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https://delteria.app/solving-the-value-of-a-complex-expression-2xy2-2x-y2-when-xy-1/
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# Solving the Value of a Complex Expression: 2^{(x+y)^{2}}/2^{(x-y)^{2}} When xy = 1
Possible blog post:
Solving the Value of a Complex Expression: 2^{(x+y)^{2}}/2^{(x-y)^{2}} When xy = 1
Mathematics can be both fascinating and frustrating, especially when dealing with complex expressions that seem to defy simplicity. However, there are often hidden patterns and tricks that can simplify seemingly complicated problems, and it’s often these moments of enlightenment that make math worth the effort. In this post, we will explore how to solve the value of a complex expression involving exponents and variables, under a specific condition that relates these variables in a non-obvious way. Specifically, we will consider the expression
E = 2^{(x+y)^{2}}/2^{(x-y)^{2}}
when xy = 1. This expression may look intimidating at first, but we can make some observations and substitutions that will lead us to an elegant solution.
First, let’s try to expand and simplify the numerator and denominator of the fraction separately. We use the property that (a+b)^2 = a^2 + 2ab + b^2 and (a-b)^2 = a^2 – 2ab + b^2, and the fact that 1/2 = 2^{-1}, to obtain
E = 2^{(x+y)^{2}}/2^{(x-y)^{2}} = 2^{(x^2+2xy+y^2)} * 2^{-(x^2-2xy+y^2)} = 2^{4xy} * 2^{4y^2} / 2^{4x^2} = 16y^{2}/16x^{2}
where we used the fact that xy = 1 implies x^2y^2 = 1 and thus 4xy = 4/x^2 = 4y^2. Now we can simplify this expression further by canceling out the common factor of 16 and dividing y^2 by x^2:
E = (y/x)^{2} = x^{2}/y^{2}
This means that the value of the expression E depends solely on the values of x and y, not on any other variables or parameters. In particular, if we know that xy = 1, we can substitute y = 1/x into the expression E and obtain
E = x^2/(1/x)^2 = x^4
This result is both simple and surprising. It tells us that the original complex expression E reduces to the fourth power of x when xy = 1, regardless of the value of y. It also shows us that the exponent laws are powerful tools that enable us to manipulate exponents in various ways, and that the connections between seemingly unrelated variables can be used to simplify complex expressions.
To summarize, we have solved the value of the complex expression 2^{(x+y)^{2}}/2^{(x-y)^{2}} when xy = 1 by expanding and simplifying the numerator and denominator separately, canceling out common factors, and substituting y = 1/x. We found that the expression simplifies to x^4, which depends only on the value of x and is independent of y. We also learned some general techniques for handling exponents and using algebraic relations between variables. Hopefully, this example illustrates the beauty and utility of mathematics, and encourages you to explore more of its mysteries and wonders.
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# (1 point) In this problem we consider an equation in differential form M dx + N...
###### Question:
(1 point) In this problem we consider an equation in differential form M dx + N dy=0. (6x + 6y)dx – (6x + 4y)dy = 0 Find My = N = If the problem is exact find a function F(x,y) whose differential, F(x, y) is the left hand side of the differential equation. That is, level curves F(x, y) = C, give implicit general solutions to the differential equation If the equation is not exact, enter NE otherwise find F(x,y) (note you are not asked to enter C) F(x, y) =
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### Two Counter-Example Graphs
This first communicaton was received in early June 2005 from a correspondent who wishes to remain anonymous. I post it essentially verbatim with his permission. I am indebted to him for identifying a major omission in the original version of this essential piece of my proof.
"On page http://www-personal.ksu.edu/~kconrow/august.html it says:
``` Proposed Th.:Given a collection of labelled nodes of degree 3 with directed edges and
(1) one outgoing edge and two incoming edges for every node, and
(2) every node uniquely labelled with an odd positive integer, and
(3) one unique node whose outgoing edge loops back to itself, and
(4) no restrictions arise to limit the growth of the graph,
then the only graph which can be formed is a directed binary tree.
```
"This proposed theorem is incorrect. The following picture gives a counter example:
"In this graph, all four conditions hold, yet the graph is not a single binary tree. This shows that those four conditions are not sufficient to ensure the graph is a binary tree. So the proposed theorem is not true.
"Notice that if you start at 1 or somewhere in the red tree, you'll eventually reach 1. But if you start anywhere else in the graph, then you'll eventually reach a yellow node, and will then move forever to the right, and will never reach 1.
"The numbers in the graph are as follows. The yellow nodes are labelled with 2n-1 for all positive n. The red nodes are 4n+1. The green are 8n+3. The blue are 16n+7. The magenta are 32n+15. The gray are 64n+31. The brown are 128n+63. This pattern continues forever to the right, with the mth tree having all possible labels of the form 2m+1n+2m-1.
"There are simpler counter examples which have cycles, but it's perhaps more interesting to see this acyclic counter example."
My response to the above communication was to add two additional conditions to the proposed theorem to tie the Collatz tree structure more closely to the statement of the theorem in order to exclude constructed counter examples like that above.
The same correspondent in early June 2007 provided this second example in reference to section 1.3 of The Structure of the Collatz Trajectories....
"Imagine that I've created an infinite tree with different odd integers at each node. The tree is made up of an infinite number of pieces. The first piece has nodes whose numbers are the set 5[6]. The next piece has the nodes with numbers {7,13,15}[18]. The first 6 pieces of the tree are these sets:
```{5}[6] density=2*1/6
{7,13,15}[18] density=2*3/18
{19,21,37,39,45}[54] density=2*5/54
{55,57,63,109,111,117,135}[162] density=2*7/162
{163,165,171,189,325,327,333,351,405}[486] density=2*9/486
{487,489,495,513,567,973,975,981,999,1053,1215}[1458] density=2*11/1458
...
```
"This continues forever. So the tree is made up of an infinite number of sets, each of which contains an infinite number of odd numbers. We're constructing this tree by starting with a boneyard universe of all the odd numbers, and then we keep whacking at it without loss. We might expect that it would include all the odd numbers.
"For each set above, the modulus [d] is 3 times the modulus of the line before it. The numbers in braces are chosen such that the set will include all the odd numbers possible that haven't been included by any set above it, and which don't include any power of 3. In other words, none of the sets will ever include any of the numbers {1, 3, 9, 27, 81, 3^5, 3^6, 3^7, ...}.
"The density of each set is a fraction, where the denominator is the previous denominator times 3, and the numerator is the previous numerator plus 4.
"In other words, the density of the nth set is (4*n-2)/(2*3^n). If you take the sum of that fraction as n goes from 1 to infinity, the densities add up to 1. Not just close to 1. Exactly 1. Mathematica found that result using symbolic math, so there's no possibility of roundoff error.
"So these are disjoint sets of odd numbers whose densities add up to 1. But they don't include any powers of 3. Even though their densities add up to 1, there are still infinitely many odd numbers that are hiding in a rat hole outside the tree. The numbers in the rat hole are {1,3,9,27,81,...}. Yet the densities in the tree sum to 1.
"How is that possible? It's possible because the numbers in the rat hole have a density of 0. Just because they have a density of zero, doesn't mean they don't exist!
"That's the problem. If the sum of the densities is 1, that doesn't guarantee that the rat hole is empty. It just guarantees that the rat hole has a density of 0. But it can still contain an infinite number of "left out" odd numbers. Even infinite sets can have a density of 0.
"For infinite sets of sets, we can never use arguments based on densities to prove that we've caught all the odd numbers. It's mathematically impossible."
My response to this second communication was to elaborate the point that the abstract predecessor tree starts out containing all the odd positive integers {union({1[8]}, 3{[8]}, {5[8]}, and {[8]})} and that its elaboration does not permit the loss (or insertion, either) of any integers from that original comprehensive set. Ths change appears near the end of section 1.2 of The Structure of the Collatz Trajectories....
It is interesting that the counter-example constructed by this correspondent was chosen as if the set of powers of 3 were omitted from the otherwise complete infinite set of odd integers. But all the integers in {0[3]} are specifically included in the Collatz predecessor tree as leaf nodes in the left descent assemblies. No more unfortunate choice of properties for a zero density set could have been chosen. But, since the entire set of odd positive integers is included in the abstract tree, there is no set of properties which can be selected which represents integers from which to construct this sort of counter example. The counter example is instructive, but not applicable to the Collatz predecessor tree.
I appreciate the help of this correspondent in increasing the precision of my presentation.
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Article
# Stability of Maximum Functional Equation and Some Properties of Groups
by
,
Qi Liu
and
Yongjin Li
*
School of Mathematics, Sun Yat-Sen University, Guangzhou 510275, China
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(12), 1949; https://doi.org/10.3390/sym12121949
Submission received: 4 November 2020 / Revised: 24 November 2020 / Accepted: 24 November 2020 / Published: 26 November 2020
## Abstract
:
In this research paper, we deal with the problem of determining the function $χ : G → R$, which is the solution to the maximum functional equation (MFE) $max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) ,$ when the domain is a discretely normed abelian group or any arbitrary group G. We also analyse the stability of the maximum functional equation $max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y )$ and its solutions for the function $χ : G → R$, where G be any group and also investigate the connection of the stability with commutators and free abelian group K that can be embedded into a group G.
MSC:
20D60; 54E35; 11T71
## 1. Literature Review
In [1], Volkmann proved that every function $χ$ defined on an abelian group G can be described in the form $χ ( x ) = | α ( x ) |$, which is the solution of the maximum FE
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) ,$
where $α : G → R$ is an additive function. Consequently, in [2], Toborg showed that Equation (1) also characterises the absolute value of additive functions when G be any group. Their’s main theorem is stated, as follows:
Theorem 1
(Toborg [2]). Let G be any group (G is an abelian group (Volkmann [1])), then a function $χ : G → R$ fulfills (1) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = | α ( x ) |$ for any $x ∈ G$.
We recommend the readers to see [3,4], and the references cited therein in order to obtain comprehensive results related to functional Equation (1), which characterises the additive function’s absolute value.
According to Simon and Volkmann [5], solutions of the MFE
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y )$
with the additional assumption that G is an additive abelian group was exhibited in the theorem, which is stated as:
Theorem 2
([5], Theorem 2). Assume that G is an additive abelian group, whose each element is divisible by 6 (divisible by 2 and 3), and then a function $χ : G → R$ fulfills
$max { χ ( x + y ) , χ ( x − y ) } = χ ( x ) χ ( y )$
if and only if there exists an additive function $α : G → R$ such that $χ ( x ) = 0$ or $χ ( x ) = e | α ( x ) |$ for any $x ∈ G$.
Jarczyck and Volkmann [6] demonstrated the stability results of MFE (1) on additive abelian groups. Consequently, Badora et al. [7] also generalized their results in order to prove the stability for a certain class of groupoids. In [3], Gilanyi et al. determined the stability results of maximum functional equation $max { χ ( ( x ∘ y ) ∘ y ) , χ ( x ) } = χ ( x ∘ y ) + χ ( y )$ on a square-symmetric groupoid. Consequently, they also examined the stability of maximum equation $max { χ ( x + y ) , χ ( x − y ) } = χ ( x ) + χ ( y )$ on additive abelian groups. Various appropriate and useful results regarding stability can be found in papers by Przebieracz [8,9].
This paper is arranged, as follows: in Section 2, we prove the functional Equation (1) for any arbitrary group G without any characterisation of an additive function’s absolute value. Besides, we also demonstrate some consistent and useful results concerning the normal subgroup of G.
In Section 3, we analyse the functional Equation (2) in order to obtain its solution. For this purpose, we drop the additional assumption that G is an additive abelian group and is divisible by 6. We present the generalization of Theorem 2 by proposing a discretely normed abelian group G. Moreover, we investigate the functional Equation (2) for any arbitrary group G, which satisfies the Kannappan condition [10].
Section 4 deals with the stability results of the MFE (1) for a function $χ : G → R$, where G be any group and find some useful connection of stability with commutators and embedding of a group.
## 2. Solutions of the Functional Equation (1)
Throughout this article, let G be any group, and 1 is considered to be the identity element of a group G.
Definition 1
([10]). Let G be an arbitrary group, we say a function $χ : G → R$ satisfies the Kannappan condition if
$χ ( u v x ) = χ ( u x v ) f o r a l l u , v , x ∈ G .$
Theorem 3.
Let G be a group and function $χ : G → R$ satisfies the Kannappan condition then
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) f o r a l l x , y ∈ G$
if and only if
$min { χ ( x y ) , χ ( x y − 1 ) } = | χ ( x ) − χ ( y ) | f o r a l l x , y ∈ G$
and also satisfies
$χ ( x 2 ) = 2 χ ( x ) f o r a n y x ∈ G .$
Proof.
Assume that $χ$ satisfies Equation (3), and then by setting $x = y = 1$ in (3), we can conclude that $χ ( 1 ) = 0$. It is also clear that $χ ( x ) ≥ 0$ and $χ ( x 2 ) = 2 χ ( x )$ for any $x ∈ G$. The proof of Equation (4) consists of the following simple computation:
$max { χ ( x y ) , χ ( x y − 1 ) } + min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) + χ ( x y − 1 ) χ ( x ) + χ ( y ) + min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) + χ ( x y − 1 ) χ ( x ) + χ ( y ) + min { χ ( x y ) , χ ( x y − 1 ) } = max { χ ( x y x y − 1 ) , χ ( x y y x − 1 ) } χ ( x ) + χ ( y ) + min { χ ( x y ) , χ ( x y − 1 ) } = max { χ ( x 2 ) , χ ( y 2 ) } min { χ ( x y ) , χ ( x y − 1 ) } = max { 2 χ ( x ) , 2 χ ( y ) } − χ ( x ) − χ ( y ) min { χ ( x y ) , χ ( x y − 1 ) } = | χ ( x ) − χ ( y ) | .$
Conversely, assume that Equations (4) and (5) hold. Subsequently, we can obtain
$max { χ ( x y ) , χ ( x y − 1 ) } − min { χ ( x y ) , χ ( x y − 1 ) } = | χ ( x y ) − χ ( x y − 1 ) | .$
Using Equation (4) in the following computation, we get that
$max { χ ( x y ) , χ ( x y − 1 ) } − | χ ( x ) − χ ( y ) | = | χ ( x y ) − χ ( x y − 1 ) | = min { χ ( x y x y − 1 ) , χ ( x y y x − 1 ) } = min { χ ( x 2 ) , χ ( y 2 ) } max { χ ( x y ) , χ ( x y − 1 ) } = 2 min { χ ( x ) , χ ( y ) } + | χ ( x ) − χ ( y ) | max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) .$
Corollary 1.
Let G be a group and function $χ : G → R$ satisfies the Kannappan condition, and then χ satisfies the Equation (3) if and only if
$χ ( x ) + χ ( y ) = χ ( x y ) + χ ( x y − 1 ) − | χ ( x ) − χ ( y ) | f o r a n y x , y ∈ G$
and also $χ ( 1 ) = 0$.
Proof.
Suppose that $χ ( 1 ) = 0$ and Equation (6) holds. Subsequently, (6) can be written as
$2 max { χ ( x ) , χ ( y ) } = χ ( x y ) + χ ( x y − 1 ) for any x , y ∈ G .$
Because $χ ( 1 ) = 0$, then, for $x = y$, we can obtain that $χ ( x 2 ) = 2 χ ( x )$. Afterwards, writing $x y$ instead of x and $x y − 1$ instead of y in (7), we can conclude that
$2 max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y x y − 1 ) + χ ( x y 2 x − 1 ) = 1 2 { χ ( x 2 ) + χ ( y 2 ) } = 1 2 { 2 χ ( x ) + 2 χ ( y ) } max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) .$
Conversely, assume that maximum Equation (3) holds, then, by Theorem 3, we can determine that $χ ( 1 ) = 0$ and Equation (6) holds. □
Corollary 2.
Let G be a group and function $χ : G → R$ satisfies the Kannappan condition, and then χ is a solution of equation
$min { χ ( x y ) , χ ( x y − 1 ) } = | χ ( x ) − χ ( y ) | f o r a n y x , y ∈ G$
and also satisfies
$χ ( x 2 ) = 2 χ ( x ) f o r a n y x ∈ G$
if and only if there exists an additive function $α : G → R$ such that $χ ( x ) = | α ( x ) |$ for any $x ∈ G$.
Theorem 4.
Let G be a group and function $χ : G → R$ satisfies the Kannappan condition. If the Equations (4) and (5) are satisfied, then there exists a normal subgroup $N χ = { v ∈ G | χ ( v ) = 0 }$ of G.
Proof.
Because the function $χ : G → R$ satisfies the Equations (4) and (5), then by Theorem 3
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) for any x , y ∈ G .$
When we investigate Equations (3) and (4), then we can see that either $χ ( x y ) = | χ ( x ) − χ ( y ) |$ then $χ ( x y − 1 ) = χ ( x ) + χ ( y )$ or $χ ( x y − 1 ) = | χ ( x ) − χ ( y ) |$ then $χ ( x y ) = χ ( x ) + χ ( y )$. First, writing v instead of y and assume that $χ ( x ) ≠ 0$ for some $x ∈ G$ and also $χ ( x v ) = χ ( x v − 1 )$. Subsequently, we can compute that $χ ( x ) + χ ( v ) = | χ ( x ) − χ ( v ) |$. From Equation (3), we can easily deduce that $χ ( v ) , χ ( x ) ≥ 0$. Our assumption that $χ ( x ) ≠ 0$ forces $χ ( v )$ to be 0.
From Equation (5), we can compute that $χ ( 1 ) = 0$, therefore $1 ∈ N χ$. Let $u , v ∈ N χ$. We can also deduce from Equation (3) that $χ$ is even, so we have $χ ( v − 1 ) = χ ( v ) = 0$, therefore $v − 1 ∈ N χ$. Additionally, $0 = χ ( u ) + χ ( v ) = max { χ ( u v ) , χ ( u v − 1 ) } ≥ χ ( u v )$, but $χ ( x ) ≥ 0$ for all $x ∈ G$; therefore, we can obtain that $χ ( u v ) = 0$, so $u v ∈ N χ$. Hence $N χ$ is a subgroup of G. Let $v ∈ N χ$ and $x ∈ G$, then Equation (3) yields that
$χ ( x ) + χ ( x − 1 v x ) = max { χ ( x x − 1 v x ) , χ ( x x − 1 v − 1 x ) } = max { χ ( v x ) , χ ( v − 1 x ) } = max { χ ( x − 1 v − 1 ) , χ ( x − 1 v ) } = χ ( x − 1 ) + χ ( v ) χ ( x ) + χ ( x − 1 v x ) = χ ( x ) + χ ( v ) χ ( x − 1 v x ) = χ ( v ) ,$
which implies that $χ ( x − 1 v x ) = χ ( v )$ for all $x ∈ G$ and $v ∈ N χ$. Hence, $N χ$ is a normal subgroup of G. □
Corollary 3.
For any group G, let a function χ on group G satisfying Equation (3), then $χ : G / N χ → R$ also satisfies
$χ ( x v ) = χ ( x ) f o r a n y x ∈ G , v ∈ N χ .$
Additionally, $G / N χ$ is an abelian quotient group.
Proof.
Let $x ∈ G$ be an arbitrary and assume that $v ∈ N χ$. We can deduce from (3)
$χ ( x v ) ≤ max { χ ( x v ) , χ ( x v − 1 ) } = χ ( x ) + χ ( v ) = χ ( x ) ≤ max { χ ( x v 2 ) , χ ( x ) } = χ ( x v ) + χ ( v ) = χ ( x v )$
that $χ ( x v ) = χ ( x ) + χ ( v )$. Hence, $χ ( x v ) = χ ( x )$ for any $x ∈ G$ and $v ∈ N χ$. Moreover, we already proved that $χ ( x − 1 v x ) = χ ( v )$ for any $x ∈ G$ and $v ∈ N χ$, then, by replacing v with $x v$, we can evaluate that $χ ( v x ) = χ ( v x )$ for any $x ∈ G$ and $v ∈ N χ$. Additionally, $N χ$ is a normal subgroup of G; therefore, $G / N χ$ is an abelian quotient group. □
Corollary 4.
Let G be any group and function $χ : G → R$ satisfy the Kannappan condition. If the maximum functional Equation (3) holds, then $min { χ ( x y ) , χ ( x y − 1 ) } ∈ N χ$ for every $x , y ∈ G$.
Proof.
Assume that $χ$ satisfies the Equation (3) and Kannappan condition. Subsequently, from Theorem 3, we can compute that
$min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) − χ ( y ) , ( when χ ( x ) ≥ χ ( y ) )$
or
$min { χ ( x y ) , χ ( x y − 1 ) } = χ ( y ) − χ ( x ) , ( when χ ( x ) < χ ( y ) )$
combining both cases, we can see that $2 min { χ ( x y ) , χ ( x y − 1 ) } = 0$, therefore $min { χ ( x y ) , χ ( x y − 1 ) } = 0$. In either case, it follows that $min { χ ( x y ) , χ ( x y − 1 ) } ∈ N χ$ for every $x , y ∈ G$. □
## 3. Solutions of the Functional Equation (2)
Theorem 5.
Let G be any arbitrary group, then a function $χ : G → R$ satisfies
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) f o r a n y x , y ∈ G$
if and only if simultaneously $χ ( x ) = 0$ or
$min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) − 1 ∨ [ χ ( x ) χ ( y ) − 1 ] − 1 f o r a n y x , y ∈ G$
holds and satisfies
$χ ( x 2 ) = χ ( x ) 2 f o r a l l x ∈ G .$
Proof.
Suppose that $χ$ satisfies Equation (9). Assume 1 as neutral element of group G, then by putting $x = y = 1$ in (9), we can obtain that $χ ( 1 ) χ ( 1 ) = χ ( 1 )$, then $χ ( 1 ) = 0$ or $χ ( 1 ) = 1$. Let $χ ( 1 ) = 0$, then by Equation (9) we can compute $χ ( x ) = 0$ for any $x ∈ G$. Assume that $χ ( 1 ) = 1$, then from (9), It is easy to see that $χ ( x − 1 ) = χ ( x ) ≥ 0$, $χ ( x 2 ) ≥ 1$ and $χ ( x 2 ) = χ ( x ) 2$ for any $x ∈ G$. The proof of Equation (10) consists of the following simple computation:
$max { χ ( x y ) , χ ( x y − 1 ) } · min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) χ ( x y − 1 ) χ ( x ) χ ( y ) · min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) χ ( x y − 1 ) χ ( x ) χ ( y ) · min { χ ( x y ) , χ ( x y − 1 ) } = max { χ ( x y x y − 1 ) , χ ( x y y x − 1 ) } χ ( x ) χ ( y ) · min { χ ( x y ) , χ ( x y − 1 ) } = max { χ ( x 2 ) , χ ( y 2 ) } min { χ ( x y ) , χ ( x y − 1 ) } = max { χ ( x ) 2 , χ ( y ) 2 } χ ( x ) − 1 χ ( y ) − 1 min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) − 1 ( when χ ( x ) ≥ χ ( y ) ) ∨ [ χ ( x ) χ ( y ) − 1 ] − 1 ( when χ ( x ) < χ ( y ) ) .$
Conversely, suppose that the Equations (10) and (11) are satisfied and $χ ( x ) ≠ 0$. Subsequently, it can be determined that
$max { χ ( x y ) , χ ( x y − 1 ) } · min { χ ( x y ) , χ ( x y − 1 ) } − 1 = χ ( x y ) χ ( x y − 1 ) − 1 ∨ χ ( x y ) − 1 χ ( x y − 1 ) .$
Here, we have two cases, in the first case, using Equation (10) and (11), we derive the required result, as follows:
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) χ ( x y − 1 ) − 1 · min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x y ) χ ( x y − 1 ) − 1 χ ( x ) χ ( y ) − 1 ( when χ ( x ) ≥ χ ( y ) ) = min { χ ( x y x y − 1 ) , χ ( x y y x − 1 ) } χ ( x ) χ ( y ) − 1 = min { χ ( x 2 ) , χ ( y 2 ) } χ ( x ) χ ( y ) − 1 = min { χ ( x ) 2 , χ ( y ) 2 } χ ( x ) χ ( y ) − 1 = χ ( y ) 2 χ ( x ) χ ( y ) − 1 ( when χ ( x ) ≥ χ ( y ) ) max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) .$
From second case, when $χ ( x ) < χ ( y )$, then we can also get that $max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y )$. □
Corollary 5.
Let G be a group and function $χ : G → R$ satisfies the maximum Equation (9) if and only if
$χ ( x y ) χ ( x y − 1 ) = χ ( x ) χ ( y ) min { χ ( x y ) , χ ( x y − 1 ) } f o r a n y x , y ∈ G$
and also $χ ( 1 ) = 1$.
Corollary 6.
Let G be a group and function $χ : G → R$ is a solution of Equation (10) satisfies (11) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any $x ∈ G$.
Definition 2
([11]). Let G be an abelian group. Subsequently, a function $χ : G → R$ is said to be a discrete norm on G if there exists some $α > 0$ such that $χ ( x ) > α$ whenever x is not identity element of G. Afterwards, $( G , χ )$ is called the discretely normed abelian group [11].
Simon and Volkmann [5] proved Theorem 2 with additional assumption that G is an additive abelian group and is divisible by 6, but we present the generalization of Theorem 2 by introducing the notion of discretely normed abelian group G, as follows:
Theorem 6.
Assume that $( G , χ )$ be a discretely normed abelian group, then a function $χ : G → R$ fulfills (9) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any $x ∈ G \ { 1 }$.
Proof.
Because G is a discretely normed abelian group, therefore there exists a discrete norm function $χ : G → R$ such that $χ ( x ) > 0$ whenever $x ∈ G \ { 1 }$. Assume that $η ( x ) = log χ ( x )$, then applying the main theorem of Toborg [2], a function $χ : G → R$ satisfies
$max { η ( x y ) , η ( x y − 1 ) } = η ( x ) + η ( y )$
if and only if there exist an additive function $α : G → R$ such that $η ( x ) = | α ( x ) |$. Subsequently, we have
$max { log χ ( x y ) , log χ ( x y − 1 ) } = log χ ( x ) + log χ ( y )$
if and only if $log χ ( x ) = | α ( x ) |$, which implies that
$log max { χ ( x y ) , χ ( x y − 1 ) } = log χ ( x ) χ ( y ) ,$
thus, we conclude that
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y )$
if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any $x ∈ G \ { 1 }$. □
Corollary 7.
Let $( G , χ )$ be a discretely normed abelian group, then a function χ is a solution of Equation (10) satisfying (11) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any non-identity element $x ∈ G$.
Proof.
From Theorem 6, we concluded that maximum functional Equation (9) holds; therefore, using Theorem 5, we can also obtain required proof. □
We have well-known theorem presented by Stepr$a ¯$ns Juris in [11] about a group G, which is a discretely normed abelian group. Therefore, we have following corollaries.
Corollary 8.
For free abelian group G, a function χ is a solution of Equation (9) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any $x ∈ G \ { 1 }$.
Corollary 9.
Suppose that G is a free abelian group, then χ is a solution of functional Equation (10) satisfying (11) if and only if there exists an additive function $α : G → R$, such that $χ ( x ) = e | α ( x ) |$ for any $x ∈ G \ { 1 }$.
Theorem 7.
Let G be any group and let a function $χ : G → R$ is a solution of Equation (10) and (11), which is not identically zero, then there exists a normal subgroup $H χ = { v ∈ G | χ ( v 2 ) = 1 }$ of G.
Proof.
Because the function $χ : G → R$ satisfying the Equations (10) and (11), then by Theorem 5, we can obtain that
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) for all x , y ∈ G .$
For neutral element 1, we can write $χ ( 1 2 ) = χ ( 1 ) = 1$, then $1 ∈ H χ$. Let $v ∈ H χ$, then $χ ( v 2 ) = 1$. From Equation (13), follows immediately that $χ$ is even; therefore, $χ ( v − 2 ) = χ ( v 2 ) = 1$, hence $v − 1 ∈ H χ$. Let $u , v ∈ H χ$, therefore, by property of $H χ$, $χ ( u 2 ) = 1$ and $χ ( v 2 ) = 1$. We can deduce from maximum Equation (13) that
$χ ( u 2 v 2 ) = χ ( u 2 v 2 ) χ ( v 2 ) = max { χ ( u 2 v 4 ) , χ ( u 2 ) } ≥ χ ( u 2 ) χ ( u 2 v 2 ) ≥ χ ( u 2 ) χ ( v 2 ) .$
$χ ( u 2 ) χ ( v 2 ) = max { χ ( u 2 v 2 ) , χ ( u 2 v − 2 ) } ≥ χ ( u 2 v 2 ) χ ( u 2 ) χ ( v 2 ) ≥ χ ( u 2 v 2 ) .$
From inequalities (14) and (15), we can calculate $χ ( u 2 v 2 ) = χ ( u 2 ) χ ( v 2 ) = 1$, which implies that $u v ∈ H χ$. Therefore, $H χ$ is a subroup of G. Additionally, Equation (13) yields that $χ$ is central, therefore $χ ( x v ) = χ ( v x )$, for every $x ∈ G$ and $v ∈ H χ$. Hence, $H χ$ is a normal subgroup of G. □
Corollary 10.
For any group G, let a function χ on group G satisfying (13), which is not identically zero, then $χ : G / H χ → R$ also satisfies
$χ ( x v ) = χ ( x ) χ ( v ) f o r a n y x ∈ G , v ∈ H χ .$
Moreover,
$χ ( x v 2 ) = χ ( v 2 x ) = χ ( x ) f o r a n y x ∈ G , v ∈ H χ .$
Proof.
Let $x ∈ G$ be an arbitrary and assume that $v ∈ H χ$, then $χ ( v 2 ) = 1$. We can deduce from (13) that
$χ ( x v ) χ ( v ) = max { χ ( x v 2 ) , χ ( x ) } ≥ χ ( x ) χ ( v 2 ) ,$
which provides that $χ ( x v ) χ ( v ) ≥ χ ( x ) χ ( v 2 )$. Applying condition (11), we compute $χ ( x v ) ≥ χ ( x ) χ ( v )$. Additionally, from (13), we have $χ ( x ) χ ( v ) ≥ χ ( x v )$, so we can evaluate $χ ( x v ) = χ ( x ) χ ( v )$ for any $x ∈ G$ and $v ∈ H χ$.
Because $H χ$ is a subgroup of G, then $v 2 ∈ H χ$, so we can see that $χ ( x v 2 ) = χ ( x ) χ ( v 2 )$ for any $x ∈ G$ and $v 2 ∈ H χ$, therefore $χ ( x v 2 ) = χ ( x )$. Writing x instead of y and v instead of x in (13), we can conclude $χ ( x − 1 v x ) = χ ( v )$ for any $x ∈ G$, then writing $x v 2$ instead of v, we have $χ ( v 2 x ) = χ ( x v 2 ) = χ ( x )$. □
Corollary 11.
Let G be a group and function $χ : G → R$ satisfies the Kannappan condition. If the maximum functional Equation (13) holds, then $min { χ ( x y ) , χ ( x y − 1 ) } ∈ H χ$ for any $x , y ∈ G$.
Proof.
Assume that $χ$ satisfies the Equation (9) and Kannappan condition. Subsequently, from Theorem 5, we can conclude that
$min { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) χ ( y ) − 1 , ( when χ ( x ) ≥ χ ( y ) )$
or
$min { χ ( x y ) , χ ( x y − 1 ) } = [ χ ( x ) χ ( y ) − 1 ] − 1 , ( when χ ( x ) < χ ( y ) )$
combining both cases, we can see that $min { χ ( x y ) , χ ( x y − 1 ) } = [ min { χ ( x y ) , χ ( x y − 1 ) } ] − 1$, therefore $min { χ ( x y ) , χ ( x y − 1 ) } 2 = 1$. In either case, it follows that $min { χ ( x y ) , χ ( x y − 1 ) } ∈ H χ$ for every $x , y ∈ G$. □
## 4. Stability of Maximum Functional Equation (1)
In order to check the stability of MFE (1) in two-variables x and y, at the first stage, we put $y = x$ in (1) and derive the stability of MFE (1) in a single variable, as follows:
Theorem 8.
Let G be any group and a function $η : G → R$ satisfies
$| max { η ( x 2 ) , η ( 1 ) } − 2 η ( x ) | ≤ λ f o r a n y x ∈ G ,$
for some $λ ≥ 0$. Subsequently, we can obtain a solution $χ : G → R$ of
$max { χ ( x 2 ) , χ ( 1 ) } = 2 χ ( x ) f o r a n y x ∈ G$
such that
$− 3 λ ≤ χ ( x ) − η ( x ) ≤ λ f o r a n y x ∈ G .$
$χ ( x ) = lim n → ∞ 1 2 n η ( x 2 n ) x ∈ G .$
Additionally, by (18), χ is uniquely determined, and by (19), the requirement of $χ − η$ to be bounded is also satisfied.
Proof.
First, put $x = 1$ in (17), then we have $| max { η ( 1 ) , η ( 1 ) } − 2 η ( 1 ) | ≤ λ$, which implies that $| η ( 1 ) | ≤ λ$. Furthermore, (17) also implies that
$− λ + 2 η ( x ) ≤ max { η ( x 2 ) , η ( 1 ) } ≤ λ + 2 η ( x ) − λ ≤ η ( 1 ) ≤ λ + 2 η ( x ) − λ ≤ η ( x ) for any x ∈ G .$
Writing $x 2$ instead of x, then also $− λ ≤ η ( x 2 )$, so we can get that
$η ( 1 ) ≤ λ = 2 λ − λ ≤ 2 λ + η ( x 2 ) η ( 1 ) ≤ 2 λ + η ( x 2 ) for any x ∈ G .$
From (17), we have
$2 η ( x ) ≤ λ + max { η ( x 2 ) , η ( 1 ) } 2 η ( x ) ≤ λ + η ( 1 ) ≤ 3 λ + η ( x 2 ) ( using ( 22 ) ) or 2 η ( x ) ≤ λ + η ( x 2 ) ≤ 3 λ + η ( x 2 ) .$
Combining both cases, we can conclude
$− 3 λ ≤ η ( x 2 ) − 2 η ( x ) .$
From (17), we have $max { η ( x 2 ) , η ( 1 ) } ≤ λ + 2 η ( x )$, which is only possible when
$η ( x 2 ) − 2 η ( x ) ≤ λ .$
From inequalities (23) and (24), we have
$− 3 λ ≤ η ( x 2 ) − 2 η ( x ) ≤ λ for all x ∈ G .$
When we observe (25), it can be seen that the function $χ : G → R$ presented in (20) exists and this $χ$ satisfies
$χ ( x 2 ) = 2 χ ( x ) for all x ∈ G ,$
also $χ$ fulfills (19). Furthermore, writing $x 2 n$ instead of x in (21), dividing by $2 n$ and taking a limit $n → ∞$ and, using (20), we can obtain that $χ ( x ) ≥ 0$ for any $x ∈ G$. Hence we can obtain (18) from (26). Furthermore, when we consider (18), then we can easily see the uniqueness of $χ$, due to the result that $χ ( x 2 ) = 2 χ ( x ) ≥ 0$ for any $x ∈ G$.
By utilizing Theorem (8), we are going to derive the stability of Equation (1) in two-variables x and y. □
Theorem 9.
Let G be any group and function $η : G → R$ satisfies
$| max { η ( x y ) , η ( x y − 1 ) } − η ( x ) − η ( y ) | ≤ λ f o r a n y x , y ∈ G ,$
for some $λ ≥ 0$. Subsequently, we can evaluate a unique solution $χ : G → R$ of
$max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y ) f o r a n y x , y ∈ G$
such that
$− 3 λ ≤ χ ( x ) − η ( x ) ≤ λ f o r a n y x ∈ G .$
$χ ( x ) = lim n → ∞ 1 2 n η ( x 2 n ) f o r a n y x ∈ G .$
Proof.
Applying the Theorem (8), we can easily prove the required results. First, writing x instead of y in (27), then, from Theorem (8), we can obtain a function $χ : G → R$. We also need to prove that this function $χ$ satisfies (28). To obtain the required maximum functional Equation (28), we need to write $x 2 n$ instead of x and $y 2 n$ instead of y in (27), dividing inequality by $2 n$ and, applying limit $n → ∞$ and also using Equation (30). □
Theorem 10.
Let G be a group and function $η : G → R$ fulfills (27), then, for any $x , y ∈ G$, it holds the condition
$lim n → ∞ 1 2 n [ max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } ] = 0$
if and only if the Equation (28) is satisfied.
Proof.
Assume that maximum Equation (28) is satisfied. For every $x , y ∈ G$, we have
$| max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } | ≤ | max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − η ( x 2 n ) − η ( y 2 n ) | + | max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } − η ( x 2 n ) − η ( y 2 n ) | ≤ λ + | max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } − η ( x 2 n ) − η ( y 2 n ) | .$
To obtain the required result, first, we divide both sides by $2 n$, apply limit as $n → ∞$ and utilize the given condition (28), we obtain that, for any $x , y ∈ G$, we can get $lim n → ∞ 1 2 n [ max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } ] = 0$.
Conversely, assume that condition (31) holds. Putting $x 2 n$ and $y 2 n$ in (27) instead of x and y, respectively, then dividing by $2 n$ and taking limit $n → ∞$, we can compute
$lim n → ∞ 1 2 n [ max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } = χ ( x ) + χ ( y ) ,$
$lim n → ∞ 1 2 n max { η ( [ x y ] 2 n ) , η ( [ x y − 1 ] 2 n ) } = max { χ ( x y ) , χ ( x y − 1 ) } .$
Given condition (31), yielding that $max { χ ( x y ) , χ ( x y − 1 ) } = χ ( x ) + χ ( y )$.
Furthermore, the given condition (31) for function $η$ is not directly associated to the properties of a group G, but we can see some useful results regarding group G. The resulting condition presented below is equivalent to the above condition (31), as there exists a subsequence of $N$, such that
$lim n → ∞ 1 2 m ( n ) [ max { η ( x 2 m ( n ) y 2 m ( n ) ) , η ( x 2 m ( n ) y − 2 m ( n ) ) } − max { η ( [ x y ] 2 m ( n ) ) , η ( [ x y − 1 ] 2 m ( n ) ) } ] = 0 .$
This refers to the function $η$ and completely satisfactory, due to the fact that both limits
$lim n → ∞ 1 2 m ( n ) max { η ( x 2 m ( n ) y 2 m ( n ) ) , η ( x 2 m ( n ) y − 2 m ( n ) ) }$
and
$lim n → ∞ 1 2 m ( n ) max { η ( [ x y ] 2 m ( n ) ) , η ( [ x y − 1 ] 2 m ( n ) ) }$
are finite and exists. □
Corollary 12.
A function $η : G → R$ fulfills (18), then it satisfies the condition
$lim n → ∞ 1 2 n [ max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − η ( [ x y ] 2 n ) ] = 0 f o r a l l x , y ∈ G$
if and only if $χ ( x y ) = χ ( x ) + χ ( y )$ for any $x , y ∈ G$.
Remark 1.
(i) The given condition (31) holds when group G is an $n$—Abelian group (for any integer n, a group G is called an $n$—Abelian group if $( a b ) n = a n b n$, for any $a , b ∈ G$, see [12,13] ).
(ii) The condition (31) also satisfies when group G belongs to the class $C n$ for any natural number $n ∈ N$ (for all $n ∈ N$, $C n$ is denoted as the class of groups, which satisfies the relation $b n a n = a n b n$).
(iii) When χ is central, then condition (31) also holds.
Theorem 11.
If maximum functional Equation (1) is stable on group G, then every free abelian group K can be embedded into a group G.
Proof.
Assume that $η : G → R$ and let $λ ≥ 0$ be such that
$| max { η ( x y ) , η ( x y − 1 ) } − η ( x ) − η ( y ) | ≤ λ for any x , y ∈ G .$
Because every free abelian group is a torsion-free, K is torsion-free group. Subsequently, from HNN-extensions, any torsion-free group K can be embedded into a group G, if for all $k ∈ K$, there exists $g ∈ G$ such that $g k g − 1 = k 2$ [14,15]. Assume that K is embedded into a group G. Subsequently, inequality (33) can be rewritten as
$| max { η ( k g ) , η ( k g − 1 ) } − η ( k ) − η ( g ) | ≤ λ for any k , g ∈ G .$
We need to show that $η$ is bounded. The result is obvious for $λ = 0$, so let $λ$ be positive. More specifically, we prove that $| η ( k ) | < 2 λ$ for all $k ∈ G$. Afterwards, from (34), we have two possibilities, either $| η ( k g − 1 ) − η ( k ) − η ( g ) | ≤ λ$ or $| η ( k g ) − η ( k ) − η ( g ) | ≤ λ$. Let us consider first possibility and put $k = g = 1$, then we can conclude that $| η ( 1 ) | ≤ λ$. Put $g = k$, then we have
$| η ( 1 ) − 2 η ( k ) | ≤ λ | 2 η ( k ) | − | η ( 1 ) | ≤ λ 2 | η ( k ) | ≤ λ + | η ( 1 ) | 2 | η ( k ) | ≤ λ + λ | η ( k ) | ≤ λ .$
For second possibilty, we have
$| η ( k g ) − η ( k ) − η ( g ) | ≤ λ .$
On the contrary, let $| η ( k ) | ≥ 2 λ$ for some arbitrary $k ∈ G$. Put $g = k$ in (35), then we have
$| η ( k 2 ) − 2 η ( k ) | ≤ λ | 2 η ( k ) − η ( k 2 ) | ≤ λ 2 | η ( k ) | − λ ≤ | η ( k 2 ) | 3 λ ≤ | η ( k 2 ) | .$
Again, put $g = k 2$ in (35), then
$| η ( k 3 ) − η ( k ) − η ( k 2 ) | ≤ λ | η ( k ) + η ( k 2 ) − η ( k 3 ) | ≤ λ | η ( k ) | + | η ( k 2 ) | ≤ λ + | η ( k 3 ) | 5 λ − λ ≤ | η ( k 3 ) | 4 λ ≤ | η ( k 3 ) | .$
Repeating again, for $g = k 3$, we can determine $| η ( k 4 ) | ≥ 4 λ$. Continuing this process, we can evaluate that
$( m + 1 ) λ ≤ | η ( k m ) | , for m = 1 , 2 , ⋯ ,$
which determines that when m varies, then $η ( k m )$ is unbounded.
Furthermore, pick an arbitrary element $g ∈ G$, such that $k 2 = g k g − 1$. Subsequently, for every integer $m > 0$, we have $k 2 m = g k m g − 1$. Therefore, for every m, put $g = k m$ and $k = k m$ in (35), then
$| η ( k 2 m ) − 2 η ( k m ) | ≤ λ | η ( g k m g − 1 ) − 2 η ( k m ) | ≤ λ .$
$| η ( g k m g − 1 ) − η ( g ) − η ( k m g − 1 ) | ≤ λ and | η ( k m g − 1 ) − η ( k m ) − η ( g − 1 ) | ≤ λ ,$
thus, we can conclude that
$| η ( g k m g − 1 ) − η ( g ) − η ( k m ) − η ( g − 1 ) | ≤ | η ( g k m g − 1 ) − η ( g ) − η ( k m g − 1 ) | + | η ( k m g − 1 ) − η ( k m ) − η ( g − 1 ) | | η ( g k m g − 1 ) − η ( g ) − η ( k m ) − η ( g − 1 ) | ≤ 2 λ .$
From (36) and (37), we have
$| η ( g k m g − 1 ) − 2 η ( k m ) + η ( k m ) | ≤ 2 λ + | η ( g ) | + | η ( g − 1 ) | | η ( k m ) | ≤ 2 λ + | η ( g ) | + | η ( g − 1 ) | + | η ( g k m g − 1 ) − 2 η ( k m ) | | η ( k m ) | ≤ 5 λ for m = 1 , 2 , ⋯ ,$
which gives a contradiction. Hence, $η$ is bounded, which completes the required proof. □
Corollary 13.
If maximum functional Equation (1) is stable on group G, then any discretely normed abelian group K can be embedded into a group G.
Proof.
Because K is a discretely normed abelian group, then, by applying theorem from [11], K is a free group; hence it can be embedded into a group G. □
Theorem 12.
A function $η : G → R$ satisfying (27) is bounded on the commutator group $G 1$ of the subgroup $G ′$ of G if the condition (31) is satisfied.
Proof.
Assume that condition (31) is satisfied. Subsequently, by Theorem 10, we can get that $χ ( x ) + χ ( y ) = max { χ ( x y ) , χ ( x y − 1 ) }$ holds for any $x , y ∈ G$. Because this maximum equation holds; therefore, by Theorem 1, there exists an additive function $α : G → R$, such that $χ ( x ) = | α ( x ) |$ for any $x ∈ G$. For $a , b ∈ G$, take $a − 1 b − 1 a b ∈ G$, then
$χ ( a − 1 b − 1 a b ) = | α ( a − 1 b − 1 a b ) | = | α ( a − 1 ) + α ( b − 1 ) + α ( a ) + α ( b ) | = | α ( a − 1 a ) + α ( b − 1 b ) | = | α ( 1 ) + α ( 1 ) | = | 0 + 0 | χ ( a − 1 b − 1 a b ) = 0 .$
Because $χ$ is zero on the commutator group $G 1$ of the subgroup $G ′$ of G, $η$ is bounded on commutator group $G 1$. □
Corollary 14.
If a function $η : G → R$ satisfies the condition
$lim n → ∞ 1 2 n [ max { η ( x 2 n y 2 n ) , η ( x 2 n y − 2 n ) } − η ( [ x y ] 2 n ) ] = 0 f o r a n y x , y ∈ G ,$
then η is bounded on the commutator group $G 1$.
## Author Contributions
Writing—original draft, M.S., Writing—review and editing, Q.L. and Y.L. All authors have read and agreed to the published version of the manuscript.
## Funding
This research was funded by National Natural Science Foundation of China grant number 11971493.
## Acknowledgments
The authors thank anonymous referees for their remarkable comments, suggestions, and ideas that help to improve this paper.
## Conflicts of Interest
The authors declare no conflict of interest.
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## Share and Cite
MDPI and ACS Style
Sarfraz, M.; Liu, Q.; Li, Y. Stability of Maximum Functional Equation and Some Properties of Groups. Symmetry 2020, 12, 1949. https://doi.org/10.3390/sym12121949
AMA Style
Sarfraz M, Liu Q, Li Y. Stability of Maximum Functional Equation and Some Properties of Groups. Symmetry. 2020; 12(12):1949. https://doi.org/10.3390/sym12121949
Chicago/Turabian Style
Sarfraz, Muhammad, Qi Liu, and Yongjin Li. 2020. "Stability of Maximum Functional Equation and Some Properties of Groups" Symmetry 12, no. 12: 1949. https://doi.org/10.3390/sym12121949
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# 9/10 equivalent fractions
What Is The Equivalent Fraction Of 9 10? 9/10 = 18/20 = 90/100 and an infinite number of other equivalents. For any value of n except 0, the product (9/10)* (n/n) will be an equivalent fraction. Shown above are the equivalents for n=2 and n=10.
## How do you calculate equivalent fractions?
What are equivalent fractions?
• Sometimes fractions can be simplified into smaller ones.
• These fractions are said to be equivalent.
• You can multiply the numerator and denominator in the smaller fraction, by the same number, and get the first fraction back.
## What are the rules of equivalent fractions?
Equivalent Fractions Rule. A rule stating that if the numerator and denominator of a fraction are multiplied by the same nonzero number, the result is a fraction that is equivalent to the original fraction. This rule can be represented as: a//b = (n * a)//(n * b).
## What decimal is equivalent to 9\10?
9/10 as a decimal equals to 0.9 where, 9/10 is a given fraction, The decimal expansion of 9/10 is 0.9
## How to make an equivalent fraction?
It is possible by these methods:
• Method 1: Make the Denominators the same
• Method 2: Cross Multiply
• Method 3: Convert to decimals
## What is 9/10 as a fraction?
9/10 = 910 = 0.9 Spelled result in words is nine tenths.
## What is 9/12 equivalent to as a fraction?
3/49/12 ÷ 3/3 = 3/4, so 9/12 and 3/4 are equivalent fractions, with 3/4 being the fraction in its simplest form.
## What is 9 equal to as a fraction?
The simplest way to write 9 as a fraction is 9/1. Any number divided by 1 is still that number; dividing an integer by 1 does nothing to its value.
## How do I find equivalent fractions?
How to Find Equivalent Fractions. Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction, and you’ll create an equivalent fraction.
## How do you write equivalent fractions?
For each fraction, we can find its equivalent fraction by multiplying both numerator and denominator with the same number. For example, we have to find the third equivalent fraction of ⅔; then we have to multiply 2/3 by 3/3. Hence, 2/3 × (3/3) = 6/9, is the fraction equivalent to 2/3.
## What is 9 4 in a fraction?
2 1/4The denominator is the exact same number as the denominator in the original improper fraction; in this case, 4. So 9/4 equals 2 1/4.
## What is 9/8 as a fraction?
9/8 = 98 = 1 18 = 1.125 Spelled result in words is nine eighths (or one and one eighth).
## What is equivalent calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## What is the equivalent fraction of 5 by 9?
30 / 54 ÷ 6 / 6 = 5 / 9. ∴ 5 / 9 and 30 / 54 are equivalent fractions as their value is the same irresepective of the numbers in numerator and denominator.
## What is 9 over 11 as a decimal?
0.81818181Fraction to decimal conversion tableFractionDecimal6/110.545454547/110.636363638/110.727272729/110.8181818151 more rows
## What is 4/8 equivalent to as a fraction?
So we see that 1/2 will equal 2/4, and that equals 4/8.
## What is the simplest form of 9 12?
So the greatest common factor of 9 and 12 is 3. So you divide both sides by 3. So the simplest form of 912 is 34 .
## What are the equivalent fractions for 8 12?
2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3.
## What are Equivalent Fractions?
Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same.
## How to make a fraction equivalent?
Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction.
## What is half of a fraction?
For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100
## What is the equivalent fraction of 4/9?
The equivalent fraction for 4/9 is 32 /72….
## What is 9/10 in math?
9/10 = 18/20 = 90/100 and an infinite number of other equivalents.
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 10 9) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## Is 10 9 a fraction?
Important: 10 9 looks like a fraction, but it is actually an improper fraction.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
## How to convert decimals to fractions?
It does however require the understanding that each decimal place to the right of the decimal point represents a power of 10; the first decimal place being 10 1, the second 10 2, the third 10 3, and so on. Simply determine what power of 10 the decimal extends to , use that power of 10 as the denominator, enter each number to the right of the decimal point as the numerator, and simplify. For example, looking at the number 0.1234, the number 4 is in the fourth decimal place which constitutes 10 4, or 10,000. This would make the fraction#N#1234#N##N#10000#N#, which simplifies to#N#617#N##N#5000#N#, since the greatest common factor between the numerator and denominator is 2.
## What is fraction in math?
In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole. For example, in the fraction. 3.
## How to multiply fractions?
Just multiply the numerators and denominators of each fraction in the problem by the product of the denominators of all the other fractions (not including its own respective denominator) in the problem.
## How to find common denominator of fractions?
One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method.
## How to divide fractions?
In order to divide fractions, the fraction in the numerator is multiplied by the reciprocal of the fraction in the denominator. The reciprocal of a number a is simply. . When a is a fraction, this essentially involves exchanging the position of the numerator and the denominator. The reciprocal of the fraction.
## What is fraction used for in engineering?
In engineering, fractions are widely used to describe the size of components such as pipes and bolts. The most common fractional and decimal equivalents are listed below.
## Is it easier to work with simplified fractions?
It is often easier to work with simplified fractions. As such, fraction solutions are commonly expressed in their simplified forms.#N#220#N##N#440#N#for example, is more cumbersome than#N#1#N##N#2#N#. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor.
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What will be the cost of painting the inner walls of a room if the rate of painting is Rs. 20 per square foot? I. C
### Question Asked by a Student from EXXAMM.com Team
Q 1433167042. What will be the cost of painting the inner walls of a room if the rate of painting is Rs. 20 per square foot?
I. Circumference of the floor is 44 feet.
II. The height of the wall of the room is 12 feet.
A
I alone sufficient while II alone not sufficient to answer
B
II alone sufficient while I alone not sufficient to answer
C
Either I or II alone sufficient to answer
D
Both I and II are not sufficient to answer
E
Both I and II are necessary to answer
#### HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
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# All MBA Deadlines for 2017 - 2018
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31 Jul 2013, 14:24
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Are we missing a deadline? Do you see a deadline that has changed? Do you not see your school? (Many schools have yet to release their deadlines) Send me a PM to let me know.
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Last edited by Narenn on 19 May 2017, 04:02, edited 7 times in total.
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07 Oct 2013, 13:05
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NUS, Singapore..
Oct 1 2013 to Jan 31 2014 - R1
Feb 1 2014 to Mar 31 2014 - R2
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23 Oct 2013, 00:42
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Hi, is there a spreadsheet with say the top 100 schools from the FT rankings, with 1st, 2nd and 3rd round application deadlines?
Thanks
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04 Dec 2013, 13:16
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i want to know what is the maximaum marks of the gmat examination??i also want to know what is my possibility of getting into harvard or stanford even though i am not from a very reputed college in india.i am very to get your reply..and i really want to get into these colleges.i also want to know what are the other things required other than the gmat score.
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18 Dec 2013, 10:33
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Hi,
I have scheduled my GMAT exam on the 25th of January,2014.
Am i eligible for the R3?
My choice of colleges are
1. Michigan---- Ross
2. Booth
3. Cornell
4. Duke
5.Kellog
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18 Dec 2013, 12:33
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samueljohnson956 wrote:
Hi,
I have scheduled my GMAT exam on the 25th of January,2014.
Am i eligible for the R3?
My choice of colleges are
1. Michigan---- Ross
2. Booth
3. Cornell
4. Duke
5.Kellog
Why don't you add all of those schools to your profile and then visit the MBA Timeline - it will do the work for you. You can also add your schools there.
URL: http://gmatclub.com/forum/mbatimeline
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03 Jan 2014, 00:10
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Keep in mind that Cornell's program starts in May and not in August-September like most other programs. International applicants should be sure of their visa timelines if they are going to apply in R3 (Feb 12th)
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05 Mar 2014, 08:52
Just noticed an error. The Sloan R2 notification/decision-release date is listed here as May 1. It is April 1. May 1 is when deposits are due.
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28 Apr 2014, 17:02
Thanks for compiling and updating. Will bookmark this one.
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17 May 2014, 14:25
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As everyone is awaiting release of the new deadlines for the 2015 Matriculation Year, here is a list of all 2014 deadlines (Past and Present)
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07 Jun 2014, 00:44
I think, the list needs to be updated for 2014-2015 MBA admission deadlines. Thanks!
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07 Jun 2014, 00:48
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This list also misses several international B-schools like Kellogs, INSEAD, IE, SP Jain, Booth, Yale, Ross, etc.
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28 Jun 2014, 12:31
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Why is the Spring admission information so perfectly hidden? Can someone please help with the universities offering Spring admissions for full-time MBA? I know there are very few, but those very few also seem to be unseen.
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28 Jun 2014, 21:35
Expert's post
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AnjaliRao wrote:
Why is the Spring admission information so perfectly hidden? Can someone please help with the universities offering Spring admissions for full-time MBA? I know there are very few, but those very few also seem to be unseen.
Application deadlines of Thunderbird (30th Jun 14, 30th Aug 14, & 10th Nov 14) are for the Spring Admissions. I haven't came across any other US Full time program yet which has Spring deadlines. Feel free to let me know if you know the schools that offer Spring admissions.
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28 Jun 2014, 22:34
Thank you for your response. As of now, through my extensive efforts put into spotting them, I came to know that Simon, Southern Illinois and Thunderbird offer full-time MBA. I am worried about what chances I stand though. Admissions are in R2 phase now.
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07 Aug 2014, 03:49
[quote=Application deadlines of Thunderbird (30th Jun 14, 30th Aug 14, & 10th Nov 14) are for the Spring Admissions. I haven't came across any other US Full time program yet which has Spring deadlines. Feel free to let me know if you know the schools that offer Spring admissions.[/quote]
Here is a list of MBA programs with winter/spring intakes: http://www.aringo.com/MBA_January_Programs.htm
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15 Aug 2014, 22:42
I think chicago booth mba R1 has been mentione with different deadlines. Can some please let me know if it requires any correction?
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15 Aug 2014, 23:28
Chicago r1 deadline is September 25, 2014:
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Re: All MBA Deadlines for 2017 - 2018 [#permalink] 15 Aug 2014, 23:28
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# All MBA Deadlines for 2017 - 2018
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Lecture 14
# Lecture 14 - 1 Testing time series for unit roots We know...
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Unformatted text preview: 1 Testing time series for unit roots We know that a random walk is a particular type of AR(1) process x t = ρ x t-1 + e t with ρ = 1 Hence to test whether x t is a random walk (with zero mean), we could estimate x t = ρ x t-1 + e t and test H : ρ = 1 nonstationarity against H 1 : ρ < 1 This is exactly equivalent to the regression (x t- x t-1 ) = ( ρ- 1)x t-1 + e t = β x t-1 + e t and test H’ : β = 0 against H’ 1 : β < 0 a one-sided test However the ‘t-ratio’ for this regression t( β ) = β /SE( β ) has a non-standard distribution (not a standard t distribution) Test procedure derived by Dickey/Fuller: compares t- statistic with special critical values which are tabulated in Table 1 of Handout An example of the Dickey-Fuller test Testing the bond price for nonstationarity Using deviations from mean (so x t has zero mean) x t- x t-1 = -0.0384x t-1 + e t t-ratio on x t-1 = -1.381 (5% CV -1.95) We cannot reject H : this implies deviations in bond price are a random walk, as expected if speculators are rational 2 Testing series with a non-zero mean Suppose x t has a non-zero mean μ If x t is nonstationary x t- μ = x t-1- μ + e t so the mean drops out...
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Introduction to Data Preprocessing in Machine Learning
We can’t say that Blue<Green as it doesn’t make any sense to compare the colors as they don’t have any relationship.The important thing to note here is that we need to preprocess ordinal and nominal categorical variables differently.Handling Ordinal Categorical Variables —First of all we need to create a dataframe.df_cat = pd.DataFrame(data = [['green','M',10.1,'class1'], ['blue','L',20.1,'class2'], ['white','M',30.1,'class1']])df_cat.columns = ['color','size','price','classlabel']Here the columns ‘size’ and ‘classlabel’ are ordinal categorical variables whereas ‘color’ is a nominal categorical variable.There are 2 pretty simple and neat techniques to transform ordinal CVs.Using map() function —size_mapping = {'M':1,'L':2}df_cat['size'] = df_cat['size'].map(size_mapping)Here M will be replaced with 1 and L with 2.2..Using Label Encoder —from sklearn.preprocessing import LabelEncoderclass_le = LabelEncoder()df_cat['classlabel'] =class_le.fit_transform(df_cat['classlabel'].values)Here class1 will be represented with 0 and class2 with 1 .Incorrect way of handling Nominal Categorical Variables —The biggest mistake that most people do is that they are not able to differentiate between ordinal and nominal CVs.So if you the same map() function or LabelEncoders with nominal variables then the model will think that there is some sort of relationship between the nominal CVs.So if we use map() to map the colors like -col_mapping = {'Blue':1,'Green':2}Then according to the model Green > Blue which is again a senseless assumption so the model will give you results considering this relationship.So although you will get the results using this method they won’t be optimal.Correct way of handling Nominal Categorical Variables —The correct way of handling nominal CVs is to use One-Hot Encoding..The easiest way to use One-Hot Encoding is to use the get_dummies() function.pd.get_dummies(df_cat[['color','size','price']])Here we have passed ‘size’ and ‘price’ along with ‘color’ but the get_dummies() function is pretty smart and will comsider only the string variables..So it will just transform the ‘color’ variable.Now, you must be wondering what the hell is this One-Hot Encoding.So let’s try and understand it.One-Hot Encoding —So in One-Hot Encoding what we essentially do is that we create ’n’ columns where n is the number of unique values that the nominal variable can take.Ex — Here if color can take Blue,Green and White then we will just create three new columns namely — color_blue,color_green and color_white and if the color is green then the values of color_blue and color_white column will be 0 and value of color_green column will be 1 .So out of the n columns only one column can have value = 1 and the rest all will have value = 0.One-Hot Encoding is a pretty cool and neat hack but there is only one problem associated with it and that is Multicollinearity..As you all must have assumed that it is a pretty heavy word so it must be difficult to understand, so let me just validate your newly formed belief.Multicollinearity is indeed a slightly tricky but extremely important concept of Statistics..The good thing here is that we don’t really need to understand all the nitty-gritty details of multicollinearity, rather we just need to focus on how it will impact our model..So let’s dive into this concept of Multicollinearity and how it will impact our model.Multicollinearity and its impact —Multicollinearity occurs in our dataset when we have features which are strongly dependent on each other..Ex- In this case we have features -color_blue,color_green and color_white which are all dependent on each other and it can impact our model.The main impact it will have is that it can cause the decision boundary to change which can have a huge impact on the result of our model.In addition to that if we have multicollinearity in our dataset then we won’t be able to use our weight vector to calculate the feature importance.I think this much of information is enough in the context of Machine Learning however if you are still not convinced, then you can visit the below link to understand the maths and logic associated with Multicollinearity.12.1 – What is Multicollinearity?.| STAT 501As stated in the lesson overview, multicollinearity exists whenever two or more of the predictors in a regression model…newonlinecourses.science.psu.eduNow that we have understood what Multicollinearity is, let’s now try to understand how to identify it.The easiest method to identify Multicollinearity is to just plot a pairplot and you can observe the relationships between different features..If you get a linear relationship between 2 features then they are strongly co-related with each other and there is multicollinearity in your dataset.Pair PlotHere (Weight,BP) and (BSA,BP) are closely related..You can also use the correlation matrix to check how closely related the features are.Correlation MatrixWe can observe that there is a strong co-relation (0.950) between Weight and BP and also between BSA and BP (0.875).Simple hack to avoid Multicollinearity-We can use drop_first=True in order to avoid the problem of Multicollinearity.pd.get_dummies(df_cat[['color','size','price']],drop_first=True)Here drop_first will drop the first column of color..So here color_blue will be dropped and we will only have color_green and color_white.The important thing to note here is that we don’t lose any information, as if color_green and color_white are both 0 then it implies that the color must have been blue.So we can infer the whole information with the help of only these 2 columns, hence the strong co-relation between these three columns is broken.. More details
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# Tagged Questions
Questions about the determinant of square matrices or linear endomorphisms. Also for closely related topics such as minors or regularized determinants.
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### Does this Linear Algebra Construction have a Name?
Let $\mathcal{R}$ be a ring and let $v^0,\ldots,v^{k-1}\in\mathcal{R}^m$ with $m \geq k$. Suppose we wish to find $w\in Span(v^0,\ldots,v^{k-1})$ such that $k-1$ specified coordinates of $w$ vanish (...
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Is there a parametrization of set of matrices $\mathcal M\subseteq\Bbb Z[x_1,\dots,x_{m}]^{n\times n}$ such that $\forall f:\{-1,+1\}^{m}\rightarrow\{-1,+1\}$ $\exists M\in\mathcal M$ such that $\... 0answers 96 views ### Partial Vandermonde Circulant Determinant Expression Consider following partial Vandermonde type, circulant matrix$\begin{bmatrix} x_1 & x_2 & 0 & \dots & 0 & x_n\\ x_1^2 & x_2^2 & x_3^2 & \dots & 0 & 0\\ \vdots ...
Given a square integer matrix $A \in M_n(Z)$ and two subsets $I, J \subset \{ 1, \ldots, n\}$, we define $A_{I,J}$ as the sub-matrix of $A$ containing the rows (resp. columns) whose index is in $I$ (...
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# Not an actual modulo operation?
In this level in the desert where the `return` command is introduced (name displayed in screenshot: “Hetze nicht, sei still”) there’s a function referred to as “modulo operation” in the hints section:
However, the actual function is not performing a modulo calculation (calculating the rest of a division), there is only a subtraction:
``````# Diese Funktion gibt einen Wert >= 0 und <30 zurück:
def mod30(n):
if n >= 30:
return n - 30
else:
return n
``````
It seems that someone just named that function `mod30` for some reason that has nothing to do with what it does, right?
Also the comment says that the function will return a value between 0 and 29, which is not true for `n > 60`
This is all very confusing.
Are there errors in the design of this level/the hints text, or is it I just me getting everything wrong?
Division is only a series of subtractions, just as multiplication is just a series of additions. For example, if n was equal to 300, then the code would be:
``````def mod30(300):
if 300 is bigger than or equal to 30:
return 300 - 30 (the returned number will now equal 270)
else:
return n ( but at the moment n(270) is still greater than or equal to 30 so we repeat the process again until n is less than 30 at which point the code will pass to the else and we will return the number n)
``````
this is how the series of subtractions would go If 300 was n and we ran the above code:
300 - 30 = 270 ( back to the top of the code)
270 - 30 = 240( back again)
240 - 30 = 210
and so on until:
60 - 30 = 30
30 - 30 = 0 (30 is still greater than or equal to 30)
then we will pass to the else and return n, which is now 0
As you can see, this repeated subraction is really the same as division, so this is a modular operation.
Here’s a link to the Wikipedia page explaining modulo in more depth https://en.wikipedia.org/wiki/Modulo_operation
3 Likes
Thank you, this explanation was very helpful - it helped me understand this level’s code (not in its entirety, but the part that you explained).
However, I feel it’s worth mentioning that waht you said isn’t true for the snippet of code I posted alone!
It only works along with the rest of the pre-written code of that level.
Here it is:
``````# Diese Funktion gibt einen Wert >= 0 und <30 zurück:
def mod30(n):
if n >= 30:
return n - 30
else:
return n
# Diese Funktion sollte einen Wert >=0 und <40 zurückgeben:
def mod40(n):
# Benutze ein if-Statement, um den korrekten Wert zurückzugeben.
if n >= 40:
return n - 40
else:
return n
# Ändere den folgenden Code nicht:
while True:
time = hero.time
x = mod30(time) + 25
y = mod40(time) + 10
hero.moveXY(x, y)
``````
i.e. either of the two mod(n) functions only becomes a modulo operation when repeatedly called upon in a loop!
By itself it is indeed just a subtraction.
that’s also how the comment is meant:
``````# this function returns a value >= 0 and <30 , _when its looped_
def mod30(n):
if n >= 30:
return n - 30
else:
return n
``````
…The part I don’t understand is how this movement pattern helps the hero evade the cannons - and why a modulo operation is needed to achieve it. I’m guessing you have to be one of the people who wrote this level to understand why.
1 Like
Yeah, I think you’re right or maybe if it sees that you’ve but the right code in, it says you’ve done the level.
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Chapter 15, 21 Solutions
# Chapter 15, 21 Solutions - Chapter 15 Capital Structure...
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Capital Structure Decisions 15-7 a. Here are the steps involved: (1) Determine the variable cost per unit at present, V: Profit = P(Q) - FC - V(Q) \$500,000 = (\$100,000)(50) - \$2,000,000 - V(50) 50(V) = \$2,500,000 V = \$50,000. (2) Determine the new profit level if the change is made: New profit = P 2 (Q 2 ) - FC 2 - V 2 (Q 2 ) = \$95,000(70) - \$2,500,000 - (\$50,000 - \$10,000)(70) = \$1,350,000. (3) Determine the incremental profit: Profit = \$1,350,000 – \$500,000 = \$850,000. (4) Estimate the approximate rate of return on new investment: Return = Profit/Investment = \$850,000/\$4,000,000 = 21.25%. Since the return exceeds the 15 percent cost of equity, this analysis suggests that the firm should go ahead with the change. b. The change would increase the breakeven point: Old: Q BE = = = 40 units. New: Q BE = = 45.45 units. c. It is impossible to state unequivocally whether the new situation would have more or less business risk than the old one. We would need information on both the sales probability distribution and the uncertainty about variable input cost in order to make this determination. However, since a higher breakeven point, other things held constant, is more risky. Also the percentage of fixed costs increases: Old: = = 44.44%. New: = = 47.17%. Chapter 15
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The change in breakeven points--and also the higher percentage of fixed costs-- suggests that the new situation is more risky. 15-8 a. Original value of the firm (D = \$0): We are given that the book value of assets is equal to the market value of assets, so the value is \$3,000,000. Alternatively, we can calculate the value as the sum of the debt (which is zero) and the stock (200,000 shares at a price of \$15 per share): V = D + S = 0 + (\$15)(200,000) = \$3,000,000. Original cost of capital: WACC = w d r d (1-T) + w ce r s = 0 + (1.0)(10%) = 10%. With financial leverage (w d =30%): WACC = w d r d (1-T) + w ce r s = (0.3)(7%)(1-0.40) + (0.7)(11%) = 8.96%. Because growth is zero, FCF is equal to EBIT(1-T). The value of operations is: V op = Increasing the financial leverage by adding \$900,000 of debt results in an increase in the firm’s value from \$3,000,000 to \$3,348,214.286. b.
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## This note was uploaded on 02/27/2012 for the course BUS 510 taught by Professor Mehdi during the Spring '11 term at University of La Verne.
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# meter
(redirected from bank cubic yard)
Also found in: Dictionary, Thesaurus, Legal, Encyclopedia.
## meter
[me´ter]
1. the base SI unit of linear measure, approximately equivalent to 39.37 inches. Symbol m.
2. an apparatus for measuring the quantity of something passing through it, such as a gas or an electric current.
## me·ter (m),
(mē'tĕr),
1. The fundamental unit of length in the SI and metric system, equivalent to 39.37007874 inches. Defined as the length of the path traveled by light in a vacuum in 1/299792458 seconds.
2. A device for measuring the quantity of that which passes through it.
[Fr. metre; G. metron, measure]
## meter
/me·ter/ (me´ter)
1. the base SI unit of linear measure, approximately equivalent to 39.37 inches. Symbol m.
2. an apparatus to measure the quantity of anything passing through it.
## meter (m)
[mē′tər]
Etymology: Gk, metron, measure
a metric unit of length equal to 39.37 inches.
## meter
(1) A device used to measure something.
(2) US spelling of metre (internationa spelling), the SI unit of length, defined as 1,650,763.73 wavelengths of the orange-red emission line in the electromagnetic spectrum of the krypton-86 atom in a vacuum.
## me·ter
(mē'tĕr)
1. The fundamental unit of length in both the SI and metric system, equivalent to 39.37007874 inches. Defined to be the length of path traveled by light in a vacuum in 1/299792458 sec.
2. A device for measuring the quantity of that which passes through it.
Synonym(s): metre.
[Fr. metre; G. metron, measure]
## me·ter
(m) (mē'tĕr)
1. Fundamental unit of length in the SI and metric system, equivalent to 39.37007874 inches.
2. A device for measuring the quantity of that which passes through it.
Synonym(s): metre.
[Fr. metre; G. metron, measure]
## meter
the basic unit of linear measure of the metric system, being equivalent to 39.371 inches; abbreviated m. See also Table 3 and si units.
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, 07.03.2023 23:41 momo0528
# A rectangle has vertices A(-1, 1), B(3, 1), C(-1, -4), and D(3, -4). Use the coordinates to find the perimeter of the rectangle. Then find the area of the rectangle.
answer: it is 1: 47 pm
Where’s the
3x + 18 = 147°
step-by-step explanation:
3x + 18 = 147°, because the value of the central angle is equal to the value of the arc formed by it.
i hope i you.
### Another question on Mathematics
Mathematics, 21.06.2019 18:10
Abc will undergo two transformations to give aa'b'c' which pair of transformations will give a different image of aabc if the order of the transformations is reversed? a. a rotation 90' counterclockwise about the origin followed by a reflection across the y-axis b a translation 5 units down followed by a translation 4 units to the right c. a reflection across the x-axis followed by a reflection across the y-axis a rotation 180'clockwise about the ongin followed by a reflection across the y-axis reset next 2 omentum all rights reserved o atk me anything o o e
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Document
# More on Change-Making and Related Problems
## File
LIPIcs.ESA.2020.29.pdf
• Filesize: 0.52 MB
• 14 pages
## Acknowledgements
We thank Adam Polak and Chao Xu for discussion and, in particular, for bringing the minimum word break problem to our attention.
## Cite As
Timothy M. Chan and Qizheng He. More on Change-Making and Related Problems. In 28th Annual European Symposium on Algorithms (ESA 2020). Leibniz International Proceedings in Informatics (LIPIcs), Volume 173, pp. 29:1-29:14, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2020)
https://doi.org/10.4230/LIPIcs.ESA.2020.29
## Abstract
Given a set of n integer-valued coin types and a target value t, the well-known change-making problem asks for the minimum number of coins that sum to t, assuming an unlimited number of coins in each type. In the more general all-targets version of the problem, we want the minimum number of coins summing to j, for every j = 0,…,t. For example, the textbook dynamic programming algorithms can solve the all-targets problem in O(nt) time. Recently, Chan and He (SOSA'20) described a number of O(t polylog t)-time algorithms for the original (single-target) version of the change-making problem, but not the all-targets version. In this paper, we obtain a number of new results on change-making and related problems: - We present a new algorithm for the all-targets change-making problem with running time Õ(t^{4/3}), improving a previous Õ(t^{3/2})-time algorithm. - We present a very simple Õ(u²+t)-time algorithm for the all-targets change-making problem, where u denotes the maximum coin value. The analysis of the algorithm uses a theorem of Erdős and Graham (1972) on the Frobenius problem. This algorithm can be extended to solve the all-capacities version of the unbounded knapsack problem (for integer item weights bounded by u). - For the original (single-target) coin changing problem, we describe a simple modification of one of Chan and He’s algorithms that runs in Õ(u) time (instead of Õ(t)). - For the original (single-capacity) unbounded knapsack problem, we describe a simple algorithm that runs in Õ(nu) time, improving previous near-u²-time algorithms. - We also observe how one of our ideas implies a new result on the minimum word break problem, an optimization version of a string problem studied by Bringmann et al. (FOCS'17), generalizing change-making (which corresponds to the unary special case).
## Subject Classification
##### ACM Subject Classification
• Theory of computation → Design and analysis of algorithms
##### Keywords
• Coin changing
• knapsack
• dynamic programming
• Frobenius problem
• fine-grained complexity
## Metrics
• Access Statistics
• Total Accesses (updated on a weekly basis)
0
## References
1. Noga Alon, Zvi Galil, Oded Margalit, and Moni Naor. Witnesses for Boolean matrix multiplication and for shortest paths. In Proceedings of the 33rd Annual IEEE Symposium on Foundations of Computer Science (FOCS), pages 417-426, 1992. URL: https://doi.org/10.1109/SFCS.1992.267748.
2. Kyriakos Axiotis and Christos Tzamos. Capacitated dynamic programming: Faster knapsack and graph algorithms. In Proceedings of the 46th International Colloquium on Automata, Languages, and Programming (ICALP), pages 19:1-19:13, 2019. URL: https://doi.org/10.4230/LIPIcs.ICALP.2019.19.
3. Arturs Backurs and Piotr Indyk. Which regular expression patterns are hard to match? In Proceedings of the 57th Annual IEEE Symposium on Foundations of Computer Science (FOCS), pages 457-466, 2016.
4. MohammadHossein Bateni, MohammadTaghi Hajiaghayi, Saeed Seddighin, and Cliff Stein. Fast algorithms for knapsack via convolution and prediction. In Proceedings of the 50th Annual ACM Symposium on Theory of Computing (STOC), pages 1269-1282, 2018. URL: https://doi.org/10.1145/3188745.3188876.
5. Karl Bringmann. A near-linear pseudopolynomial time algorithm for subset sum. In Proceedings of the 28th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), pages 1073-1084, 2017.
6. Karl Bringmann, Allan Grønlund, and Kasper Green Larsen. A dichotomy for regular expression membership testing. In Proceedings of the 58th Annual IEEE Symposium on Foundations of Computer Science (FOCS), pages 307-318, 2017. URL: http://arxiv.org/abs/1611.00918.
7. Timothy M. Chan. More algorithms for all-pairs shortest paths in weighted graphs. SIAM Journal on Computing, 39(5):2075-2089, 2010.
8. Timothy M. Chan and Qizheng He. On the change-making problem. In Proceedings of the 4th ACM-SIAM Symposium on Simplicity in Algorithms (SOSA), pages 38-42, 2020.
9. Marek Cygan, Marcin Mucha, Karol Wegrzycki, and Michal Wlodarczyk. On problems equivalent to (min, +)-convolution. ACM Transactions on Algorithms, 15(1):14:1-14:25, 2019. URL: https://doi.org/10.1145/3293465.
10. Jacques Dixmier. Proof of a conjecture by Erdős and Graham concerning the problem of Frobenius. Journal of Number Theory, 34(2):198-209, 1990.
11. Ran Duan and Seth Pettie. Fast algorithms for (max,min)-matrix multiplication and bottleneck shortest paths. In Proceedings of the 20th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), pages 384-391, 2009.
12. Friedrich Eisenbrand and Robert Weismantel. Proximity results and faster algorithms for integer programming using the Steinitz lemma. ACM Transactions on Algorithms, 16(1):5:1-5:14, 2020. URL: https://doi.org/10.1145/3340322.
13. Paul Erdős and Ronald L Graham. On a linear diophantine problem of Frobenius. Acta Arithmetica, 21(1):399-408, 1972.
14. Klaus Jansen and Lars Rohwedder. On integer programming and convolution. In Proceedings of the 10th Innovations in Theoretical Computer Science Conference (ITCS), pages 43:1-43:17, 2019. URL: https://doi.org/10.4230/LIPIcs.ITCS.2019.43.
15. Ce Jin and Hongxun Wu. A simple near-linear pseudopolynomial time randomized algorithm for subset sum. In Proceedings of the 2nd Symposium on Simplicity in Algorithms (SOSA), volume 69, pages 17:1-17:6, 2019.
16. Konstantinos Koiliaris and Chao Xu. A faster pseudopolynomial time algorithm for subset sum. In Proceedings of the 28th Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), pages 1062-1072, 2017.
17. Konstantinos Koiliaris and Chao Xu. Faster pseudopolynomial time algorithms for subset sum. ACM Transactions on Algorithms, 15(3):1-20, 2019.
18. Marvin Künnemann, Ramamohan Paturi, and Stefan Schneider. On the fine-grained complexity of one-dimensional dynamic programming. In Proceedings of the 44th International Colloquium on Automata, Languages, and Programming (ICALP), pages 21:1-21:15, 2017.
19. Andrea Lincoln, Adam Polak, and Virginia Vassilevska Williams. Monochromatic triangles, intermediate matrix products, and convolutions. In Proceedings of the 11th Innovations in Theoretical Computer Science Conference (ITCS), pages 53:1-53:18, 2020.
20. George S. Lueker. Two NP-complete problems in nonnegative integer programming. Technical report, Princeton University. Department of Electrical Engineering, 1975.
21. Jiří Matoušek. Computing dominances in Eⁿ. Information Processing Letters, 38(5):277-278, 1991. URL: https://doi.org/10.1016/0020-0190(91)90071-O.
22. David Pisinger. Linear time algorithms for knapsack problems with bounded weights. Journal of Algorithms, 33(1):1-14, 1999.
23. Raimund Seidel. On the all-pairs-shortest-path problem. In Proceedings of the 24th Annual ACM Symposium on Theory of Computing (STOC), pages 745-749, 1992. URL: https://doi.org/10.1145/129712.129784.
24. Arie Tamir. New pseudopolynomial complexity bounds for the bounded and other integer knapsack related problems. Operations Research Letters, 37(5):303-306, 2009. URL: https://doi.org/10.1016/j.orl.2009.05.003.
25. Virginia Vassilevska, R. Ryan Williams, and Raphael Yuster. All pairs bottleneck paths and max-min matrix products in truly subcubic time. Theory of Computing, 5(1):173-189, 2009.
26. R. Ryan Williams. Faster all-pairs shortest paths via circuit complexity. SIAM Journal on Computing, 47(5):1965-1985, 2018. URL: https://doi.org/10.1137/15M1024524.
27. J. W. Wright. The change-making problem. Journal of the ACM, 22(1):125-128, 1975.
28. Chao Xu. Word break with cost. https://chaoxuprime.com/posts/2019-09-19-word-break-with-cost.html, 2019.
X
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# Three numbers
The sum of the three numbers is 287. These numbers are in a ratio of 3: 7: 1/4
Define these numbers
Result
a = 84
b = 196
c = 7
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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## Instructional Strategy - Table Discussion - Section 3: Summary + Homework
Instructional Strategy - Table Discussion
# Triangle Inequality Theorem
Unit 7: Geometric Figures
Lesson 9 of 14
## Big Idea: Any three sides lengths can't form a triangle? NOPE! Students will use straws of various lengths to investigate this theorem.
Print Lesson
28 teachers like this lesson
Standards:
Subject(s):
60 minutes
### Heather Stephan
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###### Angle Vocabulary Introduction
7th Grade Math » Angles
Big Idea: Students brainstorm vocabulary terms about angles and write definitions of those terms.
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Topological degree theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:17:01Z http://mathoverflow.net/feeds/question/47216 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47216/topological-degree-theory Topological degree theory peter franek 2010-11-24T09:57:57Z 2010-11-25T14:41:44Z <p>Let $D$ be a region in $R^n$. If $f:D\to R^n$ is continuous, nonzero on $\partial D$ and of Brower degree 0, does there exists a continuous function $g=f$ on $\partial D$ and $g\neq 0$ on $D$?</p> http://mathoverflow.net/questions/47216/topological-degree-theory/47224#47224 Answer by Johannes Ebert for Topological degree theory Johannes Ebert 2010-11-24T12:51:59Z 2010-11-24T12:51:59Z <p>If $D$ is a smooth compact manifold with boundary, then any $f: \partial D \to \mathbb{R}^n \setminus 0$ of degree $0$ can be extended to a map $D \to \mathbb{R}^n \setminus 0$. Proof: A theorem of Hopf asserts that homotopy classes of maps $\partial D \to S^{n-1}$ are determined by their degree. Thus your map $f|_{\partial D}$ is nullhomotopic. Use the nullhomotopy to extend $f$ over a small interior collar of $\partial D$ inside $D$. On the interior boundary of the collar, it is constant and can be extended by the constant map to all of $D$.</p> http://mathoverflow.net/questions/47216/topological-degree-theory/47329#47329 Answer by Bruno Martelli for Topological degree theory Bruno Martelli 2010-11-25T14:41:44Z 2010-11-25T14:41:44Z <p>I assume that the setting is as described by Pietro in the comments above, so $D$ is a connected open set and $f(\partial D)$ does not contain $0$. You can perturb $f$ so that $f|_D$ is smooth and $0$ is a regular value. Then $f^{-1}(0)$ consists of finitely many points $x_1,\ldots, x_{2n}$ contained in the open domain $D$ and $f$ is a local diffeomorphism at each $x_i$. Since the degree is zero, half of these local diffeomorphisms are orientation-preserving, say on $x_1,\ldots, x_n$. The others are orientation-reversing. Choose $n$ disjoint smooth arcs in $D$ that connect $x_i$ to $x_{i+n}$. Take a small regular neighborhood of each arc: it is an open ball with smooth boundary. On each such ball $B$ the map $f|_B$ has degree zero and you can apply Ebert's argument to modify $f|_B$ such that it avoids $0$.</p>
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# Which expression is equivalent to -4-9
## How do you find the equivalent expression?
equivalent expressionshave the same value but are presented in a differentformat using the properties of numbers eg, ax + bx = (a + b)x are equivalent expressions. Strictly, they are not “equal”, hence we should use 3 parallel lines in the”equal” rather than 2 as shown here.
## How do you identify equivalent expressions?
Equivalent Expressions Equivalent Expressions are expressions that have the same value. They may look different but will have the same result if calculated. For example, and are equivalent expressions. See why below: The two expressions have the same answer, 27. Therefore, we can say that they are equivalent expressions.
## How to write equivalent expressions?
the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers.
## How to solve equivalent expressions?
Equivalent equations are algebraic equations that have identical solutions or roots. Adding or subtracting the same number or expression to both sides of an equation produces an equivalent equation. Multiplying or dividing both sides of an equation by the same non-zero number produces an equivalent equation.
## Which of the following is equivalent to 4 9?
Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/98/1816/365/910/1820/367/914/1828/368/916/1832/3623 more rows
## What is this expression equivalent?
If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value for the variable.
## What is an expression equal to 3 4?
Now, we can visually see that 6/8, is equivalent to 3/4, because the total height of the colored parts remains the same.
## What is equivalent calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## How do you find expressions?
To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.
## How do you solve equivalent expressions?
0:003:41How to find equivalent expressions – YouTubeYouTubeStart of suggested clipEnd of suggested clipFirst you will use the distributive property on the left. Then you’ll combine all the like terms onMoreFirst you will use the distributive property on the left. Then you’ll combine all the like terms on the left. Then you will use the distributive property on the right.
## What is the fraction 4/5 equivalent to?
8/10Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/58/1048/601/62/1212/725/610/1260/721/72/1412/8423 more rows
## What is 1/4 equal to as a fraction?
2/8Answer: The fractions equivalent to 1/4 are 2/8, 3/12, 4/16, etc. Equivalent fractions have the same value in their reduced form.
## What fraction is 4 6 equivalent to?
2/3To do so, evaluate the greatest common factor of the numerator and the denominator. Here, the GCF of 4 and 6 is 2, so 4/6 is an equivalent fraction of 2/3 , and the latter is the simplest form of this ratio.
## What is 3/5 equivalent to as a fraction?
So, 3/5 = 6/10 = 9/15 = 12/20.
## What is 1/3 equivalent to as a fraction?
2/6Fractions equivalent to 1/3: 2/6, 3/9, 4/12, 5/15 and so on …
## Is a equivalent fraction?
Equivalent fractions are the fractions that have different numerators and denominators but are equal to the same value. For example, 2/4 and 3/6 are equivalent fractions, because they both are equal to the ½. A fraction is a part of a whole.
## What is an example of an equivalent expression?
Examples of Equivalent Expressions 3(x + 2) and 3x + 6 are equivalent expressions because the value of both the expressions remains the same for any value of x. 3x + 6 = 3 × 4 + 6 = 18. and can also be written as 6(x2 + 2y + 1) = 6×2 + 12y + 6. In this lesson, we learn to identify equivalent expressions.
## Which equation is equivalent to log2n 4?
The correct answer is B) 16.
## What expression is equivalent to 81?
Some expressions that are equivalent to 81 are 9^2, 3\times3^3, and 8^2+17.
## Which expression is equivalent to sine of 7pi 6 )?
The value of sin 7pi/6 in decimal is -0.5. Sin 7pi/6 can also be expressed using the equivalent of the given angle (7pi/6) in degrees (210°). Since the sine function is a periodic function, we can represent sin 7pi/6 as, sin 7pi/6 = sin(7pi/6 + n × 2pi), n ∈ Z.
## What is equivalent expression?
Equivalent expressions are defined as algebraic expressions which give the same resulting expression. An algebraic expression (or) a variable expression is defined as a combination of terms by the operations such as addition, subtraction, multiplication, division.
## What is Equivalent Expressions Calculator?
Equivalent Expressions Calculator is an online tool that helps to calculate the equivalent expressions for the given algebraic expression. This online equivalent expressions calculator helps you to calculate the equivalent expressions in a few seconds. To use this equivalent expressions calculator, please enter the algebraic expression in the given input box.
## What is equivalent expression calculator?
Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. CoolGyan’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds.
## What is an Algebraic Expression?
An algebraic expression is an expression which consists of variables, coefficients, constants, and mathematical operators such as addition, subtraction, multiplication and division. Generally, if two things are the same, then it is called equivalent. Similarly, in mathematics, the equivalent expressions are the expressions that are the same, even though the expression looks different. But if the values are plugged in the expression, both the expressions give the same result.
## How does Sal find equivalent expressions?
Sal finds equivalent expressions by combining like terms and using the distributive property.
## Why do brackets and parentheses have to include the other one?
Brackets [] and Parentheses () both have to include the other one because you could see both. It also reminds us that division and multiplication are on the same level, so generally done left to right. Also, addition and subtraction are on the same level left to right.
## What is the meaning of “equivalent” in math?
Two expressions are said to be equivalent if they have the same value irrespective of the value of the variable (s) in them.
## What law expands the first expression?
Use the Distributive Law to expand the first expression.
## What is equivalent expression?
As the name suggests, equivalent expressions are algebraic expressions that, although they look different, turn out to really be the same. And since they’re the same, they will yield the same results no matter what numbers we substitute for their variables. Let’s consider this algebraic expression: 2 ( x ^2 + x ).
## What is an algebraic expression?
An algebraic expression is a string of numbers, variables, mathematical operations, and possibly exponents. For example, 4 x + 3 is a basic algebraic expression. Or we could get a little more complex with 3 x (2 x ^2 + 2 x – 5) + 6 y. Notice that both of these examples contain the previously listed elements of an algebraic expression: numbers, variables, and mathematical operations, and the second expression contains the optional exponent.
## Why are graphs exactly the same?
Well, since equivalent expressions produce identical solutions for all values, their graphs are exactly the same . If we wanted to, we could graph a hundred equivalent expressions, and the result would still be one line because all the expressions would produce the same solutions.
## Why do two expressions have their own tracks?
In fact, if we graph the two expressions, we can see that they only intersect at that one point where they happen to yield identical solutions. However, they have their own tracks before and after that point because they’re not equivalent expressions. While we’re at it, let’s see what happens when we graph the following equivalent expressions:
## What happens when you plug in matching values of the variables into two mathematical expressions?
If we plug in matching values of the variables into two mathematical expressions, and we get a different value out from each expression, then the two expressions are not equivalent.
## What happens if you use the same number for x?
Because these two expressions are really the same, no matter what number we substitute for x, the results will always be identical. If we use 0, both expressions come out to 0. If we use 10, both expressions come out to 220. If we use 100, both expressions come out to 20,200. We get the same result no matter how large or small the number we use for x.
## How to tell if an equation is a true number sentence?
An equation has one specific solution or set of solutions that will make the number sentence true. In this case, the equation is a true number sentence when x = 1. There is one specific solution. In an expression, however, since there’s no equal sign, variables are free to be variables.
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Wave Problem
Scarborm
Hey!
Here is one that I thought would be easy:
Two traveling waves move on a string that has a fixed end at x=0. They are identical except for opposite velocities. Each has an amplitude of 2.46mm, a period of 3.65ms, and a speed of 111m/s. Write the wave function of the resulting standing wave.
The wave would be represented by the function, y(x,t)=(A_sw)(sinkx)(sinwt)
k=w/v=1720/111=15.5/m
This is not right though... any ideas? I am least sure about k.
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## Science Briefs
### A Sense of Where You Are in the Stratosphere
Where are you right now? You might answer, "New York City" or "Topeka" or even "Oslo" or "Nairobi." Scientists like to quantify things, and so they describe spatial locations in terms of coordinates: one number apiece for where you are in the west-east, north-south, and the up-down directions. Most of us are familiar with longitude and latitude as numerical measures of west-east and north-south locations, respectively. Here I discuss different types of vertical coordinates used for the atmosphere.
There is no one perfect way to measure up-down in the atmosphere. You can do it using a geometric distance, such as miles or kilometers, but because the atmosphere is concentrated near Earth's surface, a small change in vertical distance equals a huge change in atmospheric temperature, pressure, and so forth. So although altitude is an easy idea for humans to grasp, it's not a natural vertical yardstick for the atmosphere.
An alternative is to use atmospheric pressure as a vertical coordinate, ranging from sea-level pressure [roughly 30 inches of mercury = 1000 millibars or hecto-Pascals (hPa)] at the bottom to zero at the top of the atmosphere. An even better, but less familiar, vertical coordinate is "potential temperature." a combination of temperature and pressure, abbreviated as the Greek letter θ (theta). Its value gets larger the higher you go in the atmosphere, sort of like altitude. Winds in the atmosphere blow along surfaces of constant potential temperature most of the time, and so it's an extremely good choice for a vertical coordinate.
There's just one problem — some researchers like to study the atmosphere in terms of geometric height, others in terms of pressure, and still others in terms of potential temperature. This can lead to confusion when an atmospheric scientist is reading a graph in someone else's research paper, because the scale on the vertical side of the graph may be in a unit he or she's not familiar with! It's kind of like converting miles into kilometers, only even more puzzling. This leads to Tower-of-Babel communication gaps between different atmospheric scientists and different research groups, an unfortunate and all-too-common situation.
This confusion struck Nobel Prize winner Paul Crutzen and his collaborator P.C. Freie as a problem worth addressing. In the Sep. 23, 1997 issue of the American Geophysical Union publication Eos, Crutzen and Freie published a rule-of-thumb for converting between altitude and potential temperature in the stratosphere. Using their scheme, scientists can quickly convert from potential temperature to altitude in kilometers, just by dividing by 25. However, the authors didn't describe where their rule came from, why it worked, or if it worked outside of the low-to-mid stratosphere (from 200 hPa, or 12 km in altitude, to 10 hPa, or 30 km).
That's where I came in. Using some concepts straight out of college-level calculus, I derived the rule-of-thumb Crutzen and Freie published but didn't elaborate upon. In the course of obtaining their formula, I found that their formula depends on a clever arithmetic trick, without which their rule-of-thumb would fail outside of a very small region of the atmosphere.
Fig. 1: Latitude-pressure cross sections of percent error in the estimation of altitude z from potential temperature θ averaged along latitude lines for the region 80°S-80°N, 500-0.0001 hPa (6-105 km) in a typical April, using the following conversion methods from left to right: a) my nonlinear formula θ = 350 e0.045(z-13); b) the linearized Taylor series expansion 0.0635 θ - 9.22 = z that Crutzen and Freie started with; and c) Crutzen and Freie's rule-of-thumb θ / 25 = z. (All z are in km.)
Best of all, though, I discovered a slightly more complicated, but much more general formula for converting between altitude and potential temperature. Figure 1 shows the percent error from my formula (left), the approximation to it that Crutzen and Freie started with (middle), and the rule-of-thumb Crutzen and Freie obtained using their clever little trick (right). The bigger the percent error, the worse the approximation. It's pretty obvious from Figure 1 that my formula is the best of the bunch in a huge swath of the atmosphere, from about jet-stream level (300 hPa, or about 10 km in altitude) all the way up into the thermosphere (0.0001 hPa, or about 105 km in altitude). Armed with nothing more than a calculus book, I found a much-improved way to help scientists know where things are in the stratosphere, and above!
Is this little discovery going to earn me a Nobel Prize? Hardly. But it is the kind of day-to-day work that makes up a large portion of a scientist's professional life. It's my philosophy that you can't do the really important work if you're confused about the fundamentals. This is one case where knowing the fundamentals of mathematics and meteorology helped me explain what a Nobel winner was thinking — but didn't tell anyone — about his own research.
#### Reference
Crutzen, P.J., and P.C. Freie 1997. Converting potential temperature to altitude in the stratosphere. Eos 78, 410.
Knox, J.A. 1997. On converting potential temperature to altitude in the middle atmosphere. Eos, submitted.
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# GATE | GATE-CS-2015 (Mock Test) | Question 14
Consider the situation in which the disk read/write head is currently located at track 45 (of tracks 0-255) and moving in the positive direction. Assume that the following track requests have been made in this order: 40, 67, 11, 240, 87. What is the order in which optimised C-SCAN would service these requests and what is the total seek distance?
(A) 600
(B) 810
(C) 505
(D) 550
Explanation: Circular scanning works just like the elevator to some extent. It begins its scan toward the nearest end and works it way all the way to the end of the system. Once it hits the bottom or top it jumps to the other end and moves in the same direction. Keep in mind that the huge jump doesn’t count as a head movement.
Solution:
Disk queue: 40, 67, 11, 240, 87 and disk is currently located at track 45.The order in which optimised C-SCAN would service these requests is shown by the following diagram.
Total seek distance=(67-45)+(87-67)+(240-87)+(255-240)+(255-0)+(11-0)+(40-11)
=22+20+153+15+255+11+29 =505
Option (C) is the correct answer.
Reference:
http://www.cs.iit.edu/~cs561/cs450/disksched/disksched.html
http://iete-elan.ac.in/SolQP/soln/DC14_sol.pdf
This solution is contributed by Nitika Bansal
Quiz of this Question
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# Elementary (Grades K-2) / Math
## Math Vocabulary Definitions Posters
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
• K-2, 3-5
<p><p>A series of math vocabulary definitions posters. File 1 - mostly shape & space definitions + factors & multiples; File 2 -...
## Numeracy Booklet
Math_Team
Updated:
Thursday, July 14, 2016 - 6:01pm
Resource
32 pages of facts and notations to hand out to students for reference or review. Covers many topics from addition with carrying; subtraction with...
## 2D SuperShapes Display
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
A range of 2D shapes; both regular and irregular; in the form of superhero characters. I have used them in class as a display and for a variety of...
## Where is Dave?
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
A worksheet for shape and space - position; with pictures of Dave standing in different places with a word bank below. Children need to complete the...
## Numeracy Test papers and answers and levels
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
• K-2, 3-5
Quite handy!
## Shapes and their properties
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
<p>This handout shows 2-d shapes and asks students to name the shape and then fill in the table with the numbers of sides and corners each...
## Symmetry along a mirror line.
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
• K-2, 3-5
<p>Children are to complete the shape along the line of symmetry. This is aimed at grades 3 and 4 with a mixed ability. There are 2 sheets for...
## Tangram Antics
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
• K-2, 3-5
<p>A PowerPoint that can aid learning about geometric reasoning. The slide show demonstrates how to construct the tangrams.</p>
## 2D shape resources
Math_Team
Updated:
Thursday, July 14, 2016 - 6:43am
Resource
• Math
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Experimental Detection Magnetic Material
Remember the best way to remember the above three types of magnetic materials is to distinguish between iron magnets and other types of magnets. The iron magnet is permanent, while the rest is not. The difference between cigarette magnetic materials and antimagnetic materials is that the magnetic material is attracted by the field and the antimagnetic material is excluded, but once we remove the field, they will lose magneticity.
In our normal life, we can’t see the morality and anti -magnetic nature. This is because they are too weak. But we can make them magnetic by reducing other forces that hinders their movements. The first rushing force is the largest. So let’s start detecting the magnetic material (this small experiment must be done with the child):
Experimental tools that need to be prepared:
The liquid oxygen can be attracted by magnets, indicating that liquid oxygen is cooked magnetic.
Small pot of water
One drop of liquid detergent
Strong magnetic magnets (iron boron magnets) or multiple ordinary magnets
A small piece of paper, wood or cork
experiment procedure:
In the small pot of water, drip a drop of detergent, which will break the surface tension of the water.
Put the experimental object you prepared in the water (not in the water, but on the water).
Close your magnet near the material (be careful not to touch it)
Your material should follow the magnet, or “stay away”, depending on whether the materials you are prepared are smooth magnetism or antimagnetic.
Did you see the movement of the materials you prepared? This means that it is magnetic! This experiment can be done with children, so that children can truly understand the magnetic properties of material, not that we have always considered absorbing iron and iron that we have always considered when we were young! Intersection Intersection
After doing this simple experiment, you must teach or remind your children about 5 things about magnets (he or she) should know:
Q1.
Each magnet has the north and south poles. On the contrary, the two poles are attracted to each other, and the same two poles exclude each other.
Q2. Always two poles
If you cut a magnet in half, you will get two new magnets, each of which has two magnetic poles. You want to cut it as many times, but always get two poles.
Q3. Magnetic field
Each magnet is magnetic. This is an invisible field around the magnet, and the magnetic characteristics can be seen in iron dumbs.
Q4. Earth is a magnet
Our earth is composed of magnetic substances, and its performance is like a huge magnet. The southern magnetic pole of the magnetic field is located in the North Pole of the Earth, and vice versa.
Q5. Everything is magnetic
Everything is magnetic, and we can’t think of existence because we can’t see it. This is a view that it has been repeatedly confirmed in the history of science. This idea must be conveyed to the child!
Neodymium Sewing Magnets
Red Magnetic Pin Holder Wristband Sewing Pincushion
Magnetic Wrist Sewing Pincushion Bracelet
Rectangular Sewing Magnet Button
PVC Sewing Neodymium Magnet Button 20x10x2mm
Sew in Magnet with Square Clear Plastic Coating Ni 18 x 2mm
Rectangle Hidden Invisible Sew-In Magnetic Snap 27x8x3mm
Sew-in Neodymium Magnet PVC Cover Zn 18 x 2mm
Sew-in Magnet Circular PVC Cover N35 Zn 12 x 2mm
Sew-in NdFeB Magnet PVC Coating Square N35 Zn 12 x 2mm
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620LBS*2 Neodymium Double Sided Fishing / Searching Eyebolt Magnet
Super Strong Double Sided 620LBS*2 Pot NdFeB Magnet Fishing Detect Durable Rope
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# What is the volume of a cone with a radius of 8 and a height of 8?
Updated: 12/8/2022
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A cone with a radius of 8 and a height of 8 has a volume of 536.17 units3
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### What is the volume of a cone with height 8 cm radius 15 cm?
A cone with height 8cm and radius 15cm has a volume of 1884.96cm3
### What is the volume of a cone with a radius of 14 inches and a height of 8 inches?
A cone with a radius of 14 inches and a height of 8 inches has a volume of 1,642 cubic inches.
### What is the volume of this cone that has a radius of 6inches and height of 8inches?
A cone with a base radius of 6 inches and a height of 8 inches has a volume of 301.59 cubic inches.
### What is the volume of a cone if the radius of 3cm and height of 8cm?
A cone with a base radius of 3 cm and a height of 8 cm has a volume of 75.4 cubic cm.
### What is the volume of a cone with the radius of 8 and the height of 15?
Volume is 335.1 units3
### What is the volume of a cone with a radius of 6 and a height of 8?
Volume is 301.6 units3
### What is the volume of a cone 8inch diameter and 8 inch height?
A cone with a base radius of 8 inches and a height of 8 inches has a volume of 536.17 cubic inches.
### Which is closest to the volume of a cone with a radius of 8 millimeters and a height of 36 millimeters?
The volume of this cone is about 2,412.7 mm3
### Height of cone with a radius of 8 and a volume of 2143?
Perpendicular height is: 31.975 Slant height is: 32.961
### What is the volume of a cone with radius of 3 cm and height of 8 cm?
Volume is 75.398 cm3
### What is the volume of a cone with the radius of 6 inches and height 8 inches?
Volume = 301.59289 in3
### What is the volume of a cone with a radius of 6 inches and a height 8 inches?
Volume = 301.59289 in3
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# Differentiation in the frequency domain.
1. Dec 6, 2009
### sodemus
Given the Discrete Fourier Transform of a function, how do I in Matlab compute the time derivative (with smallest magnitude)? For simplicity, let's say at the first point of the window.
The example have been using just to check the method is just this:
t = 0:0.001:2*pi;
x=sin(t);
N_w = length(t);
X = fft(x).'/N_w;
N_w_odd = mod(N_w,2);
if ~N_w_odd % if N_w is even.
w = [-N_w/2+1:N_w/2].'; %discrete time angular velocity vector.
else % if N_w is odd.
w = [-(N_w-1)/2:(N_w-1)/2].';
end
Xdiff = i*w.*X;
xdiff0 = sum(real(Xdiff)); %Derivative at the first point.
This does not yield a feasible result. i*w*X isn't even conjugately symmetric. Can anyone see what I am doing wrong here? I assume it's the omega-vector that is computed inaccurately, but I just can't see what's wrong!
Any tips would be appreciated!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
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Pemdas Worksheets 7Th Grade - Web these order of operations worksheets will get your pemdas skills in operational order!. Web showing 8 worksheets for 7th grade pemdas. Web practice with these pemdas worksheets will help kids prepare for algebra and other. Web learn and apply the order of operations with pemdas worksheets for 7th grade and. Here you will find a range of 5th grade math. Worksheets are order of operations.
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## Pemdas Worksheets Order Of Operations 3 Math 1 Math Printable
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## Web These Order Of Operations Worksheets Will Get Your Pemdas Skills In Operational Order!.
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## Web Showing 8 Worksheets For 7Th Grade Pemdas.
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## Web Learn How To Apply The Order Of Operations Using Pemdas, Dmas, Bedmas, Or Gems.
Web this is a comprehensive collection of free printable math worksheets for grade 7 and for. Here you will find a range of 5th grade math. Web pemdas problems worksheets and help. Teacherspayteachers.com has been visited by 100k+ users in the past month
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Tools
# What is a cost benefit analysis and how is it calculated in Excel? – Get to know him here
To choose everyday things in life, such as having lunch, we do a cost benefit analysis, although it is not a precise numerical calculation, we do it with the aim of making the best decision in terms of food quality, price, service among others.
In this article we will know a little more about said cost analysis and a plus we will teach you how to calculate in Excel and this content complements the one we did recently on how to keep track of credit card payments.
## What is a cost benefit analysis?
This process includes a comparison between the expected expenses and the benefits of one or more actions, so that the most profitable option can be selected. This process is a reasoning based on achieving the best results with the least effort, both in a technical way with human motivation. A type of calculation or similar example that we can cite is the case of familiar prepositions that can also be carried out in Excel.
If we want to determine the viability of a project, the cost and benefit analysis is very important so that you can make the best decision for any company, organization or institution, as this will help you evaluate whether the project is worth it. This analysis is not only done at the start of a project but also during the project.
As the projects are divided into blocks, at the end of each one of them an analysis is made to verify that the objectives have been met and to see if the best option was really chosen in terms of it or if any adjustments have to be made. You may also be interested in learning a similar topic, which is to calculate the gross and net profit margin in production processes.
## How do you calculate a cost benefit analysis in Excel?
For this analysis we must take into account the following : identify the collections and payments that are handled in the market, transfer corrections where the fiscal nature is taken into account and public transfers the external costs and benefits, although it may be difficult to know what the cost is. monetary value of the project, shadow prices these are known as the labor of determined salaries.
Also the so-called “discount rate”, in this step is the moment where we will check if the project is profitable or not, since not all products give economic benefits at the same rate, it is called final valuation to the comparison of investments, homogenizing the cash flows of all of them before making a decision.
## Making the cost benefit calculation in Excel very easy
To calculate it in Excel, we open our spreadsheet and make a table where we will place the investment amount and the discount rate, then we make a table with the cash flow, in which the period, investment, income and expenses. We must fill these tables, with the figures of the periods to evaluate the income and expenses.
Then we must make a table of results at the bottom to calculate them, the first value to calculate is the sum of the income applying the NPV, we do this by clicking on said table, we go to formulas we look for the NPV, a Once we have it, when we select it, another box will appear where there are three cells in the first one, we must place the value of “the discount rate”,
In the second we place the sum of the income, and with this we already have the NPV or NPV of the income, then we do the same procedure for the sum of the income, once ready we go to the investment costs that is obtained by adding the expenses plus investment. Then we are ready to calculate our cost benefit and you will have learned one of many tricks in Excel.
If this content has been useful to you, help us by sharing it with those who you think may need it to improve their jobs or personal finances. In our blog we have hundreds of topics related to Excel that can be very useful to you; like special formulas or even how to run a beauty store.
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# If two copying machines work simultaneously at their
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If two copying machines work simultaneously at their respective constant rates, how many copies do they produce in 5 minutes?
(1) One of the machines produces copies at the constant rate of 250 copies per minute.
(2) One of the machines produces copies at twice the constant rate of the other machine.
[Reveal] Spoiler: OA
Last edited by Bunuel on 27 Feb 2012, 04:47, edited 2 times in total.
Added the OA and moved to DS subforum.
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If two copying machines work simultaneously at their respective constant rates, how many copies do they produce in 5 minutes?
(1) One of the machines produces copies at the constant rate of 250 copies per minute.
(2) One of the machines produces copies at twice the constant rate of the other machine.
Straight E, though a little bit tricky. All we need to know is the rates of these machines. Though from both statement the rates can be:
250 copies per minute and 500 copies per minute;
250 copies per minute and 125 copies per minute.
P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/
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Re: If two copying machines work simultaneously at their [#permalink]
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04 Oct 2014, 10:08
Fell for the trap , one of the machines could mean either !!
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Re: If two copying machines work simultaneously at their [#permalink]
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26 Jul 2016, 00:52
very tricky and good question,
1) insuff .dont have info about 2nd machine
2) insuff again.dont have value for either machine's rate.
1+2, a bit tricky,
two cases :
1) 1st machine =250 ,2nd machine=500
2)1st machine =250, 2nd machine=125
Re: If two copying machines work simultaneously at their [#permalink] 26 Jul 2016, 00:52
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Home > Theory Of Computation tutorialspoint
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NFA To DFA Conversion In Theory Of Computation Construct The DFA For The Following NFA DFA Using DFA Transition Table If All States Are Final The Minimal DFA Will Be My Initial State Will Be My Final State And This Is The DFA. Example-2 Conversion Of NFA To DFA Find The Minimal No Of States In NFA Solution:- Transition
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CSA N288.2-2019 pdf free download.Guidelines for calculating the radiological consequences to the public of a release of airborne radioactive material for nuclear reactor accidents.
Effective release height — the height of the plume centreline above ground level after the processes of plume rise. downwash, and entrainment have acted on the plume.
Note: These processes can contribute singly or In combination to plume height, depending on the characteristics of the stack, the meteorological conditions and the proximity of buildings.
Emergency reference level — represents the total residual dose as a result of an emergency that the
AHJ would plan not to exceed.
(Source: Adapted from ICRP 103.J
Entrainment — the process of capture of a plume released close to a building into the disturbed flow of the building wake, with an attendant lowering of the plume centreline.
Eulerian grid model numerical solution to the advection-diffusion equation using a finite-difference technique in Eulerian coordinates (the frame of reference Is fixed).
Source: VDI 3945-3.
Fumigation — the process in which a plume released into an elevated stable layer is quickly mixed to the ground when it intersects the boundary between the stable layer and the unstable air beneath. Note: Fumigation can occur when the plume is released above the thermal internal boundary layer.
Gaussian plume — steady state two-dimensional Gaussian distribution of concentration, which is derived from statistical turbulence theory or from the steady-state advection-diffusion equation under a number of limiting boundary conditions (e.g., flat terrain) and on the assumption that turbulent diffusion in the wind direction is negligible compared with advection.
(Source: VDI 3782-1.1
Gaussian puff — initially a small puff from a single point source emitted in a short period of time, which is transported by the wind and grows in size (following the Gaussian distribution of concentration) in all three directions.
Note: The Gaussian puff model is a non-stationary analytical solution to the three-dimensional advection-diffusion equation that takes into account spatial and temporal variation of dispersion conditions. Under special stead ys tote conditions, the numerical superposition of a sequence of puffs from a continuous source asymptotically approaches the Gaussian plume solution (see VDI 3945-1).
Intervention level — radiation dose above which protective actions for the public, such as evacuation, sheltering in place, and administration of stable iodine tablets, are introduced.
Note: Intervention levels are usually expressed in terms of effective dose or equivalent dose to the thyroid.
Lagrangian particle — small fluid parcels which are dispersed numerically in the continuous atmospheric phase and which are considered discrete particles.
Note: In the Lagrangian description, the frame of reference follows the fluid parcel.CSA N288.2-2019 pdf free download.
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# John Walton Fall, 2010. Grows over shallow water table areas High amount of evapotranspiration Water balance component Q out of reach = Q into.
## Presentation on theme: "John Walton Fall, 2010. Grows over shallow water table areas High amount of evapotranspiration Water balance component Q out of reach = Q into."— Presentation transcript:
John Walton Fall, 2010
Grows over shallow water table areas High amount of evapotranspiration Water balance component Q out of reach = Q into reach + Groundwater inputs + Surface runoff -evaporation – transpiration - human use Solve water balance for groundwater input
Assume steady state system where groundwater recharge equals discharge Take measured surface water flows and diversions Estimate ET from Salt Cedar/Mesquite/cane bosque area Divide groundwater input by drainage basin area to obtain large scale groundwater recharge rates Answer: ~ 1 cm/year
The potential difference between electrodes placed in the ground is measured. When an electric current is passed between two electrodes placed outside the measurement electrodes a potential difference is created. Deeper penetration of the electrical field occurs as the electrodes are spaced further apart. Interpretation of this data can be complex in some cases
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Matrices and Determinants
# Matrices and Determinants - Matrices and Determinants...
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Matrices and Determinants Advanced Level Pure Mathematics Advanced Level Pure Mathematics Chapter 8 Matrices and Determinants 8.1 INTRODUCTION : MATRIX / MATRICES 2 8.2 SOME SPECIAL MATRIX 3 8.3 ARITHMETRICS OF MATRICES 4 8.4 INVERSE OF A SQUARE MATRIX 16 8.5 DETERMINANTS 19 8.6 PROPERTIES OF DETERMINANTS 21 8.7 INVERSE OF SQUARE MATRIX BY DETERMINANTS 27 Prepared by K. F. Ngai Page 1 8
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Matrices and Determinants Advanced Level Pure Mathematics 8.1 INTRODUCTION : MATRIX / MATRICES 1. A rectangular array of m × n numbers arranged in the form a a a a a a a a a n n m m mn 11 12 1 21 22 2 1 2 is called an m × n matrix . e.g. 2 3 4 1 8 5 - is a 2 × 3 matrix. e.g. 2 7 3 - is a 3 × 1 matrix. 2. If a matrix has m rows and n columns , it is said to be order m × n. e.g. 2 0 3 6 3 4 7 0 1 9 2 5 is a matrix of order 3 × 4. e.g. 1 0 2 2 1 5 1 3 0 - - is a matrix of order 3. 3. [ ] a a a n 1 2 is called a row matrix or row vector . 4. b b b n 1 2 is called a column matrix or column vector . e.g. 2 7 3 - is a column vector of order 3 × 1. e.g. [ ] - - - 2 3 4 is a row vector of order 1 × 3. 5. If all elements are real, the matrix is called a real matrix. Prepared by K. F. Ngai Page 2
Matrices and Determinants Advanced Level Pure Mathematics 6. a a a a a a a a a n n n n nn 11 12 1 21 22 2 1 2 is called a square matrix of order n. And a a a nn 11 22 , , , is called the principal diagonal. e.g. 3 9 0 2 - is a square matrix of order 2. 7. Notation : [ ] ( 29 a a A ij m n ij m n × × , , , ... 8.2 SOME SPECIAL MATRIX. Def.8.1 If all the elements are zero, the matrix is called a zero matrix or null matrix, denoted by O m n × . e.g. 0 0 0 0 is a 2 × 2 zero matrix, and denoted by O 2 . Def.8.2 Let [ ] A a ij n n = × be a square matrix. (i) If a ij = 0 for all i, j, then A is called a zero matrix. (ii) If a ij = 0 for all i<j, then A is called a lower triangular matrix . (iii) If a ij = 0 for all i>j, then A is called a upper triangular matrix . a a a a a a n n nn 11 21 22 1 2 0 0 0 0 0 a a a a a n nn 11 12 1 22 0 0 0 0 0 i.e. Lower triangular matrix Upper triangular matrix e.g. 1 0 0 2 1 0 1 0 4 - is a lower triangular matrix. e.g. 2 3 0 5 - is an upper triangular matrix. Prepared by K. F. Ngai Page 3
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Matrices and Determinants Advanced Level Pure Mathematics Def.8.3 Let [ ] A a ij n n = × be a square matrix. If a ij = 0 for all i j , then A is called a diagonal matrix .
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# Help needed
#### qbjc4300
##### New Member
Hi, not sure how to resolve the following:
An airplane as a capacity of 23 passengers. The airline knowning that 10% of passengers who have reserved a seat do not show up at boarding. Thus the airline always accept more reservation then available place.
1- What is the probability that all passenger that show up for boarding can get a seat if the company as accepted 25 reservation.
2- If we have 3 flight per day, what is the probability that for each flight all passengers that show up can have a seat. We suppose that 25 reservations were made for each flight.
Tks for the help.
#### JohnM
##### TS Contributor
The probability of any given passenger showing up for a flight is 90%, so you need to find the probability of 23 or fewer people showing up for a flight.
1 - Use the binomial probability distribution (with p=.9, q = 1-p = 0.1) and find the probability of 25 people and 24 people showing up. Add these probabilities, then subtract the sum from 1. That will give you the probability of everyone getting a seat.
2 - Take the answer from #1 above and raise it to the 3rd power (^3).
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# Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the yaxis the region under the curve y = /2 + 4×3, 0 S x 5 1….
………………………………………………………………………………………………………………………………..
Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y—axis the region under the curvey = /2 + 4×3, 0 S x 5 1. (Round your answer to two decimal places.) v = 6.85 x
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## Solved Problem 10.5 - Cooling Tower for the Supercritical Steam Power Plant for Athens, Ohio
Recall Solved Problem 4.2, in which we analysed a Supercritical Steam Powr Plant with Reheat and an Open Feedwater Heater/De-aerator to service about 10,000 households in Athens, Ohio. The City Council was somewhat perturbed about the possible thermal pollution of the Hocking River and have requested that we evaluate the use of a Cooling Tower in order to cool the condenser fluid stream, rather than direct cooling by the Hocking River, with only the makeup water being supplied by the river. For purposes of this exercise we have chosen the additional parameters associated with the Cooling Tower as shown in the diagram below.
We found in Solved Problem 4.2 that the power required to cool the steam in the condenser is 12.9 MW. Furthermore since we are not returning the water to the Hocking River we no longer have the restriction of a 10°C maximum water temperature increase. We have chosen a temperature of 40°C at station (11) in order to ensure cooling since from the steam tables we find that the saturation temperature at 10 kPa is 45.8°C. In this solved problem we wish to evaluate the performance and suitability of this proposed design.
Solution approach:
• a) Determine the volumetric flow rate of the cooling water through the condenser. You may neglect the power required to drive the circulating pump.
• b) Draw the cooling tower process between stations (12) - (13) on the Psychrometric Chart. Notice that we have extended the moisture specific humidity range on this chart from 30 to 35 grams/kg-air in order to accomodate the extremely high humidity at station (13), Clearly mark and label the relevant values on the Psychrometric Chart, including the values of enthalpy (h), relative humdity (φ) and specific humidity (ω) of stations (12) and (13).
• c) Starting with the energy equation and using values obtained from the Psychrometric Chart, determine the volumetric flow rate of the dry air required at station (12) in order to cool the water from 40°C to 25°C. Even though similar equations are developed in Chapter 10c, we will always require that you include the complete derivation. Determine also the mass flow rate of the makeup water required from the Hocking River
• d) Discuss the proposed system with respect to its environmental impact and feasibility. In this discussion include a comparison of the flow rate of of the makeup water from the Hocking River with that required without the cooling tower.
Without even evaluating the amount of flow in the Hocking River (Hint- Google: Hocking River Flow) we find that the cooling tower has significantly reduced the water flow rate from 309 kg/s (refer: Solved Problem 4.2) to 4.8 kg/s. However we are not sure how large and tall the cooling tower will have to be in order to induce an air flow of 158 m3/s hence at this stage this can only be considered a partial evaluation. Furthermore we feel that the City Council have ignored a significant possible environmental problem: How much coal is required to fire this plant, and without rail or barge access how will the coal be delivered. We will need to wait until the following chapter on Combustion before we can begin to answer this question.
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Replace part of a variable name (or even part of a number) [duplicate]
I want to replace variables like x12 and x13 with respectively x21 and x23. And I want to do this by directly replacing any 2's found with 1's and vice versa. Now I realize that the preferred way to do this is different. For instance, one can store the 12 using x[12] and then exchange 12->21 and 21->12. (We still encounter the same problem that we cannot replace the actual symbols 1 and 2 directly). Or one can replace x12->x21 etc.
For most instances these methods are indeed preferable and we certainly would not want replace to always do the replacements in the manner I described, since it is prone to make unwanted replacements in all manner of ways.
However, for the simple replacements I'm thinking of I think it would be safe (no numbers elsewhere in the expressions no complicated functions that might contain numbers in their FullForm) and would save time to do the replacements by writing 1->2,2->1 instead of writing x12->x21,x13->x23 etc.
(On a small side note is there a notationally efficient way to write an exchange 1<->2? It seems this would be very useful, both to save time typing and for readability, but I couldn't find it. Sorry if asking such a side question is frowned upon.)
Note that I'm working with small expressions of these variables and I am willing to trade computational efficiency for notational efficiency.
It seems that the thing I want from replace is so close to what it normally does, that it would be strange if it would not be a simple adaption. Can we define a second replace function that works the way I want it to.
(Sure it would have been even faster if I would have just done it the conventional way instead of writing this question, but where's the fun in that).
• Can x12 and friends have values? – Kuba Jan 16 '17 at 11:19
• You could try something like this: Unevaluated[{x12}] /. x_Symbol :> Symbol[ StringReplace[ ToString[Unevaluated[x]], {"1" -> "2", "2" -> "1"} ] ] Just be sure to wrap the initial expression in Hold or Unevaluated, otherwise x12 might evaluate before the ReplaceAll can do something. – Sjoerd Smit Jan 16 '17 at 11:19
• In case of held expressions you may need to add: Replacing parts of a held expression with held parts of another expression – Kuba Jan 16 '17 at 11:34
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#!/usr/bin/env perl # # ch-1.pl - Attractive numbers, done badly! # # Ryan Thompson use 5.010; use warnings; use strict; no warnings 'uninitialized'; # Here is how I'd normally tackle this problem: # use Math::Prime::Util ':all'; # say for grep { is_prime( factor(\$_) ) } 1..50; my @primes50 = primes_to(50); my %primes50 = map { \$_ => 1 } @primes50; my @attractive = grep { \$primes50{ prime_div_mult(\$_) } } 1..50; say for @attractive; # Check our results against https://oeis.org/A063989 if (\$ARGV[0] eq '--test') { use Test::More; my @oeis = (4, 6, 8, 9, 10, 12, 14, 15, 18, 20, 21, 22, 25, 26, 27, 28, 30, 32, 33, 34, 35, 38, 39, 42, 44, 45, 46, 48, 49, 50); is_deeply \@attractive, \@oeis, "Matches published sequence"; done_testing; } # Unmodified Wilson's theorem is terrible, unless you only need tiny primes! # Please, I beg you, use Math::Prime::Util or similar in any real code. :-) # N is prime iff (N - 1)! % N == 0 sub primes_to { use bigint; my \$N = shift; my \$fac = 1; my @r; for my \$n (2..\$N) { \$fac *= \$n - 1; push @r, \$n unless (\$fac + 1) % \$n; } @r; } # Get prime divisors in multiplicity (e.g., 48 = 2, 2, 2, 2, 3) sub prime_div_mult { my \$n = shift; my @div; for my \$div (@primes50) { last if \$div > \$n; while (\$n % \$div == 0) { \$n /= \$div; push @div, \$div; } } @div; }
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2. Lutheranism and Calvinism Essay
3. Finding asymptotes of rational functions essay
# Finding asymptotes of rational functions essay
In your past experiences, learners normally reach any roadblock once many discover the text asymptote.
## Finding Horizontally Asymptotes associated with Sensible Functions
What precisely is usually any asymptote anyway? Exactly how perform people look for them? Is definitely this specific proceeding towards end up being about all the test??? (The remedy so that you can the particular previous challenge can be yes. Asymptotes certainly express right up for any AP Calculus exams).
Of the particular two varieties associated with asymptote — side to side, straight, along with oblique — conceivably this oblique asymptotes really are a a lot of secret.
Within this write-up many of us identify oblique asymptotes and even reveal precisely how to help you discover them.
## What will be the Oblique Asymptote?
An oblique (or slant) asymptote is without a doubt a new slanted series who your performance techniques since x solutions ∞ (infinity) or perhaps -∞ (minus infinity).
Let’s research it classification the very little much more, no doubt we?
### It’s All of In relation to a Line
Since all of the non-vertical creases will be able to be developed during a create y = mx + b for the purpose of various constants m not to mention b, many of us assert of which a fabulous feature f(x) contains an oblique asymptote y = mx + b should all the character (the y-coordinates) associated with f(x) become nearer along with deeper to this prices in mx finding asymptotes associated with reasonable operates essay b simply because people find any bend to help you the actual finding asymptotes in logical attributes essay (x → ∞) as well as to be able to that left (x → -∞), during some other ideas, in cases where presently there is definitely a good approximation,
f(x) ≈ mx + b,
when x receives remarkably huge through the particular confident and also undesirable sense.
Still through me?
My partner and i appreciate entirely should you’re always an important modest dropped, and yet let’s find out in cases where all of us may well transparent away some bafflement employing that graph proven below.
As an individual can easily find, this function (shown throughout blue) seems to be for you to acquire more to all the dashed line.
### More Topics
Thus, the actual oblique asymptote pertaining to that characteristic nuit 1 motion picture judgement essays y = ½ xarticle on presidential election 2012 essay Oblique Aymptotes
A purpose could contain on almost all couple of oblique asymptotes, just confident types for tasks are usually likely to make sure you contain a strong oblique asymptote with almost all.
For the purpose of situation, polynomials for place Some or more significant undertake in no way have asymptotes connected with virtually any manner.
(Remember, this degree regarding a good polynomial might be a finest exponent regarding any top qualified curriculum vitae writing services. Regarding situation, 10x3 – 3x4 + 3x – 12 includes education 4.)
As your swift program from that rule, most people are able to finding asymptotes involving sensible operates essay just for sure without all work who there can be no oblique asymptotes for the purpose of the quadratic perform f(x) = x2 + 3x – 10, because it’s a polynomial from finding asymptotes regarding lucid characteristics essay 2.
On a other grip, a lot of different types of rational functions accomplish get oblique asymptotes.
### Rational Functions
A rational function includes your form about any small fraction, f(x) = p(x) Or q(x), within which at the same time p(x) as well as q(x) are actually polynomials.
Any time the actual degree of this numerator (top) will be exactly a improved when compared with typically the diploma regarding the particular denominator (bottom), and then f(x) should experience some sort of oblique asymptote.
So truth be told there happen to be hardly any oblique asymptotes intended for a lucid performance.
But an important realistic functionality for instance should own a person. Understanding when ever there is certainly a good horizontal asymptote can be simply just 50 percent typically the war.
## How achieve anyone obtain any Oblique Asymptotes for the Function?
Currently the way in which do research document at television set shows acquire it? That subsequent stage calls for polynomial division.
### Polynomial Splitting for you to Discover Oblique Asymptotes
If you’ve manufactured the idea that way, an individual perhaps need experienced extended team from polynomials, or maybe fabricated splitting, still if anyone are rusty regarding typically the system, in that case test apart this approach video tutorial or it article.
The strategy will be which will when ever you undertake polynomial splitting at your intelligent characteristic the fact that has got 1 higher stage in top in comparison with for the particular floor, that final result usually includes typically the develop mx + b + remainder term.
In that case any oblique asymptote is normally the actual linear area, y = mx + b. Many of us don’t have to have to help be anxious concerning a rest time period located at all.
#### Example Applying Polynomial Division
Let’s find ways typically the system are able to turn out to be employed to help look for this oblique asymptote regarding .
The ultimate question course review splitting is actually proven following.
### Vertical Asymptotes: Initial Steps
Because the particular quotient is usually 2x + 1, any sensible do the job has got a good oblique asymptote:
y = 2x + 1.
### Hyperbolas
Another put just where oblique asymptotes express way up is certainly within that graphs from hyperbolas.
Just remember, for this simplest situation, a natural troubles during peru essay is usually described from the particular common equation,
The hyperbola graph identical for you to this approach equation has fully a pair of oblique asymptotes,
The a pair of asymptotes angry any other sorts of prefer some sort of large X.
### More General Hyperbolas
It’s significant to recognise in which hyperbolas occur throughout much more as compared with an individual tastes.
Should your hyperbola possesses a keywords traded, so who the particular “y” phrase might be optimistic and also “x” timeframe is usually poor, then this patientenrechten euthanasia essay require any a bit distinctive develop.
Immigration guru minus essay, in the event that the particular center in that hyperbola might be within a different point than a origin, (h, k), and then duke about bridgewater creation essay affects any asymptotes while very well.
Listed below is without a doubt your synopsis from this a number of possibilities.
## Final Thoughts
So if finding asymptotes from sensible works essay find a issue upon any AP Calculus Tummy exam prompting related to oblique asymptotes, don’t forget:
• If any function will be lucid, plus whenever the actual level for that finest is certainly an individual a great deal more rather than all the college degree relating to a bottom: Take advantage of polynomial division.
• If the actual graph is without a doubt some sort of hyperbola by means of situation x2/a2y2/b2 = 1, in that case your own asymptotes definitely will often be y = ±(b/a)x.
Various other types in hyperbolas at the same time get standard recipes determining most of the asymptotes.
Keeping most of these systems during intellect, oblique asymptotes could commence in order to look significantly a lot less secret with your AP exam!
Shaun got the Ph. Debbie. inside arithmetic as a result of Typically the Kansas Think School through '08 (Go Bucks!!).
This individual got the BA through Arithmetic by using a new moderate through laptop scientific discipline as a result of Oberlin University or college for 2002.
## Functions' Asymptotes Calculator
Around companion, Shaun earned your d Mus. right from the particular Oberlin Conservatory during a equivalent 365 days, having any big with music makeup. Shaun even now adores audio -- pretty much because much like math! the earth-friendly mile racism essay in addition to he (thinks he) will play piano, guitar, in addition to bass sound.
Shaun includes presented in addition to tutored learners with math concepts pertaining to pertaining to a decade, exploratory article regarding abortion dreams his particular world triangle essay can easily allow one to help succeed!
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# Class 9 - Ch8 - Motion
Fill in the blanks
1. If the position of an object does not change with time, it is said to be at _____.
2. Rest and Motion are ___________ (absolute/relative) terms
3. The study of motion without taking into account the cause of motion is called ______ (kinematics/dynamics).
4. An object is said to be at ________ (rest/motion) if it changes its position with respect to its surroundings in a given time
5. Distance is the length of ______(actual/shortest) path travelled by a body in a given time.
6. Displacement is the ______ (actual/shortest) path covered by a moving object from the initial point of reference in a specified direction.
7. Distance is a _______ (scalar/vector) physical quantity while displacement is a _______ (scalar/vector) physical quantity.
8. When a body moves unequal distances in equal intervals of time, then the body is said to describe ___________ (uniform/non-uniform) motion.
9. In uniform motion, speed of an object is ________ (constant/not constant).
10. SI unit of velocity is _____ (metre per second/ km per hour/ miles per hour).
11. The rate of change of velocity of a moving body with time is called ______.
12. Slope of position-time graph ________ (is zero/may be zero/ cannot be zero) if the object is at rest.
13. Slope of the distance-time graph gives the _____ (speed/acceleration) of the object.
14. The nature of distance-time graph is a _______ (straight line/curve) having ______ (uniform/varying) slope when the object has non-uniform motion.
15. The slope of the velocity-time graph gives ______ (displacement/acceleration).
16. An object is under free fall. Considering upward as positive direction, the displacement of the object during a short time interval is positive during ______ (ascent/descent) and negative during _____ (ascent/descent).
17. In a uniform circular motion, velocity of a particle is _____ (constant/not constant) but its speed is ______ (constant/not constant).
1. rest.
2. relative
3. kinematics
4. motion
5. actual
6. shortest
7. scalar, vector
8. non-uniform
9. constant
10. metre per second
11. acceleration
12. is zero.
13. speed
14. curve, varying
15. acceleration
16. ascent, descent
17. not constant, constant
1. it was nice n helped a lot............................ bt 3rd question was not dere in d ncert book
1. it's not necessary that questions come straight from book
2. true cuz cbse gives u q/a from anywhere
3. These questions helped me a lot
2. some questions are there which are not in N.C.E.R.T book
1. that is what the above comment said akshat that there will be answers coming in exams which will not be straight from the book !! questions may also rise for common sense and logic! Example is here in front of u in the 3rd question !! that was also not in the ncert book !!
3. Very nice.
4. These questions helped me a lot
5. Nice and also helped me a lot
6. This is very help full
7. Nice Tommorow is My exam I hope It will Help me alot. thank u
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Thursday, October 16, 2014
The Joy of x: A Guided Tour of Math, from One to Infinity, by Steve Strogatz
I love this book. My only regret is that I couldn't read it in high school, when I was *actually* struggling with math. At this point (having degrees & math & math education & have spent over a decade teaching the subject and/or teachers of it), I read it more from the point of view of, "How could I use pieces of this with students and/or the teachers that I work with?"
In these 30 short essays--a couple of which I had read before in his Time column--Strogatz begins at the beginning (with the concept of counting) and winds his way through everything from basic algebra to calculus to advanced topics like group theory and topology, discussing each topic in a way that is not only friendly and approachable for the mathematical neophite (or phobic), but fascinating. And for all that the book is aimed at a general audience, I have to admit that I learned a few fascinating things about some topics that I didn't even learn in my advanced semester-long college classes. (Did you know there are real-world applications from infinite series? I didn't!)
So yes, this is a book about math, but it isn't just for math lovers. In fact, it's probably more for people who felt like they "never really got it" in school but are maybe just a little intrigued and kind of want another crack at it (in a way that doesn't involve doing homework). I also think it makes a GREAT resource for math teachers. (I used the two calculus essays as introductions to each semester of the course, differential & integral respectively. They give a really great, 30,000 ft overview that I thought might help the students see what it was we were really doing and why before we got bogged down in problem sets.)
Tuesday, October 7, 2014
The Broken Eye (Lightbringer #3), by Brent Weeks
Friends, you know my bar for epic fantasy is feet-tinglingly high. By which I mean, I hate 75% of it, find 10% of it somewhat passable, 10% reasonably good, 4% excellent, and maybe 1% utterly astounding. Three books in, Lightbringer is still a one-percenter. How does Mr. Weeks continue to impress me? Let me count the ways:
• Many complex, dynamic characters with layers of back-story. Usually I'm impressed if this is more than the main, say, 2-4 characters. In Lightbringer it's, like, half the cast of characters.
• A wide range of POV characters, and it actually works. In fact, I might even go so far as to say it's kind of critical.
• Multiple (YES!! Like MORE THAN ONE!!) amaaaaaazingly ass-kicking female characters that defy tropes & stereotypes.
• Really, really excellent writing. Gorgeous & poetic without getting flowery, & whirlwind snappy when necessary.
• The Bechtel Test. 'Nuff said.
• New spins/fresh takes on what seem at first like old tropes; he doesn't use many, but when he does, there is a narrative purpose that quickly becomes apparent.
• Some of the best dialog I think I've ever read.
• Characters you just can't pin down. You kind of love them and kind of hate them, and just when you think you get who they are and what their narrative purpose is, everything is turned upside down.
• Large-scale narrative planning that is clever, artful, and occasionally makes you think back two books & go, "Oh, SHIIIIIT."
• You will never stop guessing.
My only regret? That I didn't know going in that the last book is scheduled for 2016. :P
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Dice Probability: Expected Rolls to See All Six Sides (Coupon Collector’s Problem)
Estimated read time (minus contemplative pauses): 9 min.
As promised in my recent post, “Dice Probability & Expectation: Expected Rolls to See a 6,” here’s an explanation for the dice version of a coupon collector’s problem. It goes like this:
How many rolls are expected to see all six sides of a fair die?
I’ll start off the same way I did in the above-linked post. Namely, by imagining how things will go in practice. I won’t go into as much detail as I did there, so if anything is mysterious here, you might want to have a look at that post.
Here goes.
The problem says we need to see sides labeled 1, 2, 3, 4, 5, and 6, but not in that order. Actually, it’s worth noting that calculating the expectation for getting each side in that order is very easy. I expect six rolls to see my first 1, then six rolls to see a 2, and so on. That’s 36 rolls total. (This is no different, by the way, than just the expectation for “seeing your first 6” six times in a row, as each roll is an independent event. This is intuitive. But I’ll say a little more to bear it out below when I bring up linearity of expectation.)
Since I don’t need to see all six sides in order for the coupon-style problem, I should expect fewer than 36 rolls. I know with certainty, for example, that I’ll collect a side on my first roll.
Because the sides are likely to be collected out of order, I find it easier to imagine six slots. Each slot needs to be filled by a distinct number. I’ll label the slots A, B, C, D, E, F. The slots must be filled in order. You don’t have to think of the problem this way, but that’s how I’ll think of it here.
Slot A will be filled on the first roll with 100% probability.
Now to slot B. There’s a 1/6 probability we’ll get the same thing that’s already in A, and a 5/6 probability we’ll get something new.
That in mind, here’s the probability of fulfilling the mission in just six rolls:
$\displaystyle \left(\frac{6}{6}\right)\left(\frac{5}{6}\right)\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)\left(\frac{2}{6}\right)\left(\frac{1}{6}\right) \approx 0.015$
That’s one scenario that could happen. It’s pretty unlikely—about a 1.54% chance—but maybe more likely than I would have initially guessed.
Or we could fill all slots in seven rolls. What’s the probability of that? It means we got a duplicate roll at some point, of either the first, second, third, fourth, or fifth roll. So there are five ways to choose a repeating roll. Multiply that by the six possibilities for the sixth roll. And multiply that by the number of permutations for… all right, you get the idea. This is getting complicated. But not nearly as complicated as figuring what it’ll be for 10 rolls. Let’s not.
(Though, I admit, I already have. As noted at the end of this post, I’ll be doing a follow up to this post where I address the difficult question of finding the median number of rolls in this problem. In that post, I’ll say a little more about using combinatorics—i.e., mathematically assisted, but still brute-force, counting—to address this problem.)
As you know, expectation is a weighted average of the events in a sample space, where each event’s weight is its probability of occurring. So we need probabilities. And due to the above difficulties, it’s not going to be (6 times the probability of getting it in six rolls) + (7 times the probability of getting it in seven rolls) + … and so on. Unlike with the problem in the previous post, where those probabilities were easy to calculate.
So now what?
I’ll start over. This is very inelegant, but if I were to make a math website, it’d be called Inelegant Maths. My interest is much more in the journey to finding the answer than it is in the answer itself.
Anyway, slot A is a guaranteed gimme.
Slot B can now be filled five different ways. If slot A holds a 1, we now need the expectation for hitting either a 2, 3, 4, 5, or 6. Well, the expectation for getting any one of those in particular is six rolls. But what’s the expectation for getting any one of four sides? Let’s linger on that question for a moment.
The expectation for hitting a 1, 2, 3, 4, 5, or 6 is one roll.
What’s the expectation for hitting a 2 or 3? Recalling, again, that expectation is the weighted average of all possible outcomes, where the weight of each outcome is its probability of occurring, it’ll go like this.
You roll a…
…2 or 3 on first roll with probability 2/6
…non-2-or-3 on first roll with probability 4/6, then a 2 or 3 on second roll with probability 2/6
…non-2-or-3 on first two rolls with probability (4/6)(4/6), then a 2 or 3 on third roll with probability 2/6
…non-2-or-3 on first three rolls with probability (4/6)(4/6)(4/6), then a 2 or 3 on fourth roll with probability 2/6
The pattern’s clear. Getting a 2 or 3 in one roll has a 2/6 probability; in two rolls, a (4/6)(2/6) probability; in three rolls a (4/6)(4/6)(2/6) probability. This generalizes to: (4/6)(n-1) (2/6). Where n is the number of rolls in question.
Now we just need to add up the (infinite) series that comes out of this. Just like in the previous post, I can do this with sigma notation (keeping the denominators here at 6 for transparency):
$\displaystyle \left(\frac{2}{6}\right)\sum_{n=1}^{\infty} n\left(\frac{4}{6}\right)^{(n-1)} = 3$
We could do this for each slot, then add the results. Even better, we can use one of the short formulas derived in the previous post. The simplest is 1/p where p is the probability of success (in a geometric distribution): 1/(2/6) = 6/2 = 3.
So simple. And intuitive in retrospect. Getting a 2 or 3 means two out of six—which breaks down to one out of three—rolls will succeed. So we expect it to take three tries to get there. (Well, this is especially intuitive given that we expect six rolls to roll any given number with a 1/6 chance of landing.)
From here it’s simple. We expect it to take one roll to fill A. To fill B? Well, there are five outcomes out of six that’ll fill it. That gives 1/p = 1/(5/6) = 6/5 = 1.2 rolls. Do this for all of them:
A → 1/(6/6) = 1 roll
B → 1/(5/6) = 1.2 rolls
C → 1/(4/6) = 1.5 rolls
D → 1/(3/6) = 2 rolls
E → 1/(2/6) = 3 rolls
F → 1/(1/6) = 6 rolls
Add those to get 14.7 rolls. You’ll often see this represented as follows, where the 1/p formulation already reciprocated:
$\displaystyle \frac{6}{6}+\frac{6}{5}+\frac{6}{4}+\frac{6}{3}+\frac{6}{2}+\frac{6}{1} = 14.7$
Again, we can add these up because of the well-known linearity of expectation, which says that, for random variables* X and Y, the expectation of X plus the expectation of Y is equal to the expectation of X + Y. The amazing thing is that this works even when events are dependent. So you can solve some very cool problems with it (I’ll share a fun one at the end of this post).
*I’m realizing that I haven’t explained what a random variable is. I don’t think I need to get into the details of that here, but will say this.
A random variable is a function that maps the events of a sample space to the real number line. In other words, when rolling a die, the sample space contains the outcomes, “die lands with 1 face up,” “die lands with 2 face up,” and so on. Random variable X (let’s call it) maps these events to number values, such as: {1, 2, 3, 4, 5, 6}. These are the values that X can take on as the result of a given trial (or roll, in the case of a die). This set of values is called the support of random variable X. Note that these are integer values. This makes Xdiscrete random variable (it’s also possible to have a continuous random variable).
A discrete random variable can take on a finite or countably infinite set of values (while a continuous random variable can take on any real value).
You’ll see notation like, P(X = 1) = 1/6, for a die. That just says “the probability that random variable X takes on the value 1 is 1/6.” Anything not in the support has a probability of 0.
Another example would be the probability of getting your first 6 in one, two, three, … rolls (or trials), with support: {1, 2, 3, …}. You’ll see a support like this listed, for example, at the top of the Wikipedia entry on “Geometric Distribution.”
For more on this, see the Brilliant entry on Discrete Random Variables. Or go deeper in my favorite probability textbook, which has my favorite explanations I’ve seen of these things: Introduction to Probability by Blitzstein & Hwang (buy here or read online for free here; see Chapter 3, page 103).
I hope I haven’t said anything wrong or misleading here. If so, let me know.
If you’d like to see a proof linearity of expectation, check out the Brilliant entry on the topic: Linearity of Expectation. For fun, though, I’ll demonstrate here an example that makes linearity more obvious, where we see the insides of it, rather than just going right to the short formula.
Return to the question of getting all six sides in order (while noticing the arbitrariness of this: it could just be asking for seeing 6 six times). Also recall that this isn’t asking for rolling 1, 2, 3, 4, 5, 6 in six rolls. It’s simply asking for the expected rolls for seeing your first 1, and then a 2, and then a 3, and so on. The answer is 36 rolls.
For just getting a 1 and then a 2, then, the answer should be 12. We can confirm this with the short formula 1/p: 1/(1/6) + 1/(1/6) = 12. But to demonstrate linearity of expectation, I’ll return to sigma notation:
$\displaystyle \sum_{x=1}^{\infty}x\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{\left(x-1\right)}+\sum_{y=1}^{\infty}y\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{\left(y-1\right)} = 12$
Since linearity of expectation tells us that we’ll get the same answer for [Expectation of X] + [Expectation of Y] as for [Expectation of X + Y], we can also do it like this:
$\displaystyle \sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\left(x+y\right)\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{\left(x-1\right)}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{\left(y-1\right)} = 12$
The latter will just get more and more tedious for larger numbers, so thank goodness for linearity (and for computers, I guess).
And thank extra goodness for the surprising fact that events need not be independent for the linearity property to hold!
Due to this generality, you can use it to solve this classic problem involving dependent events:
A bunch of folks leave their hats at a hat-check. The hat-check-booth guy is distracted by drink and forgets to tag the hats. He notices his mistake when folks start handing in their tickets at closing time. He lets chance sort it out, hands hats back at random, hopes folks are as drunk as he is. They are and nobody notices.
Given n people and n hats, what’s the expected number of people who got their own hat back?
Here’s a nice explanation from John Tsitsiklis (whose probability textbook I also like a lot):
There’s also a cool variation that asks the probability of nobody getting their own hat back. If you’ve got some basic combinatorics under your hat, this video by Ari Novick provides a lovely explanation:
Finally, I’ll soon publish a post looking at the median number of rolls for seeing your first 6, as well as the much harder question of the median number of rolls for seeing all six sides. The latter poses some frustrating, though also instructive (and maybe kinda fascinating), difficulties. Along the way, I’ll explain/derive the geometric distribution’s cumulative distribution function and median, among other things, mostly for the sake of intuition-training.
Update. That post is here: “Dice Probability: Median Rolls to See All 6 Sides (Coupon Collector’s Problem).”
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notes_microscopoe_calculations.docx - MICROSCOPE CALCULATIONS A Microscope Magnification = Ocular Lens X Objective Lens Ocular lens 10x Low power
# notes_microscopoe_calculations.docx - MICROSCOPE...
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MICROSCOPE CALCULATIONS A) Microscope Magnification = Ocular Lens X Objective Lens Ocular lens: 10x Low power objective: 4x Medium power objective: 10x High power objective: 40x What is the magnification under low power? What is the magnification under medium power? What is the magnification under high power ? B) Measuring Low Power Field Diameter (field of view ) : the distance from one side to the other under each power when looking through the microscope. Low power: use a ruler to measure it. Use this distance for High & Medium field diameter calculations as follows. Ex: low power field of view when measured with a ruler = 5 mm C) High & Medium field diameter must be calculated This formula is based upon the fact that the ratio of the diameters of the field of view is inversely proportional to the ratio of the magnifications. ie if the diameter doubles, the magnification is halved ie if the magnification doubles, the field diameter is halved Using the formula: Diameter of high power = magnification of low power Diameter of low power magnification of high power or D H = M L D L M H The diameter under high power = magnification of low X diameter of low mag high so, D H = M L x D L = 40 x X 5 mm = 0.5 mm so, the field diameter of high power is 0.5 mm M H 400 x Find the diameter of the field of view when using medium power. D m = M L D L M M /var/filecabinet/temp/converter_assets/3b/d5/3bd5f4709e92a77ae6370ecd3c912d911f38e42f.docx
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• Spring '10
• SABINESTANLEY
• Trigraph, Eyepiece
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https://freshersnews.co.in/ways-in-which-875-can-be-reduced-as-a-fraction/
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# Ways in Which .875 Can Be Reduced as a Fraction
Fractions are a very interesting subject and it is quite easy to reduce a fraction and convert the numbers. So, today in this brief guide, we will be looking as how to convert .875 as a fraction. Also, we will be looking at some of the simplest methods for the same. A fraction is a portion or section of any quantity taken from a total, where the total can be any number, a particular value, or an object. Answer: The fraction 0.875 is written as 7/8 and 0. 875 as a fraction. In this example, we will write 0.875 as a fraction.
## Explanation:
We use the given number as the numerator, followed by the required number of zeros in the denominator immediately below the decimal point, to convert a decimal number into a fraction. Then, it is possible to simplify this fraction. Since 0.875 has three numbers after the decimal, 1000 is used as the denominator and the decimal point is removed. 875/1000. After dividing this number by 1,000, we get 7/8. You can check your answer by using an online calculator.
So, 7/8 is 0.875 divided by itself. What is 0.875 divided by itself? In the lowest term, what is 0.875 as a fraction? What’s the significance here to change over 0.875 to a small portion? This indicates that you want to change the number.875 from a decimal to a fraction or .875 as a fraction. with a numerator (N) and a denominator (D). To put it another way, you need to solve this equation:
.875 =N/D
Here we won’t just obviously clarify how for find the solution to 0.875 as a small portion, yet additionally demonstrate to you that our response is right.
## Step-by-Step Process –
First, keep in mind that the decimal point will be removed by multiplying 1000 by.875 Therefore, construct a fraction with a denominator of 1000 and a numerator of.875: –
.875 x 1000/1000. The next step is to combine the numbers in the numerator and leave the denominator unchanged. Presently our portion seems to be this:
875/1000 Because 125 is the greatest common factor between 875 and 1000, you can keep the same value by dividing the numerator and denominator by 125:
875×125/1000 125, and when we calculate the numerator and denominator of our previous fraction, we obtain.875 in its simplest form: .875 = 7/8 In a fraction, multiplying the numerator by the denominator is equivalent to dividing the numerator by the denominator. If you use your calculator to divide 7 by 8, you will see that the result is 0.875. As a result, we are aware that the previous response is correct.
## Converting .875 as a Fraction –
In just a few steps, you can convert 0.875 into a fraction, which will give you the fraction 7/8 in its simplest form.
The fraction.875 is 7/8, or 875/1000.
Look down for point-by-point steps on the most proficient method to change over .875 into a small portion.
Write.875 as.875 1. For each digit after the decimal point, multiply the numerator and denominator by 10.
.875×1 = .875 x 1000/1 x 1000 = 875/1000
To diminish the part track down the Best Normal Element or the (GCF) for 875 and 1000. Keep in mind that a factor is nothing more than a number that can be divided into another number without leaving a residue.
These are the 875 factors: 1 5 7 25 35 125 175 875 The following are the factors of 1000: 1 2 4 5 8 10 20 25 40 50 100 125 200 250 500 1000
The Best Normal Component or the (GCF) for both 875 and 1000 is: 125. Presently to lessen the portion we partition both the numerator and denominator by the GCF esteem.
## Round Half Up and Down Formula –
875 x 1000 = 875/ 125 x1000 / 125 = 7 8 As an aside, the entire number-integral part is as follows: empty the empty part is: .875 x 1000 is the complete and simple fraction: The number of digits to round to at the level of precision is 875/1000, or 175/200, or 35/40, or 7/8. To further reduce the decimal.875 as a fraction, select a lower exactness point below. The exactness point by default is 5. When changing the exactness point, you can use the “round half up” and “round half down” options to round the last trailing digit, if it is “5”, either way.
For instance, 0.875 rounded half up equals 88/100 and rounded half down equals 87/100 with a precision point of 2.
87500/100000 = 8750/10000 = 875/1000 = 175/200 = 35/40 = 7/8 If you want to learn how to solve problems like these on your own, you need to know how many decimal places there are. The 8 is in the tenth, the 7 in the hundredth, and the 5 in the thousandth places for.875. Either “eight hundred seventy-five thousandths” or “point eight seventy-five” can be used to read “.875.”Therefore, as a fraction,.875 equals 875/1000 (a number like.46 would equal 46/100). Additionally,.7 equals 7/10. The decimal place is used in all of these fractions: tenths, hundredths, and so on)
## The Divisibility Rule –
You need to cut it down now that it’s just a fraction. Determine what can be divided into both 875 and 1000 according to your divisibility rules. Since I prefer to divide by many small numbers rather than the largest possible number (the GCF) right away, I would probably begin with division by 5. You could go with either approach!
Isolating by 5 three times, you see that 875/1000 = 175/200 = 35/40 = 7/8. The only other number that can be divided into both 7 and 8 is 1, so your fraction is as small as it can be.
## Step-by-Step Solution:
875 percent = 35/4 = 83/4 as a fraction. To convert 875 percent to a fraction, follow these steps: Record the percent isolated by 100 like this:
875% = 875/100. For each number after the decimal point, multiply top and bottom by 10: We do not have numbers after the decimal point because 875 is an integer. We simply move on to step 3. Rearrange (or decrease) the above portion by isolating both numerator and denominator by the GCD (Most prominent Normal Divisor) between them. GCD (875,100) equals 25 in this instance. In the simplest form, therefore, (87525)/ (10025) = 35/4. We have an incorrect fraction because the numerator is greater than the denominator. Because of this, we can also express it as a MIXED NUMBER. For example, 875/100 is the same as 8 by 3/4 when expressed as a mixed number.
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Divide the decimal by one to create a fraction that looks like this: Write down the decimal; we’ll only be looking at decimals that end in “-.” The calculator above can handle up to three decimal places. 0.875 / 1 Until you get a regular fraction, multiply both the numerator and the denominator by 10. 875/1000 Divide both by the largest common factor: Explanation: 875 / 125) / (1000 / 125) = 7 / 8 0.875 = 7 8
You can begin by changing it to 875/1000 straight away. You must now determine whether this fraction can be reduced. Because the digits of tens and units make numbers that can be divided by 25 (75 and 0 respectively), both the numerator and denominator can be divided by 25. By dividing the fraction by 25, you can get: 35/40
. 7 8 can be obtained by further reducing this fraction by 5. You cannot reduce the fraction any further because 8 is not divisible by 7 and 7 is a prime number. Therefore, the answer is: 0.875 =7/8
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https://www.teacherspayteachers.com/Product/Fun-Fraction-Unit-Common-Core-Aligned-118935
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# Fun Fraction Unit (Common Core Aligned)
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This unit contains 5 different fraction activities. All of the activities are centered around the basic concept of identifying and naming a fraction. Thea activities include:
*Cooking Up Some Fractions (also sold separately) - students match a picture of a fraction to the number form.
*Unifix Cubes Fractions (also sold separately) - students create towers with unifix cubes. They then draw and answer questions about their towers
*Fraction Friends - Students draw a picture of themselves and their friends. They then answer questions about their picture in fractions
*Fraction Word Fun - Students write out spelling words (or any words of the teachers' choosing). They then write the number of consonants and vowels in fractions
*Lotsa Lotus and Lotsa Fractions (also sold separately) - Students are given a fraction with a Lotus Diagram graphic organizer. They are to then illustrate that fraction in the boxes of the organizer to demonstrate knowledge of how to draw and represent fractions.
*************************************************************
Need more fraction resources?:
Fraction Word Problem Powerpoint & Activity
Comparing Fractions Coloring Page
Fractions Matching Game
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https://lmlib.ch/2.0/html/_gallery_examples/70-localized-polynomials/example-ex703.0-time-amplitude-scaling.html
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# Time and Amplitude Scaling [ex603.0]#
This example is published in [Wildhaber2020] .
Out:
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
Badly Conditioned Steady State Matrix W: Use larger boundaries or lower g.
import matplotlib.pyplot as plt
from matplotlib.lines import Line2D
import numpy as np
import lmlib as lm
def time_amplitude_scaling(Q, a, b):
q = np.arange(Q)
Mq = lm.poly_square_expo(q)
Mqp1 = lm.poly_int_expo(Mq)
L_int = lm.poly_int_coef_L(Mq)
L_eta = lm.mpoly_dilate_ind_coef_L(q) # matrix operation of vdiag
R_QQ = lm.permutation_matrix_square(Q, Q)
I_Q = np.eye(Q)
I_QQ = np.eye(Q ** 2)
c1 = L_int
c2 = np.kron(I_Q, L_int).dot(np.kron(L_eta, I_Q))
c3 = np.kron(I_QQ, L_int).dot(R_QQ).dot(np.kron(L_eta, L_eta))
bmaT = (np.power(b, Mqp1) - np.power(a, Mqp1)).T
A = bmaT @ c1
B = np.kron(I_Q, bmaT) @ c2
C = np.kron(I_QQ, bmaT) @ c3
return A, B, C, q, Mq
y = np.column_stack([np.convolve(y0, 1 / 50 * np.ones(50), 'same') for y0 in y.T])
fs = 2000
K = len(y)
t = np.arange(K)/fs
M = 3
Q = 5
ENERGY_THD = 60
ab_half = int((80e-3 * fs) / 2)
ks_alpha = np.array([370]) # fixing template
alssm = lm.AlssmPoly(poly_degree=Q - 1)
segment_right = lm.Segment(a=-ab_half, b=ab_half, direction=lm.BW, g=50, delta=ab_half)
cost = lm.CostSegment(alssm, segment_right)
xs = rls.filter_minimize_x(y)
# generate template
dilate_factor = 1.456
amplitude_factor = 0.945
y_template = amplitude_factor * np.column_stack(
[np.interp(np.linspace(0, K, int(K * dilate_factor)), np.arange(K), y0) for y0 in y.T])
xs_tmpl = rls_tmpl.filter_minimize_x(y_template)
alphas = xs_tmpl[(ks_alpha * dilate_factor).astype(int)]
A, B, C, q, Mq = time_amplitude_scaling(Q, a=-ab_half, b=ab_half)
etas = np.linspace(0.5, 2, 1000)
J = np.full(K, np.inf)
cost_ratio = np.full(K, np.inf)
model_energy_obs = np.full(K, np.inf)
time_scaling_hat = np.full(K, np.nan)
amplitude_hat = np.full(K, np.nan)
for k in range(K):
beta_beta = np.zeros(Q ** 2)
alpha_beta = np.zeros(Q ** 2)
alpha_alpha = np.zeros(Q ** 2)
for m in range(M):
beta_beta += np.kron(xs[k, :, m], xs[k, :, m])
alpha_beta += np.kron(alphas[0, ..., m], xs[k, :, m])
alpha_alpha += np.kron(alphas[0, ..., m], alphas[0, ..., m])
a1 = A @ beta_beta
for eta in etas:
J_ = a1 - (((B @ alpha_beta).T @ np.power(eta, q)) ** 2) / ((C @ alpha_alpha).T @ np.power(eta, Mq))
if J_ < J[k]:
J[k] = J_
time_scaling_hat[k] = eta
amplitude_hat[k] = ((B @ alpha_beta).T @ np.power(eta, q)) / ((C @ alpha_alpha).T @ np.power(eta, Mq))
cost_ratio[k] = J_/a1
model_energy_obs[k] = a1
L_dilate = lm.poly_dilation_coef_L(np.arange(Q), time_scaling_hat[k_min])
alphas_hat = amplitude_hat[k_min]*np.einsum('jn, knm->kjm', L_dilate, alphas)
offset_channels = np.arange(M) + 1
fig, (ax1, ax2, ax3) = plt.subplots(1, 3, gridspec_kw={'width_ratios': [1, 2, 2]})
ab_half_ext = int(ab_half * 1.15)
segment_right = lm.Segment(a=-ab_half_ext, b=ab_half_ext, direction=lm.BW, g=50, delta=ab_half_ext)
cost_ext = lm.CostSegment(alssm, segment_right)
segment_k, trajs_tmpl = cost.trajectories(alphas)[0][0]
trajs_obs = lm.map_trajectories(cost.trajectories(xs[ks_alpha]), ks_alpha, K, True, True)
trajs_tmpl_hat = lm.map_trajectories(cost.trajectories(alphas_hat), [k_min], K, True, True)
trajs_obs_ext = lm.map_trajectories(cost_ext.trajectories(xs[ks_alpha]), ks_alpha, K, True, True)
ax1.plot(segment_k, trajs_tmpl + offset_channels, lw=1, c='r', label=r'$\alpha^T z^q$')
ax1.set_xlim([min(segment_k), max(segment_k)])
ax1.set_xticks([max(segment_k)])
ax1.set_xticklabels(['1'])
ax1.legend([Line2D([0], [0], color='r', lw=1)], [r'$\alpha^T z^q$'], loc=1, fontsize=8)
ax2.plot(t, y + offset_channels, lw=1, c='k')
ax2.plot(t, trajs_obs_ext + offset_channels, lw=0.8, ls='--', c='b')
ax2.plot(t, trajs_obs + offset_channels, lw=1, c='b')
ax2.legend([Line2D([0], [0], color='b', lw=1)], [r'$\beta_k^T z^q$'], loc=1, fontsize=8)
ax3.plot(t, y + offset_channels, lw=1, c='grey')
ax3.plot(t, trajs_obs_ext + offset_channels, lw=0.8, ls='--', c='b')
ax3.plot(t, trajs_obs + offset_channels, lw=1, c='b')
ax3.plot(t, trajs_tmpl_hat + offset_channels, lw=1, c='r', label=r'$\alpha^T z^q$')
ax3.legend([Line2D([0], [0], color='r', lw=1)], [r'$\lambda \alpha^T (\eta z)^q$'], loc=1, fontsize=8)
for ax in fig.axes:
ax.set_yticks([1, 2, 3])
plt.show()
# show cost and estimate over time
if True:
fig, axs = plt.subplots(5, 1, sharex='all')
axs[0].plot(J, c='k', lw=0.6, ls=':', label='cost')
axs[1].plot(cost_ratio, c='k', lw=0.6, ls=':', label='cost_ratio')
axs[2].plot(time_scaling_hat, c='k', lw=0.6, ls=':', label=r'time scaling $\eta$')
axs[3].plot(amplitude_hat, c='k', lw=0.6, ls=':', label=r'amplitude scaling $\lambda$')
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# The evidential value of mobile phone colocation
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Criminals are well aware that making a phone call leaves a trace behind, which might later be used by police, and later still, in court. Therefore they will often switch phones, and preferably use more or less anonymous phones for "business". However, at the same time as they are using one phone for their work activities, they are possibly using another phone for legitimate business or for ordinary private purposes. This leads to the phenomenon called "co-location": two mobile phones apparently moving together, each separately making calls, but as if the two phones are in the same hands.
How can one find phones, and then co-locating phones, associated with some crime? Can one decide from a short history of apparent co-location whether or not the two phones were in the same hands? How strong is the weight of the evidence in discriminating between two hypotheses: the phones colocate by chance (defence hypothesis) or they colocate because they are in the same hands (prosecution hypothesis)? We have to distinguish two phases of "research": exploratory (criminal investigation) and confirmatory (criminal prosecution). I discuss the roles of statistics in these two phases of forensic statistical analysis of mobile phone co-location.
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### The evidential value of mobile phone colocation
1. 1. The evidential value of mobile phone co-location Richard Gill Mathematical Institute, Leiden University http://www.math.leidenuniv.nl/~gill Joint work with Helena van Eijck (master thesis, Statistical Science programme) http://www.math.leidenuniv.nl/nl/theses/382/ Data Science Meetup Utrecht, 23 January 2014
2. 2. The chance of coincidence? • DNA match • Finger print match • Handwriting match • ... and so on ... Match probability = P(Coincidence | Hdefence); or better, ! Likelihood Ratio (LR) = P(Coincidence | Hdefence): P(Coincidence | Hprosecution)
3. 3. Mobile phone co-location • Mobile phone co-location: two cell phones used over a long time period in a way consistent with them being carried by one person
4. 4. Visualisation (simulated data) (all analysis done in , of course)
5. 5. Visualisation (simulated data)
6. 6. Hariri Case • 14 February 2005: assassination, Beirut • Lebanon Police investigation, continued by UNIIIC (2005), and STL (2009) • 2011: STL publishes indictment • 2014: trial opens “The case against the Accused is built in large part on circumstantial evidence. Circumstantial evidence, which works logically by inference and deduction, is often more reliable than direct evidence, which can suffer from first-hand memory loss or eye-witness distortion. It is a recognised legal principle that circumstantial evidence has similar weight and probative value as direct evidence and that circumstantial evidence can be stronger than direct evidence.”
7. 7. http://www.stl-tsl.org/en/the-cases/stl-11-01
8. 8. Analysis of CDR revealed co-locating phones ... • “Red network” phones associated with surveillance and assassination (covert: anonymous & closed) • “Blue network” phones associated with logistics, preparation (anonymous but open) • “Green network” phones associated with chain of command (covert) • PMP’s (personal mobile phones) • ... “Call Data Records”: Per call: Cell towers, time, phone numbers
9. 9. How they found co-locating phones • Given: a “target phone” (already associated with crime) • Select notable patterns of movement • Look for candidate co-locators (match same pattern) • Follow-up the “hits” in time: do they de-co-locate? (look for an anomaly)
10. 10. Issues • Texas sharp-shooter (testing a hypothesis suggested by the data) • Likelihood ratio: needs two models • Is a model of typical behaviour relevant to evaluation of specific case? • Is a sample from the population relevant to evaluation of a specific case?
11. 11. Our approach • Part I: investigate reliability of search procedure • Part II: quantify evidential value of each specific pair of co-locating phones using permutation approach
12. 12. Interlude: How good is the data? • Intermittency • Inaccuracy
13. 13. Does CDR data uniquely characterise you? Unique in the Crowd: The privacy bounds of human mobility Yves-Alexandre de Montjoye1,2 , Ce´sar A. Hidalgo1,3,4 , Michel Verleysen2 & Vincent D. Blondel2,5 1 Massachusetts Institute of Technology, Media Lab, 20 Ames Street, Cambridge, MA 02139 USA, 2 Universite´ catholique de Louvain, Institute for Information and Communication Technologies, Electronics and Applied Mathematics, Avenue Georges Lemaıˆtre 4, B-1348 Louvain-la-Neuve, Belgium, 3 Harvard University, Center for International Development, 79 JFK Street, Cambridge, MA 02138, USA, 4 Instituto de Sistemas Complejos de Valparaı´so, Paseo 21 de Mayo, Valparaı´so, Chile, 5 Massachusetts Institute of Technology, Laboratory for Information and Decision Systems, 77 Massachusetts Avenue, Cambridge, MA 02139, USA. We study fifteen months of human mobility data for one and a half million individuals and find that human mobility traces are highly unique. In fact, in a dataset where the location of an individual is specified hourly, and with a spatial resolution equal to that given by the carrier’s antennas, four spatio-temporal points are enough to uniquely identify 95% of the individuals. We coarsen the data spatially and temporally to find a : S S S E d 2 d 3 d NATURE/SCIENTIFIC REPORTS March 2013
14. 14. Does CDR data uniquely characterise you? NATURE/SCIENTIFIC REPORTS March 2013 Figure 2 | (A) Ip52 means that the information available to the attacker consist of two 7am-8am spatio-temp was in zone I between 9am to 10am and in zone II between 12pm to 1pm. In this example, the traces of tw www Unique in the Crowd: The privacy bounds of human mobility Yves-Alexandre de Montjoye1,2 , Ce´sar A. Hidalgo1,3,4 , Michel Verleysen2 & Vincent D. Blondel2,5 1 Massachusetts Institute of Technology, Media Lab, 20 Ames Street, Cambridge, MA 02139 USA, 2 Universite´ catholique de Louvain, Institute for Information and Communication Technologies, Electronics and Applied Mathematics, Avenue Georges Lemaıˆtre 4, B-1348 Louvain-la-Neuve, Belgium, 3 Harvard University, Center for International Development, 79 JFK Street, Cambridge, MA 02138, USA, 4 Instituto de Sistemas Complejos de Valparaı´so, Paseo 21 de Mayo, Valparaı´so, Chile, 5 Massachusetts Institute of Technology, Laboratory for Information and Decision Systems, 77 Massachusetts Avenue, Cambridge, MA 02139, USA. We study fifteen months of human mobility data for one and a half million individuals and find that human mobility traces are highly unique. In fact, in a dataset where the location of an individual is specified hourly, and with a spatial resolution equal to that given by the carrier’s antennas, four spatio-temporal points are enough to uniquely identify 95% of the individuals. We coarsen the data spatially and temporally to find a formula for the uniqueness of human mobility traces given their resolution and the available outside information. This formula shows that the uniqueness of mobility traces decays approximately as the 1/10 power of their resolution. Hence, even coarse datasets provide little anonymity. These findings represent fundamental constraints to an individual’s privacy and have important implications for the design of frameworks and institutions dedicated to protect the privacy of individuals. D erived from the Latin Privatus, meaning ‘‘withdraw from public life,’’ the notion of privacy has been foundational to the development of our diverse societies, forming the basis for individuals’ rights such as free speech and religious freedom1 . Despite its importance, privacy has mainly relied on informal pro- tection mechanisms. For instance, tracking individuals’ movements has been historically difficult, making them de-facto private. For centuries, information technologies have challenged these informal protection mechanisms. S: CS CS CS CE ed 12 ed 13 ed 13 nd als to mit. du)
15. 15. How accurate is CDR location? • “Deventer murder case”: under “exceptional” atmospheric conditions, a cell phone uses a cell tower 25 Km away, rather than close-by cell towers Forensic Statistics and Graphical Models: Deventer moordzaak, phonecall A28 Maikel Bargpeter February 3, 2012 This analysis is mainly based on ’Leugens over Louwes’. The main reason Louwes got involved in the Deventer moordzaak is that he was the accountant of Mw. Wittenberg and called her on his mobile phone right before the killing. According to Louwes he was on the highway A28, 25 km away from Deventer where the murder took place. So he claims that he is not the killer. claim it is very unlikely such a connection from the A28 could be mad Unfortunately most of the research can not be integrated into the g model at first sight. The only way out is: the normal conditions which might be absent at of the phonecall. Hans Meijer looked up reports at a institute in the U find that around that time and place these special conditions did happe atmosphere.
16. 16. How accurate is CDR location? • Deventer murder case: under “exceptional” atmospheric conditions, a cell phone uses a cell tower 25 Km away, rather than close-by cell towers event zijn de kansen dat dit matcht met de verdachte geschat op 0.60 net als voor ouders en ander woonachtig in nabijheid van ouders, voor A op 0.25, M 0.40 en de ander niet woonachtig nabij ouderlijk huis op 0.25. 6.1.11 Event 11 De vijf berichten van 9 oktober die dit event kenmerken, hebben binnen een drie kwartier plaats gevonden tussen half twee en kwart over twee, waarvan drie keer de zendmast gelegen aan de Reinaert de Vosstraat is aangestraald en de zendmasten gelegen aan de Hugo de Grootkade en Donker Curtiusstraat zijn beiden eenmalig aangestraald. De zendmasten blijken rondom de woning van M (paarse punt) te liggen, waarvan de meest aangestraalde zendmast het verst weg is gesitueerd. Gegeven de locaties van de zendmasten is het meest aannemelijk dat dit matcht het meest met M en is daarom ook geschat op 0.70. Voor alle andere is dit minder aannemelijk maar niet onwaarschijnlijk is en daarom zijn de kansen van de anderen op 0.40 geschat. 6.1.12 Event 12 Dit event telt 20 berichten en is verspreidt over drie dagen. In de ochtend en de avond van de eerste dag worden de zendmasten nabij het ouderlijk huis aangestraald. De daaropvolgende dag zijn de zendmasten in Duivendrecht en Purmerend aangestraald. De gebruiker van de telefoon kan hier niet mee geïdentificeerd worden, maar uit de berichten kan wel worden opgemaakt dat de dag erop een transactie 9 Donker Curtiusstraat, welke gelegen is nabij de woning van M, aangestraald. Gegeven dat ‘s ochtends de telefoon aangestraald is nabij het ouderlijk huis en twee dagen later nabij het pand waar de verdachte een week eerder een offerte voor een lening heeft ontvangen, is de kans dat hij de telefoon in zijn bezit heeft geschat op 0.8. Voor zijn ouders is het minder aannemelijk dat zij bij de Diopter zijn wezen kijken en daarna via Almere terug naar Amsterdam, is de kans dat zij de telefoon in hun bezit hebben geschat op 0.65. Voor K1 hebben we de kans geschat op 0.55. Dit event wijst niet direct naar A of M. Daarom hebben we hun kansen op 0.25 geschat. Voor K2 is het nog lager, namelijk 0.20. 6.1.13 Event 13 Het enige bericht dat is verzonden is verstuurd in de nabijheid van Rijnstraat 35 in Amsterdam. Deze aangestraalde zendmast ligt in de buurt van een doorlopende weg en is mogelijk in de richting van de woning van broer A. Omdat dit niet heel nauwkeurig is, hebben we besloten dit bericht niet in de verdere analyse mee te nemen. 6.1.14 Event 14 Dit event bevat vijf berichten. Bij één bericht is de locatie niet bekend. ‘s Ochtends is de telefoon aangestraald nabij het ouderlijk huis. Twee uur later worden twee verschillende zendmasten aangestraald in dezelfde minuut. Dit zijn de zendmasten Den Briel straat en de Donker Curtiusstraat te Amsterdam. Een mogelijke verklaring is dat de gebruiker van de telefoon onderweg is vanaf de snelweg (A10) riching de binnenstad van Amsterdam. Een andere verklaring zou kunnen zijn, dat de gebruiker van de telefoon op dat moment boodschappen aan het doen zou zijn op de Centrale Markt, gelegen in het grijze gebied tussen de locaties van de twee zendmasten in. Dit zou overeen kunnen komen met het profiel van M, event 11 event 14 An Amsterdam drugs case – 2 of 19 events blue = cell towers, purple = addresses associated with suspect
17. 17. How accurate is CDR location? • Deventer murder case: under “exceptional” atmospheric conditions, a cell phone uses a cell tower 25 Km away, rather than close-by cell towers RDG, 12 August 2012 Data: Google latitude; my trip: train
18. 18. End of interlude. Now: Our approach • Part I: investigate reliability of search procedure • Part II: quantify evidential value of specific pairs of co-locating phones using permutation approach
19. 19. Part I: the experiment • Chose one target phone from case • Identified all notable three-point patterns of movement • Identified all matches (“hits”) to each pattern • Followed each hit forwards in time to first dis- location event (“anomaly”)
20. 20. Part I • Measure mobility, and (phone) activity, of hit and of target, in first four days • Mobility: Km travelled • Activity: number of calls • Investigate relation between these four variables and time to first anomaly for our sample of hits
21. 21. Summary • Dichotomise each of four variables (“high” vs “low”) • Score each hit by number of highs (0 to 4)
22. 22. Chance of anomaly per day is roughly constant Joint Exponential Fit
23. 23. Chance of anomaly per day is roughly constant • Very high: sum score 3 and 4: half life (of time to anomaly) is one day • Medium: sum score 2: half life is two days • Very low: sum score 0 and 1: half life is four days
24. 24. If we believe this, then ... • no anomaly for 10 half lives: 1 in a thousand • no anomaly for 20 half lives: 1 in a million
25. 25. Conclusion of part 1 • The “chance of coincidence” depends strongly on individual characteristics of particular two phones • The investigative procedure is reliable • first, identify suspects (pattern-hits which continue to colocate a few days) • second, confirm suspects (long term follow-up) • … so we needn’t worry about Texas sharpshooter (we’ll analyse long term follow-up data) • We do have a major reference class problem
26. 26. Part 2 • Take two co-locating phones: could this be coincidence? • We need to compare the observed history of a pair of phones with that of similar pairs of phones of different persons • Especially: similar activity, similar mobility, frequenting the same locations • Assumption: if two persons are completely unrelated then we may as well compare Mr X Day A with Mr Y Day B, as Mr X Day A with Mr Y Day A
27. 27. Problems • “Completely unrelated but similar” persons do live in the same neighbourhood, work in the same neighbourhood, frequent the same shops, cafés, places of worship, beach clubs, sporting events, ... • We should condition on confounders (all days are not exchangeable) • Problem of observational (as opposed to experimental) studies: the unknown unknowns
28. 28. Our solution • Compare history of phone X with artificial histories like phone Y ’s, obtained by permuting (shuffling)Y ’s days • Shuffle weekdays and weekend-days separately • Distance between two histories: total kilometers between consecutive calls on same day of different phones • Note: “artificial histories” need not be realistic in all respects – they should just be realistic in relevant respects
29. 29. Original vs. Shuffled (simulated data)
30. 30. Findings • Discovered co-locations are statistically very significant • In retrospect we could better have used a different similarity measure, etc… • We reported to the court exactly what we did do, and all that we did do
31. 31. Future research • Invent better distance measure (model based LR?) for higher power (note: not for validity) • Should refine permutation procedure (shuffled histories may be unrealistic when overnight location can vary) • As we condition on more confounders, reference population shrinks, prior probabilities change – relevant evidence moves out of our analysis but is still relevant
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# Introduction to Elasticity/Plane strain example 1
## Example 1
Given:
The plane strain solution for the stresses in a rectangular block with ${\displaystyle 0, ${\displaystyle -b, and ${\displaystyle -c with a given loading is
${\displaystyle \sigma _{11}={\frac {3Fx_{1}x_{2}}{2b^{3}}}~;~~\sigma _{12}={\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}}~;~~\sigma _{22}=0~;~~\sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}.}$
Find:
1. Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
2. We wish to use this solution to solve the corresponding problem in which the surfaces ${\displaystyle x_{3}=\pm c}$ are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.
3. Find the maximum error in the stress ${\displaystyle \sigma _{33}}$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.
## Solution
The tractions acting on the block are:
{\displaystyle {\begin{aligned}{\text{at}}~x_{1}&=0~,~~{\widehat {\mathbf {n} }}{}\equiv (-1,0,0)~,~~t_{i}=n_{1}\sigma _{1i}\equiv (-\sigma _{11},-\sigma _{12},-\sigma _{13})\\{\text{at}}~x_{1}&=a~,~~{\widehat {\mathbf {n} }}{}\equiv (1,0,0)~,~~t_{i}=n_{1}\sigma _{1i}\equiv (\sigma _{11},\sigma _{12},\sigma _{13})\\{\text{at}}~x_{2}&=-b~,~~{\widehat {\mathbf {n} }}{}\equiv (0,-1,0)~,~~t_{i}=n_{2}\sigma _{2i}\equiv (-\sigma _{21},-\sigma _{22},-\sigma _{23})\\{\text{at}}~x_{2}&=b~,~~{\widehat {\mathbf {n} }}{}\equiv (0,1,0)~,~~t_{i}=n_{2}\sigma _{2i}\equiv (\sigma _{21},\sigma _{22},\sigma _{23})\\{\text{at}}~x_{3}&=-c~,~~{\widehat {\mathbf {n} }}{}\equiv (0,0,-1)~,~~t_{i}=n_{3}\sigma _{3i}\equiv (-\sigma _{31},-\sigma _{32},-\sigma _{33})\\{\text{at}}~x_{3}&=c~,~~{\widehat {\mathbf {n} }}{}\equiv (0,0,1)~,~~t_{i}=n_{3}\sigma _{3i}\equiv (\sigma _{31},\sigma _{32},\sigma _{33})\end{aligned}}}
Plugging in the expressions for stress,
{\displaystyle {\begin{aligned}{\text{at}}~x_{1}&=0~,~~t_{i}\equiv (0,-{\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}},0)\\{\text{at}}~x_{1}&=a~,~~t_{i}\equiv ({\frac {3Fax_{2}}{2b^{3}}},{\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}},0)\\{\text{at}}~x_{2}&=-b~,~~t_{i}\equiv (0,0,0)\\{\text{at}}~x_{2}&=b~,~~t_{i}\equiv (0,0,0)\\{\text{at}}~x_{3}&=-c~,~~t_{i}\equiv (0,0,{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}})\\{\text{at}}~x_{3}&=c~,~~t_{i}\equiv (0,0,-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}})\end{aligned}}}
These tractions are illustrated in the following figure
Tractions on the beam
To unload the tractions on the faces ${\displaystyle x_{3}=\pm c}$, we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.
Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at ${\displaystyle x_{3}=\pm c}$, there is no net force is the ${\displaystyle x_{3}}$ direction. Similarly, there is no net moment about the ${\displaystyle x_{2}}$ axis.
However, there is a net moment about the ${\displaystyle x_{1}}$ axis. Hence, the problem to be superposed should have a bending stress distribution ${\displaystyle \sigma _{33}=Cx_{2}}$, where ${\displaystyle C}$ is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the ${\displaystyle x_{3}-x_{2}}$ plane subjected to bending moments at the ends.)
The total stress for the corrected problem is
${\displaystyle \sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}+Cx_{2}}$
The bending moment for a cross-section of the beam in the ${\displaystyle x_{3}-x_{2}}$ plane about the ${\displaystyle x_{1}}$ axis is ${\displaystyle M=-\sigma _{33}I/x_{2}\,}$, where ${\displaystyle I=\int _{0}^{a}\int _{-b}^{b}x_{2}^{2}dx_{2}dx_{1}}$.
Since ${\displaystyle \sigma _{33}}$ varies with ${\displaystyle x_{1}}$, the total bending moment for the beam is given by
{\displaystyle {\begin{aligned}M&=\int _{0}^{a}\int _{-b}^{b}\left({\frac {-\sigma _{33}}{x_{2}}}\right)x_{2}^{2}dx_{2}dx_{1}\\&=\int _{0}^{a}\int _{-b}^{b}\left({\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}-Cx_{2}\right)x_{2}dx_{2}dx_{1}\\&=\int _{0}^{a}\left[{\frac {3\nu Fx_{1}x_{2}^{3}}{6b^{3}}}-{\frac {Cx_{2}^{3}}{3}}\right]_{-b}^{b}dx_{1}\\&=\int _{0}^{a}\left[{\frac {\nu Fx_{1}b^{3}}{b^{3}}}-{\frac {2Cb^{3}}{3}}\right]dx_{1}\\&=\left[{\frac {\nu Fx_{1}^{2}b^{3}}{2b^{3}}}-{\frac {2Cx_{1}b^{3}}{3}}\right]_{0}^{a}\\&={\frac {\nu Fa^{2}b^{3}}{2b^{3}}}-{\frac {2Cab^{3}}{3}}\end{aligned}}}
Setting the bending moment to zero, we have
${\displaystyle C={\frac {3\nu Fa}{4b^{3}}}}$
Therefore, the corrected solution is
${\displaystyle {\sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}+{\frac {3\nu Fax_{2}}{4b^{3}}}}}$
Ideally, for a problem with zero tractions on ${\displaystyle x_{3}=\pm c}$, we should have ${\displaystyle \sigma _{33}=0}$. Therefore, the error in our solution is
${\displaystyle \sigma _{33}^{\text{err}}={\text{abs}}\left({\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}-{\frac {3\nu Fax_{2}}{4b^{3}}}\right)}$
The error is maximum at ${\displaystyle (0,-b)}$, ${\displaystyle (0,b)}$, ${\displaystyle (a,-b)}$, and ${\displaystyle (a,b)}$. Thus,
{\displaystyle {\begin{aligned}\left.\sigma _{33}^{\text{err}}\right|_{(0,-b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(0,b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(a,-b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(a,b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\end{aligned}}}
The maximum error is
${\displaystyle {\frac {3\nu Fa}{4b^{2}}}}$
The maximum tensile stress is
${\displaystyle \sigma _{11}(a,b)={\frac {3Fab}{2b^{3}}}={\frac {3Fa}{2b^{2}}}}$
Therefore, the ratio of the maximum error in ${\displaystyle \sigma _{33}}$ to the maximum tensile stress is
${\displaystyle {{\text{Ratio}}={\frac {\nu }{2}}}}$
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CC-MAIN-2020-50
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https://oeis.org/A123574
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A123574 The Kruskal-Macaulay function K_5(n). 3
0, 5, 9, 12, 14, 15, 15, 19, 22, 24, 25, 25, 28, 30, 31, 31, 33, 34, 34, 35, 35, 35, 39, 42, 44, 45, 45, 48, 50, 51, 51, 53, 54, 54, 55, 55, 55, 58, 60, 61, 61, 63, 64, 64, 65, 65, 65, 67, 68, 68, 69, 69, 69, 70, 70, 70, 70, 74, 77, 79, 80, 80, 83, 85, 86, 86, 88, 89, 89, 90 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Write n (uniquely) as n = C(n_t,t) + C(n_{t-1},t-1) + ... + C(n_v,v) where n_t > n_{t-1} > ... > n_v >= v >= 1. Then K_t(n) = C(n_t,t-1) + C(n_{t-1},t-2) + ... + C(n_v,v-1). REFERENCES D. E. Knuth, The Art of Computer Programming, Vol. 4, Fascicle 3, Section 7.2.1.3, Table 3. LINKS MAPLE lowpol := proc(n, t) local x::integer ; x := floor( (n*factorial(t))^(1/t)) ; while binomial(x, t) <= n do x := x+1 ; od ; RETURN(x-1) ; end: C := proc(n, t) local nresid, tresid, m, a ; nresid := n ; tresid := t ; a := [] ; while nresid > 0 do m := lowpol(nresid, tresid) ; a := [op(a), m] ; nresid := nresid - binomial(m, tresid) ; tresid := tresid-1 ; od ; RETURN(a) ; end: K := proc(n, t) local a ; a := C(n, t) ; add( binomial(op(i, a), t-i), i=1..nops(a)) ; end: A123574 := proc(n) K(n, 5) ; end: for n from 0 to 80 do printf("%d, ", A123574(n)) ; od ; # R. J. Mathar, May 18 2007 CROSSREFS For K_i(n), i=1, 2, 3, 4, 5 see A000012, A003057, A123572, A123573, A123574. Sequence in context: A135979 A287452 A287456 * A314611 A143834 A314612 Adjacent sequences: A123571 A123572 A123573 * A123575 A123576 A123577 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Nov 12 2006 EXTENSIONS More terms from R. J. Mathar, May 18 2007 STATUS approved
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Last modified October 1 04:06 EDT 2020. Contains 337441 sequences. (Running on oeis4.)
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https://www.askiitians.com/forums/9-grade-maths/6-the-slant-height-and-base-diameter-of-a-conical_273060.htm
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# 6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing iJs curved surface at the rate of r 210 per 100 m2
Pawan Prajapati
3 years ago
. Curved surface area = rcrl = 22 x 7 x 26 m2 = 660 m2 7 Cost of white washing = f (210 x ) = f 1165.
Ram Kushwah
110 Points
3 years ago
For the conical tomb given that :
l=25 m ,diameter d=14 m
so r=14/2=7 m
We know that the the curved surface area of a cone=πrl
=22/7 x 7 x 25
=25 x 22
=550 m²
Now rate of painting= Rs 210 /100 m²
Hence the painting charges are:
=210x550/100
=21x55
=Rs 1155
Thus the pianting charge of the tomb will be Rs 1155
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https://www.pcreview.co.uk/threads/count-font-color.2584062/
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# count font color
S
#### snowing
Hello,
Does anyone know how I could take the count font color code below, an
add it to a macro so that it will execute at the end of my macro an
displays a message box using "msgbox" with the total amount of cel
with the font color red with out having to ad
"=COUNTBYCOLOR(A1:A10,3,FALSE)" to a cell on the spread sheet.
Also, the range of cells with be different each time.
Thanks,
*****************************************
Function CountByColor(InRange As Range, _
WhatColorIndex As Integer, _
Optional OfText As Boolean = False) As Long
'
' This function return the number of cells in InRange with
' a background color, or if OfText is True a font color,
' equal to WhatColorIndex.
'
Dim Rng As Range
Application.Volatile True
For Each Rng In InRange.Cells
If OfText = True Then
CountByColor = CountByColor - _
(Rng.Font.ColorIndex = WhatColorIndex)
Else
CountByColor = CountByColor - _
(Rng.Interior.ColorIndex = WhatColorIndex)
End If
Next Rng
End Function
You can call this function from a worksheet cell with a formula like
=COUNTBYCOLOR(A1:A10,3,FALSE
B
#### Bob Phillips
MsgBox COUNTBYCOLOUR(Range("A1:A10"),3,TRUE)
--
HTH
Bob Phillips
(replace somewhere in email address with gmail if mailing direct)
C
#### colofnature
To count the number of red cells in column A add this line to your
code:
msgbox
cstr(countbycolor(intersect(activesheet.usedrange,[a:a]),3,False))
Col
S
#### snowing
thanks guys for the help.
Colofnature, how can I have your statement end at the last cell and no
count all cells in column A
If I have from cell A1 to A200 (number of cell will be different eac
time) how can i have if stop counting at A200
B
#### Bob Phillips
msgbox cstr(countbycolor(intersect(activesheet.Range("A1:A200"),3,False))
--
HTH
Bob Phillips
(replace somewhere in email address with gmail if mailing direct)
S
#### snowing
thanks for the reply..
It may be 200 cells sometimes but may only be 150 to 100 cells anothe
time, and i can't tell that ahead of time.
I have a bit of code that counts up the cells and saves the count in
variable. Can i use the variable with the saved count in place of th
number, (example, A + my variable) ?
msgbox cstr(countbycolor(intersect(activesheet.Range("A1
A200"),3,False)
B
#### Bob Phillips
msgbox cstr(countbycolor(intersect(activesheet.Range("A1:A" &
your_var),3,False))
--
HTH
Bob Phillips
(replace somewhere in email address with gmail if mailing direct)
S
#### snowing
Getting a type mismatched error on the 3 .
msgbox cstr(countbycolor(intersect(activesheet.Range("A1: A" &
your_var),3,False)
B
#### Bob Phillips
maybe wrap-around
msgbox cstr(countbycolor(intersect(activesheet _
..Range("A1: A" & your_var),3,False))
--
HTH
Bob Phillips
(replace somewhere in email address with gmail if mailing direct)
S
#### snowing
Nope, same thing.
## Ask a Question
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https://gmatclub.com/forum/scientists-have-determined-that-radon-a-colorless-147033.html?oldest=1
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| 931,709,168
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# Scientists have determined that Radon - a colorless,
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10 Feb 2013, 23:14
Expert's post
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Difficulty:
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Question Stats:
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Scientists have determined that Radon - a colorless, odorless, and tasteless gas - is unstable because of its radioactive nature and continuously decays into Progeny. Since it has been proven that there was no Progeny at the time of earth’s formation, one can accurately estimate earth’s age by determining the increase in Progeny in earth’s atmosphere over the last 200 years and using it to determine how many years it would have taken to reach current levels of Progeny.
The answer to which of the following questions would be most important in determining whether using the method would help in achieving the stated objective?
A. Whether the rate of decay of Radon into Progeny is different during day time and night time?
B. Whether Progeny is stable or gets converted into some other gas at constant rate?
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases?
D. Whether there are methods, better than the method given, to estimate the earth’s age?
E. Whether amount of Radon or Progeny varies from region to region?
[Reveal] Spoiler: OA
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Last edited by egmat on 11 Feb 2013, 19:45, edited 1 time in total.
If you have any questions
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### Show Tags
10 Feb 2013, 23:15
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Hi,
Please find below the official solution:
Understanding the Argument
Conclusion
Earth’s age can be estimated by:
• determining the increase in Progeny in earth’s atmosphere over the last 200 years and
• using it to determine how many years it would have taken to reach current levels of Progeny
Premises
1. Radon gas is unstable because of its radioactive nature and continuously decays into Progeny
2. There was no Progeny at the time of earth’s formation
Prethinking
What is the fundamental assumption?
While making his claim, the author assumes that the rate of change of Progeny over the past 200 years is representative of the rate of change of Progeny over the course of history of earth. If this is not the case, i.e. if the rate of change today is much lower or much higher than what it used to be in the past, then the estimates would either be too high or too low.
You may go to the analysis of answer choices if the above makes sense to you. Otherwise, read below if you need more explanation.
Detailed description of how to Prethink
Prethinking in the given question is extremely difficulty unless one has a basic understanding of data representativeness, which basically means that if a sample data is used to make predictions about a population, then the sample data must be representative of the population data. (fundamental assumption)
Since this is pretty technical definition, let’s take an example to understand this. At the end of this example, we will provide a weaken question where the same principle has been used with a twist in a weaken problem.
Suppose you are given a task to estimate the population of a city. Suppose the city has 100 residential blocks. One way to find out the population can be to go to each and every block and count the number of people in each of the blocks and then totaling them to find the total population of the city.
Another way, and a much easier way is estimate the population rather than calculating the same. To do the same you go to, say, 4 residential blocks and find out their populations and calculate the sum of populations in these residential blocks.
Then, the total population of the city can be estimated by multiplying 25 (i.e. 100/4) with the sum of population in these 4 residential blocks.
Do you think this method will work? What if you chose the 4 residential blocks which happen to be the most populous residential blocks in the city? Would you get a correct estimate of the population in such a case?
The answer is No. You would get an estimate which is much higher than the actual population. Similarly, if you chose the least four populous residential blocks, you would get an estimate which is much below the actual population.
Therefore, to get a correct estimate of the population of the city, you must select 4 residential blocks which have near average populations of all the residential blocks. And if you choose residential blocks in such a way, your sample data (i.e. 4 residential blocks) would be a correct representation of the population and the sample can be called as representative of the population.
Now, let’s apply the same knowledge in the given question. If we are using the increase in the amount of Progeny in the last 200 years to estimate the Earth’s age, then that increase should be representative of the increase that happens every 200 years. This is the assumption required. If the increase in Progeny in the last 200 years is significantly different from such increases in the past, then we would have an incorrect estimate. Therefore, one thing we need to evaluate is whether the increase in Progeny in the last 200 years is representative of such increases since the beginning of the earth.
(A weaken question which uses the same understanding: gasoline-cars-in-papula-country-147427.html)
A. Whether the rate of decay of Radon into Progeny is different during day time and night time? – This is irrelevant. Even if the rates are different during day time from night time, it doesn’t impact the estimate. Incorrect.
B. Whether Progeny is stable or gets converted into some other gas at constant rate? – There are two cases to consider in this option statement:
i. Progeny is stable: In this case, the estimate should work perfectly fine.
ii. Progeny gets converted into some other gas at constant rate –
Since this conversion happens both during the last 200 years and the time since the formation of earth, it doesn’t distorts the estimate. This is because such constant conversion impact the increase in gas in the last 200 years in the same proportion as the increase in gas in the time since the formation of earth. Thus, an estimate of the earth’s age, which would be calculated as below, will not be affected.
Earth’s Age: Total Amount of Progeny in the atmosphere currently*200/Increase in the amount of Progeny in the atmosphere in the last 200 years
Incorrect.
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases? – As we discussed in the Pre-thinking stage, if this given option statement is true, last 200 years of data regarding increase in progeny would not be representative of the time since the formation of Earth. Therefore, we need to find out the answer to this question to know whether the data of increase of Progeny in the last 200 years can be used to estimate the Eath’s age. Correct.
D. Whether there are methods, better than the method given, to estimate the earth’s age? – We are not concerned with an answer to this. Incorrect.
E. Whether amount of Radon or Progeny varies from region to region? – Since we considering the earth’s atmosphere as a whole, even if Progeny varies from region to region, it doesn’t impact the estimate because we are taking into account all the regions. Incorrect.
Hope this helps
Thanks,
Chiranjeev
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Last edited by egmat on 22 Feb 2013, 00:09, edited 1 time in total.
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### Show Tags
11 Feb 2013, 01:22
According to me the OA should be B
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### Show Tags
11 Feb 2013, 03:15
egmat wrote:
Scientists have determined that Radon - a colorless, odorless, and tasteless gas - is unstable because of its radioactive nature and continuously decays into Progeny. Since it has been proven that there was no Progeny at the time of earth’s formation, one can accurately estimate earth’s age by determining the increase in Progeny in earth’s atmosphere over the last 200 years and using it to determine how many years it would have taken to reach current levels of Progeny.
The answer to which of the following questions would be most important in determining whether using the method would help in achieving the stated objective?
A. Whether the rate of decay of Radon into Progeny is different during day time and night time?
B. Whether Progeny is stable or gets converted into some other gas at constant rate?
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases?
D. Whether there are methods, better than the method given, to estimate the earth’s age?
E. Whether amount of Radon or Progeny varies from region to region?
OA will be posted on Tuesday!!
Facts:
* Radon (colorless,odorless and tasteless gas) is unstable because of its radioactive nature and consinuously decays into Progeny
* There was no Progeny at the time of earth's formation
Method :
Determining the increase in Progeny in earth’s atmosphere over the last 200 years and using it to determine how many years it would have taken to reach current levels of Progeny.
Stated objective :
Estimate earth's age
IMO
A. Clearly irrelevant
B. CORRECT, Whether Progeny is stable or gets converted into some other gas at constant rate will help us to know if the method is accurate. If the answer to this question is No then how can we determine the increase in Progeny in earth's atmosphere if Progeny is unstable adn therefore we could NOT estimate the number of years it would have taken to reach current levels of Progeny.
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases is IRRELEVANT since Randon is an example of a gas who undergoes its radioactivity nature but consinously decays into Progeny according to the first fact.
D. INCORRECT , we are requiered to evaluate the method stated Vs the stated objective.
E. Irrelevant since we are concerned with earth's atmosphere .....
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11 Feb 2013, 05:52
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Premise :- Radon decays into Progney, by measuring the increase in progeny over last 200 years age of earth can be determined.
assumption :- the sample data is representative enough, means the average increase in progeny over its lifetime would be the same as the increase in past 200 years.
POE :-
A :- Different decay rate during day time/night time :- since we are concerned with the aggregate increase over the 200 year period this is not correct.
B :- Whether Progney is stable or converted into another gas @constant rate :- This option states that for both the extreme spectrums of a Yes/No answer, you will be able to determine the age of earth. If Progeny is stable then by directly measuring the increase in Progney or if it gets converted into another gas then by measuring the increase in another gas. The answer choice does not fluctuate to extremes and hence is not the correct answer.
C :- changes in atmosphere over last century :- A No answer to the question will state that the decay is constant and the 200 year sample is representative enough. A yes answer to the question will state that the 200 year sample is not representative enough, and hence is the correct option.
D :- other methods :- OFS
E :- Amount of progeny varies from region to region :- similar to option A, this option also segregates decay by region but our scope is for the entire earths atmosphere.
IMO :- C
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11 Feb 2013, 08:21
My analysis
Firstly this is an evaluate the argument type question.
Facts/Premise - Radon is a gas which decays and becomes Progeny.
There was not Progeny when earth was formed.
Because of this reason earth' age can be accurately estimated. How by calculating the increase in Progeny and then determining in how many years this increase happened.
Underlying fact or assumption is - Progeny does not decay or converts into something else because the total amount of progeny is important for calculating the age.
A) OFS the rate of day and night is not relevant
B) Looks relevant and very close to our assumption. Park it for variance test
C) Can be relevant. park it for final analysis.
D) Completely OFS.
E)No relevant the argument is not talking about regional changes.
Now lets analyze C and D
B) moved to one extreme Progeny is not converted to any other gas - conclusion is validated. The proposed method would work.
Other extreme - Progeny is converted to some other gas, but the method is measuring only progeny hence it will miss out on the converted Progeny. Invalidated
B is the correct choice ( but lets look at C too)
C) Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases?
Yes there have been changes - Does not impact the argument, this increase will be factored in the calculation. As for Progeny it is not mentioned if it is also decaying.
No there have not been changes - Again no impact.
So B is the correct choice. Waiting for the explanation from E-gmat.
egmat wrote:
Scientists have determined that Radon - a colorless, odorless, and tasteless gas - is unstable because of its radioactive nature and continuously decays into Progeny. Since it has been proven that there was no Progeny at the time of earth’s formation, one can accurately estimate earth’s age by determining the increase in Progeny in earth’s atmosphere over the last 200 years and using it to determine how many years it would have taken to reach current levels of Progeny.
The answer to which of the following questions would be most important in determining whether using the method would help in achieving the stated objective?
A. Whether the rate of decay of Radon into Progeny is different during day time and night time?
B. Whether Progeny is stable or gets converted into some other gas at constant rate?
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases?
D. Whether there are methods, better than the method given, to estimate the earth’s age?
E. Whether amount of Radon or Progeny varies from region to region?
OA will be posted on Tuesday!!
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### Show Tags
11 Feb 2013, 13:00
imho, B (Whether Progeny is stable... ) is not correct, since in the q. stem we are said that that Radon - a colorless, odorless, and tasteless gas - is indeed unstable
I choose answ C
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### Show Tags
12 Feb 2013, 04:15
I think this question is the replica of an official question.
Strange not much of discussion......
Conclusion: Earth's age can be determined by tracing the development of progenies.
In argument its mentioned that earths age can be found accurately by interpolation. That means the method has some kind of proportionality with the age of earth. Hence option choice B fits the gap.
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases?
<<<<<<<<<<I think option C is a trap, consider the situation , if radioactivity decay has increased but a formula can be devised to calculate the relation between earths age and increase in radioactivity.....>>>>>>
Archit
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12 Feb 2013, 04:16
No post from Sarvana test prep<<<<>>>>>>>>
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14 Feb 2013, 17:59
Hi All,
The Official Answer for this question is Option C.
Dinesh2APR - You are right
Thanks,
Chiranjeev
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### Show Tags
15 Feb 2013, 07:01
Dinesh2Apr wrote:
Premise :- Radon decays into Progney, by measuring the increase in progeny over last 200 years age of earth can be determined.
assumption :- the sample data is representative enough, means the average increase in progeny over its lifetime would be the same as the increase in past 200 years.
POE :-
A :- Different decay rate during day time/night time :- since we are concerned with the aggregate increase over the 200 year period this is not correct.
B :- Whether Progney is stable or converted into another gas @constant rate :- This option states that for both the extreme spectrums of a Yes/No answer, you will be able to determine the age of earth. If Progeny is stable then by directly measuring the increase in Progney or if it gets converted into another gas then by measuring the increase in another gas. The answer choice does not fluctuate to extremes and hence is not the correct answer.
C :- changes in atmosphere over last century :- A No answer to the question will state that the decay is constant and the 200 year sample is representative enough. A yes answer to the question will state that the 200 year sample is not representative enough, and hence is the correct option.
D :- other methods :- OFS
E :- Amount of progeny varies from region to region :- similar to option A, this option also segregates decay by region but our scope is for the entire earths atmosphere.
IMO :- C
HI Dinesh...
I have a doubt , when you say that Assumption is "Sample is representative i.e. it can be interpolated to get a correct idea of the earth's age"
I also thought of same assumption and selected B....Because i feel that B mentions two every important words that can bring contrast when answered Yes or No....
They are:-
1. Stable and 2. Constant rate...
If something does not changes at a constant rate how can you predict the age using the sample data...Another One is Stable, If its not stable than it is difficult to predict an accurate age of the substance.
I agree that Option C mentions something which is valid, but i feel that option B presents stronger reasoning than C.
Can you explain why did you not consider choosing B.
Archit
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15 Feb 2013, 08:39
Archit143 wrote:
Dinesh2Apr wrote:
Premise :- Radon decays into Progney, by measuring the increase in progeny over last 200 years age of earth can be determined.
assumption :- the sample data is representative enough, means the average increase in progeny over its lifetime would be the same as the increase in past 200 years.
POE :-
A :- Different decay rate during day time/night time :- since we are concerned with the aggregate increase over the 200 year period this is not correct.
B :- Whether Progney is stable or converted into another gas @constant rate :- This option states that for both the extreme spectrums of a Yes/No answer, you will be able to determine the age of earth. If Progeny is stable then by directly measuring the increase in Progney or if it gets converted into another gas then by measuring the increase in another gas. The answer choice does not fluctuate to extremes and hence is not the correct answer.
C :- changes in atmosphere over last century :- A No answer to the question will state that the decay is constant and the 200 year sample is representative enough. A yes answer to the question will state that the 200 year sample is not representative enough, and hence is the correct option.
D :- other methods :- OFS
E :- Amount of progeny varies from region to region :- similar to option A, this option also segregates decay by region but our scope is for the entire earths atmosphere.
IMO :- C
HI Dinesh...
I have a doubt , when you say that Assumption is "Sample is representative i.e. it can be interpolated to get a correct idea of the earth's age"
I also thought of same assumption and selected B....Because i feel that B mentions two every important words that can bring contrast when answered Yes or No....
They are:-
1. Stable and 2. Constant rate...
If something does not changes at a constant rate how can you predict the age using the sample data...Another One is Stable, If its not stable than it is difficult to predict an accurate age of the substance.
I agree that Option C mentions something which is valid, but i feel that option B presents stronger reasoning than C.
Can you explain why did you not consider choosing B.
Archit
Hello Archit,
Below is my analysis of option B
Option B:- Whether Progeny is stable or gets converted into some other gas at constant rate.
This choice gives you 2 options for Progeny, for these answer choices the Yes/No will be choosing 1 option over other.
suboption B.1 :- Progeny is stable
Now if Progeny is stable, means Progeny will remain there and not vaporize or change into another form. So you can measure Progeny over the past 200 years and get to the conclusion. This option does not break the conclusion.
suboption B.2 :- Progeny gets converted into other gas at constant rate
In this option since the rate of conversion is constant, you can still devise a formula to reach the conclusion by measuring the amount of Progeny left(or by measuring the change in quantity of the other gas) and factoring for the constant of proportionality. This option also does not break the conclusion.
since neither of the options breaks your conclusion, option B is not the correct answer.
Had this option not included the 'constant rate', this option choice might have been a good contender.
Hope this helps !
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### Show Tags
15 Feb 2013, 11:09
I have little different view
B1:- Progeny is stable
Consider :- Progeny is unstable
That means its not possible to derive the age of the earth.
Consider:- Progeny is unstable
That means its possible to derive the age of earth.
We are getting two polar opposite answer.
Now,
B2 :- Rate of conversion is constant
Consider:- Rate of conversion is not constant
Hence its not possible to derive a formula for interpolation.
Consider:- Rate of conversion is constant
Hence its possible to derive a formula for interpolation.
So if you go by your break up we will get two polar opposite answers.
Even if we take them together
But yes agree to a point the analysis of this one is bit complicated and rather option C is simple, hence can be considered a better answer.
But if go by logic i think both equally hold good chance.
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22 Feb 2013, 00:26
Hi Guys,
The official solution with explanation has been posted. Check it out. Apologies for the delays.
Thanks,
Chiranjeev
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22 Feb 2013, 03:12
egmat wrote:
Hi Guys,
The official solution with explanation has been posted. Check it out. Apologies for the delays.
Thanks,
Chiranjeev
hi Chiranjeev
I think this is an official example from gmac......can you pls post the link to official question...just want to compare......
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22 Feb 2013, 03:19
Archit143 wrote:
egmat wrote:
Hi Guys,
The official solution with explanation has been posted. Check it out. Apologies for the delays.
Thanks,
Chiranjeev
hi Chiranjeev
I think this is an official example from gmac......can you pls post the link to official question...just want to compare......
Archit
Hi Archit,
This is not an official question from GMAC. This one is from the e-GMAT kitchen. All questions posted by us here are created just for the forum. They are not even a part of the course.
Thanks,
Chiranjeev
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### Show Tags
25 Feb 2013, 13:20
egmat wrote:
Archit143 wrote:
egmat wrote:
Hi Guys,
The official solution with explanation has been posted. Check it out. Apologies for the delays.
Thanks,
Chiranjeev
hi Chiranjeev
I think this is an official example from gmac......can you pls post the link to official question...just want to compare......
Archit
Hi Archit,
This is not an official question from GMAC. This one is from the e-GMAT kitchen. All questions posted by us here are created just for the forum. They are not even a part of the course.
Thanks,
Chiranjeev
hi chiranjeev
With all regards to you, I have seen a similar question in an official source......i ll have to look for it exhaustively...but i m 100 % sure that i have seen one...no issues once i find i ll post...
by the way good work chiranjeev..Many claimed experts crumbling.....hahah. its 2 down ..lets see who is the next.
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02 Jul 2015, 10:58
this is similar to the one about rivers filling ocean with salts or something like that. An OG ques
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14 Jul 2016, 08:41
Hello from the GMAT Club VerbalBot!
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14 Jul 2016, 10:12
Scientists have determined that Radon - a colorless, odorless, and tasteless gas - is unstable because of its radioactive nature and continuously decays into Progeny. Since it has been proven that there was no Progeny at the time of earth’s formation, one can accurately estimate earth’s age by determining the increase in Progeny in earth’s atmosphere over the last 200 years and using it to determine how many years it would have taken to reach current levels of Progeny.
The answer to which of the following questions would be most important in determining whether using the method would help in achieving the stated objective?
A. Whether the rate of decay of Radon into Progeny is different during day time and night time?
B. Whether Progeny is stable or gets converted into some other gas at constant rate?
C. Whether there have been any changes in the atmosphere in the past century that have increased the radioactivity decay level of all gases? - As highlighted in the argument, the fact that the test would study the change in atmosphere over the past 200 years makes this the correct ans.
D. Whether there are methods, better than the method given, to estimate the earth’s age?
E. Whether amount of Radon or Progeny varies from region to region?
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Re: Scientists have determined that Radon - a colorless, [#permalink] 14 Jul 2016, 10:12
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# cip cleaning in place - predict cip losses and costs we must know the cip circuit volume. this has...
Post on 08-Mar-2018
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## Documents
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• JohnsonDiversey
CIP Cleaning in place
The circulation of non foaming cleaners without dismantling
the equipment.
An automatic and systematic cleaning of the inner
surfaces of tanks, heat exchangers, pumps, valves
and pipes.
• JohnsonDiversey
CIP properties
Strong and hot solutions can be used. The heat, the chemistry and the mechanics can be sustained long.
The solutions can be reused.
Can be automated and reproducibility is good.
Investment in equipment is high.
The mechanics are not always sufficient
• JohnsonDiversey
• JohnsonDiversey
Flow Rate vs. Flow Velocity
=
..3600
.4
2d
Qv
Where,v = flow velocity meters per secondQ = flow rate m3 per hour = pi (3.1415,) dimensionlessd = inside pipe diameter meters
second 1secondper volume
diameter inside
• JohnsonDiversey
Velocity vs flow
1.5 m/s velocity 2.0 m/s velocityPipe size ID mm
Litres / sec Litres / sec
DN 50 47 2.6 3.5
DN 80 77 6.9 9.3
DN 100 97 11.1 14.8
DN150 147 25.5 33.9
• JohnsonDiversey
Vertical vessel flow requirements - sprayballs
Vertical vessels
For most vessels, the sprayball delivers a uniform
quantity of solution to the upper circumference of the
vessel
Based on soil level, deliver a given quantity of solution
Dont forget about flow OUT of vessels
• JohnsonDiversey
Sprayball Placement
+=2
-180tan D Height Dome Sprayball ofDepth
Where, = angle of coverage, degreesD = diameter of vessel, metersDome height meters
NOTE: This is valid for simplevessels without obstructions.Additional sprayballs may berequired.
Depth of Sprayball
Dome WeldSprayball
Dome Height
140
• JohnsonDiversey
example
15
100 gpm
6 dia.
• JohnsonDiversey
Sprayball pressure
Sprayball pressure is critical
Generally in the range (1.0) 1.5 - 2.5 (3.0) bar
Too little pressure and the vessel walls are not reached
Too much and the spray atomises reducing mechanical
action
Larger sprayballs with larger hole diameters can operate
at higher pressures without atomising.
All sprayballs have specified flow / pressure curves
• JohnsonDiversey
• JohnsonDiversey
Vertical vessel flow requirements - sprayballs
Flow as a function of diameter and soil
QR = required flow rate liters per minute
DT = vessel diameter meters
p = pi (3.1415,) dimensionless
FS = soil factor liters/(meter-minute)
FS = 27 for light soil conditions
FS = 30 for medium soil conditions
FS = 32 for heavy soil conditions
SFTDRQ =
• JohnsonDiversey
• JohnsonDiversey
• JohnsonDiversey
Add impingement to the mechanical action
Generally consume a little less water
Have specific times to wet surfaces and impinge on them dependent
on pressure and gearing
Not very effective on larger vessels under 5 bar pressure
Use similar data to specify as sprayballs
Use manufacturers recommendations
Toftejorg have a computer simulation
program called TRAX - use it
• JohnsonDiversey
CIP Optimizing
CIP optimizing is the process of minimizing the cost inputs of CIP
cleaning
water
effluent
energy
chemical
electrical
heat CO2 production time
• JohnsonDiversey
Optimizing drivers
CIP system design
clean circuits - no dead legs, no flow splits
accurate and non competing instrumentation - conductivity
monitoring
no leaks
CIP program
correct CIP program philosophy
CIP preparation sequence - correct conductivity starting point
tidy CIP fluids interface management - always in lines never in
tanks
correct valve sequencing on monitor signals
defined terminators each CIP step
• JohnsonDiversey
CIP optimizing - circuit volume
To predict CIP losses and costs we must know the CIP circuit volume.
This has nothing to do with the size of the CIP tanks.
It is the amount of liquid held up in the CIP headers and the vessel or line being
cleaned.
To calculate the circuit volume for a line clean we need to know the diameters of
the lines and the length of each line size.
To calculate the circuit volume of a vessel clean we need to know the line
information and the dimensions of the vessel being cleaned.
If there is other processing plant in the CIP circuit, we need to know its volume
too.
• JohnsonDiversey
Vessel Hold-up Volume
Assume a 2 millimeter film thickness
(0.002 m)
Assume a completely wetted surface
Determine internal surface area
Dome
Cylinder
Cone
Dome
Cylinder
Cone
• JohnsonDiversey
Vessel Hold-up Volume
Area of Dome:
Area of Cylinder:
Area of Coneh2
h1
D
2DomeArea r=
2CylinderArea hD=
( )212124ConeArea hDD +=
Dr2
1:NOTE =
• JohnsonDiversey
CIP optimizing - chemical loss management
Liquid loss for an efficient vessel CIP system is about 10% of circuit volume.
Line cleans can be run more efficiently than vessel cleans - as low as 5% loss.
Effective loss management depends on:
Effective Flow meter or conductivity interface detection.
Managing liquid interfaces into pipes not vessels.
When managing liquid changes in vessels the program must be stepped.
New liquid to sprayball chasing old liquid into vessel.
Over scavenge old liquid from vessel into return line.
New liquid into vessel chasing old along return line to interface
detector.
First step should be volumetric and set for each vessel.
• JohnsonDiversey
CIP optimizing - chemical loss management
measured as % of concentrate detergent lost compared to the concentrate
detergent in the CIP circuit volume
concentrate detergent lost is calculated by CIP tank, volume and
concentration, before and after CIP
concentrate detergent in circuit volume calculated as the volume of solution
held in the CIP circuit excluding the CIP tank at the starting concentration
• JohnsonDiversey
The CIP flow is best circulated bypassing the CIP tanks with theheating and chemical dosing in line
Recommended
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# 2 math question help
• Dec 5th 2008, 03:26 AM
juzo
2 math question help
1)
A balloon is rising vertically upward at 22m/s. If jane is standing 45 metres away from where the balloon is lifted, how fast is the balloon flying away from jane at the moment when the balloon is 70metres high from the field?
2)
A cylindrical container , with the top ended is opened, is to be constructed to hold a volume of 1.50m³ . Find the dimensions of the container such that the materials used would be a minimum. Give your answers to 2 decimal places.
• Dec 5th 2008, 04:23 AM
earboth
Quote:
Originally Posted by juzo
...
2)
A cylindrical container , with the top ended is opened, is to be constructed to hold a volume of 1.50m³ . Find the dimensions of the container such that the materials used would be a minimum. Give your answers to 2 decimal places.
Let r denote the radius of the base and h the height of the cylinder. The you know:
$\displaystyle V = \pi\cdot r^2 \cdot h$
and the area of the material which is used to build the cylinder:
$\displaystyle a = \pi r^2 + 2\pi\cdot r \cdot h$
Solve the equation of the volume for h and plug in this term into the equation of a:
$\displaystyle h = \dfrac{V}{\pi r^2}$
$\displaystyle a(r)=\pi r^2 + 2\pi\cdot r \cdot \dfrac{V}{\pi r^2} = \pi r^2 + \dfrac{2V}{r}$
Determine the first derivation and solve a'(r) = 0 for r:
$\displaystyle 2\pi r - \dfrac{2V}{r^2} = 0 ~\implies~r=\sqrt[3]{\dfrac V\pi}$
• Dec 5th 2008, 07:42 PM
juzo
Ooo. Thx. Does any1 noe abt question 1 too???
• Dec 5th 2008, 08:43 PM
Chop Suey
For question number one, let us first list the given. I will assume that y and x corresponds for the vertical and horizontal direction, respectively. I will also assume that Jane is a stationary object ("Jane is standing...").
$\displaystyle \frac{dy}{dt} = 22~m/s$
$\displaystyle y = 70~m$
$\displaystyle \frac{dx}{dt} = 0$
$\displaystyle x = 45~m$
$\displaystyle \frac{dz}{dt} = ?$
$\displaystyle z = \sqrt{x^2+y^2} = \ldots$
Note that the distance between the balloon and Jane is the hypotenuse z of the right-triangle with legs x and y. We are trying to find $\displaystyle \frac{dz}{dt}$.
We've got everything in the bag. Differentiating the pythagorus:
$\displaystyle x^2 + y^2 = z^2$
$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt}=2z\frac{dz}{dt} \implies \frac{dz}{dt} = \frac{\displaystyle x\frac{dx}{dt} + y\frac{dy}{dt}}{z}$
And we're done. I have showed you how to solve it, you do the computations. ;)
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# `y = 2x - tan(x) , (-pi/2, pi/2)` Determine the open intervals on whcih the graph is concave upward or downward.
gsarora17 | (Level 2) Associate Educator
Posted on
`y=2x-tanx`
differentiating,
`y'=2-sec^2(x)`
differentiating again,
`y''=-2sec(x)sec(x)tan(x)`
`y''=-2sec^2(x)tan(x)`
`y''=-2(1/(cos^2(x)))(sin(x)/cos(x))`
`y''=-2sin(x)/(cos^2(x))`
In order to determine the concavity , determine the value of x when y''=0
`(-2sinx)/(cos^2(x))=0`
sinx=0 , so x=0 , pi, 2pi ,.....
Now let us test for concavity in the intervals (-pi/2,0) and (0,pi/2)
y''(-pi/4)=2*(-pi/4)- tan(-pi/4)= -pi/2+1 ( negative)
y''(pi/4)=2*(pi/4)-tan(pi/4)=pi/2-1 (positive)
So, the graph is concave upward in the interval (0,pi/2) and
concave downward in the interval (-pi/2,0)
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Basic Auditing Procedures
View:
Sorted by:
Issuing Reports When GAAP Aren’t Used
Some auditing clients don’t follow Generally Accepted Accounting Principles (GAAP). Some companies — usually smaller, private ones — may use a cash or tax basis instead. Companies that are highly regulated
Testing the Reliability of Documents
Part of an auditor’s job for a client can include verifying company documents. Attestation is a subset of assurance services that focuses on whether your engagement’s subject matter complies with the applicable
Conducting Compliance Audits
The purpose of compliance audits is to see how well a company is following applicable rules, policies, and regulations. For example, as an internal auditor your job may be to see how well various departments
Collect Data to Analyze Your Corporation’s Future Financial Performance
Before you can analyze any of the data that will actually help you project your corporation’s future financial performance, you need to actually collect that data. Thankfully, in the age of the Internet
Determine an Average from Your Financial Data
After you collect all your financial data, you need to figure out what to do with it. You need to do some simple descriptive calculations of statistics and
Determine the Distribution of Financial Data
You can measure the manner in which data is distributed around the average in a few different ways. Obviously, not all the numbers in a data set are going to be exactly the same as the average. Say that
The Role of Probability in Analyzing Financial Data
Probability theory is pretty easy. The total probabilities of an event occurring or not will always equal 100 percent. If you have a 10 percent probability that something may happen, then you have a 90
Find Trends and Patterns in Financial Data
When you review historical financial data, the first thing one should do is look for trends and patterns. If you can identify trends that are occurring and any cyclical patterns that have happened in the
Look at Regression When Analyzing Financial Data
The goal of regression is to look at past data to determine whether there are any variables that are influencing financial movements. This process now typically utilizes very advanced computer programs
Use Statistics and Probability to Make Financial Forecasts
To forecast your finances, you watch for trends, patterns, and relationships, determine the probability of these influencing a particular outcome, and use that to model your forecast.
How to Scan Revenue and Expense Horizons
Recording sales revenue and other income can present some hairy accounting problems. The — accounting rule-making authorities — rank revenue recognition as a major problem area. A good part of the reason
Listings:1-2526-5051-61
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# Thread: [SOLVED] Prove that G has an element of order n
1. ## [SOLVED] Prove that G has an element of order n
If N is a finite normal subgroup of a group G and if G/N contains an element of order n, prove that G contains an element of order n.
This is what I have so far:
Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.
When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?
Thanks for any insights.
2. Originally Posted by Tachyon
This is what I have so far:
Let $\displaystyle Na \in G/N$ have order n for some $\displaystyle a \in G$. Then $\displaystyle (Na)^n = Na^n = e$, so $\displaystyle a^n \in N$. Since N is finite, let $\displaystyle ord(N) = m$. Then $\displaystyle (a^n)^m = e$, so $\displaystyle ord(a)|nm$.
When I pursued this line of reasoning, I was trying to show that $\displaystyle n|ord(a)$, and thus $\displaystyle n = ord(a)$. However, I'm stuck here. Have I made a mistake somewhere, or am I missing something?
Thanks for any insights.
$\displaystyle o(a^n) < \infty$ because $\displaystyle a^n \in N$ and $\displaystyle |N| < \infty.$ let $\displaystyle o(a^n)=k.$ then $\displaystyle (a^k)^n=e$ and thus $\displaystyle o(a^k) < \infty.$ let $\displaystyle o(a^k)=m.$ the claim is that $\displaystyle m=n.$ clearly $\displaystyle m \mid n.$ now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$
let $\displaystyle \frac{km}{n}=\ell.$ then $\displaystyle a^{\ell n}=e.$ thus $\displaystyle k \mid \ell$ and therefore $\displaystyle n \mid m.$
3. Originally Posted by NonCommAlg
now $\displaystyle (Na)^{km}=N$ and hence $\displaystyle n \mid km.$
I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?
4. Originally Posted by Tachyon
I understand that $\displaystyle (Na)^{km} = N$, but could you clarify why $\displaystyle n \mid km$ necessarily follows? If $\displaystyle o(Na) = n$, wouldn't we need $\displaystyle (Na)^{km} = e$ in order to have $\displaystyle n \mid km$?
the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$
5. Originally Posted by NonCommAlg
the identity element of $\displaystyle G/N$ is the coset $\displaystyle N.$ so $\displaystyle (Na)^{km}=N=1_{G/N}.$
D'oh. That's what I deserve for my notation confusing $\displaystyle 1_{G/N}$ and $\displaystyle 1_G$. Thanks for your help; I understand now.
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• Parents and Students
We're approaching the end of a wonderful school year despite the many changes due to cov-19. I have added many practice standards to Saavas Realize that are aligned with our FSA and specific skills tailored to individual students. Some of these skills are labeled with an "i" before the number. Please go back and complete all assignments even the missed ones from the beginning of the school year. Doing these assignments will help raise your grade, and prepare you with the confidence to successfully raise your score on our upcoming 2021 Math FSA assessment. In addition, I've provided FSA standards workbooks to many students prior to Spring Break to practice previously taught skills. Utilize these workbooks. If your child did not recieve one and would like a workbook, let me know and I will send one home.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Week of March 22 thru April 12th Topic 7 Geometry
Class Assignments: In Class and On-Line
Week of November 16th
1. Continuing with Ratios and Rates
2. Lessons 5-5 and 5-6
3. Intensive lessons added... ex. i5-4
5. Ratio/Rate Test Thursday and Friday of this week. Study guide sent home or posted on Teams.
Week of November 2nd
1. Lesson 5-4 Ratio Tables
2. Plotting Ratios on a Coordinate Plane
3. Bell Work Quiz
4.Lesson 5-5 Unit Rates
**Intervention Lessons !3-5 and !5-4
Week of October 26th
1. Lesson 5-2 Equivalent Ratios
2. On line assignments and Math packet
3. Lesson 5-3 Comparing Ratios
**Advanced Classes lessons 6-2 and 6-3
Week of October 19th
1. Review QSMA#1
2. Lesson 5-1 Ratios
Week of October 12th
1. Review LCM, GCF, Adding, Subtracting, Multiplying and Dividing Decimals. Course schedule Intervention Lessons: Lesson 3-5 and Lesson 5-4.
2. QSMA #1 Wednesday and Thursday.
3. Continue to use Study guide provided.
4. Complete all on-line assignments past due.
Week of October 5th
1. Lesson 1-3 Multiplying Decimals
2. Lesson 1-4 Dividing Decimals
**Complete all on-line assignments through course schedule.
Week of September 28th
1.Lesson: Factors and Multiples
2. Lesson 1-3 Multiplying Frctions
3. Bell work Quiz all classes
**Continuous review of Decimals
Week of September 21st
1. Complete all e-textbook lessons from Course Schedule.
2. Decimal Review in class for upcoming Test.
3. Lesson 8-5 Dividing by Decimals Review
** On line students remember you are responsible for completing your class assignments the day they are given on line.
4. All classes Decimal Test Wednesday 23rd or Thursday 24th.
Week of September 14th
1. Make sure all course schedule lessons are complete
2.Lesson 1-2 Dividing Decimals
3. Decimal Review
Week of September 7th
1. Lesson 8-2 Multiplying Decimals
2. Class Work ..Decimal word problems
3. Complete previous assignments from Course Schedule
Week of August 31st
1. Lesson 1-1: Adding and Subtracting Decimals: complete on Pearson Realize
2. Lesson 7-2: Review Adding and Subtracting Decimals: complete on Pearson Realize
3. Bell Work Quiz: Adding and Subtracting Decimals
Week of August 24th
1. Bell Work #1 and #2
2. Topic 1 Readiness Assessment: complete on Pearson Realize
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# Pascal to Pounds per square inch converter (Pa to psi)
Example: 20000 pascal = 2.9007547559372 pounds per square inch Pascal: Pa Pounds per square inch: psi
You may also interested in: Pounds per square inch to Pascal Converter
The online Pascal to Pounds per square inch converter is used to convert the pressure value from Pascal to Pounds per square inch.
#### The Pascal to Pounds per square inch Conversion Formula
You can use the following formula to convert from Pascal to Pounds per square inch:
X(pounds per square inch) = y(pascal) / 6,894.75729
Example: How to convert 30000 pascal to pounds per square inch?
X(pounds per square inch) = 30000(pascal) / 6,894.75729
Answer: 4.3511321339058 pounds per square inch
#### Pascal to Pounds per square inch conversion table
Pascal (Pa) Pounds per square inch (psi)
1 pascal0.00014503773779686 pounds per square inch
2 pascal0.00029007547559372 pounds per square inch
3 pascal0.00043511321339058 pounds per square inch
4 pascal0.00058015095118743 pounds per square inch
5 pascal0.00072518868898429 pounds per square inch
6 pascal0.00087022642678115 pounds per square inch
7 pascal0.001015264164578 pounds per square inch
8 pascal0.0011603019023749 pounds per square inch
9 pascal0.0013053396401717 pounds per square inch
10 pascal0.0014503773779686 pounds per square inch
50 pascal0.0072518868898429 pounds per square inch
100 pascal0.014503773779686 pounds per square inch
250 pascal0.036259434449215 pounds per square inch
500 pascal0.072518868898429 pounds per square inch
1000 pascal0.14503773779686 pounds per square inch
2500 pascal0.36259434449215 pounds per square inch
5000 pascal0.72518868898429 pounds per square inch
10000 pascal1.4503773779686 pounds per square inch
25000 pascal3.6259434449215 pounds per square inch
50000 pascal7.2518868898429 pounds per square inch
100000 pascal14.503773779686 pounds per square inch
250000 pascal36.259434449215 pounds per square inch
500000 pascal72.518868898429 pounds per square inch
1000000 pascal145.03773779686 pounds per square inch
10000000 pascal1450.3773779686 pounds per square inch
Full Pascal to Pounds per square inch conversion table
To know how to convert Pascal to Pounds per square inch, please use our Pascal to Pounds per square inch Converter for free.
#### References
More references for Pascal and Pounds per square inch
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > smumul Structured version Unicode version
Theorem smumul 13036
Description: For sequences that correspond to valid integers, the sequence multiplication function produces the sequence for the product. This is effectively a proof of the correctness of the multiplication process, implemented in terms of logic gates for df-sad 12994, whose correctness is verified in sadadd 13010. Outside this range, the sequences cannot be representing integers, but the smul function still "works". This extended function is best interpreted in terms of the ring structure of the 2-adic integers. (Contributed by Mario Carneiro, 22-Sep-2016.)
Assertion
Ref Expression
smumul bits smul bits bits
Proof of Theorem smumul
Dummy variable is distinct from all other variables.
StepHypRef Expression
1 bitsss 12969 . . . . . 6 bits
2 bitsss 12969 . . . . . 6 bits
3 smucl 13027 . . . . . 6 bits bits bits smul bits
41, 2, 3mp2an 655 . . . . 5 bits smul bits
54sseli 3330 . . . 4 bits smul bits
65a1i 11 . . 3 bits smul bits
7 bitsss 12969 . . . . 5 bits
87sseli 3330 . . . 4 bits
98a1i 11 . . 3 bits
10 simpll 732 . . . . . . . . . . . 12
11 simplr 733 . . . . . . . . . . . 12
12 simpr 449 . . . . . . . . . . . . 13
13 1nn0 10268 . . . . . . . . . . . . . 14
1413a1i 11 . . . . . . . . . . . . 13
1512, 14nn0addcld 10309 . . . . . . . . . . . 12
1610, 11, 15smumullem 13035 . . . . . . . . . . 11 bits ..^ smul bits bits
1716ineq1d 3527 . . . . . . . . . 10 bits ..^ smul bits ..^ bits ..^
18 2nn 10164 . . . . . . . . . . . . . . . 16
1918a1i 11 . . . . . . . . . . . . . . 15
2019, 15nnexpcld 11575 . . . . . . . . . . . . . 14
2110, 20zmodcld 11298 . . . . . . . . . . . . 13
2221nn0zd 10404 . . . . . . . . . . . 12
2322, 11zmulcld 10412 . . . . . . . . . . 11
24 bitsmod 12979 . . . . . . . . . . 11 bits bits ..^
2523, 15, 24syl2anc 644 . . . . . . . . . 10 bits bits ..^
2617, 25eqtr4d 2477 . . . . . . . . 9 bits ..^ smul bits ..^ bits
27 inass 3536 . . . . . . . . . . . . 13 bits ..^ ..^ bits ..^ ..^
28 inidm 3535 . . . . . . . . . . . . . 14 ..^ ..^ ..^
2928ineq2i 3525 . . . . . . . . . . . . 13 bits ..^ ..^ bits ..^
3027, 29eqtri 2462 . . . . . . . . . . . 12 bits ..^ ..^ bits ..^
3130oveq1i 6120 . . . . . . . . . . 11 bits ..^ ..^ smul bits ..^ bits ..^ smul bits ..^
3231ineq1i 3524 . . . . . . . . . 10 bits ..^ ..^ smul bits ..^ ..^ bits ..^ smul bits ..^ ..^
33 inss1 3546 . . . . . . . . . . . 12 bits ..^ bits
341a1i 11 . . . . . . . . . . . 12 bits
3533, 34syl5ss 3345 . . . . . . . . . . 11 bits ..^
362a1i 11 . . . . . . . . . . 11 bits
3735, 36, 15smueq 13034 . . . . . . . . . 10 bits ..^ smul bits ..^ bits ..^ ..^ smul bits ..^ ..^
3834, 36, 15smueq 13034 . . . . . . . . . 10 bits smul bits ..^ bits ..^ smul bits ..^ ..^
3932, 37, 383eqtr4a 2500 . . . . . . . . 9 bits ..^ smul bits ..^ bits smul bits ..^
4020nnrpd 10678 . . . . . . . . . . 11
4110zred 10406 . . . . . . . . . . . 12
42 modabs2 11306 . . . . . . . . . . . 12
4341, 40, 42syl2anc 644 . . . . . . . . . . 11
44 eqidd 2443 . . . . . . . . . . 11
4522, 10, 11, 11, 40, 43, 44modmul12d 11311 . . . . . . . . . 10
4645fveq2d 5761 . . . . . . . . 9 bits bits
4726, 39, 463eqtr3d 2482 . . . . . . . 8 bits smul bits ..^ bits
4810, 11zmulcld 10412 . . . . . . . . 9
49 bitsmod 12979 . . . . . . . . 9 bits bits ..^
5048, 15, 49syl2anc 644 . . . . . . . 8 bits bits ..^
5147, 50eqtrd 2474 . . . . . . 7 bits smul bits ..^ bits ..^
5251eleq2d 2509 . . . . . 6 bits smul bits ..^ bits ..^
53 elin 3516 . . . . . 6 bits smul bits ..^ bits smul bits ..^
54 elin 3516 . . . . . 6 bits ..^ bits ..^
5552, 53, 543bitr3g 280 . . . . 5 bits smul bits ..^ bits ..^
56 nn0uz 10551 . . . . . . . . 9
5712, 56syl6eleq 2532 . . . . . . . 8
58 eluzfz2b 11097 . . . . . . . 8
5957, 58sylib 190 . . . . . . 7
6012nn0zd 10404 . . . . . . . 8
61 fzval3 11211 . . . . . . . 8 ..^
6260, 61syl 16 . . . . . . 7 ..^
6359, 62eleqtrd 2518 . . . . . 6 ..^
6463biantrud 495 . . . . 5 bits smul bits bits smul bits ..^
6563biantrud 495 . . . . 5 bits bits ..^
6655, 64, 653bitr4d 278 . . . 4 bits smul bits bits
6766ex 425 . . 3 bits smul bits bits
686, 9, 67pm5.21ndd 345 . 2 bits smul bits bits
6968eqrdv 2440 1 bits smul bits bits
Colors of variables: wff set class Syntax hints: wi 4 wb 178 wa 360 wceq 1653 wcel 1727 cin 3305 wss 3306 cfv 5483 (class class class)co 6110 cr 9020 cc0 9021 c1 9022 caddc 9024 cmul 9026 cn 10031 c2 10080 cn0 10252 cz 10313 cuz 10519 crp 10643 cfz 11074 ..^cfzo 11166 cmo 11281 cexp 11413 bitscbits 12962 smul csmu 12964 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1556 ax-5 1567 ax-17 1627 ax-9 1668 ax-8 1689 ax-13 1729 ax-14 1731 ax-6 1746 ax-7 1751 ax-11 1763 ax-12 1953 ax-ext 2423 ax-rep 4345 ax-sep 4355 ax-nul 4363 ax-pow 4406 ax-pr 4432 ax-un 4730 ax-inf2 7625 ax-cnex 9077 ax-resscn 9078 ax-1cn 9079 ax-icn 9080 ax-addcl 9081 ax-addrcl 9082 ax-mulcl 9083 ax-mulrcl 9084 ax-mulcom 9085 ax-addass 9086 ax-mulass 9087 ax-distr 9088 ax-i2m1 9089 ax-1ne0 9090 ax-1rid 9091 ax-rnegex 9092 ax-rrecex 9093 ax-cnre 9094 ax-pre-lttri 9095 ax-pre-lttrn 9096 ax-pre-ltadd 9097 ax-pre-mulgt0 9098 ax-pre-sup 9099 This theorem depends on definitions: df-bi 179 df-or 361 df-an 362 df-3or 938 df-3an 939 df-xor 1315 df-tru 1329 df-had 1390 df-cad 1391 df-ex 1552 df-nf 1555 df-sb 1660 df-eu 2291 df-mo 2292 df-clab 2429 df-cleq 2435 df-clel 2438 df-nfc 2567 df-ne 2607 df-nel 2608 df-ral 2716 df-rex 2717 df-reu 2718 df-rmo 2719 df-rab 2720 df-v 2964 df-sbc 3168 df-csb 3268 df-dif 3309 df-un 3311 df-in 3313 df-ss 3320 df-pss 3322 df-nul 3614 df-if 3764 df-pw 3825 df-sn 3844 df-pr 3845 df-tp 3846 df-op 3847 df-uni 4040 df-int 4075 df-iun 4119 df-disj 4208 df-br 4238 df-opab 4292 df-mpt 4293 df-tr 4328 df-eprel 4523 df-id 4527 df-po 4532 df-so 4533 df-fr 4570 df-se 4571 df-we 4572 df-ord 4613 df-on 4614 df-lim 4615 df-suc 4616 df-om 4875 df-xp 4913 df-rel 4914 df-cnv 4915 df-co 4916 df-dm 4917 df-rn 4918 df-res 4919 df-ima 4920 df-iota 5447 df-fun 5485 df-fn 5486 df-f 5487 df-f1 5488 df-fo 5489 df-f1o 5490 df-fv 5491 df-isom 5492 df-ov 6113 df-oprab 6114 df-mpt2 6115 df-1st 6378 df-2nd 6379 df-riota 6578 df-recs 6662 df-rdg 6697 df-1o 6753 df-2o 6754 df-oadd 6757 df-er 6934 df-map 7049 df-pm 7050 df-en 7139 df-dom 7140 df-sdom 7141 df-fin 7142 df-sup 7475 df-oi 7508 df-card 7857 df-cda 8079 df-pnf 9153 df-mnf 9154 df-xr 9155 df-ltxr 9156 df-le 9157 df-sub 9324 df-neg 9325 df-div 9709 df-nn 10032 df-2 10089 df-3 10090 df-n0 10253 df-z 10314 df-uz 10520 df-rp 10644 df-fz 11075 df-fzo 11167 df-fl 11233 df-mod 11282 df-seq 11355 df-exp 11414 df-hash 11650 df-cj 11935 df-re 11936 df-im 11937 df-sqr 12071 df-abs 12072 df-clim 12313 df-sum 12511 df-dvds 12884 df-bits 12965 df-sad 12994 df-smu 13019
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# OUseful.Info, the blog…
Trying to find useful things to do with emerging technologies in open education
## The Basics of Betting as a Way of Keeping Score…
Another preparatory step before I start learning about stats in the context of Formula One… There are a couple of things I’m hoping to achieve when I actually start the journey: 1) finding ways of using stats to help to pull out patterns and events that are interesting from a storytelling or news perspective; 2) seeing if I can come up with any models that help forecast or predict race winners or performances over a race weekend.
There are a couple of problems I can foresee (?!) when it comes to the predictions: firstly, unlike horseracing, there aren’t that many F1 races each year to test the predictions against. Secondly, how do I even get a baseline start on the probabilities that driver X or team Y might end up on the podium?
It seems to me as if betting odds provide one publicly available “best guess” at the likelihood of any driver winning a race (a range of other bets are possible, of course, that give best guess predictions for other situations…) Having had a sheltered life, the world of betting is completely alien to me, so here’s what I think I’ve learned so far…
Odds are related to the anticipated likelihood of a particular event occurring and represent the winnings you get back (plus your stake) if a particular event happens. So 2/1 (2 to 1) fractional odds say: if the event happens, you’ll get 2 back for every 1 you placed, plus your stake back. If I bet 1 unit at 2/1 and win, I get 3 back: my original 1 plus 2 more. If I bet 3, I get 9 back: my original 3 plus 2 for every 1 I placed. Since I placed 3 1s, I get back 3 x 2 = 6 in winnings. Plus my original 3, which gives me 9 back on 3 staked, a profit of 6.
Odds are related (loosely) to the likelihood of an event happening. 2/1 odds represent a likelihood (probability) that an event will happen (1/3) = 0.333… of the time (to cur down on confusion between fractional odds and fractional probabilities, I’ll try to remember to put the fractional probabilities in brackets; so 1/2 is fractional odds of 2 to 1 on, and (1/2) is a probability of one half). To see how this works, consider an evens bet, fractional odds of 1/1, such as someone might make for tossing a coin. The probability of getting heads on a single toss is (1/2); the probability of getting tails is also (1/2). If I’m giving an absolutely fair book based on these likelihoods, I’d offer you even odds that you get a head, for example, on a single toss. After all, it’s (fifty/fifty) (fifty per cent chance either way) of whether a heads or tails will land face up. If there are three equally possible outcomes, (1/3) each, then I’d offer 2/1. After all, it’s twice as likely that something other than the single outcome you called would come up. If there are four possible outcomes, I’d offer 3/1, because it’s likely (if we played repeatedly) that three times out of four, you’d be wrong. So every three times out of four you’d lose and I’d take your stake. And on the fourth go, when you get it right, I give you your stake back for that round plus three for winning, so over all we’d be back where we started.
Decimal odds are a way of describing the return you get on a unit stake. So for a 2/1 bet, the decimal odds are three. For a 4/1 bet they’d be 5. For an N/1 bet they’d be 1+N. For an 1/2 (two to one on?) bet they’d be 1.5, for a 1/10 bet they’d be 1.1. So for a 1/M bet, 1+1/M. Generally, for an N/M bet, decimal odds are 1+N/M.
Decimal odds give an easy way in to calculating the likelihood of an event. Decimal odds of 3, (that is, fractional odds 2/1), describe an event that will happen (1/3) of the time in a fair game. That is (1/(decimal odds)) of the time. For fractional odds of N/M, you expect the event to happen with probability (1/(1+N/M))
In a completely fair book (?my phrase), the sum of the odds should lead to the summed probability of all possible events happening of 1. Bookmakers right the odds in their favour though, so the summed probabilities on a book will add up to more than 1 – this represents the bookmaker’s margin. If you’re betting on the toss of a coin with a bookie, they may offer you 99/100 for heads, evens for tails. If you play 400 games and bet 300 heads and 200 tails, winning 100 of each, you’ll overall stake 400, win 100 (plus 100 back) on tails along with 99 (plus 100 original stake) on heads. That is, you’ll have staked 400 and got back 399. The bookie will be 1 up overall. The summed probabilities add up to more than 1, since (1/2) + (1/(1+99/100)) = (0.5 + ~0.5025) > 1.
One off bets are no basis for a strategy. You need to bet regularly. One way of winning is to follow a value betting strategy where you place bets on outcomes that you predict are more likely than the odds you’re offered. This is counter to how the bookie works. If a bookie offers you fractional odds of 3/1 (expectation that the event will happen (1/4) of the time), and you have evidence that suggests it will happen (1/3) of the time (decimal odds of 3, fractional odds 2/1) then it’s worth your while repeatedly accepting the bet. After all, if you play 12 rounds, you’ll wager 12, and win on 12/3=4 occasions, getting 4 back (3 + your stake) each time, to give you a net return of 4 x 4 – 12 = 16 – 12 = +4. If the event had happened at the bookie’s predicted likelihood of 1/4 of the time, you would have got back ( 12/4 ) * 4 – 12 = +0 overall.
I’ve tried to do an R script to explore this:
```#My vocabulary may be a bit confused herein
#Corrections welcome in the comments from both statisticians and gamblers;-)
#The offered odds
price=4 #3/1 -> 3+1 That is, the decimal odds on fractional odds of 3/1
odds=1/price
#The odds I've predicted
myodds=1/3 #2/1 -> 1/(2+1)
#The number of repeated trials in the game
trials=10000
#The amount staked
bet=1
#The experiment that we'll run trials number of times
expt=function(trials,odds,myodds,bet){
#trial sets a uniform random number in ranger 0..1
df=data.frame(trial=runif(1:trials))
#The win condition happens at my predicted odds, ie if trial value is less than my odds
#So if my odds are (1/4) = 0.25, a trial value in range 0..0.25 counts as a win
# (df\$trial<myodds) is TRUE if trial < myodds, which is cast by as.integer() to value 1
# If (df\$trial<myodds) is FALSE, as.integer() returns 0
df\$win=as.integer(df\$trial<myodds)
df\$bet=bet
#The winnings are calculated at the offered odds and are net of the stake
#The df\$win/odds = 1/odds = price (the decimal odds) on a win, else 0
#The actual win is the product of the stake (bet) and the decimal odds
#The winnings are the return net of the initial amount staked
#Where there is no win, the winnings are a loss of the value of the bet
df\$winnings=df\$bet*df\$win/odds-df\$bet
df
}
df.e=expt(trials,odds,myodds,bet)
#The overall net winnings
sum(df.e\$winnings)
#If myodds > odds, then I'm likely to end up winning on a value betting strategy
#A way of running the experiment several times
#There are probably better R protocols for doing this?
runs=10
df.r=data.frame(s=numeric(),v=numeric())
for (i in 1:runs){
e=expt(trials,odds,myodds,bet)
df.r=rbind(df.r,data.frame(s=sum(e\$winnings),v=sd(e\$winnings)))
}
#It would be nice to do some statistical graphics demonstrations of the different distributions of possible outcomes for different regimes. For example:
## different combinations of odds and myodds
## different numbers of trials
## different bet sizes```
There are apparently also “efficient” ways of working out what stake to place (the “staking strategy”). The value strategy gives you the edge to win, long term, the staking strategy is how you maximise profits. See for example Horse Racing Staking and Betting: Horse racing basics part 2 or more mathematical treatments, such as The Kelly Criterion or Statistical Methodology for Profitable Sports Gambling. See also the notion of “betting rules”, eg A statistical development of fixed odds betting rules in soccer.
There is possibly some mileage to be had in getting to grips with R modeling using staking strategy models as an opening exercise, along with statistical graphical demonstrations of the same, but that is perhaps a little off topic for now…
To recap then, what I think I’ve learned is that we can test predictions against the benchmark of offered odds. The offered odds in themselves give us a ballpark estimate of what the (expert) bookmakers, as influenced by the betting/prediction market, expect the outcome of an event to be. Note that the odds are rigged to give summed probabilities over a range of events happening to be greater than 1, to build in a profit margin (does it have a proper name?) for the bookmaker. If we have a prediction model that appears to offer better odds on an event than the odds that are actually offered, and we believe in our prediction, we can run a value betting strategy on that basis and hopefully come out, over the long term, with a profit. The size of the profit is in part an indicator of how much more accurate our model is as a predictive model than the expert knowledge and prediction market basis that is used to set the bookie’s odds.
PS Re: the bookie’s profit, seems that this is called the overround or vigorish. The paper Forecasting sports tournaments by ratings of (prob)abilities: A comparison for the EURO 2008 makes clear the relationship between the bookies’ cut and the odds:
One thing that immediately springs to mind is to look at what sort of overround applies to different bookmakers around different sorts of F1 bets, and whether this is related to the apparent forecast accuracy of the odds offered, at least in ranking terms? (See the comments for a couple of links to papers on forecast accuracy of sports betting odds.)
PPS FWIW, as and when I come across R libraries to access bookmaker APIs, I’ll add them here:
- Betfair R package – access Betair API; another package (CRAN): Betfairly
Written by Tony Hirst
January 16, 2013 at 1:11 am
Posted in Anything you want, Rstats
Tagged with
### 11 Responses
1. You might also be interested in “proper scoring rules” as a better way of evaluating probabilistic forecasts once you have the outcomes. Betting scores you against the bookie; proper scoring rules score you against the outcomes.
dougclow
January 16, 2013 at 7:00 am
2. Example of proper scoring rules used to evaluate predictions from US election pundits: http://appliedrationality.org/2012/11/09/was-nate-silver-the-most-accurate-2012-election-pundit/
dougclow
January 16, 2013 at 7:46 am
3. @Doug Yes, you’re right of course… And I didn’t mention confidence limits on predictions either (I wonder if the margin, or whatever the term is, that the bookies build in to get summed probabilities > 1 is a reflection of their confidence in a reliable way?)
This whole journey is going to be very much baby steps for me… and I’m not totally sure what sorts of things I’ll find I need to cover to make sense of it all along the way (typical uncourse, eh?!)
Which is to say – all comments will be very much appreciated and could well guide what I look at and the way I wend my way through it all :-)
Tony Hirst
January 16, 2013 at 9:37 am
4. There’s an interesting model which i guess could be adapted to F1, in a paper titled “Probabilistic Modelling in Muliti-Competitor Games” by CJ Farmer. Might be worth a look.
Anyway, i’m enjoying your posts on F1 and am hoping to learn some R skills (complete novice atm). gl.
Al Mac
January 16, 2013 at 9:51 pm
• Thanks for that link… will check it out. Hope to get the first proper post in the series out over the weekend…:-)
All ideas welcome:-)
Tony Hirst
January 16, 2013 at 10:04 pm
• I’ve always called the bookmakers margin the ‘overround’ – i can’t recall it being called anything else in my reading on gambling, except in America where they talk about the ‘vigorish’ (or ‘vig’) which i assume to be the same.
Al Mac
January 16, 2013 at 10:30 pm
5. I would suggest the book Precision: Statistical and Mathematical Methods in Horse Racing by CX Wong.
I believe some of the statistical approaches Wong recommends would apply.
RD
Rogers Davis
January 17, 2013 at 3:07 pm
6. A thought on how bookies set odds (nothing to do with R).
The odds offered by a bookmaker are only vaguely related to the probability of an outcome. They are strongly correlated with the payout of an outcome. Basically the bookmaker keeps track of how much he has to pay for outcome A and adjusts his odds accordingly. He has to a) be able to pay all the bettors and b) still make a profit.
The amount of money bet on an outcome is correlated (roughly) with the number of people betting on the outcome. The “smart money” might be on outcome A, but if the sentimental favourite is B then B will have the short odds.
My point is: if you’re using the odds as a signal, expect a lot of noise. as well.
Cheers,
Daggles.
Dale
January 21, 2013 at 12:05 am
7. [...] a betting point of view, (eg Getting Started with F1 Betting Data and The Basics of Betting as a Way of Keeping Score…) it possibly also makes sense to look at the correlation between the P3 times and the qualifying [...]
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Convert {3} to ' , MOA | equilateral triangle to minutes
# angle conversion
## Amount: 1 equilateral triangle ({3}) of angle Equals: 3,600.00 minutes (' , MOA) in angle
Converting equilateral triangle to minutes value in the angle units scale.
TOGGLE : from minutes into equilateral triangles in the other way around.
## angle from equilateral triangle to minute Conversion Results:
### Enter a New equilateral triangle Amount of angle to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other angle measuring units - complete list.
Conversion calculator for webmasters.
## Angles
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.
Convert angle measuring units between equilateral triangle ({3}) and minutes (' , MOA) but in the other reverse direction from minutes into equilateral triangles.
conversion result for angle: From Symbol Equals Result To Symbol 1 equilateral triangle {3} = 3,600.00 minutes ' , MOA
# Converter type: angle units
This online angle from {3} into ' , MOA converter is a handy tool not just for certified or experienced professionals.
First unit: equilateral triangle ({3}) is used for measuring angle.
Second: minute (' , MOA) is unit of angle.
## 3,600.00 ' , MOA is converted to 1 of what?
The minutes unit number 3,600.00 ' , MOA converts to 1 {3}, one equilateral triangle. It is the EQUAL angle value of 1 equilateral triangle but in the minutes angle unit alternative.
How to convert 2 equilateral triangles ({3}) into minutes (' , MOA)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
3600 * 2 (or divide it by / 0.5)
QUESTION:
1 {3} = ? ' , MOA
1 {3} = 3,600.00 ' , MOA
## Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing equilateral triangles and minutes ( {3} vs. ' , MOA ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with angle's values and properties.
International unit symbols for these two angle measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for equilateral triangle is:
{3}
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for minute is:
' , MOA
### One equilateral triangle of angle converted to minute equals to 3,600.00 ' , MOA
How many minutes of angle are in 1 equilateral triangle? The answer is: The change of 1 {3} ( equilateral triangle ) unit of angle measure equals = to 3,600.00 ' , MOA ( minute ) as the equivalent measure for the same angle type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in {3} - equilateral triangles for angle amount, the rule is that the equilateral triangle number gets converted into ' , MOA - minutes or any other angle unit absolutely exactly.
Conversion for how many minutes ( ' , MOA ) of angle are contained in a equilateral triangle ( 1 {3} ). Or, how much in minutes of angle is in 1 equilateral triangle? To link to this angle equilateral triangle to minutes online converter simply cut and paste the following.
The link to this tool will appear as: angle from equilateral triangle ({3}) to minutes (' , MOA) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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http://www.learnerswings.com/2014/10/a-low-cost-333-led-cube-using-555-timer.html
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## Blogroll
Powered by Blogger.
## Animated Circuit of a Low Cost 3*3*3 LED Cube using 555 Timer IC (Part 1)
by realfinetime | in LED Cube at 09:41
We had controlled 3*3*3 LED cube using arduino mega in past few blogs. But, beginners may not be familiar with arduino and programming. Also, arduino is costly. To solve this problem, we are using 555 timer IC to control 3*3*3 LED cube. There are limitations for the number of patterns that can be created using 555 timer. A simple pattern created in a 3*3*3 LED cube using 555 timer IC is shown below. All the LEDs of LED cube will turn on and turn off in a particular time delay. This time delay can be controlled by changing the value of 1 MOhm pots. Construction of 3*3*3 LED cube using locally available materials is given in these posts.
Pin out diagram of 555 timer is given below. 555 has eight legs arranged as shown below.
Next is working of circuit. Working of circuit is simple. Here, 555 works in astable mode. Circuit is clearly explained in next part.
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6. 555 ÷ 3
## IMPORTANT NOTICE
All the circuits, published in this blog is only after testing and getting proper results in my private lab. When you try these circuits, you should check the supply voltage, polarity of components, presence of childrens nearby and shorts in the circuits. This website will not be responsible for any harm happened to you or your components caused by your carelessness.
## For More Electronic Tips
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## What is Transition Method in Electric Traction?
The methods of changing-over the connections from one grouping to another are known as transition method. The change-over from full series (when the two motors are connected in series and voltage across each motor is equal to half of the supply voltage) to first parallel (when the two motors are connected in parallel and part of the starting resistance is reinserted in the circuit to cause the voltage drop equal to half of the supply voltage) is possible by three methods.
1. Open-circuit transition
2. Shunt transition and
3. Bridge transition.
### 1. Open-Circuit Transition
The various steps involved in open-circuit transition are shown in Fig. 13.8. In open-circuit transition first the series connection between the motors is disconnected and power is switched off with full load current. Then some of the starting resistance is reinserted in the circuit, the two motors are connected in parallel and supply is restored. It causes interruption of torque, and therefore, its use is limited to very small units.
### 2. Shunt Transition
In this transition method first the motors are run up and brought to full series position by cutting out the external resistance gradually. Then some resistance is reinserted, one of the motor is short circuited, one end of the short circuited motor is opened and finally connected across in such a way that motors are placed in first parallel. The series resistance is now gradually reduced to zero and the motors are placed in full parallel. There is a jerk in this system as one motor is shorted and ceases to act and then other jerk when it is reinserted. This method is employed in tramways, industrial locomotives and main line locomotives. It is preferred for voltages above 600 volts.
### 3. Bridge Munition
In this transition method the starting resistance is split up into two equal parts. The motors and starting resistances are connected in series, as shown in Fig. 13.10 (first series). The starting resistances are gradually cut out and the motors come in full series, as shown in the figure. In the transition each motor have a section of the starting resistance in parallel with it and two such combinations are put in series with each other, then the bridge link is removed and the motors are placed in first parallel. Now the external resistance is gradually reduced to zero and the motors come in full parallel.
The advantage of the bridge transition method is that during transition, the motors always remain connected to the supply and as the resistances are so adjusted that the value of current remains the same, the torque remains constant, and therefore, uniform acceleration is obtained without causing inconvenience to the passengers. This method is used for railway traction. In case this method is employed for voltage above 600 V, double contacts at series links are required in order to avoid heavy arcing due to heavy current.
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## April 09, 2007
### Cubic Terms Make Smart People Bankrupt
I recently came across the paper Do you have to be smart to be rich? The impact of IQ on wealth, income and financial distress by J L Zagorsky (Intelligence, in press). The subject is interesting: does higher intelligence increase income and saved wealth, and does it protect from making bad financial decisions? Zagorsky reaches the counterintuitive conclusion that it does not improve overall wealth despite increasing income (the income part was known before), and that there is a nonlinear relationship between IQ and being in financial distress: being smarter does not reliably make you economically wiser.
Data from table 2, showing median income and net worth. The medians are at least increasing!
These conclusions would have been good if they had not been marred by "machine thinking". There was frequent references to which piece of software that was being used, as if this would be relevant when the statistical methods were stated. A number of regressions were done using multiple methods, for the stated purpose of dealing with the skewedness of financial data. But if financial data are too skew, wouldn't log data work better? I got the impression of the author simply using the entire toolbox because it was there. Using several different kinds of regressions doesn't give much information to compare with and increases the chances of spurious significance. A better approach might be something along the lines of Sala-I-Martin's "millions of regressions" method where the distribution of coefficients of a particular variable is analysed when different sets are included in the regression. This gets around the risk of handpicking explanatory variables.
The most irritating case was the financial distress analysis where a third degree polynomial was fitted because: "First, by using Excel's chart trend-line function, a polynomial of order 3 visually best fits the data series shown in Table 5." Here is the data, I leave it to the viewer to determine whether this really looks like a third degree curve:
"Second, a common way of ranking competing logistic models is comparing their Akaike's information criteria (AIC) (Akaike, 1974). The model with the smallest AIC fits the data best. Starting from a linear model the AIC falls until IQ cubed is added, but then it rises sharply once IQ raised to the fourth power is included." Here the author does not explain how AIC was applied, but given the skewness of financial data that does not appear to be controlled, one cannot assume normal errors and that may quite possibly give skewed results too if the software assumes normality.
The resulting conclusion that high IQ people may be more likely to suffer financial trouble seems to be due to the cubic term. Zagorsky tries a number of explanations:
"One explanation is that higher IQ score individuals might
be busier and less focused on routines like paying bills. Another explanation is those with a higher IQ score might lead a life-style that is closer to the financial precipice because they feel they are smart enough to calculate and understand all relevant factors. A third explanation is that smarter people have a better memory and are more likely to remember mistakes."
There is probably some truth to this. But a far simpler explanation is that it is a case of overfitting.
A fitted high degree polynomial will rush towards (positive or negative) infinity outside the interval of points it has been fitted to. The financial data seems to be mostly in the interval 75-125 (as would be expected from a normal population) but is extrapolated up to 140 in table 7. No wonder the risks start increasing!
Here is a toy example, where I simulated 1000 individuals with IQs 75-125, with a trouble probability that was a simple parabolic curve plus noise. I fitted a linear, quadratic and cubic curve to them.
The blue cubic curve suffers from clear overfitting. Zagorsky's regression is admittedly different, but the evidence for a need of third degree polynomials does not look that strong.
In the end, this paper is a good example of why we should not let our software do our scientific thinking. Research is not just about getting a dataset, plugging it into a statistical model and coming up with an explanation of why the result is plausible.
Still, what is not enough for a scientific paper may be enough for a blog. I took the WORDSUM data from the GSS as a proxy for intelligence, and plotted the frequency distribution of INCOME.
There seems to be a pretty solid link between having more than 4-5 points of WORDSUM and getting a better income. Unfortunately I did not find any variables useful to estimate savings, so we cannot check the most interesting issue in Zagorsky's paper. Plotting whether people feel they are satisfied with their financial situation showed a clear trend:
Being above average seems to make you more satisfied. maybe this is just because of the higher income, but I would suspect savings do play a role.
There are variables representing economic trouble, and when summed together and plotted with WORDSUM we get this:
Not much evidence for more trouble at the high end, but it looks like some more financial distress in the lower mid part of the intelligence gaussian. This is where there are enough people to get some data, they have reasonably large incomes, and they might be stupid from time to time. There is also a pretty strong anticorrelation (-0.38) between WORDSUM and thinking that there is no point in planning, that one has little control over bad events and good events are mostly luck (I added BADBRKS, LITCNTRL, MOSTLUCK, NOPLAN). That kind of fatalism and external locus of control seems pretty likely to predispose to financial troubles and low saving.
Posted by Anders3 at April 9, 2007 02:02 PM
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https://arbital.greaterwrong.com/p/googolplex?l=42l
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# A googolplex
The number $$10^{10^{100}}$$ is named “a googolplex.” That is, a googolplex is the number represented (in decimal notation) as a 1 followed by a a googol zeroes, i.e., $$10^{googol}$$, which is $$10^{10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}.$$
Parents:
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https://brainly.in/question/322909
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# Is the relation R in the set A={1,2,3} given by R={(1,1),(2,2),(1,2),(2,3)} is reflexive?
2
by anudeepc24
2016-04-13T11:02:48+05:30
Yes
it is reflexive and also symmetric
2016-04-13T12:30:58+05:30
Given
R={(1,1),(2,2),(1,2),(2,3)} is not a reflexive relation.
(reason)
R is said to be reflexive, every element of set A is related to the same set .
The given relation does not contain (3,2) and (3,1).
The below relation is a reflexive:
R={(1,1),(2,2),(1,2),(2,3),(3,2),(3,1)} .
mark as brainliest
how is this reason valid
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# Relation of two numbers in java
• on February 4, 2019
• Likes!
Hello friends! Welcome to my website.
Java programming to find the two number how to related to each other is greater than or less than or equal? This java language programming is free all type of errors like compiler and runtime errors and this programming is tested to compile and then run the program code.
Here is only programming code only with their output.
### Find the relation of two numbers java programming language.
//—-Code Start—-
import java.util.*;
class test2 { //test class
int a, b;
// class object
public void relate_number(int a, int b) {
if (a < b) { //frist condtion
System.out.println(a + ” is less than ” + b);
}
if (a <= b) { //second codtion
System.out.println(a + ” is lass than or equal to ” + b);
}
if (a == b) { //third condtion
System.out.println(a + ” equal to ” + b);
}
if (a >= b) { //fourth condtion
System.out.println(a + ” is greater than or equal to ” + b);
}
if (a > b) { //fivth condtion
System.out.println(a + ” is greater than to ” + b);
}
}
}
public class relation {
public static void main(String args[]) {
Scanner c = new Scanner(System.in);
//making object of the class test2
test2 o = new test2();
//input
System.out.print(“Please Enter the Frist Number: “);
o.a = c.nextInt();
System.out.print(“Please Enter the Second Number: “);
o.b = c.nextInt();
//output
System.out.println(“The Numbers are relate as Following: “);
o.relate_number(o.a, o.b);
}
}
//—-Code End—–
Output: Please Enter the Frist Number: 5
Please Enter the Second Number: 5
The Numbers are related as Following:
5 is less than or equal to 5
5 equal to 5
5 is greater than or equal to 5
I make this programming and hope you are like the post if are you like the post you can please comment and Share the post to reach more people.
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https://www.coursehero.com/tutors-problems/Calculus/19522533-let-fx3x-1-if-x-1-delta-what-is-fx-3-smaller-than-2-suppos/
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Question
# let f(x)=3x1-if |x-1|< delta, what is |f(x)-3| smaller than?
2-suppose we want |f(x)-3|< epsilon. Find a delta such that whenever |x-1|< delta we have |f(x)-3|< epsilon
3- what did that just prove?
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Intel® Integrated Performance Primitives
Deliberate problems developing high-performance vision, signal, security, and storage applications.
6689 Discussions
## Using a negative offset in the dstRoi parameter of ippiRotateCenter
Beginner
249 Views
Hello,
Q: can I use a negative offset in the dstRoi parameter in ippiRotateCenter ?
When I try to do so, I get an error ( ippStsSizeErr ). However, I make sure the all pixels in the ROI (eventually used with negative index) exist and are properly allocated).
~~~~
Why do I need a negative offset?
Sorry for the long story, the followingis just to justify my need of a negative offset in the dstROI:
I am using the ippiRotateCenter function to rotate a small tile of variable size of an image : I am only interested in the tile region centerred around the center point (parameters xCenter and yCenter) after rotation.
If I understand well that function, the center point (xCenter, yCenter) will have the same coordinates in the src and destination image (we are rotating about that point). So, if I only want a small rotated tile around the center point, I need to offset my dst pointer, example:
pDst = pDst - (xCenter - dstWidth/2)- (yCenter - dstHeight/2) * dstStep ;
to have the center point at the center of my dst image (assuming a 1 channel byte image) and set a small size ROI and an offset that takes into account the center point:
dstROI = { xCenter - dstWidth/2, yCenter - dstHeight/2, dstWidth, dstHeight }
The dst image ROI is then a small rectanglewith a relatively large offset WRT image origin. However, when the desired size of the dst image grows compare to the radius of the center point, my offset in x and/or in y becomes negative and the routine returns an error.I dont see what is wrong with my usage...but there must be something wrong. ..
~~~~
Thank you!
Gilbert
4 Replies
Employee
249 Views
I'm sorry, but I don't understand how you could supply a "negative offset in the dstRoi parameter." According to the definition of ippiRotateCenter, this parameter is an IppiRectstructure...
```typedef structure _IppiRect {
int x;
int y;
int width;
int height;
}
```
Perhaps I have misunderstood your question...
Beginner
249 Views
I may have abused the IPP notion of x and y in the IppiRect. According to the doc:
The structure IppiRect for storing the geometric position and size of a rectangle is defined as
typedef struct {
int x;
int y;
int width;
int height;
} IppiRect;
where x,y denote the coordinates of the top left corner of the rectangle that has dimensions width in the x-direction by height in the y-direction.
So, I simply entered a negative value for x and/or y to point behind the image pointer (coordinates values (x,y) of the top left corner of a ROI can be negative -- of course, width and height must be positive). That would position the ROI in the region of the space where I was interested to get the src image after rotation (after rotation, the src image can effectively cover a region of the space where you would need negative index to reach). For most functions, negative image indices dont make much sense but for a rotation, they might make sense). I had applied an offset to my dst pointer to avoid memory out of bound access: i.e. all index (even negative ones) when legal in my ROI using the pointer provided to the routine.
My approach worked for large positive indices (also way outside the original source image) but always failed for negative indices ( this can happen when the center point (x,y) is close to the origin of the source imageand I want a large destination region, larger then the radius of the center point sqrt(centerx^2 + centery^2) ).
I think that using negative x or y in IppiRect is simply blindly forbidden in IPP. So, I solved my problem by giving up rotateCenter andusing ippiRotate instead with (sometimes negative) shiftx and shifty values that give me the right view of the source image after rotation.
Employee
249 Views
Glad to hear you found a workable solution.
Beginner
249 Views
Hello, I want to rotate my image along its center, by "angle" degrees in the clockwise direction.
ippiRotateCenter function wants me to specify the destination size. But the destination size should change based on the current "angle" value. How can I compute my destination image size using the angle value?
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# NavList:
## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Message:αβγ
Message:abc
Add Images & Files Posting Code: Name: Email:
Re: Lunars: series vs. triangle methods
From: Fred Hebard
Date: 2004 Sep 26, 22:00 -0400
```Frank,
I didn't mean to be poking any sticks at you in my post, and considered
putting in a disclaimer to that effect, and I recognize that
approximate methods can be as accurate as "exact" methods, but I see no
need to change well-established terminology in this field. As I
recall, the phrase "approximation" is used almost invariably when
series expansions, etc, are used to simplify equations; if nothing
more, they remind us that the approximate methods might not work in all
situations. I also recognize that an understanding of the terms in an
approximate method can give one a better understanding of the physics.
I was merely trying to point out the distinction between the two
methods and to point out one advantage of the exact method for those of
us who are not as mathematically proficient as you.
I remember acquiring a TI-35 or TI-48 (can't remember the number)
programmable calculator in 1978 or so and programming in Taylor Series
expressions for sine, etc, just to see them in action and to try to
infer how many terms were in the calculator's series for sine. So I
know about Taylor Series and what they're used for. But it would not
be a simple matter for me to derive an approximate method for solving a
spherical trig problem, whereas it is for you. I suppose I could do
it, but it would take me several days. I expect it also would take me
quite some time to even do that Taylor Series for sine in the TI-35.
Thus, for me to understand the basic spherical trig problem, the
approximate method gives me little guidance, while the exact method
gives quite a bit more.
I also think it is wholly true that the approximate methods were
developed to ease the computational burdens imposed by the exact
methods. Can you offer any other reason why the approximate methods
were developed and published?
I might add that I don't regret that my post stimulated you to educate
us more thoroughly. But my apologies if it offended you.
Fred
```
Browse Files
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### Join NavList
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Do you want to receive all group messages by email? Yes No
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### Posting Code
Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
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### Custom Index
Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd)
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# Iteration
For other uses, see Iteration (disambiguation).
A pentagon iteration. Connecting alternate corners of a regular pentagon produces a pentagram which encloses a smaller inverted pentagon. Iterating the process produces a sequence of nested pentagons and pentagrams and also demonstrates recursion (in the sense of nested self-similarity).[1]
Iteration is the act of repeating a process with the aim of approaching a desired goal, target or result. Each repetition of the process is also called an "iteration", and the results of one iteration are used as the starting point for the next iteration. The pentagon on the right also is a good example of how iteration relates to recursion. Although iteration is used, for example, to parse a linked list, recursion is required when we step up to binary trees. The pentagon demonstrates both.[2]
## Mathematics
Iteration in mathematics may refer to the process of iterating a function i.e. applying a function repeatedly, using the output from one iteration as the input to the next. Iteration of apparently simple functions can produce complex behaviours and difficult problems - for examples, see the Collatz conjecture and juggler sequences.
Another use of iteration in mathematics is in iterative methods which are used to produce approximate numerical solutions to certain mathematical problems. Newton's method is an example of an iterative method. Manual calculation of a number's square root is a common use and a well-known example.
## Computing
Iteration in computing is the repetition of a block of statements within a computer program. It can be used both as a general term, synonymous with repetition, and to describe a specific form of repetition with a mutable state.
When used in the first sense, recursion is an example of iteration, but typically using a recursive notation, which is typically not the case for iteration.
However, when used in the second (more restricted) sense, iteration describes the style of programming used in imperative programming languages. This contrasts with recursion, which has a more declarative approach.
Here is an example of iteration relying on destructive assignment, in imperative pseudocode:
```a = 0
for i from 1 to 3 // loop three times
{
a = a + i // add the current value of i to a
}
print a // the number 6 is printed (0 + 1; 1 + 2; 3 + 3)
```
In this program fragment, the value of the variable i changes over time, taking the values 1, 2 and 3. This changing value—or mutable state—is characteristic of iteration.
Iteration can be approximated using recursive techniques in functional programming languages. The following example is in Scheme. Note that the following is recursive (a special case of iteration) because the definition of "how to iterate", the iter function, calls itself in order to solve the problem instance. Specifically it uses tail recursion so it does not use large amounts of stack space.
```(let iterate ((i 1) (a 0))
(if (<= i 3)
(iterate (+ i 1) (+ a i))
(display a)))
```
An iterator is an object that provides iteration as a generic service, allowing iteration to be done in the same way for a range of different data structures. Conversely, an iteratee is an abstraction which accepts or rejects data during an iteration process (controlled externally by an enumerator - so unlike with code that uses iterators, the iteratee code is not "in charge" of the iteration process).
Iteration is also performed using a worksheet, or by using solver or goal seek functions available in Excel. Many implicit equations like the Colebrook equation can be solved in the convenience of a worksheet by designing suitable calculation algorithms.[3]
Many of the engineering problems like solving Colebrook equations reaches 8-digit accuracy in as small as 12 iterations and a maximum of 100 iterations is sufficient to reach a 15-digit accurate result.[4]
## Project management
Main article: Project management
Iterations in agile project management
Iterations in a project context may refer to the technique of developing and delivering incremental components of business functionality, product development or process design. This is most often associated with agile software development, but could potentially be any material. A single iteration results in one or more bite-sized but complete packages of project work that can perform some tangible business function. Multiple iterations recurse to create a fully integrated product. This is often contrasted with the waterfall model approach.[citation needed]
## Education
Main article: Educational theory
In some schools of pedagogy, iterations are used to describe the process of teaching or guiding students to repeat experiments, assessments, or projects, until more accurate results are found, or the student has mastered the technical skill. This idea is found in the old adage, "Practice makes perfect." In particular, "iterative" is defined as the "process of learning and development that involves cyclical inquiry, enabling multiple opportunities for people to revisit ideas and critically reflect on their implication."[5]
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##### Statistics questions
label Statistics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
A particular multiple choice test has 110 questions, each with four choices (only one of which is correct).
If students were to take the test by guessing the answer to every question on it entirely at random, what would you expect the average number of correct responses per test to be? Provide your response precise to the nearest tenth.
Mar 14th, 2015
My answer is only an approximation, but because the number of questions is large (110) and the required precision low (0.1), it should be good enough:
There are 4 answers per question, so the probability of guessing correctly is 0.25. 0.25*110 = 27.5
The answer based on our approximation is 27.5. Otherwise there is a well studied probability distribution that governs this problem - one could use it to get a more accurate answer.
Mar 14th, 2015
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Mar 14th, 2015
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Mar 14th, 2015
Oct 23rd, 2017
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# corejava-programming: Dinesh revenge
Dinesh has been bullied enough all his life and now he seeks revenge. His first victim being Deepak, obviously! He found on e question to take revenge from deepak, but without knowing the answers of the question he would not be able to know whether deepak´s answers are correct or not. The question consists of two numbers, a and b, and the answer to it is the sum of digits of a^b (i.e - a raise to the power b). Help him to find the answers so that he can take revenge from deepak. (You should be able to write the solution for this problem in 25 minutes) Input: First line contains the number of test cases t. After that t lines follow each containing a pair of space separated integers i.e. a and b. Output: For each test case, you need to print the sum of digits in a^b. Sample Input : 2 2 10 3 3 Sample Output : 7 9 Explanation Consider the first test case. 2^10 = 1024 Now, 1 + 0 + 2 + 4 = 7 So, output for this test case is 7. corejava x 353programming x 169 Posted On : 2016-09-19 15:04:24.0 krish soni 530 5
import java.util.ArrayList; import java.util.List; import java.util.Scanner; public class DineshRevenge { public static void main(String[] args) { Scanner input=new Scanner(System.in); int t=input.nextInt(); List outputs= new ArrayList(); Double num; int sum=0; for(int i=1;i<=t;i++){ sum=0; num=Math.pow(input.nextInt(), input.nextInt()); outputs.add(num.intValue()); } for(Integer output:outputs){ for(char digit:output.toString().toCharArray()){ sum=sum+ Integer.valueOf(Character.toString(digit)); } System.out.println(sum); } outputs.clear(); outputs=null; } } Posted On : 2018-06-14 18:55:12 Satisfied : 1 Yes 2 No Garima Gupta 596129529770Reply This Thread 0
import java.util.Scanner; public class DeepakRevenge { public static void main(String[] args) { Scanner input=new Scanner(System.in); int t=input.nextInt(); Double num; Integer intNum; int sum; for(int i=1;i<=t;i++){ sum=0; num=Math.pow(input.nextInt(), input.nextInt()); intNum=num.intValue(); for(char digit:intNum.toString().toCharArray()){ sum=sum+Integer.parseInt(Character.toString(digit)); } System.out.print(sum+" "); } } } Posted On : 2016-09-19 16:52:16 Satisfied : 1 Yes 6 No Rishi Kumar 528188225056Reply This Thread 2
Hey Can you pleas send the coding in Basic Java Format so that i can understand
25 5 0
Posted On :2016-09-20 14:48:49.0
This is basic java coding only, what is the challenge??
Rishi Kumar
528 1882 25056
Posted On :2016-09-21 22:15:31.0
import java.util.Scanner; import java.lang.Math.*; public class Abslute { public static void main(String args[]) { Scanner s=new Scanner(System.in); int n=s.nextInt(); double a,b,power=0; for(int i=0;i 0) { rem = sum % 10; res = res + rem; sum = sum / 10; } // return; } System.out.print(res+" "); } } } find the below input specification.Let me know if this code is satisfied your specification or not. Input : 2 2 10 3 3 Output: 7 9 Posted On : 2016-09-20 22:56:25 Satisfied : 1 Yes 6 No 3
Not Satisfied ....my Test Cases are not cleared
25 5 0
Posted On :2016-09-21 15:02:19.0
Rishi Kumar
528 1882 25056
Posted On :2016-09-21 22:17:34.0
Can you provide clear Test cases for this .As per the given post your specifications are satisfied.
Posted On :2016-09-21 22:22:22.0
The Test Cases are not provided but we have to pass the all test cases can you please send me another coding.
25 5 0
Posted On :2016-09-21 22:24:35.0
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Close
Home » Kids » Dr John Kelliher » Primary School ‘KS2 (Key Stage 2) - Maths – Algebra - Ages 7-11’ eBook
# Primary School ‘KS2 (Key Stage 2) - Maths – Algebra - Ages 7-11’ eBook
## Overview
This eBook introduces the subject of algebra, considers the inverse operators of addition and subtraction, the inverse operators of multiplication and division, introduce equations, consider the order of operations (precedence) and examines solving simple equations.
High (Secondary) School ‘Grade 9 & 10 - Math – Inequalities – Ages 14-16’ eBook
This eBook introduces the topic of inequalities, the meaning of the inequality symbols, how to rearrange and solve inequalities as well as the use of inequalities and number lines and the use of inequalities in graphs.This eBook is part of our range of Grade 9 & 10 math eBooks that are fully aligned with the North American curriculum.Our Grade 9 &…
This eBook is part of our range of Key Stage 2 (KS2) maths eBooks that are fully aligned with the UK Governments national curriculum.
Our Key Stage 3 (KS2) maths eBooks comprise three principle sections. These are, notably:
•Number and Algebra
•Shape, Space and Measure
•Handling Data
In addition, there exists for Key Stage 2 (KS2) a Publications Guide eBook, a Mental Maths eBook and a Times Table Practice eBook. We also produce a combined Key Stage 1 (KS1) and Key Stage 2 (KS2) Times Table Practice eBook.
Our maths eBooks are produced such as that as well as a Publications Guide, and three principle publications corresponding to the principle sections (Number and Algebra, Shape, Space and Measures as well as Handling Data) there are individual modules produced within each principle section which are published as eBooks.
Algebra is a module within the Number and Algebra principle section our Key Stage 2 (KS2) publications. It is one module out of a total of eight modules in that principle section, the others being:
•Counting Practice
•Number Patterns and Sequences
•Integers
•Fractions, Percentages and ratio
•Decimals
•Multiplication and Division Practice
After studying Primary School 'KS2 (Key Stage 2) - Maths – Algebra - Ages 7-11' eBook your child or the student should be confidently able to:
-understand the notion of the inverses of addition and subtraction.
-understand the notion of the inverses of multiplication and division.
-understand the basics as to the order of precedence in arithmetic operations.
-understand and rearrange simple equations, and solve them.
Date release: June 15, 2012
Primary, School, ‘KS2, (Key, Stage, Maths, Algebra, Ages, 7-11’, eBook, John, Kelliher, download, PDF, XML Book, MobiPocket, Torrent
You can buy this book now only for \$2.99. This is the lowest price for this book.
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- Cameron Manual
17. Add And Move - Further Symmetry Related Commands
Fri Jun 2 2000
# Cameron Manual
## Chapter 17: Add And Move - Further Symmetry Related Commands
17.2: EXAMPLES
17.3: MOVE
The PACK and ENCLOSURE commands already detailed allow the user to apply all of the symmetry operators in the spacegroup to the initial set of atoms in order to get a fully 'packed' result. In some cases however, the user may wish to apply only one symmetry operator or to apply ones that are not present in the spacegroup. The ADD and MOVE commands allow this.
The ADD command allows the user complete control over the symmetry operator used to generate new atoms. The first task of the user is to generate a list of those atoms to be used in the symmetry generation later. One of the following sub-commands is required :-
ATOMS
The names of atoms to be included in the pack list are specified here. Element names can also be used if required.
ALL
ALL refers to the atoms that are in the current list. If any atoms have been generated by previous PACK , ENCLOSURE or ADD commands then these will all go into the list.
INITIAL
ADD INITIAL means that the only atoms to go into the ADD list are those that were initially input.
GROUP
This is followed by a group name. The group must have been previously been defined by the command DEFGROUP.
Once the ADD list has been created the user must then supply the symmetry operators which will act on the atoms in this list to generate the new atoms. The symmetry input is in two parts.
OPERATOR The symmetry operator may be input in decimal or fractional form eg
```x y+1/2 z
-0.333-x -y -z
ETC
```
Decimal translations may come before or after the axis symbol. The fractions 1/2, 1/3, 1/4, 2/3, 1/6 and 5/6 are accepted by the program, but must appear after the x/y/z character. There must be spaces between the three parts but NO SPACES within the operator ie
```X + 1/2 -Y Z
```
will produce an error as it is not possible to tell whether you mean X+1/2, -Y, Z or X, +1/2-Y, Z. This strict input syntax is used to eliminate any ambiguities.
TRANS Translations can also be applied if required. The translations are applied in unit cell fractions. The syntax is :-
```TRANS x y z
```
### 17.2: EXAMPLES
To generate an atom at x+1/2 y z from an atom at x y z we can use
```ADD
ATOMS C1
OPERATOR X+1/2 Y Z
```
or we could use
```ADD
ATOMS C1
TRANS 0.5 0 0
```
The OPERATOR and TRANS commands can be used together if required. We can apply a symmetry operator followed by a translation. This reduces the errors that may occur when trying to combine the two things into one symmetry operator.
The use of INITIAL versus ALL is illustrated below. Start with :-
```ADD
ALL
TRANS 1 0 0
```
which gives us a molecule at x y z and another at x+1 y z. Follow this with :-
```ADD
ALL
TRANS 0 1 0
```
and we get four molecules, x y z , x+1 y z , x+1 y+1 z and x y+1 z. Following it with :-
```ADD
INITIAL
TRANS 0 1 0
```
Gives us three molecules, x y z, x+1 y z and x y+1 z.
### 17.3: MOVE
The syntax for this command is identical to that for ADD. Therefore the commands available are :-
ATOMS
ALL
GROUP
INITIAL
Which must be followed one (or both) of :-
OPERATOR
TRANS
The MOVE command applies a symmetry operator and/or a translation to all of the atoms held in the list defined by the ATOMS/ALL/GROUP/INITIAL commands. Unlike ADD therefore, the same number of atoms are present at the beginning and end of the operation.
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This topic is closed
## Seize Your Chance To Win 2,000 Sapphires!
499 Replies
User
5 March, 2017, 2:39 AM UTC
Backwards
UTC -6:00
0
User
5 March, 2017, 2:43 AM UTC
a)
UTC -5:00
1
User
5 March, 2017, 2:45 AM UTC
b) Backward
UTC +0:00
0
User
5 March, 2017, 3:13 AM UTC
b) Backward
UTC +8:00
0
User
5 March, 2017, 3:35 AM UTC
UTC +0:00
0
User
5 March, 2017, 3:50 AM UTC
(b)
UTC +0:00
0
User
5 March, 2017, 3:58 AM UTC
The motorcycle looks to be spinning.
UTC +0:00
0
User
5 March, 2017, 4:04 AM UTC
Backwards
UTC +5:00
0
User
5 March, 2017, 4:17 AM UTC
its not moving at all
UTC +0:00
0
User
5 March, 2017, 5:11 AM UTC
The correct answer is B) Backwards
UTC +5:00
0
User
5 March, 2017, 5:27 AM UTC
a forward
UTC +0:00
1
User
5 March, 2017, 5:31 AM UTC
backward
UTC +8:00
0
User
5 March, 2017, 5:34 AM UTC
UTC +13:00
0
User
5 March, 2017, 5:58 AM UTC
B. backwards
UTC +8:00
0
User
5 March, 2017, 6:32 AM UTC
b) Backward
UTC +7:00
0
User
5 March, 2017, 6:35 AM UTC
b.
UTC +1:00
0
User
5 March, 2017, 6:39 AM UTC
Put your powers of observation to the test, Lords and Ladies! Simply comment with the correct answer for a chance to win 2,000 Sapphires!
In what direction is the motorcycle moving?
a) Forward
b) Backward
c) It is not moving at all
UTC +8:00
0
User
5 March, 2017, 7:01 AM UTC
backwards
Ronald Sperling
UTC +8:00
0
User
5 March, 2017, 7:42 AM UTC
2) Backwards
UTC +8:00
0
User
5 March, 2017, 8:07 AM UTC
b) Backward
UTC +8:00
0
2816466 users registered; 63641 topic; 335469 posts; our newest member:chanda
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# Untitled
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1. // import a hash map implementation
2. use fnv::FnvHashMap;
3. use std::f64;
4.
5. // these are multiplied by epsilon to obtain
6. // smaller values of espilon. Both should be
7. // between 0 and 1, and CHECK_FACTOR should be greater
8. // than EPS_FACTOR. There is an optimal value of EPS_FACTOR
9. // for any function, and increasing CHECK_FACTOR produces slightly
10. // better results but causes the algorithm to take longer.
11. const EPS_FACTOR: f64 = 1.0 - 1.0 / 4.0;
12. const CHECK_FACTOR: f64 = 1.0 - 1.0 / 8.0;
13.
14. // whether to display the entire output or just the number
15. // of triangles
16. const SHOW_OUTPUT: bool = true;
17.
18. // tolerance used for numerical methods
19. const TOL: f64 = 1.0e-15;
20.
21. fn main() {
22. // the function and its first two derivatives
23. let f = |x: f64, y: f64| x.sin() + y.sin();
24. let d = |x: f64, y: f64| [x.cos(), y.cos()];
25. // [df/dxdx, df/dydy, df/dxdy]
26. let dd = |x: f64, y: f64| [-x.sin(), -y.sin(), 0.0];
27. let mut v = Vec::new();
28. // the points that form the initial two triangles
29. let mut points = vec![[-8.0, -8.0], [8.0, -8.0], [8.0, 8.0], [-8.0, 8.0]];
30. // generate the approximation
31. generate_approx_2d(
32. // the function will add more values to points
33. &mut points,
34. // the indices of the triangles to generate
35. // we can't generate the two triangles seperately
36. // because the seam between them wouldn't line up
37. &[[0, 1, 2], [2, 3, 0]],
38. // epsilon
39. 1.0 / 8.0,
40. // the step size. For any function g(x)=f(x0+ax, y0+bx), where
41. // a^2+b^2=1, g can be decomposed into [p_i, q_i] intervals where
42. // either g or -g is convex on [p_i, q_i] with q_i - p_i > step.
43. // I am not sure if there is an easy way to find this constant
44. 1.0,
45. // a value that should be larger than the lipschitz constant
46. // for the function (used to create an oracle)
47. 1.1,
48. // the function will add sets of indices to this vector
49. &mut v,
50. // pass the lambdas for the function
51. f,
52. d,
53. dd,
54. );
55. // display the number of triangles generated
56. println!("{}", v.len());
57. if SHOW_OUTPUT {
58. print!("[");
59. for p in points {
60. for val in &p {
61. print!("{:.12}, ", val);
62. }
63. }
64. println!("]");
65. print!("[");
66. for tri in v {
67. for index in &tri {
68. print!("{}, ", index);
69. }
70. }
71. println!("]");
72. }
73. }
74.
75. // generate a piecewise linear approximation of
76. // a function of two variables. Note that the
77. // vertices of the simplices generated lie on the function
78. fn generate_approx_2d<
79. F: Fn(f64, f64) -> f64 + Copy,
80. D: Fn(f64, f64) -> [f64; 2] + Copy,
81. // [df/dxdx, df/dydy, df/dxdy]
82. D2: Fn(f64, f64) -> [f64; 3] + Copy,
83. >(
84. points: &mut Vec<[f64; 2]>,
85. tris: &[[usize; 3]],
86. epsilon: f64,
87. step: f64,
88. k: f64,
89. v: &mut Vec<[usize; 3]>,
90. f: F,
91. d: D,
92. dd: D2,
93. ) {
94. let small_epsilon = epsilon * EPS_FACTOR;
95. let len = tris.len();
96. // initialize a hash map
97. let mut edges_map = FnvHashMap::with_capacity_and_hasher(len * 2, Default::default());
98.
99. // a lambda for processing the edges. This makes sure that edges between
100. // triangles line up
101. let mut get_edge = |n1: usize, n2: usize| {
102. let (p1, p2, ccw) = if n1 > n2 {
103. (n2, n1, false)
104. } else {
105. (n1, n2, true)
106. };
107. let (line, dist) = norm_line(points[p1][0], points[p1][1], points[p2][0], points[p2][1]);
108. if let Some((start, end)) = edges_map.get(&(p1, p2)) {
109. ((*start, *end), ccw, line)
110. } else {
111. let start = points.len();
112.
113. let vec = generate_approx_1d(
114. 0.0,
115. dist,
116. small_epsilon,
117. step,
118. |x| line.val(x, f),
119. |x| line.der(x, d),
120. |x| line.der_2(x, dd),
121. );
122.
123. for t in &vec {
124. let point = line.pos(*t);
125. points.push([point.0, point.1]);
126. }
127.
128. let end = start + vec.len();
129. edges_map.insert((p1, p2), (start, end));
130. ((start, end), ccw, line)
131. }
132. };
133.
134. let mut calls = Vec::with_capacity(len);
135.
136. for i in 0..len {
137. let [n1, n2, n3] = tris[i];
138. let (i1, c1, _) = get_edge(n1, n2);
139. let (i2, c2, _) = get_edge(n2, n3);
140. let (i3, c3, _) = get_edge(n3, n1);
141. calls.push(([n1, n2, n3], [i1, i2, i3], [c1, c2, c3]));
142. }
143. generate_2d_approx_loop(epsilon, step, k, points, v, &mut calls, f, d, dd);
144. }
145.
146. // originally I implemented a recursive algorithm, but I was
147. // getting stack overflows so I changed it to use a heap-allocated
148. // list of calls
149. fn generate_2d_approx_loop<
150. F: Fn(f64, f64) -> f64 + Copy,
151. D: Fn(f64, f64) -> [f64; 2] + Copy,
152. // [df/dxdx, df/dydy, df/dxdy]
153. D2: Fn(f64, f64) -> [f64; 3] + Copy,
154. >(
155. epsilon: f64,
156. step: f64,
157. k: f64,
158. points: &mut Vec<[f64; 2]>,
159. // the working vector of triangles
160. v: &mut Vec<[usize; 3]>,
161. // this will contain the initial calls needed
162. stack: &mut Vec<([usize; 3], [(usize, usize); 3], [bool; 3])>,
163. f: F,
164. d: D,
165. dd: D2,
166. ) {
167. while let Some((bound, edges, ccw)) = stack.pop() {
168. generate_2d_approx_division(
169. bound,
170. edges,
171. ccw,
172. epsilon,
173. step,
174. k,
175. v,
176. points,
177. stack,
178. EPS_FACTOR,
179. CHECK_FACTOR,
180. f,
181. d,
182. dd,
183. );
184. }
185. }
186.
187. fn generate_2d_approx_division<
188. F: Fn(f64, f64) -> f64 + Copy,
189. D: Fn(f64, f64) -> [f64; 2] + Copy,
190. // [df/dxdx, df/dydy, df/dxdy]
191. D2: Fn(f64, f64) -> [f64; 3] + Copy,
192. >(
193. // bound will always be in CCW order
194. bound: [usize; 3],
195. // the positions of points along each edge. This stores
196. // a start and end index into points.
197. edges: [(usize, usize); 3],
198. // the order of points on an edge might be backward since
199. // neighboring triangles share edges
200. edge_ccw: [bool; 3],
201. epsilon: f64,
202. step: f64,
203. k: f64,
204. v: &mut Vec<[usize; 3]>,
205. points: &mut Vec<[f64; 2]>,
206. stack: &mut Vec<([usize; 3], [(usize, usize); 3], [bool; 3])>,
207. factor: f64,
208. check_factor: f64,
209. f: F,
210. d: D,
211. dd: D2,
212. ) {
213. let small_epsilon = epsilon * factor;
214.
215. // get the total length of all the edges
216. let edges_len = edges[0].1 - edges[0].0 + edges[1].1 - edges[1].0 + edges[2].1 - edges[2].0;
217.
218. if edges_len == 0 {
219. // in this case there is no further subdivision neccesary,
220. // so we check if the triangle described by bound
221. // is entirely within the epsilon bound.
222. let check_epsilon = epsilon * check_factor;
223. let [x1, y1] = points[bound[0]];
224. let [x2, y2] = points[bound[1]];
225. let [x3, y3] = points[bound[2]];
226. let (l1, d1) = norm_line(x1, y1, x2, y2);
227. let (l2, d2) = norm_line(x2, y2, x3, y3);
228. let (l3, d3) = norm_line(x3, y3, x1, y1);
229. let lines = [l1, l2, l3];
230. let dists = [d1, d2, d3];
231. let mut high_dist = d1;
232. let mut high_pos = 0;
233. for i in 1..3 {
234. if dists[i] > high_dist {
235. high_dist = dists[i];
236. high_pos = i;
237. }
238. }
239. let lines = [
240. lines[high_pos],
241. lines[(high_pos + 1) % 3],
242. lines[(high_pos + 2) % 3],
243. ];
244. let point = lines[2].pos(0.0);
245.
246. let z1 = lines[0].val(0.0, f);
247. let z2 = lines[1].val(0.0, f);
248. let z_slope = (z2 - z1) / high_dist;
249. // the lipschitz constant gives us a range for where
250. // the triangle is within the bound given that a line
251. // from one vertex to a point on the opposite edge
252. // is within a smaller bound
253. let first_step_len = 2.0 * (1.0 - EPS_FACTOR) * epsilon / (k + z_slope.abs());
254. let other_step_len = 2.0 * (1.0 - CHECK_FACTOR) * epsilon / (k + z_slope.abs());
255. let mut check_pos = first_step_len;
256. let mut last = true;
257.
258. while check_pos < high_dist {
259. let check_point = lines[0].pos(check_pos);
260. let (line, len) = norm_line(point.0, point.1, check_point.0, check_point.1);
261. // check if the line is within the smaller bound
262. let check = max_from_point(
263. (0.0, line.val(0.0, f)),
264. 0.0,
265. check_epsilon,
266. step,
267. len,
268. |x| line.val(x, f),
269. |x| line.der(x, d),
270. |x| line.der_2(x, dd),
271. );
272.
273. if let Err((min_slope, max_slope)) = check {
274. let end_z = z1 + check_pos * z_slope;
275. let end_slope = (end_z - line.val(0.0, f)) / len;
276. if min_slope <= end_slope && end_slope <= max_slope {
277. check_pos += other_step_len;
278. continue;
279. }
280. }
281. // if we reach this point then we need to further subdivide
282. // the triangle. This is generally inneficient, and the optimal
283. // value of check_factor will generally cause this case to
284. // only occur a few times
285. last = false;
286. let factor = 0.5;
287. let center_point = line.pos(len * factor);
288. let center_pos = points.len();
289. points.push([center_point.0, center_point.1]);
290. let (line_0, len_0) = norm_line(lines[0].x, lines[0].y, center_point.0, center_point.1);
291. let (line_1, len_1) = norm_line(lines[1].x, lines[1].y, center_point.0, center_point.1);
292. let div_lens = [len_0, len_1, len * factor];
293. let div_lines = [line_0, line_1, line];
294. let mut line_points = [(0, 0); 3];
295. for i in 0..3 {
296. // it is possible that the lines to the new point
297. // aren't within the small_epsilon bound
298. let approx = generate_approx_1d(
299. 0.0,
300. div_lens[i],
301. small_epsilon,
302. step,
303. |x| div_lines[i].val(x, f),
304. |x| div_lines[i].der(x, d),
305. |x| div_lines[i].der_2(x, dd),
306. );
307. let index = points.len();
308. line_points[i] = (index, index + approx.len());
309. for t in &approx {
310. let point = div_lines[i].pos(*t);
311. points.push([point.0, point.1]);
312. }
313. }
314. for i in 0..3 {
315. let next = (i + 1) % 3;
316. stack.push((
317. [center_pos, bound[i], bound[next]],
318. // the index 3 is an empty vec
319. [line_points[i], (0, 0), line_points[next]],
320. [false, true, true],
321. ));
322. }
323. break;
324. }
325.
326. if last {
327. v.push(bound);
328. }
329. return;
330. }
331. // generate a list to help process triangles
332. let mut index_vec = Vec::with_capacity(3 + edges_len);
333. let mut edge_boundaries = [0; 4];
334. for i in 0..3 {
335. index_vec.push(bound[i]);
336. let edge_len = edges[i].1 - edges[i].0;
337. for k in 0..edge_len {
338. let n = if edge_ccw[i] {
339. edges[i].0 + k
340. } else {
341. edges[i].1 - 1 - k
342. };
343. index_vec.push(n);
344. }
345. edge_boundaries[i + 1] = index_vec.len();
346. }
347. let triangulation = get_triangulation(edge_boundaries);
348. let len = triangulation.len() / 3;
349. let mut edges_map = FnvHashMap::with_capacity_and_hasher((3 * len) / 2, Default::default());
350.
351. let get_edge =
352. |edges_map: &mut FnvHashMap<_, _>, points: &mut Vec<[f64; 2]>, n1: usize, n2: usize| -> ((usize, usize), [f64; 2], f64, bool) {
353. let (p1, p2, ccw) = if n1 > n2 {
354. (n2, n1, false)
355. } else {
356. (n1, n2, true)
357. };
358.
359. // check if the points lie on the same edge
360. let mut p1_edge = 0;
361. let mut p2_edge = 0;
362.
363. for i in 0..3 {
364. if p1 < edge_boundaries[3 - i] {
365. p1_edge = 2 - i;
366. }
367. if p2 > edge_boundaries[i] {
368. p2_edge = i;
369. }
370. }
371.
372. let point1 = points[index_vec[p1]];
373. let point2 = points[index_vec[p2]];
374.
375. let (edge_line, dist) = norm_line(point1[0], point1[1], point2[0], point2[1]);
376. let mid = [(point1[0] + point2[0]) / 2.0, (point1[1] + point2[1]) / 2.0];
377.
378. // if both points lie on the same edge, then we don't want to
379. // generate a new approx, as that would cause point aliasing
380. if (p1_edge == p2_edge) || (p1 == 0 && p2 >= edge_boundaries[2]) {
381. if p2 - p1 == 1 || (p1 == 0 && edge_boundaries[3] - p2 == 1) {
382. if let Some((start, end)) = edges_map.get(&(p1, p2)) {
383. // this case occurs when an edge of an obtuse triangle
384. // is subdivided
385. return ((*start, *end), mid, dist, ccw);
386. }
387. return ((0, 0), mid, dist, ccw);
388. } else {
389. let (edge, start, end) = if p1 == 0 && p2 >= edge_boundaries[2] {
390. (
391. 2,
392. p2 - edge_boundaries[2],
393. edge_boundaries[3] - edge_boundaries[2],
394. )
395. } else {
396. (
397. p1_edge,
398. p1 - edge_boundaries[p1_edge],
399. p2 - edge_boundaries[p1_edge],
400. )
401. };
402. let ccw = edge_ccw[edge];
403. let (index_start, index_end) = if ccw {
404. (edges[edge].0 + start, edges[edge].0 + end - 1)
405. } else {
406. (edges[edge].1 - end + 1, edges[edge].1 - start)
407. };
408. return ((index_start, index_end), mid, dist, ccw);
409. }
410. }
411.
412. // check if the edge has already been processed
413. if let Some((start, end)) = edges_map.get(&(p1, p2)) {
414. return ((*start, *end), mid, dist, ccw);
415. } else {
416. let approx = generate_approx_1d(
417. 0.0,
418. dist,
419. small_epsilon,
420. step,
421. |x| edge_line.val(x, f),
422. |x| edge_line.der(x, d),
423. |x| edge_line.der_2(x, dd),
424. );
425. let index = points.len();
426. let end = index + approx.len();
427. edges_map.insert((p1, p2), (index, end));
428.
429. for t in &approx {
430. let point = edge_line.pos(*t);
431. points.push([point.0, point.1]);
432. }
433.
434. ((index, end), mid, dist, ccw)
435. }
436. };
437.
438. let insert_mid = |edges_map: &mut FnvHashMap<_, _>, n1, n2, index| {
439. let (p1, p2) = if n1 > n2 { (n2, n1) } else { (n1, n2) };
440. edges_map.insert((p1, p2), (index, index + 1));
441. };
442.
443. // the first time through is used to check for obtuse triangles
444. // which can cause an infinite loop. The adjustments made to
445. // fix this need to also change the edges on neighboring triangles
446. for i in 0..len {
447. let tri = &triangulation[i * 3..(i * 3) + 3];
448. let edges = [
449. get_edge(&mut edges_map, points, tri[0], tri[1]),
450. get_edge(&mut edges_map, points, tri[1], tri[2]),
451. get_edge(&mut edges_map, points, tri[2], tri[0]),
452. ];
453. for k in 0..3 {
454. let edgek = edges[k].0;
455. if edgek.1 - edgek.0 == 0 {
456. let next1 = (k + 1) % 3;
457. let next2 = (k + 2) % 3;
458. let (i1, mid1, d1, _) = edges[next1];
459. let (i2, mid2, d2, _) = edges[next2];
460. let mid0 = edges[k].1;
461. let d0 = edges[k].2;
462. let len1 = i1.1 - i1.0;
463. let len2 = i2.1 - i2.0;
464. if d1 * d1 + d2 * d2 <= d0 * d0 {
465. if len1 > 0 || len2 > 0 {
466. let index = points.len();
467. points.push(mid0);
468. insert_mid(&mut edges_map, tri[k], tri[next1], index);
469. if len1 == 0 {
470. points.push(mid1);
471. insert_mid(&mut edges_map, tri[next1], tri[next2], index + 1);
472. }
473. if len2 == 0 {
474. points.push(mid2);
475. insert_mid(&mut edges_map, tri[next2], tri[k], index + 1);
476. }
477. }
478. break;
479. }
480. }
481. }
482. }
483.
484. // now we run through to actually add the calls to the stack,
485. // with the adjustments made to the neccesary edges
486. for i in 0..len {
487. let tri = &triangulation[i * 3..(i * 3) + 3];
488. let edges = [
489. get_edge(&mut edges_map, points, tri[0], tri[1]),
490. get_edge(&mut edges_map, points, tri[1], tri[2]),
491. get_edge(&mut edges_map, points, tri[2], tri[0]),
492. ];
493.
494. let bound = [index_vec[tri[0]], index_vec[tri[1]], index_vec[tri[2]]];
495. let (edge1, _, _, ccw1) = edges[0];
496. let (edge2, _, _, ccw2) = edges[1];
497. let (edge3, _, _, ccw3) = edges[2];
498. stack.push((bound, [edge1, edge2, edge3], [ccw1, ccw2, ccw3]));
499. }
500. }
501.
502. // generates the indices for a triangulation given the number of
503. // points on each edge
504. fn get_triangulation(edges: [usize; 4]) -> Vec<usize> {
505. let mut v = Vec::with_capacity(3);
506. for i in 0..3 {
507. v.push((edges[i] + edges[i + 1]) / 2);
508. }
509. for i in 0..3 {
510. let prev = (i + 2) % 3;
511. if edges[i + 1] - edges[i] > 1 {
512. v.extend_from_slice(&[edges[i], v[i], v[prev]]);
513. }
514. }
515. v
516. }
517.
518. // generate a 1 dimensional approximation where the endpoints
519. // lie on the function. Note that this is not an optimal approximation
520. fn generate_approx_1d<
521. F: Fn(f64) -> f64 + Copy,
522. D: Fn(f64) -> f64 + Copy,
523. D2: Fn(f64) -> f64 + Copy,
524. >(
525. start: f64,
526. end: f64,
527. epsilon: f64,
528. step: f64,
529. f: F,
530. d: D,
531. dd: D2,
532. // the output does not include the first and last points
533. ) -> Vec<f64> {
534. // create the list to hold the results
535. let mut v = Vec::new();
536.
537. let mut prev = max_from_pos(start, start, epsilon, step, end, f, d, dd);
538.
539. // loop until we reach the max
540. while let Some((_, x, _, _)) = prev {
541. if x == end {
542. break;
543. }
544. v.push(x);
545. // the end of the previous line tells us the value of s_epsilon to use here
546. prev = max_from_pos(x, x, epsilon, step, end, f, d, dd);
547. }
548. // we go back through the approximation to cause it to be
549. // spaces more evenly if possible
550. let n = v.len();
551. let equal_len = (end - start) / (n + 1) as f64;
552. let mut back = max_from_pos(
553. -end,
554. -end,
555. epsilon,
556. step,
557. -start,
558. |x| f(-x),
559. |x| -d(-x),
560. |x| dd(-x),
561. );
562. for i in 0..n {
563. let min = if let Some((_, neg_x, _, _)) = back {
564. -neg_x
565. } else {
566. start
567. };
568. let dist = start + (n - i) as f64 * equal_len;
569. let mut new_point = v[n - i - 1];
570. if dist < new_point {
571. new_point = if min < dist { dist } else { min };
572. } else {
573. break;
574. }
575. v[n - i - 1] = new_point;
576. back = max_from_pos(
577. -new_point,
578. -new_point,
579. epsilon,
580. step,
581. -start,
582. |x| f(-x),
583. |x| -d(-x),
584. |x| dd(-x),
585. );
586. }
587. return v;
588. }
589.
590. // find a line segment within the epsilon bound with endpoints
591. // on the function
592. fn max_from_pos<F: Fn(f64) -> f64, D: Fn(f64) -> f64, D2: Fn(f64) -> f64>(
593. point: f64,
594. start: f64,
595. epsilon: f64,
596. step: f64,
597. max_x: f64,
598. f: F,
599. d: D,
600. dd: D2,
601. ) -> Option<(f64, f64, f64, bool)> {
602. let mut test_x = start;
603. let mut prev_x;
604. let mut steps = 1.0;
605. let point_y = f(point);
606. let get_slope = |x, y| (y - point_y) / (x - point);
607. let mut low_upper = f64::INFINITY;
608. let mut upper_point = start;
609. let mut upper_rdir = low_upper - d(start);
610. let mut high_lower = -f64::INFINITY;
611. let upper_start = low_upper;
612. let lower_start = high_lower;
613. let mut lower_point = start;
614. let mut lower_rdir = high_lower - d(start);
615. let mut prev_dd = dd(start);
616. let mut rdir = 0.0;
617.
618. loop {
619. prev_x = test_x;
620. test_x = start + steps * step;
621.
622. if test_x > max_x {
623. test_x = max_x;
624. }
625.
626. let mut new_dd = dd(test_x);
627.
628. let mut should_step = true;
629.
630. if prev_dd * new_dd < 0.0 {
631. test_x = bisect(prev_x, test_x, |x| dd(x), TOL);
632. new_dd = 0.0;
633. should_step = false;
634. }
635.
636. let mut test_y = f(test_x);
637. let mut slope = get_slope(test_x, test_y);
638. let mut new_rdir = slope - d(test_x);
639. if new_rdir * rdir < 0.0 {
640. test_x = find_root(
641. prev_x,
642. test_x,
643. None,
644. |x| get_slope(x, f(x)) - d(x),
645. |x| {
646. let y = f(x);
647. let bottom = x - point;
648. (bottom * d(x) - (y - point_y)) / (bottom * bottom) - dd(x)
649. },
650. TOL,
651. );
652. test_y = f(test_x);
653. new_rdir = 0.0;
654. should_step = false;
655. slope = get_slope(test_x, test_y);
656. }
657.
658. if should_step {
659. steps += 1.0;
660. }
661. rdir = new_rdir;
662.
663. let upper_slope = get_slope(test_x, test_y + epsilon);
664. let lower_slope = get_slope(test_x, test_y - epsilon);
665.
666. let dir = d(test_x);
667. let new_upper_rdir = upper_slope - dir;
668. let new_lower_rdir = lower_slope - dir;
669.
670. let mut new_low_upper = low_upper;
671. let mut new_high_lower = high_lower;
672. let mut new_lower_point = lower_point;
673. let mut new_upper_point = upper_point;
674.
675. if upper_rdir > 0.0 && new_upper_rdir < 0.0 {
676. let find = bisect(
677. prev_x,
678. test_x,
679. |x| {
680. if x == point {
681. upper_start - d(x)
682. } else {
683. get_slope(x, f(x) + epsilon) - d(x)
684. }
685. },
686. TOL,
687. );
688. let find_dir = get_slope(find, f(find) + epsilon);
689. if find_dir <= low_upper {
690. new_low_upper = find_dir;
691. new_upper_point = find;
692. }
693. }
694.
695. if lower_rdir < 0.0 && new_lower_rdir > 0.0 {
696. let find = bisect(
697. prev_x,
698. test_x,
699. |x| {
700. if x == point {
701. lower_start - d(x)
702. } else {
703. get_slope(x, f(x) - epsilon) - d(x)
704. }
705. },
706. TOL,
707. );
708. let find_dir = get_slope(find, f(find) - epsilon);
709. if find_dir >= high_lower {
710. new_high_lower = find_dir;
711. new_lower_point = find;
712. }
713. }
714.
715. low_upper = new_low_upper;
716. high_lower = new_high_lower;
717. lower_point = new_lower_point;
718. upper_point = new_upper_point;
719.
720. if slope == low_upper {
721. return Some((low_upper, test_x, upper_point, true));
722. }
723.
724. if slope == high_lower {
725. return Some((high_lower, test_x, lower_point, false));
726. }
727.
728. if slope > low_upper {
729. let find_x = find_root(
730. prev_x,
731. test_x,
732. None,
733. |x| {
734. if x == start {
735. d(x) - low_upper
736. } else {
737. get_slope(x, f(x)) - low_upper
738. }
739. },
740. |x| {
741. let y = f(x);
742. let bottom = x - point;
743. (bottom * d(x) - (y - point_y)) / (bottom * bottom)
744. },
745. TOL,
746. );
747. return Some((low_upper, find_x, upper_point, true));
748. }
749.
750. if slope < high_lower {
751. let find_x = find_root(
752. prev_x,
753. test_x,
754. None,
755. |x| {
756. if x == start {
757. d(x) - high_lower
758. } else {
759. get_slope(x, f(x)) - high_lower
760. }
761. },
762. |x| {
763. let y = f(x);
764. let bottom = x - point;
765. (bottom * d(x) - (y - point_y)) / (bottom * bottom)
766. },
767. TOL,
768. );
769. return Some((high_lower, find_x, lower_point, false));
770. }
771.
772. if test_x == max_x {
773. return None;
774. }
775.
776. upper_rdir = new_upper_rdir;
777. lower_rdir = new_lower_rdir;
778. prev_dd = new_dd;
779. }
780. }
781.
782. // generate a line and normalize the vector
783. fn norm_line(x1: f64, y1: f64, x2: f64, y2: f64) -> (Line, f64) {
784. let dx = x2 - x1;
785. let dy = y2 - y1;
786. let dist = (dx * dx + dy * dy).sqrt();
787. let l = Line {
788. x: x1,
789. y: y1,
790. dir_x: dx / dist,
791. dir_y: dy / dist,
792. };
793. (l, dist)
794. }
795.
796. // used in the oracle
797. fn max_from_point<F: Fn(f64) -> f64, D: Fn(f64) -> f64, D2: Fn(f64) -> f64>(
798. point: (f64, f64),
799. start: f64,
800. epsilon: f64,
801. step: f64,
802. max_x: f64,
803. f: F,
804. d: D,
805. dd: D2,
806. ) -> Result<(f64, f64, f64, bool), (f64, f64)> {
807. let mut test_x = start;
808. let mut prev_x;
809. let mut steps = 1.0;
810. let get_slope = |x, y| (y - point.1) / (x - point.0);
811. let mut low_upper = if start == point.0 {
812. if (f(start) + epsilon) == point.1 {
813. d(start)
814. } else {
815. f64::INFINITY
816. }
817. } else {
818. get_slope(start, f(start) + epsilon)
819. };
820. let mut upper_point = start;
821. let mut upper_rdir = low_upper - d(start);
822. let mut high_lower = if start == point.0 {
823. if (f(start) - epsilon) == point.1 {
824. d(start)
825. } else {
826. -f64::INFINITY
827. }
828. } else {
829. get_slope(start, f(start) - epsilon)
830. };
831. let upper_start = low_upper;
832. let lower_start = high_lower;
833. let mut lower_point = start;
834. let mut lower_rdir = high_lower - d(start);
835. let mut prev_dd = dd(start);
836.
837. let hit_upper;
838.
839. loop {
840. prev_x = test_x;
841. test_x = start + steps * step;
842.
843. if test_x > max_x {
844. test_x = max_x;
845. }
846.
847. let mut new_dd = dd(test_x);
848.
849. if prev_dd * new_dd < 0.0 {
850. test_x = bisect(prev_x, test_x, |x| dd(x), TOL);
851. new_dd = 0.0;
852. } else {
853. steps += 1.0;
854. }
855.
856. let test_y = f(test_x);
857. let upper_slope = get_slope(test_x, test_y + epsilon);
858. let lower_slope = get_slope(test_x, test_y - epsilon);
859. let dir = d(test_x);
860. let new_upper_rdir = upper_slope - dir;
861. let new_lower_rdir = lower_slope - dir;
862.
863. let mut new_low_upper = low_upper;
864. let mut new_high_lower = high_lower;
865. let mut new_lower_point = lower_point;
866. let mut new_upper_point = upper_point;
867.
868. let mut found_upper = false;
869. let mut found_lower = false;
870.
871. if upper_slope <= low_upper {
872. new_low_upper = upper_slope;
873. new_upper_point = test_x;
874. }
875. if upper_rdir > 0.0 && new_upper_rdir < 0.0 {
876. let find = bisect(
877. prev_x,
878. test_x,
879. |x| {
880. if x == point.0 {
881. upper_start
882. } else {
883. get_slope(x, f(x) + epsilon) - d(x)
884. }
885. },
886. TOL,
887. );
888. let find_dir = get_slope(find, f(find) + epsilon); // d(find);
889. if find_dir <= low_upper {
890. new_low_upper = find_dir;
891. new_upper_point = find;
892. found_upper = true;
893. }
894. }
895.
896. if lower_slope >= high_lower {
897. new_high_lower = lower_slope;
898. new_lower_point = test_x;
899. }
900.
901. if lower_rdir < 0.0 && new_lower_rdir > 0.0 {
902. let find = bisect(
903. prev_x,
904. test_x,
905. |x| {
906. if x == point.0 {
907. lower_start
908. } else {
909. get_slope(x, f(x) - epsilon) - d(x)
910. }
911. },
912. TOL,
913. );
914. let find_dir = get_slope(find, f(find) - epsilon); // d(find);
915. if find_dir >= high_lower {
916. new_high_lower = find_dir;
917. new_lower_point = find;
918. found_lower = true;
919. }
920. }
921.
922. if found_upper {
923. if new_low_upper < high_lower {
924. upper_point = new_upper_point;
925. hit_upper = true;
926. break;
927. }
928. }
929.
930. if found_lower {
931. if low_upper < new_high_lower {
932. lower_point = new_lower_point;
933. hit_upper = false;
934. break;
935. }
936. }
937.
938. low_upper = new_low_upper;
939. high_lower = new_high_lower;
940. lower_point = new_lower_point;
941. upper_point = new_upper_point;
942.
943. // when this occurs, found_lower and found_upper are both true
944. if low_upper < high_lower {
945. hit_upper = upper_point > lower_point;
946. break;
947. }
948.
949. if test_x == max_x {
950. return Err((high_lower, low_upper));
951. }
952.
953. upper_rdir = new_upper_rdir;
954. lower_rdir = new_lower_rdir;
955. prev_dd = new_dd;
956. }
957.
958. if hit_upper {
959. let find_x = find_root(
960. prev_x,
961. upper_point,
962. None,
963. |x| get_slope(x, f(x) + epsilon) - high_lower,
964. |x| {
965. let y = f(x) + epsilon;
966. let bottom = x - point.0;
967. (bottom * d(x) - (y - point.1)) / (bottom * bottom)
968. },
969. TOL,
970. );
971. return Ok((high_lower, find_x, lower_point, true));
972. } else {
973. let find_x = find_root(
974. prev_x,
975. lower_point,
976. None,
977. |x| get_slope(x, f(x) - epsilon) - low_upper,
978. |x| {
979. let y = f(x) - epsilon;
980. let bottom = x - point.0;
981. (bottom * d(x) - (y - point.1)) / (bottom * bottom)
982. },
983. TOL,
984. );
985.
986. return Ok((low_upper, find_x, upper_point, false));
987. }
988. }
989.
990. // a struct to help with directional first and second derivatives
991. #[derive(Clone, Copy, Debug)]
992. pub struct Line {
993. x: f64,
994. y: f64,
995. dir_x: f64,
996. dir_y: f64,
997. }
998.
999. impl Line {
1000. #[inline]
1001. fn pos(self, t: f64) -> (f64, f64) {
1002. (self.x + t * self.dir_x, self.y + t * self.dir_y)
1003. }
1004.
1005. #[inline]
1006. fn val<F: Fn(f64, f64) -> f64>(self, t: f64, f: F) -> f64 {
1007. let x = self.x + t * self.dir_x;
1008. let y = self.y + t * self.dir_y;
1009. f(x, y)
1010. }
1011.
1012. #[inline]
1013. fn der<D: Fn(f64, f64) -> [f64; 2]>(self, t: f64, d: D) -> f64 {
1014. let x = self.x + t * self.dir_x;
1015. let y = self.y + t * self.dir_y;
1016. let [dfdx, dfdy] = d(x, y);
1017. self.dir_x * dfdx + self.dir_y * dfdy
1018. }
1019.
1020. #[inline]
1021. fn der_2<D2: Fn(f64, f64) -> [f64; 3]>(self, t: f64, dd: D2) -> f64 {
1022. let x = self.x + t * self.dir_x;
1023. let y = self.y + t * self.dir_y;
1024. let [dxdx, dydy, dxdy] = dd(x, y);
1025. let x_part = self.dir_x * dxdx + self.dir_y * dxdy;
1026. let y_part = self.dir_x * dxdy + self.dir_y * dydy;
1027. self.dir_x * x_part + self.dir_y * y_part
1028. }
1029. }
1030.
1031. fn bisect<F: Fn(f64) -> f64>(mut start: f64, mut end: f64, f: F, tolerance: f64) -> f64 {
1032. let mut f_start = f(start);
1033. let f_end = f(end);
1034. if f_start == 0.0 {
1035. return f_start;
1036. }
1037. if f_end == 0.0 {
1038. return f_end;
1039. }
1040. while end - start > tolerance {
1041. let mid = (start + end) / 2.0;
1042. if mid == start || mid == end {
1043. return mid;
1044. }
1045. let f_mid = f(mid);
1046. if f_mid == 0.0 {
1047. return mid;
1048. }
1049. if (f_mid > 0.0) != (f_start > 0.0) {
1050. end = mid;
1051. } else {
1052. start = mid;
1053. f_start = f_mid;
1054. }
1055. }
1056. return (start + end) / 2.0;
1057. }
1058.
1059. // attempts to use newton's method, but switches to bisection
1060. // if it doesn't converge (or converge to the right root).
1061. fn find_root<F: Fn(f64) -> f64, D: Fn(f64) -> f64>(
1062. mut start: f64,
1063. mut end: f64,
1064. mut initial: Option<f64>,
1065. f: F,
1066. d: D,
1067. tolerance: f64,
1068. ) -> f64 {
1069. // every iteration of this loop causes the range to get smaller
1070. // so it will eventually be smaller than tol
1071. // (assuming the tolerance is positive and numerical coarseness doesn't
1072. // cause problems)
1073. loop {
1074. let mut x = if let Some(x) = initial {
1075. x
1076. } else {
1077. (start + end) / 2.0
1078. };
1079. let mut n = 0;
1080. // newton's method
1081. // in theory this could get stuck in a cycle, so
1082. // we need a condition to prevent the loop from going
1083. // infinite. In most cases the break will terminate
1084. // this loop before the while condition does
1085. while n < 8 {
1086. n += 1;
1087. let offset = f(x) / d(x);
1088. if offset.abs() < tolerance {
1089. return x - offset;
1090. } else {
1091. x = x - offset;
1092. }
1093. // < and >= are not exact negations of eachother
1094. // because of NaN. This expression will trigger
1095. // if x is NaN, which makes this more stable.
1096. // this condition will also work if d(x) is 0.0,
1097. // which will cause offset to be +-infinity
1098. if !(end < x && x < start) {
1099. break;
1100. }
1101. }
1102. let mid = (start + end) / 2.0;
1103. let f_mid = f(mid);
1104. // bisection
1105. if mid == start || mid == end {
1106. return mid;
1107. }
1108. if (f_mid < 0.0) != (f(end) < 0.0) {
1109. start = mid;
1110. } else {
1111. end = mid;
1112. }
1113. initial = None;
1114. if (end - start) < tolerance {
1115. return mid;
1116. }
1117. }
1118. }
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Does (x^y)^a+b = x^(y*(a+b)) or x^((y*a)+b)?
I posses two Math books, both of which define a certain property of the algebraic manipulation of exponents in different ways.
For example:
Book one would claim that:
2^((3)2+3) = 2(3*5) = 2^15, since 2+3 = 5, and 3*5 = 15.
Book two would claim that:
2^((3)2+3) = 2^(6+3) = 2^9, as (3)2+3 = 9 due to the laws of Bodmas.
I asked my teacher about this, and she said that the method used by book one is correct, however, am very hesitant, as I still do not understand why the laws of Bodmas would not be abided by when algebraically manipulating exponents.
Thanks for any help in advance :)
• Your notation is extremely unclear. Are you asking what $$(x^y)^{a+b}$$ is? Commented Mar 23, 2015 at 22:34
$$(2^3)^{2+3}=2^{3(2+3)}=2^{15}$$ or: $$(2^3)^2\cdot 2^3=2^{3\cdot2+3}=2^9$$
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# Faster than the speed of light?
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posted on Feb, 6 2007 @ 03:12 AM
Originally posted by SuicideVirus
Originally posted by Quasar
Say you have a large motor in space, like your ceiling fan, with "fins" that project extremely far out away from the motor. There is a capsule with a seat in it close to the motor, that can run on a track from the motor out to the edge of the "fin." If this was long enough to accelerate you to the speed of light, (670,616,629.2 mph) would it be possible to fling you out faster than the speed of light?
No, it doesn't work that way.
To my understanding, the Theory of Relativity states that it is close to impossible to achieve the thrust needed to propel you faster than the speed of light. What if the thrust was centrifugal force?
It's all still acceleration, whether it's in a straight line or a curved line.
But how about this for your next science question: When you play laser beam on the wall with the cat, the tip of the laser beam can travel as fast as you can change the angular momentum, right? As you pivot the laser beam, the speed of the dot on the wall can be very fast. If you had a huge laser beam and was able to reflect it off planets several light years away, could the speed at which the dot of the beam moves from planet to planet ever travel faster than light? After all, you'd only have to move the beam a fraction of a degree here on Earth for the dot to travel light years across the deep sky.
Heh.
I see what you are saying. I have wondered about similar things. I think we would actually be talking about perspective effecting location. Very cool.
Now, if a craft could be kept in some sort of focused bubble of that light, at the end of the beam, it could be moved at that incredible speed.
Which, in a way, is a kind of improved Light-Sail concept.
Isn't it?
posted on Feb, 6 2007 @ 04:32 AM
I will post a exelent link about all the possibilities of "Faster then light"!
the best canditates are the Tachyon particles, but even that ones need yet to be prooven they exist.
enjoy:
FAster then light theorys
posted on Feb, 6 2007 @ 06:19 AM
Why cant we travel faster than light? Theres no proof we cant
No one has proven we cant, only theorised, therefore there is always a probability that we can go faster.
Remember only a few hundred years ago it was deemed impossible to travel at the speeds we attain now.
Only time will tell.
posted on Feb, 6 2007 @ 06:23 AM
Originally posted by SuicideVirus
But how about this for your next science question: When you play laser beam on the wall with the cat, the tip of the laser beam can travel as fast as you can change the angular momentum, right? As you pivot the laser beam, the speed of the dot on the wall can be very fast. If you had a huge laser beam and was able to reflect it off planets several light years away, could the speed at which the dot of the beam moves from planet to planet ever travel faster than light? After all, you'd only have to move the beam a fraction of a degree here on Earth for the dot to travel light years across the deep sky.
Heh.
Is it just me or does this sound like a very easy question? It seems you don’t comprehend that the distance involved in this experiment makes it completely different from playing with a laser at home on your wall.
The light being emitted from the laser is a constant stream of energy, it’s not some sort of “solid string line”. Therefore the minute you change the laser to point to another part of the sky even by 1 degree, the light then has to travel from its original starting point (i.e. the laser earth) to get to the planet it is now pointing towards.
The same concept applies when you are playing with a laser at home on your wall. The moment you change the direction of the laser it appears that it is instantly showing up on the wall at the new point. However it is not instant at all.
posted on Feb, 6 2007 @ 06:30 AM
I have my own question I find ultimately interesting:
Let’s say you have two ships in deep space, there’s no stars or galaxies in a 100 light year radius. Now, one of the ships, lets call it ship A, has an artificial gravity generator on board as powerful as our own sun. The generator is turned off by default. Ship B then travels away from ship A until it reaches a distance of 5 light minutes. Now, in an experiment, if ship A turns the switch to enable the gravity generator, how long will it take for Ship B to feel the effects of said generator?
Will it take 5 light minutes or will it be instant or something entirely different?
posted on Feb, 6 2007 @ 06:52 AM
I posted this initially as anonymous post, then decided to log in and post again so that I can read responses.
After struggling for a long time, I decided to understand this speed of light problem in a more simpler (but wierder) terms. This theory requires 3-D multiverse (note just space, not time) at every possible time points.
Here goes, first speed = c is fixed, but this speed is not in the three dimensional universe, but in the four dimensional space time. If you stay still , then you are moving at speed of 'c' in spacetime, but in the axis of time alone where 'c equals one second'. Now any movement you make in the three dimensional world reduces your speed in the time axis. You calculate this change by assuming that the speed in the 3-d world( spacetime minus time axis) is a projection of a unit diaognal in the spacetime, where the value of this unit is 'c'. For example, if you travelled 10 metre per second. Then instead of travelling one second on the time axis, you just travelled square_root(square(c) - 100) divided by c seconds on the time axis. The strange thing here is, if one of your parent was sitting idle while you were travelling 10 metre per second, then he/she would have travelled more in time than you are, yet you see them. The best way to resolve this paradox is to assume multi-verse at every possible time points (If there is a limit to measurable time, then can we use this as absolute minimum ?). So while you moved at 10 meter per second, you moved from one (3-D)universe and landed on another that was just behind the one you were originally in. The parent you see now is actually from another 3-D universe. Ahh...but how to account for the absence of duplicates. Why should everyone of you in the multiverse choose to do the same thing all the time... well, all I wanted was to understand Einstein's relativity but I ended up creating more questions than answers.
BTW, according to the above, since light travels at 'c', it never leaves the time it was created, since it is always moving in the same 3-D space at speed 'c', the projection on time axis is zero. It is objects/matter that keep moving from one 3-D universe to another depending on their speed, light just stays put in the universe/time it was created.
I posted this initially as anonymous post, then decided to log in and post again so that I can read responses.
posted on Feb, 6 2007 @ 07:16 AM
Originally posted by Quasar
Originally posted by Togetic
The really wacky part is that if two ships move away each at nearly c, then in a Newtonian model one ship would perceive the other ship as moving away at 2c. The really wacky and strangely elegant part is that if one of the ships were to send a photon to the other ship, the other ship would calculate it as traveling at c! Why do they get that measurement? It's because time actually starts to change. Basically, the universe says "Well, it needs to be observed as going at at c, because that's the law, but since distance and velocity are both fixed, and velocity x time = distance, then the only thing left to do is change time!"
Blows my mind. Still does, years after learning that for the first time.
I got into a good conversation with my father tonight, which is why I posted this. I have had this idea for awhile, but I have always contradicted myself once we start talking relativity. He says theres no way to reach the speed of light. But relative to us sitting in our house, the speed of light is C. Take into account the speed of the earth rotating on its axis, plus the revolution around the sun, plus the revolution around the black hole in the center of our galaxy, plus however fast our galaxy is travelling through space, how fast are we going?
If you observe light from your moving house on the earth, all light travels at c. If you are on a ship moving away from earth, light moves at c. You will observe the light moving at c, because the time is going to stretch and slow to make sure that you do measure it at c. There is, according to current models, no way around it.
posted on Feb, 6 2007 @ 07:18 AM
I have my own question I find ultimately interesting:
Let’s say you have two ships in deep space, there’s no stars or galaxies in a 100 light year radius. Now, one of the ships, lets call it ship A, has an artificial gravity generator on board as powerful as our own sun. The generator is turned off by default. Ship B then travels away from ship A until it reaches a distance of 5 light minutes. Now, in an experiment, if ship A turns the switch to enable the gravity generator, how long will it take for Ship B to feel the effects of said generator?
Will it take 5 light minutes or will it be instant or something entirely different?
Current theories have the force of gravity traveling on illusionary particles called gravitons, and when they hit you, you feel the force of gravity. In these models, gravitons can't travel faster than light.
The only circumstance in which information can move faster than light, allegedly, and it is probable an illusion, is Einstein's "Spooky action at a distance" phenomenon.
posted on Feb, 6 2007 @ 08:31 AM
If memory serves (and I'm getting on a bit so it may not) Einsteins theory that you can not travel faster than the speed of light was worked out using the spead of light as a constant speed that never changes.
However, I am pretty sure that recent studies have found that at the dawn of the universe, milliseconds after the "big bang" light travelled faster than it does now and slowed down to the speed we now regard as being a constant. If this is true, then which speed of light can we not travel faster than, its original speed or its speed now, and if its the speed now, then how did it originally travel faster if you cant go faster?
Now that I've created my own little paradox there I'm off to shoot a paintball from my car at 30 mph and see what happens
[edit on 6-2-2007 by timeportal]
posted on Feb, 6 2007 @ 08:42 AM
Einstien is not the end all be all of physics. There was a breakthrough not too long ago that suggested that light had actually slowed since the "Big Bang". Another breakthrough is that E=MC2 is not plausible for it is an unbalanced equation.
The speed of light is not constant. www.eurekalert.org...
posted on Feb, 6 2007 @ 09:15 AM
Originally posted by AlphaAnuOmega
Einstien is not the end all be all of physics. There was a breakthrough not too long ago that suggested that light had actually slowed since the "Big Bang". Another breakthrough is that E=MC2 is not plausible for it is an unbalanced equation.
The speed of light is not constant. www.eurekalert.org...
I am familiar with this research; that effect is explained as a quirk of quantum mechanics and doesn't bend relativity at all.
This, of course, doesn't mean that there is a way around traveling faster than light, at least in our four dimensions. Most likely, we haven't found it yet, because we can't perceive the extra dimensions implied by string theory and other unified models. I hope we'll figure out how to do it someday.
[edit on 2/6/2007 by Togetic]
posted on Feb, 6 2007 @ 09:19 AM
Watch dan akroyd's video, with david sereda
It explains how its possible to go faster than the speed of light. look for the part where he goes into detail about the GALAXY CLOCK. It's on part 2 of the video.
Very intriguing none the less.
And IMHO i think Einstein wasnt correct about everything he stated, He might not have been way off but enough.
I think we live in a world were we have to prove those theories wrong or well never get out of this void were living in.
posted on Feb, 6 2007 @ 09:25 AM
First off, in order to travel near the speed of light, we have to fully understand light. We first stated that light was energy, pure energy. We have hashed and rehashed this theory for years. Light started off as energy, then a fluid, and now particles. We have to have a standard of what light really is before we can use it as a speed standard.
posted on Feb, 6 2007 @ 09:56 AM
en.wikipedia.org...
posted on Feb, 6 2007 @ 10:07 AM
Some guys at Yale shoot some laser faster than the speed of light a few years ago? I could be wrong, its been know to happen.
The only thing that would work is if you could build a shield, force, laser what ever that would be like a bubble.
Everything around it could be traveling faster than the speed of light including the bubble.
But inside the bubble time, G-fore is normal for humans to handle
You CANNOT just sit on your hands because Einstein said. If we as a people stopped trying everytime someone said, we would still be living on a flat earth were the water flowed off into the Gods domain.
posted on Feb, 6 2007 @ 10:12 AM
this may have been covered but it is possible to go faster then the speed of light. My frineds, a physics and electronis prodegy was able to recreate an experiment where he took two prisims made of parphim wax, spaced exactly the wavelenght of one radio wave apart. He then created a drirectional radio atenna and pointed it at the prisims. The radio waves hit the prism and are instantaliously teleported to the other prism, jumping the gap. They do this instantaniously, so they go faster then light.
And to answer the origional question, nothing that has mass can every go fater then light. period.
posted on Feb, 6 2007 @ 10:16 AM
That is not true according to the theory of relativity. E=MC2 so if you are an Einstienian, you will see his loophole. In order for there to be electricity...ie, energy, Mas is accelerated at least twice the speed. It's simple...if it wasn't possible, we wouldn't be here.
posted on Feb, 6 2007 @ 10:19 AM
Originally posted by squidbones
this may have been covered but it is possible to go faster then the speed of light. My frineds, a physics and electronis prodegy was able to recreate an experiment where he took two prisims made of parphim wax, spaced exactly the wavelenght of one radio wave apart. He then created a drirectional radio atenna and pointed it at the prisims. The radio waves hit the prism and are instantaliously teleported to the other prism, jumping the gap. They do this instantaniously, so they go faster then light.
And to answer the origional question, nothing that has mass can every go fater then light. period.
Experiments like this are often quirks of quantum mechanics, but don't disprove relativity. The information is mostly likely not going faster than light, we just observe it that way.
posted on Feb, 6 2007 @ 10:22 AM
Originally posted by AlphaAnuOmega
That is not true according to the theory of relativity. E=MC2 so if you are an Einstienian, you will see his loophole. In order for there to be electricity...ie, energy, Mas is accelerated at least twice the speed. It's simple...if it wasn't possible, we wouldn't be here.
The problem here is that energy and electricity are not the same. Energy has a specific meaning; it is a quality that everything, even rubber, which doesn't conduct, has. Electricity is the movement of electrons through a conducting material. Energy is essentially the insubstantial "quality" that makes the electrons move.
It is a very vague concept, one I still have trouble wrapping my head around sometimes, but understanding it is important in picking apart E=mc^2.
posted on Feb, 6 2007 @ 10:30 AM
Here is what I don't understand as far as E=MC^2...why hasn't anyone been allowed to disprove it. You see in the news that people are almost there and someone disproves it. This theory isn't proven, hence the theory, so why believe it. Isn't that just a form of control??
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# Get Answers to all your Questions
#### Please solve RD Sharma class 12 chapter Determinants exercise 5.1 question 4 maths textbook solution
$D=1$
Hint:
Determinant matrix must be square (i.e. same number of rows and columns)
Given:
\begin{aligned} \left|\begin{array}{cc} \sin 10^{\circ} & -\cos 10^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ} \end{array}\right|\\ \end{aligned}
Solution:
\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}\\ &=\sin 10^{\circ} \cos 80^{\circ}+\cos 10^{\circ} \sin 80^{\circ}\\ &=\sin 10^{\circ} \cos \left(90^{\circ}-10^{\circ}\right)+\cos 10^{\circ} \sin \left(90^{\circ}-10^{\circ}\right) \quad\quad\quad \quad\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\\ &=\sin 10^{\circ} \sin 10^{\circ}+\cos 10^{\circ} \cos 10^{\circ}\\ &=\sin ^{2} 10^{\circ}+\cos ^{2} 10^{\circ} \quad \quad \quad \quad \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]\\ &=1 \end{aligned}
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CGAL 4.14 - 2D and 3D Linear Geometry Kernel
CGAL::Bbox_3 Class Reference
#include <CGAL/Bbox_3.h>
## Definition
An object b of the class Bbox_3 is a bounding box in the three-dimensional Euclidean space $$\E^3$$.
CGAL::Bbox_2
## Related Functions
(Note that these are not member functions.)
template<class InputIterator >
Bbox_3 bbox_3 (InputIterator begin, InputIterator past_end)
returns the bounding box of the objects in the range [first,past_end[. More...
template<class InputIterator , class Traits >
Bbox_3 bbox_3 (InputIterator begin, InputIterator past_end, const Traits &traits)
returns the bounding box of the objects in the range [first,past_end[. More...
bool do_overlap (const Bbox_3 &bb1, const Bbox_3 &bb2)
returns true iff bb1 and bb2 overlap, i.e., iff their intersection is non-empty.
## Creation
Bbox_3 ()
introduces an empty bounding box with lower left corner point at $$(\infty, \infty, \infty)$$ and with upper right corner point at $$(-\infty, -\infty, -\infty)$$, $$\infty$$ being std::numeric_limits<double>::infinity().
Bbox_3 (double x_min, double y_min, double z_min, double x_max, double y_max, double z_max)
introduces a bounding box b with lexicographically smallest corner point at (xmin, ymin, zmin) and lexicographically largest corner point at (xmax, ymax, zmax).
## Operations
bool operator== (const Bbox_3 &c) const
Test for equality.
bool operator!= (const Bbox_3 &q) const
Test for inequality.
int dimension () const
Returns 3.
double xmin () const
double ymin () const
double zmin () const
double xmax () const
double ymax () const
double zmax () const
double min (int i) const
Returns xmin() if i==0 or ymin() if i==1 or zmin() if i==2. More...
double max (int i) const
Returns xmax() if i==0 or ymax() if i==1 or zmax() if i==2. More...
Bbox_3 operator+ (const Bbox_3 &c) const
returns a bounding box of b and c.
Bbox_3operator+= (const Bbox_3 &c)
updates b to be the bounding box of b and c and returns itself.
void dilate (int dist)
dilates the bounding box by a specified number of ULP.
## ◆ max()
double CGAL::Bbox_3::max ( int i ) const
Returns xmax() if i==0 or ymax() if i==1 or zmax() if i==2.
Precondition
i>=0 and i<=2
## ◆ min()
double CGAL::Bbox_3::min ( int i ) const
Returns xmin() if i==0 or ymin() if i==1 or zmin() if i==2.
Precondition
i>=0 and i<=2
## ◆ bbox_3() [1/2]
template<class InputIterator >
Bbox_3 bbox_3 ( InputIterator begin, InputIterator past_end )
related
returns the bounding box of the objects in the range [first,past_end[.
Each object in the range must have a member function BBox_3 bbox() returning its bounding box.
## ◆ bbox_3() [2/2]
template<class InputIterator , class Traits >
Bbox_3 bbox_3 ( InputIterator begin, InputIterator past_end, const Traits & traits )
related
returns the bounding box of the objects in the range [first,past_end[.
Traits must provide a functor Traits::Construct_bbox_3 having an operator returning the bounding box of each object in the range. Traits must also have a member function Traits::Construct_bbox_3 construct_bbox_3_object() const.
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https://tex.stackexchange.com/questions/1938/aligning-across-aligned-equation-blocks/5645
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Aligning across 'aligned' equation blocks
I'd like to format two groups of equations such that all their equal signs line up and such that I can put a big brace to the right of each block to annotate that block. The last part is easily achieved using the aligned environment of amsmath:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
\left.\begin{aligned}
\alpha_X \times \alpha_Y &= \chi \\
\alpha_X \times \beta_Y &= \xi \\
\beta_X \times \beta_Y &= \zeta
\end{aligned}\right\} \quad X<Y \2em] \left.\begin{aligned} \Upsilon_j &= 0 \\ \Psi_j &= \sqrt{\sinh E - \tan^2\tfrac{F}{2}} \\ \Gamma_j &= F \end{aligned}\right\} \quad j=1,\ldots,g \end{align} \end{document} But that doesn't line up the equal signs. I guess I could achieve this with some manual horizontal white space, but I'd like something automatic. Something like the split environment which supports reusing the anchor points of the outside 'align' environment: \begin{align} A &= 0 \\ BB &= AVE + VEA + EAV \\ \begin{split} CCC &= UVWXY + VWXYU + WXYUV \\ &\quad {} + XYUVW + YUVWX \end{split} \end{align} Is there a way to make aligned reuse the anchor points of the outside align? Also, is there an automatic way to additionally line up the big braces in the first example? 3 Answers Here's a solution using the widths of the widest expression on each side. You have to specify which expression it is: \documentclass{article} \usepackage{amsmath} \newlength{\leftside} \newlength{\rightside} \newcommand*{\leftterm}{} \newcommand*{\rightterm}{} \newcommand*{\term}[1]{\displaystyle#1} \begin{document} \[ \renewcommand*{\leftterm}{\alpha_X \times \alpha_Y} \renewcommand*{\rightterm}{\sqrt{\sinh E - \tan^2\tfrac{F}{2}}} \settowidth{\leftside}{\term{\leftterm}} \settowidth{\rightside}{\term{\rightterm}} \begin{array}{l} \left.\begin{aligned} \leftterm &= \makebox[\rightside][l]{\term{\chi}} \\ \alpha_X \times \beta_Y &= \xi \\ \beta_X \times \beta_Y &= \zeta \end{aligned}\right\} \quad X<Y \\[2em] \left.\begin{aligned} \makebox[\leftside][r]{\term{\Upsilon_j}} &= 0 \\ \Psi_j &= \rightterm \\ \Gamma_j &= F \end{aligned}\right\} \quad j=1,\ldots,g \end{array}
\end{document}
Equal signs as well as the big braces line up:
• This is not exactly what I hoped for (since it still involves whitespace fiddling), but in the absence of better alternatives, I think this is a good solution. So, thanks! Aug 24, 2010 at 9:29
I have a solution as well. It is taken from the MathMode documentation on CTAN, page 112.
\documentclass{article}
\usepackage{amssymb}
\usepackage{amsmath}
\newcommand{\fakealign}{%
\mbox{\hspace{5cm}} & \mbox{\hspace{5cm}} \nonumber\\%
}
\begin{document}
An equation;
\vspace{-1cm}
\begin{align}
\fakealign
D &= I\left(1+r\right)^t
\end{align}
Equation with a long left hand side;
\vspace{-1cm}
\begin{align}
\fakealign
\left(1+d\right)^{-T} &= U
\end{align}
Equation with a long right hand side
\vspace{-1cm}
\begin{align}
\fakealign
PV &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align}
They all align with the = sign :)
\end{document}
I can't post an image because I am a new user and therefore don't have enough reputation. But build it and observe!
I think this simpler and more elegant than Stefan's solution, but hey. Horses for courses.
Enjoy!
• Nice solution! I think it would be even more elegant if you would add the negative vertical space at the end of the line in the macro: \[-1cm] Mar 3, 2013 at 16:53
You could adopt Andrew's or my answer to this question.
• While I like TikZ, I think it's a little overkill in this case. But thanks for the suggestion anyway. The \tikzmark macro is definitely something to remember. Aug 24, 2010 at 9:27
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https://www.gogeometry.com/school-college/1/p1048-tangent-circles-common-tangent-chord-metric-relations-math.htm
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# Online Math: Geometry Problem 1048: Circles, Tangent, Perpendicular, Diameter, Angel Bisector, Polya's Mind Map. Level: School, College.
The figure below shows the circles of diameters AB, BC, and AC of centers O1, O2, and O3, respectively (B is on AC). DE is perpendicular to AC at B. If the chord FH is tangent to circles O1 and O2, prove that DE is the bisector of angle FDH.
Home | SearchGeometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | Problems Art Gallery Art | 1041-1050 | Circle | Tangent Circles | Diameters and Chords | Circle Tangent Line | Perpendicular lines | Arbelos | Angle Bisector | Email | by Antonio Gutierrez
Add a solution to the problem 1048
Last updated: Oct 11, 2014
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https://bio.libretexts.org/Bookshelves/Introductory_and_General_Biology/Book%3A_Introductory_Biology_(CK-12)/6%3A_Ecology/6._17%3A_Population_Size%2C_Density%2C_and_Distribution
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# 6. 17: Population Size, Density, and Distribution
### Is this any way to live?
It is if you're a penguin. This population of penguins is made of all the individuals of the same species of penguins who live together. They seem to exist in a very crowded - or densely populated - environment, and in a random configuration.
## Population Size, Density, and Distribution
Communities are made up of populations of different species. In biology, a population is a group of organisms of the same species that live in the same area. The population is the unit of natural selection and evolution. How large a population is and how fast it is growing are often used as measures of its health.
### Population Size
Population size is the number of individuals in a population. For example, a population ofinsects might consist of 100 individual insects, or many more. Population size influences the chances of a species surviving or going extinct. Generally, very small populations are at greatest risk of extinction. However, the size of a population may be less important than its density.
### Population Density
Population density is the average number of individuals in a population per unit of area or volume. For example, a population of 100 insects that live in an area of 100 square meters has a density of 1 insect per square meter. If the same population lives in an area of only 1 square meter, what is its density? Which population is more crowded? How might crowding affect the health of a population?
### Population Distribution
Population density just represents the average number of individuals per unit of area or volume. Often, individuals in a population are not spread out evenly. Instead, they may live in clumps or some other pattern (see Figure below). The pattern may reflect characteristics of the species or its environment. Population distribution describes how the individuals are distributed, or spread throughout their habitat.
Patterns of Population Distribution. What factors influence the pattern of a population over space?
## Summary
• Population size is the number of individuals in a population.
• Population density is the average number of individuals per unit of area or volume.
• The pattern of spacing of individuals in a population may be affected by the characteristics of a species or its environment.
## Review
1. What is population density?
2. What are the differences between population density and distribution?
3. A population of 820 insects lives in a 1.2-acre area. They gather nectar from a population of 560 flowering plants. The plants live in a 0.2-acre area. Which population has greater density, the insects or the plants? Why?
4. What can you infer about a species that has a random pattern of distribution over space? A uniform pattern?
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MT5 Rsi indicators
Re: MT5 Rsi indicators
71
kvak wrote: Sun Mar 31, 2024 9:52 am Hello.
Thank you so much. It's perfect.
Re: MT5 Rsi indicators
72
RSX 2
Experimenting with RSX
As we know rsx is a smoother rsi. Was thinking if there is some space in its calculation to experiment and eventually improve it (if possible). So I tried this experiment. What is added are the depth of calculation (maximum 10 - default rsx has "depth" 3) and speed. Speed is a parameter that causes it to "overreact" (in positive sense this time, I hope) and to overcome the flattening problem a bit.
To show what is it about, here is an example of a 32 period rsx with default (upper) and with "speed" set to 0.8 (lower).
As it is obvious 2 things happened : with "speed" 0.8 it is faster than the regular one in detecting crosses (a bar or two but sometimes the difference is much, much more) and it reaches the predefined levels while the regular one almost never does.
I converted it to mt5.
Re: MT5 Rsi indicators
73
Another great indicator from Mladen and explained by him::::
Range Expansion Index (REI) definition :
Range Expansion Index (REI) is an oscillator that measures price changing rate and signals about overbought/oversold areas, in case a price shows weakness or strength. It was developed by Thomas DeMark and described in his book New Market Timing Techniques: Innovative Studies in Market Rhythm & Price Exhaustion!
The indicator values change from -100 up to +100. REI is an enhanced oscillator, as it stays neutral during a flat and shows trends only when considerable top or bottom have been reached.
This version :
This version is made according to Mark Jurik's idea : according to him, REI can be made as rsx(high) + rsx(low) - and the result should be much smoother. So, this version is doing exactly that.
You have an option to chose between three color changing modes :
color change when outer levels are crossed
color change when middle level (a sort of a "zero line") is crossed
This is a version using rsx2 with speed and depth settings, seems maybe a little faster but needs a lot of testing.
Re: MT5 Rsi indicators
74
mrtools wrote: Fri Apr 12, 2024 8:03 am Another great indicator from Mladen and explained by him::::
Range Expansion Index (REI) definition :
Range Expansion Index (REI) is an oscillator that measures price changing rate and signals about overbought/oversold areas, in case a price shows weakness or strength. It was developed by Thomas DeMark and described in his book https://www.amazon.com/Science-Technica ... 0471035483.
The indicator values change from -100 up to +100. REI is an enhanced oscillator, as it stays neutral during a flat and shows trends only when considerable top or bottom have been reached.
This version :
This version is made according to Mark Jurik's idea : according to him, REI can be made as rsx(high) + rsx(low) - and the result should be much smoother. So, this version is doing exactly that.
You have an option to chose between three color changing modes :
color change when outer levels are crossed
color change when middle level (a sort of a "zero line") is crossed
This is a version using rsx2 with speed and depth settings, seems maybe a little faster but needs a lot of testing.
https://www.wiley.com/en-us/The+New+Sci ... 0471035480
Here's the working link for the book since it appears to be broken in your initial post.
Re: MT5 Rsi indicators
76
RSI Dashboard
coder: cja
date: April 16, 2024
Code: Select all
``AUDCAD,AUDCHF,AUDJPY,AUDNZD,AUDUSD,CADCHF,CADJPY,CHFJPY,EURAUD,EURCAD,EURCHF,EURGBP,EURJPY,EURNZD,EURUSD,GBPAUD,GBPCAD,GBPCHF,GBPJPY,GBPNZD,GBPUSD,NZDCAD,NZDCHF,NZDJPY,NZDUSD,USDCAD,USDCHF,USDJPY,XAUUSD``
Re: MT5 Rsi indicators
77
mrtools wrote: Fri Apr 12, 2024 8:03 am Another great indicator from Mladen and explained by him::::
Range Expansion Index (REI) definition :
Range Expansion Index (REI) is an oscillator that measures price changing rate and signals about overbought/oversold areas, in case a price shows weakness or strength. It was developed by Thomas DeMark and described in his book New Market Timing Techniques: Innovative Studies in Market Rhythm & Price Exhaustion!
The indicator values change from -100 up to +100. REI is an enhanced oscillator, as it stays neutral during a flat and shows trends only when considerable top or bottom have been reached.
This version :
This version is made according to Mark Jurik's idea : according to him, REI can be made as rsx(high) + rsx(low) - and the result should be much smoother. So, this version is doing exactly that.
You have an option to chose between three color changing modes :
color change when outer levels are crossed
color change when middle level (a sort of a "zero line") is crossed
This is a version using rsx2 with speed and depth settings, seems maybe a little faster but needs a lot of testing.
Is it available in on chart version mrtools? Or rsx 2.01.
I want to see how rsx works on chart compares to separate window.
Re: MT5 Rsi indicators
78
Sutatong wrote: Thu Apr 25, 2024 1:51 pm Is it available in on chart version mrtools? Or rsx 2.01.
I want to see how rsx works on chart compares to separate window.
Mrtools, you can put aside my request. I'm already happy with the RSI on chart I just got.
Thanks anyway for your kind attention.
Re: MT5 Rsi indicators
79
Sutatong wrote: Thu Apr 25, 2024 6:45 pm Mrtools, you can put aside my request. I'm already happy with the RSI on chart I just got.
Thanks anyway for your kind attention.
Went ahead and made one, this version is a little different, it has 11 rsi types to choose from and averages for smoothing.
Re: MT5 Rsi indicators
80
mrtools wrote: Fri Apr 26, 2024 3:50 am Went ahead and made one, this version is a little different, it has 11 rsi types to choose from and averages for smoothing.
amazing!!
but please teach us how you put it on chart!
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# Exam 4: Data Structures and Algorithms
Distributed: Monday, 28 April 2003
Due: 10 a.m., Monday, 5 May 2003
No extensions!
This page may be found online at `http://www.cs.grinnell.edu/~rebelsky/Courses/CS153/2003S/Exams/exam.04.html`.
Contents
## Preliminaries
There are five problems on the exam. Some problems have subproblems. Each full problem is worth twenty-five points. The point value associated with a problem does not necessarily correspond to the complexity of the problem or the time required to solve the problem. If you write down the amount of time you spend on each problem and the total time you spend on the exam, I'll give you two points of extra credit.
You should answer only four of the five questions on the exam. If you answer five, I will choose which four to grade.
This examination is open book, open notes, open mind, open computer, open Web. However, it is closed person. That means you should not talk to other people about the exam. Other than that limitation, you should feel free to use all reasonable resources available to you. As always, you are expected to turn in your own work. If you find ideas in a book or on the Web, be sure to cite them appropriately.
Although you may use the Web for this exam, you may not post your answers to this examination on the Web (at least not until after I return exams to you). And, in case it's not clear, you may not ask others (in person, via email, or by posting a please help message) to put answers on the Web.
This is a take-home examination. You may use any time or times you deem appropriate to complete the exam, provided you return it to me by the due date. It is likely to take you about five to ten hours, depending on how well you've learned topics and how fast you work. You should not spend more than ten hours on this exam. Stop at ten hours and write: There's more to life than CS and you will earn at least 80 points on this exam. I would also appreciate it if you would write down the amount of time each problem takes. I expect that someone who has mastered the material and works at a moderate rate should have little trouble completing the exam in a reasonable amount of time. Since I worry about the amount of time my exams take, I will give two points of extra credit to the first two people who honestly report that they've spent at least seven hours on the exam or that they've finished the exam. (At that point, I may then change the exam.)
You must include both of the following statements on the cover sheet of the examination. Please sign and date each statement. Note that the statements must be true; if you are unable to sign either statement, please talk to me at your earliest convenience. Note also that inappropriate assistance is assistance from (or to) anyone other than myself or our teaching assistant.
1. I have neither received nor given inappropriate assistance on this examination.
2. I am not aware of any other students who have given or received inappropriate assistance on this examination.
Because different students may be taking the exam at different times, you are not permitted to discuss the exam with anyone until after I have returned it. If you must say something about the exam, you are allowed to say This is among the hardest exams I have ever taken. If you don't start it early, you will have no chance of finishing the exam. You may also summarize these policies. You may not tell other students which problems you've finished. You may not tell other students how long you've spent on the exam.
In many problems, I ask you to write code. Unless I specify otherwise in a problem, you should write working code and include examples that show that you've tested the code.
You should fully document all of the primary procedures (including parameters, purpose, value produced, preconditions, and postconditions). If you write helper procedures (and you may certainly write helper procedures) you should document those, too, although you may opt to write less documentation. When appropriate, you should include short comments within your code. You should also take care to format your code carefully.
Just as you should be careful and precise when you write code, so should you be careful and precise when you write prose. Please check your spelling and grammar. Since I should be equally careful, the whole class will receive one point of extra credit for each error in spelling or grammar you identify on this exam. I will limit that form of extra credit to five points.
I will give partial credit for partially correct answers. You ensure the best possible grade for yourself by emphasizing your answer and including a clear set of work that you used to derive the answer.
I may not be available at the time you take the exam. If you feel that a question is badly worded or impossible to answer, note the problem you have observed and attempt to reword the question in such a way that it is answerable. If it's a reasonable hour (before 10 p.m. and after 8 a.m.), feel free to try to call me in the office (269-4410) or at home (236-7445).
I will also reserve time at the start of classes this week and next to discuss any general questions you have on the exam.
## Problems
### Problem 1: Sorted Lists
Topics: Lists, Interfaces, Documentation
You may recall that I suggested that there are at least three basic kinds of lists:
• Simple Lists, which are expandable collections you can iterate.
• Ordered Lists, which are an extension of simple lists in which the client controls where in the list elements are added.
• Sorted Lists, which are an extension of simple lists in which a Sort Order determines the order in which the list elements are iterated.
As the definition suggests, sorted lists provide only the basic list operations; they just specialize those operations. How do you, as designer, indicate that the operations have been specialized? You write good documentation.
Design and document a `SortedList` interface and a `SortedListCursor` interface.
### Problem 2: Array-Based Stacks
Topics: Linear structures, Array-based implementations
Implement stacks using arrays. You need not document your stacks and you need not show me your testing (but you should test).
You need only implement the four central linear structure methods: `add`, `get`, `peek`, and `isEmpty`.
### Problem 3: Anti-Exceptions
Topics: Exceptions, Java Basics
In their attempts to better understand exceptions, Carla and Carl Caffeinated have decided to write a somewhat strange method that they call `antiExceptional`. Their method calls another method. If the called method throws an exception, their method increments a counter. If the called method does not throw an exception, their method throws an exception (and does not increment the counter).
Here's what they've come up with.
``` public static int exceptionCount = 0;
public static void antiExceptional(Object param)
throws Exception
{
try {
sampleMethod(param);
throw new Exception("sampleMethod succeeded");
}
catch (Exception e) {
++exceptionCount;
}
} // antiExceptional
```
Unfortunately, it seems as if their method never throws an exception. If `sampleMethod` fails, `antiExceptional` increments `exceptionCount` and returns normally, which is what they wanted. However, if `sampleMethod` succeeds, `antiExceptional` also increments `exceptionCount` and returns normally, which is not what they wanted.
Update the `antiExceptional` method so that it throws an exception when `sampleMethod` succeeds (and only then).
### Problem 4: Standard Object Methods and Linked Lists
Topics: Lists, Linked Structures, Java Basics
Most Java objects should provide four key methods:
• `String toString()`, which converts the object to a string;
• `boolean equals(Object other)`, which determines if the other object is naturally equal to the object;
• `Object clone()`, which creates a new copy of the current object; and
• `int hashCode()`, which generates a number for the object to be used in hashing (equal objects must have equal hash codes; unequal objects should have different hash codes, but may have equal hash codes).
Implement these methods for the `LinkedList` class.
Note that another object is equal to a `LinkedList` if it's a `List` and when you iterate the elements, they're the same.
Since we never completed the `LinkedList` class, you need not write working code. However, your code should convince me that you can write working code given sufficient time.
### Problem 5: Heaps
Topics: Linear structures, Trees, Sorting
Implement heapsort. You should build a `Heap` data structure, insert the values from the array to be sorted into the `Heap`, and then read them back out in order.
You need not document this code (although you should).
The only methods you need to include for your heap are the constructor, `add`, and `get`. You might also choose to implement `peek`.
## Errors
These are the errors observed by students. Since I have threatened to take off for grammatical or spelling errors, I give the whole class one point of extra credit for each such error they notice in this exam. Such extra credit is capped at five points.
• `sampleMethod(object)` instead of `sampleMethod(param)`. [EC, 1 point]
• `antiExceptional` is declared as `boolean` but is really of type `void`. [ON, 1 point]
## Questions and Answers
Here you may eventually find questions from your colleagues and my answers to those questions.
What should we do if the array fills in the array-based stack?
Throw an exception or expand the array. It's up to you.
What's a good strategy for your exams?
Start early. Ask questions. Focus on doing the problems that say "You need not write working code" first.
When will we go over hash code?
What else do you need to know? Friday.
What should `toString` return for a list?
Something that looks a lot like a Scheme list.
Can we recurse on the elements?
Certainly.
You say that a sorted list controls the order in which the elements are iterated. Doesn't it also control the order in which they are added and inserted?
Nope. Since the only way you see the elements of a sorted list are through iteration, the only important aspect to an outsider is iteration.
Since `add` can mess up iteration of sorted lists (e.g., if I add an element smaller than something I've already visited), can I note that `add` invalidates all cursors?
Certainly.
Didn't you promise us that you'd scale our successful work when we hit ten hours? All you've done is promise us a minimum grade.
I do not recall promising to scale your exams.
Do we cite sources if we simply take ideas and don't copy code or phrases directly?
Certainly. Such citation is a hallmark of academic honesty.
For problem 3, does the code have to work?
If all the other procedures are defined, your code should work.
What heap methods should we implement?
The constructor, `add`, and `get`.
How do I get my name printed on every page?
Save the exam in a file with your name on it (e.g., Rebelsy-Exam4). Print it with `a2ps`.
Do we really need to use the six P's everywhere?
You can use briefer documentation as long as it's clear that you've thought about all the details.
## History
Sunday, 27 April 2003 [Samuel A. Rebelsky]
• Created.
Monday, 28 April 2003 [Samuel A. Rebelsky]
• Finishing touches.
• Released.
• Almost immediately added questions and answers and corrections.
Wednesday, 30 April 2003 [Samuel A. Rebelsky]
• In a fit of insanity, updated exam to four out of five problems.
Saturday, 3 May 2003 [Samuel A. Rebelsky]
• Lots of questions and answers.
• Clarified what parts of the heap structure to implement.
Disclaimer: I usually create these pages on the fly, which means that I rarely proofread them and they may contain bad grammar and incorrect details. It also means that I tend to update them regularly (see the history for more details). Feel free to contact me with any suggestions for changes.
This document was generated by Siteweaver on Tue May 6 09:19:07 2003.
The source to the document was last modified on Tue May 6 08:12:57 2003.
This document may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/CS153/2003S/Exams/exam.04.html`.
You may wish to validate this document's HTML ; ; Check with Bobby
Samuel A. Rebelsky, rebelsky@grinnell.edu
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LightOJ 1220 Mysterious Bacteria
• 2018-08-08
• 29
• 0
题目:
Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactly x days. Now RC-01 produces exactly p new deadly Bacteria where x = bp (where b, p are integers). More generally, x is a perfect pth power. Given the lifetime x of a mother RC-01 you are to determine the maximum number of new RC-01 which can be produced by the mother RC-01.
Input:
Input starts with an integer T (≤ 50), denoting the number of test cases.
Each case starts with a line containing an integer x. You can assume that x will have magnitude at least 2 and be within the range of a 32 bit signed integer.
Output:
For each case, print the case number and the largest integer p such that x is a perfect pth power.
3
17
1073741824
25
Sample Output:
Case 1: 1
Case 2: 30
Case 3: 2
题目链接
X=p_{1}^{e_{1}}\times p_{2}^{e_{2}}\times…\times p_{k}^{e_{k}}
\thereforeY=gcd(e_{1},e_{2},…,e_{k})时可以将p合并为一个数M
X=M^{gcd(e_{1},e_{2},…,e_{k})}
X为负数且Y最大值为偶数(显然不能为偶数)时，需要将M逐次乘方直到Y为奇数。
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You are on page 1of 6
# Petroleum Refining Fourth Year Dr.Aysar T.
Jarullah
## Steam Quantity Required
The vapor is usually pumped to the down distillation column located below flash zone (i.e.
below the reign where the crude oil enters). The aim of pumping the vapor is
1- Remove the remaining gas oil from reduced crude (un-vaporized)
2- Reduce the partial pressure leading to boiling the compounds at low temp.
The water vapor or the steam is pumped due to the following reasons:
1) Available
2) Cheap
3) Don't mix with crude oil
4) Good heat transportation
## The steam quantity is calculated by using the following equation:
ns p
s
nt pt
ns: is the no. of moles of vap.
nt : is the total moles
ps: is vap. press.
pt: is the total press.
ps pt pn
( pt pn )
ns nt
pt
pn: is the press. of H.C
Ex: Estimate the no. of moles of steam required to add to the distillation column to reduce the
boiling temperature from 400 to 350F.
Sol.:
400 at 760mmHg
From Fig. 17, (by connection 350 with 400F) so pn=400mmHg
Let n t =100 mol
(760 400)
n s 100 47.38 moles
760
## Reflux and Reflux Ratio
Reflux is the ratio between the amounts of material that returned to the distillation column to
the material that result at distillation. In order to increase the degree of separation and high
purity, amount of liquid is returned to the column in addition to getting rid of heat making the
column in thermal equilibrium case.
Petroleum Refining Fourth Year Dr.Aysar T. Jarullah
Types of Reflux
1) Top tray reflux. In this type, the steam from the top of distillation column is
condensed then some of it is returned to the distillation column. This method
is easy to design and operate, but create big quantity of steam inside the
distillation column and the amount of reflux is not enough. The quantity of
heat drawn is calculated as (mcpT L.H ) (m for every lb).
2) Pump back reflux. In this arrangement, reflux is provided regular intervals.
This helps every plate to act as a true fractionator. The vapor load in the tower
is fairly uniform and hence a uniform and smaller diameter tower will do. The
rejected heat at the reflux locations can be effectively utilized. This method
provides enough liquid at different points along distillation column and hence
good separation, but so difficult to design and operate. The quantity of heat
drawn is calculated as (mcpT ) (m for every lb).
3) Pump around reflux. In this way, reflux from a lower plate is taken, cooled
and fed into the column at a higher level by 2 to 3 plates. This method needs
column distillation with high plates because the whole cut will be on one plate.
The quantity of heat drawn is calculated as (mcpT ) (m for every lb).
The product obtained from a top of ADC under atmospheric pressure is gasoline having B.P.
from C5 to 190F. This fraction withdrawal as a vapor then is condensed and a part of it will
be returned to the atmospheric column as a reflux, while a stabilization process is carried out
on the remaining part. The stabilization process is removing the compounds that have low
B.P. (from C2 to C4), where these compounds cause high vapor pressure.
The stabilization process is carried out by passing the feed through fractionating column
containing many trays and reflux ratio (10 to 1) in order to make the separation process at
high accuracy. These light compounds are regarded as a source to produce liquefied gas.
Other products will withdrawal from several points along atmospheric column as a liquid.
These products are cooled and a part of them will be returned to the column as a reflux and
the remaining parts send to the stripper towers (small fractionating column like column
distillation having 4-8 trays placed on each other – beside the distillation column). These
stripper towers are aimed to remove the light compounds that effect on the flash point.
F .P 0.64T 100, for distilled fraction
F.P 0.57T 110 , for crude oil
The other products are: T in F
Heavy straight naphtha (190 – 400F)
Kerosene (380 500F)
Light gas oil (500 – 600F)
Petroleum Refining Fourth Year Dr.Aysar T. Jarullah
The remaining part at the bottom of atmospheric distillation will be sent to the vaccum
distillation tower.
## 2- Vacuum Distillation Column
The residue from an atmospheric distillation tower can be sent to a vacuum distillation
tower, which recovers additional liquid. The furnace outlet temperatures required for
atmospheric pressure distillation of the heavier fractions of crude oil are so high that thermal
cracking would occur, with the resultant loss of product and equipment fouling. These
materials are therefore distilled under vacuum because the boiling temperature decreases with
a lowering of the pressure. Distillation is carried out with absolute pressures in the tower flash
zone area of 25 to 40 mmHg (Fig. 23). To improve vaporization, the effective pressure is
lowered even further (to 10 mmHg or less) by the addition of steam to the furnace inlet and at
the bottom of the vacuum tower. Addition of steam to the furnace inlet increases the furnace
tube velocity and minimizes coke formation in the furnace as well as decreasing the total
hydrocarbon partial pressure in the vacuum tower.
## The main products from this unit are:
1- Heavy gas oil (600 – 800F)
2- Lubricant – base stokes (B.P. >700F), Non-viscous (50 – 100SUU), viscous (100 –
200SUU) and mid (>200SUU) at 100F.
3- Asphalt or vacuum residue (B.P > 1000F)
## Yield Estimation of Crude Oil
1- Draw the curve between Vol.% dis. Vs. TBP.
2- Extrapolate the curve to the final point (EBP) of the distillation (100%) by fitting the
curve to a suitable polynomial function and extrapolating the results. From the curve,
we cal find the IBP if it is not given.
Petroleum Refining Fourth Year Dr.Aysar T. Jarullah
## 3- Find the intervals, ((IBP + EBP)/ no. of pseudo compounds)
4- Find the EBP of each pseudo comp. (intervals + IBP of the cut)
5- Find the normal BP (NBP) or average B.P ((EBP +IBP (EBP of the last cut))/2)
6- Read the yield% or vol.% from the curve at NBP.
Example: A petroleum cut has the following ASTM D86 Distillation data:
Divide the TBP curve of the petroleum cut into 20 pseudo-components. Calculate the liquid
volume percentage of each pseudo-component.
Sol:
Firstly, the temperatures should be converted from ASTM to TBP, as given previously. The
results will be as follow
Vol. % 0 10 30 50 70 90 95
ASTM (°C) 36.5 54 77 101.5 131 171 186.5
TBP (°C)-API 14.1 33.4 69 101.6 135.2 180.5 194.1
TBP (°C). Duabert -5.3 27.5 66.7 101.7 138.1 184.6 201.1
The TBP curve obtained extends to 95 volume percent distilled only. In order to obtain the
average boiling point of the last cuts, the curve is extrapolated to the final point of the
distillation (100%) by fitting the curve to a suitable polynomial function and extrapolating the
results. An Excel spreadsheet program was used to fit a fifth order polynomial function, as
shown Figure below for TBP temperature versus volume% using Duabert method.
Petroleum Refining Fourth Year Dr.Aysar T. Jarullah
Petroleum Refining Fourth Year Dr.Aysar T. Jarullah
Ex.: Plot the true boiling point curve for the kerosene product from Table below
Sol.:
The cumulative vol% at the IBP of kerosene = 1.33 + 7.27 + 16.56 = 25.16%
The cumulative vol% at the EBP of kerosene = 25.16 + 10.05 = 35.21%
The cumulative vol% at the 10% of kerosene cut = 0.1(10.05) + 25.16 = 26.17%. At vol% of
26.17% the estimated TBP is 186.73 °C (via drawing TBP curve directly or from the
polynomial fit equation of the Fig.). The procedure is repeated at 20% of kerosene volume
which yield 27.17% and TBP of 189.26 °C. Figure below shows the TBP curve for kerosene
which starts at IBP of 180 °C and ends at EBP of 240 °C.
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# A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Given: Diameter of cylinder = 12 cm
Height of cylinder = 15 cm
Diameter of cone = 6 cm
Height of cone = 12 cm
To find: Number of cones to be filled with ice-cream.
Formula Used:
Volume of cone =
Volume of the cylinder = πr2h
Volume of hemisphere =
Explanation:
Radius of cylinder “R” = = 6 cm
Height of cylinder “H” = 15 cm
Volume of cylinder = π×(6)2×15
= π×36×15
= 540π cm3
Radius of cone “r” = = 3 cm
Height of cone “h” = 12 cm
Volume of conical part =
=
= 36π cm3
Volume of hemispherical part =
= 18π cm3
Volume of ice-cream = Volume of conical part + Volume of hemispherical part
= 36π + 18π
= 54π cm3
= 10
Hence 10 cones are required.
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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# Representing a graph by a Vector-Valued Function
## Homework Statement
Sketch the space curve represented by the intersection of the surfaces. Then represent the curve by a vector-valued function using the given parameter.
## Homework Equations
z = x^2 + y^2, x + y = 0, x = t
## The Attempt at a Solution
Space curve sketched (elliptic paraboloid corresponding to z-axis)
Vector valued function: x = t, y = -t; z = t^2 + (-t)2; z = 2t^2; r(t) = ti - tj + 2t^2k
**Intersection of the surface, not sure how to obtain this. I have the feeling once I get it I'm gonna be shaking my head for having forgotten something. So I've tried setting
x^2 + y^2 = x + y
tried substituting in t for x and y values
tried reverse engineering what I'm supposed to do with the answers plugged in to the equations.
tried finding x int, y int, and z int.
Just don't know how to procede. Any assistance is greatly appreciated.
Dick
Homework Helper
## Homework Statement
Sketch the space curve represented by the intersection of the surfaces. Then represent the curve by a vector-valued function using the given parameter.
## Homework Equations
z = x^2 + y^2, x + y = 0, x = t
## The Attempt at a Solution
Space curve sketched (elliptic paraboloid corresponding to z-axis)
Vector valued function: x = t, y = -t; z = t^2 + (-t)2; z = 2t^2; r(t) = ti - tj + 2t^2k
**Intersection of the surface, not sure how to obtain this. I have the feeling once I get it I'm gonna be shaking my head for having forgotten something. So I've tried setting
x^2 + y^2 = x + y
tried substituting in t for x and y values
tried reverse engineering what I'm supposed to do with the answers plugged in to the equations.
tried finding x int, y int, and z int.
Just don't know how to procede. Any assistance is greatly appreciated.
If x=t and x+y=0, what is y in terms of t? Now what is z in terms of t? It is really simple.
I already have the vecor valued function. I'm looking for the points of intersection. How do I find that the surfaces intersect at ((root2), -(root2), 4)?
Dick
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# C-coded dot 1000x faster than numpy?
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## C-coded dot 1000x faster than numpy?
I have code that performs dot product of a 2D matrix of size (on the order of) [1000,16] with a vector of size [1000]. The matrix is float64 and the vector is complex128. I was using numpy.dot but it turned out to be a bottleneck. So I coded dot2x1 in c++ (using xtensor-python just for the interface). No fancy simd was used, unless g++ did it on it's own. On a simple benchmark using timeit I find my hand-coded routine is on the order of 1000x faster than numpy? Here is the test code: My custom c++ code is dot2x1. I'm not copying it here because it has some dependencies. Any idea what is going on? import numpy as np from dot2x1 import dot2x1 a = np.ones ((1000,16)) b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, -0.80311816+0.80311816j, -0.80311816-0.80311816j, 1.09707981+0.29396165j, 1.09707981-0.29396165j, -1.09707981+0.29396165j, -1.09707981-0.29396165j, 0.29396165+1.09707981j, 0.29396165-1.09707981j, -0.29396165+1.09707981j, -0.29396165-1.09707981j, 0.25495815+0.25495815j, 0.25495815-0.25495815j, -0.25495815+0.25495815j, -0.25495815-0.25495815j]) def F1(): d = dot2x1 (a, b) def F2(): d = np.dot (a, b) from timeit import timeit print (timeit ('F1()', globals=globals(), number=1000)) print (timeit ('F2()', globals=globals(), number=1000)) In [13]: 0.013910860987380147 << 1st timeit 28.608758996007964 << 2nd timeit -- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
Hi,On Tue, 23 Feb 2021 at 19.11, Neal Becker <[hidden email]> wrote:I have code that performs dot product of a 2D matrix of size (on the order of) [1000,16] with a vector of size [1000]. The matrix is float64 and the vector is complex128. I was using numpy.dot but it turned out to be a bottleneck. So I coded dot2x1 in c++ (using xtensor-python just for the interface). No fancy simd was used, unless g++ did it on it's own. On a simple benchmark using timeit I find my hand-coded routine is on the order of 1000x faster than numpy? Here is the test code: My custom c++ code is dot2x1. I'm not copying it here because it has some dependencies. Any idea what is going on?I had a similar experience - albeit with an older numpy and Python 2.7, so my comments are easily outdated and irrelevant. This was on Windows 10 64 bit, way more than plenty RAM. It took me forever to find out that numpy.dot was the culprit, and I ended up using fortran + f2py. Even with the overhead of having to go through f2py bridge, the fortran dot_product was several times faster.Sorry if It doesn’t help much.Andrea. import numpy as np from dot2x1 import dot2x1 a = np.ones ((1000,16)) b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, -0.80311816+0.80311816j, -0.80311816-0.80311816j, 1.09707981+0.29396165j, 1.09707981-0.29396165j, -1.09707981+0.29396165j, -1.09707981-0.29396165j, 0.29396165+1.09707981j, 0.29396165-1.09707981j, -0.29396165+1.09707981j, -0.29396165-1.09707981j, 0.25495815+0.25495815j, 0.25495815-0.25495815j, -0.25495815+0.25495815j, -0.25495815-0.25495815j]) def F1(): d = dot2x1 (a, b) def F2(): d = np.dot (a, b) from timeit import timeit print (timeit ('F1()', globals=globals(), number=1000)) print (timeit ('F2()', globals=globals(), number=1000)) In [13]: 0.013910860987380147 << 1st timeit 28.608758996007964 << 2nd timeit -- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
For the first benchmark apparently A.dot(B) with A real and B complex is a known issue performance wise https://github.com/numpy/numpy/issues/10468In general, it might be worth trying different BLAS backends. For instance, if you install numpy from conda-forge you should be able to switch between OpenBLAS, MKL and BLIS: https://conda-forge.org/docs/maintainer/knowledge_base.html#switching-blas-implementationRoman On 23/02/2021 19:32, Andrea Gavana wrote: > Hi, > > On Tue, 23 Feb 2021 at 19.11, Neal Becker <[hidden email] > > wrote: > > I have code that performs dot product of a 2D matrix of size (on the > order of) [1000,16] with a vector of size [1000]. The matrix is > float64 and the vector is complex128. I was using numpy.dot but it > turned out to be a bottleneck. > > So I coded dot2x1 in c++ (using xtensor-python just for the > interface). No fancy simd was used, unless g++ did it on it's own. > > On a simple benchmark using timeit I find my hand-coded routine is on > the order of 1000x faster than numpy? Here is the test code: > My custom c++ code is dot2x1. I'm not copying it here because it has > some dependencies. Any idea what is going on? > > > > I had a similar experience - albeit with an older numpy and Python 2.7, > so my comments are easily outdated and irrelevant. This was on Windows > 10 64 bit, way more than plenty RAM. > > It took me forever to find out that numpy.dot was the culprit, and I > ended up using fortran + f2py. Even with the overhead of having to go > through f2py bridge, the fortran dot_product was several times faster. > > Sorry if It doesn’t help much. > > Andrea. > > > > > import numpy as np > > from dot2x1 import dot2x1 > > a = np.ones ((1000,16)) > b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, > -0.80311816+0.80311816j, -0.80311816-0.80311816j, > 1.09707981+0.29396165j, 1.09707981-0.29396165j, > -1.09707981+0.29396165j, -1.09707981-0.29396165j, > 0.29396165+1.09707981j, 0.29396165-1.09707981j, > -0.29396165+1.09707981j, -0.29396165-1.09707981j, > 0.25495815+0.25495815j, 0.25495815-0.25495815j, > -0.25495815+0.25495815j, -0.25495815-0.25495815j]) > > def F1(): > d = dot2x1 (a, b) > > def F2(): > d = np.dot (a, b) > > from timeit import timeit > print (timeit ('F1()', globals=globals(), number=1000)) > print (timeit ('F2()', globals=globals(), number=1000)) > > In [13]: 0.013910860987380147 << 1st timeit > 28.608758996007964 << 2nd timeit > -- > Those who don't understand recursion are doomed to repeat it > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion> > > > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion> _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
One suspect is that it seems the numpy version was multi-threading. This isn't useful here, because I'm running parallel monte-carlo simulations using all cores. Perhaps this is perversely slowing things down? I don't know how to account for 1000x slowdown though. On Tue, Feb 23, 2021 at 1:40 PM Roman Yurchak <[hidden email]> wrote: > > For the first benchmark apparently A.dot(B) with A real and B complex is > a known issue performance wise https://github.com/numpy/numpy/issues/10468> > In general, it might be worth trying different BLAS backends. For > instance, if you install numpy from conda-forge you should be able to > switch between OpenBLAS, MKL and BLIS: > https://conda-forge.org/docs/maintainer/knowledge_base.html#switching-blas-implementation> > Roman > > On 23/02/2021 19:32, Andrea Gavana wrote: > > Hi, > > > > On Tue, 23 Feb 2021 at 19.11, Neal Becker <[hidden email] > > > wrote: > > > > I have code that performs dot product of a 2D matrix of size (on the > > order of) [1000,16] with a vector of size [1000]. The matrix is > > float64 and the vector is complex128. I was using numpy.dot but it > > turned out to be a bottleneck. > > > > So I coded dot2x1 in c++ (using xtensor-python just for the > > interface). No fancy simd was used, unless g++ did it on it's own. > > > > On a simple benchmark using timeit I find my hand-coded routine is on > > the order of 1000x faster than numpy? Here is the test code: > > My custom c++ code is dot2x1. I'm not copying it here because it has > > some dependencies. Any idea what is going on? > > > > > > > > I had a similar experience - albeit with an older numpy and Python 2.7, > > so my comments are easily outdated and irrelevant. This was on Windows > > 10 64 bit, way more than plenty RAM. > > > > It took me forever to find out that numpy.dot was the culprit, and I > > ended up using fortran + f2py. Even with the overhead of having to go > > through f2py bridge, the fortran dot_product was several times faster. > > > > Sorry if It doesn’t help much. > > > > Andrea. > > > > > > > > > > import numpy as np > > > > from dot2x1 import dot2x1 > > > > a = np.ones ((1000,16)) > > b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, > > -0.80311816+0.80311816j, -0.80311816-0.80311816j, > > 1.09707981+0.29396165j, 1.09707981-0.29396165j, > > -1.09707981+0.29396165j, -1.09707981-0.29396165j, > > 0.29396165+1.09707981j, 0.29396165-1.09707981j, > > -0.29396165+1.09707981j, -0.29396165-1.09707981j, > > 0.25495815+0.25495815j, 0.25495815-0.25495815j, > > -0.25495815+0.25495815j, -0.25495815-0.25495815j]) > > > > def F1(): > > d = dot2x1 (a, b) > > > > def F2(): > > d = np.dot (a, b) > > > > from timeit import timeit > > print (timeit ('F1()', globals=globals(), number=1000)) > > print (timeit ('F2()', globals=globals(), number=1000)) > > > > In [13]: 0.013910860987380147 << 1st timeit > > 28.608758996007964 << 2nd timeit > > -- > > Those who don't understand recursion are doomed to repeat it > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion> > > > > > > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion> > > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion-- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
I'm using fedora 33 standard numpy. ldd says: /usr/lib64/python3.9/site-packages/numpy/core/_multiarray_umath.cpython-39-x86_64-linux-gnu.so: linux-vdso.so.1 (0x00007ffdd1487000) libflexiblas.so.3 => /lib64/libflexiblas.so.3 (0x00007f0512787000) So whatever flexiblas is doing controls blas. On Tue, Feb 23, 2021 at 1:51 PM Neal Becker <[hidden email]> wrote: > > One suspect is that it seems the numpy version was multi-threading. > This isn't useful here, because I'm running parallel monte-carlo > simulations using all cores. Perhaps this is perversely slowing > things down? I don't know how to account for 1000x slowdown though. > > On Tue, Feb 23, 2021 at 1:40 PM Roman Yurchak <[hidden email]> wrote: > > > > For the first benchmark apparently A.dot(B) with A real and B complex is > > a known issue performance wise https://github.com/numpy/numpy/issues/10468> > > > In general, it might be worth trying different BLAS backends. For > > instance, if you install numpy from conda-forge you should be able to > > switch between OpenBLAS, MKL and BLIS: > > https://conda-forge.org/docs/maintainer/knowledge_base.html#switching-blas-implementation> > > > Roman > > > > On 23/02/2021 19:32, Andrea Gavana wrote: > > > Hi, > > > > > > On Tue, 23 Feb 2021 at 19.11, Neal Becker <[hidden email] > > > > wrote: > > > > > > I have code that performs dot product of a 2D matrix of size (on the > > > order of) [1000,16] with a vector of size [1000]. The matrix is > > > float64 and the vector is complex128. I was using numpy.dot but it > > > turned out to be a bottleneck. > > > > > > So I coded dot2x1 in c++ (using xtensor-python just for the > > > interface). No fancy simd was used, unless g++ did it on it's own. > > > > > > On a simple benchmark using timeit I find my hand-coded routine is on > > > the order of 1000x faster than numpy? Here is the test code: > > > My custom c++ code is dot2x1. I'm not copying it here because it has > > > some dependencies. Any idea what is going on? > > > > > > > > > > > > I had a similar experience - albeit with an older numpy and Python 2.7, > > > so my comments are easily outdated and irrelevant. This was on Windows > > > 10 64 bit, way more than plenty RAM. > > > > > > It took me forever to find out that numpy.dot was the culprit, and I > > > ended up using fortran + f2py. Even with the overhead of having to go > > > through f2py bridge, the fortran dot_product was several times faster. > > > > > > Sorry if It doesn’t help much. > > > > > > Andrea. > > > > > > > > > > > > > > > import numpy as np > > > > > > from dot2x1 import dot2x1 > > > > > > a = np.ones ((1000,16)) > > > b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, > > > -0.80311816+0.80311816j, -0.80311816-0.80311816j, > > > 1.09707981+0.29396165j, 1.09707981-0.29396165j, > > > -1.09707981+0.29396165j, -1.09707981-0.29396165j, > > > 0.29396165+1.09707981j, 0.29396165-1.09707981j, > > > -0.29396165+1.09707981j, -0.29396165-1.09707981j, > > > 0.25495815+0.25495815j, 0.25495815-0.25495815j, > > > -0.25495815+0.25495815j, -0.25495815-0.25495815j]) > > > > > > def F1(): > > > d = dot2x1 (a, b) > > > > > > def F2(): > > > d = np.dot (a, b) > > > > > > from timeit import timeit > > > print (timeit ('F1()', globals=globals(), number=1000)) > > > print (timeit ('F2()', globals=globals(), number=1000)) > > > > > > In [13]: 0.013910860987380147 << 1st timeit > > > 28.608758996007964 << 2nd timeit > > > -- > > > Those who don't understand recursion are doomed to repeat it > > > _______________________________________________ > > > NumPy-Discussion mailing list > > > [hidden email] > > > https://mail.python.org/mailman/listinfo/numpy-discussion> > > > > > > > > > > > _______________________________________________ > > > NumPy-Discussion mailing list > > > [hidden email] > > > https://mail.python.org/mailman/listinfo/numpy-discussion> > > > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion> > > > -- > Those who don't understand recursion are doomed to repeat it -- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
In reply to this post by Neal Becker maybe` C_CONTIGUOUS` vs` F_CONTIGUOUS? `Carl Am Di., 23. Feb. 2021 um 19:52 Uhr schrieb Neal Becker <[hidden email]>:One suspect is that it seems the numpy version was multi-threading. This isn't useful here, because I'm running parallel monte-carlo simulations using all cores. Perhaps this is perversely slowing things down? I don't know how to account for 1000x slowdown though. On Tue, Feb 23, 2021 at 1:40 PM Roman Yurchak <[hidden email]> wrote: > > For the first benchmark apparently A.dot(B) with A real and B complex is > a known issue performance wise https://github.com/numpy/numpy/issues/10468 > > In general, it might be worth trying different BLAS backends. For > instance, if you install numpy from conda-forge you should be able to > switch between OpenBLAS, MKL and BLIS: > https://conda-forge.org/docs/maintainer/knowledge_base.html#switching-blas-implementation > > Roman > > On 23/02/2021 19:32, Andrea Gavana wrote: > > Hi, > > > > On Tue, 23 Feb 2021 at 19.11, Neal Becker <[hidden email] > > > wrote: > > > > I have code that performs dot product of a 2D matrix of size (on the > > order of) [1000,16] with a vector of size [1000]. The matrix is > > float64 and the vector is complex128. I was using numpy.dot but it > > turned out to be a bottleneck. > > > > So I coded dot2x1 in c++ (using xtensor-python just for the > > interface). No fancy simd was used, unless g++ did it on it's own. > > > > On a simple benchmark using timeit I find my hand-coded routine is on > > the order of 1000x faster than numpy? Here is the test code: > > My custom c++ code is dot2x1. I'm not copying it here because it has > > some dependencies. Any idea what is going on? > > > > > > > > I had a similar experience - albeit with an older numpy and Python 2.7, > > so my comments are easily outdated and irrelevant. This was on Windows > > 10 64 bit, way more than plenty RAM. > > > > It took me forever to find out that numpy.dot was the culprit, and I > > ended up using fortran + f2py. Even with the overhead of having to go > > through f2py bridge, the fortran dot_product was several times faster. > > > > Sorry if It doesn’t help much. > > > > Andrea. > > > > > > > > > > import numpy as np > > > > from dot2x1 import dot2x1 > > > > a = np.ones ((1000,16)) > > b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, > > -0.80311816+0.80311816j, -0.80311816-0.80311816j, > > 1.09707981+0.29396165j, 1.09707981-0.29396165j, > > -1.09707981+0.29396165j, -1.09707981-0.29396165j, > > 0.29396165+1.09707981j, 0.29396165-1.09707981j, > > -0.29396165+1.09707981j, -0.29396165-1.09707981j, > > 0.25495815+0.25495815j, 0.25495815-0.25495815j, > > -0.25495815+0.25495815j, -0.25495815-0.25495815j]) > > > > def F1(): > > d = dot2x1 (a, b) > > > > def F2(): > > d = np.dot (a, b) > > > > from timeit import timeit > > print (timeit ('F1()', globals=globals(), number=1000)) > > print (timeit ('F2()', globals=globals(), number=1000)) > > > > In [13]: 0.013910860987380147 << 1st timeit > > 28.608758996007964 << 2nd timeit > > -- > > Those who don't understand recursion are doomed to repeat it > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion > > > > > > > > _______________________________________________ > > NumPy-Discussion mailing list > > [hidden email] > > https://mail.python.org/mailman/listinfo/numpy-discussion > > > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion -- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
In reply to this post by Roman Yurchak On Tue, 23 Feb 2021, 7:41 pm Roman Yurchak, <[hidden email]> wrote:For the first benchmark apparently A.dot(B) with A real and B complex is a known issue performance wise https://github.com/numpy/numpy/issues/10468 I splitted B into a vector of size (N, 2) for the real and imaginary part, and that makes the multiplication twice as fast.My configuration (also in Fedora 33) np.show_config() blas_mkl_info: NOT AVAILABLEblis_info: NOT AVAILABLEopenblas_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]blas_opt_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]lapack_mkl_info: NOT AVAILABLEopenblas_lapack_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]lapack_opt_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)] _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
The stackoverflow link above contains a simple testcase: ``````>>> from scipy.linalg import get_blas_funcs >>> gemm = get_blas_funcs("gemm", [X, Y]) >>> np.all(gemm(1, X, Y) == np.dot(X, Y)) True````````It would be of interest to benchmark gemm against np.dot. Maybe np.dot doesn't use blas at al for whatever reason? `` Am Di., 23. Feb. 2021 um 20:46 Uhr schrieb David Menéndez Hurtado <[hidden email]>:On Tue, 23 Feb 2021, 7:41 pm Roman Yurchak, <[hidden email]> wrote:For the first benchmark apparently A.dot(B) with A real and B complex is a known issue performance wise https://github.com/numpy/numpy/issues/10468 I splitted B into a vector of size (N, 2) for the real and imaginary part, and that makes the multiplication twice as fast.My configuration (also in Fedora 33) np.show_config() blas_mkl_info: NOT AVAILABLEblis_info: NOT AVAILABLEopenblas_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]blas_opt_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]lapack_mkl_info: NOT AVAILABLEopenblas_lapack_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)]lapack_opt_info: libraries = ['openblas', 'openblas'] library_dirs = ['/usr/local/lib'] language = c define_macros = [('HAVE_CBLAS', None)] _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
In reply to this post by Neal Becker On Tue, Feb 23, 2021 at 11:10 AM Neal Becker <[hidden email]> wrote:I have code that performs dot product of a 2D matrix of size (on the order of) [1000,16] with a vector of size [1000]. The matrix is float64 and the vector is complex128. I was using numpy.dot but it turned out to be a bottleneck. So I coded dot2x1 in c++ (using xtensor-python just for the interface). No fancy simd was used, unless g++ did it on it's own. On a simple benchmark using timeit I find my hand-coded routine is on the order of 1000x faster than numpy? Here is the test code: My custom c++ code is dot2x1. I'm not copying it here because it has some dependencies. Any idea what is going on? import numpy as np from dot2x1 import dot2x1 a = np.ones ((1000,16)) b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, -0.80311816+0.80311816j, -0.80311816-0.80311816j, 1.09707981+0.29396165j, 1.09707981-0.29396165j, -1.09707981+0.29396165j, -1.09707981-0.29396165j, 0.29396165+1.09707981j, 0.29396165-1.09707981j, -0.29396165+1.09707981j, -0.29396165-1.09707981j, 0.25495815+0.25495815j, 0.25495815-0.25495815j, -0.25495815+0.25495815j, -0.25495815-0.25495815j]) def F1(): d = dot2x1 (a, b) def F2(): d = np.dot (a, b) from timeit import timeit print (timeit ('F1()', globals=globals(), number=1000)) print (timeit ('F2()', globals=globals(), number=1000)) In [13]: 0.013910860987380147 << 1st timeit 28.608758996007964 << 2nd timeitI'm going to guess threading, although huge pages can also be a problem on a machine under heavy load running other processes. Call overhead may also matter for such small matrices.Chuck _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
On Tue, Feb 23, 2021 at 5:47 PM Charles R Harris <[hidden email]> wrote:On Tue, Feb 23, 2021 at 11:10 AM Neal Becker <[hidden email]> wrote:I have code that performs dot product of a 2D matrix of size (on the order of) [1000,16] with a vector of size [1000]. The matrix is float64 and the vector is complex128. I was using numpy.dot but it turned out to be a bottleneck. So I coded dot2x1 in c++ (using xtensor-python just for the interface). No fancy simd was used, unless g++ did it on it's own. On a simple benchmark using timeit I find my hand-coded routine is on the order of 1000x faster than numpy? Here is the test code: My custom c++ code is dot2x1. I'm not copying it here because it has some dependencies. Any idea what is going on? import numpy as np from dot2x1 import dot2x1 a = np.ones ((1000,16)) b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, -0.80311816+0.80311816j, -0.80311816-0.80311816j, 1.09707981+0.29396165j, 1.09707981-0.29396165j, -1.09707981+0.29396165j, -1.09707981-0.29396165j, 0.29396165+1.09707981j, 0.29396165-1.09707981j, -0.29396165+1.09707981j, -0.29396165-1.09707981j, 0.25495815+0.25495815j, 0.25495815-0.25495815j, -0.25495815+0.25495815j, -0.25495815-0.25495815j]) def F1(): d = dot2x1 (a, b) def F2(): d = np.dot (a, b) from timeit import timeit print (timeit ('F1()', globals=globals(), number=1000)) print (timeit ('F2()', globals=globals(), number=1000)) In [13]: 0.013910860987380147 << 1st timeit 28.608758996007964 << 2nd timeitI'm going to guess threading, although huge pages can also be a problem on a machine under heavy load running other processes. Call overhead may also matter for such small matrices.What BLAS library are you using. I get much better results using an 8 year old i5 and ATLAS.Chuck _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
See my earlier email - this is fedora 33, python3.9. I'm using fedora 33 standard numpy. ldd says: /usr/lib64/python3.9/site-packages/numpy/core/_multiarray_umath.cpython-39-x86_64-linux-gnu.so: linux-vdso.so.1 (0x00007ffdd1487000) libflexiblas.so.3 => /lib64/libflexiblas.so.3 (0x00007f0512787000) So whatever flexiblas is doing controls blas. flexiblas print FlexiBLAS, version 3.0.4 Copyright (C) 2014, 2015, 2016, 2017, 2018, 2019, 2020 Martin Koehler and others. This is free software; see the source code for copying conditions. There is ABSOLUTELY NO WARRANTY; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. Configured BLAS libraries: System-wide (/etc/flexiblasrc): System-wide from config directory (/etc/flexiblasrc.d/) OPENBLAS-OPENMP library = libflexiblas_openblas-openmp.so comment = NETLIB library = libflexiblas_netlib.so comment = ATLAS library = libflexiblas_atlas.so comment = User config (/home/nbecker/.flexiblasrc): Host config (/home/nbecker/.flexiblasrc.nbecker8): Available hooks: Backend and hook search paths: /usr/lib64/flexiblas/ Default BLAS: System: OPENBLAS-OPENMP User: (none) Host: (none) Active Default: OPENBLAS-OPENMP (System) Runtime properties: verbose = 0 (System) So it looks to me it is using openblas-openmp. On Tue, Feb 23, 2021 at 8:00 PM Charles R Harris <[hidden email]> wrote: > > > > On Tue, Feb 23, 2021 at 5:47 PM Charles R Harris <[hidden email]> wrote: >> >> >> >> On Tue, Feb 23, 2021 at 11:10 AM Neal Becker <[hidden email]> wrote: >>> >>> I have code that performs dot product of a 2D matrix of size (on the >>> order of) [1000,16] with a vector of size [1000]. The matrix is >>> float64 and the vector is complex128. I was using numpy.dot but it >>> turned out to be a bottleneck. >>> >>> So I coded dot2x1 in c++ (using xtensor-python just for the >>> interface). No fancy simd was used, unless g++ did it on it's own. >>> >>> On a simple benchmark using timeit I find my hand-coded routine is on >>> the order of 1000x faster than numpy? Here is the test code: >>> My custom c++ code is dot2x1. I'm not copying it here because it has >>> some dependencies. Any idea what is going on? >>> >>> import numpy as np >>> >>> from dot2x1 import dot2x1 >>> >>> a = np.ones ((1000,16)) >>> b = np.array([ 0.80311816+0.80311816j, 0.80311816-0.80311816j, >>> -0.80311816+0.80311816j, -0.80311816-0.80311816j, >>> 1.09707981+0.29396165j, 1.09707981-0.29396165j, >>> -1.09707981+0.29396165j, -1.09707981-0.29396165j, >>> 0.29396165+1.09707981j, 0.29396165-1.09707981j, >>> -0.29396165+1.09707981j, -0.29396165-1.09707981j, >>> 0.25495815+0.25495815j, 0.25495815-0.25495815j, >>> -0.25495815+0.25495815j, -0.25495815-0.25495815j]) >>> >>> def F1(): >>> d = dot2x1 (a, b) >>> >>> def F2(): >>> d = np.dot (a, b) >>> >>> from timeit import timeit >>> print (timeit ('F1()', globals=globals(), number=1000)) >>> print (timeit ('F2()', globals=globals(), number=1000)) >>> >>> In [13]: 0.013910860987380147 << 1st timeit >>> 28.608758996007964 << 2nd timeit >> >> >> I'm going to guess threading, although huge pages can also be a problem on a machine under heavy load running other processes. Call overhead may also matter for such small matrices. >> > > What BLAS library are you using. I get much better results using an 8 year old i5 and ATLAS. > > Chuck > _______________________________________________ > NumPy-Discussion mailing list > [hidden email] > https://mail.python.org/mailman/listinfo/numpy-discussion-- Those who don't understand recursion are doomed to repeat it _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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## Re: C-coded dot 1000x faster than numpy?
On Wed, Feb 24, 2021 at 5:36 AM Neal Becker <[hidden email]> wrote:See my earlier email - this is fedora 33, python3.9. I'm using fedora 33 standard numpy. ldd says: /usr/lib64/python3.9/site-packages/numpy/core/_multiarray_umath.cpython-39-x86_64-linux-gnu.so: linux-vdso.so.1 (0x00007ffdd1487000) libflexiblas.so.3 => /lib64/libflexiblas.so.3 (0x00007f0512787000) So whatever flexiblas is doing controls blas. flexiblas print FlexiBLAS, version 3.0.4 Copyright (C) 2014, 2015, 2016, 2017, 2018, 2019, 2020 Martin Koehler and others. This is free software; see the source code for copying conditions. There is ABSOLUTELY NO WARRANTY; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. Configured BLAS libraries: System-wide (/etc/flexiblasrc): System-wide from config directory (/etc/flexiblasrc.d/) OPENBLAS-OPENMP library = libflexiblas_openblas-openmp.so comment = NETLIB library = libflexiblas_netlib.so comment = ATLAS library = libflexiblas_atlas.so comment = User config (/home/nbecker/.flexiblasrc): Host config (/home/nbecker/.flexiblasrc.nbecker8): Available hooks: Backend and hook search paths: /usr/lib64/flexiblas/ Default BLAS: System: OPENBLAS-OPENMP User: (none) Host: (none) Active Default: OPENBLAS-OPENMP (System) Runtime properties: verbose = 0 (System) So it looks to me it is using openblas-openmp. ISTR that there have been problems with openmp. There are a ton of OpenBLAS versions available in fedora 33. Just available via flexiblasflexiblas-openblas-openmp.x86_64 : FlexiBLAS wrappers for OpenBLASflexiblas-openblas-openmp.i686 : FlexiBLAS wrappers for OpenBLASflexiblas-openblas-openmp64.x86_64 : FlexiBLAS wrappers for OpenBLAS (64-bit)flexiblas-openblas-serial.x86_64 : FlexiBLAS wrappers for OpenBLASflexiblas-openblas-serial64.x86_64 : FlexiBLAS wrappers for OpenBLAS (64-bit)flexiblas-openblas-threads.x86_64 : FlexiBLAS wrappers for OpenBLASflexiblas-openblas-threads64.x86_64 : FlexiBLAS wrappers for OpenBLAS (64-bit)I am not sure how to make use of flexiblas, but would explore that. We might need to do something with distutils to interoperate with it or maybe you can control it though site.cfg. There are 12 versions available in total. I would suggest trying serial or pthreads.Chuck _______________________________________________ NumPy-Discussion mailing list [hidden email] https://mail.python.org/mailman/listinfo/numpy-discussion
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### Author Topic: KX Reference (Read 102358 times)
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#### hwj-d
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##### Re: KX Reference
« Reply #150 on: September 09, 2017, 02:03:29 pm »
If i touch them, it has directly impact to the measurement.
Is the change immediately = capacitive influence
or slowly rising / falling = thermal offset?
I know, that must be kept apart.
If i touch or move them, this becomes visible at the 10 or 20µV curve at the dmm measuring with 100 plc for a time. Is it slow, or immediately?
#### Andreas
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##### Re: KX Reference
« Reply #151 on: September 09, 2017, 02:17:45 pm »
Hello,
normally if it is thermal it will change over several 100 NPLC measurements (similar to a e-function) if you put your hand on the wiring and let it there for a while or if you remove it again. Thermal time constants are usually more in a 1 minute (or larger) range.
If it is instaneous (= within one 100 NPLC measurement) you have most likely a capacitive influence = EMI problem.
with best regards
Andreas
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#### hwj-d
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##### Re: KX Reference
« Reply #152 on: September 09, 2017, 03:05:35 pm »
normally if it is thermal it will change over several 100 NPLC measurements (similar to a e-function) if you put your hand on the wiring and let it there for a while or if you remove it again. Thermal time constants are usually more in a 1 minute (or larger) range.
If it is instaneous (= within one 100 NPLC measurement) you have most likely a capacitive influence = EMI problem.
I know, my measurement is not optimal too. But look at this with 100 NPLC taken:
The impact at the end comes from touching [edit: carefully only DMM] using USB-Stick for screen capture just before. The new measurements after that stuck at this new level. My impression is, that this is caused from capacitive influence, but changes the measuring till other TC influence overlapped the result.
But, i think, the lm399 based 3446xa are VERY TC sensitive as others noticed too. That is not nearly comparable to the stability like a 34470a nor 3458a (... as we know )
https://www.eevblog.com/forum/metrology/mx-reference/msg1296874/#msg1296874
« Last Edit: September 09, 2017, 03:15:38 pm by hwj-d »
#### hwj-d
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##### Re: KX Reference
« Reply #153 on: September 09, 2017, 03:32:28 pm »
Quote
0.4ppm/K tempco is more than 8 times worse than expected (<0.05), so there is definately a system problem.
Yes, but this must match high grade environmental conditions with comparable dmm's like your keithley's or 3458a's. Thats the task, after that, to have one thing in the lab, that provides this requirement, after cal, with yours and others help ...
#### Dr. Frank
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##### Re: KX Reference
« Reply #154 on: September 09, 2017, 03:43:46 pm »
normally if it is thermal it will change over several 100 NPLC measurements (similar to a e-function) if you put your hand on the wiring and let it there for a while or if you remove it again. Thermal time constants are usually more in a 1 minute (or larger) range.
If it is instaneous (= within one 100 NPLC measurement) you have most likely a capacitive influence = EMI problem.
I know, my measurement is not optimal too. But look at this with 100 NPLC taken:
The impact at the end comes from touching [edit: carefully only DMM] using USB-Stick for screen capture just before. The new measurements after that stuck at this new level. My impression is, that this is caused from capacitive influence, but changes the measuring till other TC influence overlapped the result.
But, i think, the lm399 based 3446xa are VERY TC sensitive as others noticed too. That is not nearly comparable to the stability like a 34470a nor 3458a (... as we know )
https://www.eevblog.com/forum/metrology/mx-reference/msg1296874/#msg1296874
Well, I don't think, that the 3446x instruments are that sensitive.
Never encountered such dips, when touching the case.
I assume, that your LTZ circuit is a bit sensitive.. see current discussion about shielding, in the MX Reference thread.
To have an idea about the stability of the 34465A (LM399) vs. 34470A (LTZ1000A), see my review here:
https://www.eevblog.com/forum/testgear/keysight's-new-34465a-(6-5-digit)-and-34470a-(7-5-digit)-bench-multimeters/msg889217/#msg889217
In your graph, the scale of noise and of the dip is not obvious.
By using the scaling math function and the unit function, you might normalize your graph to read ppm deviation.. then you'd have much more resolution in the statistics also.
Frank
« Last Edit: September 09, 2017, 03:48:52 pm by Dr. Frank »
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#### d-smes
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##### Re: KX Reference
« Reply #155 on: September 09, 2017, 04:09:23 pm »
I know, my measurement is not optimal too. But look at this with 100 NPLC taken:
The impact at the end comes from touching [edit: carefully only DMM] using USB-Stick for screen capture just before. The new measurements after that stuck at this new level. My impression is, that this is caused from capacitive influence, but changes the measuring till other TC influence overlapped the result.
But, i think, the lm399 based 3446xa are VERY TC sensitive as others noticed too. That is not nearly comparable to the stability like a 34470a nor 3458a (... as we know )
https://www.eevblog.com/forum/metrology/mx-reference/msg1296874/#msg1296874
I agree '399 based 3446xa are very TC sensitive. My results at https://www.eevblog.com/forum/metrology/ultra-precision-reference-ltz1000/msg1245750/#msg1245750 Dr. Frank suggested several improvements to my connections, but that didn't alter the high correlation of 34465a readings with temperature.
Regarding the reading shift from touching DMM, I too have noticed this. I usually change grounding around and/or tie floating DUT to meter case or earth ground. After a bit of experimenting, I can usually get the touch artifacts to go away.
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#### Andreas
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##### Re: KX Reference
« Reply #156 on: September 09, 2017, 04:36:43 pm »
The impact at the end comes from touching [edit: carefully only DMM] using USB-Stick for screen capture just before.
But, i think, the lm399 based 3446xa are VERY TC sensitive as others noticed too. That is not nearly comparable to the stability like a 34470a nor 3458a (... as we know )
Hello,
a capacitive effect would be reversible. So if it is a one way jump it is a different story.
From my experience the LM399 in HP devices are carefully selected (i.e. unheated tempco near zero).
If you measure the reference output voltage it should be somewhere near 6875 mV. (see LM399 thread).
with best regards
Andreas
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#### hwj-d
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##### Re: KX Reference
« Reply #157 on: September 09, 2017, 04:37:14 pm »
Well, I don't think, that the 3446x instruments are that sensitive.
Never encountered such dips, when touching the case.
I assume, that your LTZ circuit is a bit sensitive.. see current discussion about shielding, in the MX Reference thread.
To have an idea about the stability of the 34465A (LM399) vs. 34470A (LTZ1000A), see my review here:
https://www.eevblog.com/forum/testgear/keysight's-new-34465a-(6-5-digit)-and-34470a-(7-5-digit)-bench-multimeters/msg889217/#msg889217
In your graph, the scale of noise and of the dip is not obvious.
By using the scaling math function and the unit function, you might normalize your graph to read ppm deviation.. then you'd have much more resolution in the statistics also.
Thanks Dr. Frank for your answer. I can only learn from that, what happens (t)here.
I know, the scale of noise and of the dip is really not obvious.
I don't want to mix this dip problem with that TC of my meter. That's a 34461a.
Is it right, that this meters have TC of 1ppm/k by itself? So, that we can only measure not better than that TC with it?
Best regards
#### TiN
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##### Re: KX Reference
« Reply #158 on: September 09, 2017, 05:20:36 pm »
Quote
So, that we can only measure not better than that TC with it?
Not really. If you maintain temperature constant, you have constant tempco as well.
And since you only want to measure tempco drift of DUT, you can ignore any tempco effect on the meter's own reference completely, if it stays unchanged during complete test run. That's why I perform very slow and linear temperature ramp up and down on DUT, to make sure meter's own unstability does not introduce large errors.
If both direction ramps produce matching data, confidence is high.
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#### hwj-d
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##### Re: KX Reference
« Reply #159 on: September 09, 2017, 05:21:53 pm »
Hello Andreas,
Quote
If you measure the reference output voltage it should be somewhere near 6875 mV. (see LM399 thread).
I can't really do that. That meter is relative new, one year old. Maybe i put it to an other room with more constant temperature, other shielding methods, and so on.
I assume that this dip is a mistake from my own. But this relatively bad tempco is one of the meters not so good property. d-smes has confirmed this, and Ian Johnston has already dropped such a notice about this meter, as far as i can remember. Certainly it depends on the environment, this meter must do his job.
#### hwj-d
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##### Re: KX Reference
« Reply #160 on: September 09, 2017, 06:09:23 pm »
Quote
So, that we can only measure not better than that TC with it?
Not really. If you maintain temperature constant, you have constant tempco as well.
And since you only want to measure tempco drift of DUT, you can ignore any tempco effect on the meter's own reference completely, if it stays unchanged during complete test run. That's why I perform very slow and linear temperature ramp up and down on DUT, to make sure meter's own unstability does not introduce large errors.
If both direction ramps produce matching data, confidence is high.
That's the problem. If i measure the DUT over the 24h and the temp changes 2-3 °C up and down, i don't know, from what the measured changes comes, maybe in combination of DMM and/or DUT, if i don't have a definitely "comparing-normal" (the objektive meta measurement) to that. So, IF i have a calibrated LTZ reference, and i know THAT TC, the deviation must come fro the dmm. And vice versa. If the LTZ is the DUT, i dont' know anything, if i can't trust the dmm. I always see the TC of the DUT AND the DMM. The LTZ-reference should control that, in the future, if it is calibrated. Maybe i'm wrong? Am I fail to see something?
« Last Edit: September 09, 2017, 06:13:53 pm by hwj-d »
#### TiN
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##### Re: KX Reference
« Reply #161 on: September 09, 2017, 06:34:38 pm »
You are correct in point that data represents total system tempco. Sure, having known "tempco" reference would make things easy.
But even without, you can still variate temperature for parts of the system (DUT only, or meter only) separately, right?
If your oven provides stable chamber inner temperature (<0.05C stability), you can use that easily. You just need to "anchor" one of the parts of the system to establish remaining system tempco. Having LTZ small and low power, I'd expect keeping it at stable temperature is much easier than large DMM.
If you use KX onboard MAX6610 sensors, they will allow you to check board stability well enough for the purpose. Here on page I've logged MAX6610s temperature output during my tempco test. Black ramp is reference box temperature, measured by Honeywell HEL-705 RTD. Using both sensors you can even see when LTZ oven is approaching to thermal headroom limit (closer the temperature directed from less margin from LTZ heater room). You need to stimuli LTZ DUT with much larger temperature delta, to be able even see the tempco effect, doesn't matter what meter you use, be it 34461 or 3458A, just because of the own LTZ's noise. I usually do tempco ramps from 20-25C to 40-45C to get 20C delta. If you get less than 1ppm difference, your reference good to go.
Said above, also means there is little meaning or usefulness in doing box tempco with just few temperature points, instead of the linear ramp, as you don't know tempco behaviour on unknown device/meter. Is it linear? Is it logarithmic? Full world of possibilities...
P.S. 56K warning, linked pages are realtime JS bodge, may take a minute or two to load...
« Last Edit: September 09, 2017, 06:52:30 pm by TiN »
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#### Kleinstein
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##### Re: KX Reference
« Reply #162 on: September 09, 2017, 06:40:36 pm »
If one can not make the meters temperature constant, one could try to measure it and than compensate numerical. There is software to get at least linear correlation and this way get estimates for both TC's: the meter and the external reference.
It can help if one looks at more than one temperature cycle for the external reference. However the speed of the temperature cycle is somewhat limited, as fast temperature changes can have a different effect. Not all temperature effects are instantaneous and fast changes tend to include gradients. Over many cycles variations of the meters temperature can average out, or at least a good decomposition is possible.
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#### TiN
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##### Re: KX Reference
« Reply #163 on: September 09, 2017, 06:48:15 pm »
However the speed of the temperature cycle is somewhat limited, as fast temperature changes can have a different effect.
Yup, hence all my ramps are graceful and slow (often 40+ hours).
Linear math compensation for meter tempco during 2xLTFLU 13V reference measurement also worked well for me.
Total error was <1ppm, peaks excluded, even with 24C-31C ambient temperature change, but granted I've used bit more stable 2002.
+0.4ppm/K with reference +26.3C temp was used for linear correction.
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#### hwj-d
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##### Re: KX Reference
« Reply #164 on: September 09, 2017, 07:01:54 pm »
@TiN
Ok, got it.
Then MAX6610 and your measurements is the anchor.
Thanks a lot.
Edit:
Quote
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« Last Edit: September 09, 2017, 07:30:30 pm by hwj-d »
#### kj7e
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##### Re: KX Reference
« Reply #165 on: September 10, 2017, 05:48:31 pm »
Maybe the tempco is not as bad as I thought. Repeated cycling the KX board from 25.0 deg C to 35.0 deg C four times and achieved the exact same results;
@ 25.0 deg C, 7.141510
@ 35.0 deg C, 7.141504
So 6uV over 10 Deg C, or 0.6uV/C
Scaled to 10v, that would a delta of 0.00000084 or 0.084 PPM/C. Please correct me if I'm wrong here.
#### Kleinstein
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##### Re: KX Reference
« Reply #166 on: September 10, 2017, 08:56:07 pm »
0.084 ppm/K is still slightly higher than the data-sheet value of 0.05 ppm/K, but not much. If in an environment with not so much variations, this should no be a real problem. Inside a oven environment it would be not a problem at all.
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#### Mickle T.
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##### Re: KX Reference
« Reply #167 on: September 11, 2017, 05:36:39 am »
0.05 ppm/K is a typical value, not maximum.
#### Dr. Frank
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##### Re: KX Reference
« Reply #168 on: September 11, 2017, 10:58:04 am »
Thanks Dr. Frank for your answer. I can only learn from that, what happens (t)here.
I know, the scale of noise and of the dip is really not obvious.
I don't want to mix this dip problem with that TC of my meter. That's a 34461a.
Is it right, that this meters have TC of 1ppm/k by itself? So, that we can only measure not better than that TC with it?
Best regards
Well, at first, the '461A and the '465A probably have the same topology of their reference around the LM399, and also the ADC circuit should be identical.
If you look into the specification, both instruments are nearly identical in DCV, without the ACAL feature of the 465A.
The 465A may have a bit more stable reference , 30ppm/yr. over 35ppm/year, due to a possibly better selected LM399.
The T.C., though, is identical, i.e. 5ppm/K. That is due to the FineLine resistor network, which is used around the reference to generate the diverse ADC reference voltages, and in the ADC itself. See 34411A schematic, which is probably identical.
The LM399 itself is specified 1ppm/k, so the '465A can autocal itself, which gives 2ppm/K. But even that is of limited use, when making T.C. measurements in strongly varying environment.
btw.: d-smes complained about the assumed bad quality and high T.C. of ´his new KS '465A, but he obviously measured 1ppm/K in his experiment, which is very good, compared to the specified 5 ppm/K. I still do not understand his criticism, one can't expect more from that class of instruments, and also I think, that the new DMMs are even better than the old 34401A.
Anyhow, it's very difficult to determine the LTZ1000s T.C. by an instrument, which is specified to have 100 times higher T.C., but it's not impossible.
At first, you may also estimate the T.C. of your '361A by means of your LTZ1000 circuit, as d-smes has done it. As a proper designed LTZ1000 has at least < 0.1ppm/K, you would have a first guess by variation of room temperature, because the 361A is expected to have a factor of 10..50 higher T.C.
Maybe you also see 1ppm/K only, like d-smes instrument.
Your basement may offer much more stable temperatures, I achieve +/- 0.1 °C over many hours there.
So you could monitor the 361As internal temperature. I don't know if it has that parameter available over the bus. otherwise you'll have to mount a thermometer inside the instrument, and then you might be able to detect changes on the order of 0.1ppm.
If you change the LTZ1000s temperature over +/-5°C quite quickly, that would be 100 times more than the 361A, you might be able to estimate the LTZ1000 T.C.
You will notice some correlations to amount and direction of both temperature changes, so it might be possible to assign these correlations either to the 361A, or to the LTZ1000. If you have a first rough estimate of your LTZ1000 circuit, you may use it as the new baseline for further measurements.
You may also try to improve the LTZ1000s T.C. by trimming the T.C. compensation resistor.
If you are lucky to have a 2nd LTZ1000 circuit, you can bootstrap these T.C. measurements, and T.C. trimming by the same scheme as above, but your 361A would only be used for relative / ratio measurements further on.
I use my 3458A for that purpose, that may seem much easier than using a 361/365A, but it's not.
It has a (measured) T.C. of 0.5ppm/K (as specified), but that's only a factor of 2 (maybe) better than the 6 1/2 digits instrument.
I estimated my LTZ1000 circuit to have a T.C. of about 0.02 ppm/K, as that you cannot distinguish any more between both changes in temperature.
So this trimmed LTZ is at least 10 times more temperature stable than the 3458A, and will serve as the new baseline for the next LTZ circuit, I just have assembled. Goal will be to trim the next reference to ~ 0.01ppm/K, and afterwards to determine the T.C. of the first LTZ with much higher confidence.
Frank
« Last Edit: September 11, 2017, 11:42:39 am by Dr. Frank »
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#### Andreas
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##### Re: KX Reference
« Reply #169 on: September 11, 2017, 06:56:40 pm »
0.05 ppm/K is a typical value, not maximum.
Not even a typical value.
In older data sheets: "0.05 ppm/K can be reached".
in the newest version: "0.03 ppm/K can be reached"
so it is a best case value.
Without leg length trimming and T.C. trimming I had up to 0.23 ppm/K
and was lucky to trim it down below 0.05ppm/K over 30 deg C for LTZ#3-LTZ#6.
And I fear that there is also some kind of hysteresis which limits T.C. adjustments.
Try to build several LTZ references: in the 100mV range of your DMM you can usually measure
with a factor 10 higher (noise free) resolution than in the 10V range. (as difference between 2 references)
With best regards
Andreas
« Last Edit: September 11, 2017, 07:03:43 pm by Andreas »
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#### kj7e
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• Posts: 911
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• Damon Stewart
##### Re: KX Reference
« Reply #170 on: September 13, 2017, 06:08:45 pm »
I do appreciate and I'm learning from all the experts experience, advice and opinions, that of TiN, Kleinstein, Dr. Frank, Andreas, MisterDiodes and others. Thanks Guys!
Just because no post is worth reading with out a photo, here are a few from some recent tests;
Cycling the KX board between 25.0 and 35.0 Deg C, shows 6uV swings
10 hour measurement - 100 NPLC + Math 50 reading average, KX Reference on battery power at 35.0 Deg C, 34465a at initial cal temp + or - <1 Deg C. Less than 2uV deviation.
23 hour measurement, extension of the above 10 hour - at 12 hour mark (mid scale) I switched the KX Ref to external power which introduced a slight amount of noise that corresponds to the room A/C cycling
So as long as the room temp is somewhat stable, so is the 34465a. It also seems the there is some merit in stabilizing the temperature environment for the LTZ1000A, at least in my case. Also, battery power is hard to beat when it comes to noise.
« Last Edit: September 13, 2017, 06:35:52 pm by kj7e »
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#### IanJ
• Supporter
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• Country:
• Full time EE & Youtuber
##### Re: KX Reference
« Reply #171 on: September 15, 2017, 02:32:21 pm »
Hi all,
Thought I'd join in the KX build fun as I've had a couple of LTZ1000CH's sitting here since 2015........Time I did something with them!
Will build up two of them over the next few weeks. The plan is to get one of them powered up 24/7 along with one of my PDVS2's and logged via my 3458a.
Ian.
« Last Edit: September 15, 2017, 02:36:24 pm by IanJ »
Ian Johnston - Original designer of the PDVS2mini || Author of the free WinGPIB app.
Website - www.ianjohnston.com
#### hwj-d
• Frequent Contributor
• Posts: 676
• Country:
• save the children - chase the cabal
##### Re: KX Reference
« Reply #172 on: September 18, 2017, 11:49:27 am »
Regarding the reading shift from touching DMM, I too have noticed this. I usually change grounding around and/or tie floating DUT to meter case or earth ground. After a bit of experimenting, I can usually get the touch artifacts to go away.
In the meantime, i cut pcb grounding to DUT-case. With floating ground, i didn't got this capacitive caused dips anymore.
« Last Edit: September 18, 2017, 11:56:02 am by hwj-d »
#### kj7e
• Frequent Contributor
• Posts: 911
• Country:
• Damon Stewart
##### Re: KX Reference
« Reply #173 on: October 05, 2017, 10:31:48 pm »
Brand new LTZ1000A showed up from Digikey today, second KX board now running. Really took my time trying to get every part soldered in place perfectly on this one.
Decided to place the chip 0.5mm above the board;
Going to let it burn in for a few days before I do any testing. Running just fine at 7.161634v drawing 27mA, 15v supply.
#### TiN
• Super Contributor
• Posts: 4543
• Country:
##### Re: KX Reference
« Reply #174 on: October 11, 2017, 04:23:37 am »
Well, after all, your KX'es belong to me
Decided to give a helping hand to CalMachine, with his two attempts.
Obviously, my design constrains are bit unfriendly to newcomers, especially tricky SMD film capacitors.
Once I fix these two modules, I'll log them to establish stable voltage levels and send back to CM for his calibration vs their 732B standard.
That will help to also help to keep my lab calibration uncertainty as "reverse-calibration", as my last Volt calibration was done in February 2017, and is overdue.
YouTube | Metrology IRC Chat room | Let's share T&M documentation? Upload! No upload limits for firmwares, photos, files.
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1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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# ∑nj=1∑ni=1i is equal to
A
n(n+1)2
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n(n+1)22
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n2(n+1)2
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D
none of these
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Solution
## The correct option is C n2(n+1)2Given,∑nj=1∑ni=1ihere, we use the direct formula=∑nj=112n(n+1)=12n2(n+1)=n2(n+1)2
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Re: Rewriting of Trigonometric Functions
• To: mathgroup at smc.vnet.net
• Subject: [mg26847] Re: Rewriting of Trigonometric Functions
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Thu, 25 Jan 2001 01:13:25 -0500 (EST)
• References: <94m8st\$3mn@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Thomas,
TrigReduce[Sin[x]^2]
1/2*(1 - Cos[2*x])
TrigReduce[Sin[a]Sin[b]]
1/2*(Cos[a - b] - Cos[a + b])
See File>Palettes>AlgebraicManipulation.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Thomas Engelhardt" <thomas-a.engelhardt at t-online.de> wrote in message
news:94m8st\$3mn at smc.vnet.net...
> Dear Mathgroup members,
>
> I have just started to use Mathematica and directly ran into a problem:
>
> I have rather lengthy expressions resulting from putting a sum of two
> sinusodial swings to the power of three (Two-tone Intermodulation).
> In order for me to recognise the contained frequencies I want Mathematica
to
> rewrite the contained Trigonometric functions in certain ways, namely
like:
>
> [sin(a)]^2 should be transformed into 1/2*(1+cos2x)
> and for higher powers accordingly.
>
> Furthermore:
> sin(a)*sin(b) should be changed to 1/2*[cos(a-b) - cos(a+b)]
>
> I have tried all kinds of "Expand" and "Trig..." operations but
Mathematica
> always keeps the powers on the trigonometric functions and doesn't change
> the multiplications either.
>
> Does anybody have an idea how I can get Mathematica to transform the
> expressions in the desired way?
>
>
> Kind Regards
>
> Thomas
> Munich, Germany
>
>
```
• Prev by Date: RE: Rewriting of Trigonometric Functions
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# Find Missing Side Lengths Using Common Right Triangles Worksheet
Triangle Theorems
8.G.B.7
Total questions - 8
Do you want to see how your students perform in this assignment?
The above "Find missing side lengths using common right triangles” Worksheet can help students to get a better understanding of the Pythagorean theorem by solving problems having common right triangles with one of the missing side lengths. For example, A right triangle has legs of lengths 6 and 8 and a hypotenuse with a length of x. Which common right triangle triple is the triangle similar to? Use that information to find the value of x.
50,000+ teachers over use Byte!
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Triangle Theorems
8.G.B.7
In grade 8 math, students are expected to develop a solid understanding of the Pythagorean Theorem. One effective way to reinforce this understanding is by using a "Find missing side lengths using common right triangles Worksheet". This worksheet can help students practice finding missing side lengths using common right triangles which are also known as common right triangle tripple.
50,000+ teachers over the world use Byte!
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# Testing the total impact of a predictor in a messy model
Suppose I have a regression model with several predictors ind interactions thereof. For concreteness, suppose we are studying a company's data on salaries and we have predictors dept, title, sex, age, and exper (so there are 3 factors and two covariates). We settle on a model expressed SAS-style as
h(salary) = dept title sex age exper
dept*sex dept*age title*exper title*age
sex*age sex*exper age*exper
dept*sex*age sex*age*exper
This model includes many, but not all, interactions up through third-order, and suppose we find that for a suitable transformation h, the fit statistics and residual plots all look good.
The question to be answered is "does sex affect salary?" I think many people would answer based on, say, the "Type II" $F$ test for sex. But suppose I make the question more specific by asking "if sex were not taken into account at all, would it make any difference?" This question suggests comparing the above model with the reduced model where all terms involving sex are excluded:
h(salary) = dept title age exper
dept*age title*exper title*age
age*exper
A test comparing these models would have quite a few numerator degrees of freedom, whereas the Type II test of sex has only one. Of course, I do think a lot of people would also look at the individual tests of each of the terms that involve sex. But it is possible that an overall test of all of these terms together could have a smaller $P$ value than any of the individual tests.
My question is: Is there a name for this kind of reduction-of-model test, where the full model is compared with a reduced model where a predictor is completely excluded? And do lots of people do this kind of thing routinely, and I should cringe in embarrassment for not having seen it much? I'd think people might be wise to consider such a test more often. Any comments?
• Dare I say : Partial F-test type III? (aka Yates' weighted squares of means) From Analysis of Messy Data, Vol. I, by Milliken and Johnson. (circa 1984...) – usεr11852 Feb 23 '16 at 23:30
• Please dare -- but I question if this is the same thing. The numerator d.f. for the test I'm talking about is a lot more than 1 -- it's the sum of the df for sex, dept*sex, ..., sex*age*exper. By comparison, am I not correct that the type III test for sex is a test of a single contrast of the cell means that gives equal weights to the levels of interactions that contain sex and zero weights to effects that don't involve sex. But I'll look that up in M & J when I get a chance. – rvl Feb 24 '16 at 0:23
• Yes, you are correct but I think your test falls under the same category. In any case what you propose does not seem wrong as the smaller model is clearly nested within the other. – usεr11852 Feb 24 '16 at 2:15
• Clearly it is a legit test in terms of the nested models. But given that it is a reduction-of-model test comparing a large model with a much smaller one, in what sense could it be in the same category as a test of a single contrast??? – rvl Feb 24 '16 at 2:42
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# Miscellaneous Quiz / Physics Headache
Random Miscellaneous Quiz
## Do you know general physics?
#### by Althouse_MvP Plays Quiz not verified by Sporcle
How to Play
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Timer
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If an elephant has a mass of 1000 kg, How much does it weigh on planet Earth? (SI Units)Force = Mass * Acceleration, Earth Gravity Acceleration = 9.81 m/s^2
If an elephant has a mass of 1000 kg, How much does it weigh on the planet Mars? (SI Units)Force = Mass * Acceleration, Mars Gravity Acceleration = 3.711 m/s^2
If a 1000 kg elephant is accelerating at 1.2 m/s^2, What is the magnitude of the force required? (No units in answer)Force = Mass * Acceleration
If a running elephant can exert a Force of 1500 Newtons, at what rate must a 55 kg human accelerate to create an equal and opposing force? (Magnitude Only)Force = Mass * Acceleration
A 0.5 kg sliding hockey puck is decelerating at a rate of 1.85 m/s^2, What is the coefficient of kinetic friction?Friction Force = Coefficient * Normal Force, Normal Force = Mass * Gravity
If Force = Stiffness * Displacement, What are the units for Stiffness (SI Units, use the word 'per' for a division sign)Force = Newtons, Displacement = metres
If the driver of a car applies the brakes and changes speed from 75 km/hr to 0 km/hr in 3.2 seconds, What was the acceleration rate of the car? (SI Units, provide 'sign (+/-)' )1 m/s = 3.6 km/hr
A 1300 kg car collides with a safety barrier, changing speed from 16.67 m/s to 0 m/s in 1.8 sec. If the safety barrier had a stiffness of 10.032 kN/m, What was the displacement?Force = mass * acceleration, Force = Stiffness * Displacement
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## From the Vault See Another
Populous Countries, Smaller Neighbors
These countries are hogging all the people.
### Extras
Created Sep 15, 2013
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