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Maths Presentation Ppt
maths presentation
Presentation Description
this presentation is based on theorem , and many other things
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Results 1 - 10 of 50
, 2004
"... Modular SOS (MSOS) is a variant of conventional Structural Operational Semantics (SOS). Using MSOS, the transition rules for each construct of a programming language can be given incrementally,
once and for all, and do not need reformulation when further constructs are added to the language. MSOS th ..."
Cited by 55 (4 self)
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Modular SOS (MSOS) is a variant of conventional Structural Operational Semantics (SOS). Using MSOS, the transition rules for each construct of a programming language can be given incrementally, once
and for all, and do not need reformulation when further constructs are added to the language. MSOS thus provides an exceptionally high degree of modularity in language descriptions, removing a
shortcoming of the original SOS framework. After sketching the background and reviewing the main features of SOS, the paper explains the crucial differences between SOS and MSOS, and illustrates how
MSOS descriptions are written. It also discusses standard notions of semantic equivalence based on MSOS. Appendix A shows how the illustrative MSOS rules given in the paper would be formulated in
conventional SOS.
- UML 2000 - The Unified Modeling Language. Advancing the Standard, vol. 1939 of LNCS , 1999
"... The UML meta model [3] captures the abstract syntax... ..."
- SOS 2005 PRELIMINARY VERSION , 2005
"... Rewriting logic is a flexible and expressive logical framework that unifies denotational semantics and SOS in a novel way, avoiding their respective limitations and allowing very succinct
semantic definitions. The fact that a rewrite theory’s axioms include both equations and rewrite rules provides ..."
Cited by 39 (11 self)
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Rewriting logic is a flexible and expressive logical framework that unifies denotational semantics and SOS in a novel way, avoiding their respective limitations and allowing very succinct semantic
definitions. The fact that a rewrite theory’s axioms include both equations and rewrite rules provides a very useful “abstraction knob” to find the right balance between abstraction and observability
in semantic definitions. Such semantic definitions are directly executable as interpreters in a rewriting logic language such as Maude, whose generic formal tools can be used to endow those
interpreters with powerful program analysis capabilities.
- Applied Categorical Structures , 1999
"... . We present a categorical characterisation of term graphs (i.e., finite, directed acyclic graphs labeled over a signature) that parallels the well-known characterisation of terms as arrows of
the algebraic theory of a given signature (i.e., the free Cartesian category generated by it). In particula ..."
Cited by 37 (24 self)
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. We present a categorical characterisation of term graphs (i.e., finite, directed acyclic graphs labeled over a signature) that parallels the well-known characterisation of terms as arrows of the
algebraic theory of a given signature (i.e., the free Cartesian category generated by it). In particular, we show that term graphs over a signature \Sigma are one-to-one with the arrows of the free
gs-monoidal category generated by \Sigma. Such a category satisfies all the axioms for Cartesian categories but for the naturality of two transformations (the discharger ! and the duplicator r),
providing in this way an abstract and clear relationship between terms and term graphs. In particular, the absence of the naturality of r and ! has a precise interpretation in terms of explicit
sharing and of loss of implicit garbage collection, respectively. Keywords: algebraic theories, directed acyclic graphs, gs-monoidal categories, symmetric monoidal categories, term graphs.
Mathematical Subject Clas...
- In Proceedings of the IJCAR 2004. LNCS , 2004
"... Abstract. Formal semantic definitions of concurrent languages, when specified in a well-suited semantic framework and supported by generic and efficient formal tools, can be the basis of
powerful software analysis tools. Such tools can be obtained for free from the semantic definitions; in our exper ..."
Cited by 35 (9 self)
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Abstract. Formal semantic definitions of concurrent languages, when specified in a well-suited semantic framework and supported by generic and efficient formal tools, can be the basis of powerful
software analysis tools. Such tools can be obtained for free from the semantic definitions; in our experience in just the few weeks required to define a language’s semantics even for large languages
like Java. By combining, yet distinguishing, both equations and rules, rewriting logic semantic definitions unify both the semantic equations of equational semantics (in their higher-order
denotational version or their first-order algebraic counterpart) and the semantic rules of SOS. Several limitations of both SOS and equational semantics are thus overcome within this unified
framework. By using a high-performance implementation of rewriting logic such as Maude, a language’s formal specification can be automatically transformed into an efficient interpreter. Furthermore,
by using Maude’s breadth first search command, we also obtain for free a semi-decision procedure for finding failures of safety properties; and by using Maude’s LTL model checker, we obtain, also for
free, a decision procedure for LTL properties of finite-state programs. These possibilities, and the competitive performance of the analysis tools thus obtained, are illustrated by means of a
concurrent Caml-like language; similar experience with Java (source and JVM) programs is also summarized. 1
, 1999
"... A novel form of labelled transition system is proposed, where the labels are the arrows of a category, and adjacent labels in computations are required to be composable. Such transition systems
provide the foundations for modular SOS descriptions of programming languages. Three ..."
Cited by 27 (6 self)
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A novel form of labelled transition system is proposed, where the labels are the arrows of a category, and adjacent labels in computations are required to be composable. Such transition systems
provide the foundations for modular SOS descriptions of programming languages. Three
, 1999
"... We introduce the notion of cartesian closed double category to provide mobile calculi for communicating systems with specific semantic models: One dimension is dedicated to compose systems and
the other to compose their computations and their observations. Also, inspired by the connection between s ..."
Cited by 20 (12 self)
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We introduce the notion of cartesian closed double category to provide mobile calculi for communicating systems with specific semantic models: One dimension is dedicated to compose systems and the
other to compose their computations and their observations. Also, inspired by the connection between simply typed -calculus and cartesian closed categories, we define a new typed framework, called
double -notation, which is able to express the abstraction /application and pairing/projection operations in all dimensions. In this development, we take the categorical presentation as a guidance in
the interpretation of the formalism. A case study of the ß-calculus, where the double - notation straightforwardly handles name passing and creation, concludes the presentation.
- In MFCS'99, Proc. 24th Intl. Symp. on Mathematical Foundations of Computer Science, Szklarska-Poreba , 1999
"... ) Peter D. Mosses 1;2 1 BRICS and Department of Computer Science, University of Aarhus, Denmark 2 Visiting SRI International and Stanford University, USA Abstract. A novel form of labelled
transition system is proposed, where the labels are the arrows of a category, and adjacent labels in c ..."
Cited by 17 (16 self)
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) Peter D. Mosses 1;2 1 BRICS and Department of Computer Science, University of Aarhus, Denmark 2 Visiting SRI International and Stanford University, USA Abstract. A novel form of labelled transition
system is proposed, where the labels are the arrows of a category, and adjacent labels in computations are required to be composable. Such transition systems provide the foundations for modular SOS
descriptions of programming languages. Three fundamental ways of transforming label categories, analogous to monad transformers, are provided, and it is shown that their applications preserve
computations in modular SOS. The approach is illustrated with fragments taken from a modular SOS for ML concurrency primitives. 1 Introduction SOS (structural operational semantics) is a widely-used
framework for defining process algebras [12, e.g.] and programming languages [13, e.g.]. Following Plotkin [22], SOS has often been preferred to the more abstract framework of denotational seman...
- PROC. OF EXPRESS’97 , 1997
"... In addition to ordinary places, called stable, zero-safe nets are equipped with zero places, which in a stable marking cannot contain any token. An evolution between two stable markings,
instead, can be a complex computation called stable transaction, which may use zero places, but which is atomic w ..."
Cited by 17 (13 self)
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In addition to ordinary places, called stable, zero-safe nets are equipped with zero places, which in a stable marking cannot contain any token. An evolution between two stable markings, instead, can
be a complex computation called stable transaction, which may use zero places, but which is atomic when seen from stable places: no stable token generated in a transaction can be reused in the same
transaction. Every zero-safe net has an ordinary Place-Transition net as its abstract counterpart, where only stable places are maintained, and where every transaction becomes a transition. The two
nets allow us to look at the same system from both an abstract and a refined viewpoint. To achieve this result no new interaction mechanism is used, besides the ordinary token-pushing rules of nets.
The refined zero-safe nets can be much smaller than their corresponding abstract P/T nets, since they take advantage of a transition synchronization mechanism. For instance, when transactions of
unlimited l...
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Is there an nontrivial function whose 'period paralellograms' are Gosper Islands?
up vote 2 down vote favorite
The Gosper island tiles the plane, so I'm curious if a nontrivial elliptic? function exists which would have a 'period gosper-island' instead of a period parallelogram. In this case, I'm using
'trivial' to denote a function that's got the right kind of lattice, but the Gosper island is overkill and unnecessary -- for example the derivative of the equianharmonic case: $d\wp(z;0,1)/dz$ (last
tiling ca.analysis-and-odes fractals
Dear Deoxy, your source for learning elliptic functions is definitely not the best one. But already there it's explained that fundamental domains of elliptic functions are parallelograms
(quotients of $\mathbb C$ by lattices), no snowflakes. More exotic fundamental domains can be constructed for automorphic rather than elliptic functions. I wonder whether one can get fractals but
tiling pictures of more regular geometric structure can be seen for example in M.Yoshida's "Hypergeometric functions, my love" (1997). – Wadim Zudilin May 23 '10 at 11:54
5 I don't think the question makes sense. An elliptic function has a period lattice $\Omega$, which is a subgroup of $\mathbb{C}$. There is nothing unique about “the” period parallellogram, though;
it is just a convenient labeling of the equivalence classes $\mathbb{C}/\Omega$ by taking one element from each equivalence class. There is no reason, apart from convenience, why it has to be a
parallellogram. Gosper islands should do just fine, though to go and pick a fractal one seems slightly perverse. – Harald Hanche-Olsen May 23 '10 at 11:56
1 Harald -- the crux of my question is whether any functions exist whose 'canonical' fundamental domain is a Gosper island -- if you were to look at domain colored pictures of these hypothetical
functions, one would say "aha, Gosper islands" instead of "aha, period parallelograms". – deoxygerbe May 23 '10 at 13:55
The point of Harald's comment is that there is no 'canonical' fundamental domain, at least until you provide extra data. – Qiaochu Yuan May 23 '10 at 20:09
I wonder how the choice of fundamental domain alters the discussion of elliptic curves – john mangual Jul 10 '13 at 11:47
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1 Answer
active oldest votes
The Gosper islands are a fundamental domain for the translation action of the Eisenstein integers $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, since the shapes can be constructed by deforming a
Voronoi decomposition of the plane with respect to that lattice. It is therefore reasonable to look for functions in the field generated by the Weierstrass $\wp$ function for the Eisenstein
integers and its first derivative $\wp'$, since this field comprises all of the meromorphic functions that are periodic with respect to the lattice.
It is not clear what selection rule you want to apply to favor one function over another. Elliptic functions do not have canonical fundamental domains, and one has to choose extra data
up vote 2 (e.g., a basis of the lattice, and a pair of paths in the homotopy class representing the basis) to write down boundaries in the usual theory.
down vote
I suppose you may want to find a function $f(z)$ such that the boundary configuration is equal to $\{ z : |f(z)| = 1 \}$, but I am somewhat doubtful that such a function exists, simply by
degree considerations.
add comment
Not the answer you're looking for? Browse other questions tagged tiling ca.analysis-and-odes fractals or ask your own question.
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Asymptotical growth of a class of random trees, Annals Of Probability 13
Results 1 - 10 of 13
, 1991
"... We survey two indices for text, with emphasis on Pat arrays (also called suffix arrays). A Pat array is an index based on a new model of text which does not use the concept of word and does not
need to know the structure of the text. to appear in Information Retrieval: Data Structures and Algori ..."
Cited by 23 (0 self)
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We survey two indices for text, with emphasis on Pat arrays (also called suffix arrays). A Pat array is an index based on a new model of text which does not use the concept of word and does not need
to know the structure of the text. to appear in Information Retrieval: Data Structures and Algorithms, R.A. Baeza-Yates and W. Frakes, eds., Prentice-Hall. 1 1 Introduction Text searching methods may
be classified as lexicographical indices (indices that are sorted), clustering techniques, and indices based on hashing (for example, signature files [FC87]). In this report we discuss
lexicographical indices, in particular, two main data structures: inverted files and Pat trees. Our aim is to build an index for the text of size similar to or smaller than the text. Briefly, the
traditional model of text used in information retrieval is that of a set of documents. Each document is assigned a list of keywords (attributes), with optional relevance weights associated to each
keyword. This ...
"... There is an upsurge in interest in the Markov model and also more general stationary ergodic stochastic distributions in theoretical computer science community recently (e.g. see
[Vitter,KrishnanSl], [Karlin,Philips,Raghavan92], [Raghavan9 for use of Markov models for on-line algorithms, e.g., cashi ..."
Cited by 17 (4 self)
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There is an upsurge in interest in the Markov model and also more general stationary ergodic stochastic distributions in theoretical computer science community recently (e.g. see [Vitter,KrishnanSl],
[Karlin,Philips,Raghavan92], [Raghavan9 for use of Markov models for on-line algorithms, e.g., cashing and prefetching). Their results used the fact that compressible sources are predictable (and
vise versa), and showed that on-line algorithms can improve their performance by prediction. Actual page access sequences are in fact somewhat compressible, so their predictive methods can be of
benefit. This paper investigates the interesting idea of decreasing computation by using learning in the opposite way, namely to determine the difficulty of prediction. That is, we will ap
proximately learn the input distribution, and then improve the performance of the computation when the input is not too predictable, rather than the reverse. To our knowledge,
"... We analyze the behavior of the level-compressed trie, LC-trie, a compact version of the standard trie data structure. Based on this analysis, we argue that level compression improves the
performance of both tries and quadtrees considerably in many practical situations. In particular, we show that ..."
Cited by 11 (6 self)
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We analyze the behavior of the level-compressed trie, LC-trie, a compact version of the standard trie data structure. Based on this analysis, we argue that level compression improves the performance
of both tries and quadtrees considerably in many practical situations. In particular, we show that LC-tries can be of great use for string searching in compressed text. Both tries and quadtrees are
extensively used and much effort has been spent obtaining detailed analyses. Since the LC-trie performs significantly better than standard tries, for a large class of common distributions, while
still being easy to implement, we believe that the LC-trie is a strong candidate for inclusion in the standard repertoire of basic data structures.
"... We study an order-preserving general purpose data structure for binary data, the LPC-trie. The structure is a compressed trie, using both level and path compression. The memory usage is similar
to that of a balanced binary search tree, but the expected average depth is smaller. The LPC-trie is well ..."
Cited by 9 (1 self)
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We study an order-preserving general purpose data structure for binary data, the LPC-trie. The structure is a compressed trie, using both level and path compression. The memory usage is similar to
that of a balanced binary search tree, but the expected average depth is smaller. The LPC-trie is well suited to modern language environments with ecient memory allocation and garbage collection. We
present an implementation in the Java programming language and show that the structure compares favorably to a balanced binary search tree.
, 1995
"... ii Abstract This thesis presents three trie organizations for various binary tries. The new trie structures have two distinctive features: (1) they store no pointers and require two bits per
node in the worst case, and (2) they partition tries into pages and are suitable for secondary storage. We ap ..."
Cited by 7 (2 self)
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ii Abstract This thesis presents three trie organizations for various binary tries. The new trie structures have two distinctive features: (1) they store no pointers and require two bits per node in
the worst case, and (2) they partition tries into pages and are suitable for secondary storage. We apply trie structures to indexing, storing and querying both text and spatial data on secondary
storage. We are interested in practical problems such as storage compactness, I/O efficiency, and large trie construction. We use our tries to index and search arbitrary substrings of a text. For an
index of 100 million keys, our trie is 10 %- 25 % smaller than the best known method. This difference is important since the index size is crucial for trie methods. We provide methods for dynamic
tries and allow texts to be changed. We also use our tries to compress and approximately search large dictionaries. Our algorithm can find strings with k mismatches in sublinear time. To our
knowledge, no other published sublinear algorithm is known for this problem. Besides, we use our tries to store and query spatial data such as maps. A trie structure is proposed to permit querying
and retrieving spatial data at arbitrary levels of resolution, without reading from secondary storage any more data than is needed for the specified resolution. The trie structure also compresses
spatial data substantially. The performance results on map data have confirmed our expectations: the querying cost is linear in the amount of data needed and independent of the data size in practice.
We give algorithms for a set of sample queries including geometrical selection, geometrical join and the nearest neighbour. We also show how to control query cost by specifying an acceptable
- Journal of Statistical Mechanics: Theory and Experiment
"... Abstract. A hierarchical model for the growth of planar arch structures for RNA secondary structures is presented, and shown to be equivalent to a tree-growth model. Both models can be solved
analytically, giving access to scaling functions for large molecules, and corrections to scaling, checked by ..."
Cited by 7 (2 self)
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Abstract. A hierarchical model for the growth of planar arch structures for RNA secondary structures is presented, and shown to be equivalent to a tree-growth model. Both models can be solved
analytically, giving access to scaling functions for large molecules, and corrections to scaling, checked by numerical simulations of up to 6500 bases. The equivalence of both models should be
helpful in understanding more general tree-growth processes. PACS numbers: 87.14.gn, 87.15.bd, 02.10.Ox, 02.50.EyA growth model for RNA secondary structures 2 1.
- Information Processing Letters , 2005
"... Andersson and Nilsson have already shown that the average depth Dn of random LC-tries is only Θ (log ∗ n) when the keys are produced by a symmetric memoryless process, and that Dn = O (log log
n) when the process is asymmetric. In this paper we refine the second estimate by showing that asymptotical ..."
Cited by 2 (2 self)
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Andersson and Nilsson have already shown that the average depth Dn of random LC-tries is only Θ (log ∗ n) when the keys are produced by a symmetric memoryless process, and that Dn = O (log log n)
when the process is asymmetric. In this paper we refine the second estimate by showing that asymptotically (with n → ∞): Dn ∼ 1 log log n, where n is the number of η keys inserted in a trie, η = −
log (1 − h/h−∞), h = −p log p − q log q is the entropy of a binary memoryless source with probabilities p, q = 1 − p (p ̸= q), and h− ∞ = − log min(p, q). Key words: average case analysis of
algorithms, trie, LC-trie. 1
, 2000
"... We apply the trie structures to indexing, storing and querying structured data on secondary storage. We are interested in the storage compactness, the I/O efficiency, the order-preserving
properties, the general orthogonal range queries and the exact match queries for very large files and databases. ..."
Cited by 1 (0 self)
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We apply the trie structures to indexing, storing and querying structured data on secondary storage. We are interested in the storage compactness, the I/O efficiency, the order-preserving properties,
the general orthogonal range queries and the exact match queries for very large files and databases. We also apply the trie structures to relational joins (set operations). We compare trie structures
to various data structures on secondary storage: multipaging and grid files in the direct access method category, R-trees/R*-trees and X-trees in the logarithmic access cost category, as well as some
representative join algorithms for performing join operations. Our results show that range queries by trie method are superior to these competitors in search cost when queries return more than a few
records and are competitive to direct access methods for exact match queries. Furthermore, as the trie structure compresses data, it is the winner in terms of storage compared to all other methods
mentioned above. We also present a new tidy function for order-preserving key-to-address transformation. Our tidy function is easy to construct and cheaper in access time and storage cost compared to
its closest competitor.
"... There is an upsurge in interest in the Markov model and also more general stationary ergodic stochastic distributions in theoretical computer science community recently, (e.g. see
[Vitter,Krishnan,FOCS91], [Karlin,Philips,Raghavan,FOCS92] [Raghavan92]) for use of Markov models for on-line algorithms ..."
Add to MetaCart
There is an upsurge in interest in the Markov model and also more general stationary ergodic stochastic distributions in theoretical computer science community recently, (e.g. see
[Vitter,Krishnan,FOCS91], [Karlin,Philips,Raghavan,FOCS92] [Raghavan92]) for use of Markov models for on-line algorithms e.g., cashing and prefetching). Their results used the fact that compressible
sources are predictable (and vise versa), and show that on-line algorithms can improve their performance by prediction. Actual page access sequences are in fact somewhat compressible, so their
predictive methods can be of benefit. This paper investigates the interesting idea of decreasing computation by using learning in the opposite way, namely to determine the difficulty of prediction.
That is, we will approximately learn the input distribution, and then improve the performance of the computation when the input is not too predictable, rather than the reverse. To our knowledge, this
is first case of a computational problem where we do not assume any particular fixed input distribution and yet computation is decreased when the input is less predictable, rather than the reverse.
We concentrate our investigation on a basic computational problem: sorting and a basic data structure problem: maintaining a priority queue. We present the first known case of sorting and priority
queue algorithms whose complexity depends on the binary entropy H ≤ 1 of input keys where assume that input keys are generated from an unknown but arbitrary stationary ergodic source. This is, we
assume that each of the input keys can be each arbitrarily long, but have entropy H. Note that H
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How to Keep Your Child From Failing Math
Don't let the title scare you!
When I tell people that DH is a math teacher I get a couple reactions, from admiration to the more common confession that they are terrified of math. I know many of you struggle to help your kids,
and even wonder how you can keep them doing well in math as they progress.
The first thing I'd tell you is to stop telling yourself that you're "bad" at math. It would be unacceptable for you to go around telling people that you're "bad" at reading, "bad" at writing, and
"bad" at driving. You keep that to yourself, because just like negative body image, your kid will pick that up right away.
The second thing that we see is preventable; it's something so simple, so basic, that if you make sure your kids knows it, they will be equipped to handle the math coming down the road...story first:
DH had a long conversation with a parent recently, a parent who was extremely upset at him because their child had 100% on all their math homework, yet scored 25% on the tests. The parent was
furious, demanding an answer from him on why their child was failing despite having perfect homework.
DH pulled out the homework; it was full of correct answers--answers...but no work shown. He said he didn’t want to accuse any student of cheating when there’s always room for divine
intervention--intervention that happened to desert the child during tests. So, DH made an offer: if the child could set up a problem from any of the homework, not solve, just set up the work, he’d be
happy to grant the child an “A.” The child burst into tears. The parent soon grasped the picture.
What’s more, DH said, can you tell me what 7 times 8 is? The child thought for a few minutes and answered “42” then after another minute “56.” Here’s the problem, DH said, you never learned your
times tables when you were supposed to, and I bet that every teacher after that allowed you to use a calculator until now, is that right? The child nodded. I suspect that you and your child will have
a different type of summer activity planned soon, DH said to the parent.
So how do you help your kids with math, even if you feel inadequate to do so? The very first step is to make sure they have memorized the addition and multiplication tables. Every double digit
addition and subtraction in their head. Every multiplication and division in their head. They should be able to answer instantaneously, with zero lag time. They should be able to do this by the 4th
grade. If your child can't, you know where to start. Your kid should also be able to to count up to the 12th multiple for the first 12 counting numbers--i.e. 12, 24, 36, 48, etc, all the way up to
144 and back again without an “uhm” or a breath.
How do these things help? Instead of trying to figure out how the numbers all fit together, your kids will be able to concentrate on the problem in front of them. If you can make sure your child
knows their tables they’ll be prepared for the advanced stuff. You would not believe how many kids reach high school without basic addition, subtraction, and multiplication--it is the fundamental
root of so many kids failing math.
Now you know. And you can prevent it.
39 comments:
My mom taught 7th and 8th grade math for many years and said the biggest problem she found was kids who didn't know their "math facts". Consequently, my children are "required" to play a game
from Fun4thebrain.com every single school day--addition up until 2nd grade, multiplication after that.
I SO agree with everything you've said!
I hate to be a brat but...when I was in high school I showed my work, never copied. (Also, none of my math teachers even accepted hw without all the work shown.) I knew my multiplication tables
like the back of my hand. I got the highest grade in my class in geometry (the only A in the class)...and I was still just plain bad at Algebra (I and II). Like, bad bad bad. Even though I came
in at lunch and after school for help...And for me, high school was only like 5 years ago...
Sarahlucy- There are many reasons why someone might have a rough time in their math classes, but the number one problem teachers see that contributes to outright failure are kids who don't know
their tables. I'm hoping to cover some of the other reasons soon, but at least this gives parents a where-to-start benchmark.
I've never understood why people think it's ok to say they are bad at math. It's like Barbie saying "math is hard". Thank you! Now let's go read some Danica McKellar books.
I'm putting in a shameless plug for a little electronic device called the flashmaster, you can find it and testimonials from amazon and it links to their website. Not only do they have fabulous
customer service, but the product is great for drilling the add/subtract/multiply/divide basics up throug the 12 tables. I have NO connection with them other than being a very happy customer.
This thing help my triplets get drilled on and learn their multiplication tables super fast. I just didn't have the bandwidth to do it on my own. I tell everyone abut it because I love it so
okay, back to regular programming!
I wish I'd have read this about five years ago.
This should be required reading for parents.
You have no idea how this hit me Carina. Like a punch in the gut.
This describes my school life to a T. I never grasped the fundamentals, and therefore struggled (and continue to) with these concepts. It still hurts. Emotionally and practically.
Every parent of young children needs this advice. Make sure your kids KNOW the basics.
Well said C.
What are te different ways to learn those facts? My kids are at the perfect ages for this step. My oldest is in 5th and (I guess) the way they taught him to memorize didn't work well for him
because he still struggles. He's had charts, flash cards, and we've tried some apps. I think we need to start from scratch :-/
My hubby is such a believer in this-- he makes our kids do math worksheets every day. In some ways, schools have pushed the 'ready 20 minutes every night' too much while neglecting math.
Great post, Carina.
Absolutely!!! I taught 3rd grade for ten years prior to moving into the library and the one thing I stressed to parents and students above all else was to memorize those math facts. "They will
haunt you from here on out if you don't", I would tell them. All it takes is a little nightly practice. Still, many parents would not carry through on this. Quite sad.
That is sooo true, thank you! I'm a primary school teacher and I couldn't agree more - if the kids don't know their tables by heart by the end of 4th grade, they run into big problems.
Love it. I've always loved math (was a math major in fact) and I'm so glad my kids' elementary school makes them do timed 100 math fact pages from 1st grade on up. It gets them to the point of
not having to think about it; they just know those facts!
Good tip, thanks. You should write more about this.
Yes! My kids have monthly tests on their math facts and they are in 1st, 4th and 6th grades. The goal is to master them and I get to see how many facts they get right each month and then I know
what we need to work on. The math program is one of the main reasons I love my kids' school.
So many successes or failures later in life can be traced back to the basics! Math tables, typing, basic spelling and grammar, etc. It isn't "busywork", it's foundational!
I love this post. I really do. My niece is having a little trouble (in 3rd grade) with her math because the school/teachers never required them to learn the basics. So now at the end of third
grade, she is suddenly have to memorize her times tables at the same time she is learning how to multiply fractions. Crazy.
In my son's school, they are fond of telling parents that it isn't the same math the we learned. I have cringed when I heard this and am prepared to be a smart mouth the next time someone at the
school says that. It is the same math - the facts don't change - just the way it is taught changes. When statements like that from teachers and administrators though, some parents are justifiably
confused as to what they should be doing to help.
I love you, your husband, and this blog post!
True story: I made flashcards last night for my first grader because he took too long to tell me what 7+7 was.... and that's just not acceptable to me.
In grade school, I was sick and had to stay home for one whole week. The week we learned our times tables. To this day I still get nervous if someone asks me a multiplication question. I know the
answers and have a fairly good aptitude for math, but that week of missing those drills and songs has forever left me pausing or flustered when faced with 8 x 7 (for example).
You can't stress enough how important those skills are! Thank you for this post. I hope it's one of a series.
I am living proof of the truth of this post. I am 29 years old, graduated second in my class (of 38, SNORT!), and have struggled my entire life with math anxiety mostly because of multiplication.
My ACT score would have been off the chart and near perfect if I hadn't scored so low on the math portion.
I could do the work but it took me so long,I had difficulty understanding concepts because I spent so much time figuring out the multiplication and division without a calculator. I even got sent
to the principal's office once after staying behind to ask for help when my teacher became concerned that I could not answer a simple multiplication problem when he asked me (I froze). I was 18
and that still gives me anxiety when I think about it.
Needless to say, I don't want my kids to have the same problem. Paul is great at math. I'm going to start working on better memorizing my times tables. It's never too late.
I so agree with this. I was sick for like a week in elementary school, and when I got back, they were doing these challeges where you had to write down all the multiplications from 0 to 12
dealing with the number 8. And you had to do it all in 1 minute. If you didn't pass it, you had to do it again the next day. I felt so behind because I had to start at 0, but I'm so glad I made
the effort to try to catch up and learn it with everyone else! (I still suck at 12s because they stopped doing it before I got there.)
My BS is in maths and I've tutored college students who have really ingrained the "Math is scary" idea. I think that some, as children, are told math is hard and then they give up when anything
that might resemble math appears. Of course, not all are like this: my sister has brain-related learning issues and tutoring her was completely different. I had one tutee who put all her math
learning into short-term memory: enough to finish the homework and even show work, but by the time the midterm came around, it wasn't there anymore. Her view of math was that it was hard, she
hated it, and she wanted to get the homework done ASAP and forget the class ASAP. It's hard as a tutor to deal with that. Where do you even start?!
The worst thing about say "I'm bad at math" is that it sometimes becomes a badge of honor (several editorialists have talked about this).
It's basically like bragging to someone "I'm not a nerd".
... although, I do want to say calculus got really hard all of a sudden when I got to college (the worst grade on my BYU transcript).
But I would never say I'm can't do math.
Che happens to be the product developer for Marbles the Brain store. The focal point of the store is to help people keep their brain young so that they may live longer. The majority of those
games are math related or solving puzzles.
Math keeps you young.
AMEN!!!!!!!!!!!!!!!!!!! If that could be font size 500 it would be. amen amen amen amen amen. AMEN!!!! This is why when I taught 6th grade and realized that my sixth graders didn't know their
multiplication tables I made them do times tests every week all year long. I had a standing offer that anyone that could beat me in a 100 problem multiplication worksheet would get an entire
pizza of their choice delivered to them at lunch. In five years only one student did it. Johnny Liao. By 2 seconds, after trying half the year. I got the idea from my own 3rd grade teacher who
took a group of us to McDonald's if we got under 3 minutes on our timing. I was always flabbergasted that by 6th grade they didn't know their tables. I was dead set on them knowing them before
middle school. It is sort of ridiculous.
This comment is getting too long. But you should tell your husband to do that with his students. It's silly for HS, but they will thank him forever for making them learn them.
*timed tests.
Right on.
Likely - J doesn't usually have time to administer tables tests, he has to help his students with his curricula. He does give parents and students resources and urges them to work on the issue at
My husband is a math major and a math tutor, and he will tell you that math *is* a hard thing, but not impossible to learn. Unless you're my brother and have a diagnosed math learning disability.
Math wasn't super hard for me until about fifth or sixth grade. I'm not horrible at math, but I'm not great at it either.
Oh by the way, people have the same reaction when I tell them that I studied English Language and Editing. They'll say, "Oh! I hate English!" I almost want to say, "Then why do you speak it?"
And about the brother with the math learning disability: my mom never told him until he finished all of his math requirements. She didn't want him to use it as an excuse. He did have to retake
math classes, but he worked his tail off to get the grades he needed.
Do you have any suggestions on how to help someone who has working memory issues? My son tested in the bottom 2 percentile in working memory function. Basically he has trouble memorizing things
and any time he has to use his short term memory and his long term memory at the same time he gets overwhelmed and just kinda shuts down. He did great in math until it came time to memorize
multiplication and now he struggles all the time.
Oh, I wouldn't say he should do that every week like I did -- I had my students ALL day--- big difference. I just think he could tell them to practice at home and dish out the challenge and then
once a month, or every couple month or if he has a few minutes to kill at the end of class he could say, "Anyone want to challenge me?". It takes all but three minutes. When he finishes, it's
Melissa -- if I were you I would post multiplication problem flashcards all over his room. Well, I would start with 1's, then 2's and on up to not overwhelm him. With my students who had a hard
time with certain facts or even with spelling, I would tape the problem on their desk (or the words they were struggling with) and because they saw it day after day after day, it stuck. Not
saying that it's a cure all, but i's a try and it can't hurt.
Oh, my friend...is it allowable to blame my frequent "I'm bad at math" declaration on the fact that I am actually quite bad at math? Can I blame it on my Canadian math education? Something?
And what do I do when my child's school teaches math in a completely different way than I learned (what little I managed to learn)? They don't learn their tables. It's actually NOT encouraged.
They've learned (sort of) the combinations to get 10..but that's IT.
It's the blind leading the blind over here. And, while I may be "not so talented" at math, my daughter could be great at it...I just don't know how to help her.
Amen to memorizing those math tables! In our school district (in Boise) they use the "Investigations" math curriculum. My daughter is learning how to multiply 4 times 5 by drawing four circles
with 5 dots in them, then counting them. Or she calculates 8 times 6 by counting by eights - 8, 16, 24... until she gets the right answer. She cannot tell you a single multiplication fact past
the 1's without some serious hesitation or just a blank stare. I fear for any of her generation that is being taught with this curriculum.
Needless to say, her summer will be filled with flash cards! Nothing can replace memorization.
Amy- that it's frowned upon is crazy. Like Angie, your best bet is flash cards or online games that are like flash cards to get the memorization done. There's no substitute.
So what do I do with a kid that aces all of his math tests, yet is failing the class because he hasn't done (or handed in) the homework??!
I feel I should preface my response with "I'm not a mom," which is true, but I was a student. I agree with your basic premise, that kids need to learn to do those four basic operations
automatically, but the very reason they need to is exactly so that they understand how the numbers work together, so that they learn how the numbers play together. But I agree that this should be
something that every child learns sooner rather than later.
I'm glad you posted this. Brady has been getting by doing the math by adding on his fingers, and I realize now that he just needs to have the sums memorized. Great advice!
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Potra-Pták Iterative Method with Memory
ISRN Mathematical Analysis
Volume 2014 (2014), Article ID 697642, 6 pages
Research Article
Potra-Pták Iterative Method with Memory
^1Department of Mathematics, Hamedan Branch, Islamic Azad University, Hamedan 65138, Iran
^2Department of Mathematics, University of Venda, Private Bag X5050, Thohoyandou 0950, South Africa
Received 8 September 2013; Accepted 5 November 2013; Published 22 January 2014
Academic Editors: I. Straškraba and C. Zhu
Copyright © 2014 Taher Lotfi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
The problem is to extend the method proposed by Soleymani et al. (2012) to a method with memory. Following this aim, a free parameter is calculated using Newton’s interpolatory polynomial of the
third degree. So the R-order of convergence is increased from 4 to 6 without any new function evaluations. Numerically the extended method is examined along with comparison to some existing methods
with the similar properties.
1. Introduction
Root finding is a great task in mathematics, both historically and practically. It has attracted attention of great mathematicians like Gauss and Newton. It has real major applications and because of
these real features it is still alive as a research field.
Kung and Traub's conjecture is the basic fact to construct optimal multipoint methods without memory [1]. On the other hand, multipoint methods with memory can increase efficiency index of an optimal
method without memory without consuming any new functional evaluations and merely using accelerator parameter(s). This great power of methods with memory has not been well considered until very
recently. So we have been motivated to extend modified Potra-Pták [2] to its with memory method.
Traub in his book [3] introduced methods with and without memory for the first time. Moreover, he constructed a Steffensen-type method with memory using secant approach. In fact, he increased the
order of convergence of the Steffensen method [4] from 2 to 2.41. This is the first method with memory based on our best knowledge. In other words, Traub changed Steffensen's method slightly as
follows (see [3, pages 185–187]): The parameter is called self-accelerator and method (1) has convergence order of 2.41. It is still possible to increase the convergence order using better
self-accelerator parameter based on better Newton interpolation. Free derivative can be considered as another virtue of (1).
We use the symbols , , and according to the following conventions [3]. If , we write or . If , we write or . If , where is a nonzero constant, we write or . Let be a function defined on an interval ,
where is the smallest interval containing distinct nodes . The divided difference with th-order is defined as follows: , Moreover, we recall the definition of efficiency index (EI) as , where is the
order of convergence and is the total number of function evaluations per iteration.
This paper is organized as follows. Section 2 reviews modified Potra-Pták's method and we try to remodify it slightly too. Error equation for our modification is provided. In Section 3, development
to with memory is carried out along with the discussion of its -order. Numerical examinations and comparisons are presented in the last section.
2. Remodified Optimal Derivative-Free Potra-Pták's Method
In this section, our primal goal is to modify Soleymani et al. method slightly so that its error equation can provide better form in the case with memory. In fact, we prove that our modified method
can generate order of convergence of 6 while theirs has order of convergence of 5.2 in the case of with memory.
Derivative-free iterative methods for solving nonlinear equation are important in the sense that in many practical situation it is preferable to avoid calculation of derivative of . One such scheme
is which is obtained from Newton's method by approximating the derivative by the quotient . Scheme (3) defines a one-parameter family of methods and has the same order and efficiency index as that of
Newton's method [3, 4].
Recently, based on scheme (3), Soleymani et al. [2] have extended the idea of this family and presented Potra-Pták's derivative free families of two-point methods without memory as follows Moreover,
they have proved.
Theorem 1 (see [2]). Let be a simple root of the sufficiently differentiable function in an open interval . If is sufficiently close to , then (5) is of local forth order and satisfies the error
equation below, where , , ., and
As you can see, the order of convergence is . It is clear that error equation (6) has linear factor ; it is better to correct approach (5) in such a way that its error equation has the quadratic
factor . So, as we can prove later, this factor increases convergence order up to 6. To this end, it is just enough to correct second step in (5) as follows:
Hence, method without memory (8) is still optimal and in the following theorem we establish its error equation.
Theorem 2. Let be a simple root of the sufficiently differentiable function in an open interval . If is sufficiently close to , then (8) is of local forth order and satisfies the error equation below
where , , , , and
Proof. We provide the Taylor expansion of any term involved in (8). By Taylor expanding around the simple root in the nth iterate, we have By considering this relation and the first step of (8), we
obtain At this time, we should expand around the root by taking into consideration (12). Accordingly, we have Additionally, we obtain Similarly, the same Taylor expansion results in Using (12)–(15)
in the last step of (8) provides finally which shows that (8) is a derivative-free family of two-step methods with optimal convergence rate of 4. This completes the proof.
3. Development and Construction with Memory Family
This section concerns with extension of (8) to a method with memory since its error equation contains the parameter which can be approximated in such a way that increase the local order of
convergence. So we set as the iteration proceeds by the formula for , where is an approximation of . We have a method through the following forms of : where is Newton's interpolatory polynomial of
third degree, set through four available approximations , , , and and By using Taylor's expansion of around the root , we have where . By using (18) and (19), we calculate According to this and (17
) we find For general case one can consult [3].
In order to obtain the order of convergence of the family of two-point methods with memory (8), where is calculated using the formula (17), we will use the concept of the -order of convergence [3].
Now, we can state the following convergence theorem.
Theorem 3. If an initial approximation is sufficiently close to the zero of and the parameter in the iterative scheme (8) is recursively calculated by the forms given in (17), then the -order of
convergence is at least 6.
Proof. Let be a sequence of approximations generated by an iterative method with memory (IM). If this sequence converges to the zero of with the -order (≥) of IM, then we write where tends to the
asymptotic error constant of IM when . Thus Let , , then we have where . In the sequel, we obtain the -order of convergence of family (8) for approach (17) applied to the calculation of .
Assume that the iterative sequences and have the -orders; and , respectively, then, bearing in mind (22) we obtain and then, we obtain Assume that the iterative sequence has the -order ; then
bearing in mind (22) we obtain and then, we obtain Combining the exponents of on the right-hand sides of (27)-(28), (29)-(30), and (23)–(31), we form the nonlinear system of three equations in , ,
and :
Nontrivial solution of this system is , , and , and we conclude that the lower bound of the -order of the method with memory is .
Similarly, one can prove the following.
Theorem 4. If an initial approximation is sufficiently close to the zero of and the parameter in the iterative scheme (5) is recursively calculated by the forms given in (17), then the -order of
convergence is at least 5.2.
4. Numerical Examples
To examine practical aspects of the proposed modified Potra-Pták's without and with memory we implement it here in action. In other words, we demonstrate the convergence behavior of the method with
memory (8), where is calculated by (17). For comparison purposes, we pick up Kung and Traub [1] and Zheng et al. [5] with and without memories. We use these notations. The errors denote
approximations to the sought zeros. stands for . Moreover, indicates computational order of convergence and is computed [2] The software Mathematica 8, with 1000 arbitrary precision arithmetic, has
been used in our computations. The results alongside the test functions are given in Tables 1 and 2, while [3]. From Tables 1 and 2, we can conclude that our methods work numerically well and are
successfully competing with the existing methods. Indeed the last columns of these tables show that both numerical and theoretical aspects support each other.
For comparison purposes, we consider the following methods.
Two-Point Method by Kung and Traub [1]:
Two-Point Method by Zheng et al. [5]:
From Tables 1 and 2, it can be seen that our modified method without memory works truly; moreover, its with memory competes the existing methods. To sum up, Potra and Pták [6] constructed two-point
method without memory with convergence order of 3; it is not optimal in the sense of Kung and Traub. Cordero et al. [7] could make it optimal. In other words, they introduce optimal two- and
three-point methods with order of convergence of 4 and 8, respectively. Though their methods are optimal, they are not derivative-free. Freshly, Soleymani et al. [2] have drawn two point methods
without memory from Potra and pták method. One is derivative-free and the other is not. In addition their derivative method results two steps method by Cordero et al. [7] for (See (5)). In this work,
we modified their derivative-free method at first. Then, we generalized it to method with memory with efficiency index ; see more about efficiency index in [7]. Therefore, a two-step method with
memory can obtain performance even better than four-step methods without memory with efficiency index .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This research was supported by the University of Venda and the Islamic Azad University, Hamedan Branch.
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2. F. Soleymani, R. Sharma, X. Li, and E. Tohidi, “An optimized derivative-free form of the Potra-Pták method,” Mathematical and Computer Modelling, vol. 56, no. 5-6, pp. 97–104, 2012. View at
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3. J. F. Traub, Iterative Methods for the Solution of Equations, Prentice Hall, New York, NY, USA, 1964. View at MathSciNet
4. J. F. Steffensen, “Remarks on iteration,” Scandinavian Actuarial Journal, vol. 1933, no. 1, pp. 64–72, 1933. View at Publisher · View at Google Scholar
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6. F. A. Potra and V. Pták, “Nondiscrete introduction and iterative processes,” in Research Notes in Mathematics, vol. 103, Pitman, Boston, Mass, USA, 1984.
7. A. Cordero, J. L. Hueso, E. Martínez, and J. R. Torregrosa, “New modifications of Potra-Pták's method with optimal fourth and eighth orders of convergence,” Journal of Computational and Applied
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A. Multiple quantum (MQ) and spin state selection
B. Three spin system of the type AMX: 2,3-dibromopropionic acid. Spin selected double quantum experiment
C. Spin selected zero quantum experiment
D. Spin system of the type AFKPX: The spectrum of 2-fluoropyridine
E. New spin state selected DQ-J resolved sequence
F. Spin system of the type AFKPX: The spectrum of 1-chloro-2-fluorobenzene
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Article discusses the effect of subtle messages communicated to children through adults' anxiety about math and science abilities
John Evans on 04 Sep 13
"Contrary to what you've probably read, you don't have to be engaging to be a great teacher-at least not in any charismatic and charming sense of the word.
You can be relatively "boring" and lead students to outstanding academic progress, mainly by staying organized, being reflective, flexible, and in constant contact with an active and ambitious
professional learning networking. Teaching differently requires work."
Garrett Eastman on 18 May 13
abstract: "A large number of studies carried out on pupils aged 8-14 have shown
that teachable agent (TA) based games are beneficial for learning. The present pi-
oneering study aimed to initiate research looking at whether TA based games can
be used as far down as preschool age. Around the age of four, theory of mind
(ToM) is under development and it is not unlikely that a fully developed ToM is
necessary to benefit from a TA's socially engaging characteristics. 10 preschool
children participated in an experiment of playing a mathematics game. The partic-
ipants playing a TA-version of the game engaged socially with the TA and were
not disturbed by his presence. Thus, this study unveil exciting possibilities for
further research of the hypothesised educational benefits in store for preschoolers
with regard to play-and-learn games employing TAs."
Garrett Eastman on 29 Jan 13
Abstract: "In this paper, students provide insight into their
use of Quick Response (QR) codes and mobile
devices to assist in mathematics homework efforts.
These QR codes were directly linked to instructional
videos related to their unit on fraction algorithms
and were hosted on YouTube. In particular, through
focus-group interviews, the students identified many
strengths associated with the implementation of this
research. The strengths include the manner in which
the YouTube clips of currently accepted instructional
strategies worked to reinforce their classroom
learning, how the mobile devices motivated students
to complete homework in a variety of non-traditional
settings, increased their communication with their
classroom teacher, and how these devices engaged
parents and siblings in the learning process."
Garrett Eastman on 10 Dec 12
A college institutional researcher reveals some data concerning math placement tests and the pressures impacting students to opt for more difficult math courses without adequate preparation and
unsatisfactory results. He writes: "In my mind, this disconnect exemplifies the degree to which incoming students and families don't grasp the difference between going to college to acquire
content knowledge and going to college to develop skills and dispositions. ... [I]f students understand that college is about developing skills and dispositions, I think that they might be more
likely to appreciate the chance to start at the beginning that is appropriate for them, savoring each experience like a slow cooked, seven course meal because they know that the culmination of
college is made exponentially better by the particular ordering and integrating of the flavors that have come before."
Leena Helttula on 05 Dec 12
"126 page free downloadable PDF book for grade 4-7 teachers showing how KenKen puzzles can turn mathematical problem solving into fun and how student motivation can be dramatically increased.
Garrett Eastman on 14 Oct 12
"This paper describes a research project on Year 3
primary school students in Malaysia in their use of computer-based
video game to enhance learning of multiplication facts (tables) in the
Mathematics subject. This study attempts to investigate whether
video games could actually contribute to positive effect on children's
learning or otherwise. In conducting this study, the researchers
assume a neutral stand in the investigation as an unbiased outcome of
the study would render reliable response to the impact of video
games in education which would contribute to the literature of
technology-based education as well as impact to the pedagogical
aspect of formal education. In order to conduct the study, a subject
(Mathematics) with a specific topic area in the subject (multiplication
facts) is chosen. The study adopts a causal-comparative research to
investigate the impact of the inclusion of a computer-based video
game designed to teach multiplication facts to primary level students.
Sample size is 100 students divided into two i.e., A: conventional
group and B conventional group aided by video games. The
conventional group (A) would be taught multiplication facts (timetables)
and skills conventionally. The other group (B) underwent the
same lessons but with supplementary activity: a computer-based
video game on multiplication which is called Timez-Attack. Analysis
of marks accrued from pre-test will be compared to post- test using
comparisons of means, t tests, and ANOVA tests to investigate the
impact of computer games as an added learning activity. The findings
revealed that video games as a supplementary activity to classroom
learning brings significant and positive effect on students' retention
and mastery of multiplication tables as compared to students who rely
only upon formal classroom instructions."
Garrett Eastman on 26 Sep 12
first mathematics education seminar: Understanding Abstract Concepts in the Context of Abstract Algebra.
Mathematics is a science of numbers, quantity, and space. All of the listed components are abstract ideas. How do we learn abstraction?"
Garrett Eastman on 16 Sep 12
New York Times story of Neil Heffernan's creation of a computerized tutor designed to emulate actual teachers, eventually becoming ASSISTments
Garrett Eastman on 21 May 12
Abstract "This article reports on a case study of the web-based educational maths application, Mathletics. The findings are drawn from an ethnographic study of children's technology use in
Melbourne, Australia. We explore the experience, governance and commerce of children's Mathletics use, and offer insights into the developing possibilities and challenges emerging through the
adoption of Web 2.0 applications for learning and education." (Full text requires subscription or purchase)
Garrett Eastman on 02 May 12
See in particular, Digital Learning Dot #3, Personalized Learning in Math Class
Garrett Eastman on 13 Apr 12
by Penta, Michael K., M.S., UNIVERSITY OF MASSACHUSETTS LOWELL, 2011, 83 pages; 1507796 Abstract record for a thesis completed in December 2011, fulltext requires subscription or purchase. A news
story on the departmental blog http://blog.uml.edu/cs/2011/12/penta_ms_video_games_for_math_learning.html describes the kinds of games involved, the participants and the results.
Garrett Eastman on 07 Apr 12
Discusses theory of motivational learning in educational games and presents a research study involving 115 elementary school students in math class using digital game-based learning
Garrett Eastman on 22 Mar 12
Abstract: "A mobile learning research project was conducted in Trinidad and Tobago to determine if mobile learning can assist high school students in learning mathematics. Several innovative
techniques were used in this research to address the problem of high failure rates of mathematics in high schools in the Caribbean. A mobile learning application was developed based on a subset
of the high school mathematics curriculum used in the English-speaking Caribbean. Game-based learning, personalization and multiple learning strategies were used in conjunction with mobile
learning to assist students in improving their performance in mathematics. Three evaluation studies were conducted with the mobile learning application. During the studies, usage data was
captured automatically by the system and this was used to determine the extent to which the students actually used the mobile application. At the end of each study, a questionnaire was used to
capture student opinions of the mobile learning application. Questionnaire data is based solely on student responses and there is no guarantee of its accuracy and reliability. This paper focuses
on the responses of the students to the questionnaire and seeks to determine if the usage data can increase the reliability of the questionnaire data. It summarizes the behaviour patterns of the
students gleaned from the usage logs and compares this to the students' responses to the questionnaire. Generally it was found that the students' responses agreed with the usage data, though
there were occasions when the responses diverged."
Garrett Eastman on 31 Jan 12
(Self-directed-Learning) "SDL model for planning, managing, and
directing the development of student progress when using the
educational games while learning math."
Garrett Eastman on 27 Jan 12
Using the MathMind tutorial and observing advantages and disadvantages for students and teachers
Garrett Eastman on 26 Jan 12
"Broadening participation among women and racial/ethnic minorities in science, technology, engineering and maths" , book chapter excerpt
Garrett Eastman on 16 Dec 11
Describes Motion Math, an iPad game, shown in a study to improve 5th graders' facility with fractions
Garrett Eastman on 16 Dec 11
"Math Quest is a role playing game that could be used as a tool to learn numbers and basic
mathematic operations. The Math Quest package consists of two main modules; learning and game modules
that can be executed separately. The use of the learning module as a tool in learning will allow for a highly
individualized and interactive environment. This paper presents the design of the learning module for
numbers and their mathematics operation. Due to its interactive and stimulating nature, the module is suitable
for school children age 9 to 12 years old to learn the subject. The development takes into consideration of
constructivism learning theories where learning is based on students' active participation in problem solving
and critical thinking regarding activity that they are involved in. The framework for each of module is as
follows: objectives, concept, examples, exercises, quizzes. A heuristic evaluation on the design was
conducted and positive feedback was obtained."
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
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You haven't written a testimonial for Owlfred.
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Statistical Comparison Framework and Visualization Scheme for Ranking-Based Algorithms in High-Throughput Genome-Wide Studies
As a first step in analyzing high-throughput data in genome-wide studies, several algorithms are available to identify and prioritize candidates lists for downstream fine-mapping. The prioritized
candidates could be differentially expressed genes, aberrations in comparative genomics hybridization studies, or single nucleotide polymorphisms (SNPs) in association studies. Different analysis
algorithms are subject to various experimental artifacts and analytical features that lead to different candidate lists. However, little research has been carried out to theoretically quantify the
consensus between different candidate lists and to compare the study specific accuracy of the analytical methods based on a known reference candidate list. Within the context of genome-wide studies,
we propose a generic mathematical framework to statistically compare ranked lists of candidates from different algorithms with each other or, if available, with a reference candidate list. To cope
with the growing need for intuitive visualization of high-throughput data in genome-wide studies, we describe a complementary customizable visualization tool. As a case study, we demonstrate
application of our framework to the comparison and visualization of candidate lists generated in a DNA-pooling based genome-wide association study of CEPH data in the HapMap project, where prior
knowledge from individual genotyping can be used to generate a true reference candidate list. The results provide a theoretical basis to compare the accuracy of various methods and to identify
redundant methods, thus providing guidance for selecting the most suitable analysis method in genome-wide studies.
Key words: genome-wide association studies, candidate lists
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Probability: Infinite Convergent Series and Random Variables
I have a random variable problem. I need to prove that my equation I came up with is a valid probability mass function.
In the problem, I came up with this for my probability mass function:
[tex]\Sigma[/tex] [tex]12/(k+4)(k+3)(k+2)[/tex]
Maple says that this does in fact converge to 1, so it's valid; however...I can't use "Maple said so" as an answer.
My attempt was to break it up using partial fraction decomposition:
([tex]6/(k+4)[/tex]) - ([tex]12/(k+3)[/tex]) + ([tex]6/(k+2)[/tex])
I was hoping that this would be telescoping, but it is not. Does anyone have an idea on how I can prove that this converges to 1?
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[NBI] Week four of the Math Blogging Initiation
September 16, 2012
It’s the final week of the Math Blogging Initiation, and I’ve got 7 awesome bloggers who deserve a huge congratulations for submitting a weekly (or more) blog post for the past month. Congrats to
everyone who participated in this outstanding project—the math twitterblogoverse is much richer thanks to your contributions.
Joe B (@forumjoe) has a blog named lim joe→∞. The fourth post for the Blogging Initiation is titled Everything is Mathematical” and the author sums it up as follows: I provide a link to a new
mathematics challenge site, Everything is Mathematical and discuss the content. I post my solution to the first puzzle and improve my Latex skills in the process. A memorable quotation from the post
is: I’m really impressed by the way Marcus du Sautoy presents the problem in an easy-to-understand way. There’s no pseudocontext here, there’s no anyqs.
My notes: joe is a Maths Teacher in the UK, who’s actually been sporadically blogging for quite some time (since 2009). He’s recently been developing is chops with LaTeX, and he is uncovering some
great resources like Everything is Mathematical that I had not seen before.
Nutter Buttersmith (@reminoodle) has a blog named The MathSmith. The fourth post for the Blogging Initiation is titled Mathemes” and the author sums it up as follows: I had a lot of fun creating
these math memes. A memorable quotation from the post is: there were no sentences.
My notes: These are some great math memes, and if you check the archives, you’ll see Nutter is fully plugged into the math twitterblogoverse; she’s posted a Made4Math Monday post on foldables, as
well as a number of My Favorite Friday posts, including this post on student math puns.
Christy Wood (@wyldbirman) has a blog named Hands on Math in High School. The fourth post for the Blogging Initiation is titled Beautiful Dance Moves” and the author sums it up as follows: My blog is
about the beautiful dance moves I do when I describe functions. I have a picture and descriptions. A memorable quotation from the post is: When I started teaching, I found myself using my arms – and
not for writing on the whiteboard.
My Notes: A nice quick tip on helping students to visualize functions through dance. Heck, this technique has even gotten some kids into college (and profiled in the NYT). Christy teaches Algebra,
Algebra II and Geometry at a school in Florida.
Jonathan Newman | Hilbert’s Hotel
Jonathan Newman has a blog named Hilbert’s Hotel. The fourth post for the Blogging Initiation is titled Tiered Assessment in Chemistry” and the author sums it up as follows: I used another tiered
assessment (stolen from Steve Grossberg) and tried to perfect it. The quiz is on scientific notation and significant figures for my Chemistry class, but is certainly usable in a math class. I will
try to post more of my tiered assessment quizzes online as I work out the kinks. A memorable quotation from the post is: And since we get bonus points for embedding a video, here’s one reason it’s
fun to be a Chemistry teacher (yes, I celebrate Pi day in my math classes even though I don’t teach Geometry…)
My Notes: Jonathan teaches precalculus, chemistry, physics and general science at Rehoboth high school in New Mexico. I think the idea of tiered quizzing is fairly interesting—students can simply
work their way up to the highest level of mastery. He also seems to be dipping his toes into Standards based grading.
Valerie Higgins @Valerie1121 has a blog named Crafty Math. The fourth post for the Blogging Initiation is titled How big is my classroom?” and the author sums it up as follows: My students worked in
groups to calculate the perimeter, area, and volume of my classroom. Then they summarized how they did this in written form. A memorable quotation from the post is: I was tired of the worksheets.
My Notes: OMG OMG OMG: Valerie did the super cool project based on Ana Soler’s incredible artwork where she creates sculptures of tennis balls appearing to bounce through the air, and she wrote a
3-part post about how she did it: part 1, part 2, part 3. I am so going to do this in my classroom.
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The generalisation of ordered pair to something having more positions is usually called a tuple (or ordered tuple). More particularly one gets the term $n$-tuple, which refers to a list, $(x_1, \
ldots, x_n)$, with $n$ entries from some set, $X$; here $n$ is a natural number. It thus corresponds to an element in the $n$-fold product set, $X^n$. The various elements $x_i$ of the $n$-tuple are
usually called its components and sometimes it is useful to call the set of components the support or range of the tuple.
• An ordered pair is a $2$-tuple. A $3$-tuple is a triple, a $4$-tuple is a quadruple, a $5$-tuple is a quintuple etc. The notions of $1$-tuple and $0$-tuple are trivial.
• The term ‘tuple’ is usually used for an $n$-tuple for a specific number $n$. If we wish to speak of an $n$-tuple for an arbitrary $n$ (particularly without specifying that $n$), then we may speak
of a list (which has other terminology, described on that page). Then the set of lists is the disjoint union over $n$ of the sets of $n$-tuples.
• The term ‘tuple’ is usually used for an $n$-tuple for a finite number $n$. If we wish to speak of an $n$-tuple for an infinite (or possibly infinite) $n$, then we may speak of a sequence.
See ordered pair for methods of formalising ordered pairs (which are $2$-tuples) in various foundations of mathematics. Some of these generalise immediately to $n$-tuples for arbitrary $n$;
otherwise, we may define $n$-tuples recursively: a triple is an ordered pair whose (say) first component is an ordered pair; a quadruple is an ordered pair whose first component is a triple, etc.
Revised on May 4, 2013 00:12:50 by
Urs Schreiber
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Let A and B be points on a circle. If the line that goes through A and B goes through the center of the circle, we say that A and B are diametically opposed.
If A and B are diametrically opposed, then the line through A and B divides the circle into two half circles.
Let A and B be two points on a circle centered at O. The arc between A and B is the set of points on the circle which are in the angle between OA and OB.
If A and B are diametrically opposed, we will have to specify which half circle we mean.
Let A and B be two points on a circle. A partition of the arc between A and B is a set of points
A = P[0], P[1], P[2], . . . , P[n] = B
on the arc between A and B ordered so that P[i ]is in the angle between OP[i-1 ]and OP[i][+1 ]where O is the center of the circle.
The inside approximation of the arc length from A to B corresponding to the partition A = P[0] , P[1], . . . , P[n] = B is the sum of the distances between consecutive points
|A, P[1 ]| + | P[1], P[2 ]| + | P[2], P[3]| + . . . + |P[n][-1], B|
Let A and B be two points on a circle. The outside approximation of the arc length from A to B corresponding to the partition A = P[0] , P[1], . . . , P[n] = B is the sum of the distances starting at
A and going through all of the intersections of tangent lines at consecutive points.
|A, T[1 ]| + | T[1], T[2 ]| + | T[2], T[3]| + . . . + |T[n], B|
Where T[i] is the point where the tangent at P[i-1] meets the tangent at P[i].
One partition is called a refinement of another partition if every point in one partition is also a point in the other one.
Let A and B be two points on a circle. The length of the arc between A and B or the arc length between A and B is the least upper bound of all of the inside approximations.^1
The unit circle is the circle of radius 1 centered at the origin.
In the angle between AB and AC, the ray from A thorough B and the ray from A through C are called the arms of the angle.
The angle illustrated above is called / BAC. The middle letter indicates the vertex of the angle, and the other two letters are points on the two arms of the angle.
For any angle, the ratio of the arc length of the arc of a circle centered at the vertex of the angle bewtween the arms of the angle to the radius of the circle is called the number of radians in the
angle , or the radian measure of the angle.^2 The notation for the number of radians in an angle / BAC will be m/ BAC.
To get the number of degrees in an angle, multiply the number of radians by 180^o/pi^3
A straight angle is one whose two rays are opposite rays of the same straight line.
If the arms of an angle are perpendicular, the angle is called a right angle.
The number of radians in a straight angle is denoted by pi. Pi is a Greek letter that looks like
A straight angle has 180 degrees.
An angle which is smaller than a right angle is called an acute angle .
An angle which is between a right angle and a straight angle in size is called an obtuse angle .
If two angles add up to a straight angle, they are called supplementary.
Supplementary angles
If two angles add up to a right angle they are called complementary.
Complementary angles
If two lines cross in a plane as in the figure below, the angles which are opposite each other called vertical angles.
A rotation is accomplished by fixing one point of the plane and moving all the points through a fixed angle about the fixed point.
In the figure, A is moved to A', B is moved to B', and C is moved to C'. In order for the movement to be a rotation we must have
|OA| = |OA'|
|OB| = |OB'|
|OC| = |OC'|
and / AOA', / BOB', and / COC' all have the same size.
If you start at a point on a circle where a horizontal line through the center meets the circle in the negative x direction and proceed in the positive y direction, you are going clockwise.
The opposite direction along a circle from clockwise is called counterclockwise.
Analytic Foundations of Geometry
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Pattern Drafting 101 - The Men's Shirt Sleeve Block
Pattern Drafting 101 – The Men’s Shirt Sleeve Block
18 February, 2011
Thank you all for your support- it makes me incredibly happy to see all those hits on my drafting posts! Let me know if there’s anything you want to see and feel free to post comments or questions!
Also, my fiancé pointed out to me today that I should include a glossary of terms on this site, so starting with this post, I’m including hyperlinks to the new glossary page. I’ll be going back and
adding them to the older posts soon.
For the men’s shirt sleeve, you’re going to need the following:
• Pencil
• Ruler
• Newsprint Paper (It’s easier to see through than thicker papers and will be useful in transferring the markings from the shirt to the sleeve. It’s also fairly inexpensive and readily available at
most craft stores.)
• A copy of your completed Men’s Shirt block – you’ll be using markings from this to create the sleeve, so make sure you haven’t erased your marked points.
Step 1: Rename the Points- For the sake of simplicity, we’re going to rename some of the points on our completed Men’s Shirt block to use on our Sleeve Block. Rename the following points according to
the chart below.
│Shirt Block │Replace with…. │
│Point K │Point (a) │
│Point L │Point (b) │
│Point Y │Point (f) │
│Point U │Point (g) │
│Point M │Point (h) │
Step 2: Measure the Armscye- Lay a flexible measuring tape or a length of string along the armscye curve. Record this measurement. Extend the line (ab) upwards a distance of 1/3 the armscye
measurement and mark the end point (c).
Step 3: Drawing in the Construction Lines- Extend the Armscye Line out to the left, through point B. It doesn’t really matter how far, just far enough. You can always make it longer if you need to.
Do the same from Point (c). Mark point (d) halfway along line (ac), and extend a line out to the left of point (d) approximately the same length as your other two construction lines. These lines
don’t need to be drawn in very heavily; you’ll only be erasing them at the end.
Step 4: Mark Some New Points- Where the construction line from (d) intersects with the left side of the armscye, mark the point (e).
Step 5: Measure and Pivot- Measure the distance from (f) to (b) in a straight line. Add 1.5cm to this and record. Pivot your ruler on point (b) until it intersects with the topmost construction line
at the distance you just recorded, to the left of point (b). Draw the line and mark the endpoint (i).
Measure the distance again from (g) and (e) in a straight line. Add 1.25cm and record. Pivot your ruler on point (i) until it intersects with the middle construction line at the distance you just
recorded, to the left of point (i). Draw the line and mark the endpoint (j).
Step 6: Measure and Pivot again- Measure the distance along the curve from (e) to (h). Add 1.25cm and record. Pivot your ruler around point (j) until it intersects with the Armscye Line at the
distance you just recorded, to the left of point (j). Draw the line and mark the endpoint (k).
Measure the distance again along the curve from (b) to (h). Add .75cm and record. Pivot your ruler on point (b) until it intersects with the Armscye Line at the distance you just recorded, to the
right of point (b). Draw the line and mark the endpoint (l).
Step 7: Sleeve Length- Draw a vertical line straight down from point (i) to the desired sleeve length (This block creates a long dress-shirt sleeve. I’ll go over pattern editing for shorter sleeves
at a later date. For now, it’s good to have a longer base to work from). Mark the endpoint (m).
Step 8: The Hem Line- Draw a horizontal line to the right of point (m) until the end lines up with point (l). Mark the end point (n) and connect (n) to (l) with a straight line. (mnl) should be a
right angle. Do the same for the left side, marking the end point (o) and connecting (o) to (k)
Step 9: Suppressing the Shirt Block- Go ahead and get rid of your shirt block lines at this point. You won’t need them after this. Make sure to keep everything you’ve done on the sleeve so far.
Step 10: Shaping the Sleeve- Measure 5cm inward from (n). Mark the point (p), and connect it to point (l). Do the same on the other side, marking the resulting point (q) and connecting it to point
Step 11: Sleeve Slit- Mark point (r) midway between (m) and (q). Draw a short, 1cm line straight down from point (r), and mark the end point (s). Draw a line straight up from (s) 10cm long. Mark the
end point (t). The line (st) will become the sleeve plaquette in the finished pattern.
Step 12: The Sleeve Hem- Draw a curved line connecting points (m), (s), and (q).
Step 13: Drawing the Sleeve Cap- Connect the points at the top of the sleeve with curved lines, following the chart below.
│From point…│To point…│Direction of Curve │Maximum Deviation│
│k │j │Downward │0.75cm │
│j │i │Upward │1.5cm │
│i │x │Upward │2cm │
│x │b │None (straight) │0cm │
│b │l │Downward │0.75cm │
Step 14: Marking the Elbow Line- Find the halfway point between (n) and (t); 2.5cm above that, mark a point (z). Draw a horizontal line through (z) until you reach the line (kq). This is the elbow
line of the sleeve.
Step 15: Cleaning up- erase all the construction lines, and you should be left with something that looks like this:
Tada! You’re done! You now have a completed men’s shirt sleeve block ready to be turned into something amazing!
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Strong Convergence Theorems for the Generalized Split Common Fixed Point Problem
Journal of Applied Mathematics
Volume 2012 (2012), Article ID 575014, 13 pages
Research Article
Strong Convergence Theorems for the Generalized Split Common Fixed Point Problem
College of Science, Civil Aviation University of China, Tianjin 300300, China
Received 9 January 2012; Accepted 17 February 2012
Academic Editor: Rudong Chen
Copyright © 2012 Cuijie Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium,
provided the original work is properly cited.
We introduce the generalized split common fixed point problem (GSCFPP) and show that the GSCFPP for nonexpansive operators is equivalent to the common fixed point problem. Moreover, we introduce a
new iterative algorithm for finding a solution of the GSCFPP and obtain some strong convergence theorems under suitable assumptions.
1. Introduction
Let and be real Hilbert spaces and let be a bounded linear operator. Given intergers , let us recall that the multiple-set split feasibility problem (MSSFP) was recently introduced [1] and is to find
a point: where and are nonempty closed convex subsets of and , respectively. If , the MSSFP (1.1) becomes the so-called split feasibility problem (SFP) [2] which is to find a point: where and are
nonempty closed convex subsets of and , respectively. Recently, the SFP (1.2) and MSSFP (1.1) have been investigated by many researchers; see, [3–10].
Since every closed convex subset in a Hilbert space is looked as the fixed point set of its associating projection, the MSSFP (1.1) becomes a special case of the split common fixed point problem
(SCFPP), which is to find a point: where and are nonlinear operators. If , the problem (1.3) reduces to the so-called two-set SCFPP, which is to find a point:
Censor and Segal in [11] firstly introduced the concept of SCFPP in finite-dimensional Hilbert spaces and considered the following iterative algorithm for the two-set SCFPP for Class- operators:
where , and is the identity operator. They proved the convergence of the algorithm (1.5) to a solution of problem (1.4). Moreover, they introduced a parallel iterative algorithm, which converges to a
solution of the SCFPP (1.3). However, the parallel iterative algorithm does not include the algorithm (1.5) as a special case.
Very recently, Wang and Xu in [12] considered the SCFPP (1.3) for Class- operators and introduced the following iterative algorithm for solving the SCFPP (1.3): Under some mild conditions, they
proved some weak and strong convergence theorems. Their iterative algorithm (1.6) includes Censor and Segal’s algorithm (1.5) as a special case for the two-set SCFPP (1.4). Moreover, they prove that
the SCFPP (1.3) for the Class- operators is equivalent to a common fixed point problem. This is also a classical method. Many problems eventually converted to a common fixed point problem; see [13–15
Motivated and inspired by the aforementioned research works, we introduce a generalized split common fixed point problem (GSCFPP) which is to find a point: Then, we show that the GSCFPP (1.7) for
nonexpansive operators is equivalent to the following common fixed point problem: where for every . Moreover, we give a new iterative algorithm for solving the GSCFPP (1.7) for nonexpansive operators
and obtain some strong convergence theorems.
2. Preliminaries
Throughout this paper, we write and to indicate that converges weakly to and converges strongly to , respectively.
An operator is said to be nonexpansive if for all . The set of fixed points of is denoted by . It is known that is closed and convex. An operator is called contraction if there exists a constant such
that for all . Let be a nonempty closed convex subset of . For each , there exists a unique nearest point in , denoted by , such that for every . is called a metric projection of onto . It is known
that for each , for all .
Let be a sequence of operators of into itself. The set of common fixed points of is denoted by , that is, . A sequence is said to be strongly nonexpansive if each is nonexpansive and whenever and are
sequences in such that is bounded and ; see [16, 17]. A sequence in is said to be an approximate fixed point sequence of if . The set of all bounded approximate fixed point sequences of is denoted by
; see [16, 17]. We know that if has a common fixed point, then is nonempty; that is, every bounded sequence in the common fixed point set is an approximate fixed point sequence. A sequence with a
common fixed point is said to satisfy the condition () if every weak cluster point of is a common fixed point whenever . A sequence of nonexpansive mappings of into itself is said to satisfy the
condition () if for every nonempty bounded subset of ; see [18].
In order to prove our main results, we collect the following lemmas in this section.
Lemma 2.1 (see [16]). Let be a nonempty subset of a Hilbert space . Let be a sequence of nonexpansive mappings of into . Let be a sequence in such that . Let be a sequence of mappings of into defined
by for , where is the identity mapping on . Then is a strongly nonexpansive sequence.
Lemma 2.2 (see [16]). Let be a Hilbert space, a nonempty subset of , and and sequences of nonexpansive self-mappings of . Suppose that or is a strongly nonexpansive sequence and is nonempty. Then .
Lemma 2.3 (see [17]). Let be a Hilbert space, and a nonempty subset of . Both and satisfy the condition and is bounded for any bounded subset of . Then satisfies the condition .
Lemma 2.4 (see [19]). Let and be bounded sequences in a Banach space and let be a sequence in with . Suppose for all integers and Then .
Lemma 2.5 (see [20]). Assume that is a sequence of nonnegative real numbers such that where is a sequence in and is a sequence such that(i),(ii) or . Then .
3. Main Results
Now we state and prove our main results of this paper.
Lemma 3.1. Let be a given bounded linear operator and let be a sequence of nonexpansive operators. Assume For each constant , is defined by the following: Then . Moreover, for , is nonexpansive on
for .
Proof. Since the inclusion is evident, now we only need to show the converse inclusion. If , then we have . Since , we take an arbitrary . Hence It follows that , then for every , hence . Next we
turn to show that is a nonexpansive operator for . Since is nonexpansive, we have Hence For , we can immediately obtain that is a nonexpansive operator for every .
From Lemma 3.1, we can obtain that the solution set of GSCFPP (1.7) is identical to the solution set of problem (1.8).
Theorem 3.2. Let and be sequences of nonexpansive operators on Hilbert space . Both and satisfy the conditions and . Let be a contraction with coefficient . Suppose . Take an initial guess and define
a sequence by the following algorithm: where , , , and are sequences in . If the following conditions are satisfied:(i);(ii)and ;(iii);(iv);(v), then converges strongly to where .
Proof. We proceed with the following steps.
Step 1. First show that there exists such that .
In fact, since is a contraction with coefficient , we have for every , . Hence is also a contraction. Therefore, there exists a unique such that .Step 2. Now we show that is bounded.
Let , then and . Hence Then By induction on , for every . This shows that and are bounded, and hence, , , and are also bounded.Step 3. We claim that and , where .
We first show the former equality. Let be a bounded sequence in . If , then Hence . On the other hand, if , combining (3.11) and , we obtain that . Hence . Therefore, .
Next, we show the latter equality. Using Lemma 2.1, we know that is a strongly nonexpansive sequence. Thus, since , from Lemma 2.2 we have Step 4. satisfies the condition , where .
Let be a nonempty bounded subset of . From the definition of , we have, for all , It follows that Since satisfies the condition and , we have that is, satisfies the condition . Since is bounded for
any bounded subset of , by using Lemma 2.3, we have that satisfies the condition , that is, satisfies the condition .Step 5. We show .
We can write (3.6) as where . It follows that From Step 2, we may assume that , where is a bounded set of . Then from (3.16), we obtain It follows that Since satisfies the condition , combining as ,
we have Hence by Lemma 2.4, we get as . Consequently, Step 6. We claim that .
From (3.6), we have and hence Since , and , we derive Thus (3.23) and Steps 2 and 3 imply that Step 7. Show , where .
Since is bounded, there exist a point and a subsequence of such that and . Since and satisfy the condition , from Step 6, we have . Using (2.1), we get Step 8. Show .
Since , using (3.8), we have which implies that for every . Consequently, according to Step 7, , and Lemma 2.5, we deduce that converges strongly to . This completes the proof.
Combining Lemma 3.1 and Theorem 3.2, we can obtain the following strong convergence theorem for solving the GSCFPP (1.7).
Theorem 3.3. Let and be sequences of nonexpansive operators on Hilbert space and , respectively. Both and satisfy the conditions and . Let be a contraction with coefficient . Suppose that the
solution set of GSCFPP (1.7) is nonempty. Take an initial guess and define a sequence by the following algorithm: where , and , , , are sequences in . If the following conditions are satisfied:(i);
(ii)and;(iii);(iv);(v), then converges strongly to where .
Proof. Set . By Lemma 3.1, is a nonexpansive operator for every . We can rewrite (3.29) as
We only need to prove that satisfies the conditions and . Assume that is a nonempty bounded subset of . For every , we have Since satisfies the condition , and is bounded, it follows from (3.31) that
Therefore, satisfies the condition .
Assume that and ; we next show that . By using , we have . Since , we choose an arbitrary point ; then for every , Hence Then we get . Since satisfies the condition and , we have . From Lemma 3.1, we
have .
Let be a nonexpansive mapping with a fixed point, and define for all . Then satisfies the conditions and . Thus, one obtains the algorithm for solving the two-set SCFPP (1.4).
Corollary 3.4. Let and be nonexpansive operators on Hilbert space and , respectively. Let be a contraction with coefficient . Suppose that the solution set of SCFPP (1.4) is nonempty. Take an initial
guess and define a sequence by the following algorithm in (3.29), where , and , , , are sequences in . If the following conditions are satisfied:(i);(ii) and ;(iii);(iv);(v). Then converges strongly
to where .
Remark 3.5. By adding more operators to the families and by setting for and for , the SCFPP (1.3) can be viewed as a special case of the GSCFPP (1.7).
This research is supported by the science research foundation program in Civil Aviation University of China (07kys09), the Fundamental Research Funds for the Central Universities (Program No.
ZXH2011D005), and the NSFC Tianyuan Youth Foundation of Mathematics of China (No. 11126136).
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MATLAB for Systems Analysis
We know how to model systems, and MATLAB. Let’s put these together. We will consider the use of MATLAB for obtaining Response in time domain Response in frequency domain For the former you will be given help to allow you to write code to Find the theoretical response to a step input Find the simulated response to a step input For the latter you will be given help to allow you to develop code to Plot Nyquist and Bode diagrams of the loop gain (strictly -loop gain) Plot Frequency response of whole, or closed loop system The assignment associated with this work will involve various systems. These systems will be represented using transfer functions, which are in turn represented using polynomials.
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.1
Example System
I controller plant 1 1+s15 9 (1+s15)(1+s8) 1+s10 O
(Minus) Loop transfer function is 1 s15 1 9 9 9 * 1 s10 (1 s15)(1 s8) (1 s10)(1 s8) 80s 2 18s 1 Closed Loop transfer function 9 9 9 80s 2 18s 1 9 80s 2 18s 1 9 80s 2 18s 10 1 80s 2 18s 1
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.2
In MATLAB, represent transfer function with polynomials: c = tf (9*[15 1], [10 1]); p = tf (1, conv([15 1],[8 1]) ); For some frequency response need loop transfer function: series (c,p); % multiplies two transfer functions in series This does not do cancelling, so simplify by: minreal (series(c,p)); For closed loop analysis, want closed loop transfer function: minreal (feedback (series (c,p), 1) ); % feedback (t1,t2) is t1/(1+t1*t2). MATLAB code to generate transfer functions of other components Trans Func
3s 2 2s 1 MATLAB tf([5 0],1); tf(2, [3 2 1]); tf([4 5], [1 0]); tf((1,4,5),[1 0]) Note, for those without the control toolbox, I have provide versions of the above MATLAB functions: these are on a separate sheet.
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.3
5 4 s
1 4 5s s
Theoretical Response
The Laplace transform lectures mean you should be able to find the K theoretical step response for a system with transfer function . 2 bs c as The following is part of a MATLAB function to return such a step response function ans = sotheory(k,a,b,c,t); % returns theoretical response of system with tf k/(as^2+bs+c) % at all the values of time in the vector t. if (a == 0) & (b ~= 0) & (c ~= 0) ans = (k/c)* (1 - exp(-t*c/b)); elseif (a ~= 0) & (b == 0) & (c ~= 0) w = sqrt(c/a); ans = k*(1-cos(w*t))/c; end NB to evaluate exp(dt)sin(wt) do exp(d*t).*sin(w*t) Note the '.*' operator
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.4
Step Response of Such a System
The command step returns the simulated step response and plots it. It needs the numerator and denominator polynomials, with optional time. Can call with output arguments, to get data but not plot it. The following code sets values for t, uses sotheory to get the theoretical response, plots that response, and then calls step to add the simulated response. It assumes values for variables k, a, b and c exist. t = [0:0.1:50]; % set range of values of t plot (t, sotheory(k, a, b, c, t) ); % calc and plot theoretical resp hold on; % hold data for the graph step (k,[a b c],t, ‘*’); % calc and plot step response Alternative method [op, dummy, t] = step(k, [a b c]); % use step to calc values and t plot (t, op, t, sotheory (k, a, b, c, t), ‘*’); % plot step & calc & plot theory % note dummy variable (it is the 'state variable', which we can ignore here) % note also, plot one graph in * so can see agreement of two methods
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.5
Doing your own simulation
O 1 sTe As an intro, consider simulating system with transfer function I 1 sTa x 1 O Could do by: and 1 sTe Here x is ‘state’ variable. I 1 sTa x Conceivably do by:
s I 1 Ta 1 X s
X(n) - X(n - 1) But, this involves differentiating, done by t n - t n -1 X values have errors, and tn-tn-1 is small, so errors amplified. So avoid.
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.6
Te I 1 Ta 1 X s
Uses differential of X we have already got. Thus, algorithm to find O at a given time is: Sx = (I – X)/Ta; % find input to integrator X = Integrate (Sx); % integrate this to get X O = X + Sx * Te; % compute O Can extend concept to system with transfer function O z n s n z n -1s n -1 .. z1s z 0 Note, pn <> 0. I p s n p s n -1 .. p s p n n -1 1 0
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.7
Zn Zn-1 Z1 Z0
1 Pn
1 s
1 s
1 s
Po P1 Pn-2 Pn-1
2/CY/I6 Modelling project
MATLAB System Analysis
RJM p 5.8
We again define state variable x, saying x 1 O and z n s n z n -1s n -1 .. z1s z 0 I p s n p s n -1 .. p s p x
n n -1
From x/I we get
And O/x gives Note, we find nth differential of x and by integrating get n-1th differential, etc. To find O and x we need these differentials, but we already have them. Note can find values of O, x and its diffs at any time with vectors: num = numerator poly, den = den poly, and x is [s nx sn-1x …sx x] % if a=[a1 a2 a3] and b=[b1 b2 b3], then dot(a,b) = a1*b1 + a2*b2 + a3*b3 x(1) = (Input - dot (den(2:n), x(2:n)) / den(1); % find nth diff of x Integrate x(2:n) % integrate to find n-1th, etc. Output = dot(x,num) % compute output
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.9
1 0 n x 1 I - p s n -1x .. p sx p x s n -1 1 0 pn O z n s n x z n -1s n -1x .. z1sx z 0 x
Problem - how to integrate
Recall integration is area under a curve
Previous Input 1 s Current Input
Output Previous Time Current Time
Assume we know integral up to previous time. Then Euler's (poor) rectangular method is Area = Area + Area of Rectangle due to current input = Area + (CurrentInput) * (CurrentTime - Previous Time) Note Runge-Kutta is better integration method. Hence we have the following MATLAB function to find the step response.
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.10
function resp = RJMStep (num, den, t) % Euler Simulation of system with transfer function num/den, at times in t % store output response in resp. Note den(1) must not be zero. % Dr Richard Mitchell, 19/9/00 x = zeros (size(den)); % x is array of state variables: set to 0 n = length(den); % find length of den array while length(num) < n % add leading zeros to num, so same length as den num = [0 num]; end; resp = [0]; % initialise response array to 0 for ct = 2:length(t), % for all time x(1) = (1 - dot(den(2:n),x(2:n)))/den(1); % calc input to first integrator x(2:n) = x(2:n) + (t(ct)-t(ct-1))*x(1:n-1); % integrate all state vars resp = [resp dot(num,x)]; % add output to resp array end; Note to be reasonably accurate, t(ct) - t(ct-1) must be sufficiently small.
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.11
5 s 2 2s 5
1.4 1.2 1 0.8 0.6 0.4 0.2 0
if t = [0:0.1:10], t=[0:0.02:10] and SOTheory
Not bad if Tstep = 0.02, but more calcs. Need better integration method
2/CY/I6 Modelling project MATLAB System Analysis RJM p 5.12
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Who is Emmy Noether?
My graduate education in mathematics at Michigan State University (more long ago than I care to remember) must somehow have been more broadminded than I thought at the time, since I can recall being
made aware of the mathematician Amalie (Emmy) Noether and studying her work. Natalie Angier’s column in the New York Times this week introduces her readers to Emmy Noether with the observation that
while “Scientists are a famously anonymous lot, but few can match in the depths of her perverse and unmerited obscurity the 20th-century mathematical genius Amalie Noether.”
Albert Einstein called her the most “significant” and “creative” female mathematician of all time, and others of her contemporaries were inclined to drop the modification by sex. She invented a
theorem that united with magisterial concision two conceptual pillars of physics: symmetry in nature and the universal laws of conservation. Some consider Noether’s theorem, as it is now called,
as important as Einstein’s theory of relativity. (New York Times)
Perhaps it was in Lee Sonneborn’s class in Abstract Algebra, studying rings and groups, the subject of some of Noether’s work that I first heard of Emmy Noether. Angier notes that Noether was born in
Erlangen, Germany, 130 years ago this month, and Angier seeks to correct the mistreatment science and mathematics have given to Noether by having seemingly forgotten her. What I want to know is why
does it seem thus so often, why does it seem that we either forget about the contributions of women mathematicians and physicists; or if, as in the case of Emmy Noether, when the contributions are so
important we can’t forget them, we apparently forget who made them.
Noether was a highly prolific mathematician, publishing groundbreaking papers, sometimes under a man’s name, in rarefied fields of abstract algebra and ring theory. And when she applied her
equations to the universe around her, she discovered some of its basic rules, like how time and energy are related, and why it is, as the physicist Lee Smolin of the Perimeter Institute put it,
“that riding a bicycle is safe.”
Ransom Stephens, a physicist and novelist who has lectured widely on Noether, said, “You can make a strong case that her theorem is the backbone on which all of modern physics is built.”
What is Noether’s Theorem? Angier describes it especially well.
What the revolutionary theorem says, in cartoon essence, is the following: Wherever you find some sort of symmetry in nature, some predictability or homogeneity of parts, you’ll find lurking in
the background a corresponding conservation — of momentum, electric charge, energy or the like. If a bicycle wheel is radially symmetric, if you can spin it on its axis and it still looks the
same in all directions, well, then, that symmetric translation must yield a corresponding conservation. By applying the principles and calculations embodied in Noether’s theorem, you’ll see that
it is angular momentum, the Newtonian impulse that keeps bicyclists upright and on the move.
Some of the relationships to pop out of the theorem are startling, the most profound one linking time and energy. Noether’s theorem shows that a symmetry of time — like the fact that whether you
throw a ball in the air tomorrow or make the same toss next week will have no effect on the ball’s trajectory — is directly related to the conservation of energy, our old homily that energy can
be neither created nor destroyed but merely changes form.
Noether’s Wikipedia entry is appropriately lengthy and adds to the list of mathematicians and physicists who considered her the most important female scientist of all time. Wikipedia discusses all of
Noether’s work, throughout her life, in consderable detail and is very well worth reading.
I particularly noted this passage about Noether’s influence on the revolution in mathematics that occurred in the 19th and the 20th Centuries.
In the century from 1832 to Noether’s death in 1935, the field of mathematics—specifically algebra—underwent a profound revolution, whose reverberations are still being felt. Mathematicians of
previous centuries had worked on practical methods for solving specific types of equations… [Beginning with work by Gauss, Gal,ois, and others in 1830...] research turned to determining the
properties of ever-more-abstract systems defined by ever-more-universal rules. Noether’s most important contributions to mathematics were to the development of this new field, abstract algebra. (
Perhaps this is why I think it must have been in Professor Sonneborn’s Abstract Algebra that I first learned of Emmy Noether.
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Independent and Common Pathways Models Example Scripts for OpenMx in R
Tue, 05/10/2011 - 09:22
Common and independent
Common and independent pathway models are the joining of factor models and ACE-type models, both of which are covered in the documentation. Let's say that you're interested in a common pathway for a
single factor solution on some set of twins. Your factor model would look very much like the two-factor model example in the documentation, with the factor for twin 1 (x in the example) constrained
to be equal to the factor for twin 2 (y in the example) through equality constraints on loadings and residual variances. The covariance between the factors is then given by the ACE model in the
genetic epidemiology example.
Look at this thread about ACE models for latent variables for more information: http://openmx.psyc.virginia.edu/thread/789
Sun, 05/22/2011 - 21:40
example scripts
Thanks for the thoughts. I am quite a novice at this so I was hoping for something more practical, as I saw that there are example scripts for classic mx so I thought maybe there was a place for
example scripts for openmx. I'll see what I can make of the references you provided but if anyone has an example of these models in openmx I'd greatly appreciate it.
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The Problem
Symbol manipulation doesn’t receive near the emphasis it did 20 years ago in the high school curriculum. Proofs too are becoming increasingly rare in the high schools.
A professional mathematician may think of himself as an inspired artist (and he might be), but proofs and tedious symbol manipulation are still 90% of the game.
More Problems
Career-oriented, scientifically-inclined students can major in any number of different fields and don’t have to “explain” their abilities to a recruiter.
A math major must be able to answer the question, “What can you do for me?” A math major must bear a relatively huge burden of self-marketing.
B.S. Mathematics
Pre-calculus (admission price)
Calculus I, II, III
Foundations (Proofs)
Linear Algebra
Two of DE, Number Theory, Geometry, Complex Variables, Modeling
One year of Modern Algebra, Analysis, or Statistics plus one semester of one of the others
Nine additional hours of upper division
Total = 41 hours
B.A. Philosophy
Phil 110-120 (admission price)
Phil 160
Phil 210
Phil 314, 315
18 additional hours in upper division
Total = 30 hours
B.A. History
History 121-122, History 201-202 (admission price)
History 301
History 499
21 additional hours in upper division
Minimum score on exit exam
Total = 24 hours
What should we do?
Neighboring B.A.’s
Ole Miss (30 hrs math)
Calc Sequence + Proofs + Linear Algebra + 12 hrs + Programming
University of Arkansas (34 hrs math)
Calc Sequence + Discrete Math + Linear Algebra + 15 hrs + Paper
Miami University (34 hrs math)
Calc Sequence + Linear Algebra + 19 hrs + Programming + 12 “related hours”
Some Numbers
Miami U has 306 “math” majors, 15000 undergrads
153 BSEd, 28 BSM, 2 BSS, 58 BSM/S, 65 BA.
At least 1/3 of BA students are double majors
U Arkansas has 141 math majors, 12800 undergrads.
95 BS, 46 BA
Traditional (33 hrs math)
Calc Sequence (11 hrs) + Math 241 + Math 310 + 16 hrs
The Arts Model
There is a clear difference between an artist and an art historian, between a musician and a music historian.
Colleges have long recognized the intellectual value of both fields by credentialing majors in both areas of study.
Speaking of Arts
Algebra + Trig? [Math 140, 141?] (admission price)
Statistics [Math 210]
One year of Calculus for Humanities [Math 220-221?]
Proofs [Math 241]
Linear Algebra [Math 310]
18 additional hours of upper division courses
require double major?
minimum score on exit exam?
Total = 33 hours
18 additional hours ?!?
Math 350, 410, 420
None of these courses requires the traditional calculus ramp
Math 340, 365, 455, 461, 462, 465
These courses claim to need Math 252
Math 451, 491, 492
These courses claim to need Math 320
We could aim for breadth: Maple + Mathematica + Derive + Excel
We could specialize in just one package, and use the course as a lens through which to view a single mathematical topic, say, differential equations.
Other Possibilities
Mathematical Literature (Gardner, Stewart, Hofstadter, etc)
Philosophy of Mathematics
Symbolic Logic (Phil 360)
Math History II
Mathematics of Art
Math “Recital” or Paper
Exit exam prep
Yet other possibilities
Total New Courses
Math 141 Trig?
Math 221 Calculus for Humanities II
Math 380 Mathematical Software
Two Math Humanities Upper Division
Baccalaureate Standards
The mathematical sciences bachelor's degree program should be consistent with the current recommendations of the MAA Committee on the Undergraduate Program in Mathematics (CUPM) Guidelines. …
Programs with no curricular track that conforms to the CUPM guidelines should be justified by a detailed and persuasive rationale for departing from those guidelines. A summary of the CUPM Report
comprises Appendix B of this document.
Guidelines for Programs and Departments in Undergraduate Mathematical Sciences
Appendix B
Calculus (with Differential Equations)
Linear Algebra
Probability and Statistics
Proof-based courses
An in-depth experience in mathematics
Applications and connections
Track courses, departmental requirements and electives
Catalog Descriptions
B.A. Option 1 – Traditional (33 hrs)
This is an excellent option for students who enjoy mathematics for the “purity” of the discipline and are not terribly interested in applications. Also an excellent choice for students wishing to
pursue a double major in a discipline of the humanities. Students hoping to pursue grad study or employment in mathematics will be better served by a B.S. degree.
B.A. Option 2 – Nontraditional (33 hrs)
This is an excellent option for students who are primarily interested in the study of mathematics as a cultural artifact. Though students will pick up some mathematical skills as a matter of
course, students taking this degree should do so primarily to become conversant in the philosophical/historical issues of the discipline. This is an excellent choice for students hoping to double
major in a discipline of the humanities. Students hoping to pursue grad study or employment in mathematics will be better served by a B.S. degree.
Q&A 1
Where in the world are these new courses going to come from?
If we are serious about the answer to this question being “release time”, we must also be serious about going after external grants to fund the development of any new courses.
Q&A 2
Could B.S. majors use any of these new courses to satisfy their nine additional hours of upper division courses?
This would be one sure way of guaranteeing that our exit exam scores fall closer to the 50th percentile.
Q&A 3
Would B.A. folks leave with any career competencies?
Who knows. An exit exam minimum might be some insurance. So would requiring a double major. In any case, UTM offering a B.A. in Mathematics would be no more irresponsible than UTM offering a B.A.
in Philosophy.
Q&A 4
Wouldn’t it be misleading to graduate people with a B.A. in Mathematics when they would in fact have little in the way of true competency in the field?
As regards option 1, we have seen three schools not overcome by such moral qualms. More of a worry for option 2 grads, I would argue this concern is not realistic: the kind of students attracted to
this option are unlikely to be seeking future employment in mathematics.
Q&A 5
What makes you think THEC is going to let us add a new B.A., when we haven’t been able to award the old B.A.?
I’m not proposing a new degree. I’m proposing we renovate the old one “down to the girders”. If we are troubled by the prospect that we are falsely credentialing our graduates in mathematics, we
could apply to rename the degree in the years following the renovation (e.g. B.A. in Mathematical Arts).
Q&A 6
Who would major?
The Miami data suggests that quite a few might take B.A. Option 1. Praxis refugees certainly. At least four of these refugees took BUS degrees with 32 hours of math to their credit. They would be
just a capstone paper away from a B.A. in math.
Q&A 7
But who would take that silly B.A. Option 2?
If the humanities calc ramp is sufficient to do upper level statistics, a few “statistics majors” might come from there. But who knows? Maybe no one. If we’re serious though about finding new majors
we need to fish in fresh ponds. The sciences pond appears to have all but dried up nationwide. The humanities pond is one the mathematical community has not yet seriously tried to farm.
Speaking of fishing
Q&A 8
Wouldn’t this destroy our B.S. in Mathematics?
The Miami data suggests that offering a more easily obtained B.A. is not a zero sum game. Far from stealing from the pool of B.S. majors, a more easily obtained B.A. appears to add to the total pool
of majors. There may even be some unexpected synergies between the nontraditional B.A. and the traditional B.S.; this could add a fresh dynamic to the department.
Luckiest Man Alive
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: compan
compan program
Documentation for the compan program is below, with links to related programs in the "see also" section.
{version = 3.25; (* of compan.p 2001 Mar 26}
(* begin module describe.compan *)
compan: composition analysis.
compan(cmp: in, anal: out, companp: in, output: out)
cmp: the input composition, which is the output of program comp;
anal: the output analysis of this program;
companp: for parameters; should contain a single integer which specifies
the maximum level for which the composition is analyzed. the
maximum allowed level is 4, or the maximum level for which the
composition was determined.
output: for user messages;
calculates chi squared from a composition using:
1) assumption of equal frequencies to calculate mono, di, tri
and tet expecteds;
2) mono frequencies to calculate di, tri and tet expecteds;
3) di frequencies to calculate tri and tet expecteds;
4) tri frequencies to calculate tet expecteds;
the partial chi squared values are given for each oligo.
the 'information' content of the composition is also calculated,
using the standard information theory definition:
information = -sum(frequency * log(frequency)),
where the sum is over each oligonucleotide of a given length
and the log is taken to the base 2. this gives the information
in bits.
see also
gary stormo
the program cannot do calculations for compositions larger than 4
(* end module describe.compan *)
{This manual page was created by makman 1.44}
{created by htmlink 1.55}
Viewing Files Accessibility
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Copyright © University of Cambridge. All rights reserved.
Well done to Daniel from Kings School, New Zealand, Josh from Chatham Grammar School for Boys, and Anna from the Sandon School, who all correctly worked out the dimensions of the running track. Here
is Josh's solution:
Radius of the curved sections:
As the straight sections of the track are both 85m long, 170m of the track is straight. That means that the remaining 230m of the track is comprised of the two semicircular sections. As there are two
semicircular sections, each of equal radii, the sum of their inside edges (of the inside lanes) is equal to the circumference of a circle that would fit neatly into the inside curves of the track.
The circumference of a circle, $c$, is given by the equation: $c = 2 \times \pi \times r$, where $r$ is the radius.
Therefore, as we have deduced that the circumference of this imaginary circle is equal to the sum of the inside curves of the track, we can write: $230 = 2 \times \pi \times r$, so $r = \frac{230}{2\
pi} = 36.606$ metres (to 3 d.p.)
This measurement is only the measurement of the radii of the semicircles which comprise the inside edge of the inside lane on the curved section. We want to find a measurement for the radii of the
semicircles which comprise the whole of both curved sections; there are 8 lanes that we must consider in order to do this. As each lane is a constant 1.25 metres, the sum of the thickness of the
lanes is equal to 8 x 1.25 = 10 metres. Therefore, the radius of the curved sections, in order to produce a scale drawing, is equal to: 36.606 + 10 = 46.606 metres (to 3 d.p.)
The 200m Staggered Start:
As it is mentioned in the problem for the 400m start that the measurement for the length of a given lane is that of its inside edge, I will assume that it is the inside edges of each lane that must
be made constant in the 200m race in order to make the race fair. The runner in lane 1 starts at the curved section on the bottom right of the track, so I will use that lane for comparison with the
others. The straights do not need to be compensated for in the staggered start, as the runners would all run the same distance here. Therefore, we only need to consider the 115m of track that the
runners will run on before they hit the straight; in this time, all runners will be running on the curved section of track.
The runner in lane 2 is running on a track with an inside edge that is 1.25m further outwards than that of lane 1. Therefore, the radius of runner 2's track is equal to $36.606 + 1.25 = 37.856$
metres. Thus, the circumference of the semicircle that makes up the curved section is equal to: $\frac{2 \pi r}{2} = \pi r = 118.927$ metres This is 3.927 metres longer than the inside edge of lane
1, so the rune in lane 2 will start 3.927 metres in front of the runner in lane 1.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $(36.606 + 1.25(n-1)) \times \pi$ I arrived at this conclusion because the formula for the arc of a
semicircle is pi x r, hence the reason for multiplying the sequence by pi, and the section in brackets is the way to determine the radius of the inside edge of any given track. Each runner, as we
move from lane 1 to lane 8, will start 3.927 metres (3 d.p.) in front of the previous runner.
The 400m Staggered Start:
The obvious conclusion for this problem would be to say that each runner starts twice as far behind the runner in the next inside lane as they did in the 200m race, but I will investigate this
Each runner is running 400 metres, but 170 metres of the track they will run (85 x 2) is made up of straight sections, where the runners will run the same distance regardless of the lane they are in,
so this does not need to be compensated for. Thus, 230m of the track they will run on is comprised of a curved section, which must be compensated for. When the runner in lane 2 reaches a curved
section, the radius of the inner edge of his/her lane is equal to $36.606 + 1.25 = 37.856$ metres. Thus, the arc lengths of both semicircles together is equal to: $2 \times \pi \times 37.856 =
237.854$ metres (3 d.p.) This is 7.854 metres longer than the length of curved track in lane 1, so the runner in lane 2 will start 7.854 metres in front of the runner in lane 2.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $2 \times ((36.606 + 1.25(n-1)) \times pi)$. I arrived at this conclusion using the nth term formula I
generated before to calculate the arc length of one of the semicircular pieces of track, then multiplied it by 2 (for the two semicircles involved in the 400m. As we move from lane 1 to lane 8, each
runner starts 7.854 metres in front of the runner on the next inside lane to them.
Rajeev, from Haberdashers' Aske's Boys' School, sent us his calculations for the staggers presented clearly in a table. You can see it here.
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matrix, in mathematics, a rectangular array of elements (e.g., numbers) considered as a single entity. A matrix is distinguished by the number of rows and columns it contains. The matrix;e15;none;1;
e15;;;block;;;;no;1;139392n;140158n;;;;;eq15;comptd;;center;stack;;;;;CE5is a 2×3 (read "2 by 3") matrix, because it contains 2 rows and 3 columns. A matrix having the same number of rows as columns
is called a square matrix. The matrix;e16;none;1;e16;;;block;;;;no;1;139392n;164586n;;;;;eq16;comptd;;center;stack;;;;;CE5is a 2×2 matrix, or square matrix of order 2; a square matrix of order n
contains n rows and n columns. Definitions are made for certain operations with matrices; for example, a matrix may be multiplied by a number, and two matrices of the same order may be added or
multiplied using an algebra of matrices that has been developed. Matrices find application in such fields as vector analysis and the solution of systems of linear equations by means of electronic
See R. C. Dorfi, Matrix Algebra (1969).
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
More on matrix from Fact Monster:
See more Encyclopedia articles on: Mathematics
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MathGroup Archive: October 2005 [00760]
[Date Index] [Thread Index] [Author Index]
Re: Importing from Adobe Illustrator
• To: mathgroup at smc.vnet.net
• Subject: [mg61619] Re: [mg61615] Importing from Adobe Illustrator
• From: "David Park" <djmp at earthlink.net>
• Date: Mon, 24 Oct 2005 01:43:59 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
It certainly is easier to work entirely within Mathematica if you can. In
this case I think you can.
The idea is to use the Rectangle graphic primitive with a third argument.
The third argument is your basic plot. You put the Rectangle on a larger
piece of paper and then use the Arrow and Text routines to add the callouts
and to add other text you may want. Here is a somewhat crude example.
{Rectangle[{0, 0}, {1, 1}, Plot[Sin[x], {x, 0, 2Pi},
DisplayFunction -> Identity,
AspectRatio -> 1,
Axes -> {True, False},
Ticks -> None]],
Arrow[ {0.5, 1.20}, {0.28, 1.0}, HeadLength -> 0.03,
HeadCenter -> 0.5],
Text["peak", {0.5, 1.22}, {-1.3, 0}],
Arrow[ {0.8, -0.15}, {0.75, -0.0}, HeadLength -> 0.03,
HeadCenter -> 0.5],
Text["trough", {0.8, -0.17}, {-1, 0}],
Arrow[ {0.6, 0.75}, {0.5, 0.5}, HeadLength -> 0.03,
HeadCenter -> 0.5],
Text["null point", {0.6, 0.78}, {-1.2, 0}],
Text[StyleForm["Wave Terms", FontSize -> 20], {0.5, 1.4}]}],
AspectRatio -> Automatic,
PlotRange -> {{-0.1, 1.1}, {-0.25, 1.5}},
Frame -> False,
Background -> Linen,
ImageSize -> 500];
If you click on the plot you will see the Rectangle on the larger plot. You
can draw arrows from outside the Rectangle to inside the Rectangle. You will
usually want to control the AspectRatio of the inner plot with respect to
the Rectangle. If you have axes, a frame and tick labeling this can change
the relative location of the items in the inner plot to the overall plot. So
you will usually want to get that settled before you start drawing callout
You can click off coordinates from the plot, but if you are within the
Rectangle the coordinates are different than if you are outside the
Rectangle. Therefore I added a Frame to the overall plot to estimate
coordinates and turned it off when I was finished. You can use 'outside'
coordinates to draw arrows to 'inside' locations.
David Park
djmp at earthlink.net
From: Ben Kovitz [mailto:bkovitz at acm.org]
To: mathgroup at smc.vnet.net
BTW, what I *really* want to do is take some Mathematica output, draw
little "callout" arrows and remarks around it, and put the result back
into a Mathematica notebook. Is there some much superior way to do
this, that bypasses exporting and importing?
Ben Kovitz
Humboldt State University
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Kelvin function of the first kind: Series representations (subsection 06/01)
Bessel-Type Functions KelvinBei[z] Series representations
Generalized power series
Expansions at generic point z==z[0]
Expansions at z==0
For the function itself
For small integer powers of the function
© 1998-2014 Wolfram Research, Inc.
Bessel-Type Functions KelvinBei[z] Series representations
Generalized power series
Expansions at generic point z==z[0]
Expansions at z==0
For the function itself
For small integer powers of the function
Generalized power series
Expansions at generic point z==z[0]
Expansions at z==0
For the function itself
For small integer powers of the function
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William P. Thurston 1946 - 2012
William P. Thurston died August 21 at the age of 65. After many groundbreaking results in foliation theory, Thurston became a pioneer in the field of low-dimensional topology. Thurston's geometric
ideas developed initially in connection with three-dimensional topology have been extremely influential in the last few decades, both in topology and geometry, and in other fields such as group
theory and complexity. His geometrization conjecture for 3-manifolds, implying the Poincaré conjecture, was proved by Perelman in 2003. Thurston is
also one of the fathers of geometric group theory. His honors include the Fields Medal (1982), the AMS Book Prize (2005), and the AMS Steele Prize for a Seminal Contribution to Research (2012).
Information: http://www.ams.org/news?news_id=1602
August 23, 2012 - 14:36 — ems_site
A tribute to Bill Thurston posted by Terence Tao
Post new comment
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Physics homework hanging stone
1. The problem statement, all variables and given/known data
A stone hangs by a fine thread from the ceiling and a section of the same thread dangles from the bottom of the stone.
2. Relevant equations
a. If a person gives a sharp pull on the dangling thread, where is the thread likely to break: below the stone or above it?
b. What if the person gives a slow and steady pull?
Explain your answer.
3. The attempt at a solution
I think that the answer for a is below while for b the answer is above, but I don't know exactly how. Can someone please explain this in detail. Thanks
have a look at the tensions in the string, if the stone has a mass of 10kg, and a second 2 kg stone was attached to the bottom of the lower string.
What happens to those tensions is the 2 kg stone is replaced by a 4kg stone? 6kg? 8kg? 12kg? 14kg? ......
That should give you a clue for part (b)
Answer me that and I will give you a hint about (a)
EDIT: for speed of calculation, and because each stone could have been a little heavier any way, use g=10 m/s^2
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How many solution sets do systems of linear inequalities have?
It can have none, one, or infinitely many. Here's a system with no solutions: x 5. Here's a system with one solution: x =3. [ ( = mean "less than or equal" and "greater than or equal") And here's a
system with infinitely many solutions: x < 3 and x < 2. ]
Not a good answer? Get an answer now. (FREE)
There are no new answers.
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C++ Programming/Code/Standard C Library/Functions/pow
Syntax #include <cmath>
double pow( double base, double exp );
The pow() function returns base raised to the expth power. There's a domain error if base is zero and exp is less than or equal to zero. There's also a domain error if base is negative and exp is not
an integer. There's a range error if an overflow occurs.
Related topics
exp - log - sqrt
Last modified on 14 June 2009, at 16:48
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High Energy Particle Physics
Previous months:
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2009 - 0904(1) - 0907(10) - 0908(5) - 0909(4) - 0910(7) - 0911(10) - 0912(4)
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2014 - 1401(14) - 1402(20) - 1403(10) - 1404(17)
Recent Submissions
Any replacements are listed further down
[485] viXra:1404.0424 [pdf] submitted on 2014-04-18 12:45:37
Complete Relativity Predicts the Recently Reported Mass of the Higgs Boson
Authors: Ramzi Suleiman
Comments: 8 Pages.
In two recent articles I have shown that a relativity theory without the Lorentz Invariance Principle, termed Complete Relativity, conforms to quantum mechanics and cosmology. Here I demonstrate that
it also conforms to the Standard Model. Using the energy expression derived from the theory, I demonstrate that the theory predicts the recently reported mass (≈125 GeV) of the Higgs boson.
Category: High Energy Particle Physics
[484] viXra:1404.0407 [pdf] submitted on 2014-04-17 08:30:46
The Resonant Nature of the Negative Z with a Mass of 4430 MeV
Authors: Sylwester Kornowski
Comments: 2 Pages.
Here, within the lacking part of ultimate theory, i.e. the Everlasting Theory, I calculated mass (4423 MeV), quantum numbers (unitary spin and positive parity) and estimated the full width (about 100
MeV; such mesons have the resonant character) of the negative resonance Z(4430). It is not a new form of matter called a tetraquark.
Category: High Energy Particle Physics
[483] viXra:1404.0369 [pdf] submitted on 2014-04-17 01:31:59
Neutrino Oscillations
Authors: Lubomir Vlcek
Comments: 5 Pages. The article is ready for publication.
In the present paper we show, that leptons ( electron, muon, tau ), W + - Z bosons and neutrinos ( electron neutrino , muon neutrino, tau neutrino) can be replaced with electron moving at different
speeds from 0.1c up to 0.999.. c . Keywords: mass, kinetic energy, electron . leptons, neutrino. PACS number: 14.20., 14.40., 14.60. 14.65. 14.70.
Category: High Energy Particle Physics
[482] viXra:1404.0279 [pdf] submitted on 2014-04-16 03:58:50
Physics is Easy
Authors: Lubomir Vlcek
Comments: 13 Pages. The article is ready for publication
In the present paper we show, that leptons ( electron, muon, tau ), W + - Z bosons and neutrinos ( electron neutrino , muon neutrino, tau neutrino) can be replaced with electron moving at different
speeds from 0.1c up to 0.999.. c . Similarly hyperons, mesons and quarks can be replaced by proton and neutron (or alpha particle respectively ) moving at different speeds from 0.1c up to 0.999.. c .
While, the neutron is composed of proton and electron orbiting around it. Thus, all particles, which are currently known, can be replaced by the various fast moving electron or proton (D,He 3 or
alpha particle respectively ).Electron and proton are the only stable fundamental elementary particles. Higgs Boson. Keywords: mass, kinetic energy, potential energy. leptons. hyperons, mesons,
quarks PACS number: 14.20., 14.40., 14.60. 14.65. 14.70.
Category: High Energy Particle Physics
[481] viXra:1404.0273 [pdf] submitted on 2014-04-16 04:09:57
Particles, Waves and Trends in Physics
Authors: Lubomir Vlcek
Comments: 35 Pages. The article is ready for publication
Speeds of electrons and protons in atoms are small. For example: An electron moving at a speed ve= 0,003c creates spectral line Hα. Accurate electron speeds are given in the table in this article.
Confirmation of Doppler´s principle in hydrogen for Balmer line Hα. Comparison the official and of the author´s results. Theory of particles, waves and heat. Accompanying activity of reaction on
movement of stable particles in the transmission medium are waves. Neutron, β electron , gamma rays – calculations. Discussion to Cobalt-60 decay. Stable electrons moving with speeds (0,99 c – c )
creates leptons (μ−, τ−), neutrinos (νe, νμ, ντ) and bosons W +, W-, Z (= β electrons). Weak interactions are caused with stable electrons, which creates leptons (μ−, τ−) = ( particles = electrons
different speeds), neutrinos νe, νμ, ντ (= waves) , bosons W +, W-, Z (= particles = β electrons moving at nearly the speed of light ) and gamma rays (=waves of extremely high frequency >1019 Hz ).
Stable particles (p +, n0, D, He-3, α) moving with speeds ( 0,3 c – 0,99 c ) creates baryons and mesons. The strong interactions are caused with stable particles (p +, n0, D, He-3, α ), which creates
baryons and mesons. Therefore creation and annihilation operators in physics are irrelevant.
Category: High Energy Particle Physics
[480] viXra:1404.0268 [pdf] submitted on 2014-04-16 04:16:15
Physics is Beautiful
Authors: Lubomir Vlcek
Comments: 25 Pages. The article is ready for publication
In the paper „ Physics is easy“ we showed, that leptons ( electron, muon, tau ), W + - Z bosons and neutrinos ( electron neutrino, muon neutrino, tau neutrino) can be replaced with electron moving at
different speeds from 0.1c up to 0.999.. c . Similarly hyperons, mesons and quarks can be replaced by proton and neutron (or alpha particle respectively ) moving at different speeds from 0.1c up to
0.999.. c . While, the neutron is composed of proton and electron orbiting around it. Thus, all particles, which are currently known, can be replaced by the various fast moving electron or proton.
Electron and proton are the only stable fundamental elementary particles. We show, that neutron is source β rays - β electrons ( bosons Zo, W+- too) , γ rays, electron neutrinos, muon neutrinos,
tauon neutrinos. Keywords: mass, kinetic energy, β rays, bosons Zo, W+- , γ rays, wave–particle duality. PACS number: 14.20., 14.40., 14.60., 14.65., 14.70.
Category: High Energy Particle Physics
[479] viXra:1404.0261 [pdf] submitted on 2014-04-16 04:23:08
Introduction to my Two Articles Physics is Easy and Physics is Beautiful
Authors: Lubomir Vlcek
Comments: 29 Pages. The article is ready for publication
. In the present paper we show, that accurate measurement kinetical energy in high-energy physics will use to determine the exact value of the speed of particles. The exact value of the speed of
particles allows us to establish the precise momentum of particles. Keywords: mass, kinetic energy, potential energy. neutron, proton, electron, leptons, quarks PACS number: 14.20., 14.40., 14.60.
14.65. 14.70.
Category: High Energy Particle Physics
[478] viXra:1404.0246 [pdf] submitted on 2014-04-16 04:47:42
Shortened Great Table of Elementary Particles
Authors: Lubomir Vlcek
Comments: 23 Pages. The article is ready for publication
Stable particles (p +, n0, D, He-3, α) moving with speeds ( 0,3 c – 0,99 c ) creates baryons and mesons. Stable electrons moving with speeds ( 0,99 c – c ) creates leptons (μ−, τ−), neutrinos (νe,
νμ, ντ) and bosons W +, W-, Z. Speeds of electrons and protons in atoms are smaller. For example: An electron moving at a speed ve= 0,003c creates spectral line Hα. Weak interactions are caused with
stable electrons, which creates leptons, neutrinos and bosons W +, W-, Z. The strong interactions are caused with stable particles (p +, n0, D, He-3, α ), which creates baryons and mesons. Therefore
creation and annihilation operators in physics are irrelevant.
Category: High Energy Particle Physics
[477] viXra:1404.0243 [pdf] submitted on 2014-04-16 04:53:01
Great Table of Elementary Particles
Authors: Lubomir Vlcek
Comments: 72 Pages. The article is ready for publication
Stable particles (p +, n0, D, He-3, α) moving with speeds ( 0,3 c – 0,99 c ) creates baryons and mesons. Stable electrons moving with speeds ( 0,99 c – c ) creates leptons (μ−, τ−), neutrinos (νe,
νμ, ντ) and bosons W +, W-, Z. Speeds of electrons and protons in atoms are smaller. For example: An electron moving at a speed ve= 0,003c creates spectral line Hα. Weak interactions are caused with
stable electrons, which creates leptons, neutrinos and bosons W +, W-, Z. The strong interactions are caused with stable particles (p +, n0, D, He-3, α ), which creates baryons and mesons. Therefore
creation and annihilation operators in physics are irrelevant.
Category: High Energy Particle Physics
[476] viXra:1404.0238 [pdf] submitted on 2014-04-16 05:03:33
Movement Principles of the Fast-Spinning Bodies
Authors: Lubomir Vlcek
Comments: 40 Pages. The article is ready for publication
In this article are the 5 tables for calculating the average translational velocities of all milisecond pulsars and others fast spinning bodies and shift the center of gravity due to the rotation.
Corrections theory ideal spining circle for real bodies. Extreme values for neutron stars. Summary values for neutron stars. A new perspective on neutronization and nuclear fusion. Consent theory
with the real Universe. Keywords: stars: kinematics, pulsars: general, stellar dynamics, , stars: neutron, PACS number: 97.60.Gb, 97.60.Jd,
Category: High Energy Particle Physics
[475] viXra:1404.0130 [pdf] submitted on 2014-04-16 03:43:54
Nuclear Fusion
Authors: Lubomir Vlcek
Comments: 7 Pages. The article is ready for publication
In the present paper we show, that leptons ( electron, muon, tau ), W + - Z bosons and neutrinos ( electron neutrino , muon neutrino, tau neutrino) can be replaced with electron moving at different
speeds from 0.1c up to 0.999.. c . When four protons fuse to become one helium nucleus, two of which must be converted into neutrons, and each such transition depends on the penetration of the two
electrons from the Universe, to the interior of the star . Penetration 1038 to 1058 of high energy electrons from the Universe to the interior of the star, transferred huge amounts of energy from the
Universe into a small space of the star. This huge cosmic energy is responsible for thermonuclear fusion. Currently prevailing opinion that the star itself is the source of the nuclear fusion
powering the star. In fact, without a high-energy electrons from other stars of the Universe, single star can not be able to a nuclear fusion, because without a high-energy electrons from other
stars, her stellar protons cannot be transform into her neutrons.
Category: High Energy Particle Physics
[474] viXra:1404.0110 [pdf] submitted on 2014-04-13 11:30:05
Quarks of 18 Types in Vedic Physics
Authors: John Frederick Sweeney
Comments: 13 Pages. several charts
Western nuclear physics contains six types of Quarks, while Vedic Nuclear Physics accounts for 18 types of Quarks and 18 types of anti – Quarks. In addition, Vedic Nuclear Physics posits three types
of matter, which carries implications for Quark charges. Vedic Nuclear Physics posits Giant Quarks, something unknown in Western Physics. This paper describes the concept of Hyper – Circles, which
are essential to Vedic Nuclear Physics and which may resemble Octonions, Sedenions or Exceptional Lie Algebras such as E7 and E8.
Category: High Energy Particle Physics
[473] viXra:1404.0100 [pdf] submitted on 2014-04-12 07:42:28
The Numerical Analysis of Beta Decay Stimulation by the High Thermal Spike of Photon Incidence to Valence Nucleons
Authors: Stefan Mehedinteanu
Comments: The work contain 60 pages
Since, at RHIC and LHC heavy-ion colliders the classical color field play an important role to study production of quark-gluon plasma, we propose a theory to describe strong-field inside the nucleons
based on Dual Ginzburg-Landau-Pitaevski (DGL) theory . We provide a detailed analysis of physically important quantities as regarding the nucleons substructure as: the uniform chromoelectric field
(vortex) strength inside a nucleon, the mass of monopole viewed as gluons which are the propagators of the QCD and carry colour and anti-colour, with an hedgehog-like configuration, or as a results
of interaction of spin-orbit of the monopole current , or of Rashba field interaction, all giving the same result; the quantification of the interaction energies of one vortex ( ) and of a giant
vortex ( ), as to be encapsulated by the Abrikosov triangular lattice generated by the coalescence of the flux lines. Therefore, it is proved for the first time, that in the nucleon exist
sufficiently high electromagnetic fields that permit to continue extract (with a rate of pair) from vacuum of pairs (virtual) of high energy electrons, of , Higgs bosons, quarks, by a Schwinger
effect, etc, to transform its into real one of very short time life, just like in a veritable laboratory. Thus, it was discovered for the first time that is in fact the Schwinger critical field for
the pair creation from vacuum. These pairs decay or annihilate into electrons, which passes the monopole condensate barrier as beta-electrons by quantum tunneling due of the phase slip with and of a
energy release, the entire model is proved for a free neutron decay life-time. Equally, the same Schwinger pairs-production rates are enhanced by a thermal Boltzmann factor in place of quantum
tunneling, when this thermalization due of the incidence of an high thermal spike of a photon with nucleons destroys the superconductivity. This effect is proved in the case of , through its β-decay
to 1.809 MeV γ-ray, when at high temperatures ( ) equilibrium is reached between and which is relevant to some high temperature astrophysical events such as novae. In the applications, as based on
these data, there are calculated: the Higgs boson energy release due of two gluons fusion during the collision at LHC, gluon pair production from space-time dependent chromofield due of the collision
of and of heavy nuclei; the pairs creation due of the thermally-induced vacuum instability as induced by a laser pulse in a crossed field of a single plane wave generated by a single high energy
photon. A proposal to use a laser pulse to reduce the half life of beta decay nuclides is discussed.
Category: High Energy Particle Physics
[472] viXra:1404.0091 [pdf] submitted on 2014-04-11 10:16:04
Boson Charge Explained
Authors: John Frederick Sweeney
Comments: 14 Pages.
Bosons come in positive, negative and zero charges, W+, W− and Z, but western science to date fails to explain why this is so. The W+, W−, and Z0 bosons, together with the photon (γ), comprise the
four gauge bosons of the electroweak interaction..Vedic Physics offers the genuine explanation of why bosons have three distinct types of charge and this paper definitively explains the reason, which
has to do with the three types of matter in the Universe. The paper concludes with a new typology of bosons based on the three different types of matter found in the universe.
Category: High Energy Particle Physics
[471] viXra:1404.0065 [pdf] submitted on 2014-04-08 13:43:39
Calculating the Trace of Products of Gamma Matrices Using an Easy Trick
Authors: Björn Schemmann
Comments: 5 Pages. The current version of the text is in German
The author is showing a memory hook to calculate the traces of products of Dirac matrices which might be useful for the average student of physics during his QED-class. Further the author proves the
universal validity of this technique applied to an arbitrary even number of gamma matrices using complete induction.
Category: High Energy Particle Physics
[470] viXra:1404.0007 [pdf] submitted on 2014-04-02 02:38:30
The Supersymmetric Planck Model
Authors: Bernard Riley
Comments: 21 Pages.
In the Planck Model, broken symmetries are manifested on the mass levels of the three Planck sequences. Superpartners, resulting from the breaking of supersymmetry, are arranged symmetrically about
superlevels within each sequence. There are two types of superlevel: every third level in each sequence (Type 1) and every fifth level in each sequence (Type 2). The superpartners of each charged
lepton and quark are identified; the superpartnerships are centred on superlevels of Type 1. The electron, the up quark and, intriguingly, the proton participate in superpartnerships that are
symmetrically arranged about coincident superlevels of Type 1. The W and Z weak gauge bosons and the 126 GeV Higgs boson participate in a trio of superpartnerships about three coincident superlevels
of Type 2. Coincident superlevels are rare, concentrated at known particle mass scales, and lie in a symmetric pattern that results from the interplay of the Planck sequences. The pattern of
coincident superlevels extends from the GUT scale to meV scale, forming the background to the see-saw mechanism of neutrino mass generation.
Category: High Energy Particle Physics
[469] viXra:1404.0005 [pdf] submitted on 2014-04-01 09:51:19
Hierarchy of Matter-Particles
Authors: Nainan K. Varghese
Comments: 12 Pages.
In material world matter provides substance to all real entities. Hence, it is logical to conceive development-hierarchy of various matter-particles, starting from unstructured matter, rather than
bifurcating superior matter-particles into inferior ones on the basis of noticed properties. Diverse properties of various matter-particles are derived from immediate inferior matter-particles and
depend on their physical structures. Very brief account of development of various matter-particles, from unstructured matter, as conceived in alternative concept presented in book ‘MATTER
(Re-examined)’, is presented in this article.
Category: High Energy Particle Physics
[468] viXra:1403.0971 [pdf] submitted on 2014-03-30 08:46:50
A Preliminary to a Unified Field Theory
Authors: Keith Maxwell Hardy
Comments: 28 Pages.
All forces emerge from a simple field based on the analogy of waves rippling out from a source. New ideas are applied to this field such as plus and minus space and time, an absolute reference frame
of nothingness, and origins of influence. This approach results in a basic unified field theory.
Category: High Energy Particle Physics
[467] viXra:1403.0300 [pdf] submitted on 2014-03-19 12:55:30
B-mode Octonionic Inflation of E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 9 Pages.
B-mode CMB polarization observed by BICEP2 is consistent with the Octonionic Inflation of E8 Physics that does not use a conventional inflaton field but instead uses NonAssociative Non-Unitary
Octonionic Quantum Processes.
Category: High Energy Particle Physics
[466] viXra:1403.0272 [pdf] submitted on 2014-03-17 00:05:24
Why the Composite Magnetic Monopoles of Yang-Mills Gauge Theory Have All the Required Chromodynamic and Confinement Symmetries of Baryons, How These May be Developed Into Topologically-Stable Protons
and Neutrons, and How to Path Integrate in Yang-Mills
Authors: Jay R. Yablon
Comments: 85 Pages.
We develop in detail, the classical magnetic monopoles of non-abelian Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine
their gauge fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. After developing inverse
solutions for the non-abelian electric charge densities while carefully examining uniqueness and gauge fixing, we use these solutions together with Dirac theory to “populate” these classical
monopoles with fermions. Applying the Fermi-Dirac-Pauli Exclusion Principle to these fermions forces the selection of a rank-3 gauge group initially chosen to be SU(3). We then find that these
non-abelian magnetic monopoles have the exact chromodynamic symmetries of baryons and interact via colored magnetic fields with the exact chromodynamic symmetries of mesons. We show that these
monopoles are also topologically stable, and that a required U(1) factor which ensures this stability also “flavors” these monopole as protons and neutrons. Because this exposition is classical, we
also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement. We finally point out how a recursive aspect of the non-abelian electric charge
solution may be used to perform an analytically-exact quantum path integration for Yang-Mills theory, proving the existence of a non-trivial quantum Yang–Mills theory on R4 for any simple gauge group
Category: High Energy Particle Physics
[465] viXra:1403.0267 [pdf] submitted on 2014-03-15 20:58:17
Photographs of a Ferrofluid Cell
Authors: Michael Snyder
Comments: 8 Pages.
This document is an update to my paper "Photonic Dipole Contours of Ferrofluid Hele-Shaw Cell" arXiv:0805.4364v2 . Photographs are the primary datasets and makeup the majority of the document. The
overall argument is that the ferrofluid cells are basically spin based radar systems that show photon scatter off of electron flows inside of the ferrofluid cell; induced by external magnetic fields.
In other words, we are watching sparks cascading through an irregular lattice like lighting jumping cloud to cloud. They build up static electricity like a solar cell.
Category: High Energy Particle Physics
[464] viXra:1403.0118 [pdf] submitted on 2014-03-13 06:54:12
Confinement of Charge Creation and Annihilation Centers by Nakanishi-Lautrup Field
Authors: Hideki Mutoh
Comments: 4 Pages.
The electromagnetic field model including Nakanishi-Lautrup (NL) field of quantum electrodynamics (QED) can easily treat creation and annihilation of positive and negative charge pairs, although it
is difficult for Maxwell's equations to treat them. However, the model does not directly satisfy the charge conservation equation and permits single charge creation and annihilation. It is shown that
the potential energy of NL field for a pair of charge creation and annihilation centers is proportional to their distance. It causes the confinement of charge creation and annihilation centers, which
means the charge conservation for this model. The quark confinement might be also explained by the energy of NL field.
Category: High Energy Particle Physics
[463] viXra:1403.0078 [pdf] submitted on 2014-03-11 17:04:04
A Tau Particle Model Based on the Sternglass Theory
Authors: Ray Fleming
Comments: 5 Pages.
Ernest Sternglass determined that a neutral meson, the π0 could be modeled as a relativistic electron-positron pair, and later determined that the muon could be modeled as an electron rotating around
a similar electron-positron pair. The author noticed that there is a second higher-energy orbital solution not previously published by Sternglass where the electron-positron pair orbits around the
electron’s center. A simple computation shows that the mass-energy of this second solution is consistent with the tau particle. Based on these models the mu and tau leptons are not fundamental
particles as described in currently popular theories but are instead two excited meta-stable states of an electron and an electron-positron pair.
Category: High Energy Particle Physics
[462] viXra:1403.0073 [pdf] submitted on 2014-03-10 18:38:18
Illustrations of a Modified Standard Model: Part 2-the Pion/muon Decays and the Neutrino Detector Nuclear Reactions
Authors: Roger N. Weller
Comments: 13 Pages.
The concepts of a proposed Modified Standard Model are applied to explain the charged pion and muon decays along with an analysis of the nuclear reactions involved in major neutrino detection
Category: High Energy Particle Physics
[461] viXra:1403.0016 [pdf] submitted on 2014-03-03 15:15:57
Introduction to the Expanded Rishon Model
Authors: Luke Kenneth Casson Leighton
Comments: 16 Pages. Many thanks to Vixra for providing this service as a means for those exploring particle physics to publish their work. I look forward to when future revisions of this paper are
peer-reviewed, but did not wish to delay sharing it in an open forum.
We introduce an expansion of the Rishon Model to cover generations, (including a previously undiscovered one), a brief explanation for how T and V exist at all, and explain particle decay in terms of
simple "phase transform" rules. We identify all current particles (with the exception of "Top") including the gluon, the Bosons and the Higgs, purely in terms of the underlying mechanism which
topologically can be considered to be Rishons.
Category: High Energy Particle Physics
[460] viXra:1403.0015 [pdf] submitted on 2014-03-03 15:53:20
Proof of Massless Particles Only Having Two Helicity One-particle States
Authors: LIU Changli, GE Fengjun
Comments: 4 Pages.
Why massless particles, for example photons, can only have two helicity one-particle states is the main subject of this work. As we know, the little group which describes massive particle
one-particle states' transformations under the Lorentz transformation is SO(3), while the little group describing massless states is ISO(2). In this paper, a new method is proposed to contract SO(3)
group to ISO(2) group. We use this contraction method to prove that the particle can only have two helicity one-particle states from the perspective of \emph{kinematics}, when the particle mass
trends to zero. Our proof is different from the dynamic explanation in the existing theories.
Category: High Energy Particle Physics
[459] viXra:1403.0013 [pdf] submitted on 2014-03-03 13:39:30
Solution of the Horizon and Flatness Problem in Cosmology Without Inflation
Authors: Friedwardt Winterberg
Comments: 9 Pages.
It is shown that the hypothesis of an inflationary universe to solve the horizon and flatness problem can be avoided by Lorentzian relativity assuming the existence of a preferred reference system at
rest with the zero point vacuum energy cut off at the Planck energy. Under this assumption there will be a “firewall” at the event horizon of an expanding universe where all matter disintegrates into
radiation having everywhere the Unruh black body radiation temperature. The replacement of the Einsteinian relativity by the Lorentzian relativity to solve the problem of quantum gravity was
suggested to the author by Heisenberg.
Category: High Energy Particle Physics
[458] viXra:1402.0178 [pdf] submitted on 2014-02-28 14:42:16
Unimodular SL(n,R) Gravity and E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 2 Pages.
Unimodular SL(n,R) Gravity in n-dimensional SpaceTime is related to E8 Physics in two ways: to 8-dim SpaceTime and to 4-dim Physical Minkowski M4 SpaceTime. The Unimodular SL(8,R) represents 8-dim
SpaceTime as a generalized checkerboard of SpaceTime HyperVolume Elements. The Unimodular SL(4,R) has conformal structure effectively equivalent to that of MacDowell-Mansouri SU(2,2) Conformal
Gravity and, further, is useful in resolving the strong CP problem.
Category: High Energy Particle Physics
[457] viXra:1402.0164 [pdf] submitted on 2014-02-26 23:02:38
Divergence-Free Versus Cutoff Quantum Field Theory
Authors: N.S. Baaklini
Comments: 24 pages, 148 equations, 10 references
We review the fundamental rules for constructing the regular and the gauge-invariant quantum field action both in the divergence-free approach and in the cutoff approach. Loop computations in quantum
electrodynamics of fermionic spinor matter, and also in quantum gravity of fermionic spinor matter, are presented in both approaches. We explain how the results of the divergence-free method
correspond to those of the cutoff method. We argue that in a fundamental theory that contains quantum gravity, the cutoff framework might be necessary, whereby the cutoff parameter and the
gravitational coupling could be related to each other quite consistently.
Category: High Energy Particle Physics
[456] viXra:1402.0157 [pdf] submitted on 2014-02-25 07:59:50
Reformulation of Relativistic Quantum Mechanics and Field Theory Equations with Space Time Sedenions
Authors: Victor L. Mironov, Sergey V. Mironov
Comments: 11 Pages.
We present an alternative type of sixteen-component hypercomplex scalar-vector values named "space-time sedenions", generating associative noncommutative space-time Clifford algebra. The
generalization of relativistic quantum mechanics and field theory on the basis of sedenionic space-time operators and hypercomplex wave functions is discussed.
Category: High Energy Particle Physics
[455] viXra:1402.0154 [pdf] submitted on 2014-02-23 10:30:53
Illustrations of a Modified Standard Model: Part 1-The Solar Proton-Proton Cycle
Authors: Roger N. Weller
Comments: 12 Pages.
A proposed modification of the Standard Model, when applied to the Solar Proton-Proton Cycle, provides a more in-depth explanation of the nuclear reactions within each step of the process.
Category: High Energy Particle Physics
[454] viXra:1402.0150 [pdf] submitted on 2014-02-22 06:41:58
Minimal Math Structures Needed for E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 4 Pages.
This is a rough outline of the minimal math structure needed for my E8 Physics model. It makes clear that a major barrier to understanding it is the amount of not-well-known mathematics of Real
Clifford Algebras, Lie Algebras, Bounded Complex Domains, etc. ... No details are given here as they can be found in viXra 1312.0036 and my other viXra papers and my web sites at tony5m17h.net and
valdostamuseum.com/hamsmith/ As to why anyone should expend the effort to understand the necessary math, the payoff is the substantially realistic results of E8 Physics calculations set out at the
end of this paper. The calculations are mostly tree-level with a few first-order results so further calculation work might bring even closer correspondence with observations.
Category: High Energy Particle Physics
[453] viXra:1402.0144 [pdf] submitted on 2014-02-21 06:36:26
Fundamental Unification Theory with the Electron, the Neutrino, and Their Antiparticles
Authors: N.S. Baaklini
Comments: 22 pages, 208 equations, 11 references
An SU3 unification theory with the electron, the positron, and the neutrino is reviewed. A 10-spacetime gravidynamic unification of the internal charges and the spin is formulated, with a
16-component Majorana-Weyl fermion that consolidates the foregoing three Weyl particles with the antineutrino, and their Dirac conjugates. Vector bosons and scalar (Higgs) particles are consolidated
in an antisymmetric tensor of 3rd rank, being the only tensor, apart from the graviton of 10-spacetime, that can couple to the unifying fermion. We write the Lorentz algebra of 10-spacetime in terms
of the 4-spacetime Lorentz algebra and the internal O6 factor, the latter expressed via its U3 subalgebra, and construct the pertinent operator representations. We exhibit the complete structure of
the unified gauge-Higgs couplings, indicating the source terms of particle masses. On the basis of this simple unification model, we propose the radical idea that all observed bosonic and fermionic
particles, whether leptonic or hadronic, may be composed of just the underlying four fundamental fermions.
Category: High Energy Particle Physics
[452] viXra:1402.0134 [pdf] submitted on 2014-02-20 11:21:36
A Wave Equation for Ultra-Relativistic Particles
Authors: Ginés R. Pérez Teruel
Comments: 5 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. In addition, we introduce the Hamiltonian form of the wave equation, and discuss another aspects such as the equation of the
probability conservation.
Category: High Energy Particle Physics
[451] viXra:1402.0108 [pdf] submitted on 2014-02-16 15:11:59
Why Yang-Mills Magnetic Monopoles Appear to Confine their Gauge Fields and have Composite Features, Similarly to Baryons
Authors: Jay R. Yablon
Comments: 15 Pages.
We develop in detail, the classical magnetic monopoles of Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. Because this exposition is
classical, we also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement, and what would need to also be considered in a complete quantum field
Category: High Energy Particle Physics
[450] viXra:1402.0107 [pdf] submitted on 2014-02-16 09:44:21
Higgs Energy Spheres and Higgs Two-Step Mass Generation
Authors: Malcolm H, Mac Gregor
Comments: 19 pages
Elementary particles can be represented as localized concentrations of energy, which act as Higgs energy spheres S. The sphere radius specifies its energy, and hence its coupling to the scalar Higgs
field. Massless photon energy spheres are defined by the Planck equation E = hν. Massive particle energy spheres are defined by the fine structure electrostatic energy equation and the fine structure
inertial mass equation, which are derived from the fine structure constant α~1/137. Higgs particle generation occurs in two steps: (1) an electrostatic-energy sphere is created; (2) it adiabatically
expands radially by a factor of 137 and transforms into a Compton-sized inertial-mass sphere, which represents a Higgs unit energy quantum. Four Higgs energy channels (electron, boson, fermion and
gauge boson) accurately reproduce lepton, constituent-quark, hadron, and average-gauge-boson masses.
Category: High Energy Particle Physics
[449] viXra:1402.0091 [pdf] submitted on 2014-02-14 05:56:39
The Topological Skyrme Model and the Current and the Constituent Quarks in Quantum Chromodynamics
Authors: Syed Afsar Abbas
Comments: 8 Pages.
It is well known that a quark manifests its current-character and simultaneously its constituent-character, depending upon what energy scale or length scale one is interested in for a particular
phenomenon. Here we show that there is a fundamental conflict between these two concepts, when one looks carefully as to how the electric charge for a quark is defined in these two, current and
constituent, structures. This is a crisis for quantum chromodynamics. We show then that the topological Skyrme Model comes to the rescue. We prove in this paper as to how unambiguously and uniquely,
it is the Skyrme Model which solves this conundrum in Quantum Chromodynamics.
Category: High Energy Particle Physics
[448] viXra:1402.0085 [pdf] submitted on 2014-02-12 23:03:50
Graphs and Expressions for Higher-Loop Effective Quantum Action
Authors: N.S. Baaklini
Comments: 11 pages, 8 references, 408 KB
We present the Feynman graphs and the corresponding expressions, up to 4th loop order, for a generic effective quantum field theory action. Whereas there are 2 graphs in the 2-loop order, and 8
graphs in the 3-loop order, we obtain 43 irreducible graphs in the 4-loop order. These results are obtained using Mathematica programming, where the underlying code is capable of generating graphs
and expressions to any desired loop order. We explain the associated programming strategy.
Category: High Energy Particle Physics
Recent Replacements
[388] viXra:1403.0300 [pdf] replaced on 2014-03-24 11:01:31
B-mode Octonionic Inflation of E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 18 Pages.
B-mode CMB polarization observed by BICEP2 is consistent with the Octonionic Inflation of E8 Physics that does not use a conventional inflaton field but instead uses NonAssociative Non-Unitary
Octonionic Quantum Processes. V2 adds details such as inflation producing a matter-dominated Universe. V3 corrects and expands material about matter-antimatter and generation-antigeneration.
Category: High Energy Particle Physics
[387] viXra:1403.0300 [pdf] replaced on 2014-03-24 00:22:43
B-mode Octonionic Inflation of E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 18 Pages.
B-mode CMB polarization observed by BICEP2 is consistent with the Octonionic Inflation of E8 Physics that does not use a conventional inflaton field but instead uses NonAssociative Non-Unitary
Octonionic Quantum Processes. V2 adds details such as inflation producing a matter-dominated Universe.
Category: High Energy Particle Physics
[386] viXra:1403.0272 [pdf] replaced on 2014-04-01 22:03:34
Why the Composite Magnetic Monopoles of Yang-Mills Gauge Theory Have All the Required Chromodynamic and Confinement Symmetries of Baryons, How These May be Developed Into Topologically-Stable Protons
and Neutrons, and How to Path Integrate in Yang-Mills
Authors: Jay R. Yablon
Comments: 101 Pages. Revision 2 adds two new sections 12 and 13
We develop in detail, the classical magnetic monopoles of non-abelian Yang-Mills gauge theory and show how these monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. After developing inverse
solutions for the non-abelian electric charge densities while carefully examining uniqueness and gauge fixing, we use these solutions together with Dirac theory to “populate” these classical
monopoles with fermions. Applying the Fermi-Dirac-Pauli Exclusion Principle to these fermions forces the selection of a rank-3 gauge group initially chosen to be SU(3). We then find that these
non-abelian magnetic monopoles have the exact chromodynamic symmetries of baryons and interact via colored magnetic fields with the exact chromodynamic symmetries of mesons. We show that a required U
(1) factor ensures that these monopoles are topologically stable, and also “flavors” these monopole as protons and neutrons. Because this exposition is classical, we also discuss the extent to which
classical field theory can be used to effectively analyze baryons and confinement. We finally point out how a recursive aspect of the non-abelian electric charge solution may be used to perform an
analytically-exact quantum path integration for Yang-Mills theory, proving the existence of a non-trivial quantum Yang–Mills theory on R4 for any simple gauge group G.
Category: High Energy Particle Physics
[385] viXra:1403.0016 [pdf] replaced on 2014-04-09 15:55:35
Introduction to the Expanded Rishon Model
Authors: Luke Kenneth Casson Leighton
Comments: 16 Pages.
We introduce an expansion of the Rishon Model to cover quark generations, (including a previously unnoticed one), provide an explanation for T and V as a topologically convenient moniker representing
aspects of phase and polarity within knots (of for example String Theory), and explain particle decay in terms of simple ``phase transform" rules. We identify all current particles (with the
exception of ``Top") including the gluon, the Bosons and the Higgs, purely in terms of the underlying mechanism which topologically can be considered to be Rishons. All this is predicated on the
simple assumption that all particles in effect photons phase-locked in a repeating pattern inherently obeying Maxwell's equations, in symbiotic support of their own outwardly-propagating
electro-magnetic synchotronic radiation, and that Rishons represent a phase ``measurement" (real or imaginary) at key strategic points on the photon's path.
Category: High Energy Particle Physics
[384] viXra:1403.0013 [pdf] replaced on 2014-03-04 11:32:13
Solution of the Horizon and Flatness Problem in Cosmology Without Inflation
Authors: Friedwardt Winterberg
Comments: 9 Pages.
It is shown that the hypothesis of an inflationary universe to solve the horizon and flatness problem can be avoided by Lorentzian relativity assuming the existence of a preferred reference system at
rest with the zero point vacuum energy cut off at the Planck energy. Under this assumption there will be a “firewall” at the event horizon of an expanding universe where all matter disintegrates into
radiation having everywhere the Unruh black body radiation temperature. The replacement of the Einsteinian relativity by the Lorentzian relativity to solve the problem of quantum gravity was
suggested to the author by Heisenberg.
Category: High Energy Particle Physics
[383] viXra:1402.0150 [pdf] replaced on 2014-03-05 08:56:46
Minimal Math Structures Needed for E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 4 Pages.
This is a rough outline of the minimal math structure needed for my E8 Physics model. It makes clear that a major barrier to understanding it is the amount of not-well-known mathematics of Real
Clifford Algebras, Lie Algebras, Bounded Complex Domains, etc. ... No details are given here as they can be found in viXra 1312.0036 and my other viXra papers and my web sites at tony5m17h.net and
valdostamuseum.com/hamsmith/ As to why anyone should expend the effort to understand the necessary math, the payoff is the substantially realistic results of E8 Physics calculations set out at the
end of this paper. The calculations are mostly tree-level with a few first-order results so further calculation work might bring even closer correspondence with observations. V2 and V3 correct and
clarify some typography re gauge groups and spinor fermions.
Category: High Energy Particle Physics
[382] viXra:1402.0150 [pdf] replaced on 2014-02-24 13:51:46
Minimal Math Structures Needed for E8 Physics
Authors: Frank Dodd Tony Smith Jr
Comments: 4 Pages.
This is a rough outline of the minimal math structure needed for my E8 Physics model. It makes clear that a major barrier to understanding it is the amount of not-well-known mathematics of Real
Clifford Algebras, Lie Algebras, Bounded Complex Domains, etc. ... No details are given here as they can be found in viXra 1312.0036 and my other viXra papers and my web sites at tony5m17h.net and
valdostamuseum.com/hamsmith/ As to why anyone should expend the effort to understand the necessary math, the payoff is the substantially realistic results of E8 Physics calculations set out at the
end of this paper. The calculations are mostly tree-level with a few first-order results so further calculation work might bring even closer correspondence with observations. V2 corrects and
clarifies some typography re gauge groups.
Category: High Energy Particle Physics
[381] viXra:1402.0134 [pdf] replaced on 2014-03-19 07:28:52
New Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 8 Pages.
In this letter we present a novel wave equation for ultra-relativistic matter, particles that satisfy $p>>m$, and whose energy-momentum relation can therefore be approximated by $E\simeq p+\frac{m^2}
{2p}$. We discuss in detail the implications of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino
oscillations. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another aspects, such as a conservation law analogue to the continuity equation. In addition,when
the wave equation is solved for an infinite potential well, it is found not only the quantization of the energy, but also the quantization of the mass. Finally, we point out that in the electron
case, the energy regime to test the model is $E\sim 500 $GeV, accessible at the next generation lepton collider (ILC).
Category: High Energy Particle Physics
[380] viXra:1402.0134 [pdf] replaced on 2014-03-16 11:42:31
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 7 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, particles that satisfy $p>>m$, and whose energy-momentum relation can therefore be approximated by $E\simeq p+\frac{m^2}
{2p}$. We discuss in detail the implications of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino
oscillations.The non-trivial dispersion relation is also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of
these waves satisfies $V_{ph}\geq c$, the associated group velocity can never surpass the speed of light. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another
aspects, such as a conservation law analogue to the continuity equation. In addition,when the wave equation is solved for an infinite potential well, it is found not only the quantization of the
energy, but also the quantization of the mass. Finally, we point out that in the electron case, the energy regime to test the model is $E\sim 500 $GeV, accessible at the next generation lepton
collider (ILC).
Category: High Energy Particle Physics
[379] viXra:1402.0134 [pdf] replaced on 2014-03-14 13:04:10
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 7 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another aspects, such as a conservation law
analogue to the continuity equation.In addition,when the wave equation is solved for an infinite potential well, it is found not only the quantization of the energy, but also the quantization of the
Category: High Energy Particle Physics
[378] viXra:1402.0134 [pdf] replaced on 2014-03-12 18:07:03
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 6 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another aspects, such as a conservation law
analogue to the continuity equation.In addition,when the wave equation is solved for an infinite potential well, it is found not only the quantization of the energy, but also the quantization of the
Category: High Energy Particle Physics
[377] viXra:1402.0134 [pdf] replaced on 2014-03-10 13:40:25
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 6 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another aspects, such as a conservation law
analogue to the continuity equation.In addition,when the wave equation is solved for an infinite potential well, it is found not only the quantization of the energy, but also the quantization of the
Category: High Energy Particle Physics
[376] viXra:1402.0134 [pdf] replaced on 2014-03-09 17:22:15
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 5 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. On the other hand,we introduce the Hamiltonian form of the wave equation, and discuss another aspects, such as a conservation law
analogue to the probability conservation theorem.In addition,when the wave equation is solved for an infinite potential well, it is found not only the quantization of the energy, but also the
quantization of the mass.
Category: High Energy Particle Physics
[375] viXra:1402.0134 [pdf] replaced on 2014-03-04 00:34:23
A Wave Equation For Ultra-relativistic Particles
Authors: Ginés R.Pérez Teruel
Comments: 5 Pages.
In this letter we present a new wave equation for ultra-relativistic matter, matter whose energy can be approximated by the relation, $E\simeq p+\frac{m^2}{2p}$. We discuss in detail the implications
of this wave equation and analyse their possible solutions, such as the plane-wave, which is completely consistent with the theory of neutrino oscillations.The non-trivial dispersion relation it is
also derived from the wave equation, allowing to obtain the phase and group velocities of the ultra-relativistic waves; while the phase velocity of these waves is always greater than $c$, the
associated group velocity can never surpass the speed of light. In addition, we introduce the Hamiltonian form of the wave equation, and discuss another aspects such as the equation of the
probability conservation. When the wave equation is solved for an infinite potential well, it is found that not only the energy in quantized, but also the rest mass of the particle.
Category: High Energy Particle Physics
[374] viXra:1402.0108 [pdf] replaced on 2014-02-22 07:28:57
Why Yang-Mills Magnetic Monopoles Appear to Confine their Gauge Fields and have Composite Features, Similarly to Baryons
Authors: Jay R. Yablon
Comments: 15 Pages.
We develop in detail, the classical magnetic monopoles of Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. Because this exposition is
classical, we also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement, and what would need to also be considered in a complete quantum field
Category: High Energy Particle Physics
[373] viXra:1402.0108 [pdf] replaced on 2014-02-21 15:58:32
Why Yang-Mills Magnetic Monopoles Appear to Confine their Gauge Fields and have Composite Features, Similarly to Baryons
Authors: Jay R. Yablon
Comments: 15 Pages.
We develop in detail, the classical magnetic monopoles of Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. Because this exposition is
classical, we also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement, and what would need to also be considered in a complete quantum field
Category: High Energy Particle Physics
[372] viXra:1402.0108 [pdf] replaced on 2014-02-17 17:21:12
Why Yang-Mills Magnetic Monopoles Appear to Confine their Gauge Fields and have Composite Features, Similarly to Baryons
Authors: Jay R. Yablon
Comments: 15 Pages.
We develop in detail, the classical magnetic monopoles of Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. Because this exposition is
classical, we also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement, and what would need to also be considered in a complete quantum field
Category: High Energy Particle Physics
[371] viXra:1402.0108 [pdf] replaced on 2014-02-16 16:02:31
Why Yang-Mills Magnetic Monopoles Appear to Confine their Gauge Fields and have Composite Features, Similarly to Baryons
Authors: Jay R. Yablon
Comments: 15 Pages.
We develop in detail, the classical magnetic monopoles of Yang-Mills gauge theory, and show how these classical monopoles, when analyzed using Gauss’ / Stokes’ theorem, appear to confine their gauge
fields, and also, appear to be composite objects. Of course, baryons, which include the protons and neutrons at the heart of nuclear physics, also confine their gauge fields and are
similarly-composite objects. This raises the question whether the magnetic monopoles of Yang-Mills theory are in some fashion related to the observed physical baryons. Because this exposition is
classical, we also discuss the extent to which classical field theory can be used to effectively analyze baryons and confinement, and what would need to also be considered in a complete quantum field
Category: High Energy Particle Physics
|
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The Absence of Absolutely Continuous Spectra for Radial Tree Graphs
Seminar Room 1, Newton Institute
We will introduce a family of Schrödinger operators on tree graphs with coupling conditions given by (b_n-1)^2+4 real parameters where b_n is the branching number. We will show the unitary
equivalence of the Hamiltonian on the tree graph and the orthogonal sum of the Hamiltonians on the halflines. We will use this unitary equivalence to prove that for a large family of coupling
conditions there is no absolutely continuous spectrum of the Hamiltonian on the sparse tree. On the other hand, we will show nontrivial examples of trees with the spectrum which is purely absolutely
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible.
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|
in QFT Lagrangian what is the potential energy exactly?
if I have χψ*ψ term in the Lagrangian, the scattering ψψ→ψψ behaves as if we had a potential (non-rel. limit) Exp(-mr)/r.
but when I try to calculate the expectation value of χψ*ψ in any state I get zero.
is that a different kind of potential energy? does it have to do with re-normalization?
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Brier, WA Science Tutor
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Topic: how to get the detailed values from a linear model class?
Replies: 2 Last Post: Mar 6, 2013 1:26 PM
Messages: [ Previous | Next ]
James how to get the detailed values from a linear model class?
Posted: Mar 5, 2013 4:00 PM
Posts: 77
Registered: 10/24/05 I am using the example on the mathworks website:
X = randn(100,5);
y = X*[1;2;3;4;5] + 6 + randn(100,1);
mdl = LinearModel.fit(X,y);
one silly question is that how to get the values of those model details from mdl? I can easily see them, but not sure how to output them or assign to a variable in matlab.
Date Subject Author
3/5/13 how to get the detailed values from a linear model class? James
3/6/13 Re: how to get the detailed values from a linear model class? Alan Weiss
3/6/13 Re: how to get the detailed values from a linear model class? James
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Sugar Hill, GA Math Tutor
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Reginald Brooks
Contact Email:
From biology, to medicine, to physics, and to numbers and geometry, my pursuit has always been visually based...though not necessarily on retinal imagery, but on any visual association which enhances
the mystery and thus engages the mind...as a painter and sculptor for the past 35 years. _______
Past solo exhibitions:_________________
"~breathe~ A Celebration of Women", 2008
"A Void", 2007
"Infinite Dimensions...without borders", 2005
"The E-P Series: digital-encaustic with scratchboard", 2001;____
"Dance Number Appendage: A Surrealistic Look At Patterns in Dance and Numbers", 1997;____
"Lucy:Contemporary Reflections on Paleonathropologic
"A Search for the Multicultural Mythology of Our Times", 1991.__________
Writings based on visual art, math, music and physics______________________
2014- MathspeedST: The Ubiquitous Information of Numbers (or, How Fractals Exceed the Speed Limit of Light) The General Ubiquity Rule (GUR) in effect, asserts that if you have a fixed numerical
pattern that is ubiquitously present across all space and time (spacetime, ST), that information is present across the same, passing any need to actually travel or transports said information at or
above the speed of light.
The Ubiquitous Information of Numbers, a new media net.art project, presents a summary of this idea in a somewhat playful form. The gist of the argument is as stated in the GUR above. A more local
example of a checkered tablecloth is used to demonstrate the point. No matter how folded (or not), of whatever size, once you know the pattern of the cloth in any one grid square you know the the
pattern of grids at any other spot on the cloth. Substituting the BBS-ISL (Brooks Base Square-Inverse Square Law) matrix, with its fixed numerical pattern that expands across any and all sizes,
spaces, and times of distribution, gives us exactly that: The Ubiquitous Information of Numbers.
2013-"MathspeedST: Leapfrogging LightspeedST FASTER Than The Speed of Light" The fundamental, Underlying Question: How can information be ubiquitous...effectively traveling FASTER than the speed of
light? is examined in this ebook that brings together the white papers and new medial net.art projects on Brooks (Base) Square (BBS-ISL matrix), A Fresh Piece of Pi(e), Numbers of Inevitability, and
more, examining the fundamental, underlying nature of SpaceTime (ST) itself as defined below the Observable Universe. Is there a connection? This free ebook is the complement to LightspeedST.
"LightspeedST: Leapfrogging @ The Speed of Light" The Ultimate Question: Why is the speed of light constant? is laid out in this ebook containing a detailed look at the 8+-part series (below)
examining the fundamental nature of SpaceTime (ST) and the role the constant "the speed of light" plays. Is there a connection? This ebook is a complementary work to MathspeedST.
"LightspeedST: 0. "The Ultimate Question"
The Ultimate Question Intro to this 7+-part series examining the fundamental nature of SpaceTime (ST) and the role the constant "the speed of light" plays. Is there a connection?
"LightspeedST: 8. "Triad"
The triad of velocity, space and time ... ruminations.
"LightspeedST: 7. Quark Portraits"
Part 7 of the now expanded series. Quark Portraits:
A quick reference to the I- and L-quarks and the composite particles that they make. Includes reference charts, a quick look at L.U.F.E. and The LUFE Matrix, and a whole section on Spinor Rotation.
2012-"LightspeedST: 1. 20 Questions"
Part 1 of the 5-part series. Twenty plus question examining the fundamental nature of SpaceTime (ST) and the role the constant "the speed of light" plays. Is there a connection?
"LightspeedST: 2. 20 Answers"
Part 2 of the 5-part series. Twenty plus answers to the 20+ questions, framed in bite-sized chunks...the better to swallow and digest these tender new morsels. Lightspeed=ST suggests a new view of
our Universe(s).
"LightspeedST: 3. The Mechanics of SpaceTime (ST) ~A Portrait of SpaceTime~"
Part 3 of the 5-part series. Actually, this "Portrait of SpaceTime" was written first. (Actually...really...this is all based on L.U.F.E., the Layman's Unified Field Expose´ from 1985. See Links.)
The goal: in the most concise and succinct manner...and always with an eye on simplification...lay out the case for tying together the quantum nature of light with the gravitational dependence of
spacetime. LightspeedST Parts 1,2,4,and 5 quickly followed as the antidote.
"LightspeedST: 4. The Movie Script"
Part 4 of the 5-part series. All text and no play doesn't mean anything more than the notion that every one's individual imagination will play out differently...because that is a good thing! And yet,
there is something especially apropos about seeing the original intent...the roadmap to infinity. Don't miss this one! (IN PRODUCTION)
"LightspeedST: 5. The Movie"
Part 5 of the 5-part series. As a visual journey, "LightspeedST: 5. The Movie" will quickly swoosh you...through moving images and text...up and over, down and through, the highlights of
LightspeedST...leapfrogging @ the speed of light. (IN PRODUCTION)
"LightspeedST: 6: Links & References"
A quick References and Links page.
"AFPOP: A Fresh Piece of Pi(e)...and the √2, too...Fractal - Fractal - Fractal"
"Circles and squares...as well as other geometric forms...have always been inter-related. Inherent in their geometry is the relationship of their linear elements to their areas...forming the basis of
the Inverse Square Law (ISL), as formally presented in the Brooks Base Square (BBS) matrix. At their core, these geometric forms are based on three irrational, infinity-based numbers of
proportion...π, √2 and φ...pi, the square root of two, and the golden mean. These core proportion-based relationships suggest that they may be acting as fractals in the formation of The
Architecture Of SpaceTime (TAOST). Fractals of infinity.!"
"GoMAC: The Geometry of Music And Color, Parts II and III." How the natural, geometric divisions of the musical monochord and the color spectrum are inherently derived from the same universal law
that informs how light, sound, electromagnetism, gravity...indeed, "The Architecture of SpaceTime" itself...behaves. That law is the Inverse Square Law (ISL) and it has a specific relationship to all
the quantitative, whole integer numbers as shown in the Brooks (Base) Square matrix (BBS).
Part II of the three-part Geometry of Music And Art (GoMAC) series focuses on a visual journey of how the musical spectrum...and specifically the chromatic and diatonic scales...is based on the ISL.
A visual walk-through and summary of key concepts relating the ISL-derived Brooks (Base) Square, BBS, matrix with a geometrically derived division of both the musical monochord...resulting in the
chromatic and diatonic major scale...and the color spectrum (as seen in GoMAC part III). Modulations around the Circle of Fifths and the Golden Rectangle are...key.
Based on "The Geometry of Music, Art and Structure...linking science, art and esthetics" white paper, "Geometry of Music And Color, I" new media net.art project, and the BBS matrix. Original
soundtracks (A and B).
Part III focuses on a visual journey of how the color spectrum...and specifically the complementary colors of each hue, is based on the ISL.
A visual walk-through and summary of key concepts relating the ISL-derived Brooks (Base) Square, BBS, matrix with a geometrically derived division of both the musical monochord...resulting in the
chromatic and diatonic major scale (part II)...and the color spectrum, in which color complements inside and outside of the 400-750 nm visible spectrum dictate the colors that we perceive.
"Numbers of Inevitability ~Let’s do a number on that!~" Number relationships are like the best and worst...the highest highs, lowest lows...of the most significant human relationships you have ever
known. When they are really wrong it really shows and pretty much everyone can see it. And, when they are right...really right...they seem, well, almost inevitable.
2011-"The Architecture Of SpaceTime (TAOST) as defined by the Brooks (Base) Square matrix and the Inverse Square Law (ISL)"
"The single most important concept to take from here is that not only is ST not rigid and static...indeed it is more like a fluid made of light, each quanta of which flashes on and
off...pulse-propagates into and out of existence...at a frequency (and wavelength) according to its energy...and this fluid light ST unit resonates off a wave of energy at the peak of each
pulse-flash...that wave, like a pond ripple, fans out to infinity (or the edge of the Universe...which ever comes first) where it is absorbed back into the sub-Planck event horizon (Singularity) to
fuel the next round...communicating...passing on information...as to the location source and energy density...the amount and degree of its presence...its curvature...to every other ST unit throughout
the Universe(s), BUT that the ST formation and curvature information propagation is informed, dictated and quantified by the matrix of BBS and the ISL...the bean counters par excellence!"
"Brooks (Base) Square interactive (BBSi) matrix: Part I: BASICS
This "BASIC" introduction to the matrix grid and the simplest, most fundamental pattern of the numbers that are formed from the main, Prime Diagonal are presented in a step by step fashion in five
different interactive, multimedia formats. The Prime Diagonal is none other than square of the axial numbers (1,2,3,...) forming the core of the Inverse Square Law sequence (1,4, 9,...). All the
numbers...and their many exquisite patterns...are derived directly from the Prime Diagonal numbers. This is Part I. Part II will follow and will present some of the more interesting patterns. The
Inverse Square Law defines how a force or influence (like energy or light) falls off in strength as the distance is increased. It is one of the fundamental relationship laws of the universe effecting
everything from gravity to electromagnetism. It can not but help to also be part of the spacetime fabric. It's part of "The Architecture Of SpaceTime" (TAOST).
2010-"Brooks (Base) Square (BS): The Architecture of Space-Time (TAOST) and The Conspicuous Absence of Primes (TCAOP) -the complete work (Rules 1-177)
The table of numbers that, while excluding the *prime numbers ... and only the *prime numbers ... reveals the number patterns that are primary ... primordial ... to the structure of the Universe.
This primordial matrix defines the Inverse Square Law (ISL) that informs gravity, light (electromagnetism) and atomic structure ... the underpinnings of that which we see as the observable Universe.
It is the architecture of space-time itself ... that which ultimately informs ... that is, reveals its disposition that in turn presents the physical laws, and their expression, of light, gravity and
the structure of matter ... that fundamentally defines the true structure of the Universe ... seen and unseen ... encompassing all matter (light and dark) and all energy (light and dark). The ISL is
a key relationship describing the interplay of space and time. It is postulated here that the true underpinnings of space-time itself are inherently defined by numerical relationships ... numbers ...
the numbers of Brooks (Base) Square (BS).
2009-"GoMAS: The Geometry of Music, Art and Structure ...linking science, art and esthetics" ~ That simple geometric principles can be responsible for the manifestation of the multitudes of beauty,
diversity and richness of Nature is further supported here in this paper. The Inverse-Square-Law (ISL), a simple geometric principle, is shown to be the common link between physics, chemistry,
biology, and the science and esthetics of music and art. Nature has incorporated the ISL as the major game rule to building the cosmos. Our perception of pleasing form, color and sound is
prophetically built into our sensory being, our biologic evolution, and our structural past.
2008- "The Conservation of SpaceTime ~The role of the Higgs, graviton and photon bosons in defining Dark Matter = Dark Energy (the inverse of) ~
2008-"Quantum Gravity ...by the book" ~Merging Quantum Mechanics with Gravity~ a simple, non-mathematical insight.
2007-"The History of the Universe - update - 'The Big Void'" ~A short, update supplement to "Dark Matter = Dark Energy (the inverse of): The Conservation of Spacetime by The Conservation of Force"
and "The History of the Universe in Scalar Graphics."
2006-"The Butterfly Prime Determinant Number Array (DNA)~conspicuous abstinence~". ~The role of the highly ordered positive space pattern of the odd non-primes in revealing the reverse, negative
space pattern of the primes as a determined number array is revealed.
-"The Butterfly Prime Directive ... ~metamorphosis~". ~There is a logic behind the ordered and predictable pattern of prime numbers.
2005- "Butterfly Primes ... let the beauty seep in". ~An ordered and predictable pattern for the prime numbers emerges.
-"The LUFE Matrix: E=mc2 ... and then some". The focus is on energy. From Einstein's initial insight into the transformation between mass and energy, we now have a large number of similar
transformations between energy and the other fundamental parameters of matter ... i.e., charge and spin ... and their activities within their space-time environments such as velocity and
acceleration, frequency and temperature, momentum, pressure and force, potential difference, heat and work, etc.
2004-"The History of the Universe in Scalar Graphics" A one-page visual supplement to "Dark-Dark-Light" below.__
-"The LUFE Matrix: Infinite Dimensions ~a room with a view...within a room with a view...within a room with a view~" A new, speculative journey to account for the multiple dimensions that have
informed The LUFE Matrix since its introduction in 1985. FAQ-graphic format.
2003-Dark-Dark-Light: Dark Matter = Dark Energy (the inverse of): The Conservation of Spacetime by The Conservation of Force_____
-The LUFE Matrix: the distillation of SI units into more fundamentally base units of space and time dimensions (Short Title: The Multi-dimensional LUFE Matrix_____
-The LUFE Matrix Supplement: Examples and Proofs_____
2001-Part II below, from 2000, was rewritten and recast for online publication as a three part series:___
Part 1: "PIN: Pattern in Number...from primes to DNA" A brief primer on visual patterns found in simple, well known number systems leading into the remarkable number magic and esthetic beauty of the
DNA double helix (axial view) found in Part 2.GoDNA and continued in Part 3.SCoDNA._____
Part 2: "GoDNA: the Geometry of DNA (axial view) revealed"_____
Which came first? The self replicating underlying geometry or the form itself with its dependence on enzymes, hydrogen bonding and a matching code of bases?_____
Part 3: "SCoDNA: the Structure and Chemistry of DNA (axial view)"_____
Extending GoDNA into the realm of structural detail and chemistry forming the composite design._____
2000-"The Geometry of Music, Art and Structure?linking science, art and esthetics, Part II" self-published. The original material for this paper was presented (self-published) at the 1997 solo
exhibition at the Montage Gallery, Portland, OR under the title "Patterns In Numbers: From Prime Numbers to DNA".____
1998-"The Geometry of Music, Art and Structure?linking science, art and esthetics, Part I" presented at the Eighth International Conference on Engineering Computer Graphics and Descriptive Geometry
in Austin, TX, sponsored and published by the International Society of Geometry and Graphics. The original material for this was presented at the 1996 solo exhibition at the Montage Gallery,
Portland, OR.______
1991-"The LUFE Matrix: the distillation of SI units into more fundamentally base units of space and time dimension (Short Title: The Multi-dimensional LUFE Matrix)" with supplement, paper,
1987-"Fundamental Constants of Nature: Mass", paper, self-published_____
1987-"Fundamental Constants of Nature: Ratios", paper, self-published_____
1987-"Fundamental Constants of Nature: Charge", paper, self-published_____
1987-"Fundamental Constants of Nature: The LUFE Matrix", paper, self-published_____
1986-"Mass Variation in the Electron by Photon Absorption & The Inverse Square Law", paper, self-published_____
1986-"Expose on Music, Color and the Inverse Square Law", paper, self-published_____
1985-"L.U.F.E.: The Layman?s Unified Field Expose" (a visual guide), textbook-manuscript, self-published_____2008-placed on line
Contact and additional links to this work:_______________
Reg Brooks___
Brooks Design___
535 NW 107th Ave___
Portland, OR 97229-6211___
503.646.1548 Phone___
Portfolio at:
(drawings, new media, paintings & sculpture)
All New Media net.art projects and their White Papers (Art Theory 101) found here:
Brooks Design-Contemporary Graphics:
"RealSurReal...aClone, 2001" net art project
"Hey! U funk'n with my DNA?" net art project
"PIN: Pattern in Number...from primes to DNA"
"GoDNA: the Geometry of DNA (axial view)"
"SCoDNA: the Structure and Chemistry of DNA (axial view)"
"GoDNA: The Geometry of DNA (composite axial view image variations)"
"Naughty Physics (a.k.a. The LUFE Matrix)" net.art project
"The LUFE Matrix: the distillation of SI units into more fundamentally base units of space and time dimensions"
"The LUFE Matrix Supplement: Examples and Proofs"
"Dark Matter=Dark Energy (the inverse of): The Conservation of Spacetime by the Conservation of Force"
"The LUFE Matrix: Infinite Dimensions ~a room with a view...within a room with a view...within a room with a view~" has recently been added. A new, speculative journey to account for the multiple
dimensions that have informed The LUFE Matrix since its introduction in 1985. It is presented in a FAQ-graphic format.
"Your sFace or Mine?" net.art project.
"The History of the Universe in Scalar Graphics"
"Butterfly Primes ... let the beauty seep in"
"The Butterfly Prime Directive ... ~metamorphosis~"
"The Butterfly Prime Determinant Number Arrray (DNA) ... ~conspicuous abstinence~"
"Butterfly Primes ... ~Prejudicial Numbers~" net.art project
"Geometry of Music & Color: ~the role of the ISL~" net.art project
"Brooks (Base) Square & The Inverse Square Law (ISL)" net.art project
"BBSi Matrix: Deconstructed" net.art project
"Sunspots and Solar Flares" net.art project
"Geometry of Music And Color, II-III" net.art project
"AFPOP: A Fresh Piece of Pi(e) ... and the √2, too ... Fractal - Fractal - Fractal" net.art project
"LightspeedST: LeapFrogging ODOU @ The Speed Of Light" net.art project
"Your sFace or Mine? Anonymity" net.art project.
"MathspeedST: Leapfrogging LightspeedST FASTER Than The Speed of Light" The fundamental, Underlying Question: How can information be ubiquitous...effectively traveling FASTER than the speed of light?
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How to save money on chicken feed?
post #11 of 60
4/21/12 at 7:59pm
Thread Starter
Originally Posted by
Why is it costing you so much? How much are you buying at a time? How many and how old are your chickens?
I just checked my math as I went from buying 50ld bags of Purina layer to local grain company's layer in 100lb bags. The cost of producing an egg for me went from .15 cents to just under .10 cents
per egg. I was paying $15.50 for a 50 lb bag of Purina and now buy 100lb bags of Ventura layer for $22.32.
So for a dozen eggs is $1.20 plus the box @.22 is $1.44 cost total.
I am buying 50 pounds at a time for $15 dollars. I have 4 layers w/ one silkie which will be 1 year old at the end of June. of 2012. Sounds like I need to find a place where I can buy 100 pounds of
feed at a time. I'll ask my TSC and see if they ever sell larger quanities and check around a bit.
Originally Posted by Scooter&Suzie
I am buying 50 pounds at a time for $15 dollars. I have 4 layers w/ one silkie which will be 1 year old at the end of June. of 2012. Sounds like I need to find a place where I can buy 100 pounds of
feed at a time. I'll ask my TSC and see if they ever sell larger quanities and check around a bit.
Look for local mills. TSC is retail, of course. Find out where the "real" farmers buy feed.
With 4 layers and a broody banty you should only being through a pound of food or so a day which would be about 45/50 days of food from a bag. I was just running your numbers,do you only get one egg
a day? One egg a day for 45 days into $15 bag of food is .33 cents per egg equals $4 a dozen. I am thinking creatures other than chickens might be helping themselves to the feed.
"The difference between being involved and being committed is the same as the difference between eggs and bacon. The chicken is involved. But the pig is committed" Anonymous
"The difference between being involved and being committed is the same as the difference between eggs and bacon. The chicken is involved. But the pig is committed" Anonymous
Don't know if you are in SE PA, but there are several places in Berks County & many more in Lancaster & Chester County to get cheaper local feed.
Originally Posted by Scooter&Suzie
The best price I can find is $15 per 50 pounds at TSC. I buy the Dumor feed. I cannot find a better price where I live. Maybe I should try looking around at all the other possible stores. I'll
make some calls on Monday, I guess. I live in Pennsylvania, and I don't know of any locally milled feed around here. I was thinking, could I get together some of my friends who have chickens and
together buy some feed in bulk in order to save money? I just have a few back yard hens (4 layers, and 1 bantam with 8 eggs under her). Although, I would like to get 3 more layers and a couple more
bantams. I just can't do it yet with feed costing so much :/
Originally Posted by duckinnut
With 4 layers and a broody banty you should only being through a pound of food or so a day which would be about 45/50 days of food from a bag. I was just running your numbers,do you only get one egg
a day? One egg a day for 45 days into $15 bag of food is .33 cents per egg equals $4 a dozen. I am thinking creatures other than chickens might be helping themselves to the feed.
Thanks for doing the math, duckinnut. I was thinking along the same lines, too.
It sounds like a good price because it reproduces so quickly. Just let it multiply to a larger number before you start feeding it to the chickens. I dont know if chickens can eat snails, you might
have to google that one.
-Put the duckweed in a seperate container as your fish because the fish will eat it. I grow duckweed and everyday I give some to my chickens and some to my fish. Both eat it up very quickly. If you
put 1/4 cup in your fish tank it will be gone in a few minutes.
Hmmm.... Maybe I did something wrong there. I read in a chicken book that every 3.5 pounds of food a chicken takes in it produces 1 dozen eggs. So when I did the math on how much I was paying, I
got $4.29. Do they not need 3.5 pounds to make a dozen eggs then?
You have to go by the numbers you get not by some book. Take the number of eggs per week(4 hens @ 5eggs each =20), a bag of food should last at least 6 weeks with 4 layers and a broody. (20 eggs a wk
times 6 wks =120 or 10 dz) $15 for food divided by 10 dz should be $1.50 dz. Those numbers put you at about .12/.13 cents per egg. These numbers are moveable I am going just on what would be average.
"The difference between being involved and being committed is the same as the difference between eggs and bacon. The chicken is involved. But the pig is committed" Anonymous
"The difference between being involved and being committed is the same as the difference between eggs and bacon. The chicken is involved. But the pig is committed" Anonymous
Originally Posted by duckinnut
You have to go by the numbers you get not by some book. Take the number of eggs per week(4 hens @ 5eggs each =20), a bag of food should last at least 6 weeks with 4 layers and a broody. (20 eggs a wk
times 6 wks =120 or 10 dz) $15 for food divided by 10 dz should be $1.50 dz. Those numbers put you at about .12/.13 cents per egg. These numbers are moveable I am going just on what would be average.
Oh wow, thanks so much!
post #12 of 60
4/21/12 at 8:29pm
post #13 of 60
4/22/12 at 5:42am
post #14 of 60
4/22/12 at 6:46am
post #15 of 60
4/22/12 at 6:57am
post #16 of 60
4/22/12 at 8:57am
post #17 of 60
4/22/12 at 10:35am
Thread Starter
post #18 of 60
4/22/12 at 4:30pm
post #19 of 60
4/23/12 at 4:37pm
Thread Starter
post #20 of 60
4/23/12 at 4:57pm
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Math Help
July 26th 2008, 12:38 AM #1
Junior Member
Dec 2007
the curve f(x) for which f'(x)= 16x+k, where k is a constant, has a stationary point at (2,1), find the value of f(x) when x=7
the answers actually 201 i just can't get the working
We know two things :
- That x=2 is a stationary point hence f'(2)=0. This equation can be solved for k.
- That f(2)=1. As f'(x)=16x+k, integrating you get f(x)=... + constant and the equation f(2)=1 will give you the value of the constant.
Then simply compute f(7) and you are done.
July 26th 2008, 01:12 AM #2
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i can't wrap my head around how to do this rational inequalities question :( help?
solve and show work, help please, i would really really really appreciate it )': test after lunch! 1-x/x+5>=2-x/x+6
You first need to get a common denominator, such as $\left ( x+5 \right )\left ( x+6 \right )$ Can you go from there?
i tried to get common denominators and added up the numerators, i just became stuck on where to go from there... i guess i just don't understand what the question is asking
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Math Forum Discussions - Re: Matheology � 224
Date: Mar 22, 2013 3:40 PM
Author: Virgil
Subject: Re: Matheology � 224
In article
WM <mueckenh@rz.fh-augsburg.de> wrote:
> But I don't want to remove a set. When has it been prohibited to
> handle elements of a set?
One cannot remove an element from a set of elements without removing the
singleton set containing that element from the set of all subsets of the
set of elements.
One cannot remove a line from the set of all lines without removing the
set containing only that line from the set of all sets of lines.
At least not outside Wolkenmuekenheim.
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A mysterious Heisenberg algebra identity from Sylvester, 1867
up vote 21 down vote favorite
I am trying to understand two papers by James Joseph Sylvester:
P92: "Note on the properties of the test operators which occur in the calculus of invariants, their derivatives, analogues, and laws of combination; with an incidental application to the development
in a Maclaurinian series of any power of the logarithm of an augmented variable."
P95: "On the multiplication of partial differential operators."
[The numbering is from volume 2 of Sylvester's Collected Works. These, incidentally, are spread over four volumes: volume 1, volume 2, volume 3, volume 4 (first two courtesy of anonymous book
scanners), with some duplicates on the Internet Archive. All of them are out of copyright.]
On the first page of P95 (aka page 11 of the linked djvu), Sylvester states that
"If $\phi$ be any such function [i. e., a polynomial or power series in infinitely many commuting variables $x$, $y$, $z$, ..., $\delta_x$, $\delta_y$, $\delta_z$, ... (here, $\delta_x$, $\delta_y$,
$\delta_z$, ... are just symbols, not differential operators!) which is multilinear with respect to $\left(\delta_x,\delta_y,\delta_z,...\right)$], [we have]
$e^{\displaystyle t\phi\star} = \left[e^{\displaystyle \left(e^{\displaystyle t\phi\star}-1\right)\phi}\right]\star$."
Here, as far as I understand, the $\star$ operation is defined as follows (see page 1 of P92, aka page 1 of the linked djvu): If $\psi$ is any polynomial or power series in infinitely many variables
$x$, $y$, $z$, ..., $\delta_x$, $\delta_y$, $\delta_z$, ..., then $\psi\star$ means the differential operator we obtain if we collect all the $\delta_x$, $\delta_y$, $\delta_z$, ... variables at the
right end of every monomial and replace them by the partial derivative operators $\frac{\delta}{\delta x}$, $\frac{\delta}{\delta y}$, $\frac{\delta}{\delta z}$, .... I cannot say that I am sure
about this, though, because no matter how I try to obtain a small, verifiable example for the formula, I get some nonsense which is either wrong or I am not able to check.
Sylvester studied these in the context of classical invariant theory, but nowadays quantum field theorists are interested in these differential operators as elements of the Heisenberg algebra. Is
there any modern (readable) reformulation of the above identity? Has anyone else tried to comprehend its meaning? Is it related to the identity $\left(\exp a\right)\left(\exp b\right)\left(\exp a\
right)^{-1} = \exp\left(\left(\exp\left(\mathrm{ad} a\right)\right)\left(b\right)\right)$ which holds for any two elements $a$ and $b$ of a ring for which these exponentials make sense? (This is
speculation based on nothing more than the appearance of nested exponentials in both identities.)
noncommutative-algebra lie-algebras combinatorial-identities qa.quantum-algebra
I don't have it with me now but it seems to me that the book of Ben-Zvi and Frenkel on vertex algebras might help you. – DamienC Jul 15 '12 at 11:53
Hmm, thanks. This is certainly related to the normally ordered product in the universal enveloping algebra of the Heisenberg algebra; but I don't yet see much connection to vertex algebras. (That
said, I barely know the definition of the latter.) – darij grinberg Jul 15 '12 at 12:48
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4 Answers
active oldest votes
The following instance of your identity is well known in physics (and is sometimes called an "operator disentangling" identity)
$:\exp\left[\left(e^W-1\right)_{ij}a_i^\dagger a_j\right]:\;=\exp\left(W_{ij}a^\dagger_i a_j\right)$
where $::$ denotes normal ordering, $a_i$ and $a^\dagger_j$ are canonical Bose annhilation and creation operators satisfying $\[a_i,a^\dagger_j\]=\delta_{ij}$, and W is an arbitrary
matrix (summation implied).
For general (rather than just quadratic) $\phi$ the formula is completely new (indeed remarkable) to me.
up vote 9 Frustratingly, it's hard to track down the origins of the above formula. Here's a recent discussion that includes the above version for a single boson mode (Eq. 30):
down vote
Combinatorics and Boson normal ordering: A gentle introduction American Journal of Physics, 75 (7), pp. 639 (2007)
The authors' comments after Eq. 30 seem to imply that the formula doesn't generalize simply.
EDIT: I realized that Sylvester initially states the quadratic form above, and then limits his generalization to functions $\phi$ "linear quantic in $\delta_x$, $\delta_y$, $\
delta_z$,...". Still, this generalization appears to contradict Eq. 31 of the above article.
1 According to Harold Davis in The Theory of Linear Operators (Principia Press, 1936, pg. 199), the commutator [q,p]=1 "was apparently first studied by Charles Graves as early as 1857."
Davis goes on to use the commutator to get some "normal ordering" results obtained by Graves and to expand on them. – Tom Copeland Mar 7 at 10:32
1 Very interesting! Graves essentially had the form of the star product. – Austen Mar 7 at 16:08
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Thanks, Austen; I didn't think a particular case could be that strong. (I can't say it's an immediate particular case, though. It took me some transformations to get the $e^W-1$ term.)
Meanwhile I have understood the claim (and found a proof; more about it later today or tomorrow).
The formula I quoted above is slightly wrong, at least as I understand it. It should be $e^{\displaystyle t\phi\star} = \left[e^{\displaystyle \left(\left(e^{\displaystyle t\phi\star}-1\
right) / \left(\phi\star\right)\right) \phi}\right]\star$, where $\left(e^{\displaystyle t\phi\star}-1\right) / \left(\phi\star\right)$ is to be understood "formally" (as in, "plug in $\phi\
star$ as $x$ in the power series $\left(e^{tx}-1\right)/x$"). Sylvester gets this right in his first paper (P92) even though he misses an assumption (multilinearity with respect to the $\
delta$ variables) there. Apparently, the wrong formula is due to an off-by-$1$ error.
Let me rewrite the correct formula in some more modern notations.
Theorem 1 (Sylvester). Let $K$ be a commutative ring.
Let $K\left[a,b,c,...\right]$ be the ring of polynomials in the commuting indeterminates $a$, $b$, $c$, ..., and let $K\left[\delta_a,\delta_b,\delta_c,...\right]$ be the ring of polynomials
in the commuting indeterminates $\delta_a$, $\delta_b$, $\delta_c$, ... (which, as for now, have nothing to do with $a$, $b$, $c$, ... except being similarly labelled). Let $\mathrm{Diff}\
left(a,b,c,...\right)$ be the ring of polynomial differential operators on $K\left[a,b,c,...\right]$. Then, we can define a $K$-module isomorphism
$M : K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right] \to \mathrm{Diff}\left(a,b,c,...\right),$
$P \otimes Q \mapsto P \cdot Q\left(\dfrac{\partial}{\partial a},\dfrac{\partial}{\partial b},\dfrac{\partial}{\partial c},...\right)$.
(Only $Q$, not $P\cdot Q$, is being evaluated at $\left(\dfrac{\partial}{\partial a},\dfrac{\partial}{\partial b},\dfrac{\partial}{\partial c},...\right)$ here.)
Let $\left(K\left[\delta_a,\delta_b,\delta_c,...\right]\right)_1$ be the $K$-submodule of $K\left[\delta_a,\delta_b,\delta_c,...\right]$ spanned by $\delta_a$, $\delta_b$, $\delta_c$, ...
(that is, the degree-$1$ part of $K\left[\delta_a,\delta_b,\delta_c,...\right]$).
up vote
7 down The ring $\mathrm{Diff}\left(a,b,c,...\right)$ acts on the tensor product $K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]$ by acting on the first tensorand only
vote . Denote this action by $\rightharpoonup$. In other words, for any differential operator $R\in \mathrm{Diff}\left(a,b,c,...\right)$, any $P\in K\left[a,b,c,...\right]$ and any $d\in K\left[\
delta_a,\delta_b,\delta_c,...\right]$, set $R\rightharpoonup \left(P\otimes d\right) = R\left(P\right)\otimes d$, and extend this by linearity to an action of $\mathrm{Diff}\left(a,b,c,...\
right)$ on the whole tensor product.
Let $D \in K\left[a,b,c,...\right] \otimes \left(K\left[\delta_a,\delta_b,\delta_c,...\right]\right)_1$ be arbitrary. Let $T$ be the power series
$\sum\limits_{i\geq 1} \left(M\left(D\right)\right)^{i-1} \rightharpoonup D \dfrac{t^i}{i!} \in \left(K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]\right)\left
Then, $M\left(\exp T\right) = \exp\left(tM\left(D\right)\right)$, where $M\left(\exp T\right)$ is shorthand for "the image of $T$ under the canonical map $M : \left(K\left[a,b,c,...\right] \
otimes K\left[\delta_a,\delta_b,\delta_c,...\right]\right)\left[\left[t\right]\right] \to \left(\mathrm{Diff}\left(a,b,c,...\right)\right)\left[\left[t\right]\right]$ induced by $M$".
Note that this is closely related to normal ordered products, but I find the $:...:$ notation for normal ordered products annoyingly restrictive (as it forces everything between the colons
to be normal ordered, but the proof of Theorem 1 needs a normal ordered product with a non-normal ordered product inside). Maybe this is because I don't really understand the subtleties of
this notation. I would personally just use some different symbol for the commutative multiplication map on $\mathrm{Diff}\left(a,b,c,...\right)$ transferred from $\left[a,b,c,...\right] \
otimes K\left[\delta_a,\delta_b,\delta_c,...\right]$ by the isomorphism $M$. What about $\boxdot$?
Theorem 1 is by no means the whole content of the two papers I've linked, and I'd welcome any readable "translations" of the other results. (I'm going to try that myself, too.)
Note that there are two typos in the second equation on page 568 of P92. It says
$\phi_1\star\phi_1\star\phi_1\star = \left(\phi_1^s + 2 \phi_1\phi_2 + \phi_3\right)\star$.
First, the $s$ should be a $3$ (this is probably because whoever made the djvu set the threshold for identity of shapes too liberally); second, the $2$ coefficient should be a $3$.
So you've ended up with a rep of the Bell partition polynomials oeis.org/A036040 as Sylvester's equation $e^{t\phi _1*}=(e^{T})*$ suggests, with $T=e^{t\phi_{.} }$ umbrally? – Tom
Copeland Jul 17 '12 at 14:07
1 I interpret Sylvester's $e^{tx\frac{\mathrm{d} }{\mathrm{d} x}*}=\left ( e^{\left ( e^t-1 \right )x\frac{\mathrm{d} }{\mathrm{d} x}} \right )*$ as $e^{tx\frac{\mathrm{d} }{\mathrm{d} x}}=
e^{tB_{.}(\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}})}=e^{\left ( e^t-1 \right )\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}}}$ where $(B_{.}(x))^n=B_{n}(x)$ are the regular Bell
polynomials and $(\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}})^n=x^{n}\frac{\mathrm{d^n} }{\mathrm{d} x^n}$. – Tom Copeland Jul 17 '12 at 21:47
Your phrase "by acting on the first tensor only" reminds me of a pre-Lie algebra. – Tom Copeland Mar 4 at 7:10
Tom, thanks for reviving this thread. Unfortunately I'm lacking time these days, but I hopefully will come back to it. Which links are broken? (NB: I really don't understand umbral
notation and anything remotely similar to it -- Young's raising operators, the symbolic method etc.; sorry.) – darij grinberg Mar 5 at 5:50
Too bad about the symbolic method-- that's what Sylvester was woking at in the British tradition. Easier to relate to trees and general Tylor operator shifts (normal ordering), making
relations more transparent than Hopf algebra and operadic notations in many cases, to me at least. They seem to climb back into the Cayley trees to get a clear view of the landscape
anyway. – Tom Copeland Mar 5 at 7:23
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This can be interpreted cleanly using the notion of Pre-Lie algebra.
Indeed, vector fields on an affine space form a Pre-Lie algebra. To prove the desired identity, it is enough to consider the free Pre-Lie algebra on one generator.
This identity is known to be related to a pre-Lie version of the Baker-Campbell-Hausdorff formula. This is written in several places in the literature.
Here are a few references:
up vote 5
down vote • Agracev, A. A. and Gamkrelidze, R. V., Chronological algebras and nonstationary vector fields
• Dominique Manchon, A short survey on pre-Lie algebras, E. Schrödinger Institut Lectures in Math. Phys., Eur. Math. Soc.
• Pierre Cartier, Vinberg algebras, Lie groups and combinatorics. (English summary) Quanta of maths, 107–126, Clay Math. Proc., 11,
Thank you! I'll try to read Manchon's paper ASAP (although ASAP doesn't mean soon with my current schedule). Just some links: www2.maths.ox.ac.uk/cmi/library/proceedings/cmip011c.pdf
for the third paper, and people.sissa.it/~agrachev/agrachev_files/chrono.pdf for the first one. – darij grinberg Mar 5 at 5:53
John Baez's math.ucr.edu/home/baez/week299.html provides a nice intro to pre-Lie algebra, and F.C.'s "Rooted trees and an exponential-like series" arxiv.org/abs/math/0209104. – Tom
Copeland Mar 16 at 2:34
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A formula similar to Sylvester's:
$$exp(t\;g(z) D)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]=exp[:<z^{-1}exp(t\;g(z) D)z-1>\;z D:].$$
First, given a fct. $\omega=h(z)$, its compostional inverse $z=h^{-1}(\omega)$, and $g(z)=\frac{1}{h^\prime(z)},$ then
$$exp(t\;g(z) \frac{d}{dz})f(z)=exp(t\; \frac{d}{dh(x)})f(x)=exp(t\; \frac{d}{d\omega})f(h^{-1}(\omega))=f(h^{-1}(t+\omega))=f(h^{-1}(t+h(z))).$$
(This expression for the Lie derivative was studied as early as 1857 by Charles Graves.)
In particular with $f(z)=z,$
$$exp(t\;g(z) \frac{d}{dz})z=h^{-1}(t+h(z)).$$
(Note that with $f(z)=z$, $h(0)=0$, $g(0)=1$ and $t=h(y)$, the operator generates a formal group law (FGL), so the operator can be used to simply explore the properties of the FGL in
general. For an easy intro to FGLs, see Olver, Applications of Lie Groups to Differential Equations.)
But also using the generalized shift operator (generalized Taylor-Maclaurin series),
$$exp[:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:]f(z)=f(h^{-1}(t+h(z))),$$
and, in particular,
$$exp[:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:]z=h^{-1}(t+h(z)),$$
where $:ABC:^n=A^nB^nC^n,$ i.e., the power distributes over the operators maintaining their order. This notation allows more succinct and suggestive formulas and shouldn't be confused
with normal ordering although it can give the same result. In this case, $:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:^n=[h^{-1}(t+h(z))-z]^n\frac{d^n}{dz^n}.$
So recursively, using $D=d/dz,$
$$exp(t\;g(z) D)f(z)=f(h^{-1}(t+h(z)))=exp[:[h^{-1}(t+h(z))-z]\; D:]f(z)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]f(z),$$
where <...> denotes evaluation within the symbols.
up vote 2 Removing the intervening steps, we obtain a formula similar to Sylvester's:
down vote
$$exp(t\;g(z) D)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]=exp[:<z^{-1}exp(t\;g(z) D)z-1>\;z D:].$$
A) Let $g(z)=z$, $h(z)=ln(z)$, $h^{-1}(z)=e^z$.
$$exp(t\;z D)=exp[(e^t -1) :zD:],$$
$$exp(t\;z D)f(z)=f((e^t-1)z+z)=f(e^tz).$$
B) With $g(z)=z^2$, $h(z)=-1/z$, $h^{-1}(z)=-1/z$,
$$z^{-1}h^{-1}(t+h(z))-1=\frac{1}{1-t \cdot z}-1,$$
$$exp(t\;z^2 D)=exp[:[\frac{1}{1-t \cdot z}-1] zD:],$$
$$exp(t\;z^2 D)f(z)=f([\frac{1}{1-t \cdot z}-1]z+z)=f[\frac{z}{1-t \cdot z}].$$
Explicit Expansion of LHS:
For expansion of the LHS of the formula in terms of rooted trees, see references in OEIS A139605.
Explicitly for the third order term with $g^j_i=[D^i g(z)]^j$,
$$(g(z)D)^3=(g^1_0 g^2_1+g^2_0g^1_2)D+3g^2_0g^1_1 D^2+g^3_0D^3.$$
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North Hills Calculus Tutor
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I Get Helped!
I was trying to solve this problem and stumbled on a subproblem. What is the probability that one person does NOT see their name when it is in one of 100 identical boxes and they can open 50 of them?
My gut told me there is a 50% chance I don’t see my name, but another part of me said, there is a 99/100 chance I don’t see my name on the first try, a 98/99 chance I don’t see it on the second try,
etc. Should I add or multiply all those numbers? How?
So, I turned to Twitter. Alexander Bogomolny put an explanation on his Facebook page that convinced me my initial 50% hunch was right and helped me see that I wasn’t accounting for the probability
that I have to even try a 2nd, 3rd, 4th, etc. time.
Shawn Urban helped me realize why I wanted to be adding, not multiplying, all those probabilities (when adjusted with Alexander’s suggestion).
Earl Samuelson showed another way to confirm that 50% is the correct answer that fit into my most basic understanding of probability: # of favorable outcomes / # of possible outcomes. The total
number of possible outcomes is all the ways to choose 50 boxes out of 100 (100 C 50 on a graphing calculator). The total number of successful outcomes are all the outcomes where you pick the 1 box
with your name in it from its 1 location (1 C 1) and 49 other wrong boxes out of the remaining 99 (99 C 49). So the probability of picking your name is (1 C 1 * 99 C 49) / (100 C 50) = .5, and the
probability of not picking your name is 1 – .5 = .5
I really, really love it when multiple solutions and ways of reasoning through a problem yield the same result. I often learn something about deeper patterns in math when that happens.
Thanks Twitter!
1. ishan says:
100C50 + 99C50=
2. ishan says:
100C50 +99C49+98C48………………..50C0=
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1006 Submissions
[7] viXra:1006.0069 [pdf] submitted on 30 Jun 2010
An Application of Sondat's Theorem Regarding the Orthohomological Triangles
Authors: Ion Pătraşcu, Florentin Smarandache
Comments: 4 pages.
In this article we prove the Sodat's theorem regarding the orthohomological triangle and then we use this theorem and Smarandache-Patrascu's theorem in order to obtain another theorem regarding the
orthohomological triangles.
Category: Geometry
[6] viXra:1006.0059 [pdf] submitted on 13 Mar 2010
Properties of a Hexagon Circumscribed to a Circle
Authors: Ion Pătraşcu, Florentin Smarandache
Comments: 3 pages.
In this paper we analyze and prove two properties of a hexagon circumscribed to a circle
Category: Geometry
[5] viXra:1006.0058 [pdf] submitted on 13 Mar 2010
[4] viXra:1006.0024 [pdf] submitted on 13 Mar 2010
A Theorem about Simultaneous Orthological and Homological Triangles
Authors: Ion Pătraşcu, Florentin Smarandache
Comments: 13 pages.
In this paper we prove that if P[1],P[2] are isogonal points in the triangle ABC , and if A[1]B[1]C[1] and A[2]B[2]C[2] are their ponder triangle such that the triangles ABC and A[1]B[1]C[1] are
homological (the lines AA[1] , BB[1] , CC[1] are concurrent), then the triangles ABC and A[2]B[2]C[2] are also homological.
Category: Geometry
[3] viXra:1006.0015 [pdf] submitted on 11 Mar 2010
An Economics Model for the Smarandache Anti-Geometry
Authors: Roberto Torretti
Comments: 3 pages
The Smarandache anti-geometry is a non-euclidean geometry that denies all Hilbert's twenty axioms, each axiom being denied in many ways in the same space. In this paper one finds an economics model
to this geometry by making the following correlations: (i) A point is the balance in a particular checking account, expressed in U.S. currency. (Points are denoted by capital letters). (ii) A line is
a person, who can be a human being. (Lines are denoted by lower case italics). (iii) A plane is a U.S. bank, affiliated to the FDIC. (Planes are denoted by lower case boldface letters).
Category: Geometry
[2] viXra:1006.0004 [pdf] submitted on 3 Jun 2010
A Class of Orthohomological Triangles
Authors: Claudiu Coandă, Florentin Smarandache, Ion Pătraşcu
Comments: 5 pages
In this article we propose to determine the triangles' class... (see paper for full abstract)
Category: Geometry
[1] viXra:1006.0003 [pdf] submitted on 3 Jun 2010
The Hyperbolic Menelaus Theorem in The Poincaré Disc Model of Hyperbolic Geometry
Authors: Florentin Smarandache, Catalin Barbu
Comments: 4 pages
In this note, we present the hyperbolic Menelaus theorem in the Poincaré disc of hyperbolic geometry.
Category: Geometry
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Braingle: 'Overlapping Squares' Brain Teaser
Overlapping Squares
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #29898
Category: Probability
Submitted By: cms271828
Corrected By: DaleGriffin
Two squares each 17cm by 17cm are drawn randomly inside a square measuring 1m by 1m.
The edges of the two smaller squares are both parallel/perpendicular to the edges of the large square.
What is the exact probability the two smaller squares overlap (or touch)?
Since the centers of the two smaller squares each lie inside a square of side 83cm, we can simplify the problem to:
Find the probability that the horizontal and vertical distances between 2 points in a square of side 83 cm are less than or equal to 17 cm.
Considering the horizontal distances:
Let x and y be the horizontal distances of the two points from the left edge of the 83 by 83 square.
Clearly x and y can take any value from 0 to 83.
We can represent this space of possibilities using a grid where the lower left corner is (0,0), the horizontal axis represents x, and the vertical axis represents y.
We must work out what part of this space represents the two points being a distance of 17cm or less apart, we can do this by plotting a point where it is true.
Firstly assume y>x, then we must have y<=x+17.
This inequality is simply all the points on and below the straight line from (0,17) to (66,83).
Now assume, y<=x, then we must have y>=x-17.
This inequality is simply all the points on and above the straight line from (17,0) to (83,66).
Combining both inequalities gives a shaded area that lies between 2 right-angled triangles.
The smaller sides of each triangle will be 66 cm each. So the area of both triangles together is 66^2.
Hence the shaded area=83^2-66^2=2533
Hence the probability the distance between x and y is less than or equal to 17 is 2533/83^2=2533/6889.
Now Prob(any 2 points in 83 by 83 square having vertical AND horizontal distance <=17) =
Prob(horizontal distance between points is <=17) * Prob(vertical distance between points is <=17)
=2533/6889 * 2533/6889 (Due to symmetry)
=6,416,089/47,458,321 Hide
What Next?
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A Quantum Approach to Complex Networks’ Structure
Friday, October 12, 2012 at 4:30 pm
DA 5th fl
: Silvano Garnerone, PhD
: University of Waterloo, Canada
I will discuss two recent works about the application of quantum information tools in the context of complex networks. In the first part of the talk I will present a quantum algorithm generating a
quantum state encoding of the PageRank vector. Extensive numerical simulations provide evidences that this algorithm can prepare the quantum PageRank state in a time which typically scales
polylogarithmically in the number of web pages, allowing for a polynomial quantum speed-up in the estimation of the most important entries of the PageRank. We argue that the main topological feature
of the underlying web graph allowing for such a scaling is the out-degree distribution. [This part is based on Phys. Rev. Lett. 108, 230506 (2012)].
In the second part of the talk I will consider quantum walks on directed graphs. Using a large deviation approach I show how to construct a thermodynamic formalism allowing us to define a dynamical
order parameter, and to identify transitions between dynamical regimes. For a particular class of dissipative quantum walks I propose a new quantum generalization of the classical pagerank vector.
[This part is based on arXiv:1205.5744].
Host: Assistant Professor Ginestra Bianconi
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In this tutorial, we will use GeoGebra to approximate the area of a circle. These strategies of approximating the area of a circle was used by the Greek mathematician Archimedes.
Since this tutorial is very long I have split it into two parts. In Part I, we will inscribe a regular polygon in a circle, increase its number of sides, and investigate the relationship between
their areas. In Part II (to be posted next week), we will use the polygon circumscribing the same circle and increase its number of sides to approximate the circle’s area. Before following the
tutorial step-by-step, click here to view the final output.
1. Open GeoGebra. We will need the Algebra window and the Axes so be sure that they are displayed. If not, use the View menu from the menu bar to show them.
2. We will only need to label of the points, so we will not GeoGebra automatically label other objects. To do this, click the Options menu, click Labeling, then click All New Points Only.
3. To create a slider r for the radius of our circle, select the Slider tool, then click anywhere on the drawing pad to display the Slider dialog box.
4. In the Slider dialog box, type r in the Name box, type 0.1 in the min box, and leave the max value as 5 and increment as 0.1, then click the Apply button.
5. Create another slider name it n, set the minimum to 3, maximum to 30 and increment to 1. Slider n, will determine the number of sides of our inscribed polygon.
6. Next, we do not want any new objects to have labels. To do this, click the Options menu, click Labeling then click No New Objects.
7. To construct a circle with center A and radius r, type circle[A,r] . Move slider r and see what happens.
8. To construct the intersection of the circle and the x-axis, type (r,0) in the input box and press the ENTER key.
9. Now, we compute for the central angle of our inscribed polygon. To do this, we divide 360 by n. For example, if we want to have an equilateral triangle, we must divide 360 by 3, which will be our
central angle. To do this, type a = (360/n)° then press the ENTER key. The degree sign, tells GeoGebra that a is an angle measure. You can display the degree sign by selecting it at the drop down
list box right next to the input box.
10. To create angle BAB’, click the Angle with Given Size tool, click point B and click point A. This will display the Angle with Given Size dialog box.
11. In the Angle dialog box, type a in the Angle text box and choose the counter clockwise button and then click the OK button. Your drawing should look like the one shown in Figure 2.
12. To hide the angle measure (green sector), right click it then click Show Object.
13. To construct the inscribed polygon, select the Regular Polygon tool, click B and then click B’. This will display the Regular Polygon dialog box.
14. In the Regular Polygon dialog box, type n. This means, that we want to inscribe a polygon with n sides, then click the Ok button.. Now, drag slider n and see what happens. If you set n to 30,
your drawing should look like the one shown in Figure 3.
15. Our problem now is to hide the labels of all the points and the segments. With n set to 30, right click the polygon, then click Properties from the context menu.
16. In the Properties dialog box, select the Basic tab, click Point (be sure that the Point text is highlighted) in the Objects list, and uncheck the Show Label check box. This will hide the labels
of all the points. Now, click Segment text and uncheck the Show Label check box to hide the labels of all the sides of the polygons.
17. Now, using the text tool, we will display the area of the circle and the area of the inscribed triangle. To display the area of the circle, click the Insert Text tool and click anywhere and type
the following:
“The area of the circle with radius ” + r + “is ” + pi*r^2.
The blue text enclosed by double quotes are constants and will exactly appear as they are. The red texts r and pi*r^2 are variables and will display numbers based on the value of the slider r and
the result of the computation. GeoGebra interprets pi as the mathematical constant which approximately equals 3.1416. Note that constants are always enclosed by double quotes. Constants and
variables are always separated by the + symbol.
18. Use the text tool to display the entire polygon. In the Text tool, type
“The area of the inscribed polygon is “ + poly1
Note that poly1 is the area of our polygon (see the Algebra window). Adjust the positions of the text as needed. Move the sliders and observe what happens.
19. Your drawing should look like the Figure below. (The font of the text has been resized to make it more visible in the drawing).
20. What is the relationship between the area of the circle and the and the area of the inscribed polygon?
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[Strategy] [Sign-ups always open] Risk: Earth III [Game started!] [6 people remain!]
Forums → Forum Games →
[Strategy] [Sign-ups always open] Risk: Earth III [Game started!] [6 people remain!] (locked)
202 posts
Flag Post Risk: Earth 3
Jaskaran2000 The traditional risk game, forum style!
8157 posts
As the game begins, you start out with 5 territories each (No, I won’t be giving bonuses if you sign first). You can sign whenever you want, but starting at a later stage is
disadvantageous. Only exception to this is when there are like 1-3 nations left. In a turn, you can expand your territory, or attack someone and capture their territory, the quantity
of the territories you capture and the battles are all dependent on random.org, much like the real board game, I will be rolling a D10000 dice.
If it ends with something not following this, you get 3 territories.
If it ends with 00, 10, 20, 30, 40, or 50, you get 10 territories.
If it ends with 60, 70, 80, or 90, or a palindrome of three numbers (323, 787) you get 20 territories.
If it ends with a three-digit straight (123, 234, etc.) or triples (111, 222, etc.), you get 50 territories.
If it ends with 69, 404 or 420 you get 80 territories.
If it ends with a four-digit straight (1234, 2345, etc.) or quads (1111, 2222, etc.), you get 100 territories.
If the number is 1337 or 10000 you get 120 territories.
You must tell me in which direction you want to expand, but better be specific about it, because I won’t be making any changes if you’re vague and I’ve done something you didn’t want
me to do.
Everyone gets 15 turns per day. Keep any number of that, and 10% of that number will be removed before I add the 15 turns of the day.
Flag Post First off, I’ll roll the D10000 dice. Then,
Jaskaran2000 If you roll a number that is not below, the new number rolled will be from 1 to 100, with 1 to 50 meaning the attacking player conquers, and 51 to 100 meaning the defending player
8157 posts defends.
If you roll a number ending in 00, 10, 20, 30, 40, or 50, the new number rolled will be from 1 to 110, with 1 to 60 meaning the attacking player conquers, and 61 to 110 meaning the
defending player defends.
If you roll a number ending in 60, 70, 80, or 90, the new number rolled will be from 1 to 120, with 1 to 70 meaning the attacking player conquers, and 71 to 120 meaning the defending
player defends.
If you roll a number that is straight such as 123, 234, etc., or triples the new number rolled will be from 1 to 150, with 1 to 100 meaning the attacking player conquers, and 101 to
150 meaning the defending player defends.
If you roll a number ending in 69, 420 or 404, the new number rolled will be from 1 to 180, with 1 to 130 meaning the attacking player conquers and 131 to 180 meaning the defending
player defends.
If you roll a number that is straight such as 1234, 2345, etc., or quads, the new number rolled will be from 1 to 200, with 1 to 150 meaning the attacking player conquers, and 151 to
200 meaning the defending player defends.
If you roll the numbers 1337 or 10000, the new number rolled will be from 1 to 220, with 1 to 170 meaning the attacking player conquers, and 171 to 220 meaning the defending player
Person ~ Kingdom ~ Colo(u)r
Hamuka ~ Hamham’s Almighty Empire ~ Dark Green
Flag Post Thijsel ~ Green Nation of Doom ~ Green
Yiu113 ~ Derptopia ~ Pink
Jaskaran2000 [DEL:Helltank ~ Empire of Helltank ~ Black:DEL] Eliminated
8157 posts [DEL:Zzzip50 ~ Anonymous ~ Silver:DEL] Voluntary Resignation
Myhome16 ~ Siberian Siberia ~ Red
[DEL:2Cents ~ Kortex ~ Cyan:DEL] Eliminated
Djrockstar ~ DEMACIA ~ Brown
[DEL:Behemoth542 ~ Twin Raider Empire ~ Orange:DEL] Eliminated
Bloodyrain10001 ~ Dueling Network ~ Dark Blue
[DEL:Calebmock ~ Republic of Dread ~ Yellow:DEL] Eliminated
I) Answer the following questions (only one chance):
a) What is my sister’s age? (+2 terr)
Flag Post b) In which month was my dog born? (+1 terr)
c) Which Operating System do I use? (+1 terr)
8157 posts Want to join? Just fill this application out and post it in the thread!
Name of kingdom: (=<18 characters)
Starting place: (If you want to start randomly, say so. Don’t just leave it blank)
Flag Post
EDIT: Oops.
hamuka 4621
posts Name of kingdom: Hamham’s Almighty Empire
Color: Dark Green
Starting place: East-Central Europe (or Hungary, more accurately)
Flag Post Sign
Empire name: TieSoul’s Awesome Green Nation of Doom.
thijsel 352 Color: Green
Flag Post Empire: Derptopia
Color: Pink
yiu113 4486 Starting Place: Lower coast of Australia.
(Also you did something werid with the thing and said that a palindrome/ 60,70 etc. gives you ten territories. Gives you 20 territories.)
Flag Post Name of Kingdom:The Empire of Helltank
Helltank Starting Place:Australia
9677 posts
Flag Post Thijsel, please choose a starting area.
Jaskaran2000 @yiu: You managed to see it in the five seconds it took me to fix it.
8157 posts
Flag Post Originally posted by Jaskaran2000:
hamuka 4621 Hamuka, thijsel and yiu, please choose a starting area.
Did that 4 minutes ago :/
Flag Post Originally posted by Jaskaran2000:
yiu113 4486 @yiu: You managed to see it in the five seconds it took me to fix it.
… Winning?
Flag Post
yiu, let’s do an alliance. Agree not to attack me and I’ll agree not to attack you. If he approves, I want to make a special request to Jask so I will not attack yiu under any
Helltank circumstances even if I expand in the direction he is at.
9677 posts
Flag Post
It would be lovely if people who sign from this point on do not exceed empire name character limit of 18, sorry for any inconvenience. Oh, and thij, please shorten your empire name
Jaskaran2000 too.
8157 posts
Flag Post
thijsel 352 Empire name: Green Nation of Doom
posts Color: Green
Starting Place: Netherlands
Flag Post Name of kingdom: Anonymous
Color: Silver
Zzzip50 7124 Starting place: (’MURICA)
Flag Post
@Zip: I’ve added your kingdom to the USA, since I wasn’t sure whether you meant North America, South America, or USA. USA is in NA, so I went with that.
8157 posts
Flag Post Yeah, sorry I didn’t make that clear.
Zzzip50 7124 From this point onwards I shall address the USA as ’MURICA.
Flag Post Name of kingdom: Siberian Siberia
Color: Red for Russia
myhome16 Starting place: SIBERIA
4916 posts
Originally posted by Jaskaran2000:
Flag Post Risk: Earth 3
Firespread The traditional risk game, forum style!
7691 posts
cough this is stolen from 4chan uncough
Flag Post
@Fire: The previous game was too, I don’t see a problem with it. Can’t give credits because anonymous.
8157 posts
I did the last one that died, so…
Flag Post
Name of Kingdom: Kortex
2Cents 189 Color: Cyan
posts Starting place: Istanbul, Turkey
And I shall do the same sign-up :)
Flag Post Name of Kingdom: DEMACIA
Color: Brown
djrockstar Starting Place: Qatar
1921 posts
Flag Post
hey, we passed the amount of players of the previous game :O
thijsel 352
Flag Post The game has begun!
Jaskaran2000 Everyone: 15
8157 posts
Flag Post Spend 5 of my movement to get into South America.
yiu113 4486 If I succeed, spread north in South America with 5 more movement.
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Posts from January 2012 on Mir Books
Monthly Archives: January 2012
In the Little Mathematics Library now we come to Systems of Linear Inequalities by A. S. Solodovnikov. This booklet is one of longest in the LML series, having more than 120 pages. The back cover of
the book says the … Continue reading
After a break we being with our postings again. To start, we will see a title in the Little Mathematics Library which is a`natural continuation’ of last post, this one is called Induction in Geometry
and the authors are L. … Continue reading
In the Little Mathematics Library we now come to the book called The Method of Mathematical Induction by I. S. Sominsky (aka Sominskii). In the foreword it is said: The method of mathematical
induction, which is the subject of this … Continue reading
We now come to another book in the Little Mathematics Library titled Stereographic Projection by B. A. Rosenfeld and N. D. Sergeeva. As the title suggests the book deals with projections on planes.
The present booklet is devoted to proofs … Continue reading
In the Little Mathematics Library we now come to Method of Successive Approximations by N. Ya. Vilenkin. As the title suggests the book has to do with approximation methods, but what kind of
approximations and for what kind of use … Continue reading
Posted in little mathematics library, mathematics, mir publishers Tagged achilles, chords, Derivatives, Geometrical Meaning, iterations, mappings, mathematics, square roots, succesive approximations,
tortoise 3 Comments
After the last two posts by V. A. Uspenskii (check out his site here) which dealt with Post’s Machine, and Pascal’s Triangle, we now come to another book by him in the Little Mathematics Library
series titled Gödel’s Incompleteness Theorem. … Continue reading
In continuing from the last post on Post’s Machine by V. A. Uspenskii (sometimes Uspensky) we come to another volume by him titled Pascals’s Triangle. The book opens with an interesting note The
reader who is not familiar with Pascal’s … Continue reading
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Count within 1000; skip-count by 5s, 10s, and 100s. *TNCore Focus Standard
This particular standard expands on students’ prior knowledge of basic counting skills. Students should recognize, read, and write numerals within 1000, as well as count on in a variety of sequencing
patterns anywhere in the range of numbers from 1 to 1000. Students should be able to count on by 5s, 10s, and 100s. One motivation for this standard is that students will be able to develop stategies
for addiiton and subtraction. For example, a student who wants to add 784 + 58 can start at 784 and count five tens: 794, 804, 814, 824, 834. Then add eight ones: 835, 836, 837, ...., 842. This
technique can also be helpful in determining a missing addend. Suppose we want to solve the equation 479 + ? = 520. We can do this by counting up from 479 until we get close to 520: 489, 499, 509,
519. So far we have counted up by 10 four times. Counting up by one then produces 520, so we have added four tens + one = 41.
Understanding the Standard:
• Provide students with opportunities to count by ones, and then to skip-count by more difficult intervals (by 5, 10, 100), starting at any number less than 1000.
• Provide students with opportunities to understand and practice skip-counting by using base-ten blocks to illustrate the process of adding tens. .
• Provide students with opportunities to regroup numbers into ones, tens and hundreds to aid them in skipcounting..
Questions to Focus Instruction:
• Can students count on fluently beginning at any whole number less than 1000?
• Can students skip-count fluently beginning at any whole number less than 1000 when given a specific sequencing interval (for example, by 5, 10, or 100)?
Prior to:
Students can count, recognize and extend sequencing patterns up to 120.
Go to 1.NBT.1 to see previous skills in this progression
At Grade Level:
Students will be expected to count in a variety of sequences to 1,000. They will recognize and read the word name and digit form of these numbers as well as model them with concrete objects, such as
base 10 models.
Moving Beyond:
Mastery of this skill is further demonstrated in a student’s ability to fluently perform all four operations with numbers. This standard helps students to find additional strategies for addition and
subtraction through the development of their skip-counting skills from any number.
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Theater of Pompey - Theatrum Pompei ProjectUntitled
Pompey Resources
Primary Sources
Latin Designations
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Topographic Dictionary
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Communis Opinio
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Building Activity
PDCS Home Page
An Abridgement Based on T. S. R. Broughton's Magistrates of the Roman Republic vol. 2 and 3 (Atlanta: Scholars Press, 1952, 1986).
Return to Rome During the Fifties B.C.
Jump to:
59 B.C., A.U.C. 695 [Return to Top of Page]
Consuls: C. Iulius C. f. C. n. Caesar Pat. (Cos. 48, 46-44, Pr. 62): See MRR 2. 187-188.
M. Calpurnius C. f. Bibulus (Pr. 62)
Praetors: T. Ampius Balbus
L. Appuleius Saturninus
Cn. Cornelius Lentulus Spinther Clodianus Pat.: Presided at trial of C. Antonius, prob. in Quaestio de maiestate.
Q. Fufius Calenus (Cos. 47): Carried law to require separate reports of the votes of the three classes of jurors.
T. Vettius Sabinus: Quaestio de repetundis. Presided at trial of L. Valerius Flaccus for extortion.
Aed. Cr.: ? C. Licinius Murena
? C. Visellius Varro
Aed. Pl.: ? L. Calpurnius Bestia
Iudex Q.: P. Licinius Crassus (Pr. 57): L. Vettius was indicted before him for a breach of peace.
Tr. Pl.: C. Alfius Flavus: Supporter of Caesar, Pompeius, and Crassus.
Q. Ancharius (Pr. 56): One of 3 tribunes who opposed Caesar, Pompeius, and Crassus.
Q. Caecilius Metellus Pius Scipio Nasica (Cos. 52, Pr. 55)
C. Cosconius (Pr. 54?)
Cn. Domitius Calvinus (Cos. 53, 40, Pr. 56): One of 3 tribunes who opposed Caesar, Pompeius and Crassus.
C. Fannius (Pr. 55?): One of 3 tribunes who opposed Caesar, Pompeius, and Crassus.
? P. Nigidius Figulus (Pr. 58)
P. Vatinius (Cos. 47, Pr. 55): Chief supporter of Caesar, Pompeius, and Crassus. See MRR 2.190 for legislation.
Quaest.: L. Aemilius Lepidus Paullus Pat. (Cos. 50, Pr. 53): Served under C. Octavius in Macedonia.
M. Favonius (Pr. 49)
Promag.: ? L. Afranius (Cos. 60): Perhaps Proconsul in Cisalpine Gaul.
Cn. Cornelius Lentulus Marcellinus Pat. (Cos. 56, Pr. 60): Governor, perhaps Proconsul, of Syria.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Governor, perhaps Proconsul, of Nearer Spain.
? L. Culleolus: Governor of Illyricum, before Caesar.
C. Octavius (Pr. 61): Proconsul in Macedonia. Saluted as Imperator for his victory over the Bessi.
C. Papirius Carbo (Pr. 62): Governor of Bithynia and Pontus.
? Q. Pompeius Rufus (Pr. 63): Proconsul in Africa.
C. Pomptinus (Pr. 63): Vatinius refused to recognize validity of the supplicationes for his victory in Gaul.
Q. Tullius Cicero (Pr. 62): Proconsul in Asia.
C. Vergilius Balbus (Pr. 62): Propraetor in Sicily.
Legates: L. Aelius Tubero: Served as Legate under Q. Cicero in Asia.
A. Allienus (Pr. 49): Served as Legate under Q. Cicero in Asia.
M. Gratidius: Served as Legate under Q. Cicero in Asia.
Sp. Cm.: 1. Board of Twenty to assign land under Caesar's agrarian laws:
M. Attius Balbus (Pr. by 59)
Cn. Pompeius Magnus (Cos. 70, 55, 52)
M. Terentius Varro (Pr. -)
Cn. Tremellius Scrofa (Pr. -)
2. Board of Five with judicial also:
M. Valerius Messalla Niger Pat. (Cos. 61)
? C. Cosconius (Pr. 63): Died before July 59.
? M. Licinius Crassus (Cos.70, 55, Pr. 73?): It is uncertain as to which board he belonged, if not to both.
Augurs: Q. Caecilius Metellus Celer (Cos. 60, Pr. 63): ? - 59. Died.
Successor: Unknown.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Functioned as augur at adoption of Clodius into plebeian family.
58 B.C., A.U.C. 696 [Return to Top of Page]
Consuls: L. Calpurnius L. f. L. n. Piso Caesoninus (Pr. 61?)
A. Gabinius A. f. (Pr. 61)
Praetors: L. Cornelius Lentulus Crus Pat. (Cos. 49)
L. Domitius Ahenobarbus (Cos. 54): With C. Memmius, he attacked Caesar's acta. Discussed measure for Cicero's recall.
C. Fabius Hadrianus
L. Flavius: Came into conflict with Clodius when Tigranes was taken from his custody.
C. Memmius: With Domitius, he attacked Caesar's acta. Attempted to try Vatinius for violation of lex Iunia Licinia.
P. Nigidius Figulus
L. Villius Annalis
[The following are possibilities: See MRR 2.192 for further details.]
C. Considius Longus
Q. Minucius Thermus
P. Silius Nerva
Cn. Tremellius Scrofa
Aed. Cr.: M. Aemilius Scaurus Pat. (Pr. 56): Gave aedilician games of special magnificence.
P. Plautius Hypsaeus (Pr. 55)
Iudex Q.: ? C. Visellius Varro
Tr. Pl.: Aelius Ligus: Associated himself with Clodius & vetoed Ninnius's proposal for Cicero's recall.
L. Antistius: Attempted to prosecute Caesar for his actions while Consul, but was restrained by his colleagues.
P. Clodius Pulcher: See MRR 2.195-6.
L. Ninnius Quadratus: Supported Cicero's cause. Also attempted to prevent one of Clodius's men from celebr. Ludi Compitales.
L. Novius [Niger?]: Took cognizance of a supposed attempt by Clodius on Pompeius's life.
Q. Terentius Culleo: Proposed to annul law exiling Cicero and attempted to draw Pompeius away from his Caesarian associates.
Quaest.: C. Calpurnius Piso Frugi: Assigned to Pontus and Bithynia, he gave up his province in order to aid in Cicero's recall, his father-in-law.
Cn. Plancius: Served in Macedonia under L. Appuleius Saturninus and aided Cicero during his exile.
Promag.: T. Ampius Balbus (Pr. 59): Proconsul in Asia.
L. Appuleius Saturninus (Pr. 59): Propraetor in Macedonia.
Cn. Cornelius Lentulus Marcellinus Pat. (Cos. 56, Pr. 60): Governor of Syria for two years.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): See MRR 2.197.
C. Pomptinus (Pr. 63): Remained outside of Rome awaiting confirmation of a triumph.
M. Porcius Cato (Pr. 54): Appointed under lex Clodia as Quaestor pro Praetore toannex Cyprus and restore exiles at Byzantium.
Q. Tullius Cicero (Pr. 62): Proconsul in Asia, whence he returned in May.
C. Vergilius Balbus (Pr. 62): Propraetor in Sicily. Debarred by terms of lex Clodia from giving Cicero a refuge.
T. Vettius Sabinus (Pr. 59?): Governor of Africa.
Legates: M. Mettius: Caesar sent him to Ariovistus who made him a prisoner.
C. Valerius Procillus: Caesar sent him to Ariovistus who made him a prisoner.
? C. Claudius Pulcher Pat. (Pr. 56): Brother of P. Clodius in charge of Caesar's troops in Italy in early 58.
T. Labienus (Pr.-): Served under Caesar as Legatus pro praetore.
? L. Aurunculeius Cotta: Legate under Caesar in Gaul.
Q. Pedius (Cos. Suff. 43, Pr. 48): Legate under Caesar in Gaul.
? Ser. Sulpicius Sabinus: Legate under Caesar in Gaul.
? Q. Titurius Sabinus: Legate under Caesar in Gaul.
? P. Vatinius (Cos. 47, Pr. 55): Legate under Caesar in Gaul.
Prefects: P. Licinius Crassus: In command of cavalry against Ariovistus.
Pontif.: L. Pinarius Natta: Recently elected Pontifex who officiated at dedication of shrine to Libertas on site of Cicero's house.
Flam.: Sex. Iulius Caesar Pat.: Flamen Quirinalis.
57 B.C., A.U.C. 697 [Return to Top of Page]
Consuls: P. Cornelius P. f. Cn. n. Lentulus Spinther Pat. (Pr. 60): Worked for recall of Cicero and carried bill in comitia centuriate for his recall.
Q. Caecilius P. f. Q. n. Metellus Nepos (Pr. 60): Cooperated w/colleague in Cicero's recall & in giving Pompeius charge of grain supply.
Praetors: C. Caecilius Cornutus: Aided in securing Cicero's recall from exile.
L. Caecilius Rufus: Pr. Urbanus.
M. Calidius.
Ap. Claudius Pulcher Pat. (Cos. 54): Quaestio de repetundis. Didn't join his colleagues in supporting Cicero's recall.
P. Licinius Crassus
Sex. Quinctilius Varus Pat.
C. Septimius
Q. Valerius Orca
Aed. Cr.: ? Q. Caecilius Metellus Pius Scipio Nasica: Gave splendid gmaes in honor of his adoptive father.
Q. Fabius Maximus Pat. (Cos. Suff. 45, Pr. 48): Restored the Fornix Fabinaus.
Aed. Pl.: ? L. Calpurnius Bestia: Defended by Cicero from a charge of ambitus.
C. Cosconius (Pr. 54)
Tr. Pl.: T. Annius Milo (Pr. 55): Cooperated with Sestius, he organized gangs to fight ag. Clodius, while working for Cicero's recall.
Sex. Atilius Serranus Gavianus: Opposed Cicero's recall.
C. Cestilius: Aided in Cicero's recall.
M. Cispius (Pr. af. 54): He was later condemned for bribery.
M'. Curtius Peducaenus (Pr. 50)
Q. Fabricius: Led in attempting to carry on Jan. 25 the bill of the eight Tribunes for Cicero's recall.
T. Fadius [Gallus?]
C. Messius: Offered bill for Cicero's recall at the beg. of his tribunate & in Sept. proposed bill to give P. vast powers over grain supply.
Q. Numerius Rufus (Gracchus): With Serranus he opposed the moves for Cicero's recall.
P. Sestius: With Milo he worked most actively for Cicero's recall.
Promag.: L. Calpurnius Piso Caesonius (Cos. 58, Pr. 61): Proconsul in Macedonia with unlimited imperium. He was proclaimed imperator.
C. Fabius Hadrianus (Pr. 58): Proconsul in Asia
A. Gabinius (Cos. 58, Pr. 61?): Proconsul in Syria with unlimited imperium. He was refused a supplicatio for his successes.
C. Iulius Caesar (Cos. 59, 48, 46-44, Pr. 62): Proconsul of both Gallic provinces and Illyricum. Honored with supplicatio for 15 days.
C. Memmius (Pr. 58): Governor of Bithynia and Pontus, with Catullus and Cinna on his staff.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Received charge of grain supply with imperium pro consule for 5 years with 15 legates.
C. Pomptinus (Pr. 63): Remained outside of Rome awaiting triumph.
M. Porcius Cato (Pr. 54): Quaestor pro praetore in charge of annexation of Cyprus and the restoration of Byzantine exiles.
T. Vettiuis Sabinus (Pr. 59): He may have remained as governor of Africa.
Legates: L. Aurunculeius Cotta: Legate under Caesar in Gaul.
? Cornelius Sisenna: Stepson and probably Legate or Prefect under Gabinius in Syria.
T. Labienus (Pr. -): Legatus pro praetore under Caesar in Gaul.
? P. Licinius Crassus: Served under Caesar in Gaul.
Q. Marcius Crispus (Pr. 46): Legate under Piso in Macedonia.
Q. Pedius (Cos. Suff. 43, Pr. 48): Legate of Caesar in Gaul.
? Servilius or Servianus: Served under Gabinius in Syria.
Ser. Sulpicius Galba Pat. (Pr. 54): Legate under Caesar in Gaul.
Q. Titurius Sabinus: Legate under Caesar in Gaul.
M. Tullius Cicero (Cos. 63, Pr. 65): Legate under Pompeius.
Q. Tullius Cicero (Pr. 62): Legate under Pompeius; served in Sardinia beg. in autumn of 57.
L. Valerius Flaccus (Pr. 63): Legate under Piso in Macedonia.
P. Vatinius (Cos. 47, Pr. 55): Legate under Caesar in Gaul.
C. Vergilius Balbus (Pr. 62): Legate under Piso in Macedonia.
Prefects: M. Antonius (Cos. 44, 34, Cos. Desig. 31): Commander of cavalry under Gabinius in Syria.
Pontif.: C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Pontifex Maximus
P. Cornelius Lentulus Spinther (Cos. 57, Pr. 60)
P. Servilius Vatia Isauricus (Cos. 79, Pr. 90)
M. Terentius Varro Lucullus (Cos. 73, Pr. 76)
Q. Caecilius Metellus Creticus (Cos. 69, Pr. 74)
M'. Acilius Glabrio (Cos. 67, Pr. 70)
M. Valerius Messala Pat. (Cos. 61, Pr. 64?)
L. Cornelius Lentulus Niger Pat. (Pr. bef. 60): Flamen Martialis.
P. Sulcipius Galba Pat. (Pr. bef. 65)
Q. Caecilius Metellus Pius Scipio Nasica (Cos. 52, Pr. 55)
C. Fannius (Pr. 55?)
M. Aemilius Lepidus Pat. (Cos. 46, 42, Pr. 49)
L. Claudius Pulcher Pat.: Rex Sacrorum.
M. Aemilius Scaurus Pat. (Pr. 56)
M. Licinius Crassus (Cos. 70, 55) or M. Licinius Crassus
C. Scribonius Curio (Cos. 76, Pr. 80?)
Sex. Iulius Caesar Pat.: Flamen Quirinalis.
Q. Cornelius: Pontifex Minor.
P. Albinovanus: Pontifex Minor.
Q. Terentius Culleo: Pontifex Minor.
Augurs: P. Cornelius Lentulus Spinther Pat.: Son of Consul of 57, adopted into the family of the Manlii Torquati to make him eligible.
Faustus Cornelius Sulla Pat.: Augur before 57.
56 B.C., A.U.C. 698 [Return to Top of Page]
Consuls: Cn. Cornelius P. f. Lentulus Marcellinus Pat. (Pr. 60): He opposed candidacy of Crassus and Pompeius for consulships of 55.
L. Marcius Philippus (Pr. 62)
Praetors: M. Aemilius Scaurus Pat.: Quaestio de vi. Presided at trial of Sestius.
Q. Ancharius
C. Claudius Pulcher Pat.: Attempted to prevent removal of tablet on Capitoline under which Cicero was exiled.
Cn. Domitus Calvinus (Cos. 53, 40): Quaestio de ambitu. Presided at trial of Bestia de ambitu and perhaps Caelius de vi.
Aed. Cr.: P. Clodius Pulcher: See MRR 2.208.
? M. or C. Claudius Marcellus (Cos. 51, or 50, or 49): Probably to be identified with Consul of 50, C. Marcellinus.
Iudex Q.: ? Cn. Domitius: See MRR 2.208 for uncertainties as to his identity.
Antistius Vetus
L. Caninius Gallus: Sought to give Pompeius, rather than Lentulus Spinther, the duty of restoring Ptolemy Auletes.
M. Nonius Sufenas: Prosecuted with C. Cato in 54 for his part in obstructing elections in this year.
Cn. Plancius
A. Plautius (Pr. 51?): Read to Senate a letter from Ptolemy asking to be restored by Pompeius and two lictors.
C. Porcius Cato: Supported neither Lentulus or Pompeius with regards to restoring Ptolemy.
L. Procilius: Prosecuted and convicted in 54 for his part in delaying elections in this year.
L. Racilius: Supported optimate stand ag. Clodius.
P. Rutilius Lupus: Attacked Caesar's agrarian law and supported motion to have Pompeius restore Ptolemy.
Quaest.: ? M. Coelius Vinicianus (Pr. 48)
Promag.: C. Caecilius Cornutus: Governor of Bithynia and Pontus.
Q. Caecilius Metellus Nepos (Cos. 57, Pr. 60): Proconsul in Nearer Spain where he dealt with a rising of the Vaccaei.
L. Caecilus Rufus (Pr. 57): Proconsul of Sicily.
L. Calpurnius Piso Caesoninus (Cos. 58, Pr. 61?): Proconsul in Macedonia. It was decided in this year to send him a succesor in 55.
Ap. Claudius Pulcher Pat. (Cos. 54, Pr. 57): Governor of Sardinia.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proc. in Cilicia. Commissioned in late 57 to restore Ptolemy until oracle was read.
A. Gabinius (Cos. 58, Pr. 61?): Prconsul in Syria. Prepared for expedition against the Parthians.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proconsul in two Gallic provinces and Illyricum.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul in charge of the grain supply. See MRR 2.211.
C. Pomptinus (Pr. 63): Remained outside of Rome awaiting triumph.
M. Porcius Cato (Pr. 54): Quaestor pro praetore to annex Cyprus and restore Byzantine exiles. Returned in 56.
Sex. Quinctilius Varus Pat. (Pr. 57): Proconsul in Farther Spain.
C. Septimius (Pr. 57): Proconsul in Asia.
Q. Valerius Orca: Proconsul in Africa.
[Unknown governors for Crete and Cyrene.]
Tr. Mil.: C. Volusenus: Served under Caesar in Gaul
Legates: T. Silius: Pref. or Tr. Mil. of equestrian rank sent by Crassus to collect grain from Veneti, Esbuii, and Curiosolites. Arrested and held.
T. Terrasidius: Pref. or Tr. Mil. of equestrian rank sent by Crassus to collect grain from Veneti, Esbuii, and Curiosolites. Arrested & held.
M. Trebius Gallus: Pref. or Tr. Mil. of equestrian rank sent a Crasse to collect grain from Veneti, Esbuii, & Curiosolites. Arrested & held.
Q. Velanius: Pref. or Tr. Mil. of equestrian rank sent by Crassus to collect grain from Veneti, Esbuii, and Curiosolites. Arrested and held.
L. Aurunculeiuis Cotta: Legate under Caesar.
T. Labienus: Legatus pro praetore under Caesar. In charge of Rhine region.
? P. Licinius Crassus: Sent to Aquitania by Caesar, where he waged a successful campaign. Returned to R. with veterans for elections.
Q. Marcius Crispus (Pr. 46): Legate under Piso in Macedonia.
? Q. Pedius (Cos. Suff. 43, Pr. 48): Legate under Caesar in Gaul.
? Ser. Sulpicius Galba Pat. (Pr. 54): Legate under Caesar in Gaul. He may have returned to Rome in 56.
Q. Titurius Sabinus: Legate under Caesar in Gaul, where he crushed a rising of the Veneli, Curiosolites and the Lexovii.
Q. Tullius Cicero (Pr. 62): Legate under Pompeius, in Sardinia whence he returned in June.
L. Valerius Flaccus Pat. (Pr. 63): Legate under Piso in Macedonia.
P. Vatinius (Cos. 47, Pr. 55): Legate under Caesar, but in R. while Legate.
C. Vergilius Balbus: Legate under Piso in Macedonia.
Prefects: D. Iunius Brutus Albinus: Prefet of Caesar's fleet against the Veneti.
M. Antonius (Cos. 44, 34, Cos. Desig. 31)
Pontif.: L. Pinarius Natta Pat.: ? - 56.
Augurs: L. Licinius Lucullus Ponticus (Cos. 74, Pr. 78): ? - 56.
Flam. M.: L. Cornelius Lentulus Niger Pat. (Pr. bef. 60): bef. 69-56.
Successor: ? L. Cornelius Lentulus Pat.
Sept. Ep.: Cn. Cornelius Lentulus Marcellinus Pat. (Cos. 56)
Luperci: M. Caecilus Rufus (Pr. 48)
L. Herennius Balbus
55 B.C., A.U.C. 699 [Return to Top of Page]
Consuls: Cn. Pompeius Cn. f. Sex. n. Magnus (Cos. 70, 52): See MRR 2.214.
M. Licinius P. f. M. n. Crassus (Cos. 70, Pr. 73?): See MRR 2.214.
Censors: M. Valerius M. f. M'. f. Messalla Niger Pat. (Cos. 61, Pr. 64?): After destructive flood, he and his colleague tried to regulate the Tiber.
P. Servilius C. f. M. n. Vatia Isauricus (Cos. 79, Pr. 90): Still in office w/his colleague as of July 54, but lustrum was not completed.
Praetors: T. Annius Milo
? Q. Caecilius Metellus Pius Scipio Nasica (Cos. 52)
? P. Plautius Hypsaeus
P. Vatinius: Elected over Cato through collusion of Crassus and Pompeius.
Aed. Cr.: ? L. Aemilius Paullus Pat. (Cos. 50, Pr. 53): Undertook repair of Basilica Aemilia.
? Nonius Struma
Aed. Pl.: C. Messius: Celebrated the Floralia as Aedile.
Tr. Pl.: P. Aquilius Gallus: With Ateius Capito, he opposed plans of Pompeius and Crassus. Tried to prevent passage of lex Trebonia.
C. Ateius Capito (Pr. -): W/Gallus he opposed Pompeius & Crassus. Cursed the departure of Crassus from the city.
C. Trebonius (Cos. 45, Pr. 48): Carried against all obstruction law giving provinces to P. & Cr., each for five years.
Mamilius: Author of lex Mamilia Roscia Alleian Peducaea Fabia.
L. Roscius Fabatus (Pr. 49): Author of lex Mamilia Roscia Alleian Peducaea Fabia.
A. Allienus (Pr. 49): Author of lex Mamilia Roscia Alleian Peducaea Fabia.
Sex. Peducaeus: Author of lex Mamilia Roscia Alleian Peducaea Fabia.
C. Fabius: Author of lex Mamilia Roscia Alleian Peducaea Fabia.
Quaest.: ? P. Licinius Crassus
? L. Minucius Basilus
? C. Sallustius Crispus.
Interrex: M. Valerius Messala Pat. (Cos. 61, Pr. 64?)
Promag.: M. Aemilius Scaurus Pat. (Pr. 56): Governor of Sardinia, probably proconsul. Returned in 54 to unsuccessful prosecution.
Q. Ancharius (Pr. 58): Proconsul in Macedonia in succession to Piso.
Q. Caecilius Metellus Nepos (Cos. 57, Pr. 60): Proconsul in Nearer Spain, where he faced a rising of the Vaccaei.
L. Calpurnius Piso Caesoninus (Cos. 58, Pr. 61?): Proconsul in Macedonia, whence he returned before late summer.
C. Claudius Pulcher Pat.: Proconsul in Asia.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proconsul in Cilicia.
A. Gabinius (Cos. 58, Pr. 61?): In spring of 55 he restored Ptolemy Auletes to the throne of Egypt. Condemned for extortion in 54.
C. Iuluis Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proconsul in two Gallic provinces and Illyricum.
C. Pomptinus (Pr. 63): Continued to wait for his triumph.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul in charge of the grain supply.
? Sex. Quinctilius Varus Pat. (Pr. 57): May have remained as Proconsul in Farther Spain until Pompeius b/c governor.
Legates: L. Aurunculeius Cotta: Legate under Caesar in Gaul.
T. Labienus: Legatus pro praetore under Caesar in Gaul.
? Q. Numerius Rufus: Legate under Caesar.
? M. Plaetoriuis Cestianus (Pr. 64): Went to Lentulus Spinther in Cilicia.
P. Suplicius Rufus Pat. (Pr. 48): Legate under Caesar in Gaul.
Q. Titurius Sabinus: Legate under Caesar in Gaul.
L. Afranius (Cos. 60, Pr. 72?): Legate of Pompeius, sent to Spain bef. end of 55.
M. Petreius (Pr. bef. 63): Legate of Pompeius, sent to Spain bef. end of 55.
Prefects: M. Antonius (Cos. 44, 34, Cos. Desig. 31): Served under Gabinius in Syria and aided in restoring Ptolemy.
Augurs: P. Licinius Crassus: Successor of L. Licinius Lucullus.
54 B.C., A.U.C. 700 [Return to Top of Page]
Consuls: L. Domitius Cn. f. Cn. n. Ahenobarbus (Pr. 58): Opposed Caesar, Gabinius, Pompeius.
Ap. Claudius Ap. f. Ap. n. Pulcher Pat. (Pr. 57): Became reconciled with Cicero.
Praetors: C. Alfius Flavus: Presided over trial of Gabinius for maiestas. May have been Quaesitor, not Praetor.
? M. Claudius Marcellus (Cos. 51)
? C. ? Cosconius
? Domitius: Quaestio de vi. Presided over second prosecution of Caelius Rufus.
? Fonteius: Pr. Urbanus.
M. Porcius Cato: Quaestio de repetundis. Presided over trial of Scaurus.
P. Servilius Isuaricus (Cos. 48, 41): Brought C. Messius into court although he was legate of Caesar.
Ser. Sulpicius Galba Pat: Enabled his old commander Pomptinus to secure his triumph.
L. Aelius Tubero: Possibly praetor in this year. No conclusive evidence.
M. Aurelius Cotta: Possibly praetor in this year. No conclusive evidence.
M. Considius Nonianus: Possibly praetor in this year. No conclusive evidence.
C. Fannius: Possibly praetor in this year. No conclusive evidence.
L. Postumius: Possibly praetor in this year. No conclusive evidence.
P. Sestius: Possibly praetor in this year. No conclusive evidence.
Voconius: Possibly praetor in this year. No conclusive evidence.
Aed. Cr.: Cn. Plancius
A. Plautius Plotius (Pr. 51)
Tr. Pl.: D. Laelius: Aided Gabinius after his conviction.
C. Memmius: Pressed against Gabinius charge of extortion. Probably prosecuted Domitus Calvinus for bribery.
Q. Mucius Scaevola: Used obnuntiatio to delay elections. Opposed triumph of Pomptinus.
Terentius: Vetoed the bill to prosecute candidates for consulship who were involved in bribery scandals.
Quaest.: Fautus Cornelius Sulla Pat.
M. Licinius Crassus: Served under Caesar in Gaul.
? T. Ligarius: Quaestor urbanus. Worked with Cicero in promoting Caesar's interests.
? C. Scribonius Curio: May have served under C. Claudius Pulcher in Asia.
L. Sestius Pansa: Quaestor in Asia, probably proquaestor.
Promag.: Q. Ancharius (Pr. 56): Proconsul in Macedonia.
C. Claudius Pulcher Pat.: Proconsul in Asia.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proconsul in Cilicia.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proconsul in both of the Gallic provinces and Illyricum.
M. Licinius Crassus (Cos. 70, 55, Pr. 73?): Proconsul in Syria under lex Trebonia.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul of Spains under lex Trebonia and in charge of grain supply.
C. Pomptinus (Pr. 63): With aid of Pr. Galba and Cos. Ap. Claudius finally secured his triumph ex Allbroges.
Tr. Mil.: M Curtius Postumus (Pr. 47, 46)
Q. Laberius Durus: Killed in battle in Britian.
Petronius: Served under Crassus in Syria.
Legates: L. Afranius: Legate under Pompeius in Spain.
L. Aurunculeius Cotta: Legate under Caesar. Killed during revolt of the Euburones.
C. Fabius: Legate under Caesar in Gaul.
? A. Hirtius: Served under Caesar in Gaul.
T. Labienus (Pr. -): Legatus pro praetore unde Caesar in Gaul. In charge of Gaul during Caesar's exped. to UK.
P. Licinius Crassus: Joined with his father in Syria in winter of 54-53 with 1000 Gallic horse.
C. Messius: Legate of Caesar but returned to R. by edict of Pr. P. Servilius.
L. Munatius Plancus (Cos. 42, Pr. 45?): Legate under Caesar in Gaul.
M. Petreius (Pr. 64): Legate under Pompeius in Spain.
? L. Roscius Fabatus: Served under Caesar in Gaul.
P. Sulpicius Rufus: Served under Caesar in Gaul.
Q. Titurius Sabinus: Legate under Caesar in Gaul.
C. Trebonius (Cos. Suff. 45, Pr. 48): Legate under Caesar in Gaul.
Q. Tullius Cicero: Legate under Caesar in Gaul.
Octavius: Legate of Crassus in Syria.
Vargunteius: Legate of Crassus in Syria.
Prefects: Q. Atrius: Prefect to guard Caesar's fleet during his expedition to Britian.
53 B.C., A.U.C. 701 [Return to Top of Page]
Consuls: Cn. Domitius M. f. M. n. Calvinus (Cos. 40, Pr. 56): Entered office with colleague in July.
M. Valerius Messalla Rufus Pat. (Pr. 62?): Attempts to hold elections for 52 frustrated by Milo & Hypsaeus.
Praetors: L. Aemilius Lepidus Paullus Pat. (Cos. 50)
? P. Attius Varus
? C. Claudius Marcellus (Cos. 50)
? Q. Minucius Thermus
Aed. Cr.: ? M. Aemilius Lepidus Pat. (Cos. 46, 42, Pr. 49)
Tr. Pl.: M. Coelius Vinicianus (Pr. 48?): Together with Hirrus he proposed that Pompeius be made Dictator.
P. Licinius Crassus Dives Iunianus: Cicero dissuaded him from joining Hirrus in proposing Dictatorship for Pompeius.
C. Lucilius Hirrus: With Vinicianus proposed that Pompeius be made Dictator.
Quaest.: C. Cassius Longinus (Pr. 44): Served under Crassus in Syria as Quaestor. Escaped from Carhae, organized defense of Syria.
M. Iunius Brutus (Pr. 44): Refused to serve under Caesar in Gaul and accompanied father-in-law Ap. Claudius Pulcher to Cilicia.
Interrex: Q. Caecilius Metellus Pius Scipio Nasica (Cos. 52, Pr. 55)
M. Valerius Messalla Niger Pat. (Cos. 61, Pr. 64)
Promag.: C. Claudius Pulcher Pat. (Pr. 56): Proconsul in Asia.
Ap. Claudius Pulcher Pat. (Cos. 54, Pr. 57): Proconsul in Cilicia.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proconsul. Returned to R. and remained cum imperio while awaiting triumph.
C. Cosconius C.f. (Pr. 54?): Proconsul in Macedonia.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44): Proconsul of two Gallic provinces and Illyricum under lex Pompei Licinia.
M. Licinius Crassus (Cos. 70, 55, Pr. 73): Proconsul in Syria under lex Trebonia. His invasion of Parthia ended in disaster at Carrhae.
M. Licinius Crassus: Contined to serve under Caesar in early 53 in Gaul.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul of both Spains by lex Trebonia. Probably also remained in charge of grain supply.
C. Scribonius Curio: With C. Claudius Pulcher in Asia.
Tr. Mil.: Petronius: Served under Crassus in Syria and died defending him.
Legates: L. Afranius (Cos. 60, Pr. 72?): Legate to Pompeius in Spain.
C. Antistius Reginus: Legate of Caesar in Spain.
C. Fabius: Legate of Caesar in Gaul.
Q. Fabius Vergilianus: Legate of Ap. Claudius in Cilicia.
M. Iunius Silanus: Legate of Caesar in Gaul.
T. Labienus: Legatus pro praetore under Caesar in Gaul.
P. Licinius Crassus: Served with his father against Parthians and fell in battle at Carrhae.
? Marcus Censorinus: Senate who died with younger Crassus at Carrhae.
? Megabocchus: Served and fell with younger Crassus at Carrhae.
L. Minucius Basilus: Cavalry commander, probably Legate, under Caesar in Gaul.
? Q. Mucius Scaevola: Legate under Ap. Claudius in Cilicia.
L. Munatius Plancus: Legate under Caesar in Gaul.
Octavius: Legate under Crassus at Carrhae. Killed defending Crassus from capture.
? M. Octavius Cn. f.: On staff of Ap. Claudius in Cilicia.
M. Petreius (Pr. 64?): Legate of Pompeius in Spain.
T. Sextius (Pr. bef. 44): Legate of Caesar in Gaul.
P. Sulpicius Rufus Pat. (Pr. 48): Legate under Caesar.
C. Trebonius (Cos. Suff. 45, Pr. 48): Legate under Caesar in Gaul with special command ag. Eburones.
Q. Tullius Cicero (Pr. 62): Legate under Caesar in Gaul.
? C. Valerius Flaccus L. f.: On staff of Ap. Claudius in Cilicia.
Prefects: ? Coponius: In command of Roman garrison at Carrhae.
? Egnatius: In command of body of 300 cavalry which escape Carrhae.
? C. Trebonius: Roman equites in command of vexillum of the legion under Q. Cicero.
C. Volcatius Tullus: Placed in charge of guard at bridge over Rhine.
Pontif.: C. Scribonius Curio (Cos. 76, Pr. 80): ca. 60 - 53.
Successor: C. Scribonius Curio.
M. Licinius Crassus (Cos. 70, 55): ca. 60- 53. If he was pontifex, his spot became vacant.
Augurs: P. Licinius Crassus: ca. 56-53. Died at Carrhae.
Successor: M. Tullius Cicero (Cos. 63, Pr. 65): Nominated by Pompeius and Hortensius.
52 B.C., A.U.C. 702 [Return to Top of Page]
Consuls: Cn. Pompeius Cn. f. Sex. n. Magnus (Cos. 70, 55): See MRR 2.234.
Q. Caecilius Q. f. Q. n. Metellus Pius Scipio Nasica (Pr. 55?)
Praetors: ? C. Claudius Marcellus (Cos. 49)
? M. Nonius Sufenas
? P. Silius
Cn. Tremellius Scrofa
Aediles: ? M. Aufidius Lurco
M. Favonius
Tr. Pl.: M. Caelius Rufus (Pr. 48): Supported Milo and joined Cicero in defense of M. Saufeius.
Q. Manilius Cumans: Joined Caelius in restoring to Milo a slave being kept as witness in home of Triumvir Capitalis.
T. Munatius Plancus Byrsa: Supporter of Clodius. Prominent in disorders af. his death. Prosecuted by Cicero in 52.
Q. Pompeius Rufus: Supporter of Clodius. Prominent in disorders af. his death. Prosecuted by Caelius in 52 and sent into exile.
C. Sallustius Crispus (Pr. 46): Opposed Milo, supported colleagues Plancus and Rufus.
[All ten tribunes joined together to pass law permitting Caesar to stand for consulship in absentia.
Quaest.: ? L. Ateius Capito (Pr. -)
M. Antonius (Cos. 44, 34, Cos. Desig. 31): Chosen without the lot, he served under Caesar in Gaul.
L. Caecilius Metellus: Quaestor in Sicily.
Q. Cassius Longinus: Chosen, not allotted, he served under Pompeius in Spain.
M. Eppius
Interrex: M. Aemilius Lepidus Pat. (Cos. 46, 42, Pr. 49): First interrex.
Ser. Sulpicius Rufus Pat. (Cos. 51, Pr. 65): Interrex who presided over election of Pompeius as sole consul.
M. Valerius Messalla Niger Pat. (Cos. 61, Pr. 64?)
Quaestit.: Considius: Presided over trial of Saufeius under the lex Plautia de vi.
L. Domitius Ahenobarbus (Cos. 54, Pr. 58): Presided over the trial of Milo under the lex Pompeia de vi.
L. Fabius: Presided over the conviction in absence of Milo, under a law (the Plautian?) de vi.
M. Favonius (Pr. 49): Presided over the conviction of Milo, presumably in absentia, under the Lex Licinia de sodaliciis.
A. Manlius Torquatus Pat. (Pr. 70): Presided over the choice of prosecutor and conviction of Milo de ambitu.
Promag.: ? P. Attius Varus: Governor of Africa a few years before 49 B.C.
C. Cassius Longinus (Pr. 44): Proquaestor in Syria.
Ap. Claudius Pulcher Pat. (Cos. 54, Pr. 57): Proconsul in Cilicia.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proconsul and Imperator awaiting his triumph outside of Rome.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proconsul in both Gallic provinces and Illyricum under lex Licinia Pompeia.
? Q. Minucius Thermus (Pr. 53 or by 58?): Propraetor in Asia.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul of both Spains under lex Trebonia but his command was prorogued for 5 more years.
Tr. Mil.: M. Aristius: Served under Caesar in Gaul.
Legates: L. Afranius (Cos. 60, Pr. 72?): Continued to serve under Pompeius in Gaul.
C. Antistius Reginus: Legate under Caesar in Gaul.
? M. Antonius (Cos. 44, 34, Cos. Desig. 31): Legate of Caesar in Gaul.
C. Caninius Rebilus (Cos. Suff. 45, Pr. 48?): Legate of Caesar in Gaul.
C. Fabius: Legate under Caesar in Gaul.
Q. Fabius Vergillianus: Legate under Ap. Claudius in Cilicia.
L. Iulius Caesar Pat. (Cos. 64): Legate of Caesar in Gaul, in charge of Narbonese province.
T. Labienus (Pr. -): Legatus pro praetore under Caesar in Gaul.
L. Minucius Basilus (Pr. 45): Served under Caesar in Gaul.
L. Munatius Plancus (Cos. 42, Pr. 45?): Legate under Caesar in Gaul.
P. Nigidius Figulus (Pr. 48): Legate.
? M. Octavius: On staff of Appius Claudius in Cilicia.
M. Petreius (Pr. 64): Legate under Pompeius in Spain.
M. Sempronius Rutilius: Served under Caesar in Gaul.
T. Sextius: Legate under Caesar in Gaul.
P. Sulpicius Rufus Pat. (Pr. 48): Legate under Caesar.
C. Trebonius (Cos. Suff. 45, Pr. 48): Legate under Caesar in Gaul.
Q. Tullius Cicero (Pr. 62): Legate under Caesar in Gaul.
C. Valerius Flaccus Pat. Legate under Ap. Claudius in Cilicia.
C. Volcatius Tullus: Legate under Caesar.
Prefects: D. Iunius Brutus Albinus (Cos. Desig. 42, Pr. 48): Served under Caesar in Gaul.
M. Scaptius: Appointed by Ap. Claudius in Cilicia as Prefect of cavalry to collect debts owed to Brutus by Salamis.
? C. Volusenus Quadratus: Sent to put Commius the Atrebatian to death.
Pontif.: C. Scribonius Curio: Elected in succession to his father.
51 B.C., A.U.C. 703 [Return to Top of Page]
Consuls: Ser. Sulpiciuis Q. f. Rufus Pat. (Pr. 65)
M. Claudius M. f. M. n. Marcellus (Pr. 54?): Strongly anti-Caesarian opposed by both colleague and Pompeius.
Praetors: M. Iuventius Laterensis: Quaestio de repetundis.
A. Plautius: Praetor Urbanus.
Tr. Pl.: C. Caelius: Vetoed anti-Caesarian resolution of the Senate.
P. Cornelius: Vetoed anti-Caesarian resolution of the Senate.
C. Vibius Pansa (Cos. 43, Pr. 48): Vetoed several anti-Caesarian resolutions of the Senate.
L. Vinicius: Vetoed anti-Caesarian resolution of the Senate.
Quaest.: C. Antonius (Pr. 44): One of three brothers of whom Marcus was Quaest. in 52, and Lucius in 50.
? T. Furfanius Postumus: Quaestor in Sicily.
Furius Crassipes: Served under Silius in Bithynia and Pontus.
L. Mescinius Rufus: Served under Cicero in Cilicia.
Canini Sallustius: Served as Proquaestor in 50 under Bibulus in Syria. Often confused with Sallust the historian.
Promag.: M. Antonius (Cos. 44, 34, Cos. Desig. 31): Continued to serve under Caesar in Gaul.
M. Calpurnius Bibulus (Cos. 59, Pr. 62): Proconsul in Syria, where he arrived late in year after Cassius repelled Parthian invasion.
C. Cassius Longinus (Pr. 44): Proquaestor in Syria. Led Parthian invaders into ambush at Antioch.
Ap. Claudius Pulcher Pat. (Cos. 54, Pr. 57): Procos. in Cilicia, leaving in August. Claimed as Imperator a triumph, but backed down.
C. Considius Longus (Pr. bef. 57? or 52?): Governor of Africa in 50 and probably in 51.
P. Cornelius Lentulus Spinther Pat. (Cos. 57, Pr. 60): Proconsul and Imperator; celebrated triumph from Cilicia in latter part of 51.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proc. & Imperator in two Gallic provinces and Illyricum under lex Pompeia-Licinia.
Q. Minucius Thermus (Pr. bef 57? or 53?): Propraetor in Asia.
M. Nonius Sufenas (Pr. 52?): Governor of a province in East, probably Crete or Macedonia.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul in both Spains. Continued to remain in Italy.
P. Silius (Pr. bef. 57? or 52?): Propraetor in Bithynia and Pontus.
Cn. Tremellius Scrofa (Pr. bef. 57? or 52?): Governor of a province in the East.
M. Tullius Cicero (Cos. 63, Pr. 66): Proconsul and was acclaimed Imperator in October after victory over tribesmen of Amanus.
Tr. Mil.: Q. Fufidius: Served under Cicero in Cilicia.
Sex. Lucilius: Fell in battle while serving under Bibulus in Syria.
Legates: D. Antonius: Sent by Cicero to Ap. Claudius, the retiring governor of Cilicia.
L. Afranius: Legate under Pompeius in Spain.
M. Anneius: Legate under Cicero in Cilicia.
? C. Antistius Reginus: Legate under Caesar in Gaul.
C. Caninius Rebilus (Cos. Suff. 45, Pr. 48?): Legate under Caesar in Gaul.
C. Fabius: Legate under Caesar in Gaul.
C. Fabius Vergillianus: Legate under Caesar in Gaul.
Q. Fufius Calenus (Cos. 47, Pr. 59): Legate under Caesar in Gaul.
L. Iulius Caesar Pat. (Cos. 64): Legate under Caesar in Gaul.
T. Labienus (Pr. -): Legatus pro praetore unde Caesar in Gaul.
Q. Ligarius: Legate under Considius in Africa.
L. Minucius Basilus (Pr. 45): Legate under Caesar in Gaul.
L. Munatius Plancus (Cos. 42, Pr. 45?): Legate under Caesar in Gaul.
P. Nigidius Figulus (Pr. 58): Legate in Asia, presumably under Minucius Thermus and left Asia in July 51.
? M. Octavius: On staff of Ap. Claudius in Cilicia
M. Petreius: Legate under Pompeius in Spain.
C. Pomptinus: Legate under Cicero in Cilicia.
T. Sextius: Legate under Caesar in Gaul.
P. Sulpicius Rufus Pat. (Pr. 48): Legate under Caesar.
T. Titius: Legate in a province imp. for its grain supply, probably af. conclusion of Pompeius' term in charge of supply.
C. Trebonius (Cos. Suff. 45, Pr. 48): Legate under Caesar in Gaul.
L. Tullius: Legate under Cicero in Cilicia.
Q. Tullius Cicero (Pr. 62): Legate under brother in Cilicia.
? C. Valerius Flaccus Pat.: On the staff of Ap. Claudius in Cilicia.
P. Vatinius (Cos. 47, Pr. 55): Legate under Caesar in Gaul.
? C. Volcatius Tullus: Legate under Caesar in Gaul.
Prefects: D. Antonius: Praefectus evocatorum under Cicero in Cicilia.
Q. Atius Varus: Praefectus equitum under Caesar in Gaul.
L. Clodius: Praefectus fabrum under Appius Claudius in Cilicia.
Q. Lepta: Praefectus fabrum under Cicero in Cilicia.
Q. Volusius: Sent to Cyprus by Cicero to attend to the litigation of the Roman citizens there.
C. Volusius Quadratus: Praefectus equitum under Caesar in Gaul.
15Viri: P. Cornelius Dolabella Pat. (Cos. Suff. 44): Elected this year, defeating L. Cornelius Lentulus Crus.
Addendum: The following men were witnesses to resolution of the Senate reported by Caelius to Cic. in Fam. 8.8.5-6:
L. Domitius Ahenobarbus (Cos. 54)
Q. Caecilius Metellus Pius Scipio Nasica (Cos. 52)
L. Villius Annalis: Pr. before 57.
C. Septimius (Pr. 57)
C. Lucilius Hirrus (Tr. Pl. 53)
C. Scribonius Curio (Tr. Pl. 50)
L. Ateius Capito (Pr.-): Ex-Quaestor in 51.
M. Eppius: Ex-Quaestor in 51.
50 B.C., A.U.C. 704 [Return to Top of Page]
Consuls: L. Aemilius M. f. Q. n. Lepidus Paullus Pat. (Pr. 53): Tended to support Caesar.
C. Claudius C. f. M. n. Marcellus (Pr. by 53): Opponent of Caesar.
Censors: Ap. Claudius Ap. f. Ap. n. Pulcher Pat. (Cos. 54, Pr. 57): Zealous in his duties, removing, among others, Sallust from the Senate.
L. Calpurnius L. f. L. n. Piso Caesoninus (Cos. 58, Pr. 61?): More moderate, protected Curio from his colleague.
Praetors: ? M. Considius Nonianus
C. Curtius Peducaenus
? L. Postumius
? L. Scribonius Libo
C. Titius L. f. Rufus: Pr. Urbanus.
M. Livius Drusus Claudianus: President of a court to which cases of violation of the Lex Scantinia were brought. May have been iudex.
Aed. Cr.: M. Caelius Rufus (Pr. 48): Quarrelled with Censor Ap. Claudius. Prosecuted persons who were diverting aqueduct water.
M. Octavius
Tr. Pl.: C. Furnius (Pr. 42): Cicero relied on him to prevent prorogation of his provincial command.
C. Scribonius Curio: Elected in place of Servaeus who had been convicted of bribery in the election. See MRR 2.249.
Quaest.: T. Antistius: Quaestor in Macedonia.
L. Antonius (Cos. 41): Served under Minucius Thermus in Asia. Upon leaving the province Thermus left him in command.
C. Coelius Caldus: Served under Cicero in Cilicia. Left in command when Cicero returned to R.
Promag.: M. Aurelius Cotta (Pr. 55? or 54?): Gov. of Sardinia in early 49.
M. Calpurnius Bibuls (Cos. 59, Pr. 62): Proconsul of Syria. Honored with supplicatio. His sons were murdered in Egypt this year.
C. Considius Longus (Pr. before 57? or 52?): Returned from Africa late in 50 to convass for consulship.
T. Furfanius Postumus (Pr. 46): Proquaestor in Sicily.
C. Iulius Caesar Pat. (Cos. 59, 48, 46-44, Pr. 62): Proc. of both Gallic provinces + Illyricum. Crisis over successor erupted at year's end.
L. Mescinius Rufus: Served under Cicero in Cilicia and left provinces with him.
Q. Minucius Thermus (Pr. b/f 57? and 53?): Propraetor in Asia, whence he returned to Italy this year.
M. Nonius Sufenas (Pr. 52?): Governor of a province in East, probably Crete or Macedonia.
Cn. Pompeius Magnus (Cos. 70, 55, 52): Proconsul in both Spanish provinces. He continued to govern by legates.
Canini Sallustius: Served as Proquaestor under Bibulus in Syria.
P. Silius (Pr. 52?): Propraetor in Pontus and Bithynia.
Cn. Tremellius Scrofa (Pr. 52?): Governor of an eastern province, perhaps Macedonia.
M. Tullius Cicero (Cos. 63, Pr. 65): Proconsul in Cilicia and Imperator. Supplicatio was decreed for his victory. Left Cilicia in June 50.
Tr. Mil.: Q. Fufidius: Served under Cicero in Cilicia.
M. Scaptius: He first accepted then refused this position from Cicero while attending business for Brutus in Cappadocia.
Legates: A. Hirtius (Cos. 43, Pr. 46): Caesar sent him to Rome early in Dec. to meet Balbus and Metellus Scipio, but he hurried back w/o meeting.
L. Afranius (Cos. 60, Pr. 72?): Legate under Pompeius in Spain.
M. Anneius: Legate under Cicero in Cilicia.
? C. Antistius Reginus: Legate under Caesar.
C. Fabius: Legate under Caesar in Gaul.
Q. Fufius Calenus: Legate under Caesar in Gaul.
L. Iulius Caesar Pat. (Cos. 64): Legate under Caesar in Gaul.
T. Labienus (Pr. -): Legatus pro praetore under Caesar in Gaul who received charge of Cisalpine Gaul in 50.
Q. Ligarius: Legate under Considius Longus in Africa, left in command when Considius returned to Rome.
L. Minucius Basilus: Legate under Caesar in Gaul.
L. Munatius Plancus (Cos. 42, Pr. 45?): Legate under Caesar in Gaul.
M. Petreius (Pr. 64): Legate under Pompeius in Spain.
C. Pomptinus (Pr. 63): Legate under Cicero in Cilicia, but returned to Italy before him.
? T. Sextius (Pr. b/f 44): Legate under Caesar in Gaul.
P. Sulpicius Rufus Pat. (Pr. 48): Legate under Caesar.
M. Terentius Varro (Pr. -)
C. Trebonius (Cos. Suff. 45, Pr. 48): Legate under Caesar in Gaul, placed in charge of winter-quarters in Belgic Gaul.
L. Tullius: Legate under Cicero in Cilicia.
Q. Tullius Cicero (Pr. 62): Legate under Cicero in Cilicia.
P. Vatinius (Cos. 47, Pr. 55): Legate under Caesar in Gaul.
Veiento: Legate left in command of Syria by Bibulus.
C. Volcatius Tullus: Legate under Caesar in Gaul.
Prefects: L. Gavius: Received prefecture from Cicero in order to attend to business for Brutus in Cappadocia.
Q. Lepta: Praefectus fabrum under Cicero in Cilicia.
M. Scaptius: Received praefecture from Cicero in orrder to attend to business in Cappadocia for Brutus.
Q. Volusius: Served under Cicero in Cilicia.
Pontif.: L. Domitius Ahenobarbus (Cos. 54, Pr. 58): Elected af. 57.
M. Iunius Brutus (Pr. 44): ?-42. Colleague of Metellus Scipio.
Augurs: Q. Hortensius Hortalus (Cos. 69, Pr. 72): 67-50.
Successor: M. Antonius (Cos. 44, 34, Cos. Desig. 31)
Augurs before Death of Hortensius:
? Q. Cassius Longinus: Probably before 55.
C. Claudius Marcellus (Pr. 80)
Ap. Claudius Pulcher Pat. (Cos. 54, Pr. 57)
P. Cornelius Lentulus Spinther Pat.
Faustus Cornelius Sulla Pat.: Elected before Lentulus Spinther.
Q. Hortensius (Cos. 69, Pr. 72)
L. Iulius Caesar Pat. (Cos. 64)
? L. Marcius Philippus (Cos. Suff. 38, Pr. 44)
Cn. Pompeius Magnus (Cos. 70, 55, 52)
? P. Servilius Isauricus (Cos. 48, 41, Pr. 54)
Ser. Sulpicius Galba Pat. (Pr. 54)
Q. Mucius Scaevola
M. Tullius Cicero (Cos. 63, Pr. 65)
M. Valerius Messalla Rufus Pat. (Cos. 53, Pr. 62): Augur for 55 years.
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Yahoo Groups
An example 4x4x4 BLD solve
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Hey everyone,
So yeah the work on the webpage is not proceding very well due to my
jobs :-(
I really don't want to be the guy that says I'll write up some info
but then never does. So, here is an example solve of 4x4x4 BLD -
literally every single thing that goes through my head when I do a solve.
Below is the scramble and the description of everything about the
solve. Before you scramble please hold the cube in the same
orienation you consider to be the "solved" orientation. This will
make it much easier to follow along with what I am doing.
F B R2 b r D' B' d2 l' D2 b L' U2 L2 f d' F' b l2 U' d2 L' F' D2 B2 L
f2 F2 d' f' b d B2 b r2 D2 f U2 d l
I actually memorized and solved this scramble before typing up this
example to make sure I would write up exactly what I do, even during a
real solve. The overall time was 8:41.85 minutes, which is pretty
much right at my average lately, which has been in the 8:40's. The
memorization step took 4:33 minutes and the solving then took 3:58
After the scramble I always rotate the cube around to find the
position with the most centers already solved. After scrambling the
cube I did the rotation z' x'. If you don't like xyz notation I first
rotated the whole cube counterclockwise as if I was doing the move F'
and I then rotated the whole cube as if I was doing the move R'
After the cube rotation I now have the following cycles on the cube:
lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU -> buL -> flU -> dbL ->
urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF -> lbU
1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -> dBL
-> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
2) (rUF <-> dFL)
3) (bUL -> fDL -> dBR -> rDF -> bUL)
I always memorize and solve in this order for the 4x4x4:
1) memorize center permutation
2) memorize edge permutation
3) memorize corner permutaiton
4) memorize corner orientation
5) solve corner orientation
6) solve corner permutation
7) solve center permutation
8) solve edge permutation
During the memorization of this solve I debated memorizing and solving
with center blocks on the F and R faces, but decided against it
because I had already memorized one of the pieces I would have made a
block out of into an image, and didn't want to rewrite the first
journey location any.
MEMORIZATION STEP
center cycle: lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU -> buL ->
flU -> dbL -> urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF -> lbU
center cycle in letters: A -> I -> E -> V -> O -> W -> B -> Q -> C ->
S -> F -> K -> D -> H -> L -> X -> G -> A
The journey I was using was halfway through one of my 28 location
journeys. This particular journey is a walk from my dormitory my
junior year of college through to my class on the other side of campus.
At each location I need to put at least three images, sometimes more
which does happen on this solve during the edge memorization.
Let me walk through each location one at a time:
Location 1:
This location is outside the dining hall with a bunch of tables and
At this location I have the letters AIEVO. I take the very first
letter and give it a single letter image (famous person). This is so
I know which piece is the first piece of every one of my three cycles.
The letter is "A" so the image is "Dan Aykroyd".
Ok, so now I have to take the letters in pairs after the first image
is made. The next two are IE which is the Internet Explorer logo,
only about 6 feet tall. The next two letters are VO which is a 2 foot
tall Volcano that is spewing and erupting a huge column of ash.
I put these images in a left-to-right order. So on the left Dan
Aykroyd is rolling a giant 6 foot tall model of the Internet Explorer
logo onto the top of the two foot tall volcano to clog it up and make
it stop erupting ash everywhere.
That's it. I don't really view the motion, it is just implied. I see
the tables and chairs and the dining hall very clearly in the
background, and in the foreground I see Dan Aykroyd (motionless) in
the act of rolling the Internet Explorer logo onto the volcano. I
picture the logo about 3/4 the way up the volcano, so that the volcano
is still to the right of the Internet Explorer logo, which is to the
right of Dan Aykroyd.
Location 2:
This location is beside a wall about 20 feet away from the dining hall
on the way to a set of stairs.
At this location I have the letters WBQCSF. I now take letters in
pairs always to make double images. The letters WB form the image of
a giant 6 foot tall spider web between two large tree limbs. The
letters QC form the image of a two foot tall Quaich. The letters SF
form an image of a 4 foot tall model of the San Francisco Golden Gate
At this location I can very clearly see the wall, as well as the hand
railing leading to the stairwell that is the next location (all in the
background). In the foreground I very clearly see (in left to right
order) a giant web where a Quaich is falling into it having rolled off
the end of the miniature San Francisco Golden Gate bridge.
Location 3:
This location is at the bottom of the stair case pictured at the last
location. I view it from the top of the staircase, looking at the
space just after the last step at the bottom. There are maybe 6 steps
to this staircase.
The letters at this location are KDHLXG. KD forms the image of Billy
the Kid the famous gunslinger. HL form the image of Satan as
described by Dante in Dante's Inferno. The only difference is my
image of him is about 5 feet tall instead of the massive proportions
Dante gives him. The letters XG form the image of Gambit from the
X-men (the guy who charges stuff with energy and the object then
explodes after a few seconds).
This image is graphic, but this is exactly how I memorize. I picture,
in left to right order, Billy the kid drawing his pistol and quickly
unloading all 6 rounds into Satan, who is trapped in the ice as in
Dante's Inferno. This doesn't kill Satan, but only wounds him. To
the right of Satan Gambit charges a playing card (his usual weapon)
and places it in the mouth of Satan. When the card explodes it
explodes Satan's head and blood and guts are thrown absolutely everywhere.
Yes this image is graphic, but because of that it was very easy to
memorize. Anything funny or violent tends to be easier to hold onto.
BLAST DOOR
Yes that literally is a blast door, like for a bomb shelter. I
imagine a 200 foot tall steel blast door dropping down onto my journey
in the space between the 3rd location and the 4th. I image that in
order to get past the door I have to fly over it as if I were in a
helicopter, and I literally picture this flight over the blast door.
This is my marker that the centers are now done - i.e. when I get to
the blast door during the solving phase, I know that what follows is
now information for the edges.
edge cycles:
1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -> dBL
-> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
2) (rUF <-> dFL)
3) (bUL -> fDL -> dBR -> rDF -> bUL)
edge cycles as letters:
1) (Q -> F -> E -> S -> D -> H -> L -> M -> R -> U -> C -> B -> P -> X
-> J -> T -> A -> O -> Q)
2) (I <-> W)
3) (N -> G -> V -> K)
Location 4:
I am now memorzing the edge permutation. This location is at a "+"
shaped intersection of two very large walkways leading into one of the
quads on my school that is a very large enclosed rectangular space
formed by about 8 buildings. I can clearly see the open space of the
quad, as well as the two walkways at this location.
The letters at this location are QFESD. I need to know which piece
will be my main piece, or the one that will be the first piece in all
of my cycles, so I take the first letter "Q" and make it into an
image. The letter Q goes with the image "Queen". I now take the
letters in pairs after the first piece. The next two letters FE form
the image of a 10 foot tall Ferris Wheel. The next two letters SD
form the image of a 10 foot long Star Destroyer (like from Star Wars)
that is floating in the air about 6 feet above the ground.
In left to right order I see the Queen hiding behind a miniature
Ferris Wheel while a miniature Star Destroyer bombards it with lasers
and explosives of all kinds. The ferris wheel is on fire and so
decrepit it is about to collapse.
Location 5:
This location is part-way down a pedestrian walkway that goes into the
center of the large quad we are now in. I can picture a big tree and
a small wall to the right.
At this location I have the letters HLMRUC. HL stands for the Satan
as portrayed by Dante again. MR stands for Marvin the paranoid
android from Hitchiker's Guide to the Galaxy (as portrayed in the
movie). UC stands for a Unicorn. In left to right order I see Satan
frozen in the ice trying to frighten Marvin the android, who could not
be more unimpressed. Satan gets pissed off about this, so he posesses
a unicorn just to the right of (behind) Marvin and has the Unicorn ram
it's horn through Marvin's head, effectively killing him.
Location 6:
This location is the center of the quad, where the two largest
walkways meet at a sort of tall skinny "X" shape. I can see the
buildings on the other side of the quad as well as a very big tree to
the left at this location.
Here I have the letters BPXJTA. BP stands for a backpack which is
oversized (about 6 feet tall) and filled with 100's of regular sized
books. XJ is the image of Jubilee from the X-men. TA is the image of
a tank with it's gun turret pointed menacingly at the object next to it.
In left to right order I see a backpack lying on the ground, filled to
the brim with books. Jubilee is using her sparks to completely
destroy the backpack which is now on fire and billowing smoke
everywhere. Behind her a tank aims and takes a shot from it's huge
gun turret but Jubilee ducks out of the way just in time.
I normally stop after three images, but this time I see that the next
piece, the "O" piece, goes to the "Q" spot and completes this cycle.
Instead of starting at a new location I tack on my single letter "O"
image (Ozzy Osbourne) behind the tank, as if he were orchestrating the
whole incident of the tank firing on Jubilee.
This cycle is now finished. I know I am not done memorizing edges
though, because I have only used three journey locations, and the
average is 4 for edges. During the actual solve I looked around for
an edge I had not used yet when I spotted the two cycle (rUF <-> dFL)
or in my letters that is (I <-> W). Rather than start this image at
the next journey location, since a two cycle is always just one image,
I tack it on to the current location just behind Ozzy Osbourne, only
surrounded by curtains on both the left and right. This is my
miniature, within a journey location, form for the "BLAST DOOR". It
lets me know that the image within the curtains is a two cycle. I got
this idea from John Louis, so thank you very much! The image for IW
is a "Y-Wing" space fighter from Star Wars. Within curtains I imagine
the Y-Wing taking aim at Ozzy Osbourney (through the curtains) to take
him out and make him stop firing on Jubilee with the Tank.
OK so I know I have used a 3 locations, and a little more since there
are 5 images at my 3rd location instead of the usual 3. I know that I
am not done, but I have no idea how many edges are left to look for.
So I go over my images (just for edges). I sort of half use this as
rehearsal for my edges and half to just count them and find out how
many edges I have memorized right now. I counted 20 edges when I went
back through my edge images. I now glance quickly at the cube to see
that no edges are already in their solved locations - I saw none. So
I know that there are 4 edges somewhere to look for. I find them and
they are in one cycle, which I put at my last location.
Location 7:
This location is on the other side of the quad. I am very close to
the buildings that before were only in the background.
At this location I have the 4 cycle (N->G->V->K->N). I take the N and
make it into a single letter image, that of Chuck Norris. The GV I
put together as a giant Gavel, about 6 feet tall. Since I only have
one piece left, the K, I make it into a single letter image of Chris
Katan from Saturday Night Live.
In left to right order I see Chuck Norris roundhouse kicking a giant
Gavel so hard that it slams into Chris Katan and knocks him over.
****END OF JOURNEY****
I now know that I am done memorizing the centers and edges, since I
have accounted for all 24 edges. I now memorize the Corner
permutation visually and the Corner orientation visually. Then I put
on the blindfolded.
SOLVING PHASE
Solving the corners is pretty uninteresting I think, since you can
just use the same stuff as from the 2x2x2 or 3x3x3. The only change
is that you *HAVE* to use supercube safe algs (algs that don't rotate
center caps if done on a supercube).
Here is how I solved the corners:
Corner orientation:
y' R' U2 R U R' U R L U2 L' U' L U' L' followed by U' R U R' U' R U R'
L' U' R U R' U' R U R' L' U' R U R' U' R U R' L2 y
or if you don't prefer xyz notation the above is equivalent to F' U2 F
U F' U F B U2 B' U' B U' B' followed by U' F U F' U' F U F' B' U' F U
F' U' F U F' B' U' F U F' U' F U F' B2
Corner permutation:
1) UBL->DFL->UFL : D F2 L' B' L F2 L' B L D'
2) UBL->UFR->DBL : (L B' L' B)x3 D2 (R F' R' F)x3 D2
3) (UBL <-> DBR) and (UBR <-> DFR) : F2 D (L' B L B')x3 D' F2
Ok time to actually solve the centers now.
First location: Dan Aykroyd->Internet Explorer->Volcano
I start with Dan Aykroyd then I take the first image, Internet Explorer.
Those are the letters AIE which is the cycle lbU->ufR->ulF :
1) To solve I use the commutator: [f' U' f, u']
I am here using the same notation I learned from Stefan and Per. [f'
u' f, u'] means to perform the first part, then the second part, then
the inverse of the first part, then the inverse of the second part.
So [f' u' f, u'] = (f' u' f) (u') (f' u f) (u) = f' u' f u' f' u f u
2) After Internet Explorer comes VO. So the cycle is now A->VO or
lbU->rbD->drB : D' r' [U', b d b'] r D
3) Aykroyd->Web or A->WB or lbU->lfD->rbU : [b' R' b, l2]
4) Aykroyd->Quaich or A->QC or lbU->ubL : [U, l u' l']
5) Aykroyd->San Francisco or A->SF or lbU->dbL->urF : L2 [f U2 f', u] L2
6) Aykroyd->Billy the Kid or A->KD or lbU->dfR->rfU : [U2, b d2 b']
7) Aykroyd->Satan or A->HL or lbU->drF->bdR : [r U2 r', d']
8) Aykroyd->Gambit or A->XG or lbU->rfD->dlF : r2 [b d' b', U'] r2
Centers are now solved. Notice that only three of the three cycles
required setup moves. One of the three cycles required two. Average
number of setup moves needed before being able to perform a commutator
was 4/7 or 0.57 moves.
Ok now I realize that I have run into the blast door. So I know that
the next stuff I recall deals with edges.
The first location has me recalling Queen, then the very next image is
a Ferris Wheel.
1) Queen->Ferris Wheel or Q->FE or lBU->fUR->fUL : [R B' R', f]
2) Queen->Star Destroyer or Q->SD or lBU->lDB->uFR : R [U', B d B'] R'
3) Queen->Satan or Q->HL or lBU->fDR->rDB : [B2, U' f2 U]
4) Queen->Marvin the Paranoid Android or Q->MR or lBU->bUR->lUF : F'
[B d' B', U'] F
5) Queen->Unicorn or Q->UC or lBU->dBL->uFL : [L2, F' l2 F]
6) Queen->BackPack or Q->BP or lBU->uBR->bDL : [U b2 U', B]
7) Queen->Jubilee or Q->XJ or lBU->dFR->rUB : l [B' d B, U2] l'
8) Queen->Tank or Q->TA or lBU->lDF->uBL : F2 [U2, F' u' F] F2
performed as F2 U2 F' u' F U2 F' u F'
9) Queen<->Osbourne and Y-Wing OR (Q <-> O) and (I <-> W) or (lBU <->
bDR) and (rUF <-> dFL)
Done as two three cycles
9a) [r' D' r, U2]
9b) [U2, B d' B']
10) Chuck Norris -> Gavel or N->GV or bUL->fDL->dBR : [b' R' b, L2]
11) Chuck Norris -> Chris Katan or (N <-> K) This is edge parity. To
solve I did:
f' D' f y' r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 y f' D f
The cube is now solved.
The three cycles averaged at 4 setup moves in 11 cycles or 0.36 setup
moves per three cycle. Parity used 3 setup moves, which is probably
average. A lot of cases require 4 setup moves if you want to preserve
centers. If you don't mind not preserving centers you might average
2.5 setup moves for parity.
Hope this helps. I tried to include absolutely everything, and
looking back over it that is how a BLD solve feels to me.
I double checked this message about three times, and the solve and
scramble work. If you work through the solve and it doesn't work try
again, because I double checked and there are no typos in the scramble
alg or in the stuff about solving.
View Source
Great post... my only question is why you do not do the corners last?
--- In
, "cmhardw"
<foozman17@...> wrote:
> Hey everyone,
> So yeah the work on the webpage is not proceding very well due to my
> jobs :-(
> I really don't want to be the guy that says I'll write up some info
> but then never does. So, here is an example solve of 4x4x4 BLD -
> literally every single thing that goes through my head when I do a
> Below is the scramble and the description of everything about the
> solve. Before you scramble please hold the cube in the same
> orienation you consider to be the "solved" orientation. This will
> make it much easier to follow along with what I am doing.
> Scramble:
> F B R2 b r D' B' d2 l' D2 b L' U2 L2 f d' F' b l2 U' d2 L' F' D2 B2 L
> f2 F2 d' f' b d B2 b r2 D2 f U2 d l
> I actually memorized and solved this scramble before typing up this
> example to make sure I would write up exactly what I do, even during a
> real solve. The overall time was 8:41.85 minutes, which is pretty
> much right at my average lately, which has been in the 8:40's. The
> memorization step took 4:33 minutes and the solving then took 3:58
> minutes.
> After the scramble I always rotate the cube around to find the
> position with the most centers already solved. After scrambling the
> cube I did the rotation z' x'. If you don't like xyz notation I first
> rotated the whole cube counterclockwise as if I was doing the move F'
> and I then rotated the whole cube as if I was doing the move R'
> After the cube rotation I now have the following cycles on the cube:
> centers:
> lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU -> buL -> flU -> dbL ->
> urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF -> lbU
> edges:
> 1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -> dBL
> -> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
> 2) (rUF <-> dFL)
> 3) (bUL -> fDL -> dBR -> rDF -> bUL)
> I always memorize and solve in this order for the 4x4x4:
> 1) memorize center permutation
> 2) memorize edge permutation
> 3) memorize corner permutaiton
> 4) memorize corner orientation
> 5) solve corner orientation
> 6) solve corner permutation
> 7) solve center permutation
> 8) solve edge permutation
> During the memorization of this solve I debated memorizing and solving
> with center blocks on the F and R faces, but decided against it
> because I had already memorized one of the pieces I would have made a
> block out of into an image, and didn't want to rewrite the first
> journey location any.
> *****************
> MEMORIZATION STEP
> *****************
> center cycle: lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU -> buL ->
> flU -> dbL -> urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF -> lbU
> center cycle in letters: A -> I -> E -> V -> O -> W -> B -> Q -> C ->
> S -> F -> K -> D -> H -> L -> X -> G -> A
> The journey I was using was halfway through one of my 28 location
> journeys. This particular journey is a walk from my dormitory my
> junior year of college through to my class on the other side of campus.
> At each location I need to put at least three images, sometimes more
> which does happen on this solve during the edge memorization.
> Let me walk through each location one at a time:
> Location 1:
> -----------
> This location is outside the dining hall with a bunch of tables and
> chairs.
> At this location I have the letters AIEVO. I take the very first
> letter and give it a single letter image (famous person). This is so
> I know which piece is the first piece of every one of my three cycles.
> The letter is "A" so the image is "Dan Aykroyd".
> Ok, so now I have to take the letters in pairs after the first image
> is made. The next two are IE which is the Internet Explorer logo,
> only about 6 feet tall. The next two letters are VO which is a 2 foot
> tall Volcano that is spewing and erupting a huge column of ash.
> I put these images in a left-to-right order. So on the left Dan
> Aykroyd is rolling a giant 6 foot tall model of the Internet Explorer
> logo onto the top of the two foot tall volcano to clog it up and make
> it stop erupting ash everywhere.
> That's it. I don't really view the motion, it is just implied. I see
> the tables and chairs and the dining hall very clearly in the
> background, and in the foreground I see Dan Aykroyd (motionless) in
> the act of rolling the Internet Explorer logo onto the volcano. I
> picture the logo about 3/4 the way up the volcano, so that the volcano
> is still to the right of the Internet Explorer logo, which is to the
> right of Dan Aykroyd.
> Location 2:
> -----------
> This location is beside a wall about 20 feet away from the dining hall
> on the way to a set of stairs.
> At this location I have the letters WBQCSF. I now take letters in
> pairs always to make double images. The letters WB form the image of
> a giant 6 foot tall spider web between two large tree limbs. The
> letters QC form the image of a two foot tall Quaich. The letters SF
> form an image of a 4 foot tall model of the San Francisco Golden Gate
> bridge.
> At this location I can very clearly see the wall, as well as the hand
> railing leading to the stairwell that is the next location (all in the
> background). In the foreground I very clearly see (in left to right
> order) a giant web where a Quaich is falling into it having rolled off
> the end of the miniature San Francisco Golden Gate bridge.
> Location 3:
> -----------
> This location is at the bottom of the stair case pictured at the last
> location. I view it from the top of the staircase, looking at the
> space just after the last step at the bottom. There are maybe 6 steps
> to this staircase.
> The letters at this location are KDHLXG. KD forms the image of Billy
> the Kid the famous gunslinger. HL form the image of Satan as
> described by Dante in Dante's Inferno. The only difference is my
> image of him is about 5 feet tall instead of the massive proportions
> Dante gives him. The letters XG form the image of Gambit from the
> X-men (the guy who charges stuff with energy and the object then
> explodes after a few seconds).
> This image is graphic, but this is exactly how I memorize. I picture,
> in left to right order, Billy the kid drawing his pistol and quickly
> unloading all 6 rounds into Satan, who is trapped in the ice as in
> Dante's Inferno. This doesn't kill Satan, but only wounds him. To
> the right of Satan Gambit charges a playing card (his usual weapon)
> and places it in the mouth of Satan. When the card explodes it
> explodes Satan's head and blood and guts are thrown absolutely
> Yes this image is graphic, but because of that it was very easy to
> memorize. Anything funny or violent tends to be easier to hold onto.
> **********
> BLAST DOOR
> **********
> Yes that literally is a blast door, like for a bomb shelter. I
> imagine a 200 foot tall steel blast door dropping down onto my journey
> in the space between the 3rd location and the 4th. I image that in
> order to get past the door I have to fly over it as if I were in a
> helicopter, and I literally picture this flight over the blast door.
> This is my marker that the centers are now done - i.e. when I get to
> the blast door during the solving phase, I know that what follows is
> now information for the edges.
> edge cycles:
> 1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -> dBL
> -> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
> 2) (rUF <-> dFL)
> 3) (bUL -> fDL -> dBR -> rDF -> bUL)
> edge cycles as letters:
> 1) (Q -> F -> E -> S -> D -> H -> L -> M -> R -> U -> C -> B -> P -> X
> -> J -> T -> A -> O -> Q)
> 2) (I <-> W)
> 3) (N -> G -> V -> K)
> Location 4:
> -----------
> I am now memorzing the edge permutation. This location is at a "+"
> shaped intersection of two very large walkways leading into one of the
> quads on my school that is a very large enclosed rectangular space
> formed by about 8 buildings. I can clearly see the open space of the
> quad, as well as the two walkways at this location.
> The letters at this location are QFESD. I need to know which piece
> will be my main piece, or the one that will be the first piece in all
> of my cycles, so I take the first letter "Q" and make it into an
> image. The letter Q goes with the image "Queen". I now take the
> letters in pairs after the first piece. The next two letters FE form
> the image of a 10 foot tall Ferris Wheel. The next two letters SD
> form the image of a 10 foot long Star Destroyer (like from Star Wars)
> that is floating in the air about 6 feet above the ground.
> In left to right order I see the Queen hiding behind a miniature
> Ferris Wheel while a miniature Star Destroyer bombards it with lasers
> and explosives of all kinds. The ferris wheel is on fire and so
> decrepit it is about to collapse.
> Location 5:
> -----------
> This location is part-way down a pedestrian walkway that goes into the
> center of the large quad we are now in. I can picture a big tree and
> a small wall to the right.
> At this location I have the letters HLMRUC. HL stands for the Satan
> as portrayed by Dante again. MR stands for Marvin the paranoid
> android from Hitchiker's Guide to the Galaxy (as portrayed in the
> movie). UC stands for a Unicorn. In left to right order I see Satan
> frozen in the ice trying to frighten Marvin the android, who could not
> be more unimpressed. Satan gets pissed off about this, so he posesses
> a unicorn just to the right of (behind) Marvin and has the Unicorn ram
> it's horn through Marvin's head, effectively killing him.
> Location 6:
> -----------
> This location is the center of the quad, where the two largest
> walkways meet at a sort of tall skinny "X" shape. I can see the
> buildings on the other side of the quad as well as a very big tree to
> the left at this location.
> Here I have the letters BPXJTA. BP stands for a backpack which is
> oversized (about 6 feet tall) and filled with 100's of regular sized
> books. XJ is the image of Jubilee from the X-men. TA is the image of
> a tank with it's gun turret pointed menacingly at the object next to it.
> In left to right order I see a backpack lying on the ground, filled to
> the brim with books. Jubilee is using her sparks to completely
> destroy the backpack which is now on fire and billowing smoke
> everywhere. Behind her a tank aims and takes a shot from it's huge
> gun turret but Jubilee ducks out of the way just in time.
> I normally stop after three images, but this time I see that the next
> piece, the "O" piece, goes to the "Q" spot and completes this cycle.
> Instead of starting at a new location I tack on my single letter "O"
> image (Ozzy Osbourne) behind the tank, as if he were orchestrating the
> whole incident of the tank firing on Jubilee.
> This cycle is now finished. I know I am not done memorizing edges
> though, because I have only used three journey locations, and the
> average is 4 for edges. During the actual solve I looked around for
> an edge I had not used yet when I spotted the two cycle (rUF <-> dFL)
> or in my letters that is (I <-> W). Rather than start this image at
> the next journey location, since a two cycle is always just one image,
> I tack it on to the current location just behind Ozzy Osbourne, only
> surrounded by curtains on both the left and right. This is my
> miniature, within a journey location, form for the "BLAST DOOR". It
> lets me know that the image within the curtains is a two cycle. I got
> this idea from John Louis, so thank you very much! The image for IW
> is a "Y-Wing" space fighter from Star Wars. Within curtains I imagine
> the Y-Wing taking aim at Ozzy Osbourney (through the curtains) to take
> him out and make him stop firing on Jubilee with the Tank.
> REHEARSAL:
> ----------
> OK so I know I have used a 3 locations, and a little more since there
> are 5 images at my 3rd location instead of the usual 3. I know that I
> am not done, but I have no idea how many edges are left to look for.
> So I go over my images (just for edges). I sort of half use this as
> rehearsal for my edges and half to just count them and find out how
> many edges I have memorized right now. I counted 20 edges when I went
> back through my edge images. I now glance quickly at the cube to see
> that no edges are already in their solved locations - I saw none. So
> I know that there are 4 edges somewhere to look for. I find them and
> they are in one cycle, which I put at my last location.
> Location 7:
> -----------
> This location is on the other side of the quad. I am very close to
> the buildings that before were only in the background.
> At this location I have the 4 cycle (N->G->V->K->N). I take the N and
> make it into a single letter image, that of Chuck Norris. The GV I
> put together as a giant Gavel, about 6 feet tall. Since I only have
> one piece left, the K, I make it into a single letter image of Chris
> Katan from Saturday Night Live.
> In left to right order I see Chuck Norris roundhouse kicking a giant
> Gavel so hard that it slams into Chris Katan and knocks him over.
> ****END OF JOURNEY****
> I now know that I am done memorizing the centers and edges, since I
> have accounted for all 24 edges. I now memorize the Corner
> permutation visually and the Corner orientation visually. Then I put
> on the blindfolded.
> *************
> SOLVING PHASE
> *************
> Solving the corners is pretty uninteresting I think, since you can
> just use the same stuff as from the 2x2x2 or 3x3x3. The only change
> is that you *HAVE* to use supercube safe algs (algs that don't rotate
> center caps if done on a supercube).
> Here is how I solved the corners:
> Corner orientation:
> y' R' U2 R U R' U R L U2 L' U' L U' L' followed by U' R U R' U' R U R'
> L' U' R U R' U' R U R' L' U' R U R' U' R U R' L2 y
> or if you don't prefer xyz notation the above is equivalent to F' U2 F
> U F' U F B U2 B' U' B U' B' followed by U' F U F' U' F U F' B' U' F U
> F' U' F U F' B' U' F U F' U' F U F' B2
> Corner permutation:
> 1) UBL->DFL->UFL : D F2 L' B' L F2 L' B L D'
> 2) UBL->UFR->DBL : (L B' L' B)x3 D2 (R F' R' F)x3 D2
> 3) (UBL <-> DBR) and (UBR <-> DFR) : F2 D (L' B L B')x3 D' F2
> Centers:
> --------
> Ok time to actually solve the centers now.
> First location: Dan Aykroyd->Internet Explorer->Volcano
> A->IE->VO
> I start with Dan Aykroyd then I take the first image, Internet Explorer.
> Those are the letters AIE which is the cycle lbU->ufR->ulF :
> 1) To solve I use the commutator: [f' U' f, u']
> I am here using the same notation I learned from Stefan and Per. [f'
> u' f, u'] means to perform the first part, then the second part, then
> the inverse of the first part, then the inverse of the second part.
> So [f' u' f, u'] = (f' u' f) (u') (f' u f) (u) = f' u' f u' f' u f u
> 2) After Internet Explorer comes VO. So the cycle is now A->VO or
> lbU->rbD->drB.
> lbU->rbD->drB : D' r' [U', b d b'] r D
> 3) Aykroyd->Web or A->WB or lbU->lfD->rbU : [b' R' b, l2]
> 4) Aykroyd->Quaich or A->QC or lbU->ubL : [U, l u' l']
> 5) Aykroyd->San Francisco or A->SF or lbU->dbL->urF : L2 [f U2 f', u] L2
> 6) Aykroyd->Billy the Kid or A->KD or lbU->dfR->rfU : [U2, b d2 b']
> 7) Aykroyd->Satan or A->HL or lbU->drF->bdR : [r U2 r', d']
> 8) Aykroyd->Gambit or A->XG or lbU->rfD->dlF : r2 [b d' b', U'] r2
> Centers are now solved. Notice that only three of the three cycles
> required setup moves. One of the three cycles required two. Average
> number of setup moves needed before being able to perform a commutator
> was 4/7 or 0.57 moves.
> Ok now I realize that I have run into the blast door. So I know that
> the next stuff I recall deals with edges.
> Edges:
> ------
> The first location has me recalling Queen, then the very next image is
> a Ferris Wheel.
> 1) Queen->Ferris Wheel or Q->FE or lBU->fUR->fUL : [R B' R', f]
> 2) Queen->Star Destroyer or Q->SD or lBU->lDB->uFR : R [U', B d B'] R'
> 3) Queen->Satan or Q->HL or lBU->fDR->rDB : [B2, U' f2 U]
> 4) Queen->Marvin the Paranoid Android or Q->MR or lBU->bUR->lUF : F'
> [B d' B', U'] F
> 5) Queen->Unicorn or Q->UC or lBU->dBL->uFL : [L2, F' l2 F]
> 6) Queen->BackPack or Q->BP or lBU->uBR->bDL : [U b2 U', B]
> 7) Queen->Jubilee or Q->XJ or lBU->dFR->rUB : l [B' d B, U2] l'
> 8) Queen->Tank or Q->TA or lBU->lDF->uBL : F2 [U2, F' u' F] F2
> performed as F2 U2 F' u' F U2 F' u F'
> 9) Queen<->Osbourne and Y-Wing OR (Q <-> O) and (I <-> W) or (lBU <->
> bDR) and (rUF <-> dFL)
> Done as two three cycles
> 9a) [r' D' r, U2]
> 9b) [U2, B d' B']
> 10) Chuck Norris -> Gavel or N->GV or bUL->fDL->dBR : [b' R' b, L2]
> 11) Chuck Norris -> Chris Katan or (N <-> K) This is edge parity. To
> solve I did:
> f' D' f y' r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 y f' D f
> The cube is now solved.
> The three cycles averaged at 4 setup moves in 11 cycles or 0.36 setup
> moves per three cycle. Parity used 3 setup moves, which is probably
> average. A lot of cases require 4 setup moves if you want to preserve
> centers. If you don't mind not preserving centers you might average
> 2.5 setup moves for parity.
> -----------------------------------------------
> Hope this helps. I tried to include absolutely everything, and
> looking back over it that is how a BLD solve feels to me.
> I double checked this message about three times, and the solve and
> scramble work. If you work through the solve and it doesn't work try
> again, because I double checked and there are no typos in the scramble
> alg or in the stuff about solving.
> Chris
View Source
Excellent example. I didn't do the scramble, but I read it all the
way through. I must go and practice now. It's been a while since
I've tried memorizing the 4x4 or 5x5. I'm finally "trying" to learn
OLL and the G Perms. Once again, though, Blindfolded cubing is
taking precedence.
I agree, why solve the corners first? However, I'm not confident in
my big cube blindfolded abilities yet. So I don't know why you've
chosen this way. Perhaps it just feels better. Just as different
memorization methods feel better.
I did just realize that he's posted before about blindfold solving a
cube and he messed up the Corner Permutation, when the rest of the
cube was perfect. So perhaps he'd rather use short term memo for
his corners in BLD.
--- In
, "kyuubree"
<agent_m80@...> wrote:
> Great post... my only question is why you do not do the corners
> --- In blindfoldsolving-rubiks-cube@yahoogroups.com, "cmhardw"
> <foozman17@> wrote:
> >
> > Hey everyone,
> >
> > So yeah the work on the webpage is not proceding very well due
to my
> > jobs :-(
> >
> > I really don't want to be the guy that says I'll write up some
> > but then never does. So, here is an example solve of 4x4x4 BLD -
> > literally every single thing that goes through my head when I do
> solve.
> >
> > Below is the scramble and the description of everything about the
> > solve. Before you scramble please hold the cube in the same
> > orienation you consider to be the "solved" orientation. This
> > make it much easier to follow along with what I am doing.
> >
> > Scramble:
> > F B R2 b r D' B' d2 l' D2 b L' U2 L2 f d' F' b l2 U' d2 L' F' D2
B2 L
> > f2 F2 d' f' b d B2 b r2 D2 f U2 d l
> >
> > I actually memorized and solved this scramble before typing up
> > example to make sure I would write up exactly what I do, even
during a
> > real solve. The overall time was 8:41.85 minutes, which is
> > much right at my average lately, which has been in the 8:40's.
> > memorization step took 4:33 minutes and the solving then took
> > minutes.
> >
> > After the scramble I always rotate the cube around to find the
> > position with the most centers already solved. After scrambling
> > cube I did the rotation z' x'. If you don't like xyz notation I
> > rotated the whole cube counterclockwise as if I was doing the
move F'
> > and I then rotated the whole cube as if I was doing the move R'
> >
> > After the cube rotation I now have the following cycles on the
> >
> > centers:
> > lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU -> buL -> flU ->
dbL ->
> > urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF -> lbU
> >
> > edges:
> > 1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -
> dBL
> > -> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
> > 2) (rUF <-> dFL)
> > 3) (bUL -> fDL -> dBR -> rDF -> bUL)
> >
> > I always memorize and solve in this order for the 4x4x4:
> >
> > 1) memorize center permutation
> > 2) memorize edge permutation
> > 3) memorize corner permutaiton
> > 4) memorize corner orientation
> > 5) solve corner orientation
> > 6) solve corner permutation
> > 7) solve center permutation
> > 8) solve edge permutation
> >
> > During the memorization of this solve I debated memorizing and
> > with center blocks on the F and R faces, but decided against it
> > because I had already memorized one of the pieces I would have
made a
> > block out of into an image, and didn't want to rewrite the first
> > journey location any.
> >
> > *****************
> > MEMORIZATION STEP
> > *****************
> >
> > center cycle: lbU -> ufR -> ulF -> brD -> drB -> lfD -> rbU ->
buL ->
> > flU -> dbL -> urF -> dfR -> rfU -> drF -> dbR -> rfD -> dlF ->
> >
> > center cycle in letters: A -> I -> E -> V -> O -> W -> B -> Q ->
C ->
> > S -> F -> K -> D -> H -> L -> X -> G -> A
> >
> > The journey I was using was halfway through one of my 28 location
> > journeys. This particular journey is a walk from my dormitory my
> > junior year of college through to my class on the other side of
> >
> > At each location I need to put at least three images, sometimes
> > which does happen on this solve during the edge memorization.
> >
> > Let me walk through each location one at a time:
> >
> > Location 1:
> > -----------
> > This location is outside the dining hall with a bunch of tables
> > chairs.
> >
> > At this location I have the letters AIEVO. I take the very first
> > letter and give it a single letter image (famous person). This
is so
> > I know which piece is the first piece of every one of my three
> > The letter is "A" so the image is "Dan Aykroyd".
> >
> > Ok, so now I have to take the letters in pairs after the first
> > is made. The next two are IE which is the Internet Explorer
> > only about 6 feet tall. The next two letters are VO which is a
2 foot
> > tall Volcano that is spewing and erupting a huge column of ash.
> >
> > I put these images in a left-to-right order. So on the left Dan
> > Aykroyd is rolling a giant 6 foot tall model of the Internet
> > logo onto the top of the two foot tall volcano to clog it up and
> > it stop erupting ash everywhere.
> >
> > That's it. I don't really view the motion, it is just implied.
I see
> > the tables and chairs and the dining hall very clearly in the
> > background, and in the foreground I see Dan Aykroyd (motionless)
> > the act of rolling the Internet Explorer logo onto the volcano.
> > picture the logo about 3/4 the way up the volcano, so that the
> > is still to the right of the Internet Explorer logo, which is to
> > right of Dan Aykroyd.
> >
> > Location 2:
> > -----------
> > This location is beside a wall about 20 feet away from the
dining hall
> > on the way to a set of stairs.
> >
> > At this location I have the letters WBQCSF. I now take letters
> > pairs always to make double images. The letters WB form the
image of
> > a giant 6 foot tall spider web between two large tree limbs. The
> > letters QC form the image of a two foot tall Quaich. The
letters SF
> > form an image of a 4 foot tall model of the San Francisco Golden
> > bridge.
> >
> > At this location I can very clearly see the wall, as well as the
> > railing leading to the stairwell that is the next location (all
in the
> > background). In the foreground I very clearly see (in left to
> > order) a giant web where a Quaich is falling into it having
rolled off
> > the end of the miniature San Francisco Golden Gate bridge.
> >
> > Location 3:
> > -----------
> > This location is at the bottom of the stair case pictured at the
> > location. I view it from the top of the staircase, looking at
> > space just after the last step at the bottom. There are maybe 6
> > to this staircase.
> >
> > The letters at this location are KDHLXG. KD forms the image of
> > the Kid the famous gunslinger. HL form the image of Satan as
> > described by Dante in Dante's Inferno. The only difference is my
> > image of him is about 5 feet tall instead of the massive
> > Dante gives him. The letters XG form the image of Gambit from
> > X-men (the guy who charges stuff with energy and the object then
> > explodes after a few seconds).
> >
> > This image is graphic, but this is exactly how I memorize. I
> > in left to right order, Billy the kid drawing his pistol and
> > unloading all 6 rounds into Satan, who is trapped in the ice as
> > Dante's Inferno. This doesn't kill Satan, but only wounds him.
> > the right of Satan Gambit charges a playing card (his usual
> > and places it in the mouth of Satan. When the card explodes it
> > explodes Satan's head and blood and guts are thrown absolutely
> everywhere.
> >
> > Yes this image is graphic, but because of that it was very easy
> > memorize. Anything funny or violent tends to be easier to hold
> >
> > **********
> > BLAST DOOR
> > **********
> >
> > Yes that literally is a blast door, like for a bomb shelter. I
> > imagine a 200 foot tall steel blast door dropping down onto my
> > in the space between the 3rd location and the 4th. I image that
> > order to get past the door I have to fly over it as if I were in
> > helicopter, and I literally picture this flight over the blast
> > This is my marker that the centers are now done - i.e. when I
get to
> > the blast door during the solving phase, I know that what
follows is
> > now information for the edges.
> >
> > edge cycles:
> > 1) (lBU -> fUR -> fUL -> lDB -> uFR -> fDR -> rDB -> bUR -> lUF -
> dBL
> > -> uFL -> uBR -> bDL -> dFR -> rBU -> lDF -> uBL -> bDR -> lBU)
> > 2) (rUF <-> dFL)
> > 3) (bUL -> fDL -> dBR -> rDF -> bUL)
> >
> > edge cycles as letters:
> > 1) (Q -> F -> E -> S -> D -> H -> L -> M -> R -> U -> C -> B ->
P -> X
> > -> J -> T -> A -> O -> Q)
> > 2) (I <-> W)
> > 3) (N -> G -> V -> K)
> >
> > Location 4:
> > -----------
> > I am now memorzing the edge permutation. This location is at
a "+"
> > shaped intersection of two very large walkways leading into one
of the
> > quads on my school that is a very large enclosed rectangular
> > formed by about 8 buildings. I can clearly see the open space
of the
> > quad, as well as the two walkways at this location.
> >
> > The letters at this location are QFESD. I need to know which
> > will be my main piece, or the one that will be the first piece
in all
> > of my cycles, so I take the first letter "Q" and make it into an
> > image. The letter Q goes with the image "Queen". I now take the
> > letters in pairs after the first piece. The next two letters FE
> > the image of a 10 foot tall Ferris Wheel. The next two letters
> > form the image of a 10 foot long Star Destroyer (like from Star
> > that is floating in the air about 6 feet above the ground.
> >
> > In left to right order I see the Queen hiding behind a miniature
> > Ferris Wheel while a miniature Star Destroyer bombards it with
> > and explosives of all kinds. The ferris wheel is on fire and so
> > decrepit it is about to collapse.
> >
> > Location 5:
> > -----------
> > This location is part-way down a pedestrian walkway that goes
into the
> > center of the large quad we are now in. I can picture a big
tree and
> > a small wall to the right.
> >
> > At this location I have the letters HLMRUC. HL stands for the
> > as portrayed by Dante again. MR stands for Marvin the paranoid
> > android from Hitchiker's Guide to the Galaxy (as portrayed in the
> > movie). UC stands for a Unicorn. In left to right order I see
> > frozen in the ice trying to frighten Marvin the android, who
could not
> > be more unimpressed. Satan gets pissed off about this, so he
> > a unicorn just to the right of (behind) Marvin and has the
Unicorn ram
> > it's horn through Marvin's head, effectively killing him.
> >
> > Location 6:
> > -----------
> > This location is the center of the quad, where the two largest
> > walkways meet at a sort of tall skinny "X" shape. I can see the
> > buildings on the other side of the quad as well as a very big
tree to
> > the left at this location.
> >
> > Here I have the letters BPXJTA. BP stands for a backpack which
> > oversized (about 6 feet tall) and filled with 100's of regular
> > books. XJ is the image of Jubilee from the X-men. TA is the
image of
> > a tank with it's gun turret pointed menacingly at the object
next to it.
> >
> > In left to right order I see a backpack lying on the ground,
filled to
> > the brim with books. Jubilee is using her sparks to completely
> > destroy the backpack which is now on fire and billowing smoke
> > everywhere. Behind her a tank aims and takes a shot from it's
> > gun turret but Jubilee ducks out of the way just in time.
> >
> > I normally stop after three images, but this time I see that the
> > piece, the "O" piece, goes to the "Q" spot and completes this
> > Instead of starting at a new location I tack on my single
letter "O"
> > image (Ozzy Osbourne) behind the tank, as if he were
orchestrating the
> > whole incident of the tank firing on Jubilee.
> >
> > This cycle is now finished. I know I am not done memorizing
> > though, because I have only used three journey locations, and the
> > average is 4 for edges. During the actual solve I looked around
> > an edge I had not used yet when I spotted the two cycle (rUF <->
> > or in my letters that is (I <-> W). Rather than start this
image at
> > the next journey location, since a two cycle is always just one
> > I tack it on to the current location just behind Ozzy Osbourne,
> > surrounded by curtains on both the left and right. This is my
> > miniature, within a journey location, form for the "BLAST
DOOR". It
> > lets me know that the image within the curtains is a two cycle.
I got
> > this idea from John Louis, so thank you very much! The image
for IW
> > is a "Y-Wing" space fighter from Star Wars. Within curtains I
> > the Y-Wing taking aim at Ozzy Osbourney (through the curtains)
to take
> > him out and make him stop firing on Jubilee with the Tank.
> >
> > REHEARSAL:
> > ----------
> > OK so I know I have used a 3 locations, and a little more since
> > are 5 images at my 3rd location instead of the usual 3. I know
that I
> > am not done, but I have no idea how many edges are left to look
> > So I go over my images (just for edges). I sort of half use
this as
> > rehearsal for my edges and half to just count them and find out
> > many edges I have memorized right now. I counted 20 edges when
I went
> > back through my edge images. I now glance quickly at the cube
to see
> > that no edges are already in their solved locations - I saw
none. So
> > I know that there are 4 edges somewhere to look for. I find
them and
> > they are in one cycle, which I put at my last location.
> >
> > Location 7:
> > -----------
> > This location is on the other side of the quad. I am very close
> > the buildings that before were only in the background.
> >
> > At this location I have the 4 cycle (N->G->V->K->N). I take the
N and
> > make it into a single letter image, that of Chuck Norris. The
GV I
> > put together as a giant Gavel, about 6 feet tall. Since I only
> > one piece left, the K, I make it into a single letter image of
> > Katan from Saturday Night Live.
> >
> > In left to right order I see Chuck Norris roundhouse kicking a
> > Gavel so hard that it slams into Chris Katan and knocks him over.
> >
> > ****END OF JOURNEY****
> >
> > I now know that I am done memorizing the centers and edges,
since I
> > have accounted for all 24 edges. I now memorize the Corner
> > permutation visually and the Corner orientation visually. Then
I put
> > on the blindfolded.
> >
> > *************
> > SOLVING PHASE
> > *************
> >
> > Solving the corners is pretty uninteresting I think, since you
> > just use the same stuff as from the 2x2x2 or 3x3x3. The only
> > is that you *HAVE* to use supercube safe algs (algs that don't
> > center caps if done on a supercube).
> >
> > Here is how I solved the corners:
> >
> > Corner orientation:
> > y' R' U2 R U R' U R L U2 L' U' L U' L' followed by U' R U R' U'
R U R'
> > L' U' R U R' U' R U R' L' U' R U R' U' R U R' L2 y
> >
> > or if you don't prefer xyz notation the above is equivalent to
F' U2 F
> > U F' U F B U2 B' U' B U' B' followed by U' F U F' U' F U F' B'
U' F U
> > F' U' F U F' B' U' F U F' U' F U F' B2
> >
> > Corner permutation:
> > 1) UBL->DFL->UFL : D F2 L' B' L F2 L' B L D'
> > 2) UBL->UFR->DBL : (L B' L' B)x3 D2 (R F' R' F)x3 D2
> > 3) (UBL <-> DBR) and (UBR <-> DFR) : F2 D (L' B L B')x3 D' F2
> >
> > Centers:
> > --------
> > Ok time to actually solve the centers now.
> >
> > First location: Dan Aykroyd->Internet Explorer->Volcano
> > A->IE->VO
> >
> > I start with Dan Aykroyd then I take the first image, Internet
> >
> > Those are the letters AIE which is the cycle lbU->ufR->ulF :
> > 1) To solve I use the commutator: [f' U' f, u']
> >
> > I am here using the same notation I learned from Stefan and
Per. [f'
> > u' f, u'] means to perform the first part, then the second part,
> > the inverse of the first part, then the inverse of the second
> >
> > So [f' u' f, u'] = (f' u' f) (u') (f' u f) (u) = f' u' f u' f' u
f u
> >
> > 2) After Internet Explorer comes VO. So the cycle is now A->VO
> > lbU->rbD->drB.
> > lbU->rbD->drB : D' r' [U', b d b'] r D
> >
> > 3) Aykroyd->Web or A->WB or lbU->lfD->rbU : [b' R' b, l2]
> > 4) Aykroyd->Quaich or A->QC or lbU->ubL : [U, l u' l']
> > 5) Aykroyd->San Francisco or A->SF or lbU->dbL->urF : L2 [f U2
f', u] L2
> > 6) Aykroyd->Billy the Kid or A->KD or lbU->dfR->rfU : [U2, b d2
> > 7) Aykroyd->Satan or A->HL or lbU->drF->bdR : [r U2 r', d']
> > 8) Aykroyd->Gambit or A->XG or lbU->rfD->dlF : r2 [b d' b', U']
> >
> > Centers are now solved. Notice that only three of the three
> > required setup moves. One of the three cycles required two.
> > number of setup moves needed before being able to perform a
> > was 4/7 or 0.57 moves.
> >
> > Ok now I realize that I have run into the blast door. So I know
> > the next stuff I recall deals with edges.
> >
> > Edges:
> > ------
> >
> > The first location has me recalling Queen, then the very next
image is
> > a Ferris Wheel.
> >
> > 1) Queen->Ferris Wheel or Q->FE or lBU->fUR->fUL : [R B' R', f]
> > 2) Queen->Star Destroyer or Q->SD or lBU->lDB->uFR : R [U', B d
B'] R'
> > 3) Queen->Satan or Q->HL or lBU->fDR->rDB : [B2, U' f2 U]
> > 4) Queen->Marvin the Paranoid Android or Q->MR or lBU->bUR-
>lUF : F'
> > [B d' B', U'] F
> > 5) Queen->Unicorn or Q->UC or lBU->dBL->uFL : [L2, F' l2 F]
> > 6) Queen->BackPack or Q->BP or lBU->uBR->bDL : [U b2 U', B]
> > 7) Queen->Jubilee or Q->XJ or lBU->dFR->rUB : l [B' d B, U2] l'
> > 8) Queen->Tank or Q->TA or lBU->lDF->uBL : F2 [U2, F' u' F] F2
> > performed as F2 U2 F' u' F U2 F' u F'
> > 9) Queen<->Osbourne and Y-Wing OR (Q <-> O) and (I <-> W) or
(lBU <->
> > bDR) and (rUF <-> dFL)
> > Done as two three cycles
> > 9a) [r' D' r, U2]
> > 9b) [U2, B d' B']
> > 10) Chuck Norris -> Gavel or N->GV or bUL->fDL->dBR : [b' R' b,
> > 11) Chuck Norris -> Chris Katan or (N <-> K) This is edge
parity. To
> > solve I did:
> > f' D' f y' r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 y f' D f
> >
> > The cube is now solved.
> >
> > The three cycles averaged at 4 setup moves in 11 cycles or 0.36
> > moves per three cycle. Parity used 3 setup moves, which is
> > average. A lot of cases require 4 setup moves if you want to
> > centers. If you don't mind not preserving centers you might
> > 2.5 setup moves for parity.
> >
> >
> > -----------------------------------------------
> >
> > Hope this helps. I tried to include absolutely everything, and
> > looking back over it that is how a BLD solve feels to me.
> >
> > I double checked this message about three times, and the solve
> > scramble work. If you work through the solve and it doesn't
work try
> > again, because I double checked and there are no typos in the
> > alg or in the stuff about solving.
> >
> > Chris
> >
View Source
I took the time yesterday to make several blindfolded attempts on
the 4x4 cage. I believe I made three. I sat down to memorize one
and I was distracted by a mediocre chore, so I stopped. My Second
attempt, I spent about 13 minutes memorizing the cage and I go for
the solve. I had the scramble perfectly memorized. I start corner
permutation first, and I messed up the exectution of my first corner
alg. I was doing the left handed Y perm. And everything just felt
wrong! I stopped then and there ); ... My third attempt I did at
my grandmother's just before work. Memo 16 minutes; Execution 13
minutes. I recalled everything perfectly. There was no 3x3 or 4x4
parity. However, I messed up a setup. Once again it felt wrong. I
tried to fix it. The error occurred in the last 6 edges. It was so
close! I have a long way to go. There was one cubie I was lost on
during the final execution. I recalled the chess game. I recalled
the accelerated math class lesson. After permuting 15 cubies I
start a new cycle. I started at lUB.
So I recalled:
"Skip to Mu; Dragoon" ... very long pause ... another pause ...
ohh, "Dragoon becomes King." "Lamba, Ace, Galanoth, Mu" Solved
The rUF was permuted by default, that's my 4x4 reference point. The
lUF was solved too. Resulting in 21 permutations + 1 new cycle = no
parity. I was glad that I was able to recall the King image.
That cycle equals:
Skip to lUB; fUL->rDB->lDB->rUB->bUL->lUB;
Haha, I just realized another life adventure. I can cause my images
to have accidents that my families experienced. Such as my 3yr old
sister cutting all of her hair off, or my getting ran off the road
last week. A brother getting injured on 4th of july.
Daniel Beyer
--- In
, "cmhardw"
<foozman17@...> wrote:
> Hey everyone,
> I normally try to keep from clogging the group here with posts
> big cubes BLD, but I had to post about this.
> I just broke my record for 4x4x4 BLD with a new best time of
> minutes! The coolest part though was the breakdown of the solve.
> The memorization took me about 4:25-4:26 I can't remember which
> exactly. Let's just say 4:26 to be safe. This means that I
solved in
> 2 minutes 57 seconds, a sub-3 minute solve while blindfolded!
> I was trying something new on this solve, and that was to memorize
> my normal pace, but to try to plan ahead what my next step would be
> while executing the current step. 4:26 memorization is fairly
> for me right now, maybe 5-10 seconds slower than my overall average
> average for memo. But 2:57 solving is by far the best I've ever
> on the solving phase, by about 23 seconds!
> What I did was while I was actually performing a three cycle to
try to
> not only see which images come up for my next three cycle, but to
> picture which pieces those are and also which three cycle I will
> I found that this meant that a couple times I did 2 three cycles
> right after the other with hardly any delay between them! I'd
like to
> get to the point where I can do this continuously throughout the
> solve, I definitely think it will be possible.
> I was thinking that to speed up I needed to work solely on memo,
> this showed me my solving has a ways to go too.
> With my record 3:20 fastest ever correct memorization and this new
> 2:57 fastest ever correct solve I predict a 6:17 solve overall
will be
> possible for the 4x4x4, and most likely even faster. I'm starting
> really think that a sub-6 minute overall solve might actually be a
> possiblity with another year or two of practice (for a best single
> time, not an average I mean).
> Yay big cubes BLD! This is so much fun! I really can see
> in this just like in regular speedcubing. You don't improve for a
> while, and then you make a big jump. Then you hit a plateau for a
> while and later have another big jump - just like speedcubing.
> I think you don't really need a phenomenal natural memory right
> the start, you can just treat it like speedcubing and practice a
> to get faster memorization times.
> Just had to post, I'm focusing almost entirely on my 4x4x4 BLD
> now, but will eventually start trying to apply what I learn to the
> 5x5x5 as well. I really want to start getting sub-20 minutes on
> 5x5x5 fairly often before RWC2007, and averaging sub-8 on 4x4x4 as
> Happy BLD'ing all!
> Chris
View Source
--- In
, "kyuubree"
<agent_m80@...> wrote:
> Great post... my only question is why you do not do the corners last?
Hey Marcus,
I've just found that I really like solving them first so during
memorization I can just barely memorize them, just long enough to put
on the blindfolded and solve them.
There is one major downside though, if I get corner parity, I have to
wait until after I solve the centers before I can fix it since an odd
permutation on the corners must also perform an odd permutation on the
centers (supercube-wise).
I've found though that if I use auditory memory to help me remember I
have parity in the corners it helps a lot. Once I know I have corner
parity I literally say out loud "corner parity" and I have found that
it usually sticks in my mind long enough for me to finish centers, and
still remember that on this solve (not the one before) I have corner
parity, since it's still ringing in my head so to speak.
It's just a personal preference, I know solving them last would be
fast too, since I wouldn't even have this corner parity problem, but I
prefer to do it this way after trying both.
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|
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Haciendas Constancia, PR
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Emergency Admissions for Cardiovascular and Respiratory Diseases and the Chemical Composition of Fine Particle Air Pollution
Population-based studies have estimated health risks of short-term exposure to fine particles using mass of PM[2.5] (particulate matter ≤ 2.5 μm in aerodynamic diameter) as the indicator. Evidence
regarding the toxicity of the chemical components of the PM[2.5] mixture is limited.
In this study we investigated the association between hospital admission for cardiovascular disease (CVD) and respiratory disease and the chemical components of PM[2.5] in the United States.
We used a national database comprising daily data for 2000–2006 on emergency hospital admissions for cardiovascular and respiratory outcomes, ambient levels of major PM[2.5] chemical components
[sulfate, nitrate, silicon, elemental carbon (EC), organic carbon matter (OCM), and sodium and ammonium ions], and weather. Using Bayesian hierarchical statistical models, we estimated the
associations between daily levels of PM[2.5] components and risk of hospital admissions in 119 U.S. urban communities for 12 million Medicare enrollees (≥ 65 years of age).
In multiple-pollutant models that adjust for the levels of other pollutants, an interquartile range (IQR) increase in EC was associated with a 0.80% [95% posterior interval (PI), 0.34–1.27%] increase
in risk of same-day cardiovascular admissions, and an IQR increase in OCM was associated with a 1.01% (95% PI, 0.04–1.98%) increase in risk of respiratory admissions on the same day. Other components
were not associated with cardiovascular or respiratory hospital admissions in multiple-pollutant models.
Ambient levels of EC and OCM, which are generated primarily from vehicle emissions, diesel, and wood burning, were associated with the largest risks of emergency hospitalization across the major
chemical constituents of PM[2.5].
Keywords: cardiovascular disease, chemical components, hospital admission, particulate matter, PM[2.5], respiratory disease, Speciation Trends Network
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08.05.02: Take Your Best Guess: Exploring 1, 10 and 100
You enter my classroom as a second grade student and are asked, "What's the difference between 1, 10 and 100? How much bigger is 100 than 10? Is it a big difference?" You try to answer, but you're
not quite sure. Then I ask, "OK, well let's look at the number 11. What digits are in the number 11 and what do they stand for?" Again, you don't understand. "What's a digit?" you think to yourself.
This is how most of my students react when we start to learn about place value.
I teach second grade in an unusual school. We are located within a school district that serves mostly underprivileged minority students. However, we are located amongst a fairly white, upper-middle
class neighborhood. We have come to operate like a "magnet" school in that we take students from all over the City of Richmond when we have open spaces. There are often open spaces because a majority
of the students in the surrounding neighborhood attend local private schools. These spaces are usually filled by students from schools in the district that are seen as "failing" because they are not
making adequate yearly progress. The implication of this for my classroom is that I have a large variety of students from different backgrounds, with sometimes vastly different support systems and
early childhood experiences. While one student may have a very supportive parent who is able to provide a lot of extra help at home, another may have a parent who works two jobs and isn't able to be
as involved. Some of my students are very well off, and others qualify for free lunch. Many of my students did not start at our school in Kindergarten, and may have just joined our school because a
slot opened up.
Just as there is a huge range in the personal backgrounds of my students, they are also often at very different levels in their academic developments. I may have a student who has a very deep
understanding of place value, has memorized the math facts to 20, can do two-digit addition, and is ready for multiplication and division when he or she walks into my room in September. On the other
hand, a different student may not have any fluency with math facts to 10 and does not have any strategies for one digit addition and subtraction. However, year after year, I have noticed that the
majority of my students would not be able to answer the questions presented above. It is the goal of this unit to create a scope and sequence that increases awareness of the meaning of decimal
notation and enable my students to use this more accurate number sense to help them estimate lengths and quantities in the real world.
This unit is intended to teach second grade students that there is a significant difference in the value of a digit when it is written in different places in decimal notation. Specifically, it seeks
to show them that tens are a lot bigger than ones, and hundreds are a lot bigger than tens. It also seeks to further enforce the difference in the "places" using digits other than 1. For example,
we'll explore the difference between 5, 50 and 500. These extensions will help solidify the idea that place value is a significant aspect of our numerical notation system. These concepts will be
practiced in a small unit on measurement and estimation in the 3^ r^ d quarter of the school year.
However, the path I would use to guide the students to this understanding of place value and order of magnitude starts at the beginning of the year and winds through lessons on digits, composing and
decomposing numbers to 10, number bonds to 20, expanded form, multiples of 10 and order of magnitude. In order to be able to successfully engage in the estimation activities described in this unit, I
will begin to teach them to have a new perspective of our number system starting at the beginning of the year. I believe very strongly that we need to strengthen our students' sense of number and
help them gain fluency in breaking apart and combining numbers. They need to readily be able to see a multi-digit number in many different ways and break it into its place value components before we
can ask them to compare large quantities and understand the usefulness of rounding larger numbers. A guide for following this path is included in this unit as an appendix.
Nevertheless, I understand that not all math curricula allow as much flexibility as the one I teach with. If you are not able to implement the full three quarters worth of lessons focusing on a
structured path toward greater number sense, I believe that you could take the estimation and measurement activities and insert them into your unit however you see them connecting to your current
curriculum. While students' skill in estimating is largely based on the strength of their foundation with the concept of ten and order of magnitude, this does not mean that there is a completely
linear progression from number sense to estimation. "Mathematics instruction…should promote numeracy by making students more sensitive to order of magnitude, and to the estimation capabilities of
place value notation."^ 1 Experiences with measurements and quantities that differ by a power of ten will strengthen each student's understanding of place value, order of magnitude and relative size.
The study of place value and order of magnitude is essential to constructing a solid understanding of our decimal notation system. A substantial grasp of decimal notation is the foundation for
rounding, comparing numbers and computing. In Virginia, the Standards of Learning and benchmark tests put a strong emphasis on students being able to successfully round a value, compare two values,
and perform the 4 basic operations with multiple numbers. I feel that a key factor for helping our students achieve success in these areas is to approach building a more thorough sense of number in
early childhood education. We must help young children start to comprehend the complexity of the way we communicate mathematically before we can expect them to manipulate and compute within that
system. A key factor in their success in simple computations is an appreciation of exactly what each digit in a number stands for and that where you put a digit can make a huge impact on the value of
the number.
In second grade, students start to explore the way we order digits to represent different values. However, it is not often presented to them from this perspective. Instead, seemingly out of the blue,
we ask them to answer questions like "Which digit is in the tens place?" or "What is the value of 4 in 402?" As adults, we see these concepts as transparent. However, from the viewpoint of a child
who has been counting groups of objects from 1-10 since he or she could talk, but has not gone much beyond ten, the fact that our numbers are communicated in specific combinations using decimal
notation is NOT readily apparent. Most of the students I have worked with do not understand the concept of "digit" when we begin mathematics in September. They were not introduced to "base-10" as a
concept before they got to my class. When they think of 10, they see it as a complete picture representing ten objects. Most often, they do not see 10 as representing 1 ten and 0 ones. I think we are
doing a disservice to our students when we don't emphasize that 10 is a two-digit number well before 2^ n^ d grade. Our students learn that there are 26 letters by learning a nifty song. Then they
learn that when you combine those letters they begin to represent specific sounds that hold meaning (words). Why do we not do the same thing in early mathematics teaching? Why is there not a digit
song that all children are encouraged to learn at a very young age? (See Appendix D.)
Once a student understands the concept of "digit," he or she can begin to look at the pieces of a multi-digit number. However, they first need to be able to explain the relationships between
single-digit numbers. Most of my students do not enter second grade with this ability. They cannot readily explain the relationship between a set of numbers. If I asked them, "How are 7, 3 and 4
related? Can you show me with a picture?" they wouldn't be able to successfully answer. This is an indication that they don't have sufficient experience composing and decomposing numbers. If they
can't pull apart and recompose a number, then they can't begin to manipulate its pieces. Before I can ask them to manipulate a two or three-digit number, they need to be able to compose and decompose
smaller numbers with accuracy and fluency. (Activities for building these skills are included in Appendix B.)
The way in which we see base-10 numbers determines our success in doing computations with them. "The fact that base ten numbers are sums has a pervasive influence on the methods for computing with
them…the procedures for carrying out the four arithmetic operations with base ten numbers are largely determined by the fact that base ten numbers are sums of their place value components."^ ^ 2^ ^
We must begin to see base 10 numbers as sums of their components. When we begin to do this, we can manipulate them with greater efficiency and accuracy.
After a second grader is able to see the relationships between pieces of numbers, the activities integrating order of magnitude can be implemented. As we start to talk about multi-digit numbers in
terms of their place value components, the activities in this unit become relevant. Subsequently, we can begin to use "very round numbers" to estimate quantities and lengths.
If we don't give students the opportunity to find the proof for the rules of math on their own, then how can we expect them to truly understand mathematical processes at the elementary level? Sure,
we could tell them the age-old adage "Because I said so," but we would be leaving them with a very superficial concept of number and our decimal notation system. Children need to repeatedly see the
difference in magnitude for each place value component so that they have an internal understanding of the significance of position of the decimal notation system. It is only when children build a
solid concept of digits, place value and order of magnitude that they will truly be successful in computation, estimation and the further manipulation of numbers.
Digit: A digit is a number (0,1,2,3,4,5,6,7,8,9) that can be used alone or in combination with other digits to communicate a value.
Place Value Notation/ Decimal Notation/ Base-10 Notation: Our way of writing numbers is a complex system based on groups of ten. The order in which we arrange digits determines the value the digit
represents. We have many "places" and each place has it's own value, hence "place value." A digit placed directly to the left of the decimal point stands for ones, the next place to the left denotes
groups of ten, the next place to the left from that denotes groups of hundreds, and so on. We can represent an infinite number of possible values using just ten digits because of our notation system.
We don't have to write 200 + 30 + 9, we can just write 239 and the previous values are assumed. This allows us to write large numbers very compactly.
Place Value Components/ Very Round Numbers: This refers to the value of each digit in a multi-digit number when it is considered on its own. It is a digit times a power of 10. For example, the number
3,461 has four place value components. They include 3,000, 400, 60 and 1.^ 3 A very round number is a number with only one non-zero digit, e.g., 3000, 400, 60 and 1. Our base-ten place value system
expresses every whole number as a sum of very round numbers. For example, the number 3,461 is the sum 3000 + 400 + 60 + 1. When very round numbers are combined like this to make some number, they are
called the place value components, or very round components, of that number.
Order of Magnitude: This term refers to the number of zeros used to write a very round number.^ 4^ For an arbitrary whole number, the order of magnitude is one less than the number of digits.
Composing/Decomposing Numbers
Chinese equivalents for the terms "compose" and "decompose" are prevalent in Chinese mathematics learning.^ 5 However, these are relatively unfamiliar terms to early childhood math teachers in the
United States.^ 6 If we begin by focusing on what these words mean, we will help ourselves and our students understand how the terms relate to mathematics. Compose is a verb that means "to make or
form by combining things, parts, or elements."^ 7 In mathematics, when we compose a number, we are making a number by combining two or more smaller values. Essentially, we are looking for ways to
group the possible parts of a number. Decompose is a verb that means "to separate or resolve into constituent parts or elements."^ 8 In mathematics, when we decompose a number, we are pulling that
number apart into smaller values. There may be a variety of ways to compose or decompose a number, depending on how many digits it contains. We can compose 3 in a few ways. (1 + 1 + 1) or (2 + 1) or
(1 + 2) or (0 + 3) are all ways of making 3. We can decompose 9 in a numerous ways. (There are 30!) Some examples are: 9 can be broken into (2 and 7), (5 and 4), (1 and 8) or (2 + 3 + 4). This
concept is integral to early childhood mathematics education, and I believe we should introduce not only these terms, but also these concepts, with young children as they begin to build concepts of
number. We can use "make" or "combine" as synonyms for compose, and we can use "unmake" or "break apart" as other ways of explaining "decompose," but we should use these terms interchangeably.
Expanded Form
Expanded form expresses a number as the sum of its place value components. Expanded form helps us visualize a number by looking at its place value components separately. Each digit is considered on
its own, as a separate value. When we consider each digit separately, its order of magnitude becomes more apparent. Then we see that none of the digits can represent the same order of magnitude.^ 9
Examples of expanded form:
│ 67 = 60 + 7 │4,532 = 4,000 + 500 + 30 + 2 │
│ 503 = 500 + 3 │ 11 = 10 + 1 │
│8030 = 8,000 + 30 │ 83 = 80 + 3 │
If we take the time to explicitly teach children to break apart numbers in this way, they will be able to utilize this strategy when doing operations. The study of numbers in expanded form should be
introduced after students have broken numbers into tens and ones first. They need to be guided to see each digit as a certain number of tens and ones (or thousands and hundreds) first. Then you can
begin to use expanded form.
Order of Magnitude
Once you understand expanded notation of multi-digit numbers, you can begin to explore the concept of order of magnitude. When you break a given number into expanded form, you can see each digit
represents a different power of ten. We can refer to the pieces of a number shown in expanded form (these multiples of 10) as "place value components." None of the place value components can
represent the same place value. I can write 1,111 in expanded form like this, 1,000 + 100 + 10 + 1, and we will see that it has 4 place value components. Then we can further analyze these components.
│1,000 =│1 x 10 x 10 x 10 = │1 x 103│
│100 = │1 x 10 x 10 = │1 x 102│
│10 = │1 x 10 = │1 x 101│
│1 = │1 x 1 = │1 x 100│
This chart helps us begin to name each order of magnitude. 1,000 is order of magnitude 3 because it is 1 x 10^ 3. Any 4-digit number is order of magnitude 3, which includes numbers from 1,000-9,999.
100 is order of magnitude 2 because it is 1 x 10^ 2. Any 3-digit number from 100-999 is order of magnitude 2. 10 is order of magnitude 1 because it's 1 x 10^ 1. Any 2-digit number, 10-99, is order of
magnitude 1. 1 is order of magnitude 0 because it is 1x 10^ 0. Any other 1-digit number is also order of magnitude 0.
The chart also helps us to see that with each successive digit to the left, the value of the digit is ten times bigger. 10 is ten times bigger than 1. 100 is ten times bigger than 10. 1,000 is ten
times bigger than 100. If you want a deeper understanding of the implications of order of magnitude in estimation, I would point you to pages 27-30 in R. Howe's Taking Place Value Seriously.^ 10
Understanding order of magnitude helps us more accurately conceive the difference between very large numbers that can often be abstract. It is hard for us to know just how significant the difference
is between 1 million and 1 billion. However, if we begin to analyze the decimal notation system in term of order of magnitude, we should at least know that 1 billion is a thousand times bigger than 1
million: ten groups of a million is ten million; ten groups of ten million is one hundred million; and ten groups of one hundred million make a billion. In all, this is 1,000 groups of a million.
Then you begin to realize that that is a huge difference, and it is this realization that can help us more accurately estimate.
Second graders do not need to have an advanced understanding of order of magnitude. We needn't try to explain the role of exponents in noting order of magnitude. However, we do want to build a
foundation by emphasizing that 10 is ten times bigger than one and 100 is ten times bigger than 10. This foundation will help students grasp the significance between each of the places in our decimal
notation system. It will also help them be able to estimate more readily by giving them experience with the difference in size between values that are different order of magnitude.
Estimation and Measurement with Number Bonds to 100
The base-10 manipulatives we use in teaching place value have a very useful attribute; they are measured in centimeters cubed. 1 ones block is 1cm long and a tens rod is 10 cm long. We could use the
ones block and a tens rod to begin to use our knowledge of order of magnitude to help us estimate lengths. We will find things that are about the same length as a ones cube or a tens rod, making a
chart in our math journals. We will talk about how the items that are about the same length as a tens rod are 10 times longer than the things we found that are about the length of 1 ones cube.^ 11
We could further integrate measurement into our exploration of multiples of 10 to use the base 10 blocks to create our own centimeter rulers. Each student will be given a tens rod and a strip of
paper about 2 inches thick and 50 cm long. They will take the tens rod and line it up on the line already drawn on the paper. They will mark the end of each rod and record 10cm. They will continue to
move down the line marking lengths of 10cm until they have measured 50cm. They will go back and use the rods to mark each cm length. This will give each child an experience measuring. (See Lesson 1
for detailed instructions.)
Additionally, these rulers will also serve as a number line to model addition of larger 2-digit numbers. I might ask the students to line up, end-to-end, 2 tens and 3 ones on the centimeter ruler.
They will see that these 2 tens and 3 ones have a length of 23cm. If they were to move the blocks around, they would still get a length of 23cm. My hope is that the students will understand that no
matter which order we add the pieces, 10 + 10 + 3 = 23. This activity provides an automatic connection between the pieces we are combining and the whole number they represent once combined. The
students can SEE that 10 + 10 + 3 = 23. In comparison, when we just model a 2-digit number with the same base-10 blocks on a "Tens & Ones" chart the two-digit representation of "23" is missing. The
visual model is there, but the number that corresponds to that visual model is not immediately connected. However, when we model a number on the centimeter ruler/number line there is an immediate
visual connection between the model and the number it represents. I can use this centimeter ruler/number line in a vast number of activities to solidify each student's understanding of how to compose
and decompose larger 2-digit numbers.
After we use our centimeter rulers as number lines for a few lessons, we will be ready to add on to them so that they measure 0-100cm. This will be the first lesson in the discussion that 100 is the
same as 10 tens. Therefore, 100 is ten times bigger than 10. All of our experiences using this ruler as a number line will help us be more familiar with length in centimeters. The exposure to lengths
from 0-100 centimeters should allow them to estimate lengths in relation to this experience. I could begin to ask questions such as "Is this piece of string closer to 10 cm or 100 cm? Is this line
segment closer to 20 cm or 50 cm.? Is a cat 10-20cm long or is it 20-50cm long? Is your leg about 40cm or about 100cm long? Is 40cm to the knee, or to the hip?"
It is here that I would begin to point out that the leading digit of a number represents most of the value of the number. I would ask "Why don't I ask, 'Is this about 25?' or 'Can you guess exactly
how many marbles are in this jar?'" The point of asking this question is to point out that until now we have been working with very round numbers (multiples of 10). We do this because usually we are
only really concerned with the leading digit when we estimate. I would then try to explain this through modeling with the base-10 blocks. I would ask them to show 78 in tens and ones on a "Tens &
Ones" chart. I would then ask, "Where are there the most blocks? Are there more blocks in the tens place, or are there more blocks in the ones place?" The answer to this question may not be readily
apparent to the class. I may need to remind them that 7 tens = 70 individual items. "If we think about the number 78 in expanded form as 70 + 8, which is more, 70 or 8? Right, 70 is more. You can see
that most of 78 is in the tens place. This means that the 7 tens is more important to us than the 8 ones." I would repeat this conversation with other 2-digit numbers. I might choose many 2-digit
numbers where the tens' digit is smaller than the ones' digit. This sets up a direct conflict with the idea that the larger digit in a number must represent the larger value. Many of my former
students have often automatically looked for the biggest digit when asked to identify the digit that represents the larger value. However, in the example 78, the eight is the bigger digit, but 8 is
not bigger than 70. In the number 78, the digit that shows the bigger value is the 7.
I would then extend this conversation to support the use of very round numbers in estimating quantities. "This is why we estimate with very round numbers. Most of the important information about a
number is in the first digit, so we can just take the first place value component of a number. If we do that, we are guaranteed to work with a very round number. And, as you know, very round numbers
are easy to work with because they end with zeros." Then we could try this with a few two-digit numbers.
I would continue by explaining that when we estimate we want to estimate numbers under 100, we use multiples of 10. However, instead of just picking a random very round number out of the air, it is
more meaningful for us to say that a quantity is between __(x)___ and __(y)__. When estimating, we want to express that we couldn't possibly be sure of an exact answer. It is easier to show the
uncertainty of our estimate if we say our guess is between two numbers. When we do this we are setting boundaries for our estimates. This allows us some flexibility with our estimate. We can be more
positive that a quantity is between two very round numbers than we can be that it is about one specific quantity. Think of 35 + 47. Since 30 35 40, and 40 47 50, then 30 + 40 = 70 35 + 47 40 + 50 =
After discussing this crucial aspect of estimation, we would discuss various collections of objects. We would work to create collections of different objects that vary by order of magnitude. I would
involve the students in the creation of these collections in order to give them the chance to manipulate collections of objects that have 10 items or 100 items or 1,000 items. They could work
together in small groups to make collections of beans that differ in size from order of magnitude 1 to order of magnitude 3. You could split the class into three groups and ask them to make a
collage. However, group one makes a collage with 10 beans, group 2 makes collages with 50 beans, and group 3 makes collages with 100 beans. They could begin to think about tackling problems such as
estimating how many people are in the cafeteria by estimating it between two very round numbers. The chances for extension are vast.
We could also further explore the relationship between centimeters and meters as an extension to the activities listed below. Following the study of 100 cm = 1 meter, you could discuss and explore
the differences between 1 meter, 10 meters and 100 meters.
Lesson 1
- To explore the difference between 1 and 10 through linear measurement using base-10 manipulatives
- To measure items around the classroom in centimeters using base-10 manipulatives
- To report lengths as estimates between two precise measurements when using non-standard measuring tools
- To record information gathered in order to informally assess understanding
- math journals
- ten base-10 cubes and one base-10 rod per student
- up to 50 base-10 cubes for teacher to display
- 1 centimeter ruler for every 2 students
The lesson begins with a whole group discussion about a tens rod. (We would have already discussed the manipulatives and used them in building two digit numbers.) Give each child a base-10 tens rod
and ask, "How many cubes are in a rod?" Show the whole group 10 ones and ask, "Which manipulatives are easier to handle, the rods or the cubes? If you had 30 or 40 cubes, would you rather have them
in separate cubes or grouped together in tens?" Let the class discuss, but make sure to reinforce that one tens rod is ten ones neatly packed into a form that is easy to handle. (The Cuisenaire rod
activity described in Appendix B under Composing and Decomposing 10 would be a good activity to do prior to this lesson.)
Now pair the students up and give each pair 1 base-10 cube and 1 rod. Ask them to measure both in centimeters using the centimeter ruler with their partner (you may need to guide them with this,
depending on their prior experience). Have them trace each one in their math journal and record the length next to each outline. After they try to measure on their own, discuss that the cube is 1cm
and the rod is 10-cm/1 decimeter.) Then ask, "If one rod is ten centimeters, how long is 2 rods? How long is 3 rods? How long is 4 rods? How long is 5 rods?" They should be able to figure out that
you can count by tens, so 2 rods is 20 cm, 3 rods are 30 cm, etc.
Lesson 2
- To explore powers of ten through linear measurement using base-10 manipulatives
- To construct a 50 cm ruler that can be used to measure items around the classroom
- 1 base-10 tens rod for each child
- base-10 cubes
- 5 base-10 tens rods taped together
- strips of cardstock precut measuring 1 inch wide and 50 cm long with a straight line pre-drawn down the middle
Show the class the extra long rod that is 5 base-10 rods taped together. Ask, "How many rods did I use to make this extra long rod? If I used 5 rods, how many centimeters long is this?" Then prove
that the rod is actually 50 centimeters by measuring it with a meter stick.
Next, explain that everyone is going to make his/her own 50 cm ruler to use for the next few weeks. Demonstrate how to make the ruler by lining up one rod at one end of the strip along top of the
pre-drawn line and marking the other end of the rod with a pencil line. Write "10 cm" under the pencil mark. Then move the rod down to the pencil line you just made and measure out another 10 cm
length. Do this until you have measured and marked 50 centimeters in 10-centimeter increments. You can then go back with the tens rod and mark each individual centimeter along the ruler using the
lines built into the rod, and then label all 50 centimeters. Then give each student a strip, a tens rod and a pencil and ask them to make it themselves. Find a safe place to store the ruler until the
next lesson (they can be folded and stored in Ziploc bags), or you can have them temporarily tape them to the top of their desks.
Lesson 3
- To compare 10 centimeters with 100 centimeters
- To introduce the terms "decimeter" and "meter"
- To visually see the difference between 10 and 100
- To associate 10 tens with 1 hundred
- To see 100 as ten times bigger than 10
- To introduce the prefix deci-
- To reinforce fractions as a pieces of a whole through a concrete exploration
- To see fractions as directly applied in everyday life
- math journals
- 2 previously made 50 cm rulers per pair of students
- a few meter sticks for display
- one extra long rod created by 10 tens rods taped together
- 10 tens rods per pair of students
Call the class to the rug to make a circle. Have each student bring his or her 50 cm ruler with him or her. Pair students up with a person sitting next to them. Give each pair 10 tens rods and ask
them to line them up end to end. Ask "How many centimeters long is one rod? (10cm). How many rods do you have all together? (ten) Can you figure out a way to quickly count how many centimeters you
have in 10 rods?" They should be able to count by tens to find that there are 100 centimeters total. Explain that 100 centimeters is equal to 1 meter. Show a meter stick and line it up underneath one
groups train of tens rods. They should be exactly the same length. Pass out a few meter sticks and have the groups take turns aligning their 10 tens rods with the meter stick. Summarize again that
100cm is equal to 1 meter. Then say, "How many rods made up a meter? (ten) So, each rod is 1 out of 10 pieces. That means each rod is 1/10 of a meter. There is a special name for 10 centimeters. We
call it a decimeter. Deci- is a prefix that means one tenth (1/10). So, a decimeter is 1/10 of a meter. Thus, each tens rod is a decimeter. A ones cube is a centimeter, a tens rod is a decimeter (or
10 cm), and 10 tens rods is a meter (which is the same as 100 cm or 10 decimeters)."
Then ask the students to line up both of their 50-centimeter rulers end to end. These two rulers should be the same length as the 10 tens rods because 50 cm + 50 cm = 100 cm. If they wanted to
measure things using a meter stick, they could just combine their 50 centimeter rulers to make one 100 centimeter ruler = 1 meter.
Lesson 4
- To relate previous experiences measuring in centimeters with opportunities for estimating lengths of objects
- To create a visual display of objects about 1 cm long, 10 cm long or 100 cm long in order to make it easier to compare relative lengths
- To give students a chance to estimate lengths of items and distances in the school
- three lengths of thick yarn, 1 cm, 10 cm and 100cm
- slips of paper to write on
- plastic bags
- digital camera(s)
- push pins
- 1 ones cube, 1 tens rod, and 2 50cm rulers taped together for each pair of students
Explain that the class will make a display by finding objects and distances in the school that are about 1 centimeter, 10 centimeters/ 1 decimeter, and 100 centimeters/ 1 meter. Pair up the students
and give each pair 1 ones cube and 1 tens rod. Have them tape each of their 50-centimeter rulers together to make a meter stick (or they could each create a line of 10 tens rods taped together to use
as a meter stick). They are going to use these manipulatives to find items around the room that are about this length. Demonstrate how to use the different measuring tools to measure objects around
the room. Then show them that they should leave the tool next to the object in order to take a picture of the object for the bulletin board. If they find portable items that are 1 cm or 10 cm, they
may opt to collect the item in it's own plastic bag instead of taking a photograph of it. They can also measure distances between two things/places. They may find that it is 1 meter between the rug
and the bookcase. They can lay a meter stick in that space and take a photo of the distance between two points.
After the class has finished, print the photographs. Ask each pair of students to make a caption for each photo using small strips of paper. The caption should say "The __________ is about ____
centimeters." or "It is about _____ cm from ____________ to __________." Pin the lengths of string on a bulletin board, leaving room to add the pictures above each length. Add a label under each yarn
declaring how many centimeters long it is (you may also want to add "decimeter" and "meter.") Pin the photos above the length of string corresponds the measured length of the object/distance. You can
continue to add to this display on later dates.
Lesson 5
- To make polygons that have perimeters of 10 cm, 20 cm, 30 cm, 40 cm, 50 cm, etc. up to 100cm
- To integrate linear measurement with a study of perimeter and plane shapes
- To provide further exposure to lengths that increase by powers of ten in order to improve student ability to estimate lengths
- Materials
- Wikki Stix (or pipe cleaners) cut or combined into lengths of 10cm, 20cm, 30cm, 40cm, 50cm, 60cm, 70cm, 80cm, 90cm, and 100cm - enough for each student to have 2 different lengths
- strips of paper to make labels
- ten 10 cm Wikki Stix for demonstration
In a whole group setting, display the different lengths of Wikki Stix (WS). Point out that they are cut into lengths that are multiples of ten. Show the 10 cm WS and then the 20 cm WS. Line up two 10
cm WS alongside of the 20 cm WS. Explain, "The 20 cm WS is two times bigger than the 10 cm WS. It is the same length as two 10 cm WS, so we say that it is two times bigger." Show the 30 cm WS next.
"This WS is the same as three of the 10 cm WS, so we say that it is three times bigger. 30 is three times bigger than 10." Continue this until you get to the 100 cm WS. "This WS is the same length as
ten 10 cm WS, so we know that this WS is ten times bigger than the 10 cm one. SO, 100 must be 10 times bigger than 10!"
Now explain that you are going to use all of these different lengths to make different polygons. Demonstrate how to bend a WS to make a triangular shape, a rectangular shape or a trapezoidal shape.
Explain that each student will get two different lengths of Wikki Stix. They should trace the length of each WS in their math journal. Then, they should use their 50 cm ruler to measure the length
and record it under the line segment they just drew for each WS. This should look something like this (not to scale):
10 cm 30 cm
After they have recorded this information, they can bend the WS to make whatever polygon they would like. Once they have made their two shapes, they will draw models of the shapes in their math
journal. They will then use their 50 cm rulers to measure the length of each side of their shapes. They should record the length of each side on the drawings like so (not to scale):
Next, they need to add up all the measurements of each side to find the perimeter of their shape. 10 cm + 5 cm + 10 cm + 5 cm = 30 cm. The perimeter should be the same as the length the Wikki Stix
were before they were bent into a polygon.
Appendix A: Singapore Math as a Model for this Unit
There has been a lot of research that the Singapore and Chinese models of structuring math in the early grades gives students a stronger foundation. (See Ginsburg, Geary and Ma.) Therefore, I think
it is very important to examine the scope and sequence of the math curriculum in Singapore.
The Singapore curriculum splits the first grade year into two volumes of study 1A and 1B. While there are other topics of study, these units have a strong focus on number sense. Additionally, the
preface of the Primary Mathematics textbooks states the following,
"The main feature of this package is the use of the Concrete a Pictorial a Abstract approach. The students are provided with the necessary learning experiences beginning with the concrete and
pictorial stages, followed by the abstract stage to enable them to learn mathematics meaningfully. This package encourages active thinking processes, communication of mathematical ideas and
problem solving."^ 12
1A starts with counting numbers 0-10 and exploring number bonds. The students are guided through number stories and visual representations of how you can combine numbers to make other numbers. They
prompt the students to find pairs of numbers that make values up to 10. They also integrate early algebraic thinking by asking students to find missing addends, instead of just focusing on
calculating sums. Then they move into subtraction, defining the operation, using word problems and pictures to give practice, and linking addition with subtraction. Once the students are proficient
with number bonds to 10, they move on to counting and comparing numbers to 20. The numbers from 11-20 are introduced pictorially by showing a group of ten plus a number of separate ones. The students
are then guided through exercises that emphasize that 13 is 10 and 3 or 15 is 10 and 5, etc. Then they are prompted to add using the strategy of making a 10 first and then counting on. When
subtracting, they are prompted to split the larger number into a 10 and its other piece first. Subtract the smaller amount from the 10 and then add on the ones that were ignored initially. Also, they
are asked to compare problems like the following 8 - 3 = 5 and 18 - 3 = 15.
1B starts with comparing numbers to 10. It quickly moves into studying numbers to 40. This is where they continue to use expanded form to represent each two-digit number. For example, 23 is 20+3 and
28 is 20+8. Then they begin to compare these larger numbers as well. I think it's important to note that with each number there is a picture that emphasizes groups of ten with some ones left over and
that these pictures/diagrams take a variety of forms. Multiplication is introduced as repeated addition and division is introduced as sharing a number of items by splitting them into equal groups. By
the end of 1B the students are working with numbers to 100. They start by counting by tens, 50, 60, 70, 80, 90 and 100. Then they add and subtract multiples of 10 to a number. They also do addition
and subtraction using the expanded forms of numbers, teaching students to look for tens to add and subtract with first.
While this is their 1^ s^ t grade curriculum, I think it frames a great beginning for my 2^ n^ d graders. Our Houghton Mifflin curriculum is sequenced in a very different manner, and it does not
place such an emphasis on the decimal structure of our number system. I have designed this unit to bring elements of Singapore's curriculum into my classroom.
Appendix B: Extended Scope and Sequence
Number Bonds to 10
My students come to second grade with a large deficit in their understanding of the relationships between numbers contained in what we usually refer to as "fact families." I think it is important to
concretely show students the relationship between the numbers in a fact family. "How are 7, 3 and 4 related? Can you show me with a picture?" When we start to represent the relationship between the
members of a "fact family," we are dealing with pictorial representations called "number bonds." It is the manipulation of these number bonds that will help our students see and remember the
connection between the numbers in a fact family.
What is ten?
Very young children in the United States are not taught to understand that our number system is based on a decimal notation. They are not usually taught to appreciate what "base-10" means until about
2^ n^ d grade. I feel that this is entirely too late to introduce such an integral and fundamental explanation of our number system.^ 13
I start my math curriculum by focusing on what a digit is because this is the foundation for later knowledge of place value and order of magnitude. I explain that there are ten digits
(0,1,2,3,4,5,6,7,8,9) that we combine in a variety of ways to represent any number we can imagine. I further explain that the way in which we arrange these digits holds meaning. I reintroduce the
digits from our decimal system with a picture of the value they represent.
See larger image
I then show the students a number of counters pictured on the overhead. They have to choose the correct symbol to represent that number, and once we agree, we write the digit on the board. I do this
with many quantities of counters from 0-9.
Once the class seems familiar with the digits, I show them ten counters. I ask them how they would write this number using the digits. I often ask, "Is there a digit for this value? Can we write this
number using just one of the digits?" When they answer, "No!" I take this opportunity to explain that in our decimal notation system we love to work with groups of ten. "Working with groups of ten is
easy. 10 is a very round number, which makes it easy to deal with. Adding something to zero is simple because any number plus zero leaves you with the number you started with (n + 0 = n). Zero
doesn't ask us to change anything. When we get a full group of ten, it deserves a higher ranking, so we bundle that ten together and call it by a new name. Instead of having ten single items, we now
have 1 group of ten." I then explain that this is how we "compose a ten", and we show that we have one ten by putting the digit for one in the tens place on a tens and ones chart. "This is how we
communicate that we have one group of ten." Then I ask, "But do we have anything else to count? Are there any counters left over after we make a ten?" When they answer correctly, I assert that since
there aren't any counters left we need to account for that in our number, so we write a 0 in the ones place to show that we have made one group of ten and have no extra pieces left to count. I can
also ask, "If we didn't write the zero, how would we know that we are talking about 1 ten and not just 1 one?"
I would then have them break into pairs to quickly practice this with amounts from 0-10. Partner A puts a number of counters into the ones place on a "Tens & Ones" chart. Partner B counts the
counters and decides if they represent a one-digit number or a group of ten. If the counters equal 10 then Partner B must pick up all of the counters and put them in a cup. The cup is then moved to
the tens place on the chart. If he or she is correct, they switch roles and continue to play. This may seem like a very low level activity for second graders, but I feel like it is extremely worth
investing the small amount of time it would take to have this discussion and practice the process.
Composing and Decomposing 10
The foundation has now been laid for further exploration of how to compose a ten. The lessons taught in the first quarter are focused on giving the students the opportunity to compose and decompose
10 in a variety of ways. The goal is that the students be able to readily retrieve all the possible ways of combining single-digit numbers to equal ten. I want them to see 2 and automatically know
that if you added 8, you would be able to compose a ten. Conversely, I want them to think that if they started with ten and needed just 6, they would have to break apart that very round number into 6
pieces and 4 pieces. There are a number of activities that you can do to reinforce the composition and decomposition of a ten.
One such activity involves displaying 10 beans in a variety of combinations on popsicle sticks. Each child gets 5 popsicle sticks, and they are asked to glue a1 bean on the left side and nine small
beans on the right side of the stick, leaving a space in the middle. For example, the popsicle sticks might look like this, where the 0's are beans.
This is a visual representation of a number bond. They can physically see that 1+9=10. Furthermore, if they flip the popsicle stick around, they see the commutative property of addition in noticing
that it's also 9+1=10. They will do this with 2 and 8, 3 and 7, 4 and 6, and 5 and 5. The sticks can be saved in bags for further explorations.
Another way of reinforcing the composition of 10 is to use Cuisenaire rods. They are measured in cm and are in lengths from 1 unit to 12 units. Therefore, you could have each student measure each
different colored rod with a cm ruler and have him/her mark on the rod how long it is. The students could then be prompted to find all of the combinations of rods that are equal length to the 10 rod.
This might look like this:
There are many more activities that one can use to increase student confidence in composing and decomposing ten. I want to stress how important this skill is as a building block for later
Once the students can readily compose and decompose a 10 (noting that I will use these terms alongside of "make" and "unmake"), they will be ready to move to addition and subtraction facts to 20. It
does not make sense to teach students to just rely on the "count on" and "count back" methods in adding and subtracting to 20. This does not instill in them an appreciation for the efficient system
that is decimal notation. Instead, it implies that our mathematics is without a logical system for communicating numbers, when in reality our system is very sophisticated.
One possible way to avoid deficits created through just counting on and counting back, is to emphasize that the numbers from 10-20 are composed of one ten and a number of ones. For example, 17 should
automatically be seen as one ten and seven ones. We can help young children internalize this view (for immediate retrieval) by consistently pairing the combined digits with the expanded form of the
decimal notation and a visual representation. This may look like the following:
If we are consistent in showing this triplet in the same ways/forms then we are helping our students see the "teen" numbers in a more useful and meaningful way. The development of this perspective is
pertinent because it gives students a uniform view of the logic of our decimal notation system. You may want to draw their attention to the illogical way we name the numbers from 11-19 in English. We
call 1 ten and 1 one eleven. It would make more sense if this was called "tenty-one" because it would follow the pattern set later with "twenty-one" or "fifty-one," for example. It would also be
helpful to try to point out how to hear the "ten" in "fourteen." "Can you hear the "ten" in 14? It sounds a little funny, but it's there at the end. So, "four-teen" means we have four more than ten,
and if you listen carefully, you can hear the "four" and the "ten." This would have to make accommodations for the number names of 11, 12, 13 or 15. For example, in "fifteen" neither the "five" nor
the "ten" are obvious in the word "fifteen."
To reiterate, if we only began to talk about "regrouping" when we deal with numbers larger than 20, we do not give any recognition to the structure of the decimal notation system. And thus, we end up
with children who are confused and unable to successfully perform higher-level addition and subtraction computations. In an attempt to counteract this, we will spend a large amount of time doing
"teen" computations with the decimal notation system in mind. We will continue to compose and decompose a 10 whenever necessary. The process may look something like this,
We can now see that we can work with 4 + 6 first.
The new 10 plus the original 10 from 14 is equal to 2 tens or 20.
When we start to represent familiar numbers as a sum of its parts, then it becomes easier to look at all the possible relationships. We could look at 17 in a variety of ways. It could be 10+7 or 11+6
or 12+5, etc. The ability to see the teen numbers in this way will give students the opportunity to approach problems from many points of reference. They could solve 17-9 by thinking first that 17 =
10 + 7. They could then proceed to do 10 - 9 = 1, and then 1 + 7 = 8. Or a child might see that 17 = 9 + 8. They then would easily recognize that they could just subtract that 9 and be left with the
8, so they would successfully solve 17 - 9 as equaling 8.
We will start with 20, since we at this point we are focused on all computations dealing with numbers up to 20. I will talk about 20 in reference to the way we discussed 10. We will compose and
decompose 20, using base 10 blocks as a way to see the actual bundling and breaking apart of the 2 groups of 10.
It is now that I will also introduce the idea of multiples of 10. We will talk about 20 as being ten times bigger than 2. We will see a quantity of 2 alongside a quantity of 20. I will ask, "Which
group looks bigger? Do you know how much bigger this group is? It's ten times bigger than the group of 2!" We will then proceed to split the 20 pieces into 10 groups of 2. (This is an opportunity to
also reinforce counting by 2s.) In this way, they will see that if you have ten groups of 2 items, you have 20 things. We will then talk about 3 and 30 in the same fashion, making the point that 30
is 10 times bigger than three. We will continue this with 40 and 50. I have chosen to stop at 50 at this point for three reasons. First, looking at five multiples of ten should be enough to establish
a pattern. Second, I don't want to explore the concept of 100 equaling ten 10s until I have laid a stronger foundation. Third, we are going to make centimeter rulers that measure up to 50
centimeters, and this activity will be a perfect compliment to the idea that 10 is 1 ten, 20 is 2 tens, 30 is 3 tens, 40 is 4 tens, and 50 is 5 tens.
Annotated Teacher Bibliography
Burns, Marilyn, About Teaching Elementary Mathematics. Sausalito: Math Solutions Publications, 2007. A wonderful resource for methodology and pedagogy. A wealth of lesson ideas!
Geary, D.C., Children's Mathematical Development: Research and Practical Applications. Washington, D.C.: American Psychological Association, 1994. A research-based psychological perspective of early
childhood mathematics teaching and learning.
Ginsburg, Alan, Steven Leinward, Terry Anstrom, and Elizabeth Pollock. What the United States Can Learn from Singapore's World Class Mathematics System. Washington, D.C.: American Institutes for
Research, 2005. A research based comparison of the mathematics pedagogy in Singapore and the United States. It looks at grades 1, 3 5 and 6 in both countries and compares specific mathematics
curricula from the U.S. with that of Singapore.
Howe, Roger, Taking Place Value Seriously, Preparing Mathematicians to Educate Teachers, http://www.maa.org/pmet/resources.html. An extremely helpful guide for a deeper understanding of elementary
level operations and their implications in higher level mathematics.
Ma, Liping, Knowing and Teaching Elementary Mathematics. Mahwah, N.J.: Lawrence Erlbaum Associates, 1999. Provides a wonderful analysis of mathematics instruction and conceptualization in the United
States as compared to China. This comparison is thorough in it's analysis of how teachers craft their explanations. A great place to start!
Singapore primary math texts, U.S. Edition Curriculum Planning and Development Division, Ministry of Education, Singapore: Federal Publications. These texts are textbooks and workbooks for students
based on the texts used in Singapore. They align nicely with the framework of this unit and are a great resource for visual aids.
Wood, Terry, "Second-Grade Classroom: Psychological Perspective" and "Creating an Environment for Learning Mathematics: Social Interaction Perspective." Journal for Research in Mathematics Education,
Monograph Number 6, Reston, VA: National Council of Teachers of Mathematics, 1993: 7-20. This monograph is full of psychological research on mathematics pedagogy and largely focuses on classroom
Annotated Children's Bibliography
Clements, Andrew. A Million Dots. New York: Simon & Schuster, 2006. This book is for older grade levels because it deals with much larger numbers and tries to convey the magnitude of a million
through pictures. It is fun to read to second graders; I have found they love being presented with a more concrete representation of "million." They also seem excited about the largeness of the
Jenkins, Steve, Actual Size. Boston: Houghton Mifflin Company, 2004. This is a wonderful book about the size of animals from around the world. Sizes are given in inches and feet, so it would be a
good book to add as an extension of linear measurement with customary units. It is also a great motivator because children often find animals interesting. You could read the book and then ask them to
estimate the lengths of other animals not mentioned in the text.
Jenkins, Steve. Hottest, Coldest, Highest, Deepest. Boston: Houghton Mifflin, 2004. This book compares different geographical features around the world in terms of size and other measurements. It
would be a great extension on linear measurement and relative size. It tries to make large measurements more understandable by comparing them to the average height of a person. This would be a great
way to tie geography into the topics covered in the unit.
Goldstone, Bruce, Great Estimations. New York: Scholastic Inc., 2006. This is a great resource for photos of collections of objects. It has pictures of groups of 10, 100 and 1,000. The first nine
pages easily relate to the concepts covered in this unit. The photos may give students a broader reference point to answer questions like, "Is it about 10 items or about 100 items? Does this look
like 100 or 1,000?"
1. R. Howe, Taking Place Value Seriously, 3.
2. R. Howe, Taking Place Value Seriously, 5.
3. R. Howe, Taking Place Value Seriously, 3.
4. R. Howe, Taking Place Value Seriously, 3.
5. L. Ma, Knowing and Teaching Elementary Mathematics, 1-27.
6. D.C. Geary, Children's Mathematical Development, 67.
7. http://www.dictionary.com
8. http://www.dictionary.com
9. R. Howe, Taking Place Value Seriously, 4.
10. R. Howe, Taking Place Value Seriously, 27-30.
11. M. Burns, About Teaching Mathematics, 75.
12. Singapore Primary Math Texts, Curriculum Planning & Development Division, Ministry of Education, Singapore, vol. 1A, 5.
13. D.C. Geary, Children's Mathematical Development, 44-46.
14. Commonwealth of Virginia, Board of Education, Standards of Learning, http://www.doe.virginia.gov/ VDOE/Superintendent/Sols/math2.pdf
15. National Council of Teachers of Mathematics. (2000) Principles and standards for school mathematics. Reston, VA: National Council of Teachers of Mathematics.
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Bel Tiburon, CA Math Tutor
Find a Bel Tiburon, CA Math Tutor
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Talmo SAT Math Tutor
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Synthetic Division Calculator
April 5th 2008, 03:23 PM
Synthetic Division Calculator
I recently put together a calculator to perform synthetic division for a polynomial of maximum degree of 6 or less divided by (x - c) where c is a constant that the user enters.
Located on the algebra page, and searchable by synthetic, division, quotient, polynomial, this calculator will show you each step in the synthetic division process. It provides the quotient
answer including a remainder amount if applicable. I've done some moderate testing of this program. Please let me know if anybody wants enhancements, or sees corrections that can be made.
Calculator instructions are located inside the link at the top of the page.
For the math work section, I wanted to follow how they do it on Wikipedia here ---> Ruffini's rule - Wikipedia, the free encyclopedia
The calculator is located here, enjoy:
Synthetic Division
November 10th 2008, 07:35 PM
Enhancement update. This calculator has been written in a new language to better line up columns and rows as well as color code entries for each step in the synthetic division process to better
help you follow along.
A few other design items were added. The search engine keywords are the same as well as the link to the program.
I've received no feedback on expanding the polynomial entries past a degree of 6, so I'll keep it that way unless I hear otherwise.
As always, let me know if you want enhancements or see any errors.
January 28th 2009, 01:01 PM
Hi, I love your synthetic calculator. It's freaking amazing...however I think this is a mistake although I know I'm not a genius at math...
Synthetic Division
and you come to the part where -3 is supposed to be added to -1 and to me that seems like it would be -4 but your calculator solves it as -1. Therefore throwing the rest of the equation off.
Please let me know if I'm just a moron...or is that not correct?(Bow)
January 28th 2009, 03:19 PM
Hi, I love your synthetic calculator. It's freaking amazing...however I think this is a mistake although I know I'm not a genius at math...
Synthetic Division
and you come to the part where -3 is supposed to be added to -1 and to me that seems like it would be -4 but your calculator solves it as -1. Therefore throwing the rest of the equation off.
Please let me know if I'm just a moron...or is that not correct?(Bow)
No, you are correct. First off, thank you for the kind words. It's thanks to you that the site becomes more useful.
Here is what happened, and I've fixed it for future users. It looks like you had a - 3 and I was only removing space to the left and the right of the numbers from what you entered.
So now, I've fixed the code to remove blank space between the negative sign and numbers, so - 7 becomes -7. If you don't see the change, refresh your browser. I tested the fix and it worked.
Thanks again, this will help other users in the future if they use the site.
Please let me know if I can help with anything else.
January 29th 2009, 03:48 AM
Hey thanks for this calculator. Taking it in step by step helped me to understand what I was doing wrong. I couldn't figure out what to do with the answer I had gotten but now I know what the
numbers are for now! Thanks again:D In the answer though where 3x+18+111/(x+6) shouldn't the sign in x+6 switch to a negative? here's the origional equation / (3x^2-10x)/(x-6)
January 29th 2009, 05:21 AM
Hey thanks for this calculator. Taking it in step by step helped me to understand what I was doing wrong. I couldn't figure out what to do with the answer I had gotten but now I know what the
numbers are for now! Thanks again:D In the answer though where 3x+18+111/(x+6) shouldn't the sign in x+6 switch to a negative? here's the origional equation / (3x^2-10x)/(x-6)
Yes, I saw that for some remainders and have fixed it. Good catch! I've made the change and saved it.
January 30th 2009, 07:06 PM
One more update for the answer portion. I've placed the remainder in a fraction format if a remainder exists as well as offer a small paragraph explanation as to how we take our results line and
form our quotient equation.
March 20th 2010, 01:06 PM
It's a very cool program (Clapping), and has helped a bunch; however, it appears you currently don't have support for fractional divisors.
As well, if using a decimal it freaks out.
ex. .666666666666666666666666666666666
It did fine till it attempted to subtract .66666666666666666666 from 1
and than it gave me 1 as an answer, which is clearly wrong.
March 20th 2010, 03:06 PM
It's a very cool program (Clapping), and has helped a bunch; however, it appears you currently don't have support for fractional divisors.
As well, if using a decimal it freaks out.
ex. .666666666666666666666666666666666
It did fine till it attempted to subtract .66666666666666666666 from 1
and than it gave me 1 as an answer, which is clearly wrong.
I've added this feature to handle fractional roots. You were the first person every to ask for this. Good idea!
I've also added a link back to our fraction lesson for any addition/multiplication step that involves a fractional root or interim math step.
The only thing we may need in the future is a GCF reduction if any of our steps are not fully reduced. Let me know what you think.
March 20th 2010, 03:53 PM
July 28th 2010, 08:06 AM
Hello Mathceleb,
I must say that this is commendable. However, is there currently no functionality for solving using trinomial divisors, or have I had just an oversight?
July 28th 2010, 05:21 PM
Yes, because a trinomial divisor is polynomial long division. Have you seen this calculator?
August 7th 2010, 10:31 PM
The thread seems to not exist. Also, if I am not mistaken, it can be done through Synthetic division, not the usual long method.
August 7th 2010, 11:42 PM
I corrected the thread URL.
Let me know if you have questions.
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An introduction to wavelets
Results 11 - 20 of 547
, 1998
"... This paper describes WBIIS (Wavelet-Based Image Indexing and Searching), a new image indexing and retrieval algorithm with partial sketch image searching capability for large image databases.
The algorithm characterizes the color variations over the spatial extent of the image in a manner that provi ..."
Cited by 91 (21 self)
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This paper describes WBIIS (Wavelet-Based Image Indexing and Searching), a new image indexing and retrieval algorithm with partial sketch image searching capability for large image databases. The
algorithm characterizes the color variations over the spatial extent of the image in a manner that provides semantically meaningful image comparisons. The indexing algorithm applies a Daubechies'
wavelet transform for each of the three opponent color components. The wavelet coefficients in the lowest few frequency bands, and their variances, are stored as feature vectors. To speed up
retrieval, a two-step procedure is used that first does a crude selection based on the variances, and then renes the search by performing a feature vector match between the selected images and the
query. For better accuracy in searching, two-level multiresolution matching may also be used. Masks are used for partial-sketch queries. This technique performs much better in capturing coherence of
image, object granular...
- IEEE Computer Graphics and Applications , 1995
"... this paper. Thanks also go to Ronen Barzel, Steven Gortler, Michael Shantzis, and the anonymous reviewers for their many helpful comments. This work was supported by NSF Presidential and
National Young Investigator awards (CCR-8957323 and CCR-9357790), by NSF grant CDA9123308, by an NSF Graduate Res ..."
Cited by 87 (1 self)
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this paper. Thanks also go to Ronen Barzel, Steven Gortler, Michael Shantzis, and the anonymous reviewers for their many helpful comments. This work was supported by NSF Presidential and National
Young Investigator awards (CCR-8957323 and CCR-9357790), by NSF grant CDA9123308, by an NSF Graduate Research Fellowship, by the University of Washington Royalty Research Fund (65-9731), and by
industrial gifts from Adobe, Aldus, Microsoft, and Xerox. References
- J. Fourier Anal. Appl , 1996
"... After briefly reviewing the interrelation between Riesz-bases, biorthogonality and a certain stability notion for multiscale basis transformations we establish a basic stability criterion for a
general Hilbert space setting. An important tool in this context is a strengthened Cauchy inequality. It i ..."
Cited by 86 (22 self)
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After briefly reviewing the interrelation between Riesz-bases, biorthogonality and a certain stability notion for multiscale basis transformations we establish a basic stability criterion for a
general Hilbert space setting. An important tool in this context is a strengthened Cauchy inequality. It is based on direct and inverse estimates for a certain scale of spaces induced by the
underlying multiresolution sequence. Furthermore, we highlight some properties of these spaces pertaining to duality, interpolation, and applications to norm equivalences for Sobolev spaces. AMS
Subject Classification: 41A17, 41A65, 46A35, 46B70, 46E35 Key Words: Riesz bases, biorthogonality, stability, projectors, interpolation theory, K-method, duality, Jackson, Bernstein inequalities 1
Background and Motivation A standard framework for approximately recapturing a function v in some infinite dimensional separable Hilbert space V , say, either from explicitly given data or as a
solution of an operator equ...
- Journal of Computational and Graphical Statistics , 1996
"... The theory of wavelets has recently undergone a period of rapid development. We introduce a software package called wavethresh that works within the statistical language S to perform one- and
two-dimensional discrete wavelet transforms. The transforms and their inverses can be computed using any par ..."
Cited by 81 (24 self)
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The theory of wavelets has recently undergone a period of rapid development. We introduce a software package called wavethresh that works within the statistical language S to perform one- and
two-dimensional discrete wavelet transforms. The transforms and their inverses can be computed using any particular wavelet selected from a range of different families of wavelets. Pictures can be
drawn of any of the one- or twodimensional wavelets available in the package. The wavelet coefficients can be presented in a variety of ways to aid in the interpretation of data. The package's
wavelet transform "engine" is written in C for speed and the object-orientated functionality of S makes wavethresh easy to use. We provide a tutorial introduction to wavelets and the wavethresh
software. We also discuss how the software may be used to carry out nonlinear regression and image compression. In particular, thresholding of wavelet coefficients is a method for attempting to
extract signal from noise and ...
- IEEE TRANS. ON SIGNAL PROCESSING , 1993
"... This paper constructs K-regular M-band orthonormal wavelet bases. K-regularity of the wavelet basis is known to be useful in numerical analysis applications and in image coding using wavelet
techniques. Several characterizations of K-regularity and their importance are described. An explicit formula ..."
Cited by 79 (6 self)
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This paper constructs K-regular M-band orthonormal wavelet bases. K-regularity of the wavelet basis is known to be useful in numerical analysis applications and in image coding using wavelet
techniques. Several characterizations of K-regularity and their importance are described. An explicit formula is obtained for all minimal length M-band scaling filters. A new state-space approach to
constructing the wavelet filters from the scaling filters is also described. When M-band wavelets are constructed from unitary filter banks they give rise to wavelet tight frames in general (not
orthonormal bases). Conditions on the scaling filter so that the wavelet bases obtained from it is orthonormal is also described.
, 1994
"... Let X be a countable fundamental set in a Hilbert space H, and let T be the operator T : ` 2 (X) ! H : c 7! X x2X c(x)x: Whenever T is well-defined and bounded, X is said to be a Bessel
sequence. If, in addition, ran T is closed, then X is a frame. Finally, a frame whose corresponding T is inje ..."
Cited by 75 (22 self)
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Let X be a countable fundamental set in a Hilbert space H, and let T be the operator T : ` 2 (X) ! H : c 7! X x2X c(x)x: Whenever T is well-defined and bounded, X is said to be a Bessel sequence. If,
in addition, ran T is closed, then X is a frame. Finally, a frame whose corresponding T is injective is a stable basis (also known as a Riesz basis). This paper considers the above three properties
for subspaces H of L 2 (IR d ), and for sets X of the form X = fOE(\Delta \Gamma ff) : OE 2 \Phi; ff 2 ZZ d g; with \Phi either a singleton, a finite set, or, more generally, a countable set. The
analysis is performed on the Fourier domain, where the two operators TT and T T are decomposed into a collection of simpler "fiber" operators. The main theme of the entire analysis is the
characterization of each of the above three properties in terms of the analogous property of these simpler operators. AMS (MOS) Subject Classifications: 42C15 Key Words: Riesz bases, stable bases,
- BIOMETRIKA , 1998
"... ..."
- Department of Mathematics, MIT, Cambridge MA , 213
"... Abstract. This note is a very basic introduction to wavelets. It starts with an orthogonal basis of piecewise constant functions, constructed by dilation and translation. The "wavelet transform
" maps each f(x) to its coefficients with respect to this basis. The mathematics is simple and the transfo ..."
Cited by 71 (2 self)
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Abstract. This note is a very basic introduction to wavelets. It starts with an orthogonal basis of piecewise constant functions, constructed by dilation and translation. The "wavelet transform "
maps each f(x) to its coefficients with respect to this basis. The mathematics is simple and the transform is fast (faster than the Fast Fourier Transform, which we briefly explain), but
approximation by piecewise constants is poor. To improve this first wavelet, we are led to dilation equations and their unusual solutions. Higher-order wavelets are constructed, and it is
surprisingly quick to compute with them — always indirectly and recursively. We comment informally on the contest between these transforms in signal processing, especially for video and image
compression (including highdefinition television). So far the Fourier Transform — or its 8 by 8 windowed version, the Discrete Cosine Transform — is often chosen. But wavelets are already
competitive, and they are ahead for fingerprints. We present a sample of this developing theory. 1. The Haar wavelet To explain wavelets we start with an example. It has every property we hope for,
except one. If that one defect is accepted, the construction is simple and the computations are fast. By trying to remove the defect, we are led to dilation equations and recursively defined
functions and a small world of fascinating new problems — many still unsolved. A sensible person would stop after the first wavelet, but fortunately mathematics goes on. The basic example is easier
to draw than to describe: W(x)
, 1995
"... We are presenting a new method for adaptive surface meshing and triangulation which controls the local level-of-detail of the surface approximation by local spectral estimates. These estimates
are figured out by a wavelet representation of the surface data. The basic idea is to decompose the initial ..."
Cited by 70 (3 self)
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We are presenting a new method for adaptive surface meshing and triangulation which controls the local level-of-detail of the surface approximation by local spectral estimates. These estimates are
figured out by a wavelet representation of the surface data. The basic idea is to decompose the initial data set by means of an orthogonal or semi-orthogonal tensor product wavelet transform (WT) and
to analyze the resulting coefficients. In surface regions, where the partial energy of the resulting coefficients is low, the polygonial approximation of the surface can be performed with larger
triangles without loosing too much fine grain details. However, since the localization of the WT is bound by the Heisenberg principle the meshing method has to be controlled by the detail signals
rather than directly by the coefficients. The dyadic scaling of the WT stimulated us to build an hierachical meshing algorithm which transforms the initially regular data grid into a quadtree
representation by...
, 2001
"... This report summarizes the master's thesis Islands of Music: Analysis, Organization, and Visualization of Music Archives, which I submitted to the Vienna University of Technology on December
11th, 2001. I wrote it at the Department of Software Technology and Interactive Systems, supervised by Dr. An ..."
Cited by 68 (15 self)
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This report summarizes the master's thesis Islands of Music: Analysis, Organization, and Visualization of Music Archives, which I submitted to the Vienna University of Technology on December 11th,
2001. I wrote it at the Department of Software Technology and Interactive Systems, supervised by Dr. Andreas Rauber, and assessed by Prof. Dr. Dieter Merkl
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Braingle: 'Hats and Feathers' Brain Teaser
Hats and Feathers
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #3711
Category: Probability
Submitted By: rayneeday
Corrected By: cnmne
You have five boxes with five hats in each and each hat has five feathers with five colored dots. Now the hats are in a rainbow of color and so are the feathers but the dots are all various random
combinations of red, yellow, and blue-green.
What is the probability of getting a hat with only red dots?
Show Hint
Show Answer
What Next?
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Playing Games in the Transfinite: An Introduction to "Ordinal Chomp"
Playing Games in the Transfinite: An Introduction to “Ordinal Chomp”
Chomp is a two-player game which is played as follows: The two players, A and B, start with a “board” which is a chocolate bar divided into $n \times m$ small squares. With Player A starting, they
take turns choosing a square and eating it together with all squares above and to the right. The catch is that the square at the lower left-hand corner is poisonous, and the player who is forced to
eat it loses.
This image from the Wikipedia article shows a typical sequence of moves for a $5\times 3$ chocolate bar:
At this point, Player A is forced to eat the poisoned square and hence loses the game.
Although the question of what the winning strategies are for this game is very much an open problem, the question of who has a winning strategy is not: On the $1\times 1$ board, Player B wins (since
Player A must eat the poison piece on his first move). But for any other board, Player A has a winning strategy.
To see why, suppose not. Then if Player A’s first move is to eat just the one square in the top right-hand corner, Player B must have a winning response (since we are supposing that Player B has a
winning response to any move that Player A makes). But if Player B’s response is winning, then Player A could have simply made that move to start with.
However, suppose we play Chomp not just on $n\times m$ boards, but on $\alpha\times\beta$ boards, where $\alpha$ and $\beta$ are arbitrary ordinals. The game still makes sense just as before, and
will always end in finite time, but Player A no longer wins all of the time (there will no longer be a top right-hand corner square if either $\alpha$ or $\beta$ is a limit ordinal).
Scott Huddleston and Jerry Shurman investigated Ordinal Chomp in this paper, and showed that it has a number of interesting properties. I’ll describe a few of them below.
First of all, to make things a bit easier to discuss, we will consider only $n\times \alpha$ games where $n$ is finite rather than $\beta\times\alpha$ games where $\beta$ is arbitrary. However,
everything said will go through fine in that case as well.
Secondly, we will use the following game which is equivalent to Chomp: At any point in the game, the board is described by a non-increasing sequence $(\alpha_0,\ldots, \alpha_n)$. On each turn, the
appropriate player picks an $i$ and a $\beta$ such that $\beta < \alpha_i$ and replaces the sequence with $(\alpha_0,\ldots, \alpha_{i-1},\beta,\ldots, \beta)$. We call this taking a bite of height $
\beta$. Playing on an $n\times m$ chocolate bar as described above corresponds to playing with the sequence $(m,\ldots, m)$ consisting of $n$$m$‘s.
For ease of discussion later, we make the convention that any sequence (not just a non-increasing one) describes a game position by stipulating that $(\alpha_1,\ldots, \alpha_n)$ is the same as the
position $(\alpha_1,\min\{\alpha_1,\alpha_2\},\ldots,\min\{\alpha_1,\ldots,\alpha_n\})$.
We saw above that Player A wins all non-trivial finite Chomp games, so let’s start by looking at a transfinite chomp game that Player B wins: $2\times \omega$, or $(\omega,\omega)$ in our new
notation. What is Player B’s winning strategy? Well, notice that Player A’s first move has to put the game either in the state $(n,n)$ for some $n$ or $(\omega, n)$ for some $n$. In either case,
Player B can then move the game into state of the form $(m+1,m)$ for some $m$. From a position of that form, whatever Player A does, Player B can again move to a position of that form. Eventually,
Player B will move to the position $(1,0)$, and Player A will be forced to eat the poison piece.
So, Player B can win at least some transfinite Chomp games, although he still loses a lot of them: for example, he loses all games of the form $2\times \alpha$ where $\alpha > \omega$. The reason is
that in such a game Player A can win by first moving to the position $2\times \omega$ and then using Player B’s winning strategy! Similarly, Player B loses all games $n \times \omega$ for $n > 2$.
In fact, for any ordinal $\alpha$, there is exactly one ordinal $\beta$ such that Player B wins $\alpha\times\beta$. This is a consequence of what Huddleston and Shurman call the Fundamental Theorem
of Transfinite Chomp in the above paper. Another interesting consequence, which illustrates the style of reasoning used in their proof of the Fundamental Theorem, is the following:
Theorem: For any sequence $\alpha_2,\ldots, \alpha_n$ of ordinals, there is exactly one ordinal $\alpha_1$ such that Player B wins the game $(\alpha_1,\ldots,\alpha_n)$. (Remember the convention
above about sequence which are not necessarily non-increasing.)
Proof: The uniqueness is similar to the argument given above: If Player B has a winning strategy on the game $(\alpha_1,\ldots,\alpha_n)$ then Player A has a winning strategy on all games $(\
alpha'_1,\alpha_2,\ldots,\alpha_n)$ where $\alpha'_1 > \alpha_1$, since Player A can just move to the position $(\alpha_1,\ldots, \alpha_n)$ and then use Player B’s winning strategy.
The existence is by induction. Fix $\alpha_2,\ldots, \alpha_n$ and suppose that for all $\beta_2,\ldots,\beta_n$ where $\beta_i \leq \alpha_i$ for all $i$ and for at least one $i$, $\beta_i < \
alpha_i$ we know that there is a $\beta_1 = h(\beta_2,\ldots,\beta_n)$ such that Player B has a winning strategy.
For each ordinal $\gamma$, let $B_\gamma$ be the set of all $(\beta_2,\ldots, \beta_n)$ obtainable from $(\alpha_2,\ldots, \alpha_n)$ by taking a bite of height $\gamma$ (this term was defined above,
if you forgot what it means).
Let $H = \{ h(\beta_2,\ldots,\beta_n)\mid (\beta_2,\ldots,\beta_n)\in B_\gamma\text{ and }h(\beta_2,\ldots,\beta_n) > \gamma \}$. Let $\alpha_1$ be the minimal ordinal not in $H$. I claim that Player
B has a winning strategy for $(\alpha_1,\ldots,\alpha_n)$. We will show this by showing that, for any move Player A makes, Player B can make a move that we know leads to a winning strategy for B.
So suppose Player A moves to $(\beta_1,\ldots, \beta_n)$. Either $\beta_1 < \alpha_1$ or $\beta_1 = \alpha_1$. First suppose that $\beta_1 = \alpha_1$. This means that Player A moved by taking a bite
of height (say) $\gamma$ out of $(\alpha_2,\ldots, \alpha_n)$ by moving to $(\alpha_1,\beta_2,\ldots, \beta_n)$. But by construction, we know that $\alpha_1 e h(\beta_2,\ldots, \beta_n)$, which means
that the player who is to move (Player B in this case) has a winning strategy.
Now suppose $\beta_1 < \alpha_1$. This means by construction of $\alpha_1$ that $\beta_1 = h(\beta'_2,\ldots, \beta'_n)$ where $\beta'_i \leq \beta_i$ for all $i$ and $\beta'_i < \beta_i$ for at
least one $i$. Thus, if Player B moves to the position $(\beta_1,\beta'_2,\ldots, \beta'_n)$ he ensures himself a win. $\square$.
Note that this proof is constructive. This means that you can actually use it to compute that (as we already know), the unique $\alpha$ such that Player B wins $(\alpha,\omega)$ is $\omega$. As a
puzzle, you might like to find the unique $\alpha$ such that Player B wins $(\alpha,\omega,1)$ (or such that Player B wins $(\alpha,\omega,\omega)$ or $(\alpha,\omega^\omega,\omega^2)$ or whatever
you like, although the last one will be hard).
The much-more-general Fundamental Theorem of Transfinite Chomp in the paper linked above is also constructive. It allows you, in theory, to compute who will win the $n$-dimensional game (we have been
considering $2$-dimensional games) $\alpha_1\times \cdots \times \alpha_n$ for any ordinals $\alpha_i$. However, this is quite difficult in practice: according to the Wikipedia article, it is an open
question who wins the $3$-dimensional game $\omega\times\omega\times\omega$.
As a final note, in the book Tracking the Automatic Ant, David Gale gives a very nice non-constructive proof that for all $n$, there is a unique $\alpha$ such that Player B wins $n\times \alpha$.
4 responses to “Playing Games in the Transfinite: An Introduction to “Ordinal Chomp””
1. No doubt it is a silly point and perhaps I am just being thick here, but here goes anyway… Now, if we speak of an n x m board where n and m are finite, this uniquely picks out both 1) a board
whose highest column/row are specified by the finite ordinals n and m, and 2) a board whose column/row count has cardinality the cardinality of the ordinals n and m. You get both, because these
descriptions pick out the very same board.
But this nomenclature breaks down when you turn to speak of an alpha x beta board where alpha and/or beta are limit ordinals. The board whose highest column has ordinal alpha is just one of the
many board(s) whose column count has the cardinality of alpha.
If when you speak of an alpha x beta board, you mean to pick out board(s) by cardinality, then you will have picked out not a single board but a rather large collection of boards, and in
particular have not distinguished between the board with a column for every n < alpha and no more, and the board whose highest column has ordinal alpha. So, in that case, your nomenclature cannot
distinguish between these two rather different sorts of board.
Notice, it would then be a mistake to assert that there is no upper right square on an alpha x beta board, because one of the boards thus picked out does have a highest column/row, namely the one
whose highest column has ordinal alpha, and whose highest row has ordinal beta.
On the other hand, if your talk of an alpha x beta board is meant to speak of a board whose highest column/row have the ordinals alpha and beta respectively, then in this case it would again be a
mistake to say that there would be no upper right square. It is the square at column alpha, row beta.
Finally, maybe the nomenclature is supposed to work differently for limit and non-limit ordinals. One might say that an alpha x beta board means a board with a) columns for every n<=alpha for
non-limit alpha, and b) columns for every n<alpha for limit alpha; & ditto for rows. But this would not be quite happy either, since we would be skipping a board at every limit ordinal, namely
the one whose highest column has ordinal alpha for limit alpha (ditto for rows).
Maybe I am just trying to think too early in the morning…
2. Actually, it’s just that an $\alpha\times \beta$ board has a column for each ordinal $\alpha'$ where $0\leq \alpha' < \alpha$ (and similarly for the rows with $\beta$).
I believe that works for all cases, but I definitely should have made it more clear.
3. In your discussion of non-non-increasing sequences I think you mean “min” instead of “max”.
4. Quite right. Thanks for the correction.
Filed under Uncategorized
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Understanding Compressors Curves
Compressor Curves are generally misunderstood in day-today working of process engineers and many confusions arises out of discussion among cross functional teams. This becomes more confusing in case
of Recycle loop systems for example in Ammonia synthesis loop OR Ethylene oxide Synthesis Loop.
In case of Compressors, first thing is to understand the Head Vs Flow Curve. Please do not get confused between Head and discharge pressure which are significantly different than pumps where usually
fluid specific gravity is around 1.0.
Now Let us first note down the requirement, as below
1. Curve Readings - Head & Flow from Design Curve.
2. Speed at Which the Curve is applicable.
3. Design Suction P & T.
4. Design Molecular Weight of the Gas
These are the minimum requirements to simulate your compressor for any modification study.
Now if you need to consider any speed change it can be (As shown in attached file), otherwise new operating conditions can be specified. So the first step is to consider the new flow & head curve
based on revised speed.
Then the program in attached sheet calculates actual suction volume at design conditions and then calculates developed head at new speed at specified flow. This head can be converted to new
conditions in terms of discharge pressure.
Or if desired head is given new flow can be calculated by iterations from the given curve at revised speed.
In this way when we convert actual condition to design basis and consider converted suction volume the original curve is still applicable. This helps in avoiding any confusion due to change in
suction condition in actual plant operation compared to design. The fundamental is simple - if you convert gas volume at design condition that will be understood by the compressor as if it is
operating on the curve.
OR if compressor handles Actual M3 of flow at design condition then the developed head point will lie on the original curve or revised curve if speed is changed.
The assumptions in this sheet are
1. No speed change beyond 10% of design value because errors are larger due to change in internal flow pattern.
2. No change in efficiency. Though I have considered efficiency correction factor if you wish to apply it manually.
3. No thermodynamic calculation of discharge temperature is done.
If you have any query related to any compressor problem kindly let me know. I have also developed a BC++ program for any high pressure real gas mixture compression system which uses equation of
states for the calculation of thermodynamic properties of the gaseous mixture and its efficiency, which is tested for highly non ideal system also upto 220 bar pressure.
To Download the file , just click .... Compressor Evaluation
No of Downloads
Readymade pack is available for Hydrogen, Air, Nitrogen, Oxygen, Synthesis Gas of Ammonia plant, Recycle gas for EO plant, CO2. For any other gaseous mixture it can be easily modified.
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Subspace of $L^2$ that lies in $L^\infty$
up vote 20 down vote favorite
Let $E$ be a closed subspace of $L^2[0,1]$. Suppose that $E\subset{}L^\infty[0,1]$. Is it true that $E$ is finite dimensional?
PS. This is actually a question from the real analysis qualifier. I came across it as I was teaching qualifier preparation course, and was solving problems from old qualifiers. So, though it might
follow from some advanced theory of Banach spaces, I am most interested in the 'elementary' solution, using only methods from standard real analysis course. Note: if $E\subset{}C[0,1]$, then it is a
problem from Folland, and there is a solution there. However, it does not work for $L^\infty$, not without some trick.
fa.functional-analysis banach-spaces hilbert-spaces
Sorry, I left a reference (to Rudin's book) as an answer without properly reading your question, and in particular the fact that you wanted a proof avoiding functional-analytic tricks – Yemon Choi
Jan 19 '11 at 19:12
Since it seems that using the open mapping theorem is allowed, perhaps the reference to Rudin might still be of interest? It's Theorem 5.2 in Functional Analysis (2nd ed) – Yemon Choi Jan 20 '11 at
add comment
3 Answers
active oldest votes
Another solution: as Mikael wrote, $||f||_{\infty} \leq C ||f||_2$ for every $f \in E$. Let $f_1,\ldots,f_n$ be an orthonormal family in your subspace. Then for every $x \in [0,1]$,
$f_1(x)^2+\ldots+f_n(x)^2 \leq ||f_1(x)f_1+\ldots+f_n(x)f_n||_{\infty} \leq C \|f_1(x)f_1+\ldots+f_n(x)f_n\|_2$ $$=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2},$$
up vote 25 down
vote accepted and by squaring we get $f_1(x)^2+\ldots+f_n(x)^2 \leq C^2$, and integrating gives $n \leq C^2$.
5 I like this!!!! – Nate Eldredge Jan 19 '11 at 17:06
1 Many thanks! This is actually the trick to extend the solution from Folland from $C[0,1]$ to $L^\infty$. – Rostyslav Kravchenko Jan 19 '11 at 18:44
4 This is the "right" elementary solution and is what the makers of the qualifying exam had in mind. At a deeper conceptual level, the problem is obvious (the 2-summing norm of the
identity operator on any $n$-dimensional space is $\sqrt {n}$). – Bill Johnson Jan 19 '11 at 22:53
add comment
First note that by the closed graph theorem, there is a $C$ such that $\|f\|_\infty \leq C\|f\|_2$ for any $f$ iin $E$. (this can also be checked directly if you consider that the closed
graph theorem is advanced Banach space theory).
Assume $E$ is infinite dimensional, and take $(f_n)_{n\ge 1}$ an orthonormal basis of $E$. Let $A_n$ be the subset of $[0,1]$ on which the real part of $f_n$ is greater than $1/10$.
Replacing if necessary $f_n$ by $-f_n$ or $i f_n$ or $-if_n$, we can assume that $\|Re(f_n)^+\|_2 \ge 1/2$, so that the measure of $A_n$ is greater than some constant $\delta$ depending on
up vote 7 $C$ only ($\delta = (1/2^2-1/10^2)/C^2$ works for example).
down vote
Since the integral of $1_{A_1}+1_{A_2}+...+1_{A_n}$ is greater than $n\delta$, there exist $i_1,...i_k$ with $k$ being the integer part of $n\delta$, such that $A_{i_1},...,A_{i_k}$ have
non-trivial intersection. Then the sum $1/\sqrt k (f_{A_{i_1}}+...f_{A_{i_k}})$ has $L^2$ norm $1$, but $L^\infty$ norm greater than $\sqrt k/10$ (because its real part is greater than $\
sqrt k /10$ on a non-trivial subset). A contradiction.
3 Could you give a hint on how you can check $\|f\|_{\infty} \leq C \|f\|_{2}$ without appealing to the open mapping theorem (or closed graph theorem)? – Theo Buehler Jan 19 '11 at 16:45
1 In fact I was not very honest: the proof I had in mind was more or less a proof of the closed graph theorem (see arxiv.org/abs/1005.1585). Anyway, here it goes. A first attempt would be
to take a sequence $g_n$ in the unit ball of $E$ such that $\|g_n\|_\infty > 4^n$, consider $f=\sum_n \omega_n 2^{-n} g_n$ with a good choice of $\omega_n\in \mathbb C$ with $|\omega_n|=
1$, and hope that $f$ is not bounded. I do not believe that this works in general, but if the $g_n$ are chosen carefully this will work... – Mikael de la Salle Jan 19 '11 at 22:05
... In fact $\|g_n\|_\infty>4^n$ means that there is $A_n \subset [0,1]$ of positive measure such that the (modulus of the) average $E[g|A_n]$ of $g_n$ on $A_n$ is greater than $4^n$.
Changing $g_n$ if necessary, we can assume that $g_n$ is "the worst" such function, ie that $E[g_n|A]= c_n > 4^n$, and that $|E[g|A]| \leq (c_n+1) \|g\|_2$ for any $g \in E$. So define
$f_0=g_0$ and $f_{n+1} = f_n+ 2^{-n} e^{-i\theta_n} g_{n+1}$, where $\theta_{n}$ is an argument of $E[f_n|A_{n+1}]$, so that $|E[f_{n+1}|A_{n+1}]|\geq c_n/2^n$... – Mikael de la Salle
Jan 19 '11 at 22:15
Hum, I noticed that I already got it wrong: I should have replaced the $2^{-n}$ by $3^{-n}$ in the definition of $f_{n+1}$, and I would conclude that $|E[f_n|A_n]|\geq c_n 3^{-n+1}$.
Then $f=\lim f_n$ belongs to $E$, and $\|f-f_n\|_2 \leq 3^{-n+1}/2$, so that $|E[f|A_n]| \ge |E[f_n|A_n]|-(c_n+1)\|f_f_n\|_2 \ge 3^{-n+1}(c_n -1)/2$. This goes to $\infty$, and proves
that $f$ is not in $L^\infty$. – Mikael de la Salle Jan 19 '11 at 22:38
Thank you very much, I think I see what you're saying and I'll have to have a closer look at Sokal's paper which looks really neat! While trying to answer the original question I came up
with something similar as you're sketching now and I was afraid that a substantially simpler argument escaped me - I interpreted using the open mapping theorem as "cheating" from the PS
in the question... – Theo Buehler Jan 19 '11 at 23:11
add comment
Here's one solution. There may be cleaner ones.
Let $E$ be as supposed. The natural inclusion $T : L^\infty([0,1]) \hookrightarrow L^2([0,1])$ is bounded, so $E = T^{-1}(E)$ is therefore also closed in $L^\infty$. By the open mapping
theorem, it follows that $T^{-1}$ is bounded on $E$, so there exists $C$ such that for all $f \in E$, $||f||_\infty \le C ||f||\_2$. Now if $||f||\_2 = 1$, we have $||f||\_\infty \le C$, and
so by noting $$1 = \int |f|^2 \le C^2 m(|f| > \epsilon) + \epsilon^2$$ and taking, say, $\epsilon = 1/2$, we have $m(|f| > 1/2) \ge 1/4C^2$.
up vote
7 down Now suppose $E$ is infinite dimensional; then it contains an $L^2$-orthonormal sequence $\{f_n\}$. By replacing $f_n$ by $-f_n$ as necessary we may assume that for each $f_n$, $m(f_n > 1/2)
vote \ge 1/8C^2$. By a pigeonhole argument there is a set $A$ of positive measure where $f_{n_k} > 1/2$ for infinitely many $n_k$ (edit: actually I am not sure about this step). Now $f = \sum_k k
^{-1} f_{n_k}$ converges in $L^2$ and so $f \in E$; since the $L^2$ and $L^\infty$ norms are equivalent on $E$, the sum also converges uniformly to $f$ on a set of full measure. But this
implies that $f = +\infty$ a.e. on $A$, which is absurd.
1 Oops, Mikael beat me. – Nate Eldredge Jan 19 '11 at 15:44
It is funny that we had exactly the same construction. I just do not follow your use of the pigeohole principle. – Mikael de la Salle Jan 19 '11 at 16:29
1 Hm, in retrospect, neither do I. – Nate Eldredge Jan 19 '11 at 17:07
@Nate, I think that you are taking $A = \limsup \{f_n > 1/2\}$ and observing that $m(A) \ge \limsup m(f_n > 1/2) \ge 1/8C^2$. – L Spice Jan 19 '11 at 22:35
@L Spice: That's what I was thinking, but unfortunately, $A = \limsup \\{ f_n > 1/2\\}$ doesn't quite do the trick. It gives us that for each $x \in A$, there is some sequence $f_{n_k}$
1 with $f_{n_k}(x) > 1/2$ for all $n_k$; unfortunately, the particular sequence that works may vary with $x$, and it is not clear to me that there is a single sequence that works for all $x
\in A$ (or even a positive-measure subset). – Nate Eldredge Jan 19 '11 at 23:06
add comment
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Vector Addition with Integer Components
written by Andrew Duffy
supported by the National Science Foundation
The Vector Addition with Integer Components model allows the user to split a vector into its components, and practice finding the magnitude and direction of a vector if you know the components. In
this simulation the x and y components of each vector are all integers. In the "Find components" mode, you are given the magnitude and direction of the vector, and your goal is to find the
x-component and the y-component of the vector. In the "Find magnitude and direction" mode, you are given the two components, and you need to find the magnitude and direction of the vector.
The Vector Addition with Integer Components model was created using the Easy Java Simulations (EJS) modeling tool. It is distributed as a ready-to-run (compiled) Java archive. Double clicking the
ejs_bu_vector_components_integer.jar file will run the program if Java is installed.
Please note that this resource requires at least version 1.5 of Java (JRE).
Vector Addition with Integer Components Source Code
The source code zip archive contains an XML representation of the Vector Addition with Integer Components model. Unzip this archive in your EJS workspace to…
more... download 5kb .zip
Published: April 25, 2010
Subjects Levels Resource Types
Classical Mechanics
- Instructional Material
- Newton's Second Law
- High School = Curriculum support
= Force, Acceleration
- Lower Undergraduate = Interactive Simulation
Mathematical Tools
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- Trig and Pre-cal
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This material is released under a GNU General Public License Version 3 license.
Rights Holder:
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EJS, Easy Java Simulation, components, force, resultant, vector addition, vector simulation, vectors
Record Cloner:
Metadata instance created May 2, 2010 by Mario Belloni
Record Updated:
August 25, 2013 by Matt Mohorn
Last Update
when Cataloged:
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11. Common Themes
11B. Models
• 6-8: 11B/M2. Mathematical models can be displayed on a computer and then modified to see what happens.
• 9-12: 11B/H2. Computers have greatly improved the power and use of mathematical models by performing computations that are very long, very complicated, or repetitive. Therefore, computers can
reveal the consequences of applying complex rules or of changing the rules. The graphic capabilities of computers make them useful in the design and simulated testing of devices and structures
and in the simulation of complicated processes.
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@misc{ Author = "Andrew Duffy", Title = {Vector Addition with Integer Components}, Month = {April}, Year = {2010} }
%A Andrew Duffy
%T Vector Addition with Integer Components
%D April 16, 2010
%U http://www.compadre.org/Repository/document/ServeFile.cfm?ID=10010&DocID=1643
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%0 Computer Program
%A Duffy, Andrew
%D April 16, 2010
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%8 April 16, 2010
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Advogato: Blog for Pseudonym
harrisj: Thanks for the bouquet. It's only because I happen to be good at remembering this esoteric excreta.
You mentioned verification in P-time, and I think it's worth exploring what that means. One technical definition of the class NP is:
NP is the class of decision problems (languages) L such that there is a polynomial time function f(x,c) where x is a string, c is another string whose size is polynomial in the size of x, and f
(x, c)=True if and only if x is in L.
(taken from the comp.theory FAQ)
As an example, x might be a graph, c might be a path in that graph, and f(x, c) is the function "is c a Hamiltonian path in the graph x?"
There are at least three kinds of questions that you may want to ask about this problem, given x. They are:
• Is x in L, that is, does there exist c such that f(x, c)=True? That question corresponds to "does the graph have a Hamiltonian cycle"? Or "is the number composite"? This is called a "decision
problem", and is not necessarily verifiable in P (unless P=NP, of course), because verifying an answer is identical to finding the answer.
• Given x, give me a c which proves that x is in L. This kind of problem is trivially certifiable in P, because all you need to do is take the certificate c and compute the value of f(x, c). This
question corresponds to "find me a Hamiltonian cycle in this graph", or "find me factors of this number".
• Given x, find me the certificate c which minimises (or maximises) some cost function cost(c). This is called an "NP optimisation problem", and corresponds to something like "find me the
Hamiltonian cycle of minimal distance" (i.e. the travelling salesman problem). I don't know whether this kind of problem is certifiable in P or not. Could someone please let me know?
Now to EXPTIME. EXPTIME-hard problems are not certifiable in P-time if the certificate is polynomial in the size of the input. If they were, you could construct a nondeterministic polynomial
algorithm to solve them. (Nondeterministically guess the certificate in NP time then test it in P time.) So I guess EXPTIME-hard public-key systems and zero-knowledge proof systema aren't feasable.
If the certificate were not polynomial in the size of the input, that might be a different matter. That is, certifying EXPTIME-hard problems might be polynomial in the size of the certificate. I'm
not sure about that one.
EXPTIME-hard problems often turn up when dealing with logic and theorem proving (e.g. constraint systems, conditional logic) or formal languages and the machines which generate/recognise them (e.g.
tree automata). Here are some example EXPTIME-complete problems:
• Evaluation of datalog queries for single rule programs containing only a single clause.
• For tree automata, the intersection problem (i.e. given two tree automata, is the intersection of their languages empty) and equivalence problem (do two given tree automata accept the same
• Removing recursion from formulas in the mu calculus (i.e. given a recursive formula, give me a non-recursive equivalent if it exists)
• Satisfiability for definite set constraints
• Determining a winning strategy for chess generalised to an N*N board or go generalised to an N*N board. I'm not entirely sure how you generalise chess piece movement, but there you go.
And just for fun, here are a couple of NEXPTIME-complete problems:
• Satisfiability for two-variable first-order logic.
• Provability in first-order multiplicative additive linear logic
Naturally, NEXPTIME-complete problems are certifiable in EXPTIME.
Happy theory nerding.
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[Israel.pm] BruteForcing function
Gaal Yahas gaal at forum2.org
Mon Dec 20 08:02:33 PST 2004
On Mon, Dec 20, 2004 at 05:17:37PM +0200, amit sides wrote:
> Gaal Yahas wrote:
> > @alphabet = ( 'a' .. 'z', 'A' .. 'Z', '0' .. '9', qw/!_-/ );
> > for (0 .. $#alphabet) {
> > print $alphabet[$_];
> > }
> i thought about it.....so we solve the first problem ...now i need to
> solve the sec one
> i dont see how this helps me show all the possible option of combinations
> do you have an idea how to print @alphabet chars by loops ? how to
> arrange the loops ?
You haven't defined your problem, then. The above generates all possible
single-letter "words" from the alphabet. If you want to generate all
"words" with *more* letters, you simply want to enumerate over a larger
numerical range, in base n (where n is the size of the alphabet).
Think of it this way: you have a combination lock with three rolling
digits, that is, it offers numbers like [1][2][3] and [0][0][7]. To try
all combinations you simply need to try all numbers from 0 to 999. But
if I go and replace the *label* on each roller to say [A] to [I] instead
[0] to [9], your problem has remained essentially the same; all you need
is a presentation function to translate the output from (say) [2][0][1]
to [C][A][B].
Now your real rollers are bigger. Instead of 0-9 they accept that range
you defined earlier, which has 65 "digits". So to generate all
combinations of three letter "words", you have 65*65*65 (65^3) numbers
to visit, but what you do with each one is still the same: thinking of
it as a base-65 number, translate it to your alphabet.
foreach my $num (0 .. (@alphabet ** $width - 1)) {
print base_n(\@alphabet, $num);
You still have to write base_n, but if you Google for "base conversions
perl" you'll get plenty of ideas on how to do that.
Gaal Yahas <gaal at forum2.org>
More information about the Perl mailing list
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Standard 2: Reason abstractly and quantitatively
The Standard:
Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative
relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without
necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved.
Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute
them; and knowing and flexibly using different properties of operations and objects.
Classroom Observations:
Teachers who are developing students’ capacity to "reason abstractly and quantitatively" help their learners understand the relationships between problem scenarios and mathematical representation, as
well as how the symbols represent strategies for solution. A middle childhood teacher might ask her students to reflect on what each number in a fraction represents as parts of a whole. A different
middle childhood teacher might ask his students to discuss different sample operational strategies for a patterning problem, evaluating which is the most efficient and accurate means of finding a
solution. Visit the video excerpts below to view these teachers engaging their students in abstract and quantitative reasoning.
Connections to Classroom Practices
5th Grade
[Mathematically proficient students bring] the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the JW Player goes here
symbols involved.
See this video in the
Hillary Lewis-Wolfsen leads a re-engagement lesson on the proportions and ratios, helping students to recognize what a visual representation of a simplified fraction looks context of an entire
like. In this clip, she gives her students “think time” to jot down their ideas, then responds to a student with the correct answer by asking another student to explain her lesson.
answer. Lewis-Wolfsen comments that her “students often understand more if they hear the explanation in a variety of ways, and not just from the teacher.”
5th/6th Grade
Mathematically proficient students make sense of quantities and their relationships in problem situations... Quantitative reasoning entails... attending to the meaning of
quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. JW Player goes here
Fran Dickinson leads a lesson on numeric patterning, helping students to investigate a numeric pattern and to generalize what they see happening as the pattern grows. In this See this video in
clip, Dickinson describes the importance of individual think time before he asks his students to discuss the relative strength of two different approaches to a patterning task. the context of an
One pair discusses the numbers within the sample strategy, and Dickinson repeats back their conversation to the whole group, telling his students, “I’ve heard two really good entire lesson.
questions about Learner B’s strategy. One was, what are all these 3’s? and Kelcey’s question was, what about this 4? Where’s the 4 coming from?” This clip is also indicative of
standard 1 (make sense of problems and persevere in solving them).
6th Grade
Mathematically proficient students make sense of quantities and their relationships in problem situations... Quantitative reasoning entails habits of creating a coherent
representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using JW Player goes here
different properties of operations and objects.
See this video in the
Joe Condon works with his 6th grade students to identify strategies for comparing unit rates. In these clips, he begins with a definition of rate developed and expanded by context of an entire
the class over the two-week period. His students are asked to name familiar rates. He then presents them with interactive and real-life scenarios and asks them to calculate lesson.(Parts 1 - 6)
the fastest unit rates and justify their answers. Condon refers his students to strategies for comparing rates, asking the students to use these strategies with new data. The
students then discuss their responses with the whole group, and conclude that the best rates depend on the goal of the calculation.
Mathematically proficient students make sense of quantities and their relationships in problem situations... Quantitative reasoning entails habits of creating a coherent
representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using
different properties of operations and objects.
JW Player goes here
Joe Condon presents a lesson study for 6th grade students and observing teachers to identify strategies for comparing unit rates. In these clips, he starts by establishing
classroom norms for active listening. Then he engages the class in a whole-group experiment of the teacher power walking a 5-meter strip. The members of the class are given See this video in the
jobs for collecting data or monitoring time. Then Condon uses a ratio table to talk about rate and find equivalent rates and unit rates. Students are asked to name some context of an entire
familiar or common rates. Students then try to give their own definition of rate after looking at these examples. Finally students are given a ratio with no words and asked lesson.(Parts 1 - 6)
what it could mean.
Students conduct 3 rate experiments: stringing beads on a shoelace, picking up cubes with chopsticks, and counting rice. After the experiments the teacher debriefs the
results as a whole class. "Who won?" "Why can't we tell from the raw data?"
9th-12th Grade
Mathematically proficient students make sense of quantities and their relationships in problem situations...Quantitative reasoning entails habits of creating a coherent
representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly JW Player goes here
using different properties of operations and objects.
See this video in the context
Carlos Cabana works with his high school algebra for English language learning students on algebraic reasoning and multiple representations around parabolas. In these of an entire lesson.(Parts 8
clips, Cabana analyzes groups' work in terms of logical flow and carefulness of organization, trying to make a guess about what they really understand. Students explain & 10)
their thinking to each other as they look to find the vertex of the parabola. Cabana challenges them to move between their representations.
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BIOGRAPHY 9.3 Siméon Poisson (1781-1840)
Simeon Denis Poisson was born in Pithiviers, France. He was sent to study at the famous École Polytechnique of Paris and performed so well that he was exempted from the final exams and remained there
as a professor for 40 years. He also held a great variety of supplementary posts and wrote 300 papers on mathematics, astronomy, and physics. Most famous at the time was his Traité de Mé
canique (1811), but today he is best known for his 1837 presentation of the exponential limit of the binomial distribution. Although the same result had been reached in 1718 by de Moivre (see
Biography 8.2), the probability distribution so derived is now called the Poisson probability distribution. Important as it has turned out to be, Poisson's achievement remained, nevertheless,
practically unknown for more than 60 years. Attention was drawn to it in 1898 in a paper by Ladislaus von Bortkiewicz. He noted how deaths of Prussian soldiers, caused by horsekicks, could be
described as a Poisson process and how their number could be estimated by Poisson's formula. When examining the experience of 10 army corps over 20 years (a total of 200 observations), von
Bortkiewicz found the absolute and relative frequency distribution noted in the Chapter 9 Preview. Von Bortkiewicz calculated the weighted mean annual number of such deaths as
and he introduced this value of µ in the Poisson formula:
The result, given in the last column of text Table 9.1, closely matched the empirical probabilities shown in the relative frequency column. From then on, the Poisson probability distribution has
remained in the limelight.
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Acharya Nagarjuna University Maths-II model question papers
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Course: B.Sc Electronics University/board: Acharya Nagarjuna University
B.A. / B.Sc. DEGREE EXAMINATION, MODEL PAPER
(Examination at the end of second year, for 2009 - 2010 onwards)
Time : 3 Hours Max. Marks : 100
SECTION - A (6 X 6 = 36 Marks)
Answer any SIX questions.
1. Prove that a subgroup H of a group G is a normal subgroup of G iff every left coset
of H in G is a right coset of H in G .
2. Prove that every subgroup of a cyclic group is cyclic.
3. Prove that an ideal M of a commutative ring R with unity is maximal iff R / M is a
4. State and prove division algorithm for polynomials.
5. Prove that a monotonic sequence is convergent iff it is bounded.
6. Examine the continuity of the function f defined by f (x) =| x|+| x -1| at x = 0 and 1.
7. State and prove Cauchy's mean value theorem.
8. If f : [a, b]?R is continuous on [a, b] then prove that f is R -integrable on [a, b] .
SECTION - B (4 X 16 = 64 Marks)
Answer any ALL questions.
9.(a) Let H and K be two subgroups of a group G . Prove that H ? K is a subgroup of
G iff either H ? K or K ? H .
(b) If G = Q- {1} and * is defined on G as a *b = a + b - ab then show that (G, *) is
an abelian group.
10.(a) State and prove Lagrange's theorem on finite groups.
(b) Let G be a group. Show that the mapping f :G ?G defined by f (a) = a-1 ?a?G
is one one onto. Also show that f is an automorphism iff G is abelian.
11.(a) Define the terms integral domain and field.
Prove that every finite field is an integral domain.
(b) If Q( 2) = {a + b 2 :a,b ?Q} then show that Q( 2) is a field.
12.(a) State and prove fundamental theorem of homomorphism of rings.
(b) Find the sum and product of the polynomials f (x) = 2x3 + 4x2 + 3x + 2 and
g(x) = 3x4 + 2x + 4 over Z5 .
13(a) State and prove D'Alembert's ratio test.
(b) Test for convergence of
( )
? + + + +
? ?
? ?
n n n
14(a) If f in continuous on [a,b] then prove that f is bounded on [a,b] .
(b) If s
n n = 1+ + + +
! !
then show that {s } n converges.
15(a) State and prove Rolle's theorem.
(b) Derive the expansion of sin x by using Maclaurin's theorem.
16(a) State and prove fundamental theorem of integral calculus.
(b) Prove that
p p p
= ? sec x dx = using the first mean value theorem.
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Summer Research Program for Science Teachers
Jacqueline Kennedy Onassis High School
Summer 2000
How can we determine the concentration of an unknown solution?
Instructional Objectives:
a. Students will create a standard curve, plotting the relationship between the concentration of a solution and the amount of light that it is able to absorb. [Content Standard Unifying Concepts-
Change, constancy, and measurement] [9-12 Content Standard B- Properties of matter]
b. Students will be able to use the standard curve they have created to determine the concentration of an unknown solution. [9-12 Content Standard A- Formulate explanations using evidence]
CBL System
TI Graphing Calculator [9-12 Content Standard E- Understandings about science and technology]
Vernier Colorimeter
Vernier adapter cable
TI-Graph Link
One cuvette
Five 20 x 150 mm test tubes
Tissues (preferrably lint free)
30 mL of 0.40 M NiSO[4 ]
5 mL of NiSO[4 ] of unknown concentration
two 10 ml pipets (or graduated cylinders)
pipet pump or pipet bulb
distilled water
test tube rack
two 100mL beakers
stirring rod [Teaching Standard D- Make accessible science tools]
1. Using the table below, use the distilled water provided to dilute the NiSO[4 ] and make 5 solutions with known concentrations.
│ │0.40 M NiSO[4] │[]H[2]0│Concentration │
│Trial Number│ │ │ │
│ │(mL)[ ] │(mL) │(M) │
│1 │2 │8 │0.08 │
│2 │4 │6 │0.16 │
│3 │6 │4 │0.24 │
│4 │8 │2 │0.32 │
│5 │10 │0 │0.40 │
2. Calibrate the colorimeter. Prepare a blank by filling a cuvette ¾ full of distilled water. With the light source turned off, enter this absorbance value obtained as 0% transmittance. With the
wavelength knob in the Red LED position (635 nm), enter the absorbance value obtained as 100% transmittance.
3. In this same manner, collect absorbance data for each of the five standard solutions. When the percent transmittance value for each solution is displayed , enter the molar concentration for that
4. Using your calculator, construct a graph of absorbance vs. concentration. Then perform a linear regression on your data. [9-12 Content Standard A- Use mathematics to improve scientific
communication] If the data you have obtained are consistent with Beer s Law (a direct relationship between absorbance and concentration), the regression line should closely fit the five data points
and should pass through (or near) the origin of the graph.
5. Obtain about 5 mL of the unknown solution of NiSO[4] . Find the absorbance for the unknown solution. Then use your calculator to interpolate along the regression line on your Beer s Law curve.
6. Use the TI Graph link cable and program to transfer the graph of absorbance vs. concentration (including the interpolated unknown concentration) to a laptop computer. Print a copy of the graph.
Return to Chemistry Lesson Plans Menu
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Question and Answer thread
Posted 2011-March-05, 09:54
Deleted -- too general
Posted 2011-March-07, 14:47
Ken, is an "indestructable" number only one that is fixed under the powertrain mapping, or is a number "indesctructible under mapping M(f)" if M(x) = x? If so, does the term apply to any generic set
element under a mapping of element type to the same element type?
Posted 2011-March-08, 21:37
mycroft, on 2011-March-07, 14:47, said:
Ken, is an "indestructable" number only one that is fixed under the powertrain mapping, or is a number "indesctructible under mapping M(f)" if M(x) = x? If so, does the term apply to any generic set
element under a mapping of element type to the same element type?
So the first answer is: I don't know. I first heard of indestructible numbers by reading the "interesting numbers" thread and following up by googling Below and Conway on it all. So I don't know
But I am not sure that I understand your question. Yes I am (reasonably) sure that the term is context specific, meaning that Conway or Bello or one of those guys invented it to deal with numbers
arising in some manner from the powertrain map. I say this just on a general feel for how terms like this come into being. Sometimes it seems there is a gnome locked away somewhere making up these
Someone recently sent me a video related to this perverse invention of words:
Posted 2011-March-12, 07:28
what happens to the ships traveling the Pacific when the tsunamis come to them?
Posted 2011-March-12, 07:53
Fluffy, on 2011-March-12, 07:28, said:
what happens to the ships traveling the Pacific when the tsunamis come to them?
As I understand it, hardly noticeable in deep water--a gentle one-meter rise in water level. It's when the tsunami reaches shallow areas that the huge wall of water forms.
The growth of wisdom may be gauged exactly by the diminution of ill temper. Friedrich Nietzsche
The infliction of cruelty with a good conscience is a delight to moralists that is why they invented hell. Bertrand Russell
Posted 2011-March-12, 08:29
PassedOut, on 2011-March-12, 07:53, said:
As I understand it, hardly noticeable in deep water--a gentle one-meter rise in water level. It's when the tsunami reaches shallow areas that the huge wall of water forms.
Exactly. Tsunami (japanese for "harbor wave") waves have a large wave length (tens of kilometers), so that they are a very large and shallow bulge. In deep water, this is not much of a problem and
completely harmless to a ship.
However, the speed of the wave decreases with depth in shallower water (you can notice this at the beach when waves from the deep water come in at an angle, the part closest to the shore will show
down so it appears the wave turns its wake parallel to the coast). This means that the back of the wave starts to catch up with the front, leading to piling up of water and to the wall of water that
we know as a tsunami.
As the wave displaces such a large volume of water, before the top reaches the shore the low point of the wave reaches it first and can be observed as a sudden quick low tide.
Posted 2011-March-12, 10:35
sr but I don't see the purpose of this thread. If you have a question isn't it more practical to post it as a separate topic? Then people can decide whether to read it or not depending on the topic
title. And we don't get confusion with discussions about one topic mixed up with other topics.
For the same reason I dislike threads of the type "a couple of hands from yesterday". Just post each hand as separate topic (unless they have a common theme, e.g. both were about whether to make an
offshape 1NT opening, or both were about whether to cover an honor lead from dummy).
the Netherlands is a very small country, so they probably have room only for tables that are about 2' x 2' --- Vampyr
Posted 2011-March-12, 20:18
I agree with Helen.
On the other hand, can I get an invitation to quora?
wyman, on 2012-May-04, 09:48, said:
Also, he rates to not have a heart void when he leads the ♥3.
rbforster, on 2012-May-20, 21:04, said:
Besides playing for fun, most people also like to play bridge to win
Mi Blog
In all fields of endeavour emotion is the arch-enemy of judgement.
Posted 2011-March-13, 10:12
helene_t, on 2011-March-12, 10:35, said:
sr but I don't see the purpose of this thread. If you have a question isn't it more practical to post it as a separate topic? Then people can decide whether to read it or not depending on the topic
title. And we don't get confusion with discussions about one topic mixed up with other topics.
For the same reason I dislike threads of the type "a couple of hands from yesterday". Just post each hand as separate topic (unless they have a common theme, e.g. both were about whether to make an
offshape 1NT opening, or both were about whether to cover an honor lead from dummy).
If you want you can still post it as a separate topic, if you feel like it's maybe not such an important question, or for whatever other reasons you may have, you can post it here.
... and I can prove it with my usual, flawless logic.
George Carlin
Posted 2011-March-13, 11:38
Yes. There are various questions that occur to me that don't really seem to warrant the dignity of having their own dedicated thread but still I might like to ask. "What's the name of that paper and
pencil game with the boxes?" seems right. Anyway, I'll keep checking in.
Posted 2011-March-13, 12:25
2**1 = 2
2**2 = 4
2**3 = 8
2**4 = ..6
2**5 = ..2
2**6 = ..4
2**7 = ..8
2**8 = ..6
ad infinitum. the powers of all other last digits have the same period of 4 behaviour (in addition, 0, 1, 5 and 6 remain constant; 4 and 9 have a period of 2; 2, 3, 7, 8 are the ones where the
smallest period is 4). Is there some fancy explanation to this or is this just a combination of a coincidence and its consequences?
... and I can prove it with my usual, flawless logic.
George Carlin
Posted 2011-March-13, 15:34
helene_t, on 2011-March-12, 10:35, said:
sr but I don't see the purpose of this thread.
You can then ask for the purpose of this thread on the questions and answers post in the water cooler
Posted 2011-March-13, 15:37
gwnn, on 2011-March-13, 12:25, said:
2**1 = 2
2**2 = 4
2**3 = 8
2**4 = ..6
2**5 = ..2
2**6 = ..4
2**7 = ..8
2**8 = ..6
ad infinitum. the powers of all other last digits have the same period of 4 behaviour (in addition, 0, 1, 5 and 6 remain constant; 4 and 9 have a period of 2; 2, 3, 7, 8 are the ones where the
smallest period is 4). Is there some fancy explanation to this or is this just a combination of a coincidence and its consequences?
You already know that the last digit is dependant only in the previous last digit and the base of the power, so you already know the answer, why do you ask?
Posted 2011-March-13, 16:27
Why isn't there any last digits that have a period of 5? Or 3?
... and I can prove it with my usual, flawless logic.
George Carlin
Posted 2011-March-13, 20:44
gwnn, on 2011-March-13, 12:25, said:
2**1 = 2
2**2 = 4
2**3 = 8
2**4 = ..6
2**5 = ..2
2**6 = ..4
2**7 = ..8
2**8 = ..6
ad infinitum. the powers of all other last digits have the same period of 4 behaviour (in addition, 0, 1, 5 and 6 remain constant; 4 and 9 have a period of 2; 2, 3, 7, 8 are the ones where the
smallest period is 4). Is there some fancy explanation to this or is this just a combination of a coincidence and its consequences?
, there is a
fancy explanation
(both link to wikipedia). Given any base "b" you are correct in realizing that all that matters is the last digit. Another way of saying this is that what matters is the remainder of the number when
you divide by "b".
One can do arithmetic on the numbers from 0 to b-1 (the remainders). It's referred to as "clock arithmetic" sometimes or "modular arithmetic." If you add to numbers, you imagine "wrapping around."
For example, if "b = 5" then "2 + 4 = 6 = 1" (since the remainder of 6 and 1 is the same when divided by 5). We can also multiply "3*4 = 12 = 2" (since 12 and 2 have the same remainder when divided
by 5).
You may see what this has to do with your problem. We want to know the first time that 2*2*2....*2 = 2 (with b = 10). There is a theorem that if "b" is prime then this number must always divide b-1.
In general it will divide the Euler Totient function of "b." To define this we assume that "b" factors into products of powers of prime numbers as:
b = p_1^(e_1)*p_2^(e_2)...*p_k^(e_k)
Then the Euler Totient function of "b" is
As an example with b = 10. It factors as 2*5, so the totient function is (2-1)*(5-1) = 4. This is why all the "periods" you see divide 4.
I fear I may have given too much detail (and at once not enough), but I'm a bit too tired to edit this so please ask follow up questions if you want more details about parts of this.
Bridge Personality: 44 44 43 34
Never tell the same lie twice. - Elim Garek on the real moral of "The boy who cried wolf"
Posted 2011-March-14, 05:37
cool, I will need to read a little more on this, but this has been bugging me on and off for the last 8 years or so! I never could find a good way of googling it and when my mathematician roommate
explained to me, I didn't understand anything and I didn't want to ask twice (out of shame or laziness).
... and I can prove it with my usual, flawless logic.
George Carlin
Posted 2011-March-14, 14:10
Is it possible to "Lowjack" a thread?
I have purged myself of negative feelings and transferred them into a clone. I've named it "EddieDane", based on J.E. Freemans character in the movie "Millers Crossing".
Do not underestimate the power of the dark side. Or the ninth trumph.
Best Regards Ole Berg
We should always assume 2/1 unless otherwise stated, because:
- If the original poster didn't bother to state his system, that means that he thinks it's obvious what he's playing. The only people who think this are 2/1 players.
Posted 2011-March-19, 08:23
"I think maybe so and so was caught cheating but maybe I don't have the names right". Sure, and I think maybe your mother .... Oh yeah, that was someone else maybe. -- kenberg
"...we live off being battle-scarred veterans who manage to hate our opponents slightly more than we hate each other. -- Hamman, re: Wolff
Posted 2011-March-19, 16:09
wyman, on 2011-March-19, 08:23, said:
can I
a thread?
Posted 2011-March-21, 14:37
kenberg, on 2011-March-08, 21:37, said:
So the first answer is: I don't know. I first heard of indestructible numbers by reading the "interesting numbers" thread and following up by googling Below and Conway on it all. So I don't know
But I am not sure that I understand your question. Yes I am (reasonably) sure that the term is context specific, meaning that Conway or Bello or one of those guys invented it to deal with numbers
arising in some manner from the powertrain map. I say this just on a general feel for how terms like this come into being. Sometimes it seems there is a gnome locked away somewhere making up these
I was just wondering if the term was generic (having a name for an element that is invariant when a particular mapping is applied seems useful, for low-grade versions of useful). And if it was
generic, it needn't apply solely to numbers; a function could be indestructible when a mapping M(function)->function is applied, and the same could be of any element of any set and its mappings.
And this is why I'm an engineer and not a pure math guy - because this is interesting, but not useful, to me.
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Archives of the Caml mailing list > Message from Didier Remy
Date: -- (:)
From: Didier Remy <Didier.Remy@i...>
Subject: Re: generalization in tuples
On Mon, Oct 16, 2000 at 02:42:11PM +0200, David Monniaux wrote:
> 1/ Is it possible to do what I want to do, even if it means using a
> kludge? The above code, using multiple let's, is not good: it's not
> useable in the middle of an expression (this is for CamlP4-generated
> code).
> (acceptable kludges include the use of Obj.magic)
In principle, Obj.magic should do the job, but it does not:
Obj.magic (fun x -> x)
is treated as an application and returns a weak type ;-(
The problem is that "Obj.magic" is defined as a primitive and the above is
typed as any other application, but I don't see any reason except technical
to treat Obj.magic as a constructor. Anyway, I don't think Obj.magic is
a good fix...
> 2/ Is there a finer notion of a "generalizable" expression that
> encompasses the above code, and could the "let generalization" procedure
> in the compiler be improved so that the above code gets a polymorphic
> type?
Yes, there is a very simple generalization of the value-only polymorphism
Expressions need to be partitioned into two sets: expansive and
non-expansive expressions, such that the evaluation of non-expansive is
guaranteed not to create any storage.
For instance, non-expansive expressions may include variables, values (hence
functions), as well as constructors applied to non-expansive expressions.
Note that subexpressions of non-expansive expression are often expansive
(e.g. typically when the expression is under lambda-abstraction).
Given an expression e, we are only interested in outer expansive
sub-expressions of e, i.e. those that are not sub-expressions of a
non-expansive sub-expression of e (in which case, they are protected from
When typing an expression e, all type variables appearing in at least one
outer expansive sub-expression of e may also be the type of a store cell
allocation and should not be generalized. All other type variables can be
For instance, in (a simpler version of) your example:
let x = (ref [], fun x -> x);;
The expression (ref [], fun x -> x) has type 'a ref * ('b -> 'b);
here, "ref []" is an application, hence an (outer) expansive expression and
'a appearing in its type cannot be generalized. Conversely, "fun x -> x" is
non-expansive and since variable "'b" only appear in the type of this
non-expansive subexpression, it can be generalized.
A few more examples:
let x = (let y = fun x -> x in ref y, y)
: ('a -> 'a) ref * ('a -> 'a)
Here 'a appears both in an outer expansive expansive expression and in a
non-expansive expressions. Hence it is dangerous can cannot be generalized.
let x = fun x -> ref x
: ('a -> 'a ref)
The expression is protected, i.e. non-expansive and "'a" can be
(Note that this is a strict generalization of the actual solution.()
The implementation
This is actually quite simple: while typeckecking an expression, just keep
track of whether the expression is the outermost non-expansive part of a
let-bound expression, and if not, make its variable non-generalizable.
In fact, I experimented this in MLART a while ago:
#morgon:~/caml/camlart/src$ ./camlrun ./camltop -I lib
> Caml Light version 0.5 (modified with extensible records)
#(ref 1, fun x -> x);;
- : int ref * ('a -> 'a) = ref 1, <fun>
or, using extensible records :-)
#{!a = fun x -> x};;
> Toplevel input:
>{!a = fun x -> x};;
> Cannot generalize 'a in {a : mut. 'a -> 'a; abs. 'b}
#{!a =1; b = fun x -> x};;
- : {a : mut. int; b : pre. 'a -> 'a; abs. 'b} = {!a = 1; b = <fun>}
This has never been implemented in Ocaml, probably because, besides me,
you are one of first persons to complain about the drastic implementation of
value-only polymorphism restriction.
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My Favorite Strange Number: Ω (classic repost)
I’m away on vacation this week, taking my kids to Disney World. Since I’m not likely to
have time to write while I’m away, I’m taking the opportunity to re-run an old classic series
of posts on numbers, which were first posted in the summer of 2006. These posts are mildly
Ω is my own personal favorite transcendental number. Ω isn’t really a specific number, but rather a family of related numbers with bizarre properties. It’s the one real transcendental number that I
know of that comes from the theory of computation, that is important, and that expresses meaningful fundamental mathematical properties. It’s also deeply non-computable; meaning that not only is it
non-computable, but even computing meta-information about it is non-computable. And yet, it’s almost computable. It’s just all around awfully cool.
So. What is it Ω?
It’s sometimes called the halting probability. The idea of it is that it encodes the probability that a given infinitely long random bit string contains a prefix that represents a halting program.
It’s a strange notion, and I’m going to take a few paragraphs to try to elaborate on what that means, before I go into detail about how the number is generated, and what sorts of bizarre properties
it has.
Remember that in the theory of computation, one of the most fundamental results is the non-computability of the halting problem. The halting problem is the question “Given a program P and input I, if
I run P on I, will it ever stop?” You cannot write a program that reads an arbitrary P and I and gives you the answer to the halting problem. It’s impossible. And what’s more, the statement that the
halting problem is not computable is actually equivalent to the fundamental statement of Gödel’s incompleteness theorem.
Ω is something related to the halting problem, but stranger. The fundamental question of Ω is: if you hand me a string of 0s and 1s, and I’m allowed to look at it one bit at a time, what’s the
probability that eventually the part that I’ve seen will be a program that eventually stops?
When you look at this definition, your reaction should be “Huh? Who cares?”
The catch is that this number – this probability – is a number which is easy to define; it’s not computable; it’s completely uncompressible; it’s normal.
Let’s take a moment and look at those properties:
1. Non-computable: no program can compute Ω. In fact, beyond a certain value N (which is non-computable!), you cannot compute the value of any bit of Ω.
2. Uncompressible: there is no way to represent Ω in a non-infinite way; in fact, there is no way to represent any substring of Ω in less bits than there are in that substring.
3. Normal: the digits of Ω are completely random and unpatterned; the value of and digit in Ω is equally likely to be a zero or a one; any selected pair of digits is equally likely to be any of the
4 possibilities 00, 01, 10, 100; and so on.
So, now we know a little bit about why Ω is so strange, but we still haven’t really defined it precisely. What is Ω? How does it get these bizarre properties?
Well, as I said at the beginning, Ω isn’t a single number; it’s a family of numbers. The value of an Ω is based on two things: an effective (that is, turing equivalent) computing device; and a
prefix-free encoding of programs for that computing device as strings of bits.
(The prefix-free bit is important, and it’s also probably not familiar to most people, so let’s take a moment to define it. A prefix-free encoding is an encoding where for any given string which is
valid in the encoding, no prefix of that string is a valid string in the encoding. If you’re familiar with data compression, Huffman codes are a common example of a prefix-free encoding.)
So let’s assume we have a computing device, which we’ll call φ. We’ll write the result of running φ on a program encoding as the binary number p as &phi(p). And let’s assume that we’ve set up φ so
that it only accepts programs in a prefix-free encoding, which we’ll call ε; and the set of programs coded in ε, we’ll call Ε; and we’ll write φ(p)↓ to mean φ(p) halts. Then we can define Ω as:
Ω[φ,ε] = Σ[p ∈ Ε|p↓] 2^-len(p)
So: for each program in our prefix-free encoding, if it halts, we add 2^-length(p) to Ω.
Ω is an incredibly odd number. As I said before, it’s random, uncompressible, and has a fully normal distribution of digits. But where it gets fascinatingly strange is when you start considering its
computability properties.
Ω is definable. We can (and have) provided a specific, precise definition of it. We’ve even described a procedure by which you can conceptually generate it. But despite that, it’s deeply
uncomputable. There are procedures where we can compute a finite prefix of it. But there’s a limit: there’s a point beyond which we cannot compute any digits of it. And there is no way to compute
that point. So, for example, there’s a very interesting paper where the authors
computed the value of Ω for a Minsky machine; they were able to compute 84 bits of it; but the last 20 are unreliable, because it’s uncertain whether or not they’re actually beyond the limit, and so
they may be wrong. But we can’t tell!
What does Ω mean? It’s actually something quite meaningful. It’s a number that encodes some of the very deepest information about what it’s possible to compute. It gives a way to measure the
probability of computability. In a very real sense, it represents the overall nature and structure of computability in terms of a discrete probability.
Ω is actually even the basis of a halting oracle – that is, if you knew the
value of Ω, then you could easily write a program which solves the halting problem!
Ω is also an amazingly dense container of information – as an infinitely long number and so thoroughly non-compressible, it contains an unmeasurable quantity of information. And we can’t even figure
out what most of it is!
1. #1 Ben December 31, 2008
Bro, you should link back to the original article so that the comments can be read.
Happy new year!
2. #2 Alex Besogonov December 31, 2008
By the way, halting probability is sometimes called “Chaitin’s_constant”.
3. #3 rick December 31, 2008
Yes, Alex, you remind me of a pretty nice piece of Chaitin’s here which includes a bit about Omega, the halting problem, Godel’s incompleteness, noncomputability and the whole bit
4. #4 Dave S. December 31, 2008
In marine science Ω has a different and prosaic meaning, its the saturation state of calcium carbonate.
5. #5 John Fouhy December 31, 2008
I read a book by Chaitin — _Meta Math!_. Most of it has now left me, but one thing sticks in my mind: Apparently, if you choose a real number at random, then with probability 1, it will be:
1. Transcendental.
2. Random.
3. Uncomputable.
In other words, we can’t write it down, and neither can we describe it precisely. Even if I knew exactly what this number is, there is no way I could impart that information to you. So, in what
sense do real numbers really exist anyway?
6. #6 - December 31, 2008
I truly and deeply cannot understand this post, and no amount of italics will help that.
And that is OK.
7. #7 Jair January 1, 2009
I’ve also read that book by Chaitin, I really enjoyed it. It really makes you philosophically suspicious about the set of real numbers. Also it’s mostly non-technical from what I remember.
8. #8 The Young Linguist January 1, 2009
As an undergraduate math major, my question is this.
If I define an $\Omega$, can I determine the computing machine that it is for?
I’m just curious, because wouldn’t that make this ironically close to ‘the question’ and ‘the answer’ in HHGTTG? You can only know an $\Omega$ or a machine, not both?
(I realize it would be amazingly hard to define an \Omega, given Uncompressible and Normal)
9. #9 Jonathan Vos Post January 1, 2009
John Fouhy:
“in what sense do real numbers really exist anyway?”
You are on the edge of entering a vast and deep forest of metaphysical disagreement.
You are making assumptions about what “with probability 1″ means, which are intuitive to you, but not what you think.
You are making assumptions about what it means for mathematical objects to exist, which wise people have debated for centuries.
Here be dragons.
10. #10 Mark C. Chu-Carroll January 1, 2009
Re John #5:
Chaitin makes a very strong case that the whole idea of the real numbers is philosophically ill-founded.
It’s a damned tricky question. On the one hand, we’ve got
real observable things that at least approximate irrational, transcendental numbers – things like π and e. In that sense, they’ve got to exist. A perfect circle doesn’t exist in the real world –
but it seems strange to say that circles don’t exist. And yet, if circles exist, then π must exist; and if we accept the reality of π, then we’re stuck with the whole mess of irrationals.
But on the other hand, the moment we accept the idea of the irrational numbers, suddenly the whole concept of numbers starts to become downright nonsensical – as you point out, the overwhelming
majority of numbers can’t even be described.
(And contrary to JvP, I don’t think that you’re misunderstanding what “with probability 1″ means; I think the fact that you’re questioning whether the reals really exist in a meaningful sense
shows that you get it; the
undescribable numbers so outweigh the describables that
the describables are a vanishingly small fraction of the set of reals.)
11. #11 Jonathan Vos Post January 1, 2009
“the describables are a vanishingly small fraction of the set of reals”
Yes, but they are a particularly *meaningful* fraction, and the amount of information in each is well modeled by the Minimum Description Length paradigm. MDL is more easily calculated than the
Kolmogorov approach.
Someone who works with computers for a living may, in a sense, be a Constructivist (in my classification of the Ontology of Mathematical objects), and ONLY care about describable numbers. Because
those lead to the describable numbers on one’s paychecks.
12. #12 Valhar2000 January 2, 2009
Are irrational numbers really that problematic? After all, even if you can’t describe them, you can work with them. No matter how hard they are to describe, e^iπ=-1, and sqrt(2)^2=2.
13. #13 Jonathan Vos Post January 2, 2009
Valhar2000: The problem is with real numbers whose description intrinsically is infinitely long. For instance, a real number whose decimal expansion or continued fraction expansion has an
infinite (countable) number of digits, and no way to predict what the (n+1)-th digit is from knowing the first n digits, and no way to “compress” the number to any shorter or more efficient
That’s the essence of the problem: that “almost all” real numbers fall into that class.
I don’t consider that these problems mean that such numbers do not “exist” — but that they can’t be finitely constructed makes some people question their ontological and epistemological status —
i.e. do they exist, and what can we know about them?
This goes beyond the algebraic / irrational / transcendetal distinction to the difficulties that Chaitin discusses.
14. #14 Mark C. Chu-Carroll January 2, 2009
Re JvP #12:
It’s true that the computable reals are particularly important in a philosophical sense. But still – they are so dwarfed by the undescribables that the probability of a randomly selected real
number being undescribable is 1.
That probability statement says nothing about the philosophical or practical importance of the describable or computable numbers.
One thing I might post about at some point is the computable numbers. I saw a presentation a few years ago (I *think* it was one of Chaitin’s, but I’m not sure) that presented a construction of
the set of computable numbers. It was an interesting idea – it defined computable numbers as numbers where you could write a program that would emit the digits of that number in sequence. So, for
example, π is computable – because we can write a program that emits the digits of π. But undescribable numbers aren’t – because there’s no finite program to emit their digits. It was a very neat
idea – and the argument that surrounded the definition of the computables was that we should, perhaps, think of math in terms of the computable numbers, rather than the reals, because the reals
are philosophically ill-founded because of the undescribables.
15. #15 Jonathan Vos Post January 2, 2009
Computable numbers were defined independently by Turing, Post and Church. See The Undecidable, ed. Martin Davis, for further original papers.
Wikipedia reminds us of the following.
Informally, Minsky defines the numbers to be computed in a manner similar to those defined by Alan Turing in 1936, i.e. as “sequences of digits interpreted as decimal fractions” between 0 and 1:
“A computable number [is] one for which there is a Turing machine which, given n on its initial tape, terminates with the nth digit of that number [encoded on its tape].” (Minsky 1967:159)
The key notions in the definition are (1) that some n is specified at the start, (2) when the machine’s internal counter reaches this n the computation terminates after printing the nth decimal
digit on its tape — otherwise it would continue computing forever.
An alternate form of (2) — the machine successively prints all n of the digits on its tape, halting after printing the nth — emphasizes Minsky’s observation: (3) That by use of a Turing machine,
a finite definition — in the form of the machine’s TABLE — is being used to define what is a potentially-infinite string of decimal digits.
Marvin Minsky 1967, Computation: Finite and Infinite Machines, Prentice-Hall, Inc. Englewood Cliffs, NJ. No ISBN. Library of Congress Card Catalog No. 67-12342. See especially chapter §9 “The
Computable Real Numbers”
There are presentations floating around of key results from:
Oliver Aberth 1968, Analysis in the Computable Number Field, Journal of the Association for Computing Machinery (JACM), vol 15, iss 2, pp 276-299. This paper describes the development of the
calculus over the computable number field.
Klaus Weihrauch 2000, Computable analysis, Texts in theoretical computer science, Springer, ISBN 3540668179. (online version) §1.3.2 introduces the definition by nested sequences of intervals
converging to the singleton real. Other representations are discussed in §4.1.
Related to my classification of mathematical ontologies, one naturally asks: “is possible to disregard or throw away the full set of reals and use computable numbers for all of mathematics?” This
part of the ideocosm is appealing to Constructivists, especially by what Bishop and Richman call the Russian school of constructive mathematics. Caveat: Even though the calculus can be developed
over the computable numbers, the set of computable numbers is not closed under basic operations such as taking the supremum of a bounded sequence, so this set can NOT be used as a replacement for
the full set of real numbers in classical mathematics.
16. #16 Bob January 2, 2009
“in what sense do real numbers really exist anyway?”
Or, are real numbers “real”? Some arguments here begin to sound like the question of Berkeley: “If a tree falls in the forest…” Valhar2000 reminds us that “e^iπ=-1, and sqrt(2)^2=2″ are “real” in
the sense that they have a meaning beyond irrationality. But before the observations about the sqrt(2) and pi as “real” numbers, they surely did exist (in a “real” sense), no? Until a specific
number enters our collective conscious, what does it symbolize?
What is the nature of “Schrodinger’s Cat”? Is the cat a manifestation of a calculation that comes to an end? An observation of a cat (dead or alive) only makes sense when there is a sentient
being to do the observing. Are we here collectively observing the Omega’s out there by considering their existence? We (collectively) “use” pi, for example, whether or not we can ever write it
When a program does “halt”, the probability of halting is immediately known: it is 1. I suspect that anyone who bothers to comment here tacitly acknowledges the importance of Omega.
We all presume that our lives and our consciousness will someday come to a halt. Are we not programs?
17. #17 Sandler January 3, 2009
Fuck me, aren’t you an arrogant little Jew, preserving your own CLASSIC posts for posterity.
Skewed perspective, Jewbacca.
18. #18 Dante January 3, 2009
You cannot write a program that reads an arbitrary P and I and gives you the answer to the halting problem. It’s impossible.
How much time you have to wait for the answer? I’d write a debugger which runs P with input I and returns true when P exits, or false when P enters a previous state (and thus would loop forever).
19. #19 Bruno January 4, 2009
Dante: finite time. So your debugger itself must be proven not to ever, under any circumstances, get on an infinite loop itself. Also, returning to a previous state is only one kind of infinite
looping. There are also recursion infinite loops, forks, or legit programs which never go back to a previous state and still loop forever — like any program that computes an uncomputable number
(say, e). Your debugger would have to be able to interpret what the program is doing, and tell whether or not it eventually halts. If you did come up with one such debugger, one nice sanity check
would be to make it test itself and see if it can tell whether or not, for any given input, it eventually halts.
20. #20 Ian Calvert January 6, 2009
“I’d write a debugger which runs P with input I and returns true when P exits, or false when P enters a previous state (and thus would loop forever). ”
Ok, how about I modify your debugger, now called D, to loop infinitely if the input program halts and halt if the program would loop infinitely. [I'm pretty sure this is right, it's 5am though]
What if I call D(D)? Does it halt or not?
It’s essentially *this statement is false* in a computer progrm
21. #21 Dante January 6, 2009
I realize I’m probably wrong and people have been looking a lot at this problem, I just don’t see it.
So I have to make a proof, beside the program? I don’t think I can, other than “It’s trivial to see that D answers the halting problem, if computing resources are not a problem (memory, time)”
About P being the program for e/weirdly recursive, since there are only so many machine instructions, D is possible. Unless Marks also means “on any (future) machine architecture”.
My debugger would work since it never halts, for any program: just returns true or (after a couple of days) false.
22. #22 Dante January 6, 2009
Repost, please ignore/delete my previous entry
I realize I’m probably wrong and people have been looking a lot at this problem, I just don’t see it.
So I have to make a proof, beside the program? I don’t think I can, other than “It’s trivial to see that D answers the halting problem, if computing resources are not a problem (memory, time)”
About P being the program for e/weirdly recursive, since there are only so many machine instructions, D is possible. Unless Mark also means “on any (future) machine architecture”.
My debugger would work since it always halts, for any program: just returns true or (after a couple of days) false.
23. #23 Kyle Lahnakoski January 6, 2009
Jonathan Vos Post said:
“…the set of computable numbers is not closed under basic operations such as taking the supremum of a bounded sequence…”
Of course computable numbers are not closed under incomputable operations!
24. #24 Jonathan Vos Post January 7, 2009
Kyle: which is why I said so, in pointing out the flaw in what Bishop and Richman call the Russian school of constructive mathematics. The point seems to be that we NEED the full set of the Real
Numbers to do Calculus. We can’t throw the baby out with the ontological bathwater.
Unless you can find some position in between what is obvious to you and what is desired by those hard-core wishful-thinking constructivists. If so, you might have something worth publishing.
How much CAN we do with only computable numbers? What’s the least that we need to add to the set to do some of the other things basic to Mathematics?
25. #25 NeilF May 5, 2009
These are deep waters. (Yeah, I know this is very very late, and no-one except maybe Mark CC will read it.)
It’s true, as you’ve stated, that almost all reals are undefinable. There is no pattern or rule that picks them out, no algorithm that can generate their digits.
Even worse, the standard axioms of set theory tell us almost nothing about how *many* of these undefinable reals there are.
Using forcing one can (in a certain sense) adjoin extra real numbers to the universe. One can even add vastly more reals than there were to begin with. (This can be done without collapsing any
cardinals – see next para.)
Going the other way, one can using forcing to “collapse” a cardinal and the make the continuum hypothesis come out true: Whatever number 2^(aleph_0) is, one can graft onto the universe a new set
which is a bijection between that number and aleph_1 (and this can be done without inadvertently adding any extra reals.)
Intuitively the idea of “all subsets of the natural numbers” seems so straightforward and unambiguous that it must determine a single abstract entity just as definitively as the Peano postulates
determine the natural numbers. However, the independence phenomena of set theory make this sound naive.
Even so, I would disagree with any normative suggestion that comes out of this (e.g. “we ought to think of math in terms of computable numbers only”), for the following reasons:
Surprisingly, in “normal mathematics” the indeterminate properties of the reals (of which cardinality is the most notable, but not the only one) rarely come to the fore, and the reals do behave
like a single Platonic entity. Furthermore, in order for the basic theorems of real analysis to come out right, we *do* need the reals to be complete, which presupposes uncountability, which
requires there to be all of those mysterious uncomputable numbers floating around.
Also, the suggestion that we somehow ‘disregard’ uncomputable numbers inevitably entails that set theorists should simply put their pens down and go and do something ‘worthwhile’ instead.
However, if set theorists can prove interesting results about them (and they can) then of course they’re doing something just as ‘worthwhile’ as researchers in any other esoteric branch of pure
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Gas Laws Quiz #1
1. What will the pressure be in atm when a gas at 7 atm is heated from 15 to 30 degrees Celsius, if the volume stays at a constant 5L?
2. A gas is cooled from 350 to 300 degrees kelvin, with a constant volume of 5L. What would the final pressure of the gas be in mm-Hg if the initial pressure measures at 850 mm-hg?
3. A gas has a pressure of 630 mm-Hg at 296 degrees kelvin. At standard pressure what is the temperature of this gas in K?
4. The temperature of a 10 L sample of gas is at 500 degrees kelvin and lowers to 300 degrees kelvin, What will the volume of the gas be if the pressure remains at a constant of 150 kPa?
5. A gaseous sample of carbon dioxide occupies a volume of 3.27 L at a temperature of 20 degrees Celsius. If the pressure remains constant, what is the temperature in Kelvin if the volume lowers to
6. A sample gas occupies 0.5 liters at 30 degrees Celsius and 2000 mm - Hg. What is the temperature in Kelvin at which the gas will occupy 1 liter at 2280 mm-Hg
7. 400L of gas is stored at 100 degrees Celsius at a pressure of 709kPa. The gas is moved to a tank that is 20 degrees Celsius cooler and the pressure of the tank is 3000kPa. What is the volume of
the gas in L?
8. A gas in a balloon that has a volume of 5 L is at a pressure of 0.7atm. The gas is able to expand at a constant temp of 10 degrees Celsius until the pressure is at 0.5atm. What is the final volume
of the box?
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Torsional pendulum, logarithmic decrement
1. The problem statement, all variables and given/known data
The amplitude of a torsional vibration decreases so that the amplitude on the 100th cycle is 13% of the the amplitude of the first cycle. Determine the level of damping in terms of the logarithmic
2. Relevant equations
3. The attempt at a solution
Is this simply ln(100/13)= 2.04
or ln(13/100)= -2.04?
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[racket] static variables question
From: Joe Gilray (jgilray at gmail.com)
Date: Sun Feb 19 19:11:44 EST 2012
Thanks Gary (and Neil and Robby)
Very enlightening. I didn't know about "when". Of course what you said
about "set!" and "cond" makes perfect sense. Minor mods to your code
(using your set! elimination idea twice and caching (integer-sqrt x)) gives
the code below.
I noticed that I can also remove the define for candidate if I'm willing to
call (first candidate-primes) twice. Comments?
Thanks again!
; function to create a list of prime factors of a number
; invoke as (factor n)
(define (factor n)
(let loop-factors ([facts '()] [x n] [start 2] [end 1000]
[candidate-primes (primes-from-to 2 1000)])
(define isqrtx (integer-sqrt x))
(cond [(and (empty? candidate-primes) (>= end isqrtx))
(if (= 1 x) facts (append facts (list x)))]
[else (cond [(empty? candidate-primes)
; attempt to pull in more primes in an efficient
(set! start end)
(set! end (* 2 end))
(when (> (* 1.5 end) isqrtx)
(set! end isqrtx))
(loop-factors facts x start end (primes-from-to
start end))]
[else (define candidate (first candidate-primes))
(cond [(zero? (remainder x candidate))
(loop-factors (append facts (list
candidate)) (quotient x candidate) start end candidate-primes)]
(loop-factors facts x start end (rest
On Sun, Feb 19, 2012 at 12:50 PM, Gary Baumgartner <gfb at cs.toronto.edu>wrote:
> Applying a lot of what Neil said:
> ; function to create a list of prime factors of a number
> ; invoke as (factor n)
> (define (factor n)
> (let loop-factors ([facts '()] [x n] [start 2] [end 1000]
> [candidate-primes (primes-from-to 2 1000)])
> (cond [(and (empty? candidate-primes) (>= end (integer-sqrt x)))
> (if (= 1 x) facts (append facts (list x)))]
> [else (cond [(empty? candidate-primes)
> ; attempt to pull in more primes in an efficient
> manner
> (set! start end)
> (set! end (* 2 end))
> (when (> (* 1.25 end) (integer-sqrt x))
> (set! end (integer-sqrt x)))
> (set! candidate-primes (primes-from-to start end))
> (loop-factors facts x start end candidate-primes)]
> [else (define candidate (first candidate-primes))
> (cond [(zero? (remainder x candidate))
> (set! facts (append facts (list
> candidate)))
> (loop-factors facts (quotient x
> candidate) start end
> candidate-primes)]
> [else
> (loop-factors facts x start end (rest
> candidate-primes))])])])))
> When you're using a lot of 'begin's in 'if's, consider 'cond' which has
> imlicit 'begin'.
> And for single-branch 'if's [which must be for side-effect], 'when' or
> 'unless'.
> As for the 'set!'ing: notice, e.g., that
> (set! facts (append facts (list
> candidate)))
> (loop-factors facts (quotient x
> candidate) start end
> candidate-primes)
> is just
> (loop-factors (append facts (list
> candidate))
> (quotient x candidate)
> start end
> candidate-primes)
> ____________________
> Racket Users list:
> http://lists.racket-lang.org/users
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Weekly Problem 9 - 2014
Copyright © University of Cambridge. All rights reserved.
'Weekly Problem 9 - 2014' printed from http://nrich.maths.org/
In Matt's pocket there are 8 watermelon jellybeans, 4 vanilla jellybeans and 4 butter popcorn jellybeans. What is the smallest number of jellybeans he must take out of his pocket to be certain that
he takes at least one of each flavour?
This problem is taken from the UKMT Mathematical Challenges.
View the previous week's solutionView the current weekly problem
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To those opposed to imputing missing outcomes.
August 7, 2012
By Carlisle Rainey
(This article was originally published at Carlisle Rainey » Methods/Statistics, and syndicated at StatsBlogs.)
I've discussed imputing missing outcome variables before, but I've realized a new problem for those who oppose it. If you are opposed to multiply imputing an outcome variable, then you have to be
opposed to duration models that adjust for time censoring. You have to oppose models such as the tobit that deal with truncated observations. You also have to oppose partial observability models.
Each of these models is dealing with a specific form of missingness in the outcome variable. The goal is exactly the same as multiple imputation and the mathematics is very similar. You cannot oppose
multiple imputation of outcome variables and then use these models. Simply list-wise delete and move ahead.
Please comment on the article here: Carlisle Rainey » Methods/Statistics
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from The American Heritage® Dictionary of the English Language, 4th Edition
• n. Reversion; retrogression.
• n. Relapse to a less perfect or developed state.
• n. Psychology Reversion to an earlier or less mature pattern of feeling or behavior.
• n. Medicine A subsidence of the symptoms or process of a disease.
• n. Biology The return of a population to an earlier or less complex physical type in successive generations.
• n. Statistics The relationship between the mean value of a random variable and the corresponding values of one or more independent variables.
• n. Astronomy Retrograde motion of a celestial body.
• n. Geology A relative fall in sea level resulting in deposition of terrestrial strata over marine strata.
from Wiktionary, Creative Commons Attribution/Share-Alike License
• n. An action of regressing, a return to a previous state.
• n. A psychotherapeutic method whereby healing is facilitated by inducing the patient to act out behaviour typical of an earlier developmental stage.
• n. An analytic method to measure the association of one or more independent variables with a dependent variable.
• n. An equation using specified and associated data for two or more variables such that one variable can be estimated from the remaining variable(s).
• n. The reappearance of a bug in a piece of software that had previously been fixed.
from the GNU version of the Collaborative International Dictionary of English
• n. The act of passing back or returning; retrogression; retrogradation.
from The Century Dictionary and Cyclopedia
• n. The act of passing back or returning; retrogression.
• n. In astronomy, motion from east toward west.
• n. In geometry, contrary flexure; also, the course of a curve at a cusp.
• n. In mathematics, a discrete series which has a last element but no first.
• n. In statistics, the tendency of one variable phenomenon that is correlated with another to revert to the general type and not to equal the amount of deviation of the particular phenomenon with
which it is correlated.
• n. In breeding, the decline toward mediocrity of offspring from the mean of the two parents. Sometimes called filial regression.
from WordNet 3.0 Copyright 2006 by Princeton University. All rights reserved.
• n. an abnormal state in which development has stopped prematurely
• n. (psychiatry) a defense mechanism in which you flee from reality by assuming a more infantile state
• n. the relation between selected values of x and observed values of y (from which the most probable value of y can be predicted for any value of x)
• n. returning to a former state
regress + -ion (Wiktionary)
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[Theorycrafting] LoH and attack speed with wicked wind
• #1 10/2/2012 2:25:34 AM
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Based on a discussion going in another thread I thought I'd make a new thread solely to theorycraft and discuss how WW is affected by attack speed, both for damage and LoH returns. I'll share my
facts, previous assumptions, my new conclusions, and then give the methodolgy and results last, so people not really interested in how I came upon my conclusions can skip the end. Please feel free to
do your own testing to verify or attempt to contradict my results. I'm also planning to do similar analysis with Meteor, though anyone else who has some free time is welcome to do so.
WW LoH coefficient is 12.5%
EDIT: Someone on the official forums went through some testing similar to what I did, only moreso, and made a nice table with # WW procs vs AS.
My previous assumptions were that WW does damage 4x per second, regardless of attack speed, but displays damage every 0.5s and that life gain tics always tic every 0.5s.
My conclusions
1) Damage from WW displays every half second regardless of attack speed
2) Damage from WW is independent of attack speed
3) Life regen displays every half second, with occasional slight variations I attribute to server lag
4) Life gain from LoH is based on attack speed and the tics display at most every 0.5s
Note that (2) and (4) are somewhat contradictive, but that was what I observed. Doubling my attack speed resulted in life gain tics that were for 2x as much. I really don't understand why because the
damage clearly seems to be independent of attack speed, yet LoH returns are not.
My setup
I had a 0.9 aps white weapon with 18-19 damage range, a 1.2 aps white weapon with 9-12 damage range, a 400 LoH amulet, and a 1.5 aps weapon with 3-7 damage. I also had gloves with 7% IAS, rings with
7 and 5 IAS, an amulet with 8% IAS, and bracers with 9 IAS. With all those equipped I was able to get a 1.4 APS weapon up to 1.9 attacks per second so my range was 0.9 - 1.9 attacks per second.
To test dps I also had a ring with +11 min damage that I could pair with the white weapons to give a consistant amount of damage, in theory. It works great for single cast spells but wasn't as good
for WW.
My tests
1) Remove all gear except white weapon and +min damage ring for base case to test damage.
Then equip IAS gear and compare damage to base case. Repeat for both white weapons. Remove IAS gear and equip item with +Int to bring dps close to the dps with IAS gear.
2) Remove all gear except 0.9 APS weapon and 400 LoH amulet, then take a hit so I'm not at full life. Teleport to mob and drop a WW on mob while it's fighting my fracture clone.
Note life gains and frequency of life gains.
Repeat with IAS gear on and repeat with other white weapons.
1) Damage remained same with and without IAS for each weapon. The tics were steady at 2 per second regardless of attack speed. This test was a little difficult to perform because the tics had more
varience than I expected so I tried to focus on the min damage tic while the mob was in the middle of the WW. Just to be sure the lack of a change in damage wasn't due to the small DPS increase of
the IAS gear, by wearing no IAS gear but a pair of boots that brought my dps about the same as it was with the IAS gear, I was able to conclude that IAS does not impact the damage done by WW because
my damage DID increase with the boots on, but didn't for just adding IAS gear. With both weapons the tics were precisely equal to the damage expected from 252% weapon damage done over 6s at 0.5s
intervals, further showing that WW displays damage 0.5s apart, independent of weapon speed.
2) With 0.9 APS I saw tics of 50 every half second. When I added a couple pieces of IAS the tics varied from every 1s to every half second and most of the tics were for 100. With the 1.2 APS weapon,
pretty much every tic was for 100, but the tics weren't quite every half second. At 1.92 APS the tics displayed pretty consistantly every half second, but varied between 100 and 150.
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#2 10/2/2012 3:45:40 AM
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Isn't it the same principle behind the archon beam ? When you add attack speed the ticks have the same frequency but higher damage per tick (or life back).
I think the ticks are constant on 0.5s as long as you have more than 1 attacks/second.
When you hit 2 attacks/second you get the wired ticks where half of the LOH is gained in the first half tick, and the second half in the later tick.
I really would love to test this, but the stupid char copy is still not working for Euro region.
BTW: most people I know switched from LoH to LifeSteal, as the damage gets high enough, it just gives more life back, it seems ...
#3 10/2/2012 4:48:59 AM
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That's the usual test, however with such a test it's impossible to conclude if the WW ticks more / faster with AIS - the damage seem to add up per 0.5 seconds according to what community accepts
right now. This is primarily based on the belief that the channeled / beam spells do crit but we never get to see the crit damage as a larger yellow number because they tick faster than 0.5 seconds
and every 0.5 seconds the damage is added up. According to that post by blizzard - the tick speed scales with attack speed, assuming that the duration of WW would be the same under all conditions -
then it would be reasonable to assume that the attack speed increases the number of ticks of the WW as well. However, i really do not see how it could be tested based on the damage numbers appearing
on the screen.
"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
-- Albert Einstein
#4 10/2/2012 11:23:17 AM
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I make the assumption here that because LoH numbers did increase that number of critical hits per second were increasing as well.
I'd really like to have solid data on the under the hood pulse rate of wicked wind and meteor's DoTs. I think this can be observed via LoH since the coefficient is known to be 0.125. So my question
is this. If the coefficient is 0.125 and you have 500 life on hit, you can engineer a tooltip attacks per second of 1.0, and observe from the half second amalgamated heals how many times the spell is
triggering LoH.
With 500 LoH you should see a multiple of 62.5 per tick and that multiple is the baseline pulse rate of the spell.
If possible, doing similar tests with various meteor runes to obtain similar pulse rates for their DoTs would be helpful as well.
#5 10/2/2012 2:26:07 PM
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Based on my results, it is NOT the same principle as the Archon beam. For the beam, the tics display at the same rate as WW but when you increase attack speed the tics increase in damage. For WW the
tics did not increase in damage or frequency. I realize most people are trying to switch to Life Steal but the idea behind looking a LoH is that is should also apply to how CM procs work since they
both use the same mechanics.
How is it impossible to determine if the tics are faster with higher attack speed or not? I did tests that ranged from 0.9 to 1.92 attacks per second. It seems to me if some attack speed make the
tics faster or slower, I would have noticed it, but they were consistant at 2 tics per second. The easy way to check tic rate is to equip life regen gear since that tics pretty consistantly at 2 per
second. WW damage then lines up pretty well with the life gain tics.
I'd also like to get some solid information about Meteor tics to go along with WW, since I'd greatly prefer to use meteor in a CM build if possible.
To all who responded or have any input, could it be that the varience I saw with the tic damage was due to more frequent damage being averaged? If it was due to crits, like I expected, it shouldn't
really happen with much frequency at only 5% crit rate. Probably the best way to do the analysis is to take a vid and post it but I don't have any video recording software so I basically just dropped
my WW and saw the same damage number tic several times in a row. It really doesn't make sense that the LoH gains would depend on atatck speed but the damage done doesn't. One other thing I could try
is finding a 1.5 APS weapon with the same damage as a 0.9 APS weapon, then I can stack the attack speed to over 2x the 0.9 APS weapon and get better results for comparison. Are there any free vid
recording software programs that people use for D3?
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#6 10/2/2012 3:59:07 PM
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Some more WW info. I downloaded a trial version of Fraps and went to Belial on normal mode since he doesn't move. He also has a decent HP pool so it's a great place to test damage. I bought 2 new
weapons and a new ring:
0.9 APS 22-23 damage
1.5 APS 7-21 damage
+27 min damage ring
+12 min damage ring
0.9 APS w/12 min dmg ring => 34 - 34 damage range. DPS = 125.16 with 0% IAS, so 0.9 APS total
1.5 APS w/27 min dmg ring => 34 - 34 damage range. DPS = 254.50 with 22% IAS, or 1.83 APS total
For each weapon setup indicated I cast 1 WW at a time while recording with FRAPS. I recorded 4 casts for the 0.9 APS weapon and 6 for the 1.5 APS weapon. Since I'm using a trial version, I can only
record 30s of video at a time, but I feel the data is decent.
I replayed each cast in slow motion and recorded the tics on a spreadsheet.
0.9 APS Total Damage
Average = 353.75
1.83 APS Total Damage
Average = 357.83
Based on those results I again conclude that WW damage is independent of attack speed since the 1.83 APS setup had twice the APS but the same weapon damage as the 0.9 APS setup.
I'll repeat the experiment for LoH and post shortly. I'll probably also repeat for meteor eventually.
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#7 10/2/2012 5:32:19 PM
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LoH results are in. My results indicate that more attack speed = more LoH tics, which is similar to the results shown in my first post. I have quantified it a bit, and it seems there are breakpoints
Results (sorry for horrible formating but I haven't figured out how to put a table in these posts)
procs = total life gain / life per proc = Total Life gain / (LoH / 8) since the coefficient is 0.125 or 1/8
APS, #LoH procs
0.9 11
0.963 12
1.01 13
1.071 13
1.5 18
1.605 20
1.71 22
1.785 23
A couple things to note are that at 1.5 attacks per second, the results indicate you get about 3 procs per second. It appears that at 1 APS you get about 2. The trends indicate that at 2 APS you'd
get 4 procs per second, or twice as many as with 1 APS. Thus it seems the approximate relationship is Procs per second = 2*APS, but it is not a smooth relationship. It appears to take the number or
procs and round up, but the formula doesn't work for the 1.7+APS cases, though it works for all other cases. In other words,
Procs = Ceiling(2Base procs per attack * 6 Second duration * Attacks per second)
where Ceiling rounds a number UP to the nearest integer.
I'll work on updating the OP later with this information. My next step will be to do a similar analysis with Meteor. I'm thinking of also doing such an analysis with Archon beam to get an idea of how
frequently the LoH procs occur, but that I'll save for last if I have time.
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#8 10/2/2012 9:08:03 PM
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Are there any free vid recording software programs that people use for D3?
Bandicam is free and I use it for all my D3 videos. The only limitation on the free version (that I am aware of) is that it puts a "www.Bandicam.com" banner on the top of your video and it can only
record a video of up to 10 mins. I havn't had any problem with either of those restrictions, so I'd recommend it. It's also pretty lighweight.
Thanks for doing this testing, it's important information. Especially since I think anybody who wants to farm rings in the patch will need to be CM.
Regarding meteor, I am pretty sure you are going to find that IAS has no effect on the spell (aside from actually decreasing the casting animation, which is something), but by all means please do the
testing to confirm. My basis for this hypothesis is that IAS was demonstrated, long ago, to have no effect on Blizzard, and essentially these spells operate in the same way: they have an AP cost and
no cooldown. I would have guessed the same result for WW, frankly, but a lot of people have been saying that IAS has an effect. I guess it turns out it works on LOH but not damage, which is a bizarre
Also, take a look at this thread, although you might already be aware of all the information: http://www.reddit.com/r/Diablo3Strategy/comments/yz2jn/wizard_104_comprehensive_proc_ratios/
The proc coefficients are common knowledge, but the interesting data are the cast speed and tick speed coefficients.
#9 10/2/2012 9:17:34 PM
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This is very interesting... from here: http://eu.battle.net...08511084?page=4
Also, I don't know if some people already knew but the green numbers does not appears every 0.5 second like some people tested it on 1.02. Instead the game adapt the green number timing on your
life on hit. I learned it the hard way and I don't want people to struggle like I did so I am saying it anyway.
About that strange green number timing : I am currently retesting arc lighting... I might lack sleep because the cast animation (cosmetic) speed appears to be at max on 1.2 attacks per second
then suddenly decrease at 1.3... But i have seen green numbers pops between cosmetic cast with 1.3 attacks per second and damage seems to be dealt properly. The tick speed coef appears to be near
2.5. There is really something we don't know about this spell !
I'm having trouble following exactly what he's saying, but I think it might explain the disparity in your results re: IAS changing the LOH healing but not the damage.
#10 10/2/2012 10:08:55 PM
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0.9 APS Total Damage
Average = 353.75
1.83 APS Total Damage
Average = 357.83
Based on those results I again conclude that WW damage is independent of attack speed since the 1.83 APS setup had twice the APS but the same weapon damage as the 0.9 APS setup.
Nice work so far,
did i miss something or is there 4 ticks in the first case and 6 ticks in the second? At the same time you proclaim that WW damage is independent of APS, well the average damage seems to remain,
while the total number of ticks seems to have increased thus increasing the total damage of WW?
"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
-- Albert Einstein
#11 10/2/2012 10:56:59 PM
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Those are case with total damage done over the 12s, I just only ran 4 cases for the slower attack speed because the rest of my vids got cut off and I was too lazy to make more since the results were
pretty conclusive. Those values are the total damage done for the entire duration of a WW, i.e., 6s. For both weapon speeds and all cases I observed 12 tics total, which matches the 2 tics per second
theory. The average there is the average damage done by the WW over 6s. Sorry for any confusion about that.
Basically, the total damage done by the WW is the same for each weapon, despite the char sheet dps being almost twice as much for the faster weapon.
Thanks for the links. I'll take a look at them tonight. I actually haven't looked up any information about this stuff, so I don't really know what all people have examined. I'll also look into
Bandicam since the demo version of FRAPS only lets me record 30s at a time.
I didn't expect there to be any difference with WW and IAS because the damage did not change, but I am suprised that there is some sort of correlation with LoH, so naturally I want to examine Meteor
also, since it might behave in a similar manner. Even if there is similar results in those links I'll probably do my own testing just to be sure.
My EHP and DPS Spreadsheet, mostly useful for wizards.
My wizard, mostly useful for killing demons and collecting loot.
Wizard CM DPS Simulator, written in Matlab, release version 1.01
Comprehensive CMWW Guide, including gear and build suggestions, plus Uber strats
Arcane Mines Guide, detailed guide for the highest dps wizard build, including vids
#12 10/3/2012 12:25:47 AM
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Quote from Morphos
Are there any free vid recording software programs that people use for D3?
Bandicam is free and I use it for all my D3 videos. The only limitation on the free version (that I am aware of) is that it puts a "www.Bandicam.com" banner on the top of your video and it can
only record a video of up to 10 mins. I havn't had any problem with either of those restrictions, so I'd recommend it. It's also pretty lighweight.
Thanks for doing this testing, it's important information. Especially since I think anybody who wants to farm rings in the patch will need to be CM.
Regarding meteor, I am pretty sure you are going to find that IAS has no effect on the spell (aside from actually decreasing the casting animation, which is something), but by all means please do
the testing to confirm. My basis for this hypothesis is that IAS was demonstrated, long ago, to have no effect on Blizzard, and essentially these spells operate in the same way: they have an AP
cost and no cooldown. I would have guessed the same result for WW, frankly, but a lot of people have been saying that IAS has an effect. I guess it turns out it works on LOH but not damage, which
is a bizarre result.
Also, take a look at this thread, although you might already be aware of all the information: http://www.reddit.co...ve_proc_ratios/
The proc coefficients are common knowledge, but the interesting data are the cast speed and tick speed coefficients.
After reading those posts, it seems the LoH relationship I found is the same as that posted in the thread. Also, the thread indicates Meteor has no speed coefficient which means it shouldn't depend
on IAS, but I'll still check.
Damage isn't mentioned in the thread either, so that's something different I'm doing.
Again, thanks for the links. I wasn't aware there had been an update since the other LoH Coefficient thread I had been using before that is quicky getting out of date.
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#13 10/4/2012 9:27:01 PM
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Well my Meteor - Star Pact results are in, and it behaves exactly as expected. The blast and resulting damage tics are unaffected by attack speed. The LoH returns from the blast and tics are also
unaffected by attack speed, with the coefficients being 0.25 for blast and tics but with the first tic occuring at the same time as the blast. Note that this proc is being reduced to 0.10625 or 17/
Since I don't really have anything to do tonight, I'll probably throw together a rough CM simulator to compare WW and Meteor in the next patch to see if I can determine any sweet spots in terms of
attack speed, APoC, etc. I think it will also be interesting to cacluate the theoretical dps output of each build for say a 60s solo fight, i.e., MP10 keywarden type fight. Is the AP cost of Star
Pact still 35 on the ptr? If so, I might also look at the other runes since they only cost 50 AP, based on the PTR notes. Liquify in particular looks like a decent possiblity. With 50% crit the
average duration is something like 5.5s or 6.5 total procs, compared to the 4 of star pact. The AP cost is only 43% higher but the proc rate would be 62.5% higher, so it seems like it might be worth
My EHP and DPS Spreadsheet, mostly useful for wizards.
My wizard, mostly useful for killing demons and collecting loot.
Wizard CM DPS Simulator, written in Matlab, release version 1.01
Comprehensive CMWW Guide, including gear and build suggestions, plus Uber strats
Arcane Mines Guide, detailed guide for the highest dps wizard build, including vids
#14 10/4/2012 10:21:37 PM
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Loroese, my question here is:
Does attack speed affect the speed of ticks or ticks per second of the WW?
The problem with this is: according to what community believes as well as i recall seeing it stated by Blizzard - the damage numbers for any skill are added up at 0.5 seconds interval. So for example
if the skill ticks twice a second or the skill ticks 4 times or even 20 times a second you will only see 2 damage numbers coming up on the screen.
While the overall damage of the skill is going to remain the same, the amount of ticks it performs is...? The larger number of ticks is going to certainly provide for a larger chance to proc CM,
hence making AIS a valuable stat for the CM builds using WW.
According to your tests - the LoH values seem to increase based on AIS. That alone, suggests that the number or ticks definitely increased assuming that the coefficient remains the same no matter the
APS. Using the same model of healing done added every 0.5 seconds before being displayed - this clearly explains the reason why the numbers increased from 50 to a 100, etc.
So the final conclusion then would be that APS increases the amount of ticks WW performs? Or am i completely lost?
"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
-- Albert Einstein
#15 10/4/2012 10:59:46 PM
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In order to answer your question I have to break up the spell into Damage Tics and Proc Tics.
The Damage Tics are the displayed damage done, which occur every 0.5s regardless of attack speed. The damage displayed from the tics is also independent of attack speed, so a 1 APS weapon with 100
damage will do the same damage per tic as a 2APS weapon with 100 damage. In short, Damage Tics do not depend on attack speed.
The Proc Tics are the CM/LoH procs. It is easiest to see these tics with LoH because they show up as green numbers. The number of tics DOES depend on attack speed and the formula is #Procs =
2*AttackSpeed*6s duration, so with 1 APS you see 12 procs, with 2 APS you see 24 procs, or you gain twice as much life over the same period of time.
So basically the procs follow a different relationship than the actual damage, which makes no sense to me. While you effectively get 24 procs from 2APS, you still only see 12 damage tics.
My conclusions are that IAS increases the number of procs per WW, but doesn't actually increase the damage of WW.
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#16 10/4/2012 11:42:12 PM
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Correct, what i was saying is that with 1 APS you get 12 damage ticks of WW and with 2 APS you get 24 damage ticks as well. The thing is that - it is virtually impossible to see that due to the way
the damage numbers are shown ingame at 0.5 seconds intervals. The damage of the spell is going to be broken down into smaller more frequent ticks rather than larger bigger ticks instead.
The reason i am saying this is that it would be rather stupid to have 2 completely different formulas working for every spell, if you just use 1 formula for both the damage ticks and the proc ticks -
it would be far more sensible and easier to maintain.
"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."
-- Albert Einstein
#17 10/5/2012 12:06:49 AM
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I agree it would make more sense to have 1 system, but what I'm saying is there isn't just 1 system, or if there is, it's not what we're thinking it should be. If attack speed increased the number of
damage tics, and the displayed tics were just sums of the individual tics, then the actual damage done would increase with attack speed. As it is, you get 12 damage tics with 1 APS and 12 damage tics
with 2 APS.
Otherwise if you got 24 damage tics with 2 APS, for the damage that is displayed every 0.5s to be the same as the 1 APS case, then each of the 24 damage tics would have to do half the damage as each
of the 12 damage tics for the 1 APS case, which makes no sense if both have the same weapon damage.
I hope it doesn't make things more confusing to mention this but channeled spells are treated the way you're thinking WW should be. Specifically the displayed damage tics are averaged damage over
0.5s intervals and higher attack speed leads to larger displayed tics while the procs tic at the same rate as the actual damage is dealt. So if you have a beam spell with a speed of 3, you get 3 hits
per tic per attack speed. In other words, with APS=1, your beam deals damage 6 times a second, so every 3 hits is totaled and displayed as damage. Your proc rate is 6 times per second, which matches
the damage rate. If you have APS=2, the beam still only displays damage twice a second but now it deals damage 6 times between tics and procs 12 times per second. All channeled spells work that way,
with their own speed coefficient though.
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#18 10/5/2012 8:17:27 AM
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Well my Meteor - Star Pact results are in, and it behaves exactly as expected. The blast and resulting damage tics are unaffected by attack speed. The LoH returns from the blast and tics are also
unaffected by attack speed, with the coefficients being 0.25 for blast and tics but with the first tic occuring at the same time as the blast. Note that this proc is being reduced to 0.10625 or
Since I don't really have anything to do tonight, I'll probably throw together a rough CM simulator to compare WW and Meteor in the next patch to see if I can determine any sweet spots in terms
of attack speed, APoC, etc. I think it will also be interesting to cacluate the theoretical dps output of each build for say a 60s solo fight, i.e., MP10 keywarden type fight. Is the AP cost of
Star Pact still 35 on the ptr? If so, I might also look at the other runes since they only cost 50 AP, based on the PTR notes. Liquify in particular looks like a decent possiblity. With 50% crit
the average duration is something like 5.5s or 6.5 total procs, compared to the 4 of star pact. The AP cost is only 43% higher but the proc rate would be 62.5% higher, so it seems like it might
be worth trying.
I'm very interested to see the calculator. All of the other meteor runes are in fact 50 AP, and I think it is agreed that Liquify and Star Pact are the two best options for meteor.
#19 10/5/2012 9:19:06 AM
• Zealot
• Join Date: 8/21/2012
• Posts: 121
APS, #LoH procs
0.9 11
0.963 12
1.01 13
1.071 13
1.5 18
1.605 20
1.71 22
1.785 23
Unless I miss my guess, the third attack speed in your list should actually be 1.026 as you appear to be using a 0.9 attack speed weapon with 0%, 7%, 14%, and 19% IAS, and then a 1.5 speed weapon
with 0%, 7%, 14%, and 19% IAS.
This data doesn't quite fit the hypothosis that you simply get 2 * attack speed ticks per second but there is always a tick at the beginning and end of the spell essentially rounding up. If that were
the case your final 2 tests should have yielded 21 and 22 ticks.
You can get a slightly better fit assuming that there is rounding ocurring twice, that there is a base number of ticks for each weapon speed of 2*weapon speed rounded up, and then that creates a base
tick rate that is increased by IAS% and rounded up again at cast time to determine the total number of ticks for the spell. This yields slightly closer results but still suggests that your final case
(1.5 attack speed weapon 19% increased attack speed from gear) should have 22 ticks rather than 23.
#20 10/5/2012 2:34:59 PM
• Council Member
• Join Date: 6/4/2012
• Posts: 1,600
Quote from DisposableHeero
Unless I miss my guess, the third attack speed in your list should actually be 1.026 as you appear to be using a 0.9 attack speed weapon with 0%, 7%, 14%, and 19% IAS, and then a 1.5 speed weapon
with 0%, 7%, 14%, and 19% IAS.
This data doesn't quite fit the hypothosis that you simply get 2 * attack speed ticks per second but there is always a tick at the beginning and end of the spell essentially rounding up. If that
were the case your final 2 tests should have yielded 21 and 22 ticks.
You can get a slightly better fit assuming that there is rounding ocurring twice, that there is a base number of ticks for each weapon speed of 2*weapon speed rounded up, and then that creates a
base tick rate that is increased by IAS% and rounded up again at cast time to determine the total number of ticks for the spell. This yields slightly closer results but still suggests that your
final case (1.5 attack speed weapon 19% increased attack speed from gear) should have 22 ticks rather than 23.
You're right about the 1.01 attack speed being off a bit, though it should be 1.008 since I had a ring with 5% IAS.
Regarding the idea of a tick to start and end, I'm not completely sure what the model would be, something like 2+2*5*APS instead of 6*2*APS?
Quote from Morphos
I'm very interested to see the calculator. All of the other meteor runes are in fact 50 AP, and I think it is agreed that Liquify and Star Pact are the two best options for meteor.
I'll finish it later today. I got stuck a bit when trying to figure out how to account for stacking tics of spells but I think I figured out a good way to treat them while making as few
approximations as possible. It's basically going to be a monte carlo type simulator. Unfortunatelly it will be made in MATLAB, which is not a freeware program, but if anyone is familiar with other
programming platforms, it should be easy to convert to C++ since I'm not using any MATLAB specific functions. I'll make a new thread for that when it's done and include the code for others to look
My EHP and DPS Spreadsheet, mostly useful for wizards.
My wizard, mostly useful for killing demons and collecting loot.
Wizard CM DPS Simulator, written in Matlab, release version 1.01
Comprehensive CMWW Guide, including gear and build suggestions, plus Uber strats
Arcane Mines Guide, detailed guide for the highest dps wizard build, including vids
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East Elmhurst Calculus Tutor
...I have been living in Queens for five years. I love to teach Mathematics and it has become one of my hobbies. I am in a Junior level toward an Electrical Engineering degree at City College of
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15-317 Constructive Logic
15-317 Constructive Logic
• There are 10 homework assignments, worth a total of 400 points.
• Some assignments may offer additional problems for extra credit, which is recorded separately.
• Extra credit will be considered when determining midterm and final grades for borderline cases.
• Assignments generally are given out Thursday in lecture and are due the following Thursday.
• Homeworks may require use of the course software, or simply a write-up with pencil and paper.
• Machine-checked assignment must be submitted via the course software before the start of lecture on the due date.
• Written homeworks are to be handed in at the beginning of lecture on the due date.
• We will try our best to return graded homework during the lecture following the due date.
• For typesetting deductions in LaTeX, we use proof.sty
Date Assignment Due Solutions
Sep 3 Homework 1 Natural Deduction (Tutch requirements, LaTeX) Thu Sep 10 Written, Tutch
Sep 10 Homework 2 Quantifiers and Proof Terms (Tutch requirements, LaTeX) Thu Sep 17 Written, Tutch
Sep 17 Homework 3 Natural Numbers and Classical Reasoning Thu Sep 24 Written, Tutch
(Tutch requirements, Starter code, LaTeX)
Sep 24 Homework 4 Classical and Intuitionistic Logic (Tutch requirements, LaTeX) Thu Oct 1 Written, Tutch
Oct 8 Homework 5 Sequent Calculus for Proof Search (Starter code, Test harness, LaTeX) Thu Oct 15 Written, Code
Hard test cases
Oct 15 Homework 6 Logic Programming and Inversion (LaTeX) Thu Oct 22 Written, Prolog
Oct 22 Homework 7 Programming in Prolog (Starter code, LaTeX) Thu Oct 29 Written, Prolog
Oct 29 Homework 8 Representing Proofs in Twelf (Starter code) Thu Nov 5 Twelf
Nov 12 Homework 9 Proving Metatheorems in Twelf (Starter code, LaTeX) Thu Nov 19 Written, Twelf
Nov 19 Homework 10 Bottoms Up! (Test cases) Thu Dec 3 Lollibot
All assignments in this course are individual assignments. The work must be your own. Do not copy any parts of the solution from anyone, and do not look at other students solutions. Do not make any
parts of your solutions available to anyone and make sure noone else can read your files. We will rigorously apply the university policy on cheating and plagiarism.
We may modify this policy on some specific assignments. If so, it will be clearly stated in the assignment.
It is always permissible to clarify vague points in assignments, discuss course material from notes or lectures, and to give help or receive help in using the course software such as proof checkers,
compilers, or model checkers.
[ Home | Schedule | Assignments | Handouts | Software ]
Frank Pfenning
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Kenilworth, NJ ACT Tutor
Find a Kenilworth, NJ ACT Tutor
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Sex, evolution and parasitic wasps
Sometimes a state of affairs becomes so familiar to us that it is simply expected, and we may forget to wonder why it should be that way in the first place. Sex ratios are a good example of this: the
number of men and women in the world is roughly equal, but why should this be the case?
Many mammals have an equal sex ratio at birth.
For many 18th century thinkers the answer lay in divine providence – the goodness of God – and the equal number of boys and girls born each year was seen as an endorsement of monogamous marriage
written into the very laws of nature. This changed with the publication of Charles Darwin's Origin of Species in 1859, followed by The Descent of Man in 1872, and the focus began to shift to finding
an explanation in terms of evolution and natural selection. Darwin made an attempt but quickly withdrew it, concluding that "the whole problem is so intricate that it is safer to leave its solution
for the future".
This solution was to come in the 20th century. The reason why most species produce roughly equal numbers of male and female offspring is actually very simple, so let's see how maths can be used to
prove an intuitive biological argument.
Fisher's principle
It's disputed whether the great British geneticist Ronald Fisher was really the first to explain why evolution favours equal numbers of sons and daughters but the result still bears his name:
Fisher's principle. Begin by considering a diploid species, in which each organism has two copies of each chromosome, one coming from the mother and one from the father. This means that half of each
generation's genetic material must come from female parents and half from male parents.
Assume that mating within the population is random. This will mean that each male and female organism can expect to pass on the same quantity of genetic material to the next generation as any other
member of the same sex. In practice this probably couldn't happen, but it's a useful model. Now consider a population with more males than females. The total amount of genetic material passed on by
the males is equal to that passed on by the females, because each child contains an equal amount of both. Since there are more males, it follows that the expected share for each individual male will
be lower than that for each female, and any given female can expect to pass on more genes than any given male.
Ronald Fisher, 1890-1962.
Think of each parent as an investor, looking to get the greatest genetic return on their finite number of children. In the population considered above it's clear that there's a better "payoff"
associated with daughters than with sons, so natural selection will favour organisms which go against the population's male bias and produce more female offspring. They will have more grandchildren,
and therefore the female-biased sex ratio will become more common until the population has equal numbers of both sexes. Conversely, it pays to have more sons in female-majority populations. The equal
sex ratio is thus the only one which can't be beaten by a different strategy and hence is said to be stable.
Notice that we're talking about a stable strategy rather than an optimal one. A population's maximum possible growth is proportional to the number of females, so its members could probably all do
better if they universally adopted a somewhat daughter-biased sex ratio. However, natural selection would favour any individual in the population which bucked this trend to produce more sons so this
strategy would not be stable. Having half male children and half female may not be the optimum strategy but it is unbeatable: no strategy competing against it can do better so natural selection will
favour the status quo.
Proving that 1:1 is the best strategy
These arguments seem fairly convincing, but can it be proved mathematically that equal investment is an unbeatable strategy? Let's begin by defining some terms.
Imagine a population of wasps. Any species would do, but wasps are a nice example as the females of some varieties have the ability to choose the sex of their offspring. Assume each female can expect
to have proportion of these children which are male by
To prove Fisher’s principle, we need to build up a model for the fitness of different reproductive strategies, then show that there’s no way to beat the 1:1 ratio. Fitness is a measure of how
successful an organism is in spreading their genes through the population, and a good definition for it here is the success of an organism’s offspring in producing children of their own. To keep
things simple, assume there is one variant female, who can adopt any sex ratio, and a certain number herd females who always produce equal numbers of sons and daughters.
What will the fitness of each starting wasp be? In words the answer is clear:
Note that this answer won't be the same as the number of grandchildren. The assumption of completely random mating includes offspring between siblings, so there'll be some overlap between the
offspring of the daughters and of the sons.
Let’s start with the variant wasp and try to put these terms in numerical form. Represent the fitness as a function
The number of sons is also easy – it’s defined to be
We can get the herd values by taking the formulae we’ve got for the variant’s offspring, setting
As expected, the even sex ratio means that the number of sons and daughters are equal. Substituting all of these in gives
This is everything needed to write out the formula for the variant wasp’s fitness in full
which can be rewritten as
We’ll also need a formula for the fitness of a member of the herd, which we denote by
With these terms removed the formulae are both independent of
If Fisher’s principle is correct then the herd’s strategy can’t be defeated:
This can be simplified to
Notice that the denominator of
This can be rewritten as
Since this expression is a square, it can never be negative. It is equal to zero for
This is exactly what we wanted to prove:
In biological terms, we’ve just shown that the equal sex ratio is unbeatable by any alternative.
Except when it isn't
Well, that isn't quite true. Any deduction is only as good as its starting assumptions, and several of the ones used here could be called into question.
For now let's focus on just one assumption, that every organism is competing with every other organism in the population. This can be a useful model but it obviously doesn't hold true in reality: a
lion in South Africa's Kruger National Park won't be competing with those in London Zoo. Most organisms will only be competing with a relatively small subset of the general population at any one
Certain species of wasp take this to extremes. In these varieties several females lay eggs in a food source, such as a paralysed caterpillar, and then die. When the eggs hatch the next generation
will feed, grow and then mate indiscriminately with the other wasps hatched in the same source. The females will then leave the host to seek out a new host to lay their eggs in while the males die
There are two aspects of this system that need to be taken into account here. The first is the local fitness of different sex ratios: how successful will particular ratios be in the competition
between different wasps in one particular food source. The second is the global fitness of a sex ratio: how does the success of a particular strategy compare with the average success of all wasps in
the population. If a strategy is fitter than the average across the population, then it will spread.
The earlier proof of Fisher's principle will still hold for local fitness: if a wasp adopts a equal sex ratio then it can't do worse than any of the other wasps who laid eggs in the same food source.
The interesting thing is that this doesn't guarantee success on the global level, as doing better than local competitors doesn't necessarily mean doing better than the population average.
This poor caterpillar carries wasp larvae parasytes. Image: Stsmith.
Using similar ideas to the ones we have already encountered, it is possible to come up with a generalised model which applies to this situation. We won’t develop this model here, but only notice that
it predicts a stable sex ratio of
Two interesting aspects of this result are the answers it returns for very small and very large values of
At the other extreme, the larger
These predictions can be compared to observed sex ratios in nature with encouraging results. There are a number of factors which complicate the idealised model we've constructed here, nudging the
actual sex ratio one way or the other, but studies have shown that initially female-leaning sex ratios do move towards 1:1 as the number of wasps per host grows. It's a testament to the power of
applied mathematics that, using only a few reasonable assumptions and some A-level techniques, this result could be anticipated years before it was demonstrated experimentally.
About the author
Paul Taylor is studying towards an MSc in Biomathematics at Durham University where he previously read maths and philosophy as an undergraduate. In his spare time he plays Go and writes about science
for student media.
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SEC. 1125. TARGETED GRANTS TO LOCAL EDUCATIONAL AGENCIES.
``SEC. 1125. TARGETED GRANTS TO LOCAL EDUCATIONAL AGENCIES.
``(a) Eligibility of Local Educational Agencies.--A local educational agency in a State is eligible to receive a targeted grant under this section for any fiscal year if the number of children in the
local educational agency counted under subsection 1124(c), before application of the weighting factor described in subsection (c), is at least 10, and if the number of children counted for grants
under section 1124 is at least 5 percent of the total population aged 5 to 17 years, inclusive, in the local educational agency. Funds made available as a result of applying this subsection shall
be reallocated by the State educational agency to other eligible local educational agencies in the State in proportion to the distribution of other funds under this section.
``(b) Grants for Local Educational Agencies, the District of Columbia, and Puerto Rico.--
``(1) In general.--The amount of the grant that a local educational agency in a State or that the District of Columbia is eligible to receive under this section for any fiscal year shall be the
product of--
``(A) the weighted child count determined under subsection (c); and
``(B) the amount in the second sentence of subparagraph 1124(a)(1)(A).
``(2) Puerto Rico.--For each fiscal year, the amount of the grant for which the Commonwealth of Puerto Rico is eligible under this section shall be equal to the number of children counted under
subsection (c) for Puerto Rico, multiplied by the amount determined in subparagraph 1124(a)(3).
``(c) Weighted Child Count.--
``(1) Fiscal years 1966091998.--
``(A) In general.--The weighted child count used to determine a county's allocation under this section is the larger of the two amounts determined under clause (i) or (ii), as follows:
``(i) By percentage of children.--This amount is determined by adding--
``(I) the number of children determined under section 1124(c) for that county constituting up to 12.20 percent, inclusive, of the county's total population aged 5 to 17, inclusive,
multiplied by 1.0;
``(II) the number of such children constituting more than 12.20 percent, but not more than 17.70 percent, of such population, multiplied by 1.75;
``(III) the number of such children constituting more than 17.70 percent, but not more than 22.80 percent, of such population, multiplied by 2.5;
``(IV) the number of such children constituting more than 22.80 percent, but not more than 29.70 percent, of such population, multiplied by 3.25; and
``(V) the number of such children constituting more than 29.70 percent of such population, multiplied by 4.0.
``(ii) By number of children.--This amount is determined by adding--
``(I) the number of children determined under section 1124(c) constituting up to 1,917, inclusive, of the county's total population aged 5 to 17, inclusive, multiplied by 1.0;
``(II) the number of such children between 1,918 and 5,938, inclusive, in such population, multiplied by 1.5;
``(III) the number of such children between 5,939 and 20,199, inclusive, in such population, multiplied by 2.0;
``(IV) the number of such children between 20,200 and 77,999, inclusive, in such population, multiplied by 2.5; and
``(V) the number of such children in excess of 77,999 in such population, multiplied by 3.0.
``(B) Puerto Rico.--Notwithstanding subparagraph (A), the weighting factor for Puerto Rico under this paragraph shall not be greater than the total number of children counted under subsection
1124(c) multiplied by 1.72.
``(2) Fiscal years after 1999.--
``(A) In general.--For each fiscal year beginning with fiscal year 1999 for which the Secretary uses local educational agency data, the weighted child count used to determine a local
educational agency's grant under this section is the larger of the two amounts determined under clauses (i) and (ii), as follows:
``(i) By percentage of children.--This amount is determined by adding--
``(I) the number of children determined under section 1124(c) for that local educational agency constituting up to 14.265 percent, inclusive, of the agency's total population aged 5
to 17, inclusive, multiplied by 1.0;
``(II) the number of such children constituting more than 14.265 percent, but not more than 21.553 percent, of such population, multiplied by 1.75;
``(III) the number of such children constituting more than 21.553 percent, but not more than 29.223 percent, of such population, multiplied by 2.5;
``(IV) the number of such children constituting more than 29.223 percent, but not more than 36.538 percent, of such population, multiplied by 3.25; and
``(V) the number of such children constituting more than 36.538 percent of such population, multiplied by 4.0.
``(ii) By number of children.--This amount is determined by adding--
``(I) the number of children determined under section 1124(c) constituting up to 575, inclusive, of the agency's total population aged 5 to 17, inclusive, multiplied by 1.0;
``(II) the number of such children between 576 and 1,870, inclusive, in such population, multiplied by 1.5;
``(III) the number of such children between 1,871 and 6,910, inclusive, in such population, multiplied by 2.0;
``(IV) the number of such children between 6,911 and 42,000, inclusive, in such population, multiplied by 2.5; and
``(V) the number of such children in excess of 42,000 in such population, multiplied by 3.0.
``(B) Puerto Rico.--Notwithstanding subparagraph (A), the weighting factor for Puerto Rico under this paragraph shall not be greater than the total number of children counted under section
1124(c) multiplied by 1.72.
``(d) Local Educational Agency Allocations.--For fiscal years 1995 through 1998, grants shall be calculated by the Secretary on the basis of the number of children counted under section 1124 for
counties, and State educational agencies shall suballocate county amounts to local educational agencies, in accordance with regulations published by the Secretary. In any State in which a large
number of local educational agencies overlap county boundaries, the State educational agency may apply to the Secretary for authority during any particular fiscal year to make the allocations
under this part (other than section 1124A) directly to local educational agencies without regard to the counties. If the Secretary approves an application of a State educational agency for a
particular year under this subparagraph, the State educational agency shall provide assurances that--
``(1) such allocations will be made using precisely the same factors for determining a grant as are used under this part;
``(2) such allocations will be made using alternative data approved by the Secretary that the State determines best reflects the distribution of children in poor families and is adjusted to be
equivalent in proportion to the number of children determined in accordance with section 1124(c); or
``(3) such allocations will be made using data that the State educational agency submits to the Secretary for approval that more accurately target poverty.
In addition, the State educational agency shall provide assurances that a procedure will be established through which local educational agencies dissatisfied with the determinations made by the State
educational agency may appeal directly to the Secretary for a final determination. For fiscal years beginning in 1999, for each local educational agency serving an area with a total population of
at least 20,000 persons, the grant under this section shall be the amount determined by the Secretary. For local educational agencies serving areas with total populations of fewer than 20,000
persons, the State educational agency may either (1) distribute to such local educational agencies grants under this section equal to the amounts determined by the Secretary; or (2) use an
alternative method, approved by the Secretary, to distribute the share of the State's total grants under this section that is based on local educational agencies with total populations of fewer
than 20,000 persons. Such an alternative method of distributing grants under this section among a State's local educational agencies serving areas with total populations of fewer than 20,000
persons shall be based upon population data that the State educational agency determines best reflects the current distribution of children in poor families among the State's local educational
agencies serving areas with total populations of fewer than 20,000 persons. If a local educational agency serving an area with total populations of less than 20,000 persons is dissatisfied with
the determination of its grant by the State educational agency, then the local educational agency may appeal this determination to the Secretary. The Secretary shall respond to this appeal within
45 days of receipt.
``(e) State Minimum.--Notwithstanding any other provision of this section or subsection (b)(1) or (d) of section 1122, from the total amount available for any fiscal year to carry out this section,
each State shall be allotted at least the lesser of--
``(1) 0.25 percent of total appropriations; or
``(2) the average of--
``(A) one-quarter of 1 percent of the total amount available to carry out this section; and
``(B) 150 percent of the national average grant under this section per child described in section 1124(c), without application of a weighting factor, multiplied by the State's total number of
children described in section 1124(c), without application of a weighting factor.
-###-SEC. 1124A. CONCENTRATION GRANTS TO LOCAL EDUCATIONAL AGENCIES
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Valid joint probability density function?
September 30th 2012, 11:17 AM #1
Sep 2012
Valid joint probability density function?
I have to show that this is a valid joint probability density function. My understanding is that this is valid when the integral is equal to 1. My question is, how to do I do a joint integration
when only one of the variables is in play?
f(x,y) = 1/y for 0<x<y, 0<y<1
I appreciate the help!
Re: Valid joint probability density function?
Is the function non-negative everywhere?
What is $\int_0^1 {\int_0^y {\frac{1}{y}dxdy} }=~?$
Re: Valid joint probability density function?
They did not specifically say 0 elsewhere, but I assume it is non-negative outside of the limits. I will try it like that, thank you.
September 30th 2012, 11:27 AM #2
September 30th 2012, 12:05 PM #3
Sep 2012
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Acting on Specific Elements in a Matrix
Using MATLAB, there are several ways to identify elements from an array for which you wish to perform some action. Depending on how you've chosen the elements, you may either have the list of
elements to toss or the list if elements to retain. And you might not have much if any control yourself how the list gets presented to you since the list could be passed to you from another
calculation. The lists might be indices, subscripts, or logical arrays (often referred to as masks). Let's look at how you might arrive at such a situation and see what the code looks like to perform
one particular action, setting the desired element values to 0.
Note: I am not discussing efficiency in this article. It is highly dependent on the number of elements in the original array and how many will be retained or thrown out. This article focuses on
specifying what to keep or replace.
General Setup
Here's the setup for this investigation. I will use a fixed matrix for all the methods and always end up with the same final output. The plan is to show you multiple ways to get the result, since
different methods may be appropriate under different circumstances.
A = magic(17);
Result = A;
Result( A < mean(A(:)) ) = 0;
Let's look at the nonzero pattern of Result using spy.
Method #1 - Using Subscripts of Keepers
Here's a list of the subscripts for the elements to keep unchanged.
[rA,cA] = find(A > (17^2)/2);
Next we convert the subscripts to indices.
Result1 = zeros(size(A));
indices = sub2ind(size(A),rA,cA);
Result1(indices) = A(indices);
isequal(Result, Result1)
ans =
Why did I convert subscripts to indices? Let me illustrate with a very small example.
matrix = [ -1 1 0; 2 0 -2; 0 3 -3]
[rows,cols] = find(matrix==0)
matrix =
-1 1 0
2 0 -2
0 3 -3
rows =
cols =
Now let's see what I get if I use the subscripts to address the selected elements:
ans =
0 3 -3
2 0 -2
-1 1 0
I get the full matrix back, even though I selected only 3 elements. This definitely surprised me when I first encountered this. What's happening?
MATLAB matches each row element with each column element. matrix([1 2 3],2) returns the elements from rows 1 through 3 in column 1.
ans =
To learn more about indexing in general, you might want to read these posts or search the MATLAB documentation.
Method #2 - Using Indices of Keepers
Here we used the single output form of find which returns indices instead of subscripts.
indA = find(A > (17^2)/2);
Result2 = zeros(size(A));
Result2(indA) = A(indA);
isequal(Result, Result2)
ans =
Method #3 - Using Logical Keepers
We'll try keeping about half of the elements unchanged.
keepA = (A > (17^2)/2);
Result3 = zeros(size(A));
Result3(keepA) = A(keepA);
isequal(Result, Result3)
ans =
keepA is a logical matrix the same size as A. I use logical indexing to populate Result3 with the chosen values from A.
Method #4 - Subscripts for Elements to Set to Zero
If instead we have a list of candidates to set to 0, we have an easier time since we don't need to start off with a matrix of zeros. Instead we start with a copy of A.
Result4 = A;
[rnotA,cnotA] = find(A <= (17^2)/2);
Convert indices to subscripts, as in method #1.
indices = sub2ind(size(A),rnotA,cnotA);
Now zero out the selected matrix elements.
Result4(indices) = 0;
isequal(Result, Result4)
ans =
Method #5 - Indices for Elements to Set to Zero
If we're instead given indices, we simply skip the step of converting subscripts and follow similar logic to that in method #4.
Result5 = A;
indnotA = find(A <= (17^2)/2);
Result5(indnotA) = 0;
isequal(Result, Result5)
ans =
Method #6 - Using Logical Arrays to Specify Zero Elements
Finally, if we have a mask for the values to set to 0, we simply use it to select and set elements.
Result6 = A;
keepnotA = (A <= (17^2)/2);
Result6(keepnotA) = 0;
isequal(Result, Result6)
ans =
Which Method(s) Do You Prefer?
Which method or methods do you naturally find yourself using? Do you ever invert the logic of your algorithm to fit your way of thinking about addressing the data (the ins or the outs)? Please post
your thoughts here. I look forward to seeing them.
Get the MATLAB code
Published with MATLAB® 7.6
32 CommentsOldest to Newest
I definitely prefer the use of logical index, because the it’s more explicit. I think the code stay more readable. It’s something that I miss on other lenguajes.
It depends on the algorithm whether logical or numerical indexing is most suitable.
When I need a list of indices several times because I don’t want to create new variables, a list of numerical indices is shorter than a logical matrix. Also, when the positions of selected elements
are of interest, a list of numerical indices is better, because subscripts are easy to calculate from numerical indices.
In other cases there is the warning “logical indexing is usually faster than find”. Then I use logical indexing.
Personally, I have started to use logical indexing as much as possible. What I like about logical indexing is that I can plot the data and the mask in the same plot and clearly see what I am removing
and what I am not removing.
Thanks to all of you. For a long time, it seemed like people weren’t learning about logical indexing but now people are. I think that’s good
(but, as OkinawaDolphin says, not always optimal when the indices need to hang around for several operations and the array is large compared to the number of values of interest).
nowadays I do use logical indexing (with the exception that OkinawaDolphin pointed out). But the ONE thing which prevented me from using logical indexing from the beginning is the following.
When I saw the result of the following operations
> a = [1 2 3 4 5]
> i1 = (a > 3)
ans =
and when I realized that I can use this array in the following way (for logical indexing):
> a(i1)
ans =
I thought I can create the logical arrays by hand and use them:
> i2 = [0 0 0 1 1]
> a(i2)
??? Subscript indices must be real positive integers or logicals.
Of course, now I know that i2 created that way is real array which cannot be used as a mask and that the right way is
> a(logical(i2))
ans =
Trivial. But for me, this was the reason why I tended to use indices and ‘find’ instead of logicals for a long time.
Thanks for clarifying that the array must be of logical type for logical indexing.
I started using logical arrays a LOT more often after heeding the warnings and recommendations issued by ‘mlint’. Initially, I didn’t really believe logical arrays were that much more efficient, but
profiling my code before and after has shown direct performance improvements.
I’ve always enjoyed using array indexing of all sorts (logical or otherwise) because it can be a very compact way of expressing an operation.
Here, for example, is a “logical array” method for removing row R and column C (both scalars) from a matrix X…
X(R~=1:end, C~=1:end)
How cool is that ?
…how would you do the same thing if R and C were vectors ?
Actually, to answer Loren’s original question “Which Method(s) Do You Prefer?”, I have to say “none of the above”.
I often rely on the rather naughty mix of arithmetic and logical operations:
A .* (A > (17^2)/2)
…because it doesn’t require an assignment statement and can be used inline as part of a longer expression.
For myself, generally the answer to my preference depends on what I’m going to do with the selected elements. If I’m going to be doing something that can be vectorized (is “vectorizable”?) then I try
to use logical arrays. If not, I tend to need the actual indices of the selected elements. As a related question, is there any way to determine which of the following will be faster (or even a rule
of thumb)?
A= rand(10000,100);
A=A(1:5000,1:100); %Explicityly stating the number of columns
for n=1:size(A,2)
some other method?
When I’m dealing with very large data sets, this is one of the most frustrating slow points, since it can exceed the time requirements of many of the calculations.
With regard to removing rows R and columns C from a matrix, you can use:
A = reshape(1:49, 7, 7);
B = A; % for comparison
R = 1:2:7; % Rows to remove
C = 2:2:7;
B(R, :) = [];
B(:, C) = [];
A = reshape(1:49, 7, 7);
R = 1:2:7; % Rows to remove
C = 2:2:7;
B = A(setxor(R, 1:7), setxor(C, 1:7));
% or
B = A(setdiff(1:7, R), setdiff(1:7, C));
As for the arithmetic/logical combination, that will work … as long as all your elements of A are finite.
A = 1:1000;
B = A.*(A > (17^2)/2);
all(B == A | B == 0)
A = 1:1000;
A([5 982]) = NaN;
A([7 562]) = Inf;
A([324 870]) = -Inf;
B = A.*(A > (17^2)/2);
all(B == A | B == 0)
Note that not only are elements 5 and 982 of the second B NaN (as you would expect, since any arithmetic operation on a NaN results in a NaN result) but so are elements 324 and 870. This is because
0*Inf and 0*(-Inf) also both return NaN.
Nice reply to Tony’s challenge. You can use end’s in the expressions of your first solutions as well:
A = reshape(1:49, 7, 7);
R = 1:2:7; % Rows to remove
C = 2:2:7;
B = A(setxor(R, 1:end), setxor(C, 1:end));
% or
B = A(setdiff(1:end, R), setdiff(1:end, C));
I definitely prefer, and use as often as possible, logical indexing, possibly keeping the indices stored for future reference only if really needed.
It makes the code much more readable and, as I used to be programming with SQL languages, it exploits the semantic power of the WHERE clauses which are what make DB programming so powerful.
I tend to use all methods mentioned based on my need.
What I find interesting is how common a problem it is for (not necessarily beginner) users to run into the issue mentioned in Method #1. Often users are using subscripting and don’t realize they need
to convert to linear indices to solve a specific problem.
I find myself having to repeatedly explain the difference and am glad you have addressed the topic here.
Steve, Loren,
Thanks very much for this solution…
B = A(setdiff(1:end, R), setdiff(1:end, C));
Steve – very neat,
Loren – I had no idea I could use “end” in a function call !
…where’s that documented ?
Here’s the documentation for end:
The capability has been allowed, I believe, since its introduction as an index (and overloadable with the class system).
Hey Loren,
Logical indexing is nice, but not always practical. We produce hundreds of GB of data every day a subsequently load it into MATLAB. It’s possible to construct a short array of numerical indices,
which contains the subset of the data I may be interested in but keeping a logical index around is not really feasible, either in MATLAB or on disk.
Further, to the comparison of logical vs. numerical indexing (on a smaller footprint example)
x = rand(2 * 1e9, 1); %16GB of data
tic; l = x > 0.5; toc
Elapsed time is 4.771542 seconds.
tic; i = find(l); toc
Elapsed time is 29.321260 seconds.
>> tic; y=x(l); toc
Elapsed time is 49.782177 seconds.
>> tic; z=x(i); toc
Elapsed time is 51.138457 seconds.
As you can see, there is very little in it for a large set, since most of the computation is likely to be memory bound. I have to say that I’m disappointed with the implementation of the numeric
subscript. It is not surprising that a logical index must grow the resulting array, since it doesn’t know in advance how many elements may be in the result. However, to see reallocations on my system
monitors during the final subscript call seems unnecessary, since I know *exactly* how many elements there’ll be. Memory reallocation would actually be the reason why, for large datasets, I’d expect
the numerical index to outperform the logical index.
On a similar note, here are two features I’d like to see in MATLAB:
1) Don’t construct the logical array if you don’t need to: For example,
i = find(logical expression);
or even
b = logical expression;
i = find(b);
% b no longer used after here
There seems no reason why you have to do construct the temporary logical array first and then follow the computation by another pass over the logical array. This is slow and wastes an unnecessary
amount of memory for a temporary.
2) Again, suppose I have a large dataset (200+ GB in size). I want to perform a computation on a subset of it (say 50GB). I have to make a copy to use the data in the computation, which is a problem,
as I’m already riding hard against the amount of available memory. If I have a list of the indices already, I’d like to use an “indirect” array – an object that holds both indices and the original
data and allows me to subscript it as a regular array. To preempt objections: a) Yes, I could write a class to do it but that’s not very fast. b) The object could be read only, so that you wouldn’t
have to worry about multiple numerical indices with the same value.
Many thanks,
Thank you very much for Method #4.
I’ve been trying hard to find a way for vectorizing relational matrix manipulation in Matlab.
Maybe there is an even simpler way that I am not aware of, but your #4 works.
I used it with
indices = sub2ind(size(A),rnotA+1,cnotA+1);
Note +1. I basically needed to manipulate the i+1,j+1 neighbors of elements satisfying a condition without relying on a for loop.
Thanks again!
i want to calculate numbers of columns in a matrix on matlab, will some1 kindly help me,,,?
Perhaps the function size will help you out, especially with 2 inputs ( http://www.mathworks.com/access/helpdesk/help/techdoc/ref/size.html ).
hi i would like to ask something…I want to remove an especific NaN element from a matrix, I do not know how to do it. All the solutions that I’ve read explain how to remove rows where the NaN exists,
but not specific elements..Could you please help me? I appreciate your help
If you have a vector, you can remove NaNs like so:
a(isnan(a)) = [];
But if you have a matrix, you aren’t guaranteed that the same number of NaNs are in each row/column so you may not end up with a rectangular array.
i have a question about removing elements from an array based on their indices, not their values. please consider the following code snippet showing 3 methods for doing such an operation:
n=100; % square nxn matrices
A0=rand(n,n,n); % generate matrices
for i=1:n % for each matrix
% method 1 of dropping a matrix
% method 2 of dropping a matrix
% method 3 of dropping a matrix
% mean computation for comparison purposes
note that upon running profiler, method 3 takes about half the time of method 1, and about 1/3 of the time of method 2. but the mean computation takes far less time than any!
this quite surprising result (to me) leads me to have the following questions:
1) why does it take so long?
2) is there a way to make it go *much* faster?
many thanks,
mean takes less time since there’s no need to copy the array at all before acting on it.
What exactly is taking so long that you want to go faster? I doubt you can do better than the 3rd method, which I think makes the fewest temporary intermediate arrays (none). It only makes a copy of
the array to keep and nothing more.
hey loren, thanks for the response. so, the first line of the first two methods, where the copy is made, takes nearly no time. it is the second line, where i only discard elements of the matrix, that
takes time. that might have been confusing. in practice, what i do now is something like:
where ‘s’ is a vector listing the indices that i care about.
and the reason i care about all this, is because i am doing leave-one-out cross-validation. i have a bunch of training data, and i am fitting my model, but want to ensure that i have not overfit. so,
i loop through every data point, discard it to fit the model, and then check the model accuracy on it. the part of the code that takes by far the longest was dropping a matrix from the array. now, i
skip that by doing it all in one line, as shown above. but now that line just takes about as long as dropping the matrix did, so i didn’t get any speed up.
specifically, my code now looks like this:
n=100; % square nxn matrices
A0=rand(n,n,n); % generate matrices
for i=1:n % for each matrix
% method 1 of getting mean
% method 2 of getting mean
M2=mean(A0(:,:,2:n),3); % mean computation for comparison purposes
running the profiler shows that the line to compute M2 takes equally long as the line to compute A1 and M1 combined. and, if i look carefully, i see that actually computing M2, matlab only spends a
fraction of the time computing the mean, the rest is passing the variable to the mean function.
so my question remains: is there anyway for matlab to not take a long time when dropping elements in an array? trying to trick it by putting it all on one line did not seem to work, sadly ;(
The first line takes no time since the 2 arrays are identical and MATLAB only does lazy copy or copy on write (see blogs on memory to learn more about this). Once an element is changing, MATLAB makes
a full copy of the original array to be sure it doesn’t get modified improperly.
Again, the best way to drop elements is to create the new array and fill it, you’re 3rd way – only the copy needed for the output is created in addition to the original, I believe. I don’t think you
can do better than that.
interesting. ok, i guess i’ll have to live with that ;)
thanks for your help…
hi every body
pls help
How to find the position of a matrix element…
thanking you…
Check out the functions find and ismember.
Hi Loren,
i have red this post, but didn’t find what I was looking for. Is there a easy way to make a logical matrix from subscript indices like the function find does but the other way round. Under method
number 2 in your post you gave a small example what happens when you use subscripts to address the selected elements of a matrix. You get the full matrix back and not only the selected elements. I
think the easiest way would be a logical to adress and get back only the selected elements. Using the function sub2ind is too much bother in my opinion.
As far as I know, there is not a simple 1-liner to go from subscripts to logicals without using sub2ind. Sorry you don’t care for that solution.
wasnt aware of this before, solved a bug i had been working on for two days already.
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[Scipy-tickets] [SciPy] #869: stats.invnorm.cdf returns NaNs with low mu values
[Scipy-tickets] [SciPy] #869: stats.invnorm.cdf returns NaNs with low mu values
SciPy Trac scipy-tickets@scipy....
Fri Aug 19 21:51:50 CDT 2011
#869: stats.invnorm.cdf returns NaNs with low mu values
Reporter: jpaalasm | Owner: somebody
Type: defect | Status: needs_review
Priority: normal | Milestone: 0.10.0
Component: scipy.stats | Version: devel
Keywords: |
Comment(by josefpktd):
I have no new ideas.
If we had a numerically precise norm.logcdf then this could help, if
someone finds a proof that the 0*inf term is actually zero, we could
restrict to the first term.
mu = 0.0028 seems a bit large for breaking down already, but the variance
is already pretty small 0.0028**3=2.1952e-08 and so the pdf is quite
R library VGAM: same breakdown and nans
> pinv.gaussian(0.003, 0.0029, 1)
[1] 0.7443139
> pinv.gaussian(0.003, 0.0028, 1)
[1] NaN
Ralf: comment for docs looks good
I would add reference to this ticket as code comment. Because mu=0.0028 is
quite large and the pdf and veccdf still work, I think there is room for
improvement. If we close the tickets, then code comments would give at
least a hint that there is a problem and possible (approximate) work
Ticket URL: <http://projects.scipy.org/scipy/ticket/869#comment:6>
SciPy <http://www.scipy.org>
SciPy is open-source software for mathematics, science, and engineering.
More information about the Scipy-tickets mailing list
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P=NP Update
As we reported here, there was much flurry on the internet about a potential proof of the P=NP problem. There is still an ongoing discussion about the proof and the ideas introduced in the proof,
but the consensus seems to be that there is some serious doubt that the current paper settles the question. As Terence Tao put it:
I think there are several levels to the basic question “Is the proof correct?”:
1. Does Deolalikar’s proof, after only minor changes, give a proof that P != NP?
2. Does Deolalikar’s proof, after major changes, give a proof that P != NP?
3. Does the general proof strategy of Deolalikar (exploiting independence properties in random k-SAT or similar structures) have any hope at all of establishing non-trivial complexity separation
After all the collective efforts seen here and elsewhere, it now appears (though it is perhaps still not absolutely definitive) that the answer to #1 is “No” (as seen for instance in the issues
documented in the wiki), and the best answer to #2 we currently have is “Probably not, unless substantial new ideas are added”. But I think the question #3 is still not completely resolved, and
still worth pursuing (though not at the hectic internet speed of the last few days).
– Terence Tao
But we’ll keep you posted on any big new developments!
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Various 2D Adjoint-Based Airfoil Designs
Application #58. Various 2D Adjoint-Based Airfoil Designs
This set of results is from some past work on the continuous adjoint method on unstructured grids. This test case is for a 2-element airfoil where the objective is to modify the pressure distribution
on the front airfoil by changing the shape of the aft airfoil. Results are shown after 3 design cycles.
This image shows the grid used for the design case.
This image shows the pressures contours.
This image shows a comparison of target, initial, and final pressure distributions.
Laminar Airfoil
This result is for a compressible, viscous case. The freestream Mach number is 0.75, the Reynolds number is 5000, and the angle of attack is fixed at 5 degrees. The objective in this case is simply
to increase the lift coefficient with no attention paid to the drag or the feasibility of the design. As seen in the pressure distributions shown in the figure, the lift has been increased by about a
factor of 16. It is evident that the final airfoil is not very practical because it is extremely thin.
Contours of the pressure coefficient are shown below for the initial and the final geometries.
Turbulent Airfoil
The next 2 results are for compressible, turbulent flows over airfoils. For these calculations, the turbulence model (Spalart-Allmaras) has been differentiated and included in the solution of the
costate variables along with the density, momentum, and energy. For both cases shown, the spacing at the wall is 1E-5 and the turbulence model is integrated all the way to the wall. The first case is
for an airfoil that I just made up. The Mach number is 0.4, the angle of attack is 2 degrees, and the Reynolds number is 5 million based on the chord of the airfoil. The grid is shown in the figures
below. There are only 3 design variables for this case.
The goal of the optimization was simply to match a desired pressure distribution. There are only 3 design variables for this case so the process converges fairly quickly. As can be seen, the desired
pressure distribution is obtained very well and is indistinguishable from the original. Similarly, the airfoil shape returns to that of the original shape that was used for getting the target
This next case is more difficult because there are 75 design variables. The Mach number is 0.725, the angle-of-attack is 2.54 degrees, and the Reynolds number is 6.5 million. The grid used for this
is very similar to that for the previous case and so it is not shown. This case is not quite converged but the pressures match the desired ones reasonably well.
Turbulent Airfoil with Curvature Constraints
The last case is another viscous design case. This time, the objective is to reduce the drag. In addition, there are 152 constraints placed on the geometry. Many of the constraints deal with the
curvature so the nose radius is not too small and the curvature is positive over the upper surface. In addition, there are 38 constraints which define upper and lower bounds for the design variables.
The conditions are a freestream Mach number of 0.75, a Reynolds number of 9 million, and a lift coefficient of 0.53.
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onds with different
Pricing bonds with different cash flows and compounding frequencies
Equation (1) defines the value of a bond that pays coupons on an annual basis and a principal at maturity.
The value of a bond paying a fixed coupon interest each year (annual coupon payment) and the principal at maturity, in turn, would be:
Equation (1)
Where M = Number of years to maturity
With the coupon payment fixed each period, the C term in Equation 1 can be factored out and the bond value can be expressed as:
Bonds, of course, differ in the frequency in which they pay coupons each year, and many bonds have maturities less than one year. Also, when investors buy bonds they often do so at non-coupon dates.
Equation (1), therefore, needs to be adjusted to take these practical factors into account.
Semiannual Coupon Payments
Many bonds pay coupon interest semiannually. When bonds make semiannual payments, three adjustments to Equation (1) are necessary: (1) the number of periods is doubled; (2) the annual coupon rate is
halved; (3) the annual discount rate is halved. Thus, if our illustrative 10-year, 9% coupon bond trading at a quoted annual rate of 10% paid interest semiannually instead of annually, it would be
worth $937.69:
Note that the rule for valuing semiannual bonds is easily extended to valuing bonds paying interest even more frequently. For example, to determine the value of a bond paying interest four times a
year, we would quadruple the periods and quarter the annual coupon payment and discount rate. In general, if we let n be equal to the number of payments per year (i.e., the compoundings per year), M
be equal to the maturity in years, RA be the discount rate quoted on an annual basis (simple annual rate), and R be equal to the periodic rate, then we can express the general formula for valuing a
bond as follows:
C^A = Annual coupon = (CR)(F)
n = number of payments per year
Periodic coupon = Annual coupon/n
M = term to maturity in years
N = number of periods to maturity = (n)(M)
Required periodic rate = R = Annual rate/n = R^A/n
Thus, the value of a 20-year, 6% coupon bond, with semiannual payments, a par value of $1,000, and a required return of 8% would be $802.07:
N = number of periods = 40 [= (20 years)(2)]
F = $1,000
C = Semiannual coupon = (.06/2)($1,000) = $30
R = required semiannual rate = .08/2 = .04
Compounding Frequency
The 10% annual rate in the first example and the 8% rate in the second is a simple annual rate: It is the rate with one annualized compounding. With one annualized compounding and a 10% annual rate,
we earn 10% every year and a $100 investment would grow to equal $110 after one year: $100(1.10) $110. If the simple annual rate were expressed with semiannual compounding, then we would earn 5%
every six months with the interest being reinvested; in this case, $100 would grow to equal $110.25 after one year: $100(1.05)2 $110.25. If the rate were expressed with monthly compounding, then we
would earn 0.8333% (10%/12) every month with the interest being reinvested; in this case, $100 would grow to equal $110.47 after one year: $100[1 + (.10/12)]12 $110.47. If we extend the compounding
frequency to daily, then we would earn 0.0274% (10%/365) daily, and with the reinvestment of interest, a $100 investment would grow to equal $110.52 after one year: $100[1 + (.10/365)]365 $110.52.
Note that the rate of 10% is the simple annual rate, whereas the actual rate earned for the year is [1+(RA/n)]^n-1. This rate that includes the reinvestment of interest (or compounding) is known as
the effective rate.
When the compounding becomes large, such as daily compounding, then we are approaching continuous compounding with the n term in Equation (2.3) becoming very large. For cases in which there is
continuous compounding, the future value (FV) for an investment of A dollars M years from now is equal to
where e is the natural exponent (equal to the irrational number 2.71828). Thus, if the 10% simple rate were expressed with continuous compounding, then $100 (A) would grow to equal $110.52 after one
year: $100e(.10)(1) $110.52. (After allowing for some slight rounding differences, this is the value obtained with daily compoundings.) After two years, the $100 investment would be worth $122.14:
$100e(.10)(2) $122.14.
Note that from the FV expression, the present value (A) of a future receipt (FV)is
If R .10, a security paying $100 two years from now would be worth $81.87, given continuous compounding: PV $100e-(.10)(2) $81.87. Similarly, a security paying $100 each year for two years
would be currently worth $172.36:
Thus, if we assume continuous compounding and a discount rate of 10%, then the value of our 10-year, 9% bond would be $908.82:
It should be noted that most practitioners use interest rates with annual or semiannual compounding. Most of our examples in this book, in turn, will follow that convention. However, continuous
compounding is often used in mathematical derivations, and we will make some use of it when it is helpful.
Article copyright 2011 by R. Stafford Johnson. Reprinted and adapted from Bond Evaluation, Selection, and Management, 2nd Edition with permission from John Wiley & Sons, Inc. The statements and
opinions expressed in this article are those of the author. Fidelity Investments® cannot guarantee the accuracy or completeness of any statements or data. This reprint and the materials delivered
with it should not be construed as an offer to sell or a solicitation of an offer to buy shares of any funds mentioned in this reprint.
The data and analysis contained herein are provided "as is" and without warranty of any kind, either expressed or implied. Fidelity is not adopting, making a recommendation for or endorsing any
trading or investment strategy or particular security. All opinions expressed herein are subject to change without notice, and you should always obtain current information and perform due diligence
before trading. Consider that the provider may modify the methods it uses to evaluate investment opportunities from time to time, that model results may not impute or show the compounded adverse
effect of transaction costs or management fees or reflect actual investment results, and that investment models are necessarily constructed with the benefit of hindsight. For this and for many other
reasons, model results are not a guarantee of future results. The securities mentioned in this document may not be eligible for sale in some states or countries, nor be suitable for all types of
investors; their value and the income they produce may fluctuate and/or be adversely affected by exchange rates, interest rates or other factors.
In general, the bond market is volatile, and fixed income securities carry interest rate risk. (As interest rates rise, bond prices usually fall, and vice versa. This effect is usually more
pronounced for longer-term securities). Fixed income securities also carry inflation risk, liquidity risk, call risk and credit and default risks for both issuers and counterparties. Lower-quality
fixed income securities involve greater risk of default or price changes due to potential changes in the credit quality of the issuer. Foreign investments involve greater risks than U.S. investments,
and can decline significantly in response to adverse issuer, political, regulatory, market, and economic risks. Any fixed-income security sold or redeemed prior to maturity may be subject to loss.
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Waveguides and Cavity Resonators
Transmission line notes. Also, more links to relevant topics.
Microwaves for satellite communications.
Scattering parameters, scattering matrix, s-parameters.
Four-port S-matrix relations and directional couplers.
Problems 3, on waveguides.
Java applet of wave motion
This applet shows a visualisation of wave motion and you can add your waveguide walls with the mouse.
Waves in one, two, and three dimensions.
The number of dimensions in the section heading refers to the number of dimensions in which the wave can move.
• One dimension.
On the transmission lines discussed elsewhere the wave is restricted to one dimension only. It can travel along the line from generator to load, or indeed back again, but it can't escape from the
length of the line.
• Two dimensions.
Waves on the surface of water, for example your bath, or a pond, or the sea, have two dimensions they can travel. We might refer to these as "perpendicular to the shore" and "parallel to the
shore", or x and y. Any two orthogonal directions may be chosen as the fundamental reference directions. "Orthogonal" means taking two directions at 90 degrees to each other.
In the case of waves on a pond, we may see various patterns of wave crests. Two important patterns are
1. when the crests form concentric circles, moving radially out from a point where maybe a pebble was cast, and
2. when the crests form ranks of straight lines, moving uniformly in a direction at right angles to the crests.
• Three dimensions.
The case which concerns us in general is when waves, perhaps radiation from an antenna, can move in a direction unrestricted in the three space dimensions. Various co-ordinate systems may be used
to describe these waves. Three of the important patterns of wave crests are
1. when the crests form concentric spheres, moving radially outwards from a point. These are best described by spherical polar co-ordinates, having radial, azimuth, and elevation directions.
2. when the crests form nested cylinders, moving radially away from a straight line of source. These are usually described by cylindrical polar co-ordinates, having radial, azimuth, and linear-z
3. when the crests form parallel planes, moving in a direction normal to the plane surfaces. These are described by rectangular Cartesian coordinates, in x y and z directions. Such waves are
called "plane waves".
Electromagnetic waves are "transverse waves", that is the directions of local magnetic and electric field lie at right angles to each other and to the local direction of travel of the wave. There
is a picture here .
Generally speaking, the direction of electric and magnetic field does not change as we move along with the wave motion, but stays fixed in space. An exception to this general rule occurs in
"optically active" regions, where the plane of "polarisation" (electric field direction) can rotate as we advance with the wave. Also, it is possible to superpose two plane waves having
orthogonal electric field directions which are out of time phase with each other by +/-90 degrees. In this case, the resultant wave appears to have its resultant electric field vector rotate as
the wave advances; this has nothing to do with optical activity. It is referred to as "circular polarisation". If the electric field vector rotates clockwise as we advance with the wave, it is
"right handed circular polarisation". If the electric field vector rotates anti-clockwise as we advance with the wave it is "left handed circular polarisation".
What is a waveguide?
A waveguide restricts the three dimensional "free space" propagation of the electromagnetic wave to a single dimension. Usually waveguides are
• Low loss. That is, the wave travels along the guide without greatly attenuating as it goes.
• Routeable. This means that we can gently bend the guiding structure without losing contact with the wave, without generating reflections, and without incurring much additional loss.
Thus we see that waveguides are transmission lines, in terms of the properties described in the notes referred to above.
There are a great many different wave guiding structures. Usually they are uniform in the direction of travel of the guided wave; that is, one cannot tell where one is along the waveguide by
physically looking around one. There is an exception in the case of corrugated waveguide and other slow wave structures which we shall return to later. These guides may have spatially periodic
structure in the direction of wave travel.
The different co-ordinate systems described above are appropriate to different waveguide cross sectional shapes. Cylindrical polar co-ordinates are used to describe circular cross section waveguide,
and coaxial cables. Rectangular Cartesian co-ordinates are preferred for rectangular waveguide. In the case of more exotic structures such as microstrip, fin-line, or coplanar waveguide, it is usual
to use rectangular co-ordinates and solve approximately for the cross sectional field distributions by using "conformal mapping", a technique borrowed from complex variable theory.
Rectangular metal pipe waveguide
Boundary conditions.
If an electric field is to exist close to a conductor, we have to make sure that the conductor does not short out the electric field. In practice, this means that all electric fields meet conductors
at right angles to the surface. The sources of electric field are the charges which reside on the metal surface. If the electric field represents a wave crest, the line along the crest of the wave
meets the guide metal wall at right angles and moves laterally, parallel to the surface of the metal. The associated charges move along the guide and set up currents which are linked to magnetic
fields, parallel to the guide metal surface and perpendicular to the electric fields.
There can be no time-dependent magnetic field meeting a metal wall at right angles, since the metal would act as a shorted turn and suppress the magnetic field.
Constructing a waveguide.
Let us suppose we think about plane waves having electric field along the z direction, but moving in the x-y plane. We can clearly add a metal sheet in the x-y plane without disturbing the wave
motion. The Ez field meets the sheet at right angles, and the wave travels at an arbitrary angle to the x axis (all 360 degrees of bearing are possible) with magnetic field at right angles to the
direction of propagation, but parallel to the x-y plane. The currents in the x-y metal sheet are in the direction of propagation of the wave. If we have more than one wave, with more than one
direction of travel, we merely add up the fields due to each to get the resultant field patterns. Of course, the addition has to be done vectorially.
Now let us add another sheet aligned in the x-y plane but at a different height z. The electric field will then start on one sheet and finish on the other. It is still lined up along the z direction.
It starts on a positive charge and ends on a negative charge. Both charges are notionally moving in the direction of propagation, but they represent oppositely directed currents. The adjacent
magnetic fields are in the same sense however, as they lie on opposite sides of the oppositely directed currents. One is above the current sheet and the other below.
We have made a Transverse Electromagnetic (TEM) wave guiding structure. There is no way the waves can escape from the space between the x-y metal sheets, which act as perfect screens. The electric
field, magnetic field, and direction of wave travel are all mutually orthogonal; that is, they are each at right angles to the other two. The electric and magnetic fields are uniform (they don't
depend on z) within the gap.
We have not yet made a waveguide, however. The wave still has a choice of direction in the x-y plane. Nevertheless, our structure is strongly reminiscent of a 2 wire transmission line, if we assume
the line length to lie along the arbitrary direction (in the x-y plane) of propagation.
Exercise for the student. Draw a plan view of a few wave crests looking down along the z axis. You should see straight lines on your paper, orientated at an arbitrary angle to the x axis, but
uniformly spaced apart (by a wavelength). Measure the angle alpha between the crest lines and the x axis. Measure the spacing lambda between the crest lines along the direction of wave travel.
Measure the crest intersection distance lambda(x) along the x axis. Show by geometry, or with a calculator, that [lambda]/[lambda(x)] = cos (alpha). Similarly, measure the crest intersection distance
lambda(y) along the y axis and show that [lambda]/[lambda(y)] = sin (alpha).
Then using Pythagoras' theorem (sin^2 + cos^2 = 1) show that
1/[lambda]^2 = 1/[lambda(x)]^2 + 1/[lambda(y)]^2
We are nearly there now. If we put walls in the x-z plane spaced by
a = (1/2)lambda(y)
we have made a tube whose axis lies along x. We have made a waveguide. Two waves are needed to make a standing wave pattern across this tube. Nulls are needed at a spacing of (1/2)lambda(y) to
satisfy the boundary condition that the Ez field vanishes on these walls. We are left with the waveguide formula
1/[lambda]^2 = 1/(2a)^2 + 1/[lambda(x)]^2
and the wavelength along the guide is
lambda(x) = lambda/cos(alpha).
Lambda(x) is usually referred to as the guide wavelength lambda(g). It is longer than the free space wavelength lambda because cos(alpha) is always less than unity. The plane waves are travelling at
an angle alpha to the direction along the waveguide. The energy travels at a velocity c*cos(alpha), but the wave pattern travels at the velocity given by f*lambda(g) = c/cos(alpha).
Group and phase velocity.
The energy and the modulations on the microwave signal going down the waveguide both travel at the "group velocity" c*cos(alpha) which is necessarily less than the velocity of light c. The pattern
however travels at the "phase velocity" c/cos(alpha) which is necessarily greater than the velocity of light. The product of (group velocity)*(phase velocity) = c^2.
The field pattern is formed from the superposition of two plane waves travelling in different directions, (the two directions are plus and minus alpha with respect to the direction along the
waveguide). These two waves have the same free space wavelength, and give the standing wave pattern along y required to make the fields vanish at the side guide walls. The whole field pattern is
called a "Mode". For the TE10 mode (transverse electric), if you plot out a plan view of the field patterns carefully, you will see that the electric field is always out of the paper, but that the
magnetic field forms stadia loops with repetition distance equal to a guide wavelength; there are two stadia loops per guide wavelength, in opposite senses of magnetic field circulation.
The magnetic field is always parallel to the top and bottom guide walls, and turns to become parallel to the side guide walls where it approaches them. The guide wall currents are at right angles to
the adjacent magnetic field lines.
Other modes
The rectangular pipe has cross section a by b metres, with wall planes x-y and x-z. We chose the electric field to lie along the z direction in our first example. However, one can equally satisfy the
boundary conditions with the electric field along the y direction, and the standing wave along z. Since the electric field in these modes is entirely transverse to the direction x of propagation,
they are called "transverse electric" or TE modes. The two modes here are TE10 and TE01. The 1 refers to the number of half-wave loops across the guide. By convention, the TE10 mode has its single
loop across the largest guide lateral dimension, and the TE01 mode has its loop across the smaller guide lateral dimension. Thus the cutoff frequency for the TE10 mode is the lowest frequency at
which the waveguide will transmit without attenuation.
There are also transverse magnetic modes, TM modes. These require loops across both dimensions. The lowest TM mode is the TM11 mode. The magnetic field is entirely transverse; there is a component of
longitudinal electric field in the x direction in the centre of the guide. Both parallel electric field and normal magnetic field fall to zero at the guide walls because of the standing wave pattern.
One can have higher order modes, TE20, TE30, ... TEn0
and also TE21, TE31,....TEmn
and the corresponding transverse magnetic modes TMmn, providing there is at least one loop in each transverse direction.
There are two pictures of the field patterns for some of the higher order modes in rectangular and circular waveguide.
Rectangular guide mode patterns (148kB)
Circular guide mode patterns (118kB)
Looking at our derivation of the waveguide formula, it is not difficult to see that for a general TEmn or TMmn mode the waveguide formula is
1/[lambda]^2 = {m/2a}^2 + {n/2b)^2 + 1/[lambda(g)]^2
Waveguides are generally used in a frequency range where only the lowest mode, or exceptionally the lowest few modes, will propagate. If one wants to launch a mode in a guide which supports more than
one propagating mode, one drives it with probes or loops which have a symmetry that does not excite any of the other (unwanted) modes. Thus, for example, to excite a TE10 mode we might use a probe in
the middle of the wide face of the guide; this would not excite the TE20 mode which has a null in the middle of the guide. To excite the TE20 mode we could drive two probes, spaced 1/4 and 3/4 of the
way across the broad face, in anti-phase. This would be the wrong symmetry to excite the TE10 mode, for which the two probes thus placed would have to be driven in phase with each other. It is easy
to see how to extend this method to other scenarios.
Mode field patterns
Here we have shown a picture of the field and current patterns in the TE10 (H10) and TM11 (E11) modes in square cross section waveguide. The magnetic field lines are GREEN, the electric field lines
are RED, and the currents in the guide walls are BLUE.
Nomenclature; originally the British called a TE mode an "H mode". In a TE mode there is no longitudinal E, but there is a longitudinal H field. However you can see that in the TE mode there is also
a transverse component of magnetic field. The British thought that the unique longitudinal H field was a better descriptor than the transverse electric field with no longitudinal E component.
Thus in the diagram, H10 = TE10 and E11 = TM11.
The picture is a heavily processed image taken from an original in "The Services Textbook of Radio", vol 5, 1958 HMSO, wherein there are other pictures and a very clear exposition and discussion of
the principles behind wave transmission and propagation. This book is long out of print, but if you can find a copy in the library it is well worth a read.
A problem.
An exercise for the reader. The cutoff wavelength for these modes is lambda(c) where
1/[lambda(c)]^2 = {m/2a}^2 + {n/2b}^2, at which point the guide wavelength lambda(g) has gone to infinity, and the waves are bouncing back and forth across the guide and making no progress along it.
Calculate the cutoff frequency of a waveguide formed by the BB21 corridor, assumed to be 3 metres high and 2 metres wide. Hint. Find the lowest cutoff frequency of the mn modes; nothing can propagate
below this frequency. Of course, we assume here that some kind soul has papered the walls with kitchen foil....
My answer... 50 MHz.
Slots in waveguide walls.
If we look at the blue current lines in the waveguide walls, we find there are directions of flow where a slot, cut parallel to the flow lines, will not interrupt the currents. These slots are
"non-radiating" slots.
However, if we deliberately cut a slot such as to interrupt the current flow lines, the currents have to go round the slot and this gives rise to a distortion of the electric field and magnetic field
patterns inside the guide. One finds that there is a voltage difference between opposite edges of the slot, in the middle of the slot. The slot acts as a dipole antenna and will radiate, and the
energy leaks out from the waveguide.
This principle is used to construct slot antennas in waveguide.
An important application of slotted waveguide is to construct a "slotted line" for measurements. This is usually done for the TE10 mode pattern, which is the only mode which will propagate for
sufficiently low frequencies. The slot is cut parallel to the guide axis, in the middle of the wide face. It is a non-radiating slot, and the field pattern inside the guide is but little disturbed by
it. A probe may be put into the slot to sample the local electric field strength, and moved up and down the slot along the waveguide to see the standing wave patterns and to measure the VSWR
On the X band waveguide in the labs there is a ferrite collar around the slot on the slotted lines. This reduces residual radiation from the slot. There is some residual radiation because of the
finite width of the slot, and because the probe and carriage distort the field patterns somewhat.
Rectangular cavity resonators
Starting from a rectangular waveguide of cross section a by b metres, we can add short circuit walls in the y-z planes, along the direction of propagation.
This gives a rectangular box whose resonant frequency is given by f where (f*lambda) = c = 3*10^8, and
1/[lambda]^2 = {m/2a}^2 + {n/2b}^2 + {p/2d}^2
Here, there are m half wavelength loops along y, n half wavelength loops along z, and p half wavelength loops along x. It is possible for just one only of the loop numbers m, n, and p to take the
value zero. The spacings of the walls are
d along x, a along y, and b along z. We see there are many modes of a rectangular cavity.
Exercise for the student. List the 30 lowest resonant frequencies of my microwave oven cavity, which has dimensions 0.36 metres by 0.33 metres by 0.23 metres. Assume it is empty, (no chickens, coffee
cups, or turntables. ) Now justify the choice of operating frequency for such a microwave oven.
Other shapes of cavities
Clearly, a cavity can be many other shapes than rectangular. The field theory for calculating the modes of arbitrary shaped cavities is straightforward, but often numerical methods are needed as
there are no analytic solutions. Often, cylindrical cavities are used. It is possible to use more than one mode in a cavity filter, with tuning screws and stubs to convert energy from one mode to
Any equipment box will have microwave resonant frequencies, often in the range around 1GHz. This poses problems for the engineer faced with EMC (Electromagnetic compatibility) and interference
problems, if there is any source of signal within the box having spectral frequency components somewhere near the box modes.
Another (salutary) exercise for the student. Do you have a cellnet phone? On what frequency does it transmit? Calculate the dimensions of an enclosure which is resonant at this frequency, and compare
it with your car, the lift, and other metallic enclosed spaces in which you sometimes find yourself.
If you transmit in a screened enclosure, all the energy from the transmitter is concentrated in your body, which is very lossy to microwave energy.
Launching waves in rectangular waveguide.
Slots and apertures may be cut in waveguide walls. If these slots or apertures do not interrupt the currents flowing on the interior surface of the guide walls, very little radiation escapes from the
guide through them.
A probe or loop may be introduced through the opening, and used to excite electric fields (probes) or magnetic fields (loops).
For a waveguide supporting a TE10 mode, a non-radiating slot may be cut in the centreline of the wide face of the waveguide. This allows a probe to be inserted to sample the electric field in the
waveguide. Alternatively, a small hole may be drilled on this centreline and a longer probe introduced, formed by the extension of the centre conductor of a coaxial cable.
If the waveguide is closed by a short circuit sheet, lambda(g)/4 away from the probe, the probe will be placed at a standing wave maximum, and will form a radiating antenna transition into the fields
inside the waveguide.
As with all transitions in transmission line, if the impedance of the waveguide mode is matched to the characteristic impedance of the coaxial cable, then there is no reflection set up at the
Various parameters may be adjusted to help in the matching process. The hole diameter, the penetration depth of the probe, the probe wire diameter, the spacing from the waveguide short, the
dielectric properties and dimensions of any probe sleeving, are all adjustable. Usually it is desired to have a reasonably broadband match, over maybe 10% fractional bandwidth.
The design procedure is a job for the specialist, equipped with electromagnetic field modelling software and CAD packages.
Cavities and filters for satcoms applications
Waveguide filters and cavity resonators are larger and heavier than their microstrip, and other transmission line, equivalents. But they have the following advantages for Satcoms applications.
• Power handling capacity
• Lower losses (resistive and dielectric)
• Dimensional stability against vibration, changes in temperature, and pressure.
• Frequency and bandpass response stability.
• Ease of tuning after manufacture
• Robustness
• Outgassing advantages
• Larger so tolerance of manufacture is easier
• No dielectric, so non-linearities in high fields are less.
Copyright D.Jefferies 1996, 1997, 1999, 2002, 2005, 2007. D.Jefferies@surrey.ac.uk 6th June 2007
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Gah! More math!
Gah! More math!
I've looked everywhere!
There has to be a special rule for finding the derivative of (x^x), because (x^x) is not an exponential expression, is it constant? No... can't be....
gaah! Anyone good with math/calc help me! :)
This isn't homework BTW; I'm working on some calculus algorithms :)
the derivative of (x^x) is (ln(x) + 1) * x^x
why? my calculator told me so. :D
That doesn't help... I got the same thing on my calculator; I need to be able to get there, I cannot find any forumals, proof, identities... ANYTHING that helps me :mad: grrrr... this bites
let's start off at the basics
the derivative of a*n^r == a*r*n^(r-1)
however, 1 * x^x == 1 * x * x^(x-1) == 1 * x^x
maybe this will help
//edit: here's the definite solution:
Captain Penguin
Ahh good 'ol calculus...
had to whip out my stuff from last year, but here you go: (It's called logarithmic differentiation, btw)
y = x^x
take natural log of both sides
ln(y) = x*ln(x) < - this last part is a property of logarithms
now take the derivative of both sides...
(dy/dx)*ln(y) = (dy/dx)(x*ln(x))
because taking the derivative of y gives dy/dx....
(1/y)*(dy/dx) = 1*ln(x) + x*(1/x) < - product rule
(1/y)*(dy/dx) = ln(x) + 1 <- just simplified the right side...
multiply both sides by y...
(dy/dx) = y(ln(x) + 1)
and since we know y = x^x...
(dy/dx) = x^x * (ln(x) + 1)
and voila.
I don't know how that would be programmed easily, but thats how you get there..
Captain Penguin
Originally posted by ygfperson
maybe this will help
That gives an approximated numerical derivative... (like a calculator gives) perhaps not what he needs?
and that drmath explanation.. kinda stinks. He is making it overly complicated by doing e^(log(x^x))...
thanks guys... I figured it out before I checked back here.
I just need to build a math function for the special case of (x^x) because it is neither a power expression nor an exponential expression...
(d/dx)[x^x] = [(ln(x)+1)(x^x)] works great!
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