor systems with time-varying delay, Applied Mathematics & Computation, 337 (2018), 158-178.  doi: 10.1016/j.amc.2018.05.008. [10] Z. Liu and C. Gao, A new result on robust H$\infty$ control for uncertain time-delay singular systems via sliding mode control, Complexity, 21 (2016), 165-177.  doi: 10.1002/cplx.21793. [11] R. M. Nagarale and B. M. Patre, Exponential function based fuzzy sliding mode control of uncertain nonlinear systems, International Journal of Dynamics & Control, 4 (2016), 67-75.  doi: 10.1007/s40435-014-0117-2. [12] T. Niknam and M. H. Khooban, Fuzzy sliding mode control scheme for a class of non-linear uncertain chaotic systems, Iet Science Measurement & Technology, 7 (2013), 249-255. [13] L. Ren, S. Xie and Z. G. Miao, Fuzzy robust sliding mode control of a class of uncertain systems, Journal of Central South University, 23 (2016), 2296-2304. [14] D. B. Salem and W. Saad, Integral sliding mode control for systems with time-varying input and state delays, International Conference on Engineering & Mis, (2017), 978–982. [15] A. Si-Ammour, S. Djennoune and M. Bettayeb, A sliding mode control for linear fractional systems with input and state delays, Communications in Nonlinear Science & Numerical Simulation, 14 (2009), 2310-2318.  doi: 10.1016/j.cnsns.2008.05.011. [16] H. Wang, B. Zhou and R. Lu, New stability and stabilization criteria for a class of fuzzy singular systems with time-varying delay, Journal of the Franklin Institute, 351 (2014), 3766-3781.  doi: 10.1016/j.jfranklin.2013.02.030. [17] Y. Wang, Y. Xia and H. Li, A new integral sliding mode design method for nonlinear stochastic systems, Automatica, 90 (2018), 304-309.  doi: 10.1016/j.automatica.2017.11.029. [18] J. Wu, Robust stabilization for uncertain T-S fuzzy singular system, International Journal of Machine Learning & Cybernetics, 7 (2016), 699-706. [19] Y. Xia, H. Yang, M. Fu and P. Shi, Sliding mode control for linear systems with time-varying input and state delays, Circuits Systems, and Signal Processing, 30 (2011), 629-641.  doi: 10.1007/s00034-010-9237-x. [20] L. Xiao, H. Su and J. Chu, Sliding mode prediction tracking control design for uncertain systems, Asian Journal of Control, 9 (2010), 317-325.  doi: 10.1111/j.1934-6093.2007.tb00417.x. [21] X. G. Yan, S. Spurgeon and Y. Korlov, Output feedback control synthesis for non-linear time- delay systems using a sliding-mode observer, IMA Journal of Mathematical Control and Information, 31 (2014), 501-518.  doi: 10.1093/imamci/dnt028. [22] S. Y. Yoon and Z. Lin, Robust output regulation of linear time-delay systems: A state predictor approach, International Journal of Robust & Nonlinear Control, 26 (2016), 1686-1740.  doi: 10.1002/rnc.3374. [23] J. Yu and Z. Yi, Stability analysis and fuzzy control for uncertain delayed T-S nonlinear systems, International Journal of Fuzzy Systems, 18 (2016), 1-8.  doi: 10.1007/s40815-016-0203-z. [24] Y. Zhang, Robust stability and H$\infty$ control of discrete-time uncertain impulsive systems with time-varying delay, Circuits Systems & Signal Processing, 35 (2016), 3882-3912. [25] Y. Zhao, J. Shen and D. Chen, New stability criterion for discrete-time genetic regulatory networks with time-varying delays and stochastic disturbances, Mathematical Problems in Engineering, 2016 (2016), 1-13.  doi: 10.1155/2016/7634680.
Trajectory of state $x_{1}(t)$ before adding controller
Trajectory of state $x_{2}(t)$ before adding controller
control input signal
Trajectory of state $x_{1}(t)$ after adding controller
Trajectory of state $x_{2}(t)$ after adding controller
Trajectory of sliding mode variable
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# city-block metric

The city-block metric, defined on $\mathbb{R}^{n}$, is

 $d(a,b)=\sum_{i=1}^{n}|b_{i}-a_{i}|$

where $a$ and $b$ are vectors in $\mathbb{R}^{n}$ with $a=(a_{1},\ldots,a_{n})$ and $b=(b_{1},\ldots,b_{n})$.

In two dimensions and with discrete-valued vectors, when we can picture the set of points in $\mathbb{Z}\times\mathbb{Z}$ as a grid, this is simply the number of edges between points that must be traversed to get from $a$ to $b$ within the grid. This is the same problem as getting from corner $a$ to $b$ in a rectilinear downtown area, hence the name “city-block metric.”

Title city-block metric CityblockMetric 2013-03-22 12:12:57 2013-03-22 12:12:57 akrowne (2) akrowne (2) 9 akrowne (2) Definition msc 54E35 city-block distance taxicab metric


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# Navier-Stokes system

I have to study this system which name is Navier-Stokes. Can you explain please what means that $p$, $u$ and $(u \cdot \nabla)u$. What represents in reality? Tell me please, how should I read the factor: $(u \cdot \nabla)u$? "$u$ multiplied with gradient applied to $u$ " ?

$(N-S)\begin{cases} -\mu \Delta u +(u \cdot \nabla)u+\nabla{p}=f &\mbox{in } \Omega, \\ \mbox{div }u=0 & \mbox{in } \Omega,\\ u_{\mid{\Gamma}}=0. \end{cases}$

one more question, what happens with with the system if $(u \cdot \nabla )u=0$ ? I found that the system describe the motion of a incompressible viscous fluid and it suppose the the motion is stationary but no slow, what means that stationary and that slow?

-
This is all explained in the Wikipedia article about the Navier-Stokes equations. Can you indicate what it is, that is not clear about this? – Bernhard Apr 7 '13 at 21:02
if $(u \cdot \nabla)u=0$ then the system won't be slow ? – Iuli Apr 7 '13 at 21:04
Can you restrict this to a single answerable question? – Sklivvz Apr 7 '13 at 21:13
That not the conventional Navier-Stokes system because it's missing the time derivative. – aberration Apr 25 '13 at 19:22
The time derivative is the (u⋅∇)u term. The partial derivative with respect to time is missing, which means the flow is steady. – jjack Aug 16 '15 at 19:18

$(u \cdot \nabla)u$ is the so called advective acceleration term which arises when you consider the Navier-Stokes equations in an Eulerian frame of reference. It accounts for the effect that the we are following the particle as it moves around in the fluid, presumably to regions of the flow where the velocity is different. In contrast, if you consider the Navier-Stokes in Lagrangian coordinates, we are by definition tracking individual particles and therefore that term is not present. In large magnitudes, this term is highly-nonlinear and responsible for much of the more interesting behavior we see in fluid motion.

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The term $(u \cdot \nabla)u$ describes non-linear advective acceleration through a fixed point in the stationary frame of reference (Eulerian frame of reference). In your momentum equation, you need to multiply by this term by $\rho$ to have rate of change of momentum per unit volume.

If $(u \cdot \nabla)u = 0$ in your momentum equation, then you have Stokes flow (creeping flow). This applies for very low Reynolds numbers, i.e. $Re << 1$, where inertial forces are small compared with viscous forces.

-

The answer to the second part: if $( u. \nabla) u =0$ would mean the flow is uniform in space , i.e. uniform flow. It can still have a time varying part.

x component of : $$( u. \nabla) u = u {du \over dx} + v {du \over dy} + w {du \over dz}$$ Similarly you would have y-component (in terms of v) and z-component (in terms of w).

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Should'n be partial derivatives? – aberration Apr 25 '13 at 19:23
Sorry about that. @aberration is right. I stand corrected. – jadelord May 3 '13 at 5:55

The Navier-Stokes equations represent the flow of a fluid with friction. They consist of the following conservation equations:

1. Conservation of mass (second line):

div u=0

2. Conservation of momentum (first line)

−μΔu+(u⋅∇)u+∇p=f

3. Conservation of energy (probably, in the third line, but I don't know the nomenclature):

u∣Γ=0.

-

For the first part look up the difference between Lagrangian and Eulerian reference frames and also total derivative. Second bit read about the Reynolds number of a system.

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## protected by Qmechanic♦Aug 16 '15 at 18:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.


text:
21 Haru 13 - The Results

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I don't mind including Jejima in the list of participants this basho as he always plays "A"s. If he had different choices, I wouldn't agree to be included as you may say that he have used the stats table.

And the more participants the better.

Regards,

Achiyama

Points well taken. A ONE-TIME deadline extension for All-A Jejima is being hotly discussed by the Board of Directors during our traditional pre-basho gourmet repast of peanut butter sandwiches and Calpis. Stand by.

Edit: We are convening a conclave.

Edited by Asojima

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It was a long conclave.

At the end we lost many time to search and find the right fuel to produce white smoke.

But we have as ever no new pope, but a new special rule for a very special law case.

This new rule will enter into the history of Game 21 and runs as follows:

If a player always in the past has picked all bouts A and one time missed the deadline,

but not the first Makuuchi bout on Shonichi, again voting all bouts A,

he shall be allowed to take part with this and only this pick.

Tis new rule shall be named forever "LEX JEJIMA".

It is signed from all participants of the conclave.

In their all name

Profomisakari

Edited by Profomisakari

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On day 1 we had 3 bouts.

It's logical, to have now too three leaders, amomg them with the best tiebreaker Mmikasazuma and three Red lamp holders, among them Profomisakari with the best tiebreaker.

This Haru Basho Game 21 will be a hard job for me.

This is the ranking table:

Rk Player . . . . .Pts

.1 Mmikasazuma. . . 3

.2 Yamasanzan . . . 3

.3 Holleshoryu. . . 3

.4 Kuroimori. . . . 2

.5 Tsubame. . . . . 2

.6 Pandaazuma . . . 2

.7 Chisaiyama . . . 2

.8 Achiyama . . . . 1

.9 Tochinofuji. . . 1

10 Peeter . . . . . 1

11 Wamahada . . . . 1

12 Bugman . . . . . 1

13 Asojima. . . . . 1

14 Ronnie . . . . . 1

15 Flohru . . . . . 1

16 Screeching Owl . 1

17 Susanoo. . . . . 1

18 Takanorappa. . . 1

19 Jejima . . . . . 1

20 Profomisakari. . 0

21 Fujisan. . . . . 0

22 Terarno. . . . . 0

Profomisakari

• 1

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Now after two days we have 5 bouts and two players with the full point result.

Per tiebreaker leads now Yamasanzan before Holleshoryu and 3 players with each one point back.

Because both leaders the tomorrow's bout have different picked, we then have only one leader with the foll points.

Who will it be?

This is the ranking:

Rk Player . . . . .Pts

.1 Yamasanzan . . . 5

.2 Holleshoryu. . . 5

.3 Kuroimori. . . . 4

.4 Tsubame. . . . . 4

.5 Mmikasazuma. . . 4

.6 Wamahada . . . . 3

.7 Pandaazuma . . . 3

.8 Chisaiyama . . . 3

.9 Flohru . . . . . 3

10 Screeching Owl . 3

11 Achiyama . . . . 2

12 Tochinofuji. . . 2

13 Peeter . . . . . 2

14 Bugman . . . . . 2

15 Asojima. . . . . 2

16 Ronnie . . . . . 2

17 Profomisakari. . 2

18 Susanoo. . . . . 2

19 Takanorappa. . . 2

20 Jejima . . . . . 2

21 Fujisan. . . . . 1

22 Terarno. . . . . 1

Profomisakari

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Baruto won and now one player leads alone with 6 : 0 points before four pursuers: Holleshoryu.

Tomorrow he will hold this place, because we have no bout.

The players ranking table:

Rk Player . . . . . Pts

.1 Holleshoryu. . . 6

.2 Kuroimori. . . . 5

.3 Tsubame. . . . . 5

.4 Mmikasazuma. . . 5

.5 Yamasanzan . . . 5

.6 Wamahada . . . . 4

.7 Chisaiyama . . . 4

.8 Peeter . . . . . 3

.9 Asojima. . . . . 3

10 Ronnie . . . . . 3

11 Profomisakari. . 3

12 Pandaazuma . . . 3

13 Flohru . . . . . 3

14 Screeching Owl . 3

15 Achiyama . . . . 2

16 Tochinofuji. . . 2

17 Bugman . . . . . 2

18 Fujisan. . . . . 2

19 Susanoo. . . . . 2

20 Takanorappa. . . 2

21 Jejima . . . . . 2

22 Terarno. . . . . 1

Profomisakari

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Sorry, but I don't agree to this "Lex Jejima" - deadline is deadline and if the deadline has passed the deadline has passed and no more entries are possible, independent whether participant has always the same voting scheme. What do the others think about this?

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Sorry, but I don't agree to this "Lex Jejima" - deadline is deadline and if the deadline has passed the deadline has passed and no more entries are possible, independent whether participant has always the same voting scheme. What do the others think about this?

Before accepting his entry, we obtained a concession that, should he win, the awarded trip up the Yalu River would be by rowboat.

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Holleshoryu goes his way.

After seven bout he has as ever only correct picks.

So he leads befre three competitors with each 6 and two with 5 points.

The ranking table is as follows:

Rk Player . . . . .Pts

.1 Holleshoryu. . . 7

.2 Kuroimori. . . . 6

.3 Tsubame. . . . . 6

.4 Mmikasazuma. . . 6

.5 Wamahada . . . . 5

.6 Yamasanzan . . . 5

.7 Ronnie . . . . . 4

.8 Profomisakari. . 4

.9 Chisaiyama . . . 4

10 Flohru . . . . . 4

11 Screeching Owl . 4

12 Achiyama . . . . 3

13 Peeter . . . . . 3

14 Asojima. . . . . 3

15 Pandaazuma . . . 3

16 Fujisan. . . . . 3

17 Jejima . . . . . 3

18 Tochinofuji. . . 2

19 Bugman . . . . . 2

20 Susanoo. . . . . 2

21 Takanorappa. . . 2

22 Terarno. . . . . 2

Profomisakari

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Only one bout we had today.

Tochinoshin won and only 6 players had voted for him.

Among them was nobody of the 6 best placed of yesterday's ranking table.

So we have the same leaders as yesterday with Holleshoryu on the top.

Tomorrow is no Game 21 bout.

This is the new ranking table:

Rk Player . . . . .Pts

.1 Holleshoryu. . . 7

.2 Kuroimori. . . . 6

.3 Tsubame. . . . . 6

.4 Mmikasazuma. . . 6

.5 Wamahada . . . . 5

.6 Ronnie . . . . . 5

.7 Yamasanzan . . . 5

.8 Profomisakari. . 4

.9 Chisaiyama . . . 4

10 Flohru . . . . . 4

11 Fujisan. . . . . 4

12 Screeching Owl . 4

13 Jejima . . . . . 4

14 Achiyama . . . . 3

15 Peeter . . . . . 3

16 Bugman . . . . . 3

17 Asojima. . . . . 3

18 Pandaazuma . . . 3

19 Susanoo. . . . . 3

20 Takanorappa. . . 3

21 Tochinofuji. . . 2

22 Terarno. . . . . 2

Profomisakari

Edited by Profomisakari

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Ha...interesting to see that I am just as rubbish at this game as I am at so many others this basho!

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Ha...interesting to see that I am just as rubbish at this game as I am at so many others this basho!

Hey, some of us our proud of our three points you know. :-D

Edited by Bugman

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Two bouts today - leader Holloshoryu has both bouts picked false, all three pursuers have one correct.

So we now have 4 players with each 7 points, Holle-san with the worst tiebreaker of them, but Kuroimoi with the best.

The ranking table:

Rk Player . . . . .Pts

.1 Kuroimori. . . . 7

.2 Tsubame. . . . . 7

.3 Mmikasazuma. . . 7

.4 Holleshoryu. . . 7

.5 Yamasanzan . . . 6

.6 Fujisan. . . . . 6

.7 Wamahada . . . . 5

.8 Ronnie . . . . . 5

.9 Profomisakari. . 5

10 Chisaiyama . . . 5

11 Flohru . . . . . 5

12 Screeching Owl . 5

13 Jejima . . . . . 5

14 Achiyama . . . . 4

15 Peeter . . . . . 4

16 Bugman . . . . . 4

17 Asojima. . . . . 4

18 Pandaazuma . . . 4

19 Susanoo. . . . . 4

20 Takanorappa. . . 4

21 Tochinofuji. . . 2

22 Terarno. . . . . 2

Profomisakari

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Two buts today, all leader one point more.

The same row as yesterday at the top places.

Tomorrow is no game 21 bout.

The players ranking is now as follows:

Rk Player . . . . .Pts

.1 Kuroimori. . . . 8

.2 Tsubame. . . . . 8

.3 Mmikasazuma. . . 8

.4 Holleshoryu. . . 8

.5 Fujisan. . . . . 7

.6 Yamasanzan . . . 6

.7 Flohru . . . . . 6

.8 Screeching Owl . 6

.9 Peeter . . . . . 5

10 Wamahada . . . . 5

11 Ronnie . . . . . 5

12 Profomisakari. . 5

13 Pandaazuma . . . 5

14 Chisaiyama . . . 5

15 Susanoo. . . . . 5

16 Jejima . . . . . 5

17 Achiyama . . . . 4

18 Bugman . . . . . 4

19 Asojima. . . . . 4

20 Takanorappa. . . 4

21 Tochinofuji. . . 3

22 Terarno. . . . . 2

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Two buts today, all leader one point more.

The same row as yesterday at the top places.

Tomorrow is no game 21 bout.

The players ranking is now as follows:

Rk Player . . . . .Pts

10 Wamahada . . . . 5

Shouldn't I have 6 points now, as I had Wakanosato in my prediction?

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3 buts today, not 2! ;-)

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Thank you Peeter,

I oversaw bout 21 and Waka's win.

This bout has of the leaders group only Kuroimori correct picked.

So he leads now no more per tiebreaker, but per points (9) before four pursuers

(Fujisan came new to the group).

This is the table:

Rk Player . . . . .Pts

.1 Kuroimori. . . . 9

.2 Tsubame. . . . . 8

.3 Mmikasazuma. . . 8

.4 Fujisan. . . . . 8

.5 Holleshoryu. . . 8

.6 Yamasanzan . . . 7

.7 Flohru . . . . . 7

.8 Peeter . . . . . 6

.9 Wamahada . . . . 6

10 Ronnie . . . . . 6

11 Profomisakari. . 6

12 Pandaazuma . . . 6

13 Screeching Owl . 6

14 Susanoo. . . . . 6

15 Achiyama . . . . 5

16 Bugman . . . . . 5

17 Chisaiyama . . . 5

18 Jejima . . . . . 5

19 Tochinofuji. . . 4

20 Asojima. . . . . 4

21 Takanorappa. . . 4

22 Terarno. . . . . 2

And as ever: excuse me please!

Profomisakari

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For each basho, we rank the entrants by the number of responses that they have in agreement with the other entrants. The table shows the number of correct responses and the agreement ranking for each entrant in the last 7 bashos. For example, the winner (1.) of the Hatsu 12 basho (1/12) had 12 correct answers and was #3 in the agreement ranking ( #1 is most agreeable).

 Place The Bashos 1/12 3/12 5/12 7/12 9/12 11/12 1/13 1. 12/ 3 14/22 11/ 9 9/ 3 10/ 3 11/23 11/20 2. 12/14 13/19 10/15 9/ 4 10/ 1 10/19 11/10 3. 12/ 2 13/23 10/11 8/13 10/ 7 10/16 11/19 4. 12/ 1 11/16 10/ 3 8/ 8 10/ 8 9/11 11/ 3 5. 11/14 11/ 8 9/ 2 8/ 6 9/ 2 9/24 11/22 6. 11/15 11/21 9/ 4 8/20 9/11 8/11 11/ 2 7. 11/10 10/16 9/18 7/18 9/16 8/15 10/11 8. 11/ 6 10/ 3 9/16 7/20 8/18 8/ 7 10/ 5 9. 10/11 10/20 9/19 7/ 1 8/20 8/ 8 10/24 10. 10/ 9 10/ 1 9/ 5 7/ 7 8/ 8 7/11 9/ 6 11. 10/ 8 10/14 8/ 6 7/12 8/19 7/11 9/12 12. 10/ 7 9/ 2 8/14 7/10 8/16 7/ 5 9/ 1 13. 9/12 9/6 8/ 8 6/16 7/ 8 7/18 9/16 14. 9/23 9/10 7/20 6/11 7/ 6 7/17 8/14 15. 9/ 6 9/ 6 7/7 6/16 7/21 6/19 8/ 8 16. 9/16 8/ 3 7/17 6/ 5 7/12 6/ 8 7/15 17. 8/18 8/16 7/10 6/19 6/14 6/ 4 7/ 4 18. 8/18 8/ 9 6/13 6/ 2 5/12 6/ 2 7/18 19. 8/20 8/13 6/ 1 6/14 5/14 6/ 1 7/ 7 20. 8/13 7/16 6/21 6/22 5/ 4 6/11 7/13 21. 7/21 7/11 5/ 9 4/ 4 6/21 6/21 22. 7/17 7/ 6 4/15 6/10 6/16 23. 7/22 7/ 6 4/23 4/ 3 6/ 8 24. 3/22 4/23

Given that each entrant, potentially, can have 21 correct responses, we can reach some conclusions from the table:

1. There is no obvious correlation between agreeability and the expected number of correct responses.
2. The vast majority of us have no idea what we are doing.

Edited by Asojima
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Kuroimori holds his one point lead.

Until the finish? We'll see!

Rk Player . . . . Pts

.1 Kuroimori. . . .10

.2 Mmikasazuma. . . 9

.3 Fujisan. . . . . 9

.4 Tsubame. . . . . 8

.5 Yamasanzan . . . 8

.6 Flohru . . . . . 8

.7 Holleshoryu. . . 8

.8 Peeter . . . . . 7

.9 Wamahada . . . . 7

10 Ronnie . . . . . 7

11 Profomisakari. . 7

12 Pandaazuma . . . 7

13 Screeching Owl . 7

14 Susanoo. . . . . 7

15 Achiyama . . . . 6

16 Jejima . . . . . 6

17 Tochinofuji. . . 5

18 Bugman . . . . . 5

19 Chisaiyama . . . 5

20 Takanorappa. . . 5

21 Asojima. . . . . 4

22 Terarno. . . . . 3

Profomisakari

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Only one bout will be on Senshuraku.

Now we know too, whic h bouts take not place.

There are 6 X boutsbut they brought only all i all 6 points for 5 players.

This is the new ranking table:

Rk Player . . . . .Pts

.1 Kuroimori. . . .10

.2 Tsubame. . . . . 9

.3 Mmikasazuma. . . 9

.4 Flohru . . . . . 9

.5 Fujisan. . . . . 9

.6 Peeter . . . . . 8

.7 Yamasanzan . . . 8

.8 Holleshoryu. . . 8

.9 Susanoo. . . . . 8

10 Wamahada . . . . 7

11 Bugman . . . . . 7

12 Ronnie . . . . . 7

13 Profomisakari. . 7

14 Pandaazuma . . . 7

15 Screeching Owl . 7

16 Achiyama . . . . 6

17 Jejima . . . . . 6

18 Tochinofuji. . . 5

19 Chisaiyama . . . 5

20 Takanorappa. . . 5

21 Asojima. . . . . 4

22 Terarno. . . . . 3

And now a little exercise:

Do we have a winner or must we wait until tomorrows great Yokozuna final??

Try it to find out!

I say:

Congratulations Kuroimori!

But details tomorrow!!

Profomisakari

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The final results:

 The Results for Haru 13 A [/th][th=2]B X Day 1. Ye Harumafuji ( 8) Yw Hakuho (13) (0) -- 15 2. O1e Kisenosato (13) O1w Kotooshu ( 7) (1) -- Osh kyujo from Day 6 3. O2e Kakuryu (19) O2w Kotoshogiku ( 2) (0) -- 9 4. Se Goeido (10) Sw Baruto (11) (0) -- 3 5. Ke Tochiozan (19) Me2 Chiyotairyu ( 2) (0) -- Chiyo kyujo from Day 7 6. Kw Aminishiki (15) M3e Ikioi ( 6) (0) -- 9 7. M1e Takayasu (17) M3w Tokitenku ( 3) (1) -- 10 8. M1w Tochinoshin ( 5) M2w Myogiryu (16) (0) -- 7 9. M4e Toyonoshima (17) M5w Kaisei ( 4) (0) -- 2 10. M4w Shohozan (14) M8w Kyokutenho ( 4) (3) -- No Match 11. M5e Aran (12) M7e Okinoumi ( 8) (1) -- 10 12. M6e Gagamaru (16) M6w Kitataiki ( 5) (0) -- 1 13. M7w Toyohibiki ( 6) M8e Aoiyama (13) (2) -- No Match 14. M9e Fujiazuma (13) M9w Yoshikaze ( 8) (0) -- 1 15. M10e Takarafuji ( 9) M11w Jokoryu (12) (0) -- 2 16. M10w Tamawashi (18) M11e Takekaze ( 3) (0) -- No Match 17. M12e Shotenro (11) M12w Masunoyama (10) (0) -- 1 18. M13e Tochinowaka (13) M14e Chiyonokuni ( 8) (0) -- Tochi kyujo from Day 7 19. M13w Sadanofuji (12) M14w Sagatsukasa ( 9) (0) -- 6 20. M15e Daido (16) M16e Oiwato ( 5) (0) -- 12 21. M15w Sotairyu ( 7) M16w Wakanosato (13) (1) -- 10

No 21 matches scheduled for Days 4, 5, 8, 11, 13 and 14.

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It's reality!!

The winner was even yesterday, but now very clear,

Kuroimori.

Great congratulations!

(Applauding...) (Gyoji...) (Neener, neener...) (Clapping wildly...) (I am not worthy...) (First prize...) (Yusho winner...)

This is the final players ranking:

Rk Player . . . . Pts

.1 Kuroimori. . . .11

.2 Tsubame. . . . .10

.3 Flohru . . . . .10

.4 Fujisan. . . . .10

.5 Mmikasazuma. . . 9

.6 Susanoo. . . . . 9

.7 Peeter . . . . . 8

.8 Wamahada . . . . 8

.9 Bugman . . . . . 8

10 Ronnie . . . . . 8

11 Pandaazuma . . . 8

12 Yamasanzan . . . 8

13 Holleshoryu. . . 8

14 Achiyama . . . . 7

15 Profomisakari. . 7

16 Screeching Owl . 7

17 Chisaiyama . . . 6

18 Jejima . . . . . 6

19 Tochinofuji. . . 5

20 Asojima. . . . . 5

21 Takanorappa. . . 5

22 Terarno. . . . . 4

Thank you all for playing

and especially Asojima for Selection and daily bout listing.

Let's see together in May to play together a new game 21!!

Profomisakari

Edited by Profomisakari
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Woah 8 points, i don't think i've ever done better hehe. :)

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It's reality!!

The winner was even yesterday, but now very clear,

Kuroimori.

Great congratulations!

(Applauding...) (Gyoji...) (Neener, neener...) (Clapping wildly...) (I am not worthy...) (First prize...) (Yusho winner...)

This is the final players ranking:

Rk Player . . . . Pts

.1 Kuroimori. . . .11

.2 Tsubame. . . . .10

.3 Flohru . . . . .10

.4 Fujisan. . . . .10

.5 Mmikasazuma. . . 9

.6 Susanoo. . . . . 9

.7 Peeter . . . . . 8

.8 Wamahada . . . . 8

.9 Bugman . . . . . 8

10 Ronnie . . . . . 8

11 Pandaazuma . . . 8

12 Yamasanzan . . . 8

13 Holleshoryu. . . 8

14 Achiyama . . . . 7

15 Profomisakari. . 7

16 Screeching Owl . 7

17 Chisaiyama . . . 6

18 Jejima . . . . . 6

19 Tochinofuji. . . 5

20 Asojima. . . . . 5

21 Takanorappa. . . 5

22 Terarno. . . . . 4

Thank you all for playing

and especially Asojima for Selection and daily bout listing.

Let's see together in May to play together a new game 21!!

Profomisakari

Profomisakari

I am humbled to have won the 21 Haru 2013 Yusho. (Bow...)

It proved to be a tough bout selection, as I luckily finished just barely over the .500 mark.

I will gambarize as much as possible to come out strong for the 21 Natsu 2013 competition!

Thanks to all fellow competitors and especially to Asojima and Profomisakari for running the show!


text:
#### Number System One Mark Questions

7th Standard EM

Reg.No. :
•
•
•
•
•
•

Maths

Time : 01:00:00 Hrs
Total Marks : 20
5 x 1 = 5
1. The temperature at 12 noon at a certain place was 18° above zero. If it decreases at the rate of 3° per hour at what time it would be 12° below zero?

(a)

12 mid night

(b)

12 noon

(c)

10 am

(d)

10 pm

2. (-5)-(-18)=____

(a)

23

(b)

-13

(c)

13

(d)

-23

3. Which of the following does not represent an Integer?

(a)

0÷(-7)

(b)

20÷(−4)

(c)

(-9)÷3

(d)

(12)÷5

4. (−200)÷10is

(a)

20

(b)

-20

(c)

-190

(d)

210

5. The set of integers is not closed under

(a)

(b)

Subtraction

(c)

Multiplication

(d)

Division

6. 5 x 1 = 5
7.  -(30)+____=60

()

90

8. -44+____ =-88

()

(-44)

9. (100) $\times$ ______=-500

()

-5

10.  (−40) ÷____=40

()

-1

11. (-52)+(-52)=____

()

-104

12. 5 x 1 = 5
13. The additive inverse of (–32) is (–32)

(a) True
(b) False
14. (-675)-(-400)=-1075

(a) True
(b) False
15. 15-(-18) is the same as 15+18

(a) True
(b) False
16. (-45)-(-8)=(-8)-(-45)

(a) True
(b) False
17. (-15)×5=75

(a) True
(b) False
18. 5 x 1 = 5
19. For any two integers 72 and 108, 72+108 is an also an integer

20. (1)

Multiplicative identity

21. For any three integers 68, 25 and 99
68 × ( 25 + 99) = ( 68 × 25 ) + ( 68x 9

22. (2)

Commutative property under multiplication.

23. 0 + (−138) = (−138) = (−138)+0

24. (3)

25. For any two integers (−5) and 10
(−5) × 10 = 10 ×(−5)

26. (4)


text:
mikeymcclain1 |x| + 5 = 8 one year ago one year ago

1. twalker204

3

Notably $-x+5=8\text{ when }x<0\\x+5=8\text{ when }x\geq0$Because that's how $$|x|$$ is defined, myyess?

3. mikeymcclain1

could it be 3 or a negitive 3 ?

4. hoblos

|x| + 5 = 8 |x| = 3 x=±3


text:
# Ethanol fermentation

In ethanol fermentation, (1) one glucose molecule breaks down into two pyruvates. The energy from this exothermic reaction is used to bind the inorganic phosphates to ADP and convert NAD+ to NADH. (2) The two pyruvates are then broken down into two acetaldehydes and give off two CO2 as a by-product. (3) The two acetaldehydes are then converted to two ethanol by using the H- ions from NADH, converting NADH back into NAD+.

Ethanol fermentation, also called alcoholic fermentation, is a biological process which converts sugars such as glucose, fructose, and sucrose into cellular energy, producing ethanol and carbon dioxide as by-products. Because yeasts perform this conversion in the absence of oxygen, alcoholic fermentation is considered an anaerobic process. It also takes place in some species of fish (including goldfish and carp) where (along with lactic acid fermentation) it provides energy when oxygen is scarce.[1]

Ethanol fermentation has many uses, including the production of alcoholic beverages, the production of ethanol fuel, and bread cooking.

## Biochemical process of fermentation of sucrose

A laboratory vessel being used for the fermentation of straw.
Fermentation of sucrose by yeast.

The chemical equations below summarize the fermentation of sucrose (${\displaystyle {{\ce {C12H22O11}}}}$) into ethanol (${\displaystyle {\ce {C2H5OH}}}$). Alcoholic fermentation converts one mole of glucose into two moles of ethanol and two moles of carbon dioxide, producing two moles of ATP in the process.

The overall chemical formula for alcoholic fermentation is:

${\displaystyle {\ce {C6H12O6 -> 2 C2H5OH + 2 CO2}}}$

Sucrose is a dimer of glucose and fructose molecules. In the first step of alcoholic fermentation, the enzyme invertase cleaves the glycosidic linkage between the glucose and fructose molecules.

${\displaystyle {\ce {C12H22O11 + H2O + invertase -> 2 C6H12O6}}}$

Next, each glucose molecule is broken down into two pyruvate molecules in a process known as glycolysis.[2] Glycolysis is summarized by the equation:

${\displaystyle {\ce {C6H12O6 + 2 ADP + 2 P_i + 2 NAD+ -> 2 CH3COCOO^- + 2 ATP + 2 NADH + 2 H2O + 2 H+}}}$

CH3COCOO is pyruvate, and Pi is inorganic phosphate. Finally, pyruvate is converted to ethanol and CO2 in two steps, regenerating oxidized NAD+ needed for glycolysis:

1. CH3COCOO + H+ → CH3CHO + CO2

catalyzed by pyruvate decarboxylase

This reaction is catalyzed by alcohol dehydrogenase (ADH1 in baker's yeast).[3]

As shown by the reaction equation, glycolysis causes the reduction of two molecules of NAD+ to NADH. Two ADP molecules are also converted to two ATP and two water molecules via substrate-level phosphorylation.

### Related processes

Fermentation of sugar to ethanol and CO2 can also be done by Zymomonas mobilis, however the path is slightly different since formation of pyruvate does not happen by glycolysis but instead by the Entner–Doudoroff pathway. Other microorganisms can produce ethanol from sugars by fermentation but often only as a side product. Examples are[4]

## Effect of oxygen

Fermentation does not require oxygen. If oxygen is present, some species of yeast (e.g., Kluyveromyces lactis or Kluyveromyces lipolytica) will oxidize pyruvate completely to carbon dioxide and water in a process called cellular respiration, hence these species of yeast will produce ethanol only in an anaerobic environment (not cellular respiration).

However, many yeasts such as the commonly used baker's yeast Saccharomyces cerevisiae, or fission yeast Schizosaccharomyces pombe, prefer fermentation to respiration. These yeasts will produce ethanol even under aerobic conditions, if they are provided with the right kind of nutrition. During batch fermentation, the rate of ethanol production per milligram of cell protein is maximal for a brief period early in this process and declines progressively as ethanol accumulates in the surrounding broth. Studies demonstrate that the removal of this accumulated ethanol does not immediately restore fermentative activity, and they provide evidence that the decline in metabolic rate is due to physiological changes (including possible ethanol damage) rather than to the presence of ethanol. Several potential causes for the decline in fermentative activity have been investigated. Viability remained at or above 90%, internal pH remained near neutrality, and the specific activities of the glycolytic and alcohologenic enzymes (measured in vitro) remained high throughout batch fermentation. None of these factors appears to be causally related to the fall in fermentative activity during batch fermentation.

The formation of carbon dioxide — a byproduct of ethanol fermentation — causes bread to rise.

Ethanol fermentation causes bread dough to rise. Yeast organisms consume sugars in the dough and produce ethanol and carbon dioxide as waste products. The carbon dioxide forms bubbles in the dough, expanding it to a foam. Less than 2% ethanol remains after baking.[5][6]

## Alcoholic beverages

Primary fermentation cellar, Budweiser Brewery, Fort Collins, Colorado

All ethanol contained in alcoholic beverages (including ethanol produced by carbonic maceration) is produced by means of fermentation induced by yeast.

• Wine is produced by fermentation of the natural sugars present in grapes; cider and perry are produced by similar fermentation of natural sugar in apples and pears, respectively; and other fruit wines are produced from the fermentation of the sugars in any other kinds of fruit. Brandy and eaux de vie (e.g. slivovitz) are produced by distillation of these fruit-fermented beverages.
• Mead is produced by fermentation of the natural sugars present in honey.
• Beer, whiskey, and vodka are produced by fermentation of grain starches that have been converted to sugar by the enzyme amylase, which is present in grain kernels that have been malted (i.e. germinated). Other sources of starch (e.g. potatoes and unmalted grain) may be added to the mixture, as the amylase will act on those starches as well. Whiskey and vodka are also distilled; gin and related beverages are produced by the addition of flavoring agents to a vodka-like feedstock during distillation.
• Rice wines (including sake) are produced by the fermentation of grain starches converted to sugar by the mold Aspergillus oryzae. Baijiu, soju, and shōchū are distilled from the product of such fermentation.
• Rum and some other beverages are produced by fermentation and distillation of sugarcane. Rum is usually produced from the sugarcane product molasses.

In all cases, fermentation must take place in a vessel that allows carbon dioxide to escape but prevents outside air from coming in. This is to reduce risk of contamination of the brew by unwanted bacteria or mold and because a buildup of carbon dioxide creates a risk the vessel will rupture or fail, possibly causing injury or property damage.

## Feedstocks for fuel production

Yeast fermentation of various carbohydrate products is also used to produce the ethanol that is added to gasoline.

The dominant ethanol feedstock in warmer regions is sugarcane.[7] In temperate regions, corn or sugar beets are used.[7][8]

In the United States, the main feedstock for the production of ethanol is currently corn.[7] Approximately 2.8 gallons of ethanol are produced from one bushel of corn (0.42 liter per kilogram). While much of the corn turns into ethanol, some of the corn also yields by-products such as DDGS (distillers dried grains with solubles) that can be used as feed for livestock. A bushel of corn produces about 18 pounds of DDGS (320 kilograms of DDGS per metric ton of maize).[9] Although most of the fermentation plants have been built in corn-producing regions, sorghum is also an important feedstock for ethanol production in the Plains states. Pearl millet is showing promise as an ethanol feedstock for the southeastern U.S. and the potential of duckweed is being studied.[10]

In some parts of Europe, particularly France and Italy, grapes have become a de facto feedstock for fuel ethanol by the distillation of surplus wine.[11] Surplus sugary drinks may also be used.[12] In Japan, it has been proposed to use rice normally made into sake as an ethanol source.[13]

### Cassava as ethanol feedstock

Ethanol can be made from mineral oil or from sugars or starches. Starches are cheapest. The starchy crop with highest energy content per acre is cassava, which grows in tropical countries.

Thailand already had a large cassava industry in the 1990s, for use as cattle feed and as a cheap admixture to wheat flour. Nigeria and Ghana are already establishing cassava-to-ethanol plants. Production of ethanol from cassava is currently economically feasible when crude oil prices are above US\$120 per barrel.

New varieties of cassava are being developed, so the future situation remains uncertain. Currently, cassava can yield between 25-40 tonnes per hectare (with irrigation and fertilizer),[14] and from a tonne of cassava roots, circa 200 liters of ethanol can be produced (assuming cassava with 22% starch content). A liter of ethanol contains circa 21.46[15] MJ of energy. The overall energy efficiency of cassava-root to ethanol conversion is circa 32%.

The yeast used for processing cassava is Endomycopsis fibuligera, sometimes used together with bacterium Zymomonas mobilis.

## Byproducts of fermentation

Ethanol fermentation produces unharvested byproducts such as heat, carbon dioxide, food for livestock, water, methanol, fuels, fertilizer and alcohols.[16] The cereal unfermented solid residues from the fermentation process, which can be used as livestock feed or in the production of biogas, are referred to as Distillers grains and sold as WDG, Wet Distiller's grains, and DDGS, Dried Distiller's Grains with Solubles, respectively.

## References

1. ^ Aren van Waarde; G. Van den Thillart; Maria Verhagen (1993). "Ethanol Formation and pH-Regulation in Fish". Surviving Hypoxia. pp. 157−170. ISBN 0-8493-4226-0.
2. ^ Stryer, Lubert (1975). Biochemistry. W. H. Freeman and Company. ISBN 0-7167-0174-X.
3. ^ Raj SB, Ramaswamy S, Plapp BV. "Yeast alcohol dehydrogenase structure and catalysis". Biochemistry. 53: 5791-803. doi:10.1021/bi5006442. PMC . PMID 25157460.
4. ^ Müller, Volker (2001). "Bacterial Fermentation" (PDF). eLS. John Wiley & Sons, Ltd. doi:10.1038/npg.els.0001415. ISBN 9780470015902.
5. ^ Logan, BK; Distefano, S (1997). "Ethanol content of various foods and soft drinks and their potential for interference with a breath-alcohol test". Journal of analytical toxicology. 22 (3): 181–3. doi:10.1093/jat/22.3.181. PMID 9602932.
6. ^ "The Alcohol Content of Bread". Canadian Medical Association Journal. 16 (11): 1394–5. November 1926. PMC . PMID 20316063.
7. ^ a b c James Jacobs, Ag Economist. "Ethanol from Sugar". United States Department of Agriculture. Archived from the original on 2007-09-10. Retrieved 2007-09-04.
8. ^ "Economic Feasibility of Ethanol Production from Sugar in the United States" (PDF). United States Department of Agriculture. July 2006. Archived from the original (pdf) on 2007-08-15. Retrieved 2007-09-04.
9. ^ "Ethanol Biorefinery Locations". Renewable Fuels Association. Archived from the original on 30 April 2007. Retrieved 21 May 2007.
10. ^ "Tiny super-plant can clean up hog farms and be used for ethanol production". projects.ncsu.edu. Retrieved 2018-01-18.
11. ^ Caroline Wyatt (2006-08-10). "Draining France's 'wine lake'". BBC News. Retrieved 2007-05-21.
12. ^ Capone, John (21 November 2017). "That unsold bottle of Merlot is probably winding up in your gas tank". Quartz. Retrieved 21 November 2017.
13. ^ Japan Plans Its Own Green Fuel by Steve Inskeep. NPR Morning Edition, May 15, 2007
14. ^ Agro2: Ethanol From Cassava
15. ^ Pimentel, D. (Ed.) (1980). CRC Handbook of energy utilization in agriculture. (Boca Raton: CRC Press)
16. ^ Lynn Ellen Doxon. The Alcohol Fuel Handbook. InfinityPublishing.com. ISBN 0-7414-0646-2.
17. ^ Gil, C.; Gómez-Cordovés, C. (1986). "Tryptophol content of young wines made from Tempranillo, Garnacha, Viura and Airén grapes". Food Chemistry. 22: 59–65. doi:10.1016/0308-8146(86)90009-9.
18. ^ Szlavko, Clara M (1973). "Tryptophol, Tyrosol and Phenylethanol-The Aromatic Higher Alcohols in Beer". Journal of the Institute of Brewing. 79 (4): 283. doi:10.1002/j.2050-0416.1973.tb03541.x.
19. ^ Ribéreau-Gayon, P.; Sapis, J. C. (1965). "On the presence in wine of tyrosol, tryptophol, phenylethyl alcohol and gamma-butyrolactone, secondary products of alcoholic fermentation". Comptes Rendus de l'Académie des Sciences, Série D. 261 (8): 1915–1916. PMID 4954284. (Article in French)


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# Math Help - Unfoiling

1. ## Unfoiling

From the following example in the book

$24x^2 - 34x - 45$

Step 1
Factors of 24
1 x 24
2 x 12
4 x 6
3 x 8

Step 2
Factors of 45
1 x 45
5 x 9
3 x 15

Look at the sign of C
In this case C is negative, therefore find a value from step 1 list and another from step 2 list such that the difference of their product and the product of two remaining numbers from these steps results in b.

Choose a product from Step 1 & 2
Here is where I am having difficulty understanding
4 x 6 (e, f) form "a" and 5 x 9 (g, h) form "c"

here is where I am heavily confused, on more than one occasion the author seems to multiply the above in any combination to get the correct product. Is there any method to this madness??

Problem 1) They solve by using the product of 4 and 5 = 20. And the product of 6 and 9 = 54. 54 - 20 = 34, or our b value in this expression.

Problem 2) Then the author pairs up the binomials
(4x - 9) (6x + 5). Here the author says the binomials must be ordered as (e h) (f g). That is fine, but for the previous step I do not understand why they are mixed up.

Many thanks for any help

2. What two numbers when multiplied equal -1080 and when added equal -34?.

Hmmmmm....how about -54 and 20?. You have the right numbers. Good.

$24x^{2}-54x+20x-45$

$6x(4x-9)+5(4x-9)$

$(6x+5)(4x-9)$

There ya' go.

It can be a booger finding those two numbers, especially when they are inordinately large.


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Wilcoxon Rank-Sum Test $H_1$ notation

This notation is from Nonparametric Statistical Inference, 5th ed., by Gibbons and Chakraborti. On p. 290, it states:

Wilcoxon (1945) proposed a test where we accept the one-sided location alternative $H_L: \theta < 0$ ($X \overset{\text{ST}}{>}Y)$ if the sum of the ranks of the $X$'s is too large, or $H_L: \theta > 0$ ($X \overset{\text{ST}}{<}Y$) if the sum of the ranks of the $X$'s is too small.

What in the world does $\text{ST}$ mean? I'm also not quite sure what $\theta$ is supposed to mean either - on p. 289, it state:

The location alternative is that the populations have the same form but a different measure of central tendency. This can be expressed in symbols as follows:

$H_0: F_{Y}(x) = F_{X}(x)$ for all $x$

$H_L: F_{Y}(x) = F_{X}(x - \theta)$ for all $x$ and some $\theta \neq 0$

I think (from Pre-Calculus) that the shift $F_{X}(x) \to F_{X}(x - \theta)$ would be a right translation as long as $\theta > 0$ and a left translation as long as $\theta < 0$. So wouldn't $H_L$ be equivalent to saying $F_{X}(x) > F_{Y}(x)$ when $\theta > 0$ and $F_{X}(x) < F_{Y}(x)$ when $\theta < 0$ (in both cases, for at least one $x$?)?

• The notation is defined on page 229 (I have never seen the book aside from just now when I used Google books to locate "st" in it). – Glen_b Aug 10 '15 at 1:34
• Thank you @Glen_b, that is very helpful! :) Hopefully someday I can give this book a good read instead of just using it to look something up. – Clarinetist Aug 10 '15 at 1:36

The ST above the inequality refers to stochastic ordering. That is, we say that $X$ is stochastically larger than $Y$ if $P(X > t) > P(Y > t)$ for all $t$. $\theta$ is the location parameter of the family, so the alternative says that one distribution function is "shifted" with respect to the other. Your last observation is correct as this is the meaning of stochastic ordering.


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Browse Questions

# An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.

Toolbox:
• A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
Given that an experiement succeeds twice as often as it fails we need to find the probability that in the next six trials, there will be at least 4 successes.
Since the experiment succeed twice as often it fails $p=\large\;\frac{2}{3}$$;\;q=1\;-\;p=1\;-\large\;\frac{2}{3}=\frac{1}{3}$
Let X be the random variable that represents the number of successes in $n=6$ trials.
A random variable $X$ following binomial distribution with parameter n and p its probability of 'r' succeses $\rightarrow$ $p(X=r)=C^{n}_{r} p^{r} q^{n-r}$, where p is probability of success and $q=1-p$ and $r=0,1,\dots,n$
$P(X=r)=C^{6}_{r} \large(\frac{2}{3})^{r} \frac{1}{3}^{6-r}$,
P (at least 4 successes) = P (X $\geq$ 4) =$p(\;X\;=\;4\;)\;+\;p(\;X\;+\;5\;)\;+\;p(\;X\;=6\;)$
=$\large\;c^6_4(\frac{2}{3})^4\;(\frac{1}{3})^2\;+\;c^6_5\;(\frac{2}{3})^5\;(\frac{1}{3})+c^6_6\;(\frac{2}{3})^6\;(\frac{1}{3})^0$
=$\;15\;\large\frac{2^4}{3^6}\;$$+\;6\;\large\frac{2^5}{3^6}\;$$+\;\large\frac{2^6}{3^6}$
=$\large\;\frac{2^4}{3^6}\;$$[15\;+\;12\;+\;4]$
=$\;31\;\large\frac{2^4}{3^6}$


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### This is embarrasing...

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# CEN - EN 15359

## Solid recovered fuels - Specifications and classes

inactive, Most Current
 Organization: CEN Publication Date: 1 November 2011 Status: inactive Page Count: 28 ICS Code (Solid fuels): 75.160.10
##### scope:

This document specifies a classification system for solid recovered fuels (SRF) and a template for the specification of their properties.

SRF are produced from non-hazardous waste.

NOTE 1 Waste referred to in article 2(2)(a), points (i)-(v) of the Waste Incineration Directive (2000/76/EC) is not included in the scope of this document. This is covered by CEN/TC 335 "Solid biofuels". Waste wood from demolition of buildings and civil engineering installations is, however, included in the scope.

NOTE 2 Untreated municipal solid waste is not included in the scope of this document.

### Document History

EN 15359
November 1, 2011
Solid recovered fuels - Specifications and classes
This document specifies a classification system for solid recovered fuels (SRF) and a template for the specification of their properties. SRF are produced from non-hazardous waste. NOTE 1 Waste...


text:
# How do you solve and write the following in interval notation: |x + 5|> 18?

Aug 30, 2017

Given: $| x + 5 | > 18$

Separate into two inequalities:

$x + 5 > 18$ and $- \left(x + 5\right) > 18$

Multiply the both sides of the second inequality by -1:

$x + 5 > 18$ and $x + 5 < - 18$

Subtract 5 from both sides of both inequalities:

$x > 13$ and $x < - 23$

Interval notation:

$\left(- \infty , - 23\right)$ and $\left(13 , \infty\right)$


text:
# Suppose that f is a linear function such that f(3) = 6 and f(-2) = 1. What is f(8)?

Jan 23, 2018

$f \left(8\right) = 11$

#### Explanation:

Since it's a linear function, it must be of the form

$a x + b = 0 \text{ " " } \left(1\right)$

So

$f \left(3\right) = 3 a + b = 6$

$f \left(- 2\right) = - 2 a + b = 1$

Solving for $a$ and $b$ gives $1$ and $3$, respectively.

Therefore, substituting the values of $a$, $b$, and $x = 8$ in equation $\left(1\right)$ gives

$f \left(8\right) = 1 \cdot 8 + 3 = 11$

Jan 23, 2018

$f \left(8\right) = 11$

A lot more explanation is involved than doing the actual maths

#### Explanation:

Linear basically means 'in line'. This is implying a strait line graph situation

You read left to right on the x-axis so the first value is the least $x$
using:

$f \left(- 2\right) = {y}_{1} = 1$
$f \left(3\right) = {y}_{2} = 6$
$f \left(8\right) = {y}_{3} = \text{Unknown}$

Set point 1 as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 2 , 1\right)$
Set point 2 as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(3 , 6\right)$
Set point 2 as ${P}_{3} \to \left({x}_{3} , {y}_{3}\right) = \left(8 , {y}_{3}\right)$

The gradient (slope) of part will be the same gradient of the whole.

Gradient (slope) is the amount of up or down for a given amount of along, reading left to right.

Thus the gradient gives us: ${P}_{1} \to {P}_{2}$

$\left(\text{change in "y)/("change in } x\right) \to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{6 - 1}{3 - \left(- 2\right)} = \frac{5}{5}$

Thus we have ${P}_{1} \to {P}_{3}$ ( same ratio )

$\left(\text{change in "y)/("change in } x\right) \to \frac{{y}_{3} - {y}_{1}}{{x}_{3} - {x}_{1}} = \frac{{y}_{3} - 1}{8 - \left(- 2\right)} = \frac{5}{5}$

$\textcolor{w h i t e}{\text{dddddddd")-> color(white)("ddd")(y_3-y_1)/(x_3-x_1) =color(white)("d")(y_3-1)/10color(white)("d}} = 1$

Multiply both sides by 10

$\textcolor{w h i t e}{\text{dddddddd")->color(white)("dddddddddddddd")y_3-1color(white)("d}} = 10$

$\textcolor{w h i t e}{\text{dddddddd")->color(white)("ddddddddddddddddd")y_3color(white)("d}} = 11$


text:
# Centrifugal Force and Polar Coordinates

In Classical Mechanics, both Goldstein and Taylor (authors of different books with the same title) talk about the centrifugal force term when solving the Euler-Lagrange equation for the two body problem, and I'm a little confused about what it exactly means - is it a real centrifugal force or a mathematical consequence of using polar coordinates for solving the Euler-Lagrange equation.

Their derivations of the Langrangian $L=\frac{1}{2}\mu(\dot{r}^{2}+r^{2}\dot{\theta}^{2})-U(r)$ would lead to one motion of equation (theta) showing that angular momentum is constant and one radial equation of motion shown as $\mu\ddot{r}=-\frac{dU}{dr}+\mu r\dot{\phi}^{2}=-\frac{dU}{dr}+F_{cf}$. They call $\mu r\dot{\phi}^{2}$ the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames. The frame of reference in the two body problem was chosen such that the Center of Mass of the two bodies would be the origin so that would be an inertial frame, and I'm assuming that there are no non-inertial frames involved since neither author had talked about it in the previous chapters.

Would calling $\mu r\dot{\phi}^{2}$ an actual centrifugal force be incorrect then? Isn't it a term that describes the velocity perpendicular to the radius? From this two-body problem, it appears as though if I were to use polar coordinates when solving the Euler-Lagrange equations for any other problem, the centrifugal force term will always appear, so it would be a mathematical consequence of the choice of coordinate system rather than it being an actual fictitious force. Is that term being called a centrifugal force because it actually is a centrifugal force or is it because it has a mathematical form similar to it?

-

There are two equivalent descriptions$^1$ of the reduced two-body problem with a central potential $V(r)$:

1. In an inertial frame with no fictitious forces: Here $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the angular part of the kinetic energy.

2. In a rotating frame following the reduced particle with fictitious forces and only 1D radial kinematics: Here $-\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the centrifugal potential (notice the minus sign!).

Each description leads to the same Lagrangian

$$L~=~T-U~=~L=\frac{1}{2}\mu(\dot{r}^{2}+r^{2}\dot{\theta}^{2})-V(r).$$

$^1$ The reduced particle is confined to an orbit plane, so the problem is effectively just two-dimensional described by two coordinates $r$ and $\theta$. In both descriptions, the center-of-mass serve as the origin of the reference frame.

-
I see what happened. So there are two interpretations. The first is kinetic energy when deriving using polar coordinates. The second arises when changing to a 1-D problem. For this there are more than one way: in Goldstein one involves integration that is a hassle (Goldstein 3rd ed p.75) and the other is to consider it as a 1-D fictitious problem (p.76-77), which is easier. This fictitious problem changes the interpretation of $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ from kinetic energy to potential energy contributed by potential energy. I guess this is mostly reiterating what you've said. –  user1604449 Sep 22 '12 at 1:18

I think you skipped a step...

Once you show that angular momentum $l$ is a constant of the motion you can eliminate $\dot{\theta}$ from the two-dimensional equation of motion and arrive at an equation in 1 independent variable only ($r$): $$\mu \ddot{r} - \frac{l^2}{\mu r^3} = -\frac{dU}{dr}$$ If you think about it, this new 1-dimensional space is in fact a non-inertial frame (if it were inertial the mass would fall to $r=0$), so it's no surprise to see a new force term in addition to the original potential gradient.

Goldstein then interprets the new term as "the familiar centrifugal force" by re-writing it: $$\frac{l^2}{\mu r^3} = \mu r \dot{\theta}^2 = \frac{\mu v_\theta^2}{r}$$

And it's true: that new force is indeed one of the non-inertial forces (the centrifugal) that arise if one analyzes the problem in a rotating frame with angular velocity $\omega=\dot{\theta}$.

Aside: The other non-inertial forces in a rotating frame arise from time-varying quantities. It can be shown that the entire non-inertial potential is a velocity-dependent one analogous to that of electromagnetism: $$U_{rot} = \mu \left\{ \left[ - \frac{\left( \boldsymbol{\omega \times r} \right)^2}{2} \right] - \boldsymbol{ \dot{r} \cdot \left( \boldsymbol{\omega \times r} \right) } \right\}$$

The first term gives the centrifugal force, the second the Coriolis and Euler forces (if $\boldsymbol{\omega}$ is non-constant).

-
What a coincidence! I finally replied and it was the same time as you did. I didn't think about the inertial frame and mass falling to $r = 0$. More food for thought. Thanks! –  user1604449 Sep 22 '12 at 1:29
@user1604449: yes, it looks like you got to the same place... –  Art Brown Sep 22 '12 at 1:33