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Max – Min, Solve Using Partial Derivatives
I used this problem once in a technical writing class I taught in 2000, but no students wanted (or could) tackle it. I still think it is valuable for students to write meaningfully about math and
science, and I did not want the assignment to get lost. I doubt that I could reproduce the solution again. If anybody wants to use the assignment, please feel free. Here is the formal assignment as
I typed it up in 2000.
Formal Assignment: A long piece of tin 12 meters wide is made into a trough by bending up the sides to form equal angles relative to the base. Find the amount of tin to be bent up and the angle of
inclination of the sides that will make the carrying capacity a maximum.
Hint: The volume (carrying capacity) of the trough will be maximized if the area of the trapezoidal cross section is a maximum.
This problem can be solved using trigonometry, basic geometry, algebra, and partial derivatives.
Below is the first page of the solution.
Below is page 2 of the solution. Notice that if we get a length x equal to 0 or an angle theta equal to 0, then we have to disregard those solutions because they are physically impossible or absurd.
Below is page 3 of the solution. The beauty of this problem is that we start with just a long piece of tin and we want to maximize the throughput (for example, the maximum amount of water that can
go travel through the tin if we make it a trough) of the trough. That’s it. All we are given is the width of the material and a need to find the maximum throughput.
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LDD Detection theory Assignment Help
Select LDD Detection Theory Assignment Help And Assessment Writing Service And Grab Top Position In Your Academics!
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LDD Detection theory
Question 1: Explain how a person can choose freely, but his or her choices are caused by past events.
Answer: Free will choices are made by personal beliefs and attitudes. The connection with previous event s are not taken by people
Question 2: Draw the causal tree for Newcomb's Problem when Eve cannot perfectly detect Adam's causal history.
Answer: Eve → Ignore Events of Adam
The causal tree explains the reason for Eve's failure to find Adam's causal history and the possible alternatives to the problems. The possible cause as presented in the above diagram is theexistence
of many variables in the history which makes it complicated and difficult
Question 3: Derive the expected payoff formulas conditional on taking money from the black box only, or from both boxes. Then use them to solve for another formula that equals the smallest value of M
(denoted Mmin) required in order for Adam's average benefit from opening only the black box instead of both boxes by at least some positive multiple S of the guaranteed money L in the clear box.
What is the resulting formula for Mmin. Finally, assume (L, δ) = (400, 3), (r, w) = (.58, .43) and use this formula to calculate Mmin this case.
Answer: (M min) = (L .δ) (r + w) = (400 × 3) × (0.58 + 0.43) = 1200 x 1 = 1200
From both box = 1200/0.4 = 1500
Question 4: Suppose a CD player tries to detect whether its partner is CD player instead of a DD player, by looking for external signals that are at least as typical for DD players than for CD
players. Draw a diagram to explain how two boundaries bi, and bH are optimally determined. Show on the diagram where it is optimal to respond C versus D. Also explain what happens to the boundaries
when detection becomes more cautious by raising the minimum likelihood ratio.
Answer: Maximum likelihood ratio = P(bi) + P( bH) = r/w
The boundary of bi is less than bH for CD players than DD players
Question 5: What is the meaning of the LDD detection strategy?
Answer: LDD detection theory is a mechanism used in detecting signals from a CD or DD. It involves the identification of the detection principles that are used in the analysis of the players change
over a period
Question 6: What is the main problem with the green-beard strategy? Explain how the LDD strategy overcomes this problem.
Answer: Problem is lack of proper prediction events. LDD observes the pattern of events of DD and CD players to overcome the problem
Question 7: Assume a population contains either CD or DD players, where each player is randomly matched with a partner taken from the whole population. Also assume the fear and greed payoff
differences are equal. What are the expected payoff formulas for CD players [denoted E(CD I E" ) ] and for DD players [ E(DDIE") ] ] depending on the fraction of CD players in the population, denoted
Answer: Expected payoff = Pi€ xP(CD) + P (DD)
Question 8: Use the expected payoff formulas of Part B to algebraically derive an inequality for the signal reliability ratio r/w that determines when the CD players will outperform the DD players.
Answer: Expected pay off of CD players = P(1-0.4) = 0.6
Expected pay off DD = 0.4
Question 9: Use this inequality with Part A, to explain how CD players can always outperform DD players starting from any positive initial fraction of CD players.
Answer: CD players can predict using LDD strategy and have many positive traits than DD players
Question 10: Use the inequality from Question 3; to obtain an inequality required for *cr) = 1 to remain stable against DD invaders. Also draw the ROC diagram for visually representing this stability
Question 11: Use a diagram similar to that shown in Part A to derive a prediction of what will happen to the CD players' equilibrium probability of cooperating; if the cooperation payoff difference
increases relative to the fear and greed payoff differences.
Question 12: Again use a diagram similar to Part A to derive a prediction of what will happen to the CD players' equilibrium probability of cooperating; if they use cell phones rather than talking
Question 13: What kind of factors could further enhance or degrade the effectiveness of face-to-face communication, and if so, how would this affect the probability of cooperating ?
Answer: To enhance the face to face communication, players should be taken for recreation to create good relationship and provided with equal pay and respect for the service. This will increase the
cooperation among the team. Factors that reduce face to face communication are tit for tat, revenge and badmouthing attitudes that affect the cooperating.
Tag This :- M92066653M27, LDD Detection theory Assignment Help
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Machine Learning Demo
This 60-second short-video explores a supervised learning model, specifically a feedforward neural network, being trained to recognize handwritten digits.
The network has three layers:
• An input layer with 9 neurons, which receives data from a nine-segment sensor pad.
• A hidden layer with 9 neurons.
• An output layer with 2 neurons, which can light two bulbs to represent the digits 0 or 1.
The video shows the model being trained on labeled data points consisting of handwritten inputs and the corresponding label (i.e., correct answers). Initially, the model gets most of the
classifications wrong, and a backpropagation algorithm adjusts knobs, representing the network parameters, to improve the model’s accuracy.
The demo illustrates the training phase of a neural network. Where wrong guesses by the model result in the training algorithm calculating a loss between the correct answer and the guessed answer.
That loss is then used by the backpropagation algorithm to adjust the parameters so as to guess better in the future.
Once trained, the network can then be used in "inference" where the parameters are no longer adjusted.
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How Many Amps Does A Block Heater Draw
How Many Amps Does A Block Heater Draw - Web a typical block heater for a car or small truck may draw between 5 and 10 amps, although some larger models may draw up to 15 amps. Web when i plug in the
block heater the volts drop to 120.5. If you’re trying to wire your circuit to accommodate the needs of a block heater, aim for 1000 to 1200w. New posts new resources latest activity. Web 13,562
location westminster, md tractor john deere 3720 cab/tlb, jd455 i think they are 1000 or 1200 watts?
Web how many amps does a block heater use? If you’re trying to wire your circuit to accommodate the needs of a block heater, aim for 1000 to 1200w. It shows about 6.4 amps on the cheap ammeter that i
used this morning and it shows 7.8 amps on a little. Web generally speaking, a small semi truck may use between 10 to 15 amps, while a larger model may require up to 25 amps of power. Noralph 10385
posts · joined 2007. Web this article will explain how many amps does a block heater draw? New posts new resources latest activity.
5 Things to Know about your Block Heater... Davis GMC Buick Ltd.
The amp draw of a block heater is a critical aspect often overlooked by vehicle owners. A circuit capable of withstanding. Received 0 likes on 0 posts. Web this article will explain how many amps
does a block heater draw? So at 120 volts if it's 1200 watts that is 10. But it helps to.
How To Check Block Heater » Tacklebranch
Web i guess the block heater is making it work. The amp draw of a block heater is a critical aspect often overlooked by vehicle owners. It is stamped on to the side of the block heater but you. Web
how many amps does the block heater on a 2k2 psd draw? In fact, many.
How Many Amps Does A Space Heater Draw? Essential Home And Garden
That is a lot higher than the ratings you will encounter on the market. What is the cost of installing a block heater? If you’re trying to wire your circuit to accommodate the needs of a block
heater, aim for 1000 to 1200w. It is stamped on to the side of the block heater but.
how many amps does a cummins block heater draw
So at 120 volts if it's 1200 watts that is 10. Web a typical block heater for a car or small truck may draw between 5 and 10 amps, although some larger models may draw up to 15 amps. The wattage of
each heater also varies. Web i bought a 15 amp digital timer to.
how many amps does a cummins block heater draw outfitswithwhitesliponvans
That is a lot higher than the ratings you will encounter on the market. What is the cost of installing a block heater? Does anyone know how many watts the block heater takes? Web a typical block
heater for a car or small truck may draw between 5 and 10 amps, although some larger models.
How Does an Engine Block Heater Work? • ThreeTwoHome
So at 120 volts if it's 1200 watts that is 10. But it helps to prepare for the worst. New posts new resources latest activity. A circuit capable of withstanding. Noralph 10385 posts · joined 2007.
The normal block heaters are 8 amps or less, thats why you can plug two vehicles. Web i guess.
How Many Amps Does A Block Heater Draw? (Cummins, Ford) VehicleChef
Web as opposed to just plugging it in all night? It shows about 6.4 amps on the cheap ammeter that i used this morning and it shows 7.8 amps on a little. That is a lot higher than the ratings you
will encounter on the market. Received 0 likes on 0 posts. New posts new.
how many amps does a cummins block heater draw outfitswithwhitesliponvans
A circuit capable of withstanding. New posts new resources latest activity. If you’re trying to wire your circuit to accommodate the needs of a block heater, aim for 1000 to 1200w. So at 120 volts if
it's 1200 watts that is 10. The normal block heaters are 8 amps or less, thats why you can.
how many amps does a cummins block heater draw buildingbridgesbyjulia
New posts new resources latest activity. Web i guess the block heater is making it work. Web a typical block heater for a car or small truck may draw between 5 and 10 amps, although some larger
models may draw up to 15 amps. So at 120 volts if it's 1200 watts that is 10..
how many amps does a cummins block heater draw outfitswithwhitesliponvans
That is a lot higher than the ratings you will encounter on the market. Web i bought a 15 amp digital timer to kick my block heater on several hours before i start my truck and need to know if 15
amps is big enough. Web as opposed to just plugging it in all night?.
How Many Amps Does A Block Heater Draw What is the cost of installing a block heater? In fact, many of the heaters you will find are 400w. Web as opposed to just plugging it in all night? Web there
are different block heater wattages so the amps will vary on what you have. Web this article will explain how many amps does a block heater draw?
What Is The Cost Of Installing A Block Heater?
The wattage of each heater also varies. It shows about 6.4 amps on the cheap ammeter that i used this morning and it shows 7.8 amps on a little. Web when i plug in the block heater the volts drop to
120.5. Received 0 likes on 0 posts.
Web I Guess The Block Heater Is Making It Work.
Does anyone know how many watts the block heater takes? Web as opposed to just plugging it in all night? It is stamped on to the side of the block heater but you. A circuit capable of withstanding.
Web 13,562 Location Westminster, Md Tractor John Deere 3720 Cab/Tlb, Jd455 I Think They Are 1000 Or 1200 Watts?
But it helps to prepare for the worst. Web i bought a 15 amp digital timer to kick my block heater on several hours before i start my truck and need to know if 15 amps is big enough. Web generally
speaking, a small semi truck may use between 10 to 15 amps, while a larger model may require up to 25 amps of power. I suppose the savings is in not having the draw all night long in your electrical
Web How Many Amps Does A Block Heater Use?
In fact, many of the heaters you will find are 400w. Web a typical block heater for a car or small truck may draw between 5 and 10 amps, although some larger models may draw up to 15 amps. Web there
are different block heater wattages so the amps will vary on what you have. So at 120 volts if it's 1200 watts that is 10.
How Many Amps Does A Block Heater Draw Related Post :
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Formal grammar explained
A formal grammar describes which strings from an alphabet of a formal language are valid according to the language's syntax. A grammar does not describe the meaning of the strings or what can be done
with them in whatever context—only their form. A formal grammar is defined as a set of production rules for such strings in a formal language.
Formal language theory, the discipline that studies formal grammars and languages, is a branch of applied mathematics. Its applications are found in theoretical computer science, theoretical
linguistics, formal semantics, mathematical logic, and other areas.
A formal grammar is a set of rules for rewriting strings, along with a "start symbol" from which rewriting starts. Therefore, a grammar is usually thought of as a language generator. However, it can
also sometimes be used as the basis for a "recognizer" - a function in computing that determines whether a given string belongs to the language or is grammatically incorrect. To describe such
recognizers, formal language theory uses separate formalisms, known as automata theory. One of the interesting results of automata theory is that it is not possible to design a recognizer for certain
formal languages.^[1] Parsing is the process of recognizing an utterance (a string in natural languages) by breaking it down to a set of symbols and analyzing each one against the grammar of the
language. Most languages have the meanings of their utterances structured according to their syntax - a practice known as compositional semantics. As a result, the first step to describing the
meaning of an utterance in language is to break it down part by part and look at its analyzed form (known as its parse tree in computer science, and as its deep structure in generative grammar).
Introductory example
A grammar mainly consists of a set of production rules, rewriting rules for transforming strings. Each rule specifies a replacement of a particular string (its left-hand side) with another (its
right-hand side). A rule can be applied to each string that contains its left-hand side and produces a string in which an occurrence of that left-hand side has been replaced with its right-hand side.
Unlike a semi-Thue system, which is wholly defined by these rules, a grammar further distinguishes between two kinds of symbols: nonterminal and terminal symbols; each left-hand side must contain at
least one nonterminal symbol. It also distinguishes a special nonterminal symbol, called the start symbol.
The language generated by the grammar is defined to be the set of all strings without any nonterminal symbols that can be generated from the string consisting of a single start symbol by (possibly
repeated) application of its rules in whatever way possible.If there are essentially different ways of generating the same single string, the grammar is said to be ambiguous.
In the following examples, the terminal symbols are a and b, and the start symbol is S.
Example 1
Suppose we have the following production rules:
then we start with S, and can choose a rule to apply to it. If we choose rule 1, we obtain the string aSb. If we then choose rule 1 again, we replace S with aSb and obtain the string aaSbb. If we now
choose rule 2, we replace S with ba and obtain the string , and are done. We can write this series of choices more briefly, using symbols:
The language of the grammar is the infinite set
, where
times (and
in particular represents the number of times production rule 1 has been applied). This grammar is
(only single nonterminals appear as left-hand sides) and unambiguous.
Examples 2 and 3
Suppose the rules are these instead:
This grammar is not context-free due to rule 3 and it is ambiguous due to the multiple ways in which rule 2 can be used to generate sequences of
However, the language it generates is simply the set of all nonempty strings consisting of
s and/or
s.This is easy to see: to generate a
from an
, use rule 2 twice to generate
, then rule 1 twice and rule 3 once to produce
. This means we can generate arbitrary nonempty sequences of
s and then replace each of them with
as we please.
That same language can alternatively be generated by a context-free, nonambiguous grammar; for instance, the regular grammar with rules
Formal definition
See main article: Unrestricted grammar.
The syntax of grammars
In the classic formalization of generative grammars first proposed by Noam Chomsky in the 1950s,^[2] ^[3] a grammar G consists of the following components:
• A finite set N of nonterminal symbols, that is disjoint with the strings formed from G.
• A finite set
terminal symbols
that is
• A finite set P of production rules, each rule of the form
(\Sigma\cupN)^*N(\Sigma\cupN)^* → (\Sigma\cupN)^*
is the
Kleene star
operator and
set union
. That is, each production rule maps from one string of symbols to another, where the first string (the "head") contains an arbitrary number of symbols provided at least one of them is a nonterminal.
In the case that the second string (the "body") consists solely of the
empty string
- i.e., that it contains no symbols at all - it may be denoted with a special notation (often
) in order to avoid confusion.
that is the
start symbol
, also called the
sentence symbol
.A grammar is formally defined as the
. Such a formal grammar is often called a
rewriting system
or a
phrase structure grammar
in the literature.
^[4] ^[5]
Some mathematical constructs regarding formal grammars
The operation of a grammar can be defined in terms of relations on strings:
, the binary relation
(pronounced as "G derives in one step") on strings in
is defined by:
x\undersetG ⇒ y\iff\existsu,v,p,q\in(\Sigma\cupN)^*:(x=upv)\wedge(p → q\inP)\wedge(y=uqv)
(pronounced as
G derives in zero or more steps
) is defined as the reflexive transitive closure of
• a sentential form is a member of
that can be derived in a finite number of steps from the start symbol
; that is, a sentential form is a member of
\left\{w\in(\Sigma\cupN)^*\midS\overset*{\undersetG ⇒ }w\right\}
. A sentential form that contains no nonterminal symbols (i.e. is a member of
) is called a
, denoted as
, is defined as the set of sentences built by
The grammar
is effectively the
semi-Thue system
, rewriting strings in exactly the same way; the only difference is in that we distinguish specific
symbols, which must be rewritten in rewrite rules, and are only interested in rewritings from the designated start symbol
to strings without nonterminal symbols.
For these examples, formal languages are specified using set-builder notation.
Consider the grammar
is the start symbol, and
consists of the following production rules:
This grammar defines the language
denotes a string of
's. Thus, the language is the set of strings that consist of 1 or more
's, followed by the same number of
's, followed by the same number of
Some examples of the derivation of strings in
(On notation:
reads "String generates string by means of production ", and the generated part is each time indicated in bold type.)
The Chomsky hierarchy
When Noam Chomsky first formalized generative grammars in 1956,^[2] he classified them into types now known as the Chomsky hierarchy. The difference between these types is that they have increasingly
strict production rules and can therefore express fewer formal languages. Two important types are context-free grammars (Type 2) and regular grammars (Type 3). The languages that can be described
with such a grammar are called context-free languages and regular languages, respectively. Although much less powerful than unrestricted grammars (Type 0), which can in fact express any language that
can be accepted by a Turing machine, these two restricted types of grammars are most often used because parsers for them can be efficiently implemented.^[7] For example, all regular languages can be
recognized by a finite-state machine, and for useful subsets of context-free grammars there are well-known algorithms to generate efficient LL parsers and LR parsers to recognize the corresponding
languages those grammars generate.
Context-free grammars
A context-free grammar is a grammar in which the left-hand side of each production rule consists of only a single nonterminal symbol. This restriction is non-trivial; not all languages can be
generated by context-free grammars. Those that can are called context-free languages.
The language
defined above is not a context-free language, and this can be strictly proven using the
pumping lemma for context-free languages
, but for example the language
(at least 1
followed by the same number of
's) is context-free, as it can be defined by the grammar
the start symbol, and the following production rules:
A context-free language can be recognized in
time (
see Big O notation
) by an algorithm such as Earley's recogniser. That is, for every context-free language, a machine can be built that takes a string as input and determines in
time whether the string is a member of the language, where
is the length of the string.
^[8] Deterministic context-free language
s is a subset of context-free languages that can be recognized in linear time.
There exist various algorithms that target either this set of languages or some subset of it.
Regular grammars
In regular grammars, the left hand side is again only a single nonterminal symbol, but now the right-hand side is also restricted. The right side may be the empty string, or a single terminal symbol,
or a single terminal symbol followed by a nonterminal symbol, but nothing else. (Sometimes a broader definition is used: one can allow longer strings of terminals or single nonterminals without
anything else, making languages easier to denote while still defining the same class of languages.)
The language
defined above is not regular, but the language
(at least 1
followed by at least 1
, where the numbers may be different) is, as it can be defined by the grammar
the start symbol, and the following production rules:
All languages generated by a regular grammar can be recognized in
time by a finite-state machine. Although in practice, regular grammars are commonly expressed using
regular expression
s, some forms of regular expression used in practice do not strictly generate the regular languages and do not show linear recognitional performance due to those deviations.
Other forms of generative grammars
Many extensions and variations on Chomsky's original hierarchy of formal grammars have been developed, both by linguists and by computer scientists, usually either in order to increase their
expressive power or in order to make them easier to analyze or parse. Some forms of grammars developed include:
• Tree-adjoining grammars increase the expressiveness of conventional generative grammars by allowing rewrite rules to operate on parse trees instead of just strings.^[10]
• Affix grammars^[11] and attribute grammars^[12] ^[13] allow rewrite rules to be augmented with semantic attributes and operations, useful both for increasing grammar expressiveness and for
constructing practical language translation tools.
Recursive grammars
A recursive grammar is a grammar that contains production rules that are recursive. For example, a grammar for a context-free language is left-recursive if there exists a non-terminal symbol A that
can be put through the production rules to produce a string with A as the leftmost symbol.^[14] An example of recursive grammar is a clause within a sentence separated by two commas.^[15] All types
of grammars in the Chomsky hierarchy can be recursive.
Analytic grammars
Though there is a tremendous body of literature on parsing algorithms, most of these algorithms assume that the language to be parsed is initially described by means of a generative formal grammar,
and that the goal is to transform this generative grammar into a working parser. Strictly speaking, a generative grammar does not in any way correspond to the algorithm used to parse a language, and
various algorithms have different restrictions on the form of production rules that are considered well-formed.
An alternative approach is to formalize the language in terms of an analytic grammar in the first place, which more directly corresponds to the structure and semantics of a parser for the language.
Examples of analytic grammar formalisms include the following:
• Top-down parsing language (TDPL): a highly minimalist analytic grammar formalism developed in the early 1970s to study the behavior of top-down parsers.^[16]
• Link grammars: a form of analytic grammar designed for linguistics, which derives syntactic structure by examining the positional relationships between pairs of words.^[17] ^[18]
• Parsing expression grammars (PEGs): a more recent generalization of TDPL designed around the practical expressiveness needs of programming language and compiler writers.^[19]
See also
Notes and References
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Euler-Lagrange Equation - (History of Mathematics) - Vocab, Definition, Explanations | Fiveable
Euler-Lagrange Equation
from class:
History of Mathematics
The Euler-Lagrange equation is a fundamental result in the calculus of variations, providing a necessary condition for a function to be an extremum of a functional. This equation arises from Euler's
work in both analysis and number theory and plays a crucial role in the development of differential equations and variational principles, establishing a link between physical systems and mathematical
congrats on reading the definition of Euler-Lagrange Equation. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. The Euler-Lagrange equation is derived by applying the principle of stationary action, which states that the path taken by a system is the one that minimizes (or extremizes) the action
2. It can be expressed as $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0$$, where L is the Lagrangian, and q represents the generalized
3. The equation forms the foundation for modern physics, linking dynamics and calculus through its application in deriving motion equations in classical mechanics.
4. Euler's contributions to this area provided tools that extended beyond mechanics, influencing fields such as optics and electromagnetism through variational principles.
5. The generalization of the Euler-Lagrange equation has paved the way for modern theories, including those in quantum mechanics and general relativity.
Review Questions
• How does the Euler-Lagrange equation relate to physical principles like stationary action?
□ The Euler-Lagrange equation is directly tied to the principle of stationary action, which posits that physical systems follow a path that minimizes or extremizes their action. This principle
leads to the formulation of the Euler-Lagrange equation itself, as it provides the mathematical criteria for determining such paths. By setting up a functional representing action and
deriving conditions for its extrema, we find that the resulting equations describe the motion of physical systems accurately.
• Discuss how Euler's work contributed to both analysis and number theory through the development of the Euler-Lagrange equation.
□ Euler's work significantly advanced both analysis and number theory by introducing tools like the Euler-Lagrange equation. In analysis, this equation serves as a foundational result in
variational calculus, enabling mathematicians to determine extremal functions effectively. In number theory, it influenced methods for solving problems involving optimization and
approximation, showcasing how mathematical concepts can bridge diverse areas and enhance our understanding of numerical relationships.
• Evaluate the impact of the Euler-Lagrange equation on contemporary physics and mathematics, particularly regarding its applications in modern theories.
□ The Euler-Lagrange equation has had a profound impact on contemporary physics and mathematics by serving as a cornerstone for Lagrangian mechanics and variational principles. Its applications
extend into modern theories such as quantum mechanics and general relativity, where it provides essential tools for deriving equations of motion and analyzing complex systems. This
versatility highlights how Euler's early contributions continue to shape our understanding of physical laws and mathematical frameworks, illustrating the timeless relevance of foundational
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Direction: among given options only for this figure perpendic-Turito
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Direction : Among given options only for this figure perpendicular bisector theorem is used.
If AD = 8 units, find BD.
D. 4
Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. T
the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle.
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Definition, Examples | Properties of Exponents
What is an Exponent
In general, the exponent represents the number of times that a number is multiplied by itself, and the exponentiation operation can be used to perform operations like finding roots and logarithms.
In mathematics, an exponent is a symbol used to represent the power to which a number or expression is raised.
For example, in the expression 2^3, the number 3 is the exponent, and it represents the power to which 2 is raised. The expression 2^3 is read as “2 raised to the power of 3” or simply “2 to the
power of 3”.
The result of the expression 2^3 is 8, so 2 raised to the power of 3 is equal to 8.
or so 2^3 = 2 × 2 × 2 = 8
Some more examples:
Example: 4^3 = 4 × 4 × 4 = 64
• In words: 4^3 could be called “4 raised to the power of 3”, “4 to the power of 3” or simply “4 cubed”
Example: 7^2 = 7 × 7 = 49
Exponents are commonly used in mathematical operations and can be used to simplify complex mathematical expressions. Exponents can also be negative, fractional, or even complex numbers, depending on
the context in which they are used.
Properties of Exponents
Exponents are mathematical symbols that indicate a repeated multiplication of the same number. In mathematical notation, the expression “a^n” represents the result of multiplying “a” by itself “n”
times. Here are some of the key properties of exponents:
1. Law of product
If “a^m” and “a^n” are two exponential expressions with the same base, then the product of the two expressions are given below.
a^m × a^n = a^m + n
For example, (a^2) * (a^3) = a^(2 + 3) = a^5.
2. Law of quotient
If “a^m” and “a^n” are two exponential expressions with the same base we subtract the exponents., then the quotient of the two expressions are given below :
a^m / a^n = a^m – n
For example, (a^4) / (a^2) =a^(4-2) = a^2.
3. Law of power of a power
If “a^m” is an exponential expression, then (a^m )^n = a^(m*n).
For example, (a^2 )^3 = a^(2*3) = a^6.
4. Law of zero exponent
If “a^0” is an exponential expression, then “a^0” = 1.
5. Law of negative exponent
If “a^-n” is an exponential expression, then “a^-n” = 1 / (a^n).
For example, a^-2 = 1 / (a^2).
6. Law of a fraction
If “a^(m/n)” is an exponential expression, then “a^(m/n) = (a^m )^1/n“.
These properties of exponents are useful for simplifying and solving mathematical expressions. Understanding these properties is essential for advanced mathematical concepts such as logarithms and
exponential functions.
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Beginner Questions
I'm probably missing something simple here, but I can't figure it out on my own. I usually have statements of the form P (do_thing a s), but at some point I need to prove P (λp. do_thing a (f p) p)
instead (note how s has been replaced with f p). I have a helper lemma, which would give me the proof easily, but it requires bounded s, or in this case bounded (f p), which I can provide. So, in
short, I need something like this:
assumes "⋀p. bounded (f p)"
and "⋀s. bounded s ⟹ P (λp. do_thing a s p)"
shows "P (λp. do_thing a (f p) p)"
And I don't know how to approach this.
your assumption, ⋀s. bounded s ⟹ P (λp. do_thing a s p), is too weak. as written, s is not allowed to depend on p
I still can't figure it out. I managed to show ⋀f x. bounded (f x) ==> P (λp. do_thing a (f x) p), but replacing do_thing a (f x) with do_thing a (f p) and it fails to prove.
Here is how to get the Isabelle version of @Jakub Kądziołka 's error:
assumes "⋀p. bounded (f p)"
and "⋀s. bounded s ⟹ P (λp. do_thing a s p)"
shows "P (λp. do_thing a (f p) p)"
supply [[unify_trace_failure]]
apply (rule assms(2))
Cannot unify variable ?s (depending on bound variables )
with term f p
Term contains additional bound variable(s) p
In this version ⋀f x. bounded (f x) ==> P (λp. do_thing a (f x) p) there is still no dependency between f x and p: whatever p is, f x does not change. If you get f p instead, the proof will work.
assumes "⋀p. bounded (f p)"
and "⋀(f :: 'a ⇒ 'b) (x :: 'a). bounded (f x) ==> P (λp. do_thing a (f p) p)"
shows "P (λp :: 'a. do_thing a (f p) p)"
apply (rule assms(2))
apply (rule assms(1))
That's the issue though. I need to prove ⋀(f :: 'a ⇒ 'b) (x :: 'a). bounded (f x) ==> P (λp. do_thing a (f p) p), since I am not given that fact.
⋀(f :: 'a ⇒ 'b) (x :: 'a). bounded (f x) ==> P (λp. do_thing a (f p) p)
this statement is too strong. You're saying that if, for a specific x, f x is bounded, then P (%p. do_thing a (f p) p) holds, which depends on the boundedness of f p for many different p, I'd assume
and we are telling you: without more assumption it does not work
You'd want
⋀(f :: 'a ⇒ 'b) (⋀(x :: 'a). bounded (f x)) ==> P (λp. do_thing a (f p) p)
how is bounded and do_thing defined?
assumes "⋀p. bounded (f p)"
and "⋀s. bounded s ⟹ P (λp. do_thing a s p)"
shows "P (λp. do_thing a (f p) p)"
this inference can never work. Consider bounded x = True, do_thing a s p = s, P f = (EX c. f = %x. c), f x = x
bounded is an abbreviation: bounded r == (0 <= r /\ r <= 1), and roughly the fun do_thing a s p = (modify p and return a similar p, same type)
The classes I took never covered ZF axioms, so I have to be honest and say that this goes over my head. I suppose I simply have to change the structure of my proof so that I don't run into this
Thanks for the help though!
this is independent of ZF axioms, it's just logic
maybe the example has_a_maximum (λp. f p) with f p x = x and f p _ = p is easier to understand
What's the % syntax?
a λ I was too lazy to seach for... but that does not type for your definition of bound
Ah, gotcha
that is what I had in mind:
Mathias Fleury said:
maybe the example has_a_maximum (λp. f p) with f p x = x and f p _ = p is easier to understand
Last updated: Nov 11 2024 at 04:22 UTC
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2023 CIMPA School
Leandro Vendramin (Vrije Universiteit Brussel) Computational algebra: groups, rings and combinatorics. We present the computer algebra system GAP and use it to solve some research-level problems in
algebra and combinatorics. The topics to investigate are groups, certain finite rings and some combinatorial structures associated with the Yang-Baxter equation (with applications in knot theory).
GAP is a system for computational discrete algebra with emphasis on computational group theory.
Arne van Antwerpen was the assistant to Leandro Vendramin for this course’s tutorials.
Arne is a postdoc at Vrije Universiteit Brussel (VUB) in Brussels, Belgium. He completed his PhD in 2020 at Vrije Universiteit Brussel under the supervision of Eric Jespers. His research interests
include set-theoretic solutions of the Yang-Baxter equation, skew braces, finite group theory and finiteness conditions in ring theory.
Installing GAP
We strongly recommend that all participants have either SAGE or GAP installed before the lectures begin. To install SAGE, which comes with a GAP distribution, please see here. To install GAP, there
are general instructions available here. If you would prefer not to install GAP, you can find an online alternative here.
Reading material for the course
1. Computer algebra with GAP by K. Piterman. and L. Vendramin.
1. The file with the FRACTRAN program PRIMEGAME is here.
2. The file generating the Look and Say sequence is here.
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Type II Quantum Computing Algorithm for Computational Fluid Dynamics
Degree Name
Master of Science in Applied Physics
Department of Engineering Physics
First Advisor
David E. Weeks, PhD
An algorithm is presented to simulate fluid dynamics on a three qubit type II quantum computer: a lattice of small quantum computers that communicate classical information. The algorithm presented is
called a three qubit factorized quantum lattice gas algorithm. It is modeled after classical lattice gas algorithms which move virtual particles along an imaginary lattice and change the particles’
momentums using collision rules when they meet at a lattice node. Instead of moving particles, the quantum algorithm presented here moves probabilities, which interact via a unitary collision
operator. Probabilities are determined using ensemble measurement and are moved with classical communications channels. The lattice node spacing is defined to be a microscopic scale length. A
mesoscopic governing equation for the lattice is derived for the most general three qubit collision operator which preserves particle number. In the continuum limit of the lattice, a governing
macroscopic partial differential equation—the diffusion equation—is derived for a particular collision operator using a Chapman- Enskog expansion. A numerical simulation of the algorithm is carried
out on a conventional desktop computer and compared to the analytic solution of the diffusion equation. The simulation agrees very well with the known solution.
AFIT Designator
DTIC Accession Number
Recommended Citation
Scoville, James A., "Type II Quantum Computing Algorithm for Computational Fluid Dynamics" (2006). Theses and Dissertations. 3361.
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Gas Pool Heating Cost Calculator: Calculate Your Pool Heating Expenses Now! - Calculator Pack
Gas Pool Heating Cost Calculator
Gas Pool Heating Cost Calculator is the ultimate solution for all your pool heating needs. Our aim is to provide you with a hassle-free calculation tool that will help you estimate the cost of
heating your pool with gas. By making use of this calculator, you will be able to determine the exact amount of money you need to spend on heating your pool during the winter months or anytime you
desire. With our user-friendly interface, you can easily calculate the cost of gas pool heating without any knowledge of complex formulas or equations. We understand the importance of having a warm,
cozy and comfortable pool and are dedicated to helping you get the most out of your pool heating experience. So, get ready to dive into a warm and inviting pool by using our Gas Pool Heating Cost
Gas Pool Heating Cost Calculator
Calculate the cost to heat your pool using a gas heater.
Gas Pool Heating Cost Calculator Results
Pool Volume 0
Desired Pool Temperature 0
Current Pool Temperature 0
Heating Rate 0
Gas Price 0
Total Therms Needed 0
Total Cost 0
Share results with your friends
estimating the cost of gas pool heating is essential for managing pool heating expenses. Our gas pool heating cost calculator simplifies this process. To explore related pool-related calculations,
such as estimating fiberglass pool costs, link it with our fiberglass pool calculator. This dual approach provides comprehensive resources for pool heating cost estimation.
How to Use the Gas Pool Heating Cost Calculator
The Gas Pool Heating Cost Calculator is a tool that helps you estimate the cost of heating your pool using a gas heater. It is a handy tool for homeowners who want to keep their pools warm without
breaking the bank. In this article, we will provide instructions on how to use the Gas Pool Heating Cost Calculator, including the input fields, output fields, and formula.
Instructions for Utilizing the Calculator:
The Gas Pool Heating Cost Calculator requires the following input fields:
• Pool Volume (gallons)
• Desired Pool Temperature (°F)
• Current Pool Temperature (°F)
• Heating Rate (°F per hour)
• Gas Price ($/therm)
Pool Volume refers to the total volume of water in your pool. Desired Pool Temperature is the temperature you want your pool to be at. Current Pool Temperature is the current temperature of your
pool. Heating Rate is the rate at which your gas pool heater heats your pool. Gas Price is the cost per therm of natural gas in your area.
To use the calculator, you need to provide the required input data, and the calculator will output the following results:
• Pool Volume (gallons)
• Desired Pool Temperature (°F)
• Current Pool Temperature (°F)
• Heating Rate (°F per hour)
• Gas Price ($/therm)
• Total Therms Needed
• Total Cost
Total Therms Needed is the total amount of natural gas needed to heat your pool to the desired temperature, while Total Cost is the estimated cost of the gas needed to heat your pool.
Gas Pool Heating Cost Formula:
The Gas Pool Heating Cost formula is as follows:
• Total Therms Needed = (Pool Volume x Temperature Difference x 8.34) / (Heating Rate x 100,000)
• Total Cost = Total Therms Needed x Gas Price
• Temperature Difference = Desired Pool Temperature - Current Pool Temperature
Illustrative Example:
Suppose your pool volume is 20,000 gallons, and you want to heat it from 70°F to 80°F using a gas pool heater with a heating rate of 200,000 BTUs per hour. The current gas price in your area is $1.20
per therm. By plugging these values into the Gas Pool Heating Cost formula, we can calculate the Total Therms Needed and Total Cost as follows:
Temperature Difference = Desired Pool Temperature - Current Pool Temperature = 80°F - 70°F = 10°F
Total Therms Needed = (Pool Volume x Temperature Difference x 8.34) / (Heating Rate x 100,000) = (20,000 x 10 x 8.34) / (200,000 x 100,000) = 0.0834 therms
Total Cost = Total Therms Needed x Gas Price = 0.0834 x $1.20 = $0.10
The estimated cost to heat the pool from 70°F to 80°F using a gas pool heater with a heating rate of 200,000 BTUs per hour would be $0.10.
Example Table
To provide a more concrete understanding of how the Gas Pool Heating Cost Calculator works, the following table includes example input values and their corresponding output results.
Pool Volume (gallons) Desired Pool Temperature (°F) Current Pool Temperature (°F) Heating Rate (°F per hour) Gas Price ($/therm) Total Therms Needed Total Cost
10000 80 70 100 1.50 11.12 $16.68
15000 75 65 75 2.00 18.70 $37.40
20000 85 75 125 1.25 21.11 $26.39
The Gas Pool Heating Cost Calculator is a powerful tool that allows you to estimate the cost of heating your pool using a gas heater. By providing a few simple inputs such as pool volume, desired
temperature, current temperature, heating rate, and gas price, the calculator can accurately estimate the total therms needed and total cost to heat your pool. With this information, you can make
informed decisions about how to keep your pool warm and comfortable without breaking the bank. So, whether you're a pool owner or pool professional, the Gas Pool Heating Cost Calculator is a
must-have tool for anyone looking to maximize their pool enjoyment while minimizing their costs.
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MTH321: Real Analysis I (Spring 2023)
At the end of this course the students will be able to understand the basic set theoretic statements and emphasize the proofs’ development of various statements by induction. Define the limit of, a
function at a value, a sequence and the Cauchy criterion. Prove various theorems about limits of sequences and functions and emphasize the proofs’ development. Define continuity of a function and
uniform continuity of a function, prove various theorems about continuous functions and emphasize the proofs’ development. Define the derivative of a function, prove various theorems about the
derivatives of functions and emphasize the proofs’ development. Define a cluster point and an accumulation point, prove the Bolzano-Weierstrass theorem, Rolles’s Theorem, extreme value theorem, and
the Mean Value theorem and emphasize the proofs’ development. Define Riemann integral and Riemann sums, prove various theorems about Riemann sums and Riemann integrals and emphasize the proofs’
Resources for Terminal
Questions from Chapter 02
1. A convergent sequence of real number has one and only one limit (i.e. limit of the sequence is unique.)
2. Prove that every convergent sequence is bounded.
3. Suppose that $\left\{ {{s}_{n}} \right\}$ and $\left\{ {{t}_{n}} \right\}$ be two convergent sequences such that $\underset{n\to \infty }{\mathop{\lim }}\,{{s}_{n}}=\underset{n\to \infty }{\
mathop{\lim }}\,{{t}_{n}}=s$. If ${{s}_{n}}<{{u}_{n}}<{{t}_{n}}$ for all $n\ge {{n}_{0}}$, then the sequence $\left\{ {{u}_{n}} \right\}$ also converges to $s$.
4. For each irrational number $x$, there exists a sequence $\left\{ {{r}_{n}} \right\}$ of distinct rational numbers such that $\underset{n\to \infty }{\mathop{\lim }}\,\,{{r}_{n}}=x$.
Questions from Chapter 03
1. Prove that if $\sum\nolimits_{n=1}^{\infty }{{{a}_{n}}}$ is convergent then $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=0$ but converse is not true.
2. A series $\sum{{{a}_{n}}}$ is convergent if and only if for any real number $\varepsilon >0$, there exists a positive integer ${{n}_{0}}$ such that $\left|\sum_{i=m+1}^{n} a_i \right|<\
varepsilon$ $\forall$ $n\ge m>{{n}_{0}}$.
3. Suppose $\sum{{{a}_{n}}}$ and $\sum{{{b}_{n}}}$ are infinite series such that ${{a}_{n}}>0$, ${{b}_{n}}>0$ for all $n$. Also suppose that for a fixed positive number $\lambda $ and positive
integer $k$, $a_n<\lambda b_n$ $\forall$ $n\ge k$.
1. If $\sum{{{b}_{n}}}$ is convergent, then $\sum{{{a}_{n}}}$ is convergent.
2. If $\sum{{{a}_{n}}}$ is divergent, then $\sum{{{b}_{n}}}$ is divergent.
4. Prove that every absolute convergent series is convergent, but convers is not true in general.
Questions from Chapter 04
1. Consider the function $f:[0,1]\to \mathbb{R}$ defined as $f(x)=\left\{ \begin{matrix} 0 \text{ if $x$ is rational,} \\ 1 \text{ if $x$ is irrational.}\end{matrix}\right.$ Show that $\underset{x\
to p}{\mathop{\lim }}\,f(x)$, where $p\in [0,1]$ does not exist.
2. If $\underset{x\to c}{\mathop{\lim }}\,f(x)$ exists, then it is unique.
3. A function $f(x)={{x}^{2}}$ is continuous for all $x\in \mathbb{R}$.
4. Suppose $f$ is continuous on $\left[ a,b \right]$ and $f(a)\ne f(b)$, then given a number $\lambda $ that lies between $f(a)$ and $f(b)$, there exist a point $c\in (a,b)$ with $f(c)=\lambda $.
Questions from Chapter 05
1. Prove that a differentiable function is continuous, but the converse is not true.
2. Let $f$ be defined on $\left[ a,b \right]$ and it is differentiable on $(a,b)$. If $f$ has a local maximum at a point $x\in (a,b)$ and if ${f}'(x)$ exist, then ${f}'(x)=0$.
3. Let $f$ be continuous on $[a,b]$and differentiable on $(a,b)$. Then there exists a point $c\in (a,b)$ such that $\frac{f(b)-f(a)}{b-a}={f}'(c)$.
4. If $f$ and $g$ are continuous real valued functions on closed interval $\left[ a,b \right]$ and $f$ and $g$ are differentiable on $\left( a,b \right)$, then there is a point $c\in \left( a,b \
right)$ at which $\left[ f(b)-f(a) \right]{g}'(c)=\left[ g(b)-g(a) \right]{f}'(c)$.
Course contents
• The Real Number System: Ordered Fields. The Field of Reals. The Extended Real Number System. Euclidean Space.
• Numerical Sequences and Series. Limit of a Sequence. Bounded Sequences. Monotone Sequences. Limits Superior and Inferior. Subsequences.
• Limit of a Function and Continuous Functions. Uniform Continuity. Kinds of Discontinuities.
• Derivable and Differentiable Functions. Mean Value theorems. The Continuity of Derivatives. Taylor's theorem.
• Riemann Stieltijes Integrals: Definition, Existence and Properties of the Riemann Integrals. Integral and Differentiation.
Did you know?
• The development of calculus in the 18th century used the entire set of real numbers without having defined them cleanly. The first rigorous definition was given by Georg Cantor in 1871.
• In the 16th century, Simon Stevin created the basis for modern decimal notation, and insisted that there is no difference between rational and irrational numbers in this regard.
• Tuesday, 0830-1000
• Wednesday, 0830-1000
Notes, assignments, quizzes & handout
Please download PDF files of the notes given below. To view PDF files, there must be PDF Reader (Viewer) installed on your PC or mobile or smartphone. It can be downloaded from Software section or a
user viewing on Android Smartphone may consider EBookDroid to view PDF on their smartphone.
Resources for mid-term
There will be two questions having three parts each. First part of each question will be any definitions, second part will be from questions given below and third part will be related to application
of the theory.
Questions from Chapter 01
1. Archimedean Property
2. The Density Theorem
3. If $r$ is rational and $x$ is irrational, then $r+x$ is irrational.
4. If $r$ is non-zero rational and $x$ is irrational, then $r\,x$ is irrational.
5. Let $\underline{x}\,,\,\underline{y}\in {{\mathbb{R}}^{k}}$ then $\left| \,\underline{x}\cdot \underline{y}\, \right|\le \left\| \,\underline{x}\, \right\|\,\,\left\| \,\underline{y}\, \right\|$.
(Cauchy-Schwarz’s inequality)
6. Suppose $\underline{x}\,,\,\underline{y},\underline{z}\in {{\mathbb{R}}^{k}}$, then prove that
1. $\left\| \,\underline{x}+\underline{y}\, \right\|\le \left\| \,\underline{x}\, \right\|+\left\| \,\underline{y}\, \right\|$.
2. $\left\| \,\underline{x}-\underline{z}\, \right\|\le \left\| \,\underline{x}-\underline{y}\, \right\|+\left\| \,\underline{y}-\underline{z}\, \right\|$.
Questions from Chapter 02
1. A convergent sequence of real number has one and only one limit (i.e. limit of the sequence is unique.)
2. Prove that every convergent sequence is bounded.
3. Suppose that $\left\{ {{s}_{n}} \right\}$ and $\left\{ {{t}_{n}} \right\}$ be two convergent sequences such that $\underset{n\to \infty }{\mathop{\lim }}\,{{s}_{n}}=\underset{n\to \infty }{\
mathop{\lim }}\,{{t}_{n}}=s$. If ${{s}_{n}}<{{u}_{n}}<{{t}_{n}}$ for all $n\ge {{n}_{0}}$, then the sequence $\left\{ {{u}_{n}} \right\}$ also converges to $s$.
4. For each irrational number $x$, there exists a sequence $\left\{ {{r}_{n}} \right\}$ of distinct rational numbers such that $\underset{n\to \infty }{\mathop{\lim }}\,\,{{r}_{n}}=x$.
5. A Cauchy sequence of real numbers is bounded.
6. Cauchy’s General Principle for Convergence: A sequence of real number is convergent if and only if it is a Cauchy sequence.
Sample questions related to applications
1. Let $E=\left\{1,\dfrac{1}{2},\dfrac{1}{3},... \right\} \subseteq \mathbb{R}$. Then write the set of upper and lower bounds of $E$.
2. Write subset $E$ of $\mathbb{N}$ such that $\sup E = \inf E$.
3. By defintion, prove that $0$ is even number.
4. Prove that $\left\{1+\dfrac{1}{n} \right\}$ is monotone sequence.
5. Write two subsequence of $\{(-1)^n\}$, which converges to different limits.
6. Write a reason that $\{n+\frac{1}{n} \}$ is not Cauchy sequence.
Online resources
Recommended book
1. Rudin, W. (1976). Principle of Mathematical Analysis, McGraw Hills Inc.
2. Bartle, R.G., and D.R. Sherbert, (2011): Introduction to Real Analysis, 4th Edition, John Wiley & Sons, Inc.
3. Apostol, Tom M. (1974), Mathematical Analysis, Pearson; 2nd edition.
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Lesson 11
Side-Side-Angle (Sometimes) Congruence
Problem 1
Which of the following criteria always proves triangles congruent? Select all that apply.
Corresponding congruent Side-Angle-Side
Corresponding congruent Side-Side-Angle
Corresponding congruent Angle-Side-Angle
Problem 2
Here are some measurements for triangle \(\ ABC \) and triangle \(XYZ\):
• Angle \( ABC\) and angle \(XYZ \) are both 30°
• \(BC\) and \(YZ\) both measure 6 units
• \(CA\) and \(ZX\) both measure 4 units
Lin thinks thinks these triangles must be congruent. Priya says she knows they might not be congruent. Construct 2 triangles with the given measurements that aren't congruent. Explain why triangles
with 3 congruent parts aren't necessarily congruent.
Problem 3
Jada states that diagonal \(WY\) bisects angles \(ZWX\) and \(ZYX\). Is she correct? Explain your reasoning,
Problem 4
Select all true statements based on the diagram.
Angle \(CBE\) is congruent to angle \(DAE\).
Angle \(CEB\) is congruent to angle \(DEA\).
Segment \(DA\) is congruent to segment \(CB\).
Segment \(DC\) is congruent to segment \(AB\).
Line \(DC\) is parallel to line \(AB\).
Line \(DA\) is parallel to line \(CB\).
Problem 5
\(WXYZ\) is a kite. Angle \(WXY\) has a measure of 94 degrees and angle \(ZWX\) has a measure of 112 degrees. Find the measure of angle \(ZYW\).
Problem 6
Andre is thinking through a proof using a reflection to show that a triangle is isosceles given that its base angles are congruent. Complete the missing information for his proof.
Construct \(AB\) such that \(AB\) is the perpendicular bisector of segment \(CD\). We know angle \(ADB\) is congruent to \(\underline{\hspace{.5in}1\hspace{.5in}}\). \(DB\) is congruent to \(\
underline{\hspace{.5in}2\hspace{.5in}}\) since \(AB\) is the perpendicular bisector of \(CD\). Angle \(\underline{\hspace{.5in}3\hspace{.5in}}\) is congruent to angle \(\underline{\hspace{.5in}4\
hspace{.5in}}\) because they are both right angles. Triangle \(ABC\) is congruent to triangle \(\underline{\hspace{.5in}5\hspace{.5in}}\) because of the \(\underline{\hspace{.5in}6\hspace{.5in}}\)
Triangle Congruence Theorem. \(AD\) is congruent to \(\underline{\hspace{.5in}7\hspace{.5in}}\) because they are corresponding parts of congruent triangles. Therefore, triangle \(ADC\) is an
isosceles triangle.
Problem 7
The triangles are congruent. Which sequence of rigid motions takes triangle \(DEF\) onto triangle \(BAC\)?
Translate \(DEF\) using directed line segment \(EA\). Rotate \(D’E’F’\) using \(A\) as the center so that \(D’\) coincides with \(C\). Reflect \(D’’E’’F’’\) across line \(AC\).
Translate \(DEF\) using directed line segment \(EA\). Rotate \(D’E’F’\) using \(A\) as the center so that \(D’\) coincides with \(C\). Reflect \(D’’E’’F’’\) across line \(AB\).
Translate \(DEF\) using directed line segment \(EA\). Rotate \(D’E’F’\) using \(A\) as the center so that \(D’\) coincides with \(B\). Reflect \(D’’E’’F’’\) across line \(AC\).
Translate \(DEF\) using directed line segment \(EA\). Rotate \(D’E’F’\) using \(A\) as the center so that \(D’\) coincides with \(B\). Reflect \(D’’E’’F’’\) across line \(AB\).
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Mathematical Logic: An Introduction
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Daniel W. Cunningham Mathematical Logic
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Daniel W. Cunningham
Mathematical Logic �
An Introduction
Mathematics Subject Classification 2020 03-01, 03B05, 03B10, 03B22, 03B25, 03D10, 03D20 Author Dr. Daniel W. Cunningham Department of Mathematics Californial State University 5245 North Backer Avenue
M/S PB108 Fresno CA 93740 USA [email protected]
ISBN 978-3-11-078201-1 e-ISBN (PDF) 978-3-11-078207-3 e-ISBN (EPUB) 978-3-11-078219-6 Library of Congress Control Number: 2023931072 Bibliographic information published by the Deutsche
Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2023
Walter de Gruyter GmbH, Berlin/Boston Cover image: cicerocastro / iStock / Getty Images Plus Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com
Preface Mathematical logic is a particular branch of mathematics that applies mathematical tools to investigate the nature of mathematics. Consequently, in this book our attention will be focused on
the language of mathematics. This will be done by discussing formal languages, first-order logic, model-theoretic semantics, deductive systems, and their mathematical properties and relations. Such a
focus radiates a bright and direct light on mathematics itself. Aristotle was the first formal logician who identified several key principles of correct reasoning, namely, the syllogisms. On the
other hand, modern mathematical logic is based on the late nineteenth- and early twentieth-century innovative work of Boole, Frege, Peano, Russell, Whitehead, Hilbert, Skolem, Gödel, Tarski, Cantor,
and their followers. The material presented in this text is the result of these pioneers in mathematical logic. This textbook delivers an upper-division undergraduate course in mathematical logic.
This subject typically attracts students with various backgrounds, some of whom may be quite familiar with logical notation and mathematical proof while others may not be as familiar. The book
strives to address the needs of such a course. My primary goal was to write a book that would be accessible to all readers having a fundamental background in mathematics. Thus, I have made an effort
to write clear and complete proofs throughout the text. In addition, these proofs favor detail over brevity. This approach should be to the benefit of all students, readers, and instructors.
Topics covered The book presents the fundamental topics in mathematical logic that will lead to the statements and coherent proofs of Gödel’s completeness and incompleteness theorems. The basics of
logic and elementary set theory are first discussed in Chapter 1. Since students, typically, are acquainted with these topics, one should not necessarily begin the book by starting with this chapter.
However, Section 1.1.5 and Theorem 1.1.27 should definitely be discussed. In particular, Theorem 1.1.27 is a recursion theorem that justifies many of the key definitions and proofs that are presented
in the text. Few books in mathematical logic explicitly state and prove this often applied result. Such books usually justify their definitions and proofs with the expression “by recursion;” however,
we in such cases will cite and apply Theorem 1.1.27. Understanding the statement of this theorem is more important than reading and understanding its proof. Chapter 2 introduces the syntax and
semantics of propositional logic. The chapter also carefully discusses an induction principle that is illustrated by correctly proving fundamental results about the language of propositional logic.
This is followed by establishing the completeness of the logical connectives, the compactness theorem, and the deduction theorem. We also state and prove the associated soundness and completeness
VI � Preface theorems. Most students who are familiar with propositional logic have not yet seen a mathematical development of this logic. Therefore, these topics offer an important prerequisite for
the development of first-order logic. Chapter 3 discusses the syntax and semantics of first-order languages. First-order logic is quite a bit more subtle than propositional logic, although they do
share some common characteristics. The chapter also introduces structures, which can be viewed as vehicles for interpreting a given first-order language. Tarski’s definition of satisfaction gives a
precise meaning that yields a method for interpreting a first-order language in a given structure. This is then followed by the notion of a deduction (formal proof) in a first-order language. The
main goal of Chapter 4 is to present and prove the soundness and completeness theorems of first-order logic. For each of these proofs, I have respectively isolated the technical lemmas (Sections
4.1.1 and 4.2.1) that support the proofs of these important theorems. The compactness theorem is then presented with a proof. The chapter ends with several applications of the soundness,
completeness, and compactness theorems. In Gödel’s proof of the incompleteness theorem, he encodes formulas into natural numbers using (primitive) recursive functions. In preparation for the proof of
Gödel’s theorem, Chapter 5 covers (primitive) recursive functions and relations. Since Gödel’s encoding techniques created a link between logic and computing, we begin the chapter with an
introduction to abstract computing machines and partial recursive functions. Today, computability theory in mathematical logic is closely related to the theory of computation in computer science. In
Chapter 6 the focus is on the language of elementary number theory ℒ and the standard model 𝒩 for number theory. The chapter begins with the question: Is there a decidable set of ℒ-sentences, that
hold in 𝒩 , from which one can deduce all the sentences that are true in 𝒩 ? This is followed by an introduction to the Ω-axioms. These axioms allow one to deduce some of the basic statements that
are true in the standard model. Then representable relations and functions are discussed. Eventually it is established that a function is representable if and only if the function is recursive. Then
a technique is presented that allows one to perform a Gödel encoding of all of the formulas in the language ℒ. This is followed by a proof of the fixed-point lemma and two results of Gödel, namely,
the first and second incompleteness theorems. These two theorems are among the most important results in mathematical logic.
How to use the book It is strongly recommended that the reader be familiar with the basics of sets, functions, relations, logic, and mathematical induction. These topics are typically introduced in a
“techniques of proof” course (for example, see [1]). As the emphasis will be on theorems and their proofs, the reader should be comfortable reading and writing mathematical proofs.
How to use the book
� VII
If time is short or an instructor would like to end a one-semester course by covering Gödel’s incompleteness theorems, certain topics can be bypassed. In particular, the following sections can be
omitted without loss of continuity: – 3.2.5 Classes of structures – 3.2.6 Homomorphisms – 4.3.1 Nonstandard models – 4.3.4 Prenex normal form – 5.1 The informal concept – 5.2.1 Turing machines –
5.2.2 Register machines Furthermore, an instructor could focus on the statements of the technical lemmas in Sections 4.1.1 and 4.2.1 rather than on the proofs of these lemmas. These proofs could then
be given as assigned reading. These technical lemmas support the respective proofs of the soundness and completeness theorems in Chapter 4. Similarly, in Sections 5.3 and 5.4 one could focus
attention on the results rather than on the proofs. Of course, the material in Chapter 6 is more interesting than the proofs presented in these two sections. Perhaps, after seeing the theorems
presented in Chapter 6, one would be more interested in the meticulous proofs presented in Sections 5.3 and 5.4. Exercises are given at the end of each section in a chapter. An exercise marked with
an asterisk ∗ is one that is cited, or referenced, elsewhere in the book. Suggestions are also provided for those exercises that a newcomer to upper-division mathematics may find more challenging.
Acknowledgments This book began as a set of lecture notes for an undergraduate course in mathematical logic that I taught at SUNY Buffalo State. The textbook used for this course was A Mathematical
Introduction to Logic by Herbert B. Enderton [2]. As an undergraduate at UCLA, I also took courses in mathematical logic that used Enderton’s book. Needless to say, my respect for Enderton’s book has
influenced the topics and presentation in my own book. Sherman Page deserves to be recognized for his meticulous copy-editing and review of the manuscript. Sherman found many typos and errors that I
had overlooked. I am extremely grateful for his salient comments and corrections. I am also grateful to Steven Elliot, editor at De Gruyter, for his guidance and enthusiasm. Thank you, Marianne
Foley, for your support and tolerance.
Contents Preface � V Acknowledgments � IX 1 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.1.8 1.2 1.2.1 1.2.2 1.2.3
Basic set theory and basic logic � 1 Basic set theory � 1 Relations � 2 Equivalence classes and partitions � 4 Functions � 5 Induction and recursion � 6 Defining sets by recursion � 7 Countable sets
� 11 Cardinality � 13 The axiom of choice � 14 Basic logic � 16 Propositions and logical connectives � 16 Truth tables and truth functions � 17 Predicates and quantifiers � 19
2 2.1 2.2 2.2.1 2.2.2 2.3 2.4 2.5
Propositional logic � 27 The language � 27 Truth assignments � 32 Some tautologies � 36 Omitting parentheses � 37 Completeness of the logical connectives � 39 Compactness � 44 Deductions � 48
3 3.1 3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 3.2 3.2.1 3.2.2
First-order logic � 52 First-order languages � 52 Terms and atomic formulas � 54 Induction on terms principle � 56 Well-formed formulas � 58 Induction on wffs principle � 59 Free variables � 60
Notational abbreviations � 62 Examples of languages � 63 Truth and structures � 66 Structures for first-order languages � 66 Satisfaction (Tarski’s definition) � 68
XII � Contents 3.2.3 3.2.4 3.2.5 3.2.6 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.3.6 3.3.7
Logical implication � 74 Definability over a structure � 76 Classes of structures � 78 Homomorphisms � 80 Deductions � 90 Tautologies in first-order logic � 91 Generalization and substitution � 93
The logical axioms � 97 Formal deductions � 99 Metatheorems about deductions � 102 Equality � 109 More metatheorems about deductions � 112
4 4.1 4.1.1 4.1.2 4.2 4.2.1 4.2.2 4.2.3 4.3 4.3.1 4.3.2 4.3.3 4.3.4
Soundness and completeness � 119 The soundness theorem � 119 Technical lemmas � 119 Proof of the soundness theorem � 123 The completeness theorem � 124 Technical lemmas � 125 Proof of the
completeness theorem � 129 The compactness theorem � 134 Applications � 136 Nonstandard models � 137 Löwenheim–Skolem theorems � 139 Theories � 141 Prenex normal form � 145
5 5.1 5.1.1 5.1.2 5.2 5.2.1 5.2.2 5.2.3 5.3 5.3.1 5.4 5.4.1
Computability � 148 The informal concept � 148 Decidable sets � 149 Computable functions � 151 Formalizations—an overview � 159 Turing machines � 159 Register machines � 163 Primitive recursiveness
and partial search � 166 Recursive functions � 172 Bounded search � 187 Recursively enumerable sets and relations � 195 Decidability revisited � 201
6 6.1
Undecidability and incompleteness � 205 Introduction � 205
Contents �
6.2 6.3 6.3.1 6.4 6.4.1 6.5 6.5.1 6.5.2 6.5.3
Basic axioms for number theory � 206 Representable relations and functions � 211 Recursive relations and functions are representable � 219 Arithmetization of the formal language � 225 The logical
axioms revisited � 232 The incompleteness theorems � 238 Gödel’s first incompleteness theorem � 241 Gödel’s second incompleteness theorem � 244 Epilogue � 247
Bibliography � 249 Symbol Index � 251 Subject Index � 253
1 Basic set theory and basic logic 1.1 Basic set theory Fundamental definitions of set theory A set is a collection of objects which are called its elements. As usual, we write “t ∈ A” to say that t
is a member of A, and we write “t ∉ A”’ to say that t is not a member of A. We write A = B to mean that the sets A and B are equal, that is, they have the same elements. For sets A and B we write A ⊆
B to mean to mean that set A is a subset of set B, that is, every element of A is also an element of B. Remark. Recall the following: 1. A ⊆ B means that for all x, if x ∈ A, then x ∈ B. 2. A = B
means that for all x, x ∈ A if and only if x ∈ B. 3. One special set is the empty set ⌀, which has no members at all. Note that ⌀ ⊆ A for any set A. 4. For any object x, the set {x} is called a
singleton because it has only one element, namely, x.
1. 2. 3.
We now identify some important sets that often appear in mathematics: ℕ = {x : x is a natural number} = {0, 1, 2, 3, . . . }, ℤ = {x : x is an integer} = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . }, ℚ =
{x : x is a rational number}; for example, −2, 32 ∈ ℚ,
ℝ = {x : x is a real number}; for example, 32 , π ∈ ℝ.
Given a property P(x) we can form the set of just those elements in a set A that make P(x) true, that is, we can form the set {x ∈ A : P(x)}. For example, let ℕ be the set of natural numbers. Suppose
we want to collect just those elements in ℕ that are odd. We can easily describe this set by {n ∈ ℕ : n is odd}, that is, “the set of all n ∈ ℕ such that n is odd.” Thus, {n ∈ ℕ : n is odd} = {1, 3,
5, 7, . . . }. Definition 1.1.1. Given two sets A and B we define the following: 1. A ∩ B = {x : x ∈ A and x ∈ B} is the intersection of A and B, 2. A ∪ B = {x : x ∈ A or x ∈ B} is the union of A and
B, 3. A \ B = {x : x ∈ A and x ∉ B} is the set difference of A and B (also stated in English as A “minus” B), 4. A and B are disjoint if they have no elements in common, that is, A ∩ B = ⌀, 5. to add
one extra object t to a set A, we will write A; t to denote the set A ∪ {t}. Consider a set A whose members are themselves sets. The union of A, denoted by ⋃ A, is the set of objects that belong to
some member of A, that is, ⋃ A = {x : x belongs to some member of A}. https://doi.org/10.1515/9783110782073-001
2 � 1 Basic set theory and basic logic When A is nonempty, the intersection of A, denoted by ⋂ A, is the set of objects that belong to every member of A, that is, ⋂ A = {x : x belongs to every member
of A}. For example, if A = {{0, 1, 5}, {1, 6}, {1, 5}}, then ⋃ A = {0, 1, 5, 6} and
⋂ A = {1}.
In cases where we have a set Ai for each i ∈ I, the set {Ai : i ∈ I} is called an indexed family of sets. The union ⋃{Ai : i ∈ I} is usually denoted by ⋃i∈I Ai or ⋃i Ai . The intersection ⋂{Ai : i ∈
I} is usually denoted by ⋂i∈I Ai or ⋂i Ai . We will often be dealing with unions of the form ⋃n∈ℕ An and intersections of the form ⋂n∈ℕ An . More generally, a set having the form {xi : i ∈ I} is
called an indexed set and each i ∈ I is called an index. Every element in {xi : i ∈ I} has the form xi for some i ∈ I. Such sets appear frequently in mathematics and will also appear in this text.
Definition 1.1.2. Let A be a set. The power set of A, denoted by 𝒫 (A), is the set whose elements are all of the subsets of A, that is, 𝒫 (A) = {X : X ⊆ A}. Thus, X ∈ 𝒫 (A) if and only if X ⊆ A. If A
is a finite set with n elements, then one can show that the set 𝒫 (A) has 2n elements. The set A = {1, 2, 3} has three elements, so 𝒫 (A) has eight elements, namely, 𝒫 (A) = {⌀, {1}, {2}, {3}, {1,
2}, {1, 3}, {2, 3}, {1, 2, 3}}.
1.1.1 Relations Definition 1.1.3. An ordered pair has the form ⟨a, b⟩, where a is referred to as the first component and b is identified as the second component. Example 1.1.4. The pair ⟨2, 3⟩ is an
ordered pair, and so is ⟨3, 2⟩. Note that these are different ordered pairs, that is, ⟨2, 3⟩ ≠ ⟨3, 2⟩. Definition 1.1.5. Given sets A and B, the Cartesian product A × B is the set A × B = {⟨a, b⟩ : a
∈ A and b ∈ B}. In other words, A × B is the set of all ordered pairs with first component in A and second component in B. Definition 1.1.6. An n-sequence is an ordered list of the form ⟨a1 , a2 , .
. . , an ⟩, where n ≥ 1 is a natural number. The term a1 is called the first component, a2 is the second component, . . . , and an is the n-th component. We say that ⟨a1 , a2 , . . . , ak ⟩ is a
proper initial segment of ⟨a1 , a2 , . . . , an ⟩ when 1 ≤ k < n.
1.1 Basic set theory
� 3
Given an n-sequence s = ⟨a1 , a2 , . . . , an ⟩ and an m-sequence t = ⟨b1 , b2 , . . . , bm ⟩, the concatenation of s with t is the (n+m)-sequence given by s⌢ t = ⟨a1 , a2 , . . . , an , b1 , b2 , .
. . , bm ⟩. Example 1.1.7. The sequence ⟨3, −1, 2, 2⟩ is a 4-sequence, and so is ⟨2, −1, 3, 2⟩. Note that these are different sequences, that is, ⟨3, −1, 2, 2⟩ ≠ ⟨2, −1, 3, 2⟩. We will just say that
⟨a1 , a2 , . . . , an ⟩ is a sequence, when n is understood.
Definition 1.1.8. Let n ≥ 1 and let A be a set. Then the set An is defined to be An = {⟨a1 , a2 , . . . , an ⟩ : a1 ∈ A, a2 ∈ A, . . . , an ∈ A}.
In other words, An is the set of all n-sequences whose components are all in A. For the record, an n-sequence ⟨a1 , a2 , . . . , an ⟩ is rigorously defined to be a function f : {1, 2, . . . , n} →
{a1 , a2 , . . . , an } such that f (1) = a1 , f (2) = a2 , . . . , f (n) = an . Thus, ⟨a, b⟩ = ⟨c, d⟩ if and only if a = c and b = d. Moreover, ⟨a1 , a2 , . . . , an ⟩ = ⟨b1 , b2 , . . . , bm ⟩ if
and only if n = m and a1 = b1 , . . . , an = bm . Definition 1.1.9. A subset R of An is said to be an n-place relation on A. We will write R(a1 , a2 , . . . , an ) to indicate that ⟨a1 , a2 , . . . ,
an ⟩ ∈ R. A 1-place relation on a set A is simply a subset of A. Definition 1.1.10. A 2-place relation R on a set A is often just called a relation on A. The domain of R, dom(R), is the set {x ∈ A :
⟨x, y⟩ ∈ R for some y}. The range of R, ran(R), is the set {y ∈ A : ⟨x, y⟩ ∈ R for some x}. The union of dom(R) and ran(R) is called the field of R, fld(R). A relation R on A is a subset of A × A,
that is, R ⊆ A × A. We shall customarily write xRy or R(x, y) to denote ⟨x, y⟩ ∈ R. The equality relation is reflexive, symmetric, and transitive. Many relations also have some of these properties.
Relations that have all of these properties often appear in mathematics. Definition 1.1.11. For a relation ∼ on A, we define the following: ∼ is reflexive if and only if x ∼ x for all x ∈ A, ∼ is
symmetric if and only if whenever x ∼ y, then also y ∼ x, ∼ is transitive if and only if whenever both x ∼ y and y ∼ z, then x ∼ z. Finally, we say that a relation ∼ on A is an equivalence relation
if and only if ∼ is reflexive, symmetric, and transitive.
4 � 1 Basic set theory and basic logic 1.1.2 Equivalence classes and partitions A partition is a way of breaking up a set into nonempty disjoint parts such that each element of the set is in one of
the parts. Definition 1.1.12. Let A be a set. Let P be a collection of nonempty subsets of A. We say that P is a partition of A if the following hold: 1. For every element a ∈ A there is a set S ∈ P
such that a ∈ S. 2. For all S, T ∈ P, if S ≠ T, then S ∩ T = ⌀. Example 1.1.13. The set P = {S1 , S2 , S3 } forms a partition of the set ℤ where S1 = {. . . , −6, −3, 0, 3, 6, . . . },
S2 = {. . . , −5, −2, 1, 4, 7, . . . },
S3 = {. . . , −4, −1, 2, 5, 8, . . . }. That is, P = {S1 , S2 , S3 } breaks ℤ up into three disjoint nonempty sets and every integer is in one of these sets. Definition 1.1.14. Let ∼ be an
equivalence relation on a set A and let a ∈ A be an element of A. The equivalence class of a, denoted by [a]∼ , is defined by [a]∼ = {b ∈ A : b ∼ a}. We write [a] = [a]∼ when the relation ∼ is
clearly understood. So if a ∈ A, then a ∈ [a] as a ∼ a. Theorem 1.1.15. Let ∼ be an equivalence relation on A. Then for all a, b ∈ A, a∼b
if and only if [a] = [b].
Proof. Let ∼ be an equivalence relation on A. Let a, b ∈ A. We shall prove that a ∼ b if and only if [a] = [b]. (⇒). Assume that a ∼ b. We prove that [a] = [b]. First we prove that [a] ⊆ [b]. Let x ∈
[a]. We show that x ∈ [b]. Since x ∈ [a] and [a] = {y ∈ A : y ∼ a}, it follows that x ∼ a. By assumption, we also have a ∼ b. Hence, x ∼ a and a ∼ b. Since ∼ is transitive, we conclude that x ∼ b.
Now, because [b] = {y ∈ A : y ∼ b}, it follows that x ∈ [b]. Therefore, [a] ⊆ [b]. The proof that [b] ⊆ [a] is very similar. So [a] = [b]. (⇐). Assume that [a] = [b]. We prove that a ∼ b. Since a ∈
[a] and [a] = [b], it follows that a ∈ [b]. But [b] = {y ∈ A : y ∼ b}. Therefore, a ∼ b.
1.1 Basic set theory
� 5
Corollary 1.1.16. Let ∼ be an equivalence relation on a set A. Then for all a ∈ A and b ∈ A, a ∈ [b] if and only if [a] = [b]. Theorem 1.1.17 (Fundamental theorem on equivalence relations). Whenever
∼ is an equivalence relation on a set A, the collection P = {[a] : a ∈ A} is a partition of A. We denote this partition by P = A/∼. Proof. Let ∼ be an equivalence relation on A. We prove that the set
P = {[a] : a ∈ A} is a partition of A, that is, we prove that: (i) for every element x ∈ A we have x ∈ [x], (ii) for all x, y ∈ A, if [x] ∩ [y] ≠ ⌀, then [x] = [y]. Proof of (i). Let x ∈ A. Clearly,
[x] ∈ P and x ∈ [x]. Proof of (ii). Let x, y ∈ A. Thus, [x] ∈ P and [y] ∈ P. We must prove that if [x] ≠ [y], then [x] ∩ [y] = ⌀. So assume [x] ≠ [y]. Assume, for a contradiction, that [x] ∩ [y] ≠ ⌀.
Since [x] ∩ [y] ≠ ⌀, there exists a z ∈ A such that z ∈ [x] and z ∈ [y]. However, since [x] = {b ∈ A : b ∼ x} and [y] = {b ∈ A : b ∼ y}, it thus follows that z ∼ x and z ∼ y. Because ∼ is symmetric,
we conclude that x ∼ z and z ∼ y. As ∼ is transitive, we further conclude that x ∼ y. But Theorem 1.1.15 now implies that [x] = [y], which contradicts our assumption that [x] ≠ [y]. Therefore, [x] ∩
[y] = ⌀. 1.1.3 Functions One of the most important concepts in mathematics is the function concept. A function is a way of associating each element of a set A with exactly one element in another set
B. We will give a precise set-theoretic definition of a function using relations. Definition 1.1.18. A relation R is single-valued if for each x ∈ dom(R) there is exactly one y such that ⟨x, y⟩ ∈ R.
Thus, if R is a single-valued relation, then whenever ⟨x, y⟩ ∈ R and ⟨x, z⟩ ∈ R, we can conclude that y = z. Definition 1.1.19. A function f is any single-valued relation; in other words, for each x
∈ dom(f ) there is only one y such that ⟨x, y⟩ ∈ f . Let A and B be sets. A function f from A to B is a subset of A × B such that for each x ∈ A there is exactly one y ∈ B so that ⟨x, y⟩ ∈ f . For
example, let A = {a, b, c, d, e} and B = {5, 6, 7, 8, 9}. Then f = {⟨a, 8⟩, ⟨b, 7⟩, ⟨c, 9⟩, ⟨d, 6⟩, ⟨e, 5⟩} is a function from A to B because for each x ∈ A, there is exactly one y ∈ B such that ⟨x,
y⟩ ∈ f . We now express this notion in terms of a formal definition.
6 � 1 Basic set theory and basic logic Definition 1.1.20. Let A and B be sets and let f be a relation from A to B. Then f is said to be a function from A to B if the following two conditions hold:
(1) dom(f ) = A, that is, for each x ∈ A, there is a y ∈ B such that ⟨x, y⟩ ∈ f , (2) f is single-valued, that is, if ⟨x, y⟩ ∈ f and ⟨x, z⟩ ∈ f , then y = z. The set A is the domain of f and the set
B is called the codomain of f . We write f : A → B to indicate that f is a function from the set A to the set B. Thus, for each x ∈ A, there is exactly one y ∈ B such that ⟨x, y⟩ ∈ f . This unique y
is called “the value of f at x” and is denoted by f (x). Therefore, ⟨x, y⟩ ∈ f if and only if f (x) = y. We will say that x ∈ A is an input to the function f and f (x) is the resulting output. One
can also say that the function f maps x to f (x), denoted by x → f (x). Given a function f : A → B, the set {f (x) : x ∈ A} is called the range of f and is denoted by ran(f ). Clearly, ran(f ) is a
subset of B. The following lemma offers a useful tool for showing that two functions are equal. Lemma 1.1.21. Let f and g be functions such that dom(f ) = dom(g). Then f = g if and only if f (x) = g
(x) for all x in their common domain. Definition 1.1.22. For a natural number n ≥ 2 and a function f : X n → Y from the set of sequences X n to the set Y , we shall write f (⟨x1 , x2 , . . . , xn ⟩)
= f (x1 , x2 , . . . , xn ) for every element of ⟨x1 , x2 , . . . , xn ⟩ ∈ X n . We will say that f is an n-place function or that f has arity n. 1.1.4 Induction and recursion Let ℕ = {0, 1, 2, 3, .
. . } be the set of natural numbers. Often one can use proof by mathematical induction to establish statements of the form “for all n ∈ ℕ, P(n).” Principle of mathematical induction Let S ⊆ ℕ. If 1.
0 ∈ S and 2. for all n ∈ ℕ, if n ∈ S, then n + 1 ∈ S, then S = ℕ. Proof by mathematical induction Let P(n) be a statement concerning a natural number variable n. If 1. P(0) is true and 2. for all n ∈
ℕ, if P(n), then P(n + 1), then P(n) is true for all natural numbers n.
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Principle of strong induction Let P(n) be a statement about a natural number variable n. If 1. P(0) is true and 2. the statements P(0), P(1), . . . , P(n − 1) imply P(n), for all n ∈ ℕ, then P(n) is
true for all natural numbers n. Induction is often applied in mathematical proofs. One can also define a function by induction (recursion). A recursively defined function is one that is defined in
terms of “previously evaluated values of the function.” A function h on ℕ is defined recursively if its value at 0 is first specified and then all of the remaining values are defined by using a value
that has previously been evaluated. A proof of the following theorem is given in [3, Theorem 4.2.1]. Theorem 1.1.23 (Recursion on ℕ). Let a ∈ A and let g: A → A be a function, where A is a set. Then
there exists a unique function h: ℕ → A such that: 1. h(0) = a, 2. h(n + 1) = g(h(n)), for all n ∈ ℕ. 1.1.5 Defining sets by recursion In this section, we shall develop and state a generalization of
the above Theorem 1.1.23. First we must talk about defining sets by recursion. A recursive definition of a set C has the following form: (a) Basis: Specify the “initial” elements of C. (b) Induction:
Give one or more operations for constructing “new” elements of C from “old” elements of C. (c) Closure: The set C is the smallest set that contains the initial elements and is also closed under the
operations (see Theorem 1.1.25). Suppose that U is a set and we are given a function f : U n → U. We say that a set S ⊆ U is closed under f if and only if whenever x1 , x2 , . . . , xn ∈ S we have f
(x1 , x2 , . . . , xn ) ∈ S. For any A ⊆ U, let us define f [A] = {f (x1 , x2 , . . . , xn ) : x1 , x2 , . . . , xn ∈ A}. Now let ℱ be a set of functions ℓ which have the form ℓ: U n → U, where n is
the arity of ℓ. The functions in ℱ can have different arity. For any A ⊆ U let us define ℱ [A] = ⋃{ℓ[A] : ℓ ∈ ℱ }. Given B ⊆ U, in the next theorem, we show how to construct the smallest set C such
that B ⊆ C ⊆ U and C is closed under every function in ℱ . Theorem 1.1.24. Let U be a set and let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. Now let B ⊆ U. Define,
by recursion on ℕ, the following sets:
8 � 1 Basic set theory and basic logic (1) C0 = B, (2) Cn+1 = Cn ∪ ℱ [Cn ] for all n ∈ ℕ. Let C = ⋃n∈ℕ Cn . Then B ⊆ C ⊆ U and C is closed under all the functions in ℱ . Proof. See Exercise 5. The
set C, defined in Theorem 1.1.24, is referred to as the set generated from B by the functions in ℱ . One feature of defining the set C by the above recursion is that it yields an induction principle
that will be frequently applied in the coming pages. Theorem 1.1.25 (Induction principle). Let U be a set and let ℱ be a set of functions ℓ which have the form ℓ: U n → U, where n is the arity of ℓ.
Suppose that C is the set generated from B by the functions in ℱ . If I ⊆ C satisfies (a) B ⊆ I and (b) I is closed under all of the functions in ℱ , then I = C. Proof. One can prove by induction on
n that Cn ⊆ I using (1) and (2) of Theorem 1.1.24. Thus, C ⊆ I, and since I ⊆ C, it follows that I = C. Suppose that U is a set and that f is a function of the form f : U n → U. Let C ⊆ U and suppose
that C is closed under f . Then the function fC : C n → C is defined by fC (x1 , x2 , . . . , xn ) = f (x1 , x2 , . . . , xn ) for all x1 , x2 , . . . , xn ∈ C. Definition 1.1.26. Suppose that U is a
set and ℱ is a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. We shall say that C is freely generated from B by the functions in ℱ if the following hold: 1. the set C is
generated from B by the functions in ℱ , 2. fC is one-to-one for every f ∈ ℱ , 3. the range of fC and B are disjoint, for all f ∈ ℱ , 4. the range of fC and the range of gC are disjoint for all
distinct f , g ∈ ℱ . The following theorem shows that if a set C is freely generated from B, then a function h defined on the initial elements B can be extended to a function h defined on all of the
elements in C. This theorem will justify many results to be covered in the text. The proof requires some special set-theoretic tools and can be summarized as follows: The intersection of all the
approximations to h is in fact h. Theorem 1.1.27 (Recursion theorem). Let U be a set and let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. Let B ⊆ U and let C be freely
generated from B by the functions in ℱ . Let V be a set. Assume that: (a) we are given a function h: B → V , and (b) for each ℓ ∈ ℱ , there is an associated function Fℓ : V n → V of the same arity as
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Then there is a unique function h: C → V such that: (i) h(x) = h(x) for all x ∈ B, (ii) for each ℓ ∈ ℱ , we have h(ℓ(x1 , x2 , . . . , xn )) = Fℓ (h(x1 ), h(x2 ), . . . , h(xn )) for all x1 , x2 , .
. . , xn ∈ C, where n is the arity of ℓ. Proof. Let us call a relation R ⊆ C × V suitable if the following two conditions hold: (1) ⟨b, h(b)⟩ ∈ R, for all b ∈ B, (2) for all ℓ ∈ ℱ of arity n, if x1 ,
x2 , . . . , xn ∈ dom(R), then for each 1 ≤ i ≤ n, we have ⟨xi , yi ⟩ ∈ R for some yi ∈ V and ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ∈ R. Our goal is to construct a suitable relation
that is also a function. Every suitable relation is a subset of C × V . Let 𝒮 = {R : R is suitable}. Because C × V is a suitable relation, 𝒮 is nonempty. Let h = ⋂ 𝒮 . Clearly, h ⊆ C × V , so h is a
relation. We now prove that h is suitable and that it is a function. Claim 1. The relation h is suitable. Proof. We need to show that h satisfies items (1) and (2). Let b ∈ B. Every relation in 𝒮 is
suitable. So ⟨b, h(b)⟩ is an element in every relation in 𝒮 . Thus, ⟨b, h(b)⟩ ∈ ⋂ 𝒮 , that is, ⟨b, h(b)⟩ ∈ h for every b ∈ B. To prove (2), let ℓ ∈ ℱ be of arity n and let x1 , x2 , . . . , xn ∈ dom
(h). Therefore, for each 1 ≤ i ≤ n, ⟨xi , yi ⟩ ∈ h for some yi ∈ V . As h = ⋂ 𝒮 , it follows that each ⟨xi , yi ⟩ belongs to every relation in 𝒮 . Let R ∈ 𝒮 . Since ⟨xi , yi ⟩ ∈ R for all 1 ≤ i ≤ n
and R is suitable, item (2) implies that ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ∈ R. Thus, ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ belongs to every relation in 𝒮 and
therefore, ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ∈ h. Hence, h is suitable. (Claim 1) We now must prove that h is a function with domain C. Claim 2. The relation h is a function from
C to V . Proof. To prove that h is a function from C to V , we must show that for each x ∈ C, there is exactly one y ∈ V such that ⟨x, y⟩ ∈ h. Let I ⊆ C be defined by I = {x ∈ C : there is exactly
one y ∈ V such that ⟨x, y⟩ ∈ h}. We shall prove that I = C by applying Theorem 1.1.25. To do this, we must first prove that B ⊆ I. Let b ∈ B. We know that ⟨b, h(b)⟩ ∈ h, because h is suitable. To
prove that b ∈ I, we need to show there is no ⟨b, y⟩ ∈ h where y ≠ h(b). Suppose, for a contradiction, that ⟨b, y⟩ ∈ h, where y ≠ h(b). Consider the relation R = h \ {⟨b, y⟩}. So () ⟨b, y⟩ ∉ R.
Since C is freely generated from B by the functions in ℱ , it follows that whenever ℓ ∈ ℱ , we have
10 � 1 Basic set theory and basic logic ℓ(x1 , x2 , . . . , xn ) ∉ B for all x1 , x2 , . . . , xn ∈ C. As h is suitable, it thus follows that R is also suitable. Hence, R ∈ 𝒮 . Since h = ⋂ 𝒮 , we
conclude that h ⊆ R, and thus, ⟨b, y⟩ ∈ R, which contradicts (). Thus, h(b) is the only element in V such that ⟨b, h(b)⟩ ∈ h. Therefore, B ⊆ I. Now we show that I is closed under all of the
functions in ℱ . Let g: U k → U be a function in ℱ and let z1 , z2 , . . . , zk ∈ I. We need to show that g(z1 , z2 , . . . , zk ) ∈ I. For each 1 ≤ i ≤ k, since zi ∈ I, there is a unique wi such
that ⟨zi , wi ⟩ ∈ h. Because h is suitable, we conclude that ⟨g(z1 , z2 , . . . , zk ), Fg (w1 , w2 , . . . , wk )⟩ ∈ h. To show that g(z1 , z2 , . . . , zk ) ∈ I, we must show there is no ⟨g(z1 , z2
, . . . , zk ), y⟩ ∈ h where Fg (w1 , w2 , . . . , wk ) ≠ y. Suppose, for a contradiction, that there exists a y ∈ V such that () ⟨g(z1 , z2 , . . . , zk ), y⟩ ∈ h where () Fg (w1 , w2 , . . . , wk
) ≠ y. Consider the relation R = h \ {⟨g(z1 , z2 , . . . , zk ), y⟩}. Clearly, (⧫) ⟨g(z1 , z2 , . . . , zk ), y⟩ ∉ R. We now show that R is suitable. Let b ∈ B. Since C is freely generated from B by
the functions in ℱ , it follows that g(z1 , z2 , . . . , zk ) ≠ b. Since h is suitable, it follows that ⟨b, h(b)⟩ ∈ R. Thus, R satisfies item (1) of the above definition of suitability. To verify
that R satisfies item (2), let ℓ ∈ ℱ be of arity n and let x1 , x2 , . . . , xn ∈ dom(R). Since R ⊆ h, we have x1 , x2 , . . . , xn ∈ dom(h). Since h is suitable, it follows that ⟨ℓ(x1 , x2 , . . . ,
xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ∈ h. So if ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ≠ ⟨g(z1 , z2 , . . . , zk ), y⟩, then ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ ∈ R.
On the other hand, if ⟨ℓ(x1 , x2 , . . . , xn ), Fℓ (y1 , y2 , . . . , yn )⟩ = ⟨g(z1 , z2 , . . . , zk ), y⟩, (♣) ℓ(x1 , x2 , . . . , xn ) = g(z1 , z2 , . . . , zk ) and
() Fℓ (y1 , y2 , . . . , yn ) = y.
Because C is freely generated from B by the functions in ℱ , (♣) implies that ℓ = g, n = k, and x1 = z1 , x2 = z2 , . . . , xn = zk . Therefore, x1 , x2 , . . . , xn ∈ I and thus, y1 = w1 , y2 = w2 ,
. . . , yn = wk . Hence, Fℓ (y1 , y2 , . . . , yn ) = Fg (w1 , w2 , . . . , wk ), which contradicts () and (). Thus, R is suitable. So R ∈ 𝒮 . Since h = ⋂ 𝒮 , we conclude that h ⊆ R so ⟨g(z1 , z2 ,
. . . , zk ), y⟩ ∈ R (see ()), which contradicts (⧫). Therefore,
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g(z1 , z2 , . . . , zk ) ∈ I. Theorem 1.1.25 now implies that I = C and, as a result, h is a function from C to V . (Claim 2) Since the function h is suitable, it satisfies conditions (i) and (ii)
given in the statement of the theorem. To prove that h is unique, let g: C → V also satisfy properties (i) and (ii). Thus, g is a suitable relation, so g ∈ 𝒮 . Since h = ⋂ 𝒮 , we have h ⊆ g. As h and
g are functions with domain C, Lemma 1.1.21 implies that h = g. Hence, h is unique. (Theorem) The function h in Theorem 1.1.27, satisfying (1) and (2), is often said to be defined by recursion. If a
function is defined by recursion, then proofs of statements about this function typically use “proof by induction.” The proof of Theorem 1.1.27 requires that C be freely generated. To illustrate
this, let f : ℕ → ℕ and g: ℕ → ℕ be defined by f (n) = n2 and g(n) = n3 . Let B = {0, 1, 2}. Let C be the set generated from B by the functions in ℱ = {f , g}. The functions in ℱ are one-toone.
However, their ranges are not disjoint and their ranges are not disjoint with B. So C is not freely generated. Now let Ff : ℝ → ℝ be the identity function and let Fg : ℝ → ℝ be the cube root
function. Let h: {0, 1, 2} → ℝ be such that h(1) = 3. If h could be extended to a function h: C → ℝ as in Theorem 1.1.27, then we would have h(f (1)) = Ff (h(1)) = Ff (3) = 3, 3
h(g(1)) = Fg (h(1)) = Fg (3) = √3. 3 Since f (1) = g(1) = 1, we conclude that h(1) = 3 and h(1) = √3. Thus, no such function h exists.
1.1.6 Countable sets Let ℕ = {0, 1, 2, 3, . . . } be the set of natural numbers. A set is countable if it has the same size as some subset of ℕ. In other words, a set is countable if there is a
one-to-one correspondence between the set and a subset of ℕ. Our next definition expresses this concept in mathematical terms. Definition 1.1.28. A set A is countable if and only if there exists a
one-to-one function f : A → ℕ. The following theorem will be used to prove that the set of all finite sequences of a countable set is also countable (see Theorem 1.1.30). A proof of this theorem is
given in [1, Theorem 4.4.7]. Theorem 1.1.29 (Fundamental theorem of arithmetic). For every natural number n > 1, there exist distinct primes p1 , p2 , . . . , pk together with natural numbers a1 ≥ 1,
a2 ≥ 1, . . . , ak ≥ 1 such that
12 � 1 Basic set theory and basic logic a
n = p1 1 p2 2 ⋅ ⋅ ⋅ pkk . Furthermore, given any prime factorization into distinct primes, b
n = q1 1 q2 2 ⋅ ⋅ ⋅ qℓ ℓ we have ℓ = k, the primes qi are the same as the primes pj (except for order), and the corresponding exponents are the same. Theorem 1.1.30. Let A be a set and let S be the
set of all finite sequences of elements of A. If A is countable, then S is countable. Proof. Let A be a countable set and let f : A → ℕ be one-to-one. Let S be the set of all finite sequences of
elements of A. Thus, S = ⋃ An+1 . n∈ℕ
Let h: S → ℕ be defined as follows: Let ⟨a1 , a2 , . . . , am ⟩ ∈ S. Define h(⟨a1 , a2 , . . . , am ⟩) = 2f (a1 )+1 ⋅ 3f (a2 )+1 ⋅ 5f (a3 )+1 ⋅ ⋅ ⋅ pfm(am )+1 , where pm is the m-th prime. To prove
that h: S → ℕ is one-to-one, let ⟨a1 , a2 , . . . , am ⟩ and ⟨b1 , b2 , . . . , bn ⟩ be arbitrary elements of S and assume that h(⟨a1 , a2 , . . . , am ⟩) = h(⟨b1 , b2 , . . . , bn ⟩). Thus f (am )+1
2f (a1 )+1 ⋅ 3f (a2 )+1 ⋅ 5f (a3 )+1 ⋅ ⋅ ⋅ pm = 2f (b1 )+1 ⋅ 3f (b2 )+1 ⋅ 5f (b3 )+1 ⋅ ⋅ ⋅ pfn(bn )+1 .
By Theorem 1.1.29, we conclude that m = n and f (a1 ) + 1 = f (b1 ) + 1, f (a2 ) + 1 = f (b2 ) + 1, . . . , f (am ) + 1 = f (bn ) + 1. Because f is one-to-one, we see that a1 = b1 , a2 = b2 , . . . ,
am = bn . Hence, ⟨a1 , a2 , . . . , am ⟩ = ⟨b1 , b2 , . . . , bn ⟩, and thus h: S → ℕ is one-to-one. Therefore, S is countable.
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Definition 1.1.31. Let A be a nonempty set. We shall say that A is denumerable if and only if there is an enumeration a1 , a2 , a3 , . . . , an , . . .
of all of the elements in A, that is, every element in A appears in the above list indexed by the positive natural numbers. So a set is denumerable if we can list the elements of the set in the same
way that we list the set of nonzero natural numbers, namely, 1, 2, 3, 4, 5, . . . . However, our definition of a denumerable set allows elements in the list (1.1) to be repeated, or not to be
repeated. We will show below that a countable set is denumerable. Using an enumeration of a set will allow us to construct new sets that will be useful in Chapters 2 and 4 (for example, see the proof
of Theorem 2.4.2 on page 44). Theorem 1.1.32. Let A be a nonempty countable set. Then there is a function g: ℕ → A that is onto A. Proof. Assume that A is a nonempty countable set. Thus, there is a
function f : A → ℕ that is one-to-one. We shall use the function f to define our desired function g: ℕ → A. Let c ∈ A be some fixed element. Define g as follows: For each n ∈ ℕ, a, if n is in the
range of f and f (a) = n, g(n) = { c, otherwise. Because f : A → ℕ is one-to-one, one can show that g is a function and it is onto. Corollary 1.1.33. Let A be a nonempty countable set. Then A is
denumerable. Proof. Assume that A is a nonempty countable set. Theorem 1.1.32 implies that there is a function g: ℕ → A that is onto. For each n ≥ 1 let an = g(n − 1). Since g is onto, it follows
that the enumeration a1 , a2 , a3 , . . . , an , . . . lists every element in A. 1.1.7 Cardinality The cardinality of a set is a measure of how many elements are in the set. For example, the set A =
{1, 2, 3, . . . , 10} has 10 elements, so the cardinality of A is 10, denoted by |A| = 10. The cardinality of an infinite set X will also be denoted by |X|. In this section, we briefly discuss Georg
Cantor’s method for measuring the size of an infinite set without the use of numbers. There are two infinite sets where one of these sets has cardinality much larger than the other infinite set.
Therefore, it is possible for one infinite set to have “many more” elements than another infinite set. What does it mean to say that two sets have the same cardinality, that is, the same size? Cantor
discovered a simple answer to this question.
14 � 1 Basic set theory and basic logic Definition 1.1.34. For sets A and B, we say that A has the same cardinality as B, denoted by |A| =c |B|, if there is a bijection f : A → B. The expression |A|
=c |B| looks like an equation; however, the assertion |A| =c |B| should be viewed only as an abbreviation for the statement “A has the same cardinality as B.” In other words, |A| =c |B| means that
“there is a function f : A → B that is one-to-one and onto B.” The relationship =c , given in Definition 1.1.34, is reflexive, symmetric, and transitive. The following theorem is very useful for
proving many results about cardinality. The theorem states that if there are functions f : A → B and g: B → A that are both oneto-one, then there exists a function h: A → B that is one-to-one and
onto B. Theorem 1.1.35 (Schröder–Bernstein). Let A and B be any two sets. Suppose that |A| ≤c |B| and |B| ≤c |A|. Then |A| =c |B|.
1.1.8 The axiom of choice Suppose that a set S contains only nonempty sets. Is it possible to uniformly select exactly one element from each set in S? In other words, is there a function F so that
for each A ∈ S, we have F(A) ∈ A? The following set-theoretic principle will allow us to positively answer this question. Axiom of Choice. Let 𝒞 be a set of nonempty sets. Then there is a function H:
𝒞 → ⋃ 𝒞 such that H(A) ∈ A for all A ∈ 𝒞 . Cantor developed a theory of infinite cardinal numbers which, assuming the axiom of choice, allows one to measure the size of any infinite set. Thus, if A
is an infinite set, there is a cardinal number κ such that κ = |A|. Moreover, there is an ordering on these cardinal numbers such that for any cardinal numbers λ and κ, we have λ ≤ κ or κ ≤ λ.
Moreover, if λ ≤ κ and κ ≤ λ, then λ = κ. The cardinality of any countable infinite set is denoted by the cardinal ℵ0 . Zorn’s lemma is an important theorem about sets that is normally used to prove
the existence of a mathematical object when it cannot be explicitly identified. The lemma involves the concept of a chain. A chain is a collection of sets C such that for all sets x and y in C, we
have either x ⊆ y or y ⊆ x. Zorn’s Lemma 1.1.36. Let ℱ be a set of sets. Suppose that for every chain C ⊆ ℱ , ⋃ C is in ℱ . Then there exists an M ∈ ℱ that is maximal, that is, M is not the proper
subset of any A ∈ ℱ . Surprisingly, Zorn’s lemma is equivalent to the axiom of choice. There will be times when we may apply the axiom of choice or Zorn’s lemma. However, when working with countable
sets, Corollary 1.1.33 will allow us to avoid such applications.
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Exercises 1.1. 1. Let A be a set and let ℱ be a nonempty set of sets. Prove the following: (1) A \ ⋃ ℱ = ⋂{A \ C : C ∈ ℱ }, (2) A \ ⋂ ℱ = ⋃{A \ C : C ∈ ℱ }. 2. Consider the function h: ℕ → ℕ defined
by the recursion (a) h(0) = 0, (b) h(n + 1) = 5h(n) + 1, for all n ∈ ℕ. n Prove by induction that h(n) = 5 4−1 for all n ∈ ℕ. 3. Let A and B be countable sets. Since A is countable, there is a
one-to-one function f : A → ℕ. Also, as B is countable, there is a one-to-one function g: B → ℕ. (a) Prove that the function h: A → ℕ defined by h(a) = 2f (a)+1 is one-to-one. (b) Prove that the
function k: B → ℕ defined by k(b) = 3g(b)+1 is one-to-one. (c) Define the function ℓ: A ∪ B → ℕ by h(x),
ℓ(x) = {
if x ∈ A,
if x ∈ B \ A,
for each x ∈ A ∪ B. Prove that ℓ is one-to-one. Thus, A ∪ B is countable. 4. Let ℓ: ℕ × ℕ → ℕ be defined by ℓ(x, y) = 2x+1 ⋅ 3y+1 and let g: ℕ → ℕ be defined by g(x) = 2x+1 . Let C be the set
generated from B = {1, 2} by the set ℱ = {ℓ, g}. (a) Find C1 and then find an element in C2 . (b) Show that ℓC and gC are one-to-one (use Theorem 1.1.29). (c) Show that the set C is freely generated
by B. (d) Let h: B → ℝ be defined by h(1) = π and h(2) = e. Let Fℓ : ℝ × ℝ → ℝ and Fg : ℝ → ℝ be defined by Fℓ (x, y) = x + y and Fg (x) = 5x. Let h: C → ℝ be the function given by Theorem 1.1.27.
Find h(36), h(4), and h(72). *5. Let U be a set and let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. Let B ⊆ U and let C be generated from B by the functions in ℱ .
(a) Prove that C is closed under the functions in ℱ . (b) Let B ⊆ D, where D is closed under the functions in ℱ . Prove that C ⊆ D. 6. Let U be a set and let ℱ be a set of functions ℓ of the form ℓ:
U n → U, where n is the arity of ℓ. Let B ⊆ U and let C be freely generated from B by the functions in ℱ . Let f , g ∈ ℱ , x1 , x2 , . . . , xn ∈ C, and z1 , z2 , . . . , zk ∈ C. Suppose that f (x1 ,
x2 , . . . , xn ) = g(z1 , z2 , . . . , zk ). Explain why one can conclude that f = g, n = k, and x1 = z1 , x2 = z2 , . . . , xn = zk . 7. In the proof of Theorem 1.1.25, show that Cn ⊆ I for all n ∈
ℕ. *8. Let U be a set and let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. Let B ⊆ B′ ⊆ U. Let C be generated from B by the functions in ℱ and let C ′ be generated
from B′ by the functions in ℱ . Prove that C ⊆ C ′ . *9. Let U be a set and let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. For each ℓ ∈ ℱ , let Fℓ : V n → V be of
the same arity as ℓ. Let B ⊆ B′ ⊆ U. Let C be freely generated from B by the functions in ℱ and let C ′ be freely generated
16 � 1 Basic set theory and basic logic from B′ by the functions in ℱ . Let h: B → V and g: B′ → V be such that h(b) = g(b) for all b ∈ B. Show that h(x) = g(x) for all x ∈ C. *10. Let U be a set and
let ℱ be a set of functions ℓ of the form ℓ: U n → U, where n is the arity of ℓ. For each ℓ ∈ ℱ , let Fℓ : V n → V be of the same arity as ℓ. Let B ⊆ U, let h: B → V , and let C be freely generated
from B by the functions in ℱ as in Theorem 1.1.24, that is, C = ⋃n∈ℕ Cn , where: (1) C0 = B, (2) Cn+1 = Cn ∪ ℱ [Cn ] for all n ∈ ℕ. Suppose that: – the function h is one-to-one; – for each ℓ ∈ ℱ , Fℓ
is one-to-one and the functions h, Fℓ have disjoint ranges; – Fℓ and Fℓ′ have disjoint ranges whenever ℓ, ℓ′ ∈ ℱ are distinct. Let h: C → V be as in Theorem 1.1.27. Prove by induction on n that h is
one-to-one on Cn . Then conclude that h is one-to-one. Exercise Notes: For Exercise 8, let I = C ∩ C ′ . Using Theorem 1.1.25, prove that I = C. For Exercise 9, C ⊆ C ′ by Exercise 8. Let I = {x ∈ C
: h(x) = g(x)}. Using Theorem 1.1.25, prove that I = C. For Exercise 10, if c ∈ Cn+1 , let k be the least such that c ∈ Ck . If k > 0, then c is in the range of a function in ℱ restricted to Ck−1 .
1.2 Basic logic 1.2.1 Propositions and logical connectives A proposition is a declarative sentence that is either true or false, but not both. When discussing the logic of propositional statements in
this section, we shall use symbols to represent these statements. Capital letters, for instance, P, Q, R, are used to symbolize propositional statements which may be called propositional components.
Using the five logical connectives ∧, ∨, ¬, →, ↔ together with the components, we can form new logical sentences called compound sentences. For example: 1. P ∧ Q (means “P and Q” and is called a
conjunction), 2. P ∨ Q (means “P or Q” and is called a disjunction), 3. ¬P (means “not P” and is called a negation), 4. P → Q (means “if P, then Q” and is called a conditional), 5. P ↔ Q (means “P if
and only if Q” and is called a biconditional). Using the propositional components and the logical connectives, one can construct more complicated sentences, for example, ((P ∧ (¬Q)) ∨ (S → (¬R))). We
will more formally investigate propositional logic in Chapter 2.
1.2 Basic logic
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1.2.2 Truth tables and truth functions Given a collection of propositional components, say P, Q, and R, we can assign truth values to these components. For example, we can assign the truth values T,
F, T to P, Q, R, respectively, where T means “true” while F means “false.” The truth value of a sentence of propositional logic can be evaluated from the truth values assigned to its components. We
shall explain what this “means” by using truth tables. The logical connectives ∧, ∨, ¬ yield the natural truth values given by Table 1.1. Table 1.1: Truth tables for conjunction, disjunction, and
negation. (1) Conjunction P Q T T F F
T F T F
P∧Q T F F F
(2) Disjunction P Q T T F F
T F T F
P∨Q T T T F
(3) Negation P ¬P T F
F T
Table 1.1(1) has four rows (not including the header). The columns beneath P and Q list all the possible pairs of truth values that can be assigned to the components P and Q. For each such pair, the
corresponding truth value for P ∧ Q appears to the right. For example, consider the third pair of truth values in this table, FT. Thus, if the propositional components P and Q are assigned the
respective truth values F and T, we see that the truth value of P ∧ Q is F. Table 1.1(2) shows that if P and Q are assigned the respective truth values T and F, then the truth value of P ∨ Q is T.
Moreover, when P and Q are assigned the truth values T and T, the truth value of P ∨ Q is also T. In mathematics, the logical connective “or” has the same meaning as “and/or,” that is, P ∨ Q is true
if and only if only P is true, only Q is true, or both P and Q are true. Table 1.1(3) shows that the negation of a statement reverses the truth value of the statement. A truth function accepts truth
values as input and yields a unique truth value as output. Let V = {T, F}. The above three truth tables yield the corresponding truth functions F∧ : V 2 → V , F∨ : V 2 → V , and F¬ : V → V , defined
by T, if x = T and y = T, F∧ (x, y) = { F, otherwise,
T, if x = T or y = T, F∨ (x, y) = { F, otherwise,
F¬ (x) = {
if x = F,
18 � 1 Basic set theory and basic logic The standard truth tables for the conditional and biconditional connectives are given in Table 1.2, which states that when P and Q are assigned the respective
truth values T and F, then the truth value of P → Q is F; otherwise, it is T. In particular, when P is false, we shall say that P → Q is vacuously true. Table 1.2(5) shows that P ↔ Q is true when P
and Q are assigned the same truth value; when P and Q have different truth values, then the biconditional is false. Table 1.2: Truth tables for the conditional and biconditional. (4) Conditional P Q
T T F F
T F T F
P→Q T F T T
(5) Biconditional P Q T T F F
T F T F
P↔Q T F F T
Let V = {T, F}. The conditional and biconditional truth tables yield the two truth functions F→ : V 2 → V and F↔ : V 2 → V , defined by F,
F→ (x, y) = {
T, T,
F↔ (x, y) = {
if x = T and y = F, otherwise, if x = y,
(1.5) (1.6)
Using the truth tables for the sentences P ∧ Q, P ∨ Q, ¬P, P → Q, and P ↔ Q, one can build truth tables for more complicated compound sentences. Given a compound sentence, the “outside” connective is
the “last connective that one needs to evaluate.” After the outside connective has been determined, one can break up the sentence into its “parts.” For example, in the compound sentence ¬P ∧ (Q ∨ P)
we see that ∧ is the outside connective with two parts ¬P and Q ∨ P. Problem 1.2.1. Construct a truth table for the sentence ¬P → (Q ∧ P). Solution. The two components P and Q will each need a column
in our truth table. Since there are two components, there are four truth assignments for P and Q. We will enter these combinations in the two leftmost columns in the same order as in Table 1.1(1).
The outside connective of the propositional sentence ¬P → (Q ∧ P) is →. We can break this sentence into the two parts ¬P and Q ∧ P. So these parts will also need a column in our truth table. As we
can break the sentences ¬P and Q ∧ P only into components (namely, P and Q), we obtain the following truth table:
1.2 Basic logic
Step #
T T F F
T F T F
F F T T
T F F F
T T F F
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¬P → (Q ∧ P)
We will now describe in steps how to obtain the truth values in the above table. Step 1: Specify all of the truth values that can be assigned to the components. Step 2: In each row, use the truth
value assigned to the component P to obtain the corresponding truth value for ¬P, using Table 1.1(3). Step 3: In each row, use the truth values assigned to Q and P to determine the corresponding
truth value in the column under Q ∧ P via Table 1.1(1). Step 4: In each row, use the truth values assigned to ¬P and Q ∧ P to evaluate the matching truth value for the final column under the sentence
¬P → (Q∧P), employing Table 1.2(4).
1.2.3 Predicates and quantifiers Variables are used throughout mathematics and logic to represent unspecified values. They are used when one is interested in “properties” that may be true or false
depending on the values represented by the variables. A predicate is just a statement that asserts that certain variables satisfy a property. For example, “x is an irrational number” is a predicate.
We can symbolize this predicate as Ix, where I is called a predicate symbol. Of course, the truth or falsity of the expression Ix can be evaluated only when a value for x is given. For example, if x
is given the value √2, then Ix would be true, whereas if x is given the value 2, then Ix would be false. When our attention is focused on just the elements in a particular set, we will refer to that
set as our universe of discourse. For example, if we were just talking about real numbers, then our universe of discourse would be the set of real numbers ℝ. Every statement made in a specific
universe of discourse applies only to the elements in that universe. Given a predicate Px and variable x, we may want to assert that every element x in the universe of discourse satisfies Px. We may
also want to express the fact that at least one element x in the universe makes Px true. We can then form logical sentences using the quantifiers ∀ and ∃. The quantifier ∀ means “for all” and is
called the universal quantifier. The quantifier ∃ means “there exists” and it is referred to as the existential quantifier. For example, we can form the sentences: 1. ∀xPx (which means “for all x,
Px”), 2. ∃xPx (which means “there exists an x such that Px”).
20 � 1 Basic set theory and basic logic A statement of the form ∀xPx is called a universal statement. Any statement having the form ∃xPx is called an existential statement. Quantifiers offer a
valuable tool for clear thinking in mathematics, where many ideas begin with the expression “for every” or “there exists.” Of course, the truth or falsity of a quantified statement depends on the
particular universe of discourse. Let x be a variable that appears in a predicate Px. In the statements ∀xPx and ∃xPx, we say that x is a bound variable because x is bound by a quantifier. In other
words, when every occurrence of a variable in a statement is attached to a quantifier, then that variable is called a bound variable. If a variable appears in a statement and it is not bound by a
quantifier, then the variable is said to be a free variable. Whenever a variable is free, substitution may take place, that is, one can replace a free variable with any particular value from the
universe of discourse—perhaps 1 or 2. For example, the assertion ∀x(Px → x = y) has the one free variable y. Therefore, we can perform a substitution to obtain ∀x(Px → x = 1). In a given context, if
all of the free variables in a statement are replaced with values from the universe of discourse, then one can determine the truth or falsity of the resulting statement. Problem 1.2.2. Consider the
predicates (properties) Px, Ox, and Dxy, where the variables x and y are intended to represent natural numbers: 1. Px represents the statement “x is a prime number,” 2. Ox represents the statement “x
is an odd natural number,” 3. Dxy represents the statement “x evenly divides y.” Using the above predicates, determine whether or not the following logical formulas are true or false, where the
universe of discourse is the set of natural numbers: 1. ∀x(Px → Ox), 2. ∀y(y ≥ 2 → ∃x(Px ∧ Dxy)). Solution. The expression ∀x(Px → Ox) states that “all prime numbers are odd,” which is clearly false.
On the other hand, item 2 states that “every natural number greater than or equal to 2 is divisible by a prime number” and this is true. We will formally investigate predicates and quantifiers in
Chapter 3. The understanding of the logic of quantifiers is (one can argue) a requisite tool for the study of mathematics and logic. To help students learn the language of quantifier logic, Jon
Barwise and John Etchemendy created an innovative software program called Tarski’s World. This program presents a visual display of geometric shapes sitting on a grid, referred to as a world (or
universe). The shapes have a variety of colors and positions on the grid. The user can create logic formulas and then determine whether or not these formulas are true or false in the world. Tarski’s
World is named after Alfred Tarski, an early pioneer in mathematical logic. We end this section with our own version of Tarski’s World. In the following problem, we are given a Tarskian World and
1.2 Basic logic
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English statements. We will be given some predicates and be asked to translate these statements into logical form. Problem 1.2.3. Consider the following Tarskian World, where some of the individuals
are labeled with a name. The universe consists of all the objects in this Tarskian world.
Define the following predicates (properties): – Tx means “x is a triangle.” Cx means “x is a circle.” Sx means “x is a square.” – Ix means “x is white.” Gx means “x is gray.” Bx means “x is black.” –
Nxy means “x is on the northern side of y.” – Wxy means “x is on the western side of y.” – Kxy means “x has the same color as y.” The constants a, c, d, g, j are the names of five individuals in the
above world. Using the given predicates, write each of the following statements in logical form, looking for possible hidden quantifiers and logical connectives. 1. There is a black square. 2. Every
circle is white. 3. There are no black circles. 4. a is north of c. 5. a is not north of j. 6. Every circle is north of d. 7. Some circle has the same color as d. 8. d is west of every circle.
Solution. Statements 2, 6, and 7 are the only ones that are false in the given Tarskian world. We express the above sentences in the following logical forms: 1. The sentence means that “for some x, x
is black and x is a square.” In logical form, we have ∃x(Bx ∧ Sx). 2. The sentence means that “for all x, if x is a circle, then x is white.” In logical form, ∀x(Cx → Ix).
22 � 1 Basic set theory and basic logic 3.
4. 5. 6. 7. 8.
The sentence can be stated in two equivalent ways. First, the sentence means that “it is not the case that some circle is black,” that is, ¬(some circle is black). In logical form, we obtain ¬∃x(Cx ∧
Bx). Second, the sentence also means that “every circle is not black” and we get ∀x(Cx → ¬Bx). In logical form, the sentence becomes Nac. The logical form of this sentence is ¬Naj. Rephrasing, we
obtain “for all x, if x is a circle, then x is north of d.” In logical form, we have ∀x(Cx → Nxd). The sentence asserts that “for some x, x is a circle and x has the same color as d.” In logical
form, we have ∃x(Cx ∧ Kxd). Stated more clearly, we obtain “for all x, if x is a circle, then d is west of x.” In logical form, we have ∀x(Cx → Wdx).
There will be cases when we will have to prove that there is exactly one value that satisfies a property. There is another quantifier that is sometimes used. It is called the uniqueness quantifier.
This quantifier is written as ∃!xPx, which means that “there exists a unique x satisfying Px,” whereas ∃xPx simply asserts that “at least one x satisfies Px.” The quantifier ∃! can be expressed in
terms of the other quantifiers ∃ and ∀. In particular, the statement ∃!xPx is equivalent to ∃xPx ∧ ∀x∀y((Px ∧ Py) → x = y), because the above statement means that “there is an x such that Px holds,
and any individuals x and y that satisfy Px and Py must be the same individual.” Quantifier negation laws We now introduce logic laws that involve the negation of quantifiers. Let Px be any
predicate. The statement ∀xPx means that “for every x, Px is true.” Thus, the assertion ¬∀xPx means that “it is not the case that every x makes Px true.” Therefore, ¬∀xPx means there is an x that
does not make Px true, which can be expressed as ∃x¬Px. This reasoning is reversible, as we will now show. The assertion ∃x¬Px means that “there is an x that makes Px false.” Hence, Px is not true
for every x, that is, ¬∀xPx. Therefore, ¬∀xPx and ∃x¬Px are logically equivalent. Similar reasoning shows that ¬∃xPx and ∀x¬Px are also equivalent. We now formally present these important logic laws
that connect quantifiers with negation. Quantifier Negation Laws 1.2.4. The following logical equivalences hold for any predicate Px: 1. ¬∀xPx ⇔ ∃x¬Px, 2. ¬∃xPx ⇔ ∀x¬Px, 3. ¬∀x¬Px ⇔ ∃xPx, 4. ¬∃x¬Px ⇔
1.2 Basic logic
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We will be using the symbols ⇔ and ⇒ to abbreviate two English expressions. The symbol ⇔ denotes the phrase “is equivalent to” and ⇒ denotes the word “implies.” Quantifier interchange laws Adjacent
quantifiers have four forms: ∃x∃y, ∀x∀y, ∀x∃y, and ∃x∀y. How should one interpret statements that contain adjacent quantifiers? If a statement contains adjacent quantifiers, one should address the
quantifiers, one at a time, in the order in which they are presented. This is illustrated in our solutions of the following three problems. Problem 1.2.5. Let the universe of discourse be a group of
people and let Lxy mean “x likes y.” What do the following formulas mean in English? 1. ∃x∃yLxy, 2. ∃y∃xLxy. Solution. Note that “x likes y” also means that “y is liked by x.” We will now translate
each of these formulas from “left to right” as follows: 1. ∃x∃yLxy means “there is a person x such that ∃yLxy,” that is, “there is a person x who likes some person y.” Therefore, ∃x∃yLxy means that
“someone likes someone.” 2. ∃y∃xLxy states that “there is a person y such that ∃xLxy,” that is, “there is a person y who is liked by some person x.” Thus, ∃y∃xLxy means that “someone is liked by
someone.” Hence, the statements ∃x∃yLxy and ∃y∃xLxy mean the same thing. Problem 1.2.6. Let the universe of discourse be a group of people and let Lxy mean “x likes y.” What do the following formulas
mean in English? 1. ∀x∀yLxy, 2. ∀y∀xLxy. Solution. We will work again from “left to right” as follows: 1. ∀x∀yLxy means “for every person x, we have ∀yLxy,” that is, “for every person x, x likes
every person y”; hence, ∀x∀yLxy means that “everyone likes everyone”; 2. ∀y∀xLxy means that “for each person y, we have ∀xLxy,” that is, “for each person y, y is liked by every person x”; thus,
∀y∀xLxy means “everyone is liked by everyone.” So the statements ∀x∀yLxy and ∀y∀xLxy mean the same thing. Adjacent quantifiers of a different type are referred to as mixed quantifiers. Problem 1.2.7.
Let the universe of discourse be a group of people and let Lxy mean “x likes y.” What do the following mixed quantifier formulas mean in English? 1. ∀x∃yLxy, 2. ∃y∀xLxy.
24 � 1 Basic set theory and basic logic Solution. We will translate the formulas as follows: 1. ∀x∃yLxy asserts that “for every person x, we have ∃yLxy,” that is, “for every person x, there is a
person y such that x likes y.” Thus, ∀x∃yLxy means that “everyone likes someone.” 2. ∃y∀xLxy states that “there is a person y such that ∀xLxy,” that is, “there is a person y who is liked by every
person x.” In other words, ∃y∀xLxy means “someone is liked by everyone.” We conclude that the mixed quantifier statements ∀x∃yLxy and ∃y∀xLxy are not logically equivalent, that is, they do not mean
the same thing. To clarify the conclusion that we obtained in our solution of Problem 1.2.7, consider the universe of discourse U = {a, b, c, d}, which consists of just four individuals with names as
given. Figure 1.1 presents a world in which ∀x∃yLxy is true, where we porlikes
tray the property Lxy using the “arrow notation” x → y. In this world, “everyone likes someone.”
Figure 1.1: A world where ∀x∃yLxy is true, because everyone likes someone.
Let us focus our attention on Figure 1.1. Clearly, the statement ∀x∃yLxy is true in the world depicted in this figure. Moreover, note that ∃y∀xLxy is actually false in this world. This is the case
because there is no individual whom everyone likes. Thus, ∀x∃yLxy is true and ∃y∀xLxy is false in the world of Figure 1.1. We can now conclude that ∀x∃yLxy and ∃y∀xLxy do not mean the same thing. Our
solution to Problem 1.2.5 shows that assertions ∃x∃yLxy and ∃y∃xLxy both mean “someone likes someone.” This supports the equivalence: ∃x∃yLxy ⇔ ∃y∃xLxy. Similarly, Problem 1.2.6 confirms the
equivalence: ∀x∀yLxy ⇔ ∀y∀xLxy. Therefore, interchanging adjacent quantifiers of the same kind does not change the meaning. Problem 1.2.7, however, demonstrates that the two statements ∀y∃xLxy and
∃x∀yLxy are not equivalent. We conclude this discussion with a summary of the above observations:
1.2 Basic logic
– –
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Adjacent quantifiers of the same type are interchangeable. Adjacent quantifiers of a different type may not be interchangeable.
Quantifier Interchange Laws 1.2.8. For every predicate Pxy, the following three statements are valid: 1. ∃x∃yPxy ⇔ ∃y∃xPxy, 2. ∀x∀yPxy ⇔ ∀y∀xPxy, 3. ∃x∀yPxy ⇒ ∀y∃xPxy. It should be noted that the
implication in item 3 cannot, in general, be reversed. Quantifier distribution laws A quantifier can sometimes “distribute” over a conjunction or a disjunction. The quantifier distribution laws,
given below, express relationships that hold between a quantifier and the two logical connectives ∨ and ∧. Namely, the existential quantifier distributes over disjunction (see 1.2.9(1)), and the
universal quantifier distributes over conjunction (see 1.2.9(2)). The following four quantifier distribution laws can be useful when proving certain set identities. Quantifier Distribution Laws
1.2.9. For any predicates Px and Qx, we have the following distribution laws: 1. ∃xPx ∨ ∃xQx ⇔ ∃x(Px ∨ Qx), 2. ∀xPx ∧ ∀xQx ⇔ ∀x(Px ∧ Qx). If R is a predicate that does not involve the variable x,
then we have: 3. R ∧ ∃xQx ⇔ ∃x(R ∧ Qx), 4. R ∨ ∀xQx ⇔ ∀x(R ∨ Qx). Exercises 1.2. 1. Let C(x) represent the predicate “x is in the class” and let Mx represent “x is a mathematics major.” Let the
universe of discourse be the set of all students. Analyze the logical form of the following sentences. (a) Everyone in the class is a mathematics major. (b) Someone in the class is a mathematics
major. (c) No one in the class is a mathematics major. 2. Determine whether the following statements are true or false in the universe ℝ: (a) ∀x(x 2 + 1 > 0), (b) ∀x(x 2 + x ≥ 0), (c) ∀x(x > 21 → x1
< 3), 1 (d) ∃x( x−1 = 3), 1 (e) ∃x( x−1 = 0). 3. Given the properties and the Tarskian world in Problem 1.2.3 on page 21, determine the truth or falsity of the following statements:
26 � 1 Basic set theory and basic logic (a) ∀x(Ix → (Tx ∨ Sx)), (b) ∀x(Bx → (Tx ∨ Sx)), (c) ∃y(Cy ∧ Nyd), (d) ∃y(Cy ∧ Ndy), (e) ∃y(Cy ∧ (Nyd ∧ Ndy)). 4. Using the Tarskian predicates given in Problem
1.2.3, translate the following English sentences into logical sentences where b and d are names of individuals in some Tarskian world: (a) Something is white. (b) Some circle is white. (c) All
squares are black. (d) No squares are black. (e) All triangles are west of d. (f) A triangle is west of d. (g) Some triangle is north of d. (h) Some triangle is not gray. (i) Every triangle is north
of b. (j) No square has the same color as b.
2 Propositional logic In this chapter we will utilize mathematical tools, namely induction and recursion, to formally investigate an important form of logic called propositional logic. In particular,
we will establish a number of significant theorems concerning the syntax and semantics of propositional logic. An informal introduction to this topic is given in Section 1.2.1.
2.1 The language The syntax of the language of propositional logic is specified by first identifying the symbols (the alphabet) of the language and then defining expressions to be any finite string
of these symbols. Some expressions are meaningful while others are not. To single out the meaningful expressions requires a recursive definition (see Section 1.1.5). The meaningful expressions in
propositional logic will be called well-formed formulas (wffs). Definition 2.1.1. Our official language for propositional logic, denoted by ℒ, consists of the distinct symbols in the set ℒ = { (, ),
¬, ∧, ∨, →, ↔, A1 , A2 , . . . , An , . . . },
where the symbols in the language ℒ are described in the following table: Symbol
( ) ¬ ∧ ∨ → ↔ A� A� .. . An .. .
left parenthesis right parenthesis negation conjunction disjunction conditional biconditional first sentence symbol second sentence symbol .. . n-th sentence symbol .. .
punctuation punctuation English: not English: and English: or English: if , then English: if and only if
Remark. Several remarks are in order: 1. The distinct symbols ¬, ∧, ∨, →, ↔ are called logical connectives. 2. The infinite number of distinct symbols A1 , A2 , . . . , An , . . . are called sentence
symbols. Each Ai stands for a particular sentence. Recall that ⟨a1 , a2 , . . . , an ⟩ denotes a sequence (see Definition 1.1.6). https://doi.org/10.1515/9783110782073-002
28 � 2 Propositional logic Definition 2.1.2. An ℒ-expression is a finite string of symbols from ℒ obtained by omitting the sequence brackets ⟨ ⟩ and the commas from a finite sequence of symbols from
ℒ. If α and β are ℒ-expressions, then αβ is the ℒ-expression consisting of the symbols in the expression α followed by the symbols in the expression β. Example 2.1.3. Clearly, ⟨(, ¬, A1 , )⟩ is a
finite sequence of symbols from ℒ, and after removing the sequence brackets and commas we obtain the ℒ-expression (¬A1 ). Let α = (¬A1 ) and β = A1 . Then α and β are ℒ-expressions. In addition, we
can write (α → β) to denote the ℒ-expression ((¬A1 ) → A1 ). Also, note that ¬ →)(A8 is an ℒ-expression. Moreover, αβ is the expression (¬A1 )A1 . There is a one-to-one correspondence between each
ℒ-expression α and a finite sequence of symbols in ℒ denoted by ⟨α∗ ⟩, where α∗ denotes the result of putting commas between all of the symbols in α. So an ℒ-expression α is a proper initial segment
of the ℒ-expression β when ⟨α∗ ⟩ is a proper initial segment of ⟨β∗ ⟩ (see Definition 1.1.6). Grammatically correct expressions Is the English expression “then work not. is, Sally work is Bill if at
at” a grammatically correct sentence? No! English expressions that are not grammatically correct make no sense. This observation motivates the following question: What does it mean for an
ℒ-expression to be grammatically correct? Surely, the ℒ-expression (A2 → A5 ) is a meaningful expression. However, the expression → A2 )(A5 appears to be nonsensical. We know that English has
“correct grammar.” Can we give the language ℒ correct grammar? The answer is yes. We want to define the notion of a well-formed formula (wff ) in the language ℒ. The wffs will be the grammatically
correct ℒ-expressions. Our definition will have the following consequences: (a) Every sentence symbol is a wff. (b) If α and β are wffs, then so are (¬α), (α ∧ β), (α ∨ β), (α → β), and (α ↔ β). (c)
Every wff is obtained by applying (a) or (b). We will often use lower-case Greek characters to represent wffs and upper-case Greek characters to represent sets of wffs. Items (a)–(c) above declare
that every sentence symbol is a wff and every other wff is built from other wffs by using the logical connectives and parentheses in particular ways. For example, A1123 , (A2 → (¬A1 )), and (((¬A1 )
∧ (A1 → A7 )) → A7 ) are all wffs, but X3 , (A5 ), ()¬A41 , A5 → A7 ), and (A2 ∨ (¬A1 ) are not wffs. In order to define wffs we will need the following five formula building functions: Let α and β
be ℒ-expressions. Then ℰ¬ (α) = (¬α),
ℰ∧ (α, β) = (α ∧ β), ℰ∨ (α, β) = (α ∨ β),
2.1 The language
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ℰ→ (α, β) = (α → β), ℰ↔ (α, β) = (α ↔ β).
Using the above operations, our next definition is an application of Theorem 1.1.24. Definition 2.1.4. Let ℱ = {ℰ¬ , ℰ∧ , ℰ∨ , ℰ→ , ℰ↔ } and let 𝒮 ⊆ {A1 , A2 , A3 , . . . } be a set of sentence
symbols in the language ℒ. Let 𝒮 be the set generated from 𝒮 by the functions in ℱ . An ℒ-expression α is a well-formed formula (wff or formula) with sentence symbols in 𝒮 if and only if α ∈ 𝒮 . The
set 𝒮 is the set of wffs that can be built from the symbols in S using the five formula building functions in (2.1). When 𝒮 is the set of all sentence symbols, we shall just say that α is a
well-formed formula (wff or formula). For example, if S = {A3 , A7 }, then 𝒮 consists of the wffs whose only sentence symbols are A3 and/or A7 . Remark. Let 𝒮 and 𝒯 be sets of sentence symbols.
Exercise 8 on page 37 implies that if 𝒮 ⊆ 𝒯 , then 𝒮 ⊆ 𝒯 . Definition 2.1.4 asserts that a wff is an expression that can be built from sentence symbols by applying the formula building functions a
finite number of times (see Section 1.1.5). Thus, a wff is constructed by using the sentence symbols as building blocks and the formula building functions as mortar. This leads to the following
definition, which gives a road map for the construction of a particular wff. Definition 2.1.5. A construction sequence is a finite sequence ⟨ε1 , ε2 , . . . , εn ⟩ of ℒ-expressions such that for each
i ≤ n one of the following holds: 1. εi is a sentence symbol, 2. εi = ℰ¬ (εj ) for some j < i, 3. εi = ℰ⊡ (εj , εk ) for some j < i and k < i, where ⊡ represents any of the connectives ∧, ∨, →, ↔. An
ℒ-expression α is a wff whenever there is a construction sequence ⟨ε1 , ε2 , . . . , εn ⟩ such that α = εn , that is, the sequence ends in α. Problem 2.1.6. Let α denote the wff ((A1 ∧ A10 ) → ((¬A3
) ∨ A8 )). Identify a construction sequence that ends with α. Solution. The following “tree” gives a picture of how the above wff α is constructed from its sentence symbols using the formula building
30 � 2 Propositional logic Thus we have the following construction sequence that ends in α: ⟨A1 , A10 , (A1 ∧ A10 ), A3 , (¬A3 ), A8 , ((¬A3 ) ∨ A8 ), α⟩. Induction on wffs principle To construct a
wff, one starts with the sentence symbols and then applies the formula building functions. Consequently, Theorem 1.1.25 yields the following induction on wffs principle and associated proof strategy.
Wff Induction Principle. Let 𝕊(α) be a statement about a wff α. If 1. 𝕊(Ai ) is true for every sentence symbol Ai and 2. for all wffs α and β, if 𝕊(α) and 𝕊(β), then 𝕊((¬α)) and 𝕊((α ⊡ β)) (here ⊡
represents each of the connectives ∧, ∨, →, ↔), then 𝕊(α) is true for all wffs α. Proof Strategy. In order to prove a statement “for all wffs α, 𝕊(α)” by induction, use the following diagram: Base
step: Inductive step:
Prove 𝕊(Ai ) for all sentence symbols Ai . Let α and β be arbitrary wffs. Assume 𝕊(α) and 𝕊(β). Prove 𝕊((¬α)) and 𝕊((α ⊡ β)).
Applications of the induction principle Theorem 2.1.7. Let α be any wff. The number of left parentheses equals the number of right parentheses in α. Proof. We shall use proof by induction on wffs.
Base step: Let α = Ai be a sentence symbol. Then Ai has zero left parentheses and the same number of right parentheses. Inductive step: Let α and β be arbitrary wffs. Assume the induction hypothesis
the number of left parentheses in α equals the number of right parentheses, the number of left parentheses in β equals the number of right parentheses.
We must prove that the same holds for each of the following: (¬α), (α ∧ β), (α ∨ β), (α → β), (α ↔ β). From the induction hypothesis (IH), it immediately follows that the number of left parentheses
in (¬α) equals the number of right parentheses. A similar argument applies for the rest of the binary connectives.
2.1 The language
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Theorem 2.1.8. A proper initial segment of a wff has more left parentheses than right parentheses. Thus, no proper initial segment of a wff is itself a wff. Proof. We shall use proof by induction on
wffs. Base step: Let α = Ai be a sentence symbol. Since Ai has no proper initial segments, the result follows vacuously. Inductive step: Let α and β be arbitrary wffs. Assume the induction hypothesis
a proper initial segment of α has more left parentheses than right parentheses, a proper initial segment of β has more left parentheses than right parentheses.
We must prove that any proper initial segment of each of the following has more left parentheses than right parentheses: (¬α), (α ∧ β), (α ∨ β), (α → β), (α ↔ β). Let us first consider the wff (α ∨
β). Any proper initial segment of (α ∨ β) has one of the following forms: 1. (, 2. (α0 , where α0 is a proper initial segment of α, 3. (α, 4. (α ∨, 5. (α ∨ β0 , where β0 is a proper initial segment
of β, 6. (α ∨ β. Case 1 is clear. In cases 3, 4, and 6, Theorem 2.1.7 implies that these proper initial segments have more left parentheses than right ones. For cases 2 and 5, the induction
hypothesis implies that these two proper initial segments have more left parentheses than right parentheses. A similar argument applies for the other logical connectives. Theorem 2.1.7 now implies
that a proper initial segment of a wff is not a wff. Exercises 2.1. 1. Let B and S be sentence symbols that represent the sentences “Bill is at work” and “Sally is at work,” respectively. Translate
the following English sentences into wffs: (a) Bill is at work and Sally is not at work. (b) If Bill is at work, then Sally is not at work. (c) Bill is not at work if and only if Sally is at work. 2.
Let α be the wff ((¬A1 ) → ((A4 ∨(A3 ↔ A2 ))∧A3 )). Identify a construction sequence that ends with α. 3. Given any wff α, let s(α) = the number of occurrences of sentence symbols in α,
32 � 2 Propositional logic c(α) = the number of occurrences of binary connectives in α. Prove by induction on α the following statement: For all wffs α, s(α) = c(α) + 1. 4. In the inductive step in
the proof of Theorem 2.1.8, using the induction hypothesis, show that any proper initial segment of (¬α) is not a wff. 5. Prove by induction on α the following statement: For every wff α there is a
construction sequence that ends with α (see Definition 1.1.6). 6. Let α1 , α2 , . . . , αn be wffs, where n > 1. Show that α1 α2 ⋅ ⋅ ⋅ αn is not a wff.
2.2 Truth assignments The set of truth values {F, T} consists of two distinct items: F, called falsity, T, called truth. Clearly, there are an infinite number of sentence symbols. Suppose that we
have truth values assigned to each of these sentence symbols. Can we then identify a function that will evaluate the truth value of all the wffs? If so, is there only one such function? In order to
address these questions, we first need to establish a unique readability theorem, that is, we need to show that one can read a wff without ambiguity. Theorem 2.2.1 (Unique readability). Let ℱ = {ℰ¬ ,
ℰ∧ , ℰ∨ , ℰ→ , ℰ↔ } and let B = {A1 , A2 , A3 , . . . }. When restricted to the set of wffs, we have the following: (a) The range of each operation in ℱ is disjoint from B. (b) Any two distinct
operations in ℱ have disjoint ranges. (c) Every operation in ℱ is one-to-one. In other words, the set of all wffs is freely generated from the set of sentence symbols by the five operations ℰ∧ , ℰ∨ ,
ℰ¬ , ℰ→ , ℰ↔ . Proof. Let ℱ = {ℰ∧ , ℰ∨ , ℰ¬ , ℰ→ , ℰ↔ } and let B = {A1 , A2 , A3 , . . . }. (a) We show that the range of ℰ∧ is disjoint from B. Let α and β be wffs. Clearly, (α ∧ β) ≠ Ai , because
Ai has no parentheses. Hence, the range of ℰ∧ is disjoint from B. A similar argument applies for the other operations in ℱ . (b) We need to show that any two distinct operations in ℱ have disjoint
ranges. Let α, β, γ, and σ be wffs. Suppose that (α ∧ β) = (γ ∨ σ). Then, by dropping the first parenthesis in each expression, the resulting expressions are equal, that is, α ∧ β) = γ ∨ σ).
Since α and γ are wffs, Theorem 2.1.8 implies that α and γ cannot be proper initial segments of the other. Thus, (2.2) implies that ∧ β) = ∨ σ). So ∧ = ∨, which is a
2.2 Truth assignments
� 33
contradiction. So ℰ∧ and ℰ∨ have disjoint ranges. Similar reasoning applies in the other cases. (c) Finally, we must show that each operation in ℱ is one-to-one. So let α, β, γ, and σ be wffs.
Suppose that (α ∧ β) = (γ ∧ σ). Reasoning as in part (b), we see that α = γ and thus, β = σ. Thus, ℰ∧ is one-to-one on the wffs. Similar reasoning also shows that the other operations are one-to-one.
The unique readability theorem will allow us to positively address the questions posed at the beginning of this section (see Theorem 2.2.4 below). Definition 2.2.2. Let 𝒮 be a set of sentence
symbols. A function v: 𝒮 → {F, T} is called a truth assignment for 𝒮 . Example 2.2.3. Consider the set 𝒮 = {A3 , A6 , A8 } of sentence symbols. Then the function v: 𝒮 → {F, T} defined by v(A3 ) = T,
v(A6 ) = F,
v(A8 ) = F
is a truth assignment for 𝒮 . Let 𝒮 be a set of sentence symbols and let v: 𝒮 → {F, T} be a truth assignment for 𝒮 . Let ℱ = {ℰ∧ , ℰ∨ , ℰ¬ , ℰ→ , ℰ↔ }. Theorem 2.2.1 implies that the set 𝒮 of all
wffs generated by 𝒮 from the functions in ℱ is freely generated. For each function in ℱ we also have the corresponding truth functions F∧ , F∨ , F¬ , F→ , F↔ introduced in Section 1.2.2 (see (1.2)–
(1.6)). Theorem 1.1.27 now implies the following result. Theorem 2.2.4. Let 𝒮 be a set of sentence symbols and let v: 𝒮 → {F, T} be a truth assignment for 𝒮 . Then there is a unique function v: 𝒮 →
{F, T} satisfying the following: (i) For every A ∈ 𝒮 , v(A) = v(A). (ii) For every α, β ∈ 𝒮 , we have: T, if v(α) = F, (1) v((¬α)) = { F, if v(α) = T, T,
(2) v((α ∧ β)) = {
if v(α) = T and v(β) = T,
if v(α) = T or v(β) = T (or both),
(3) v((α ∨ β)) = {
F, (4) v((α → β)) = { T, T,
(5) v((α ↔ β)) = {
if v(α) = T and v(β) = F,
if v(α) = v(β), otherwise.
Conditions (1)–(5) of the above theorem are given in tabular form below:
34 � 2 Propositional logic α
(α ∧ β)
(α ∨ β)
(α → β)
(α ↔ β)
T T F F
T F T F
F F T T
T F F F
T T T F
T F T T
T F F T
Remark 2.2.5 (On Theorem 2.2.4). Let 𝒮 be a set of sentence symbols and v: 𝒮 → {F, T} be a truth assignment for 𝒮 . Let v: 𝒮 → {F, T} be as in Theorem 2.2.4. For all α, β ∈ 𝒮 (iff means “if and only
if”), (1) v((¬α)) = T iff v(α) = F, (2) v((α ∧ β)) = T iff v(α) = T and v(β) = T, (3) v((α ∨ β)) = T iff v(α) = T or v(β) = T, (4) v((α → β)) = T iff if v(α) = T, then v(β) = T, (5) v((α ↔ β)) = T
iff v(α) = v(β). Example 2.2.6. Consider the set 𝒮 = {A3 , A6 , A8 } of three sentence symbols and let v: 𝒮 → {F, T} be defined by v(A3 ) = T,
v(A6 ) = F,
v(A8 ) = F.
Let α, β, and γ be defined by α = (A3 → (¬A6 )),
β = (A3 → (A6 ∨ A8 )),
γ = ((A3 → (¬A6 )) ↔ (A3 → (A6 ∨ A8 ))).
So α, β, γ ∈ 𝒮 . What is v(α), v(β), v(γ)? Definition 2.2.7. Let φ be a wff. Let v be a truth assignment defined on the sentence symbols appearing in φ. Then v satisfies φ if and only if v(φ) = T.
Example 2.2.8. Let 𝒮 = {A3 , A6 , A8 }, let v: 𝒮 → {F, T}, and let α, β, γ all be as in the above Example 2.2.6. Does v satisfy α? Answer the same question for β and γ. Definition 2.2.9. Let Σ be a
set of wffs and let v be a truth assignment defined on the sentence symbols appearing in any wff from Σ. Then v satisfies Σ if and only if v(φ) = T for every φ ∈ Σ. We shall say that Σ is satisfiable
if there is a truth assignment that satisfies Σ. Given that Σ is a set of wffs and τ is a wff, our next definition will address the following question: When can we view τ as a conclusion that follows
from Σ, where Σ is regarded as a set of hypotheses?
2.2 Truth assignments
� 35
Definition 2.2.10. Let Σ be a set of wffs and let τ be a wff. Then Σ tautologically implies τ (written Σ τ) if and only if for every truth assignment v defined on the sentence symbols that appear
in wffs in Σ and in τ, if v satisfies Σ, then v satisfies τ. Exercise 9 on page 15 implies the following theorem. Theorem 2.2.11. Let v1 : 𝒮 → {F, T} and v2 : 𝒯 → {F, T} be two truth assignments,
where 𝒮 and 𝒯 are sets of sentence symbols such that 𝒮 ⊆ 𝒯 . If v1 (Ai ) = v2 (Ai ) for all Ai ∈ 𝒮 , then v1 (α) = v2 (α) for all α ∈ 𝒮 . Corollary 2.2.12. Let Σ be a set of wffs and let τ be a wff.
Let 𝒮 = {Ai : Ai appears in a wff in Σ or in τ}
and let 𝒯 be a set of sentence symbols such that 𝒮 ⊆ 𝒯 . Then Σ τ if and only if every truth assignment for 𝒯 that satisfies every member in Σ also satisfies τ. Definition 2.2.10 can be used to
identify valid forms of reasoning, or inference rules. For example, let Σ = {(α → β), α} and let v be a truth assignment that satisfies Σ. Thus, v((α → β)) = T and v(α) = T. So, by Remark 2.2.5(4),
we conclude that v(β) = T. Therefore, Σ β. This particular tautological implication justifies the formal inference rule called modus ponens, represented by the diagram α→β α , ∴β which asserts
that if (α → β) and α are true, we therefore can conclude that β is true. In other words, modus ponens offers an argument that is truth preserving. Remark 2.2.13. A few special cases concerning
Definitions 2.2.7–2.2.10 deserve some comments. (a) If Σ is the empty set ⌀, then every truth assignment satisfies Σ. (b) It follows from (a) that ⌀ τ if and only if every truth assignment
satisfies τ. In this case, we say that τ is a tautology (written τ). In other words, τ is a tautology if and only if for every truth assignment v containing the sentence symbols in τ, we have v(τ)
= T. (c) If there is no truth assignment that satisfies Σ, then for any τ it is vacuously true that Σ τ. This can occur if Σ contains a contradiction; for example, if (Ai ∧ (¬Ai )) ∈ Σ. (d) If Σ
is a singleton {σ}, then we write σ τ in place of {σ} τ. Problem 2.2.14. Let α and β be wffs. Show that (α → (β → α)) is a tautology. Solution. We apply Remark 2.2.13(b). Let v be a truth
assignment that contains the sentence symbols occurring in α and β. We must show that v((α → (β → α))) = T. Suppose,
36 � 2 Propositional logic for a contradiction, that v((α → (β → α))) = F. By Theorem 2.2.4(4), () v(α) = T and v((β → α)) = F. Since v((β → α)) = F, we conclude that v(α) = F, which contradicts
(). So v((α → (β → α))) = T. Thus, (α → (β → α)) is a tautology. Definition 2.2.15. Let σ and τ be wffs. Then σ and τ are said to be tautologically equivalent (written σ τ) if σ τ and τ σ.
Remark 2.2.16 (On Definition 2.2.15). Let σ and τ be wffs. Then one can show that σ τ if and only if for every truth assignment v containing the sentence symbols in σ and τ, we have v(σ) = v(τ).
2.2.1 Some tautologies For the remainder of this chapter, we will let A, B, C, …, Z denote arbitrary sentence symbols. We note the following. 1. Two simple tautologies: (A → A) (A ∨ (¬A)) 2. De
Morgan’s laws: ((¬(A ∨ B)) ↔ ((¬A) ∧ (¬B))) ((¬(A ∧ B)) ↔ ((¬A) ∨ (¬B))) 3. Commutative laws: ((A ∧ B) ↔ (B ∧ A)) ((A ∨ B) ↔ (B ∨ A)) 4. Associative laws: ((A ∨ (B ∨ C)) ↔ ((A ∨ B) ∨ C)) ((A ∧ (B ∧
C)) ↔ ((A ∧ B) ∧ C)) 5. Distribution laws: ((A ∧ (B ∨ C)) ↔ ((A ∧ B) ∨ (A ∧ C))) ((A ∨ (B ∧ C)) ↔ ((A ∨ B) ∧ (A ∨ C))) 6. Conditional laws: ((A → B) ↔ ((¬A) ∨ B)) ((A → B) ↔ (¬(A ∧ (¬B)))) (((¬A) →
B) ↔ (A ∨ B)) ((¬(A → (¬B))) ↔ (A ∧ B)) 7. Double negation law: ((¬(¬A)) ↔ A) 8. Contrapositive law: ((A → B) ↔ ((¬B) → (¬A))) 9. Biconditional law: ((A ↔ B) ↔ ((A → B) ∧ (B → A))) 10. Exportation
law: (((A ∧ B) → C) ↔ (A → (B → C)))
2.2 Truth assignments
� 37
2.2.2 Omitting parentheses We now describe some conventions that will allow us to reduce (without any ambiguity) the number of parentheses used in our wffs. These conventions will give us a
“shortcut” in how we can correctly express a propositional statement. 1. The outermost parentheses need not be explicitly mentioned. For example, we can write A ∧ B to denote (A ∧ B). 2. The negation
symbol shall apply to as little as possible. For example, we can write ¬A ∧ B to denote (¬A) ∧ B, that is, ((¬A) ∧ B). 3. The conjunction and disjunction symbols shall apply to as little as possible.
For example, we will write A ∧ B → ¬C ∨ D to denote ((A ∧ B) → ((¬C) ∨ D)), 4.
using conventions 1 and 2. When one logical connective is used repeatedly, the parentheses are assumed to be grouped to the right. For example, we shall write α ∧ β ∧ γ to denote α ∧ (β ∧ γ) and
α → β → γ to denote α → (β → γ). This convention allows us to observe that for n ≥ 2,
α1 ∧ α2 ∧ ⋅ ⋅ ⋅ ∧ αn ∧ αn+1 denotes α1 ∧ α2 ∧ ⋅ ⋅ ⋅ ∧ (αn ∧ αn+1 ),
α1 → α2 → ⋅ ⋅ ⋅ → αn → αn+1 denotes α1 → α2 → ⋅ ⋅ ⋅ → (αn → αn+1 ).
(2.3) (2.4)
In some of the following exercises, we will omit parentheses following these conventions. Exercises 2.2. 1. Using Theorem 2.2.11, prove Corollary 2.2.12. *2. Let Σ be a set of wffs and let τ be a
wff. Let τ be either in Σ or a tautology. Show that Σ τ. 3. Determine whether or not ((A → B) ∨ (B → A)) is a tautology. *4. Let α and β be wffs. Show that α → β → (α ∧ β) is a tautology. *5. Let
α and β be wffs. Show that (¬α → β) → (¬α → ¬β) → α is a tautology. *6. Let Σ be a set of wffs and let θ, ψ be wffs. Suppose that Σ θ and Σ (θ → ψ). Show that Σ ψ. 7. Let Σ1 and Σ2 be sets
of wffs and let α and β be wffs. Suppose that Σ1 α and Σ2 β. Show that Σ1 ∪ Σ2 (α ∧ β). 8. Let Σ = {α1 , α2 , . . . , αn } be a finite set of wffs and let τ be a wff. Prove that if Σ τ,
then (α1 ∧ α2 ∧ ⋅ ⋅ ⋅ ∧ αn ) → τ is a tautology.
38 � 2 Propositional logic *9. Let Σ be any set of wffs and let α and β be wffs. Show that the following hold: (a) Σ ∪ {α} β iff Σ (α → β), (b) α β iff (α ↔ β). *10. Suppose that (α ↔ β)
is a tautology. Show that (α → β) is a tautology. 11. Suppose that either Σ α or Σ β. Prove that Σ (α ∨ β). 12. Find Σ, α, β such that Σ (α ∨ β), and yet Σ ̸ α and Σ ̸ β. 13. Let α, β
be wffs. Suppose that α and (α ↔ β) are both tautologies. Show that β is also a tautology. *14. Let α, α′ , β, and β′ be wffs. Suppose that α α′ and β β′ . Show that (a) (¬α) (¬α′ ), (b) (α
∧ β) (α′ ∧ β′ ), (c) (α ∨ β) ¬((¬α′ ) ∧ (¬β′ )), (d) (α ∧ β) ¬(α′ → (¬β′ )). 15. The sets of wffs below are either satisfiable or not. Determine which are satisfiable by finding a truth
assignment that satisfies the given set Σ. If Σ is not satisfiable, explain (that is, give an argument) why this is the case. (a) Σ = {A, (A → B), ¬B}, (b) Σ = {(A → B), ¬A, ¬B}, (c) Σ = {(A → B),
¬A, B}, (d) Σ = {(A ∧ B), ¬A, ¬B}, (e) Σ = {(A ∧ ¬A), ¬C, B}, (f) Σ = {¬(A ↔ B), A, ¬B}, (g) Σ = {(A ↔ B), ¬A, ¬B}, (h) Σ = {A ∧ B, ¬(C ∧ D), D → ¬A, C → ¬B}, (i) Σ = {A ∧ B, ¬C ∧ D, D → ¬A, C → ¬B}.
*16. Let Σ be a set of wffs and let τ be a wff. Prove that
Σ ̸ τ
if and only if
Σ ∪ {¬τ} is satisfiable.
17. Let Σ be a set of wffs and let φ, ψ be wffs. Let v be a truth assignment that satisfies Σ. Prove the following statements: (a) If (φ ∧ ψ) ∈ Σ, then v(φ) = T. (b) If (φ → ψ) ∈ Σ and ¬ψ ∈ Σ, then v
(¬φ) = T. (c) If (ψ → φ) ∈ Σ and ¬φ ∈ Σ, then v(ψ) = F. (d) If ¬(ψ → φ) ∈ Σ, then v(ψ) = T. 18. Determine whether or not the following statements are true. Explain (that is, give an argument) why
each statement is true or explain why it is false. (a) Let Σ = {A, (A → B)}. Then Σ B. (b) Let Σ = {(A ∨ B), ¬B}. Then Σ A. (c) Let Σ = {C → (A → B), A, C}. Then Σ (C → B). (d) Let Σ = {C →
(¬A → B), C}. Then Σ (¬B → A). (e) Let Σ = {(A ↔ B), ¬C}. Then Σ (A → B).
2.3 Completeness of the logical connectives
� 39
(f) Let Σ = {(A ↔ B), ¬C}. Then Σ A. (g) Let Σ = {(A ↔ B), ¬C, ¬B}. Then Σ ¬A. (h) Let Σ = {(A ∨ B), ¬C}. Then Σ A. (i) Let Σ = {(A ∧ ¬A), ¬C, B}. Then Σ D. 19. Let 𝒮 be the set of all
sentence symbols. For a set Γ of wffs, define the truth assignment vΓ : 𝒮 → {F, T} by T,
vΓ (Ai ) = {
if Ai ∈ Γ, if Ai ∉ Γ.
Justify your answers to the following questions: (a) Let Γ = {(A ∧ B), A, ¬B}. Does vΓ satisfy Γ? (b) Let Γ = {(A ∧ ¬A), C, B}. Does vΓ satisfy Γ? (c) Let Γ = {¬(A ↔ B), A, ¬B}. Does vΓ satisfy Γ?
(d) Let Γ = {(A ↔ B), ¬A, ¬B}. Does vΓ satisfy Γ? (e) Let Γ = {A ∧ B, ¬(C ∧ D), D → ¬A, C → ¬B, A, B}. Does vΓ satisfy Γ? (f) Let Γ = {D, ¬C ∧ D, D → ¬A, C → ¬B}. Does vΓ satisfy Γ? 20.
(Substitution) Let α1 , α2 , …, αn , . . . be an infinite sequence of wffs. For each wff φ, let φ∗ be the result of replacing each occurrence of An in φ with αn . (a) Let v be a truth assignment for
the set 𝒮 of all the sentence symbols. Define u: 𝒮 → {F, T} by u(An ) = v(αn ). Prove that u(φ) = v(φ∗ ) for all wffs φ. (b) Prove that if φ is a tautology, then so is φ∗ . (c) Prove that if ψ φ,
then ψ∗ φ∗ .
Exercise Notes: For Exercise 14, apply Remark 2.2.16. For Exercise 19, note that if Γ = {(A ∧ B), A, ¬B}, then A ∈ Γ and B ∉ Γ. For Exercise 20(a), let P(φ) be the statement u(φ) = v(φ∗ ). Now prove
“for all wffs φ, P(φ)” by induction on φ. (Question: Is (φ ∧ ψ)∗ = (φ∗ ∧ ψ∗ )?)
2.3 Completeness of the logical connectives In logic, a truth function is one that accepts truth values as input and produces a unique truth value as output. Let V = {T, F}. Thus, an n-place truth
function has the form H: V n → V for some n ≥ 1. Truth functions are also called Boolean functions. Such functions are applied in a variety of different fields, namely, electrical engineering,
computer science, game theory, and combinatorics. Let α be a wff whose sentence symbols are among A1 , A2 , . . . , An . We can now define an n-place truth function Hα : V n → V by Hα (x1 , x2 , . .
. , xn ) = v(α), where v(A1 ) = x1 , v(A2 ) = x2 , …, v(An ) = xn .
40 � 2 Propositional logic So, given any wff, we can use it to define a truth function. For example, let α be the wff (A1 ∨ ¬A2 ). Then we see that Hα : V 2 → V satisfies Hα (T, T) = T, Hα (T, F) =
T, Hα (F, T) = F,
Hα (F, F) = T,
which is essentially the truth table for α. This invites an interesting question: Given any truth function G, is there a wff ψ such that G = Hψ ? Before addressing this question, let us look at some
examples. Example 2.3.1. Suppose that G: V 3 → V always has output F. Let α be the wff (A1 ∧ ¬A1 ). Since α is always false and its sentence symbols are among A1 , A2 , A3 , by applying (2.5) we see
that Hα (x1 , x2 , x3 ) = v(α) = F for all x1 , x2 , x3 ∈ V (see Theorem 2.2.11). So G = Hα . Example 2.3.2. Let G: V 3 → V be such that T,
G(x1 , x2 , x3 ) = {
if x1 = T, x2 = F, x3 = T, otherwise.
Thus, G(T, F, T) = T and this is the only input that produces the output value T. Let γ be the wff (A1 ∧ ¬A2 ∧ A3 ) which corresponds to T, F, T. Note that v(γ) = T if and only if v(A1 ) = T, v(A2 )
= F, v(A3 ) = T. Thus, by applying (2.5), we see that G(x1 , x2 , x3 ) = Hγ (x1 , x2 , x3 ) = v(γ) for all x1 , x2 , x3 ∈ V . So G = Hγ . In the above two examples, we were given relatively simple
truth functions G and we were able to find a wff ψ such that G = Hψ . Our next example gives a slightly more complicated truth function. These examples will provide a guide that will allow us to
prove that for any truth function G there is always a wff ψ such that G = Hψ . Example 2.3.3. Suppose that truth function G: V 3 → V is such that G(T, T, T) = T, G(T, T, F) = F, G(T, F, T) = F,
G(T, F, F) = T,
G(F, T, T) = F,
G(F, T, F) = T, G(F, F, T) = T, G(F, F, F) = F.
2.3 Completeness of the logical connectives
� 41
Let us focus on the cases in (2.6) that produce the value T. Also, for each such case let us identify a corresponding wff γi as was done in Example 2.3.2, namely, G(T, T, T) = T
γ1 = (A1 ∧ A2 ∧ A3 ),
G(T, F, F) = T
γ2 = (A1 ∧ ¬A2 ∧ ¬A3 ),
G(F, F, T) = T
γ4 = (¬A1 ∧ ¬A2 ∧ A3 ).
G(F, T, F) = T
γ3 = (¬A1 ∧ A2 ∧ ¬A3 ),
Now let α = γ1 ∨ γ2 ∨ γ3 ∨ γ4 , that is, let α be the following wff: (A1 ∧ A2 ∧ A3 ) ∨ (A1 ∧ ¬A2 ∧ ¬A3 ) ∨ (¬A1 ∧ A2 ∧ ¬A3 ) ∨ (¬A1 ∧ ¬A2 ∧ A3 ). Note that v(α) = T if and only if G(v(A1 ), v(A2 ), v
(A3 )) = T. By applying (2.5), we see that G(x1 , x2 , x3 ) = Hα (x1 , x2 , x3 ) = v(α) for all x1 , x2 , x3 ∈ V . So G = Hα . The above examples offer a foundation for the proof of the following
theorem. Theorem 2.3.4. Let V = {T, F} and let G: V n → V be any truth function. Then there is a wff α such that G = Hα , where Hα is defined by (2.5). Proof. If G: V n → V is always false, then let
α be the wff (A1 ∧ ¬A1 ). Thus, G = Hα (see Example 2.3.1). Suppose now that G has some output values being T, say, k many such cases. Let us list all the cases that yield the value T. For each such
case let us also identify a corresponding wff γi as was done in Example 2.3.3 (see (2.7)), specifically, G(x11 , x12 , . . . , x1n ) = T,
γ1 = (β11 ∧ β12 ∧ ⋅ ⋅ ⋅ ∧ β1n ),
G(x21 , x22 , . . . , x2n ) = T,
γ2 = (β21 ∧ β22 ∧ ⋅ ⋅ ⋅ ∧ β2n ),
G(xk1 , xk2 , . . . , xkn ) = T,
γk = (βk1 ∧ βk2 ∧ ⋅ ⋅ ⋅ ∧ βkn ),
.. .
.. .
where for 1 ≤ i ≤ k and 1 ≤ j ≤ n, we have Aj ,
βij = {
¬Aj ,
if xij = T, if xij = F.
Note that v(γi ) = T if and only if v(A1 ) = xi1 , v(A2 ) = xi2 , . . . , v(An ) = xin . Let α = γ1 ∨ γ2 ∨ ⋅ ⋅ ⋅ ∨ γk . So v(α) = T if and only if G(v(A1 ), v(A2 ), . . . , v(An )) = T. By applying
(2.5), we see that G(x1 , x2 , . . . , xn ) = Hα (x1 , x2 , . . . , xn ) = v(α) for all x1 , x2 , . . . , xn ∈ V . So G = Hα . Theorem 2.3.4 shows that every truth function is equal to a function of
the form Hα for some wff α. For this reason, the set {∧, ∨, ¬, →, ↔} of logical connectives can be said to be truth functionally complete. In fact, the proof of Theorem 2.3.4 shows that {∧, ∨, ¬}
42 � 2 Propositional logic is also truth functionally complete. To clarify this assertion, let us say that a wff α is in disjunctive normal form if α = γ1 ∨ γ2 ∨ ⋅ ⋅ ⋅ ∨ γk , where each γi is a
conjunction of the form γi = (βi1 ∧ βi2 ∧ ⋅ ⋅ ⋅ ∧ βin ) and each βij is a sentence symbol or the negation of a sentence symbol. The proof of Theorem 2.3.4 shows that every truth function is equal to
a function of the form Hα , where α is in disjunctive normal form. Thus, {∧, ∨, ¬} is truth functionally complete as well. This discussion now allows us to establish the following result. Theorem
2.3.5. Every wff φ is tautologically equivalent to a wff α in disjunctive normal form. Proof. Let φ be a wff and let Hφ be the truth function defined by (2.5). The proof of Theorem 2.3.4 shows that
Hφ = Hα , where α is in disjunctive normal form. Exercise 3 below shows that φ and α are tautologically equivalent. Definition 2.3.6. Let 𝒞 be a subset of {∧, ∨, ¬, →, ↔}. Then 𝒞 is tautologically
complete if and only if every wff α is tautologically equivalent to a wff α′ that contains only the connectives in 𝒞 . The following result follows from Theorem 2.3.5. Theorem 2.3.7. The set {∧, ∨,
¬} is tautologically complete. Theorem 2.3.8. The sets {¬, ∧} and {¬, ∨} are both tautologically complete. Proof. We will only prove that the set {¬, ∧} is tautologically complete. A similar proof
shows that {¬, ∨} is tautologically complete. By Theorem 2.3.7 we know that every wff is tautologically equivalent to a wff which contains only the connectives in {∧, ∨, ¬}. So it is sufficient to
prove by induction that every wff α containing only connectives in {∧, ∨, ¬} is tautologically equivalent to a wff α′ with connectives only in {¬, ∧}. Base step: Let α = Ai . As α has no connectives,
α′ = α as required. Inductive step: Let α and β be arbitrary wffs that contain only the connectives in the set {∧, ∨, ¬}. Assume the induction hypothesis α α′ , β β′ ,
where α′ contains only the connectives in {¬, ∧}, where β′ contains only the connectives in {¬, ∧}.
We prove that each of the wffs (¬α), (α ∧ β), (α ∨ β) is tautologically equivalent to a wff that contains only connectives in {¬, ∧}.
For (¬α), from (IH) and Exercise 14(a) on page 38, it follows that (¬α) where (¬α′ ) has connectives only in {¬, ∧}.
� 43
2.3 Completeness of the logical connectives
(¬α′ ),
For (α ∧ β), from (IH) and Exercise 14(b) on page 38, we have (α ∧ β) (α′ ∧ β′ ), where (α′ ∧ β′ ) contains only connectives in {¬, ∧}.
For (α ∨ β), from (IH) and Exercise 14(c) on page 38, we have (α ∨ β) ¬((¬α′ ) ∧ (¬β′ )),
where ¬((¬α′ ) ∧ (¬β′ )) contains only connectives in {¬, ∧}. Remark 2.3.9. Let ℱ = {ℰ¬ , ℰ∧ , ℰ∨ } and let 𝒮 = {A1 , A2 , A3 , . . . } be the set of all sentence symbols in the language ℒ. Let 𝒮 be
the set generated from 𝒮 by the functions in ℱ . Theorem 1.1.25 validates the proof by induction given in Theorem 2.3.8. Exercises 2.3. 1. For each of the following wffs α, using the tautologies in
Section 2.2.1, find a wff that is tautologically equivalent to α which contains connectives only in 𝒞 : (a) α = (A → B) ↔ (C → A) and 𝒞 = {∧, ∨, ¬}, (b) α = (A → B) ↔ (C → A) and 𝒞 = {¬, ∧}. 2. Let V
= {T, F} and define the truth function G: V 3 → V by T,
G(x1 , x2 , x3 ) = {
4. *5. 6.
Find a wff α in disjunctive normal form such that G = Hα , where Hα is defined as in (2.5). Let α and β be wffs with the same sentence symbols among A1 , A2 , . . . , An . Let Hα and Hβ be defined as
in (2.5). Prove that Hα = Hβ if and only if α β. Using the fact that {¬, ∧} is tautologically complete, prove (by induction) that {¬, ∨} is tautologically complete. Given that {¬, ∧} is
tautologically complete, prove (by induction) that {¬, →} is tautologically complete. Let ℱ = {ℰ∧ , ℰ→ } and let 𝒮 = {A1 , A2 , A3 , . . . } be the set of all sentence symbols in the language ℒ. Let
𝒮 be the set generated from 𝒮 by the functions in ℱ . Now let v: S → {F, T} be defined by v(Ai ) = T for all i = 1, 2, 3, . . . . Prove by induction on α that v(α) = T for all α ∈ 𝒮 . Is {∧, →}
tautologically complete?
if exactly two of x1 , x2 , x3 are T,
Exercise Notes: For Exercises 4 and 5, read the first paragraph of the proof of Theorem 2.3.8. For Exercise 5, see Exercise 14(d) on page 38.
44 � 2 Propositional logic
2.4 Compactness Let Σ be an infinite set of wffs. Suppose that every finite subset of Σ is satisfiable. Thus, every finite subset of Σ has a particular truth assignment that satisfies all of the wffs
in this finite subset. Is there a truth assignment that will thus satisfy all of the infinite number of wffs in Σ? The compactness theorem addresses this interesting question. In this section, we
shall give a proof of the compactness theorem, for propositional logic. We begin by formally introducing the following pertinent definition. Definition 2.4.1. A set of wffs Σ is finitely satisfiable
if and only if every finite subset of Σ is satisfiable. The proof of the following theorem shows that a sophisticated mathematical argument can establish a powerful result about propositional logic.
A similar argument will be applied in Section 4.2. Theorem 2.4.2 (Compactness theorem). Let Σ be a set of wffs. If Σ is finitely satisfiable, then Σ is satisfiable. Proof. The basic idea behind the
proof is as follows: Given that Σ is finitely satisfiable, we first construct a set of wffs Δ such that: (1) Σ ⊆ Δ; (2) Δ is finitely satisfiable; (3) for every wff α, either α ∈ Δ or (¬α) ∈ Δ. We
shall then use Δ to define a truth assignment that satisfies Σ. Note that Theorem 1.1.30 (on page 12) implies that the set of all wffs is a countable set, because the set of all wffs is a subset of
the set of all finite sequences of symbols in the countable language ℒ. Thus, let α1 , α2 , α3 , . . . , αn , . . .
be a fixed enumeration (see Corollary 1.1.33) of all the wffs in the language ℒ. Define by recursion on ℕ (see Theorem 1.1.23) the following sets: (i) Δ0 = Σ, Δn ∪ {αn+1 },
(ii) Δn+1 = {
if Δn ∪ {αn+1 } is finitely satisfiable,
Δn ∪ {¬αn+1 }, otherwise.
From the above recursive definition, we see that Δn ⊆ Δn+1 for all n ∈ ℕ. Thus, Σ = Δ0 ⊆ Δ1 ⊆ Δ2 ⊆ ⋅ ⋅ ⋅ . We shall now establish four claims. Claim 1. For all n ∈ ℕ, Δn is finitely satisfiable.
2.4 Compactness
� 45
Proof of Claim 1. We shall use induction on n. Base step: Let n = 0. Then Δ0 = Σ, which is finitely satisfiable by assumption. Inductive step: Let n ∈ ℕ be arbitrary. Assume the induction hypothesis
Δn is finitely satisfiable.
We prove that Δn+1 is finitely satisfiable. Since Δn is finitely satisfiable, Exercise 1 below implies that Δn+1 is finitely satisfiable. (Claim 1) Let Δ = ⋃n∈ℕ Δn . Clearly, Σ ⊆ Δ as Δ0 = Σ. Recall
that Δn ⊆ Δn+1 for all n ∈ ℕ. Claim 2. The set Δ is finitely satisfiable. Proof of Claim 2. We show that each finite subset of Δ is satisfiable. Let Π be a finite subset of Δ. Since Π ⊆ Δ = ⋃n∈ℕ Δn
and Π is finite, it follows that Π ⊆ Δn for some n. By Claim 1, Δn is finitely satisfiable. Thus, Π is satisfiable. (Claim 2) Claim 3. For every wff α, either α ∈ Δ or (¬α) ∈ Δ, but not both. Proof
of Claim 3. Let α be any wff. Since (2.8) enumerates all of the wffs, there is an n ∈ ℕ such that α = αn . Either αn ∈ Δn or (¬αn ) ∈ Δn by (ii) of the above construction. Since Δn ⊆ Δ, it follows
that α ∈ Δ or (¬α) ∈ Δ. If both α ∈ Δ and (¬α) ∈ Δ, then {α, (¬α)} would be a finite subset of Δ. Since Δ is finitely satisfiable, we must conclude that {α, (¬α)} is satisfiable. This is a
contradiction, as the set {α, (¬α)} is not satisfiable. Therefore, we cannot have both α ∈ Δ and (¬α) ∈ Δ. (Claim 3) We now continue with the proof of the theorem. Note that: (1) Σ ⊆ Δ; (2) Δ is
finitely satisfiable by Claim 2; (3) for every wff α, either α ∈ Δ or (¬α) ∈ Δ, but not both, by Claim 3. We will now use Δ to define a truth assignment. Let 𝒮 be the set of all the sentence symbols
and define v: 𝒮 → {F, T} by T, if Ai ∈ Δ, v(Ai ) = { F, if Ai ∉ Δ.
Let v: 𝒮 → {F, T} be the extension of v as given by Theorem 2.2.4. Claim 4. For every wff φ, we have v(φ) = T
if and only if
φ ∈ Δ.
Proof of Claim 4. We shall prove that (2.10) holds by induction on α.
46 � 2 Propositional logic Base step: Let α = Ai be any sentence symbol. Since v(Ai ) = v(Ai ), (2.9) implies that v(Ai ) = T iff Ai ∈ Δ. Inductive step: Let α and β be arbitrary wffs. Assume the
induction hypothesis v(α) = T v(β) = T
iff α ∈ Δ, iff β ∈ Δ.
We must prove that the same holds for each of the following wffs: (¬α), (α ∧ β), (α ∨ β), (α → β), (α ↔ β). Case (¬α): We prove that v((¬α)) = T iff (¬α) ∈ Δ. (⇒). Assume that v((¬α)) = T. Since v
((¬α)) = T, it follows that v(α) = F. So, α ∉ Δ by (IH). Therefore, by (3), (¬α) ∈ Δ. (⇐). Assume that (¬α) ∈ Δ. Since (¬α) ∈ Δ, it follows that α ∉ Δ; because if α ∈ Δ, then by (2) {(¬α), α} is a
satisfiable subset of Δ, which is false. Thus, α ∉ Δ and the induction hypothesis (IH) implies that v(α) = F, and therefore v((¬α)) = T. Case (α ∧ β): We must prove that v((α ∧ β)) = T iff (α ∧ β) ∈
Δ. (⇒). Assume that v((α∧β)) = T. We need to show that (α∧β) ∈ Δ. Since v((α∧β)) = T, it follows that v(α) = T and v(β) = T. So by (IH), α ∈ Δ and β ∈ Δ. Suppose, for a contradiction, that (α ∧ β) ∉
Δ. Therefore, by (3), we conclude that (¬(α ∧ β)) ∈ Δ. So {α, β, ¬(α ∧ β)} is a finite subset of Δ and by (2) it is satisfiable, which is false. Therefore, (α ∧ β) ∈ Δ. (⇐). Assume that (α ∧ β) ∈ Δ.
We will show that v((α ∧ β)) = T. As (α ∧ β) ∈ Δ, it must be the case that α ∈ Δ and β ∈ Δ. To see this, suppose that either α ∉ Δ or β ∉ Δ. If α ∉ Δ, then by (3), we have (¬α) ∈ Δ. So {(α ∧ β),
(¬α)} is a finite subset of Δ and by (2) it is satisfiable, which is false. A similar argument shows that β ∈ Δ. Hence, α ∈ Δ and β ∈ Δ and by (IH), we have v(α) = T and v(β) = T. Therefore, v((α ∧
β)) = T by Remark 2.2.4(2). Cases (α ∨ β), (α → β), and (α ↔ β): See Exercise 4 below.
(Claim 4)
Claim 4 implies that v satisfies Δ. Since Σ ⊆ Δ, it follows that v satisfies Σ. Therefore, there is a truth assignment that satisfies Σ. (Theorem) Remark 2.4.3. If we were working in a language that
had uncountably many sentence symbols, then Theorem 2.4.2 could be established by using Zorn’s lemma (Lemma 1.1.36) to obtain a set of wffs Δ that satisfies items (1), (2), and (3) in the above
proof. The remainder of the proof would then be as that after the proof of Claim 3. The next useful corollary is just the contrapositive of the Compactness Theorem 2.4.2. Corollary 2.4.4. Let Σ be a
set of wffs. If Σ is not satisfiable, then there is a finite subset Σ′ of Σ that is not satisfiable.
2.4 Compactness
� 47
Let Σ be a set of wffs and let τ be a wff. From Exercise 16 on page 38, we have Σ ̸ τ
if and only if
Σ ∪ {¬τ} is satisfiable.
The above equivalence (2.11) will be used in the proof of the following corollary. Corollary 2.4.5. If Σ τ, then there is a finite Σ0 ⊆ Σ such that Σ0 τ. Proof. Assume that Σ τ. We shall
prove that there is a finite Σ0 ⊆ Σ such that Σ0 τ. Suppose, for a contradiction, that for every finite Σ0 ⊆ Σ, we have Σ0 ̸ τ. The above (2.11) implies that for every finite Σ0 ⊆ Σ, Σ0 ∪ {¬τ} is
Consider the set of wffs Π = Σ ∪ {¬τ}. The above statement (2.12) implies that every finite subset of Π is satisfiable (why?). The compactness theorem thus asserts that Π = Σ ∪ {¬τ} is satisfiable.
Thus, by (2.11), we have that Σ ̸ τ, which contradicts our assumption. Exercises 2.4. *1. Let Σ be a set of wffs and let α be a wff. Assume that Σ is finitely satisfiable. Prove that either Σ ∪ {α}
or Σ ∪ {¬α} is finitely satisfiable. 2. Let Σ be a set of wffs. Show Σ is not satisfiable if and only if Σ (A ∧ (¬A)). 3. Using the above Exercise 2, show that Corollary 2.4.5 implies Theorem
2.4.2. *4. In the proof of Theorem 2.4.2, complete the proof of Claim 4. 5. Let Σ be the following 3-element set of wffs: Σ = {((¬A ∨ ¬B) ∧ C), ((¬A ∨ ¬C) ∧ B), ((¬C ∨ ¬B) ∧ A)}. (a) Show that every
subset of Σ containing less than three elements is satisfiable. (b) Is Σ is satisfiable? 6. Let Σ and Π be sets of wffs. Suppose that for every finite Π0 ⊆ Π, the set Σ ∪ Π0 is satisfiable. Using the
compactness theorem, prove that Σ ∪ Π is satisfiable. 7. Let Δ be a set of wffs such that: (1) Δ is finitely satisfiable and (2) for every wff α, either α ∈ Δ or (¬α) ∈ Δ. Let γ and β be wffs. (a)
Suppose that γ ∈ Δ and β is tautologically equivalent to γ. Show that β ∈ Δ. (b) Show that γ ∧ β ∈ Δ if and only if γ ∈ Δ and β ∈ Δ. (c) Show that γ ∨ β ∈ Δ if and only if either γ ∈ Δ or β ∈ Δ. (d)
Prove that (γ → β) ∈ Δ if and only if γ ∉ Δ or β ∈ Δ. Exercise Notes: For Exercise 1, if Σ∪{α} and Σ∪{¬α} are not finitely satisfiable, then there are finite subsets Σ1 and Σ2 of Σ such that Σ1 ∪ {α}
and Σ2 ∪ {¬α} are not satisfiable. Why? Now consider the finite set Σ1 ∪ Σ2 .
48 � 2 Propositional logic
2.5 Deductions Let Σ be a set of wffs and let ψ be a wff. We know that Σ ψ means that every truth assignment that satisfies every formula in Σ will also satisfy ψ. Rather than working with
all such truth assignments, is it possible to “deduce” ψ from Σ? If so, what methods of
deduction would be required to demonstrate this fact? Whenever mathematicians compose a proof, they justify a final conclusion by means of a series of intermediate steps.
Such steps typically appeal to some given assumptions and/or previously established theorems. A theorem in mathematics is a statement that is true. Therefore, mathemati-
cians use assumptions and true statements to derive new results. In propositional logic, can we use assumptions and tautologies to derive new results? In this section we will positively address this
Definition 2.5.1. Let Σ be a set of wffs and let ψ be a wff. A deduction of ψ from Σ is a finite sequence of wffs ⟨α1 , α2 , . . . , αn ⟩ such that αn = ψ and for each k ≤ n, 1.
αk ∈ Σ,
αk is obtained by modus ponens from some αi = (αj → αk ) and αj , where i, j < k.
αk is a tautology, or
When there is a deduction of ψ from Σ, we shall say that ψ is deducible from Σ or that ψ is a theorem of Σ, and we shall write Σ ⊢ ψ. Recall that the inference rule modus ponens is discussed on page
A deduction can be viewed as a “formal proof,” where each step is governed by
rules (1)–(3) of Definition 2.5.1. We use the term deduction to avoid confusion with our own mathematical proofs.
We now present two examples of deductions for which each “deduction sequence”
is written in a vertical form with a parallel justification for each step. Example 2.5.2. Let Σ = {α, β}. Show that Σ ⊢ (α ∧ β).
Solution. The following (vertical) sequence satisfies the conditions in Definition 2.5.1: 1. α 2. β 3. α → β → (α ∧ β) 4. β → (α ∧ β) 5. (α ∧ β)
in Σ, in Σ, tautology (see Exercise 4 on page 37), from 1 and 3, by modus ponens, from 2 and 4, by modus ponens.
Therefore, Σ ⊢ (α ∧ β). Example 2.5.3. Let Σ = {H → ¬B, D → B, H}. Show that Σ ⊢ (¬D).
2.5 Deductions
� 49
Solution. The (vertical) sequence below is a deduction: 1. H → ¬B 2. D → B 3. H 4. ¬B 5. (D → B) → (¬B → ¬D) 6. ¬B → ¬D 7. ¬D
in Σ, in Σ, in Σ, from 1 and 3, by modus ponens, tautology (see contrapositive law),1 from 2 and 5, by modus ponens, from 4 and 6, by modus ponens.
Hence, Σ ⊢ (¬D). The definition of Σ ψ involves truth assignments and the definition of Σ ⊢ ψ involves a “deduction sequence.” On the surface, these two concepts, Σ ψ and Σ ⊢ ψ, seem to be
unrelated. However, they are in fact intimately related. Our next two theorems will confirm this purported intimacy. The first theorem shows that if the assumptions in Σ are true and Σ ⊢ ψ, then the
deduction ψ is also true, that is, our deduction system preserves truth and for this reason, it is said to be sound. Theorem 2.5.4 (Propositional soundness). Let Σ be a set of wffs and ψ be a wff. If
Σ ⊢ ψ, then Σ ψ. Proof. Suppose that Σ ⊢ ψ. So let ⟨α1 , α2 , . . . , αn ⟩, where αn = ψ, be a deduction of ψ from Σ. For each natural number k, we shall prove that if 1 ≤ k ≤ n, then Σ αk . We
shall use strong induction on k. If k = 1, then by Definition 2.5.1, either α1 ∈ Σ or α1 is a tautology. It thus follows that Σ α1 (see Exercise 2 on page 37). Now let k be such that 1 < k ≤ n and
assume the induction hypothesis Σ αi
for all i < k.
We must show that Σ αk . Again, if αk ∈ Σ or αk is a tautology, then we have Σ αk . On the other hand, suppose that αk is obtained by modus ponens from some αi = (αj → αk ) and αj , where i, j
< k. By the induction hypothesis (IH), we have Σ (αj → αk ) and Σ αj . Hence, Σ αk (see page 35). This completes the induction, and thus, Σ ψ. Our next theorem shows that the converse of
Theorem 2.5.4 holds, that is, for every wff ψ, if Σ tautologically implies ψ, then there is a deduction of ψ from Σ. Thus, our deduction system is said to be complete. Theorem 2.5.5 (Propositional
completeness). Let Σ be a set of wffs and let ψ be a wff. If Σ ψ, then Σ ⊢ ψ. 1 See Exercise 10 on page 38.
50 � 2 Propositional logic Proof. Assume that Σ ψ. Corollary 2.4.5 implies that there is a finite Σ0 ⊆ Σ such that Σ0 τ. Let Σ0 = {α1 , α2 , . . . αn }. Exercise 6 below implies that α1 → α2 →
⋅ ⋅ ⋅ → αn → ψ is a tautology. By repeated use of modus ponens, it easily follows that Σ0 ⊢ ψ. As Σ0 ⊆ Σ, we conclude that Σ ⊢ ψ. Theorems 2.5.4 and 2.5.5 immediately establish the following result.
Theorem 2.5.6. Let Σ be a set of wffs and let ψ be a wff. Then Σ ψ if and only if Σ ⊢ ψ. Our next theorem can sometimes be used to indirectly show that a deduction exists. Our proof of this result
applies Theorem 2.5.6. Theorem 2.5.7 (Deduction theorem). Let Σ be a set of wffs and let α and β be wffs. Then Σ ⊢ (α → β) if and only if Σ ∪ {α} ⊢ β. Proof. Let Σ be a set of wffs and let α and β be
wffs. We thus have Σ ⊢ (α → β)
iff iff iff
Σ (α → β)
by Theorem 2.5.6,
Σ ∪ {α} ⊢ β
by Theorem 2.5.6.
Σ ∪ {α} β
by Exercise 9 on page 38,
Example 2.5.2 shows that {γ, σ} ⊢ (γ ∧ σ). So by Theorem 2.5.7, {γ} ⊢ (σ → (γ ∧ σ)), where Σ = {γ}, α is σ, and β is (γ ∧ σ). In the conditional α → β, α is called the hypothesis and β is called the
conclusion. In a mathematical proof of a conditional “if , then ”, one typically assumes the hypothesis and then derives the conclusion. The deduction theorem affirms that this is a valid technique
of proof in mathematics. Our final theorem of this section verifies that “proof by contradiction” is a valid proof technique. Definition 2.5.8. A set of wffs Σ is said to be inconsistent if there
exists a wff β such that Σ ⊢ β and Σ ⊢ ¬β. Moreover, Σ is consistent if for no wff β we have Σ ⊢ β and Σ ⊢ ¬β. Theorem 2.5.9. Let Σ be a set of wffs and let α be a wff. If Σ ∪ {¬α} is inconsistent,
then Σ ⊢ α. Proof. Assume that Σ ∪ {¬α} is inconsistent. Hence, for some wff β we have Σ ∪ {¬α} ⊢ β and Σ∪{¬α} ⊢ ¬β. Thus, by Theorem 2.5.7 (the deduction theorem), we have Σ ⊢ (¬α → β) and Σ ⊢ (¬α →
¬β). By Exercise 5 on page 37, (¬α → β) → (¬α → ¬β) → α is a tautology. Exercise 4 below now implies that Σ ⊢ α.
2.5 Deductions
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There are alternative deduction systems for propositional logic that are equivalent to the one presented in this section. In particular, there are deduction systems that start with a finite set of
tautologies and from this finite set one can then deduce all of the other tautologies. Consequently, the deductions in such systems can be much longer than those presented in this text. On the other
hand, there are deduction systems that consist only of a finite number of inference rules and from these rules one can deduce all of the tautologies. Again, such deductions can be longer than the
ones presented here. For each of these alternative systems, one can prove a corresponding analogue of Theorem 2.5.6, and thus these deduction systems are equivalent. In an introductory course in
mathematical logic, we believe that one should present a deduction system that is succinct and produces the desired results. Exercises 2.5. 1. Let Σ be an infinite set of wffs and let α be a wff.
Show that if Σ ⊢ α, then there is a finite Σ0 ⊆ Σ such that Σ0 ⊢ α. 2. Let Σ be a set of wffs and let α be a wff. Let Σ0 ⊆ Σ. Show that if Σ0 ⊢ α, then Σ ⊢ α. 3. Let Σ be a set of wffs and let ψ be a
tautology. Show that Σ ⊢ ψ. *4. Let Σ be a set of wffs and let θ, ψ be wffs. Suppose that Σ ⊢ θ and Σ ⊢ (θ → ψ). Show that Σ ⊢ ψ. 5. Let Σ1 and Σ2 be sets of wffs and let α and β be wffs. Suppose
that Σ1 ⊢ α and Σ2 ⊢ β. Show that Σ1 ∪ Σ2 ⊢ (α ∧ β). *6. Let n ≥ 1 be a natural number and let Σn = {α1 , α2 , . . . αn } be a set of n wffs. Prove that for all wffs θ, if Σn θ, then α1 → α2 → ⋅ ⋅
⋅ → αn → θ is a tautology. 7. Let n ≥ 1 be a natural number and let Σn = {α1 , α2 , . . . αn } be a set of n wffs. Prove that for all wffs θ, if α1 → α2 → ⋅ ⋅ ⋅ → αn → θ is a tautology, then Σn θ.
8. Give a deduction from the set Σ = {¬A ∨ B, B → C, A} whose last component is C. 9. Let Σ be an infinite set of wffs. Show that if Σ is inconsistent, then a finite subset of Σ is inconsistent.
Exercise Notes: For Exercise 6, one can repeatedly apply Exercise 9 on page 38 (see Remark 2.2.13(b)). A more formal proof of this result can be obtained by induction on n. For Exercise 7, a proof of
this result is obtained by induction on n (see (2.4) on page 37).
3 First-order logic Mathematical logic is a branch of mathematics that investigates the foundations of mathematics. In this chapter, we shall do the same. Specifically, in Section 3.1, we discuss
firstorder languages, together with some examples of first-order languages. First-order logic is rich enough to formalize virtually all of mathematics. In Section 3.2, we will investigate
mathematical structures (models) and Tarski’s definition of truth (satisfaction) in a structure. In Section 3.3, we shall examine the definition of proof (deduction) in a firstorder language.
3.1 First-order languages In this section, we will formally define the syntax of the language of first-order logic. First-order logic deals with formal statements that are expressed in terms of
predicates, logical connectives, variables, and quantifiers. A preview of such a logic is given in Section 1.2.3. We will first identify the symbols of the language. An expression will then be any
finite string of these symbols. Some expressions will be nonsensical, while others will be meaningful. Some of the meaningful expressions will denote terms which act as the nouns and pronouns of the
language; the terms can be interpreted as naming an individual object. Once we have the terms of the language, we can define the atomic formulas of the language. Atomic formulas are analogous to the
sentence symbols of propositional logic. We can then identify the correct rules of grammar and define the well-formed formulas (wffs) of the language. To specify the terms and wffs requires
definition by recursion (see Section 1.1.5). Definition 3.1.1. The alphabet of a first-order language ℒ consists of the following distinct symbols: A. Required symbols 1. Parentheses: (, ). 2.
Logical connectives: →, ¬. 3. Variables: v1 , v2 , v3 , v4 , . . . , vn , . . . . 4. Predicate symbols: For each natural number n > 0, a (possibly empty) set of n-place predicate symbols. 5.
Quantifier symbol: ∀. B. Optional symbols 1. Equality symbol: =,̇ a 2-place relation. 2. Constant symbols: A set of symbols, for example, {c1 , c2 , . . . }. 3. Function symbols: For each natural
number n > 0, a (possibly empty) set of n-place function symbols.
3.1 First-order languages
� 53
A dot over a familiar symbol is used to emphasize that the symbol has to be interpreted. This is also done to make a distinction between the symbol itself and the object that it commonly represents.
A finite string of symbols from the language ℒ will be called an ℒ-expression. For example, v3 c4 c1 )v3 is an ℒ-expression that starts with the symbol v3 . Moreover, v3 c4 is a proper initial
segment of v3 c4 c1 )v3 . The ℒ-expression v3 c4 c1 )v3 does not appear to be expressive. We will soon isolate the meaningful ℒ-expressions from those that are meaningless. There exists a one-to-one
correspondence between each ℒ-expression α and a finite sequence of symbols in ℒ denoted by ⟨α∗ ⟩, where α∗ is the result of putting commas between all of the symbols in α. So an ℒ-expression α is a
proper initial segment of the ℒ-expression β when ⟨α∗ ⟩ is a proper initial segment of ⟨β∗ ⟩ (see Definition 1.1.6). In the formal language ℒ, we have listed only the logical connectives → and ¬.
Since these two connectives are tautologically complete (see Exercise 5 on page 43), there is no need to add more. We also only identified the universal quantifier ∀. Since ¬∀x¬Px is equivalent to
∃xPx (see Quantifier Negation Laws 1.2.4(3)), the existential quantifier ∃x can be viewed as an abbreviation for ¬∀x¬. In the language ℒ, an n-place function symbol is intended to represent a
function of n relevant ℒ-expressions ε1 , ε2 , . . . , εn . Let f be a 3-place function with input ε1 , ε2 , ε3 . In mathematics f (ε1 , ε2 , ε3 ) would be the standard notation for the output value;
however, if f is a function symbol in the language ℒ, we shall represent this output value by using the Polish notation fε1 ε2 ε3 , where there are no parentheses or commas. An n-place predicate
symbol is intended to represent a property or relationship of n relevant ℒ-expressions; for example, if P is a 4-place predicate symbol in the language ℒ, then Pε1 ε2 ε3 ε4 can be viewed as an
assertion about ε1 , ε2 , ε3 , ε4 . Predicate symbols are sometimes also called relation symbols. The predicate, constant, and function symbols can be viewed as the parameters of the language. To
specify a language we must identify the particular predicate, constant, and function symbols that we wish to use. Suppose our language ℒ contains the equality symbol, the predicate symbols P1 , P2 ,
. . . , constant symbols c1 , c2 , . . . , and function symbols f1 , f2 , . . . . In the future we shall describe a language ℒ by expressing it as a set of these selected parameters, that is, we
shall say that ℒ is the set ℒ = {P1 , P2 , . . . , c1 , c2 , . . . , f1 , f2 , . . . , =}. ̇
Example 3.1.2 (Language of groups). When working with groups, one employs the language ℒ = {e, ∗, =}, ̇which has a 2-place function symbol ∗ for the group operation and a constant symbol e for the
identity element. We can write v1 ∗ v2 and v1 =̇ v2 to represent the Polish notation =v ̇1 v2 and ∗v1 v2 , respectively.
54 � 3 First-order logic Example 3.1.3 (Language of set theory). Set theory uses the language ℒ = {∈,̇ =}, ̇where ̇1 v2 . ∈̇ is a 2-place predicate symbol. We shall write v1 ∈̇ v2 to denote ∈v Example
3.1.4 (Language of elementary number theory). In number theory one can use the language ℒ = {
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Craps Betting Progressions systems
Submit your system! Share how you like to play. Contact me at starchip@live.com
For all even-money betting.
System betting means you're using a specific strategy in trying to be a winner. However, in the long-run, the odds favor the casino and no system can change that fact. That is not to say systems
can't work beautifully in the short term.
What makes the 1 - 3 - 2 - 6 System attractive is that you risk only two betting units for a chance to win ten units. You can make a nice profit with a minimal investment.
The first bet is one unit, if you win, add another unit, making the second bet a total of three units.
If you win the second bet, there are 6 units on the table. Remove four units, making your third bet two units.
If it wins add two more units, making it a total of six units for your fourth bet.
If the fourth bet wins you'll collect a total of 12 units, of which ten units are profit!
If you lose the first bet, the loss is one unit. Assuming you win the first, but lose the second bet, your net loss is two units.
If you win the second, but lose the third bet, you have a profit of two units.
If you win the first three bets but lose the fourth bet, you'll break even.
If you lose the second bet five out of six times and win four consecutive bets once, you'll be right back to even.
Use this system for even-money betting at any game.
Good Luck!
If you would like to explore other, more serious systems, and/or investigate my SUPERIOR ROULETTE
SYSTEM at http://www.letstalkwinning.com/roulys.htm.
Thank you. Tony R Frank
Another explanation of the same system:
The 1-3-2-6 System
The name of this system says it all. It is based on the premise that you can win four times in a row.
Your initial bet is 1 unit, the second 3 units, the third 2 units and the fourth 6 units.
Let's assume that each unit is $10 and the odds are 1:1 - even money.
( 1 ) The first bet is $10. ( you win )
they pay you $10 + $10=$20 you add $10 more off your money making the second bet $30 you now have $20 of your money at risk plus $10 of casino money
( 2 ) When winning again on the second bet, there would be $60 on the table Of this you take down $40
$30 + $ $30 = $60 -$40 = $20
( 3 ) The third bet is $20. The third bet wins, you will have $40 on the table to which you add $20 making a total of $60 for the fourth bet.
$20 + $20 =$40 +$20= $60
If the fourth bet wins, there would be a total of $120, of which $100 is net profit.
$60 + $60 = $120 - $20 = $100 profit
Now all the bet with the profit is taken down and you start the system all over again at $10.
If you lose the first bet, your loss is $10.
The second level loss is $20.
At the third level, a loss will give you a net profit of $20.
At the fourth level, a loss leaves you breaking even.
Each time you lose, you start all over again at $10.
The attraction of this system is that you risk $20 at a chance of making $100 net profit. This means you can lose five times, and with one win get your money back
This is a mixture of Martingale and Insurance systems. Bets are raised one unit after each losing bet and lowered one unit after each winning bet. The sequence and amount raised or lowered can be
varied to suit particular games and odds.
System description courtesy of iL Dado Advice + resources Gambling Guide. Learn how to win. Casino games and sports. Find it here: directory, online casinos, free-play games, news, and much more.
Fibonacci progression 0,1,1,2,3,5,8,13,21,34,55, etc.
Where each number is a sum of the previous two numbers.
Basic idea is to get two wins in a row.
For each loss, step up one in the progression.
For each win, make the same bet again.
If win again, start progression over. If lose, advance one step.
Unknown contributor.
BTW, in the fibonacci you don't need 2 wins in row to begin again.....WLW also gives you a win on the series. Kayjay
This is a good system as WLW shows a profit. However, I would add the following:
If PASS just won, bet on PASS. If DON'T PASS won, bet on D.P.
Play the winning side and you will catch any streak of wins. This will work on Craps, Roulette, Baccarat. James T Mays
A betting method to apply to all systems.
We all have losing bets and when we lose we want to recoup that money. Some try the Martingale. i.e. 1-2-4-8-16, etc. Seasoned players know this is a sure way to leave the table broke and with an
Try the Knockdown instead.
When you lose a one unit bet replace it with only one unit the first go. Then your progression will be 1-1-3-6-12, etc., ( I strongly suggest ending at 3 ) not 1-2-4-8-16, etc., as in a Martingale.
You still always win except on the second bet where you break even. We long time players know how to appreciate being even.
Over time this practice can improve most systems by several percentage points.
This system is also called the 'Cancellation' system. There are many variations. In its simplest form, you write down a series or a set of numbers; say, 1 2 3 4 5 6. The series can be short or long
and not necessarily sequential such as 1 1 1 3 3 5 7. The choice of a particular series depends on the type of game you want to apply it to and the odds of the bet.
Each number represents the amount in units or chips to bet. You bet the first and last of these numbers. In this example 1 and 6, which totals 7 units.
If you win, you cross out the two numbers and bet the next two 'ends' (the outside numbers). In this instance 2 and 5. If you win again you bet on the next two remaining numbers 3 and 4, and if you
win that too, you would have made a 'coup' or completed one game. Then you start all over again.
If you lose, then you add that one number to the end of the series. Say you lost your first bet of 7 units (1+6). Then you add number 7 to the end of the series to look like this: 1 2 3 4 5 6 7 and
your next bet would be 8 units ( 1+7). If you won the first bet but lost the second 2 and 5, then the series of numbers would look like this: 2 3 4 5 7.
If you work it out, you will see that when the series is completed or when you make a 'coup', there is always a profit. The negative side of this system is that you could end up betting large sums of
money even if your initial bet is small.
System description courtesy of iL Dado Advice + resources Gambling Guide. Learn how to win. Casino games and sports. Find it here: directory, online casinos, free-play games, news, and much more.
The Martingale system is a very old and extremely simple system. It is based on the probability of losing infinite times in a row and is usually applied to 'even money' bets.
You start with one bet. If you win, you start again with one bet. If you lose, you double your bet. Each time you lose, you double your last lost bet. Eventually you are bound to win. When you win
you would recover all your lost bets plus one unit (or chip) profit against your initial wager.
Although infallible in theory, the Martingale system requires a large bankroll, has a very low return and is a very risky one because of the maximum bet limits imposed by the casinos. If you run out
of money or reach the house limit, you can lose a lot with no chance to recover your losses.
System description courtesy of iL Dado Advice + resources Gambling Guide. Learn how to win. Casino games and sports. Find it here: directory, online casinos, free-play games, news, and much more.
This system is similar to the Paroli system and has the effect of 'pyramiding' your profit. Pyramiding is a parlay wager whereby the original wager plus its winnings are played on successive wagers.
It is commonly used in horse racing betting. Basically you make a bet and if you win you re-invest the winnings on the next bet. You 'let it ride'.
This method of play is by no means risk free, but it offers the least amount of risk of all wagers since the player is only concerned with either a win, place or show selection or a combination of
the three.
It is one of the oldest methods of wagering and was originally derived from the same premise that banking systems use to compound interest.
System description courtesy of iL Dado Advice + resources Gambling Guide. Learn how to win. Casino games and sports. Find it here: directory, online casinos, free-play games, news, and much more.
This system is in a way the opposite of the Martingale system. You start with one bet and you increase your bet when you win rather than when you lose.
However, you will need to plan a betting procedure whereby you know how far you will let the bet build before you take it down to the initial starting bet and how much to raise after each win. This
obviously depends on the type of game played and the odds of the bet.
The advantage of this system is that you do not require a large bankroll. It lets the profit run and cuts short the losses.
System description courtesy of iL Dado Advice + resources Gambling Guide. Learn how to win. Casino games and sports. Find it here: directory, online casinos, free-play games, news, and much more.
Pass / Don't pass betting progression
This system was first featured by Gil Stead in the 1980's newsletter THE CRAPSMEN.
First, you need 26 checks, all same denomination($1,$5,etc.). Take the first check (Bet A) and wager it on either Pass Line or Don't Pass Line.
If this bet is won, parlay it and wager both bet and win, or two units.
This is the heart of the system: one MUST back bet or parley the win.
Now if you are favored with a second win, you will of course be ahead.
Take your three extra checks and place them in the back (lock up) chip rack and start over again with one check.
Betting progression is as follows: 1,1,1,2,2,3,4,5,7 (26 checks total).
Example: Bet A (1check): lost
Bet B (1check): lost
Bet C (1check): lost
Bet D (2checks): won
Bet E parlay (2 checks plus 2 checks won): won.
You're now ahead by two checks (28 checks total). Start over again with a one check bet.
Think about it.You have a total of nine opportunities (consecutive) before you limit your losses.You don't ever have to limit your winnings.
Richard G. Poirier
Pass / Don't Pass Odds progression
This system progresses odds bets on either the Pass Line or Don't Pass based on the value of a running count.
Keep track of the count as follows:
A point that passes counts as +3
A point that misses counts as -2
Naturals and craps count as 0. If the count exceeds 55 or -37 then the progression is lost and started over.
Anytime the odds bets show a net profit, the progression is started over.
I win over and over playing this way. Please look it over.
I'm looking for your comments good or bad. Please post or e-mail them to me.
Have I developed THE winning system { the one that overcomes the odds ad puts the odds in the players favor } or have I just been getting lucky?
STREAKS -
Streaks of P and DP happening in a row, odds for each streak to continue are 50% less than the last streak.
EXAMPLE :
[ all examples will only go to a level of ten - P or DP occurring in a row - the math is the same no matter how far or near I take it - IT can not change ]
10 in a row - will occur __1 time
09 in a row - will occur __2 times
08 in a row - will occur __4 times
07 in a row - will occur __8 times
06 in a row - will occur _16 times
05 in a row - will occur _32 times
04 in a row - will occur _64 times
03 in a row - will occur 128 times
02 in a row - will occur 256 times
01 in a row - will occur 512 times
Streaks of P and DP - 2036 total decisions
50% will be P and 50% will be DP.
Doubling your bet after a loss.
No matter where you end your doubling a bet after a loss you will lose.
EXAMPLE :
[ ending the doubling and taking a loss at 5 losing wagers in a row ]
01 unit__ win = plus 1 unit_ lose = -1 unit then wager
02 unit__ win = plus 1 unit_ lose = -3 units then wager
04 unit__ win = plus 1 unit_ lose = -7 units then wager
08 unit__ win = plus 1 unit_ lose = -15 units then wager
16 unit__ win = plus 1 unit_ lose = -31 units
So playing the OPPOSITE of the prior decision, this is what will happen:
All streaks of
01 in a row-plus 1 unit
02 in a row-plus 1 unit
03 in a row-plus 1 unit
04 in a row-plus 1 unit
05 in a row minus 31 units
06 in a row minus 31 units
07 in a row minus 31 units
08 in a row minus 31 units
09 in a row minus 31 units
10 in a row minus 31 units
What happens when we look at the amount of streaks we will get when we play.
__1 streak of 10 in a row = minus _31 units 1x-31
__2 streaks of 9 in a row = minus _62 units 2x-31
__4 streaks of 8 in a row = minus 124 units 4x-
__8 streaks of 7 in a row = minus 248 units 8x-31
_16 streaks of 6 in a row = minus 496 units 16x-31
_32 streaks of 5 in a row = minus 992 units 32x-31
_64 streaks of 4 in a row = plus ___64 units 64x 1
128 streaks of 3 in a row = plus __128 units 128x 1
256 streaks of 2 in a row = plus __256 units 256x 1
512 streaks of 1 in a row = plus __512 units 512x 1
for a total of plus 960 units minus - 1953 units = minus - 993 units
As you can see betting on the opposite decision [ P / DP ] and doubling each losing wager until you win WILL lose you money.
Playing BOTH sides P and DP at the same time.
Play both sides at the same time increasing the wager on the side of P / DP that you have just lost a wager on and using the side that you won a wager on as a BACK UP BET.
Start by placing 1 unit on P AND DP [at the same time] then which ever side WINS place a wager of one unit and on the side that LOST place a wager of 2 units.
If the side with 2 units wins your ahead 1 unit then place 1 unit on the winning side and go up to 2 units on the side that lost, the one that only had 1 unit on it.
If the side that had 2 units on it lost keep only 1 unit on the winning side but increase your wager on the losing side up to 3 units.
If the side that has the 3 units on it wins and the side with 1 unit loses you are ahead 1 unit, so place 1 unit on the side that won and 2 units on the side that lost.
If the side with 3 units on it loses and the side with 1 unit wins keep one unit on the side that won and increase your wager on the losing side up to 5 units.
If the side with 5 units on it wins you are up 1 unit . Place 1 unit on the winning side and increase the losing side up to 2 units.
If the side with 5 units on it loses keep 1 unit on the side that won and increase your wager up to 9 units on the losing side.
If the side with 9 units on it wins you are up 1 unit . Place 1 unit on that winning side and increase the unit on the losing side up to 2 units.
If the side with 9 units loses we STOP HERE AND TAKE OUR LOSS! We are only down 15 units NOT 31 units like we would be after 5 losses in a row if we were only playing ONE SIDE [ P or DP ] and
doubling our wager after each loss.
WE ARE DOWN LESS THAN 50% of what we could be, talk about minimizing losses.
EXAMPLE: P wins then DP wins
__1____1___ wager/ P wins DP loss= even
__1____2___ wager/ P lost DP wins= plus 1 unit
EXAMPLE: P wins P wins DP wins
__1____1___ wager/ P wins DP lost= even
__1____2___ wager/ P wind DP lost= minus 1 unit
__1____3___wager/ P lost DP wins= plus 1 unit
EXAMPLE: P wins P wins P wins DP wins
__1____1___ wager/ P wins DP lost= even
__1____2___ wager/ P wins DP lost= minus 1 unit
__1____3___ wager/ P wins DP lost= minus 3 units
__1____5___wager/ P lost DP wins = plus 1 unit
EXAMPLE: P wins P wins P wins P wins DP wins
__1____1____wager/ P wins DP lost= even
__1____2____wager/ P wins DP lost= minus 1 unit
__1____3____wager/ P wins DP lost= minus 3 units
__1____5____wager/ P wins DP lost= minus 7 units
__1____9____wager/ P lost DP wins= plus 1 unit
EXAMPLE: P wins P wins P wins P wins P wins [ followed by a P or DP it will not make a difference -this is where we end our losing streak]
__1____1____wager/ P wins DP lost= even
__1____2____wager/ P wins DP lost= minus 1 unit
__1____3____wager/ P wins DP lost= minus 3 units
__1____5____wager/ P wins DP lost= minus 7 units
__1____9____wager/ P wins DP lost= minus 15 units
EXAMPLE: P wins DP wins DP wins P wins P wins P wins DP wins P wins P wins DP wins.
__1____1____wager/ P wins DP lost= even
__1____2____wager/ P lost DP wins= plus 1 unit
__2____1____wager/ P lost DP wins= even
__3____1____wager/ P wins DP lost= plus 2 units
__1____2____wager/ P wins DP lost= plus 1 unit
__1____3____wager/ P wins DP lost= minus 1 unit
__1____5____wager/ P lost DP wins= plus 3 units
__2____1____wager/ P wins DP lost= plus 4 units
__1____2____wager/ P wins DP lost= plus 3 units
__1____3____wager/ P lost DP wins= plus 5 units
Playing this way seems to work for every streak of P or DP 1-2-3 or 4 in a row your ahead 1 unit and for every streak of 5 or more you are down 15 units.
Let us look at what will happen in the long term [run] . When we look at the amount of streaks we will get.
Streaks of P and DP :
01 in a row = plus 1 unit
02 in a row = plus 1 unit
03 in a row = plus 1 unit
04 in a row = plus 1 unit
05 in a row = minus 15 units
06 in a row = minus 15 units
07 in a row = minus 15 units
08 in a row = minus 15 units
09 in a row = minus 15 units
10 in a row = minus 15 units
amount of streaks 50% P and 50% DP
Streaks of - in a row
01__512 ----- plus 512
02__256 ----- plus 256
03__128 ----- plus 128
04___64 ----- plus 64
total for streaks 1-4 = plus 960 units
05__32 ----- 32 x minus 15 units = minus 480
06__16 ----- 16 x minus 15 units = minus 240
07___8 ----- 8 x minus 15 units = minus 120
08___4 ----- 4 x minus 15 units = minus 60
09___2 ----- 2 x minus 15 units = minus 30
10___1 ----- 1 x minus 15 units = minus 15
total for streaks 5-10 = minus 945 units
Grand total plus 960 units - minus 945 units = plus 15 units
That seems like a lot of work just for 15 units not to talk about the time that would take, but think about this will you really bust at 15 units every time? Will the method of play really go like
this win 15 units then lose 15 units lose 15 units win 15 units and break even. NO. When you bust 15 units you have to ADD the amount of units you won to get the TRUE amount of the bust. Say you won
12 units and then bust 15 units your TRUE bust is only 3 units or say you won 22 units and then bust 15 units your TRUE bust is really a win of 7 units.
So in a 50/50 game [ craps ] you have to look at what the average is in the long run and that is your average bust WILL only be 7 1/2 [ 8 ] units playing this way and your average win will be 7 1/2 [
8 ] units. Now the great part.
Play what is called a 2-1-2-1 bankroll money management strategy along with what I have just shown you and you can not lose
A 2-1-2-1 strategy is easy what you do is break ALL your play into segments and win or lose you leave after a segment and take a break and return later to start another segment.
But here is how you play each segment.
Bankroll required
You need 30 units- each unit MUST be at least the table minimum bet [ example; at a $ 1.00 table you need $ 30.00 at a $ 5.00 table you need $150.00 ] per segment to be played also you NEED to be
able to play AT LEAST 3 segments
Start segment one by playing the 1-2-3-5-9-unit progression with DOUBLE the wager progression. Play 2-4-6-10-18 with 2 units on the other side as your back-up wager. Play this way until you win 30
units or bust [ your average bust will be 50 % of segment money- 15 units ] .IF YOU WIN 30 UNITS: walk away and take a break or find another table to play at.
When you return to play this time [ your 2nd segment ] Play 1-2-3-5-9 unit progression with 1 unit as your back-up wager. IF YOU BUST : start over playing 2-4-6- 10-18 unit progression with a 2 unit
back-up wager again.
EVERY TIME YOU BUST START OVER WITH THE 2-4-6-10-18 PROGRESSION WITH THE 2 UNIT BACK-UP WAGER.
EVERY TIME YOU WIN 15 UNITS ON YOUR 2ND SEGMENT START OVER PLAYING THE 2-4-6-10-18 PROGRESSION WITH A 2 UNIT BACK-UP WAGER.
So when you are playing the 2-4-6-10-18 progression your average bust will be 15 units and when you play the 1-2-3-5-9 progression your average bust will be 8 units.
Over ANY two segments you can either,
win and win
win and lose
lose and win
lose and lose
win and win = plus 45 units [ 30 and 15 ]
win and lose = plus 22 units [ 30 and -8 avg. ]
lose and lose = minus 30 units [ -15 avg. and -15 avg. ]
lose and win = plus 15 units [ -15 avg. and 30 ]
WHAT WILL HAPPEN IN THE LONG TERM [ RUN]
Wins AND losses in a row :
[ a 50/50 game ]
EXPECTED WINS IN A ROW [ sample only to 5 ]
1. 16 W-L [ 30 - 8 avg. = 22 ]
2. 8 W-W-L [ 30/15 - 15 avg. = 30 ]
3. 4 W-W W-L [ 30/15/30 - 8 avg. = 67 ]
4. 2 W-W-W-W-L [ 30/15/30/15 - 15 avg. = 75 ]
5. 1 W-W-W-W-W-L [ 30/15/30/15/30 - 8 avg. = 112 ]
1. 16 x 22 = 352
2. 8 x 30 = 240
3. 4 x 67 = 268
4. 2 x 75 = 150
5. 1 x 112 = 112
total = 1122
EXPECTED LOSS IN A ROW [ sample only to 5 ]
1. 16 L-W [ -15 avg. and plus 30 = 15 ]
2. 8 L-L-W [ -15 / -15 avg. and plus 30 = 0 ]
3. 4 L-L-L-W [ -15 / -15 / -15 avg. and plus 30 = - 15 ]
4. 2 L-L-L-L-W [ -15 / -15 / -15 / -15 avg. and plus 30 = - 30 ]
5. 1 L-L-L-L-L-W [ -15 / -15 / -15 / -15 / -15 avg. and plus 30 = - 45
1. 16 x 15 = 240
2. 8 x 0 = 0
3. 4 x -15 = -60
4. 2 x -30 = -60
5. 1 x -45 = -45
total = 75 [ plus 240 - 0 - 60 - 60 - 45 = plus 75
So if what is expected to happen happens and it will [ over the long run with little variation swings in the short run ] YOU CAN NOT LOSE
P.S. about the bar
When playing this way the bar will have just a slight influence on your play. When the pass line has a higher wager on it when the bar is rolled on the come out treat it as a lose and increase your
wager following the rules and when you win on the pass line you will BREAK EVEN not win 1 unit.
BUT if when the bar was rolled on the come out roll and you had a higher wager on it and did not get paid for the don't pass win just REPEAT the wagers on the pass and don't pass line and when you
win on the don't pass line you will also break even
George Herrlinger
See Bet Progressions for more methods.
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Backgammon - Doubling Weaker Players | The Gammon Press
White – Pips 4 (-2), Match -7 -3
Black – Pips 6 (+2), Match -3 -7
Black on roll. Cube action?
Black leads 4-0 in a 7-point match and owns a 2-cube.
(a) Assume you are Black, you are a very strong player, and you are playing another very strong player. Should Black double in this position? If Black doubles, should White take or drop?
(b) Same question, but now you are Black playing a weak player.
We’ll start by figuring out the correct cube action assuming you and your opponent will make correct decisions in the future. Then we can see if the theoretically correct action requires any
adjustment depending on the strength of your opponent.
To figure out the right cube action, we need to estimate three numbers:
(1) Black’s actual chance of winning the game from this point, disregarding the cube.
(2) The value of a 6-0 lead in a 7-point match (which happens if Black doubles to 4 and White drops, or Black doesn’t double and simply wins the game with the cube on 2.)
(3) The value of a 4-2 lead in a 7-point match (which happens if Black doesn’t double and White pulls the game out.)
Part (1) is pretty easy. White only wins if Black first doesn’t bear off his three checkers, which happens 31/36 of the time (note that 1-1 is not a winning double for Black), and White then rolls a
double, which happens 1/6 of the time.
31/36 * 1/6 = 31/216 = 14.4%
[You might ask, “How do players actually do these calculations over the board?” I use a lot of numerical tricks, as do most other players. After you’ve played a lot of tournament backgammon, you
realize that some tricks are particularly useful and some numbers recur constantly. In this calculation, I’d note that 31/36 is only a little larger than 5/6, so what’s 5/6 times 1/6? Well, 1/6 is
about 16.7%, so 5/6 of that will be a little less than 14%. We want a number a little bigger than that, so our answer is 14%+. That’s plenty good enough for our purposes. Trying to get exact answers
in your head is pretty hard, but close approximations are much easier and almost always good enough.]
What about part (2), the value of a 6-0 lead in a 7-point match? Since the Crawford Rule is in effect, White must win the next game, taking him to a 6-1 deficit. (It doesn’t matter if he wins a
gammon in the Crawford Game or not.) He’ll then double to 2 at the start of every game. If he then never wins a gammon, he’ll need to win three more games to win the match, for a total of four
straight wins overall. The probability of four straight wins between two equal players is
½ * ½ * ½ * ½ = 6.25%
But if White wins a gammon in either the second or third game, he wins 4 points and saves a game, so in that case he only needs to win three straight. The probability of that is 12.5%.
The probability of winning the match is therefore between 6.25% and 12.5%, and a little closer to the lower number, since you’re not favored to win a gammon in a two-game sequence. A good
approximation is 9% for White’s chances, and therefore 91% for Black’s chances.
Finally, for Part (3) we need the value of leading 4-2 in a 7-point match. Different match equity tables give slightly different numbers here, but the range is roughly 64% to 66%. I’ll use 65% as a
good average value.
Now we’re ready to figure out the optimal doubling and taking decisions at this score. Let’s start with White’s take/drop decision if he gets doubled.
> If White takes and redoubles to 8 when he can, he wins the match 14.4% of the time (from Part (1)).
> If White drops, he trails 0-6 to 7 and wins the match 9% of the time (Part (2)).
So if White gets doubled, he should take and reship when he can.
Now let’s look at Black’s doubling decision.
> If Black doubles, he must assume White takes and reships (we’re postulating correct play on both sides) so he’ll win the match 85.6% of the time.
> If Black doesn’t double, he’ll win this game 85.6%, getting to 6-0, and will lose 14.4%, getting to 4-2. His total winning chances then look like this:
85.6% of the time he’s leading 6-0, and wins 91% of those.
Wins in this variation = 85.6% * 91% = 77.9%.
14.4% of the time he’s leading 4-2, and wins 65% of those.
Wins in this variation = 14.4% * 65% = 9.4%.
Total wins from not doubling = 77.9% + 9.4% = 87.3%.
So he wins 87.3% if he doesn’t double, 85.6% if he does. So Black shouldn’t double, even in this two-roll position.
Now we’re ready to tackle the two main questions. What happens when we face real opponents of varying strengths?
(a) Against a very strong player (and assuming Black is a strong player himself) Black should just make the theoretically correct play and not double. If Black errs and doubles, White should take and
redouble to 8 if Black misses.
(b) What about if Black is a strong player and is facing a weak opponent? This is a really interesting question and in fact is the whole point of this problem.
The theory of how to play against weak players originated in the 1970s with the publication of Barclay Cooke’s The Cruelest Game. Cooke expanded on his notions in two later books, Paradoxes and
Probabilities and Championship Backgammon. Cooke’s idea was that you should be very conservative with the cube against weak players, doubling only when you were pretty sure you’d get a pass, and
aiming to grind them down in a long series of 1-point and 2-point games, giving your huge skill advantage in checker play the maximum chance to work. The worst possible disaster was to give your weak
opponent the cube in a volatile position, allowing him to rewhip to 4, win an 8-point gammon, and turn the match around.
Cooke was a well-liked fellow, a real gentleman of the old school. In addition, he was an absolutely superb writer, who was able to convey better than anyone else the glamour and excitement of
high-level backgammon. The Cruelest Game, published in 1975, was one of the big influences driving the backgammon explosion of the mid-1970s. To this day, it remains the one book I would recommend if
a friend who didn’t know the game came to me and wanted to understand why backgammon was popular and what the fuss was all about. Cooke captured the drama of the game better than anyone else, before
or since.
Cooke’s ultra-conservative approach to playing weak players soon became accepted wisdom and was echoed in other books in the 1970s and 1980s. But it is, I think, completely the wrong approach. Let’s
see why.
In Greek mythology, the gryphon was a majestic creature combining some disparate parts: the body of a lion with the wings of an eagle. Cooke’s “weak player” is a little like that: a hybrid containing
components not likely to be found in nature. He plays the checkers so poorly that you’re a huge favorite to grind him down, one or two points at a time. But he handles doubling decisions superbly; he
scoops up cubes in volatile but takeable positions that might have a strong player scratching his head, then he whips it back when the game starts to turn in his favor, applying maximum pressure.
What a tiger!
Do real “weak players” actually play like this? In my experience, almost none do. In the real world, bad players handle the checkers poorly, but they handle the cube even worse. Their cube action is
mostly tentative; they know they don’t play well, so they try to postpone decisions that might make them look foolish. They double late, or not at all, because they’re waiting for positions that are
so strong that doubling can’t be a mistake. When doubled, they’d rather drop than take, because dropping only loses a point, while taking might lose four points. Besides, they know you’re a better
player, and they’re picking up cues from you. If you’re doubling, then you must believe you have a big advantage. Who are they to argue? Better to drop.
The best way to play against weak players is to be very aggressive. Double a little early, especially in volatile positions. They’ll probably drop, but if they take, you’re better off in a number of
> Their checker errors will now be occurring with the cube on 2 rather than 1.
> They may redouble prematurely, giving you an extra edge.
> They may redouble late or not at all, giving you a huge edge.
All of these edges add up to huge vigorish over time.
Interestingly, poker players tend to handle weak opponents better than backgammon players do. Poker players understand the value of relentless aggression, and they apply it ruthlessly, raising their
limps, 3-betting their raises, and pushing the action after the flop. In part, I think, this is because poker players don’t really attach much importance to individual hands; there are many more
hands in a poker session than games in a backgammon session, so poker players find it easier to just make what their experience tells them is the best move against this particular opponent, and let
the chips fall where they may (pun intended).
Now, after this long intermezzo, back to Part (b). What do we do if we’re a strong player and our opponent is a weak player?
We double, and we do so quickly, without giving any hint that there might be something to think about. Our opponent will drop, because it’s a two-roll position, and everybody knows that’s a pass. And
we pocket our two points and our 91-9 edge without risking losing the game.
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Vedic Mathematics by W. B. Kandasamy, F. Smarandache
Vedic Mathematics
by W. B. Kandasamy, F. Smarandache
Publisher: Automaton 2006
ISBN/ASIN: 1599730049
ISBN-13: 9781599730042
Number of pages: 220
In traditional Hinduism, the Vedas are considered divine in origin and are assumed to be revelations from God. In this book the authors probe into Vedic Mathematics and explore if it is really Vedic
in origin or Mathematics in content.
Download or read it online for free here:
Download link
(790KB, PDF)
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Characteristics of Exponential Functions | Progress and Predictions
we call functions like this "exponential" because x is in the exponent
1% or 10% or 100% interestโ ฆ does it change the picture?
When you put your money in the bankโ ฆ of course it matters.
In the big scheme of things there is no difference as long as we take a number greater than one and keep multiplying it with itself repeatedly we see a similar curve:
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Evaluating Mathematica expressions in Python/Scipy | Alex Seeholzer
Evaluating Mathematica expressions in Python/Scipy
In my research I am currently using both Python/Scipy and Mathematica, the former for data evaluation and the latter for symbolic computations. I currently have to transfer symbolic expressions
computed in Mathematica to Python and evaluate them for given parameters – something I didn’t find a satisfying solution right away.
In order not to translate simple expressions by hand each time I was looking for a MathML/Latex to Python parser. Apparently Sympy offers MathML parsing, but I found a workaround, allowing you to
copy/paste lengthy Mathematica expressions into your py code and evaluate them efficiently, using scipy.weave and Mathematica’s CForm.
1. In Mathematica use CForm[Expression] to get a C version of your expression (using custom Mathematica functions, e.g. Power).
2. Either copy the header file mdefs.h from Mathematica/SystemFiles/IncludeFiles/C to /usr/local/include or change the reference in the snippet below – I like my headers in one place.
3. Then use scipy.weave.inline(code,args) to compile and execute your C code, but be sure to include the mdefs.h header. I wrote a quick wrapper for this:
[code lang=”python”]
from scipy import weave
def evalcform(code,argdict):
return weave.inline(‘#include “mdefs.h”\n’+code,
4. You can now pass code and a dictionary of variables and values to the wrapper, which will be passed on to the compiled expression and evaluated. E.g.:
[code lang=”python”]
code = “””
return_val = Power(A,2) + Power(B,0.5)*Sin(C*Pi);
vars = {‘A’:2.,’B’:2.,’C’:.5}
print mathematica.evalcform(code,vars)
which (probably) evaluates to
Note that the C code will only be compiled once, until the code (not the values of your variables) changes, which is quite fast for plotting for example. Also mind this is only a fast workaround, and
will work only for expressions involving standard trigonometrical, exponential and power functions (have a look into the mdefs.h for an overview).
2 Comments
Nice one!
Thanks – this saved me from some terrible regex/substitution/eval() nightmare!
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In mathematics, a normal map is a concept in geometric topology due to William Browder which is of fundamental importance in surgery theory. Given a Poincaré complex X (more geometrically a Poincaré
space), a normal map on X endows the space, roughly speaking, with some of the homotopy-theoretic global structure of a closed manifold. In particular, X has a good candidate for a stable normal
bundle and a Thom collapse map, which is equivalent to there being a map from a manifold M to X matching the fundamental classes and preserving normal bundle information. If the dimension of X is ${\
displaystyle \geq }$ 5 there is then only the algebraic topology surgery obstruction due to C. T. C. Wall to X actually being homotopy equivalent to a closed manifold. Normal maps also apply to the
study of the uniqueness of manifold structures within a homotopy type, which was pioneered by Sergei Novikov.
The cobordism classes of normal maps on X are called normal invariants. Depending on the category of manifolds (differentiable, piecewise-linear, or topological), there are similarly defined, but
inequivalent, concepts of normal maps and normal invariants.
It is possible to perform surgery on normal maps, meaning surgery on the domain manifold, and preserving the map. Surgery on normal maps allows one to systematically kill elements in the relative
homotopy groups by representing them as embeddings with trivial normal bundle.
There are two equivalent definitions of normal maps, depending on whether one uses normal bundles or tangent bundles of manifolds. Hence it is possible to switch between the definitions which turns
out to be quite convenient.
1. Given a Poincaré complex X (i.e. a CW-complex whose cellular chain complex satisfies Poincaré duality) of formal dimension ${\displaystyle n}$ , a normal map on X consists of
• a map ${\displaystyle f\colon M\to X}$ from some closed n-dimensional manifold M,
• a bundle ${\displaystyle \xi }$ over X, and a stable map from the stable normal bundle ${\displaystyle u _{M}}$ of ${\displaystyle M}$ to ${\displaystyle \xi }$ , and
• usually the normal map is supposed to be of degree one. That means that the fundamental class of ${\displaystyle M}$ should be mapped under ${\displaystyle f}$ to the fundamental class of ${\
displaystyle X}$ : ${\displaystyle f_{*}([M])=[X]\in H_{n}(X)}$ .
2. Given a Poincaré complex ${\displaystyle X}$ (i.e. a CW-complex whose cellular chain complex satisfies Poincaré duality) of formal dimension ${\displaystyle n}$ , a normal map on ${\displaystyle
X}$ (with respect to the tangent bundle) consists of
• a map ${\displaystyle f\colon M\to X}$ from some closed ${\displaystyle n}$ -dimensional manifold ${\displaystyle M}$ ,
• a bundle ${\displaystyle \xi }$ over ${\displaystyle X}$ , and a stable map from the stable tangent bundle ${\displaystyle \tau _{M}\oplus \varepsilon ^{k}}$ of ${\displaystyle M}$ to ${\
displaystyle \xi }$ , and
• similarly as above it is required that the fundamental class of ${\displaystyle M}$ should be mapped under ${\displaystyle f}$ to the fundamental class of ${\displaystyle X}$ : ${\displaystyle f_
{*}([M])=[X]\in H_{n}(X)}$ .
Two normal maps are equivalent if there exists a normal bordism between them.
Role in surgery theory
Surgery on maps versus surgery on normal maps
Consider the question:
Is the Poincaré complex X of formal dimension n homotopy-equivalent to a closed n-manifold?
A naive surgery approach to this question would be: start with some map ${\displaystyle M\to X}$ from some manifold ${\displaystyle M}$ to ${\displaystyle X}$ , and try to do surgery on it to make a
homotopy equivalence out of it. Notice the following: Since our starting map was arbitrarily chosen, and surgery always produces cobordant maps, this procedure has to be performed (in the worst case)
for all cobordism classes of maps ${\displaystyle M\to X}$ . This kind of cobordism theory is a homology theory whose coefficients have been calculated by Thom: therefore the cobordism classes of
such maps are computable at least in theory for all spaces ${\displaystyle X}$ .
However, it turns out that it is very difficult to decide whether it is possible to make a homotopy equivalence out of the map by means of surgery, whereas the same question is much easier when the
map comes with the extra structure of a normal map. Therefore, in the classical surgery approach to our question, one starts with a normal map ${\displaystyle f\colon M\to X}$ (suppose there exists
any), and performs surgery on it. This has several advantages:
• The map being of degree one implies that the homology of ${\displaystyle M}$ splits as a direct sum of the homology of ${\displaystyle X}$ and the so-called surgery kernel ${\displaystyle K_{*}
(M)=ker(f_{*}\colon H_{*}(M)\to H_{*}(X))}$ , that is ${\displaystyle H_{*}(M)=K_{*}(M)\oplus H_{*}(X)}$ . (Here we suppose that ${\displaystyle f}$ induces an isomorphism of fundamental groups
and use homology with local coefficients in ${\displaystyle Z[\pi _{1}(X)]}$ .)
By Whitehead's theorem, the map ${\displaystyle f}$ is a homotopy equivalence if and only if the surgery kernel is zero.
• The bundle data implies the following: Suppose that an element ${\displaystyle \alpha \in \pi _{p+1}(f)}$ (the relative homotopy group of ${\displaystyle f}$ ) can be represented by an embedding
${\displaystyle \phi :S^{p}\to M}$ (or more generally an immersion) with a null-homotopy of ${\displaystyle f\circ \phi :S^{p}\to X}$ . Then it can be represented by an embedding (or immersion)
whose normal bundle is stably trivial. This observation is important since surgery is only possible on embeddings with a trivial normal bundle. For example, if ${\displaystyle p}$ is less than
half the dimension of ${\displaystyle X}$ , every map ${\displaystyle S^{p}\to X}$ is homotopic to an embedding by a theorem of Whitney. On the other hand, every stably trivial normal bundle of
such an embedding is automatically trivial, since ${\displaystyle \pi _{p}(BO,BO_{k})=0}$ for ${\displaystyle k>p}$ . Therefore, surgery on normal maps can always be done below the middle
dimension. This is not true for arbitrary maps.
Notice that this new approach makes it necessary to classify the bordism classes of normal maps, which are the normal invariants. Contrarily to cobordism classes of maps, the normal invariants are a
cohomology theory. Its coefficients are known in the case of topological manifolds. For the case of smooth manifolds, the coefficients of the theory are much more complicated.
Normal invariants versus structure set
There are two reasons why it is important to study the set ${\displaystyle {\mathcal {N}}(X)}$ . Recall that the main goal of surgery theory is to answer the questions:
1. Given a finite Poincaré complex ${\displaystyle X}$ is there an ${\displaystyle n}$ -manifold homotopy equivalent to ${\displaystyle X}$ ?
2. Given two homotopy equivalences ${\displaystyle f_{i}\colon M_{i}\rightarrow X}$ , where ${\displaystyle i=0,1}$ is there a diffeomorphism ${\displaystyle h\colon M_{0}\rightarrow M_{1}}$ such
that ${\displaystyle f_{1}\circ h\simeq f_{0}}$ ?
Notice that if the answer to these questions should be positive then it is a necessary condition that the answer to the following two questions is positive
1.' Given a finite Poincaré complex ${\displaystyle X}$ is there a degree one normal map ${\displaystyle (f,b)\colon M\rightarrow X}$ ?
2.' Given two homotopy equivalences ${\displaystyle f_{i}\colon M_{i}\rightarrow X}$ , where ${\displaystyle i=0,1}$ is there a normal cobordism ${\displaystyle (F,B)\colon (W,M_{0},M_{1})\to (X\
times I,X\times 0,X\times 1)}$ such that ${\displaystyle \partial _{0}F=f_{0}}$ and ${\displaystyle \partial _{1}F=f_{1}}$ ?
This is of course an almost trivial observation, but it is important because it turns out that there is an effective theory which answers question 1.' and also an effective theory which answers
question 1. provided the answer to 1.' is yes. Similarly for questions 2. and 2.' Notice also that we can phrase the questions as follows:
1.' Is ${\displaystyle {\mathcal {N}}(X)eq \emptyset }$ ?
2.' Is ${\displaystyle f_{0}=f_{1}}$ in ${\displaystyle {\mathcal {N}}(X)}$ ?
Hence studying ${\displaystyle {\mathcal {N}}(X)}$ is really a first step in trying to understand the surgery structure set ${\displaystyle {\mathcal {S}}(X)}$ which is the main goal in surgery
theory. The point is that ${\displaystyle {\mathcal {N}}(X)}$ is much more accessible from the point of view of algebraic topology as is explained below.
Homotopy theory
1.' Let X be a finite n-dimensional Poincaré complex. It is useful to use the definition of ${\displaystyle {\mathcal {N}}(X)}$ with normal bundles. Recall that a (smooth) manifold has a unique
tangent bundle and a unique stable normal bundle. But a finite Poincaré complex does not possess such a unique bundle. Nevertheless, it possesses a substitute - a unique in some sense spherical
fibration - the so-called Spivak normal fibration. This has a property that if ${\displaystyle X}$ is homotopy equivalent to a manifold then the spherical fibration associated to the pullback of the
normal bundle of that manifold is isomorphic to the Spivak normal fibration. So it follows that if ${\displaystyle {\mathcal {N}}(X)eq \emptyset }$ then the Spivak normal fibration has a bundle
reduction. By the Pontrjagin-Thom construction the converse is also true.
This can be formulated in terms of homotopy theory. Recall ${\displaystyle BG}$ the classifying space for stable spherical fibrations, ${\displaystyle BO}$ the classifying space for stable vector
bundles and the map ${\displaystyle J\colon BO\rightarrow BG}$ which is induced by the inclusion ${\displaystyle O\hookrightarrow G}$ and which corresponds to taking the associated spherical
fibration of a vector bundle. In fact we have a fibration sequence ${\displaystyle BO\rightarrow BG\rightarrow B(G/O)}$ . The Spivak normal fibration is classified by a map ${\displaystyle u _{X}\
colon X\rightarrow BG}$ . It has a vector bundle reduction if and only if ${\displaystyle u _{X}}$ has a lift ${\displaystyle {\tilde {u }}_{X}\colon X\rightarrow BO}$ . This is equivalent to
requiring that the composition ${\displaystyle X\rightarrow BG\rightarrow B(G/O)}$ is null-homotopic.
Note that the homotopy groups of ${\displaystyle B(G/O)}$ are known in certain low-dimensions and are non-trivial which suggests the possibility that the above condition can fail for some ${\
displaystyle X}$ . There are in fact such finite Poincaré complexes, and the first example was obtained by Gitler and Stasheff, yielding thus an example of a Poincaré complex not homotopy equivalent
to a manifold.
2.' Relativizing the above considerations one obtains an (unnatural) bijection
${\displaystyle {\mathcal {N}}(X)\cong [X,G/O].}$
Different categories
The above bijection gives ${\displaystyle {\mathcal {N}}(X)}$ a structure of an abelian group since the space ${\displaystyle G/O}$ is a loop space and in fact an infinite loop space so the normal
invariants are a zeroth cohomology group of an extraordinary cohomology theory defined by that infinite loop space. Note that similar ideas apply in the other categories of manifolds and one has
${\displaystyle {\mathcal {N}}(X)\cong [X,G/O]}$ , and ${\displaystyle {\mathcal {N}}^{PL}(X)\cong [X,G/PL]}$ , and ${\displaystyle {\mathcal {N}}^{TOP}(X)\cong [X,G/TOP].}$
It is well known that the spaces
${\displaystyle G/O}$ , ${\displaystyle G/PL}$ and ${\displaystyle G/TOP}$
are mutually not homotopy equivalent and hence one obtains three different cohomology theories.
Sullivan analyzed the cases ${\displaystyle G/PL}$ and ${\displaystyle G/TOP}$ . He showed that these spaces possess alternative infinite loop space structures which are in fact better from the
following point of view: Recall that there is a surgery obstruction map from normal invariants to the L-group. With the above described groups structure on the normal invariants this map is NOT a
homomorphism. However, with the group structure from Sullivan's theorem it becomes a homomorphism in the categories ${\displaystyle CAT=PL}$ , and ${\displaystyle TOP}$ . His theorem also links these
new group structures to the well-known cohomology theories: the singular cohomology and real K-theory.
• Browder, William (1972), Surgery on simply-connected manifolds, Berlin, New York: Springer-Verlag, MR 0358813
• Gitler, Samule; Stasheff, James D. (November 1965), "The first exotic class of BF", Topology, 4 (3): 257–266, doi:10.1016/0040-9383(65)90010-8
• Lück, Wolfgang (2002), A basic introduction to surgery theory (PDF), ICTP Lecture Notes Series 9, Band 1, of the school "High-dimensional manifold theory" in Trieste, May/June 2001, Abdus Salam
International Centre for Theoretical Physics, Trieste 1-224
• Ranicki, Andrew (2002), Algebraic and Geometric Surgery, Oxford Mathematical Monographs, Clarendon Press, CiteSeerX 10.1.1.309.8886, doi:10.1093/acprof:oso/9780198509240.001.0001, ISBN
978-0-19-850924-0, MR 2061749
• Wall, C. T. C. (1999), Surgery on compact manifolds, Mathematical Surveys and Monographs, vol. 69 (2nd ed.), Providence, R.I.: American Mathematical Society, CiteSeerX 10.1.1.309.8451, doi
:10.1090/surv/069, ISBN 978-0-8218-0942-6, MR 1687388
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Math 302A - For Elementary Education Teaching Majors
Math Placement, Transfer Credit/Credit by Exam, PPL (Math Placement Test) → First and Second Year Math Courses → Math 302A - For Elementary Education Teaching Majors
Description: Development of a basis for understanding the common processes in elementary mathematics related to whole numbers, fractions, integers, and probability. This course is for elementary
education majors only.
Placement Level: ALEKS PPL score of 60-100% required. Test scores expire after one year.
If my scores are lower than the Placement Level: Math 100, then Math 106, then Math 302A
Comments: For majors within the College of Education only. This course cannot be used as a prerequisite for any other mathematics courses (except Math 302B for Elementary Education.)
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How do we determine the plate cost of the Biryaaniwala menu?
Determining the plate cost of the Biryaaniwala menu involves calculating the ingredients used for each dish and adding associated costs to understand the overall expense. Here’s a step-by-step guide:
Let’s assume we are calculating the plate cost for a Chicken Biryani serving:
Total Ingredient Cost = ₹10 + ₹30 + ₹5 + ₹2 + ₹1.5 = ₹48.5
Therefore, the plate cost for a Chicken Biryani serving is ₹80.17.
By accurately calculating the plate cost for each menu item, Biryaaniwala can price their dishes appropriately to ensure profitability while maintaining quality and customer satisfaction.
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Improved bounds in the metric cotype inequality for Banach spaces
It is shown that if (X,||·||X) is a Banach space with Rademacher cotype q then for every integer n there exists an even integer m≲n^1+1/q such that for every f:Z[m]^n→X we have. where the
expectations are with respect to uniformly chosen x∈Z[m]^n and ε∈{-1,0,1}n, and all the implied constants may depend only on q and the Rademacher cotype q constant of X. This improves the bound of
m≲n2+1q from Mendel and Naor (2008) [13]. The proof of (1) is based on a "smoothing and approximation" procedure which simplifies the proof of the metric characterization of Rademacher cotype of
Mendel and Naor (2008) [13]. We also show that any such "smoothing and approximation" approach to metric cotype inequalities must require m≳n^1/2 + 1/q.
All Science Journal Classification (ASJC) codes
• Bi-Lipschitz embeddings
• Coarse embeddings
• Metric cotype
Dive into the research topics of 'Improved bounds in the metric cotype inequality for Banach spaces'. Together they form a unique fingerprint.
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What does the Pi symbol mean? Because we know it's (sadly) not about pie
What does the Pi symbol mean? Because we know it’s (sadly) not about pie
Every year on March 14th, teachers, scientists, and those who just love math and numbers celebrate Pi Day, a special day that pays tribute to Pi… otherwise known as the number 3.14 (plus a bunch of
other digits that are really hard to memorize). But unless you’re a math teacher, you might not know why that number is represented by a certain symbol. So what does the Pi symbol mean?
In case it’s been awhile since you sat down in a math class, Pi is the ratio of a circle’s circumference to its diameter, and it’s represented by the symbol π, which, just by looking at it, doesn’t
seem to relate much to the definition or the number that it represents.
But the meaning behind that symbol goes a little further. According to Live Science, the Pi symbol is meant to represent the idea of Pi as it pertains to a river.
"A river's windiness is determined by its 'meandering ratio,' or the ratio of the river's actual length to the distance from its source to its mouth as the crow flies," the site says. "Rivers
that flow straight from source to mouth have small meandering ratios, while ones that lollygag along the way have high ones. Turns out, the average meandering ratio of rivers approaches — you
guessed it — Pi."
So the symbol itself is showing the straight line connection between two bends in a river, which is actually pretty cool — and that connecting line is also showing the diameter. Mind = blown.
Pi is, by definition, an irrational number, so it makes sense that it would relate to something else you can’t control, like a river. Math and science can often seem pretty intimidating, especially
if they’re not your best subjects. But even something as simple as a Pi symbol is kind of beautiful, isn’t it?
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The history of physics: the construction of the theory of relativity - Daily Info Blog
In his electronic theory, Lorentz formulated the fundamentals of electrodynamics of moving media and proved that the laws of electromagnetism are the same in all inertial reference frames.
The most general transformations of spatial coordinates and time found by him in 1904 — the so—called Lorentz transformations – played an important role in the preparation of the theory of
relativity, which was formulated in 1905 by A. Einstein in the work “On the electrodynamics of moving media”.
This theory replaced the classical views on space and time, which clearly contradicted the facts when it came to movements, where the speed of bodies could no longer be neglected in comparison with
the speed of light.
Maxwell’s equations did not change their form during Lorentz transformations, but the formulas of classical mechanics turned out to be non-invariant, so Lorentz’s theory did not solve the differences
between classical mechanics and Maxwell’s laws.
Experiments of Fizeau and Michelson
The French physicist Hippolyte Louis Fizeau (1819-1896), who was the first to determine the speed of light in terrestrial conditions, established the influence of the motion
of the medium on the speed of light propagation by measuring in 1851 the speed of light in moving water (Fizeau’s experiment), this experience proved that light is partially captured by a moving
The American experimental physicist Albert Abraham Michelson (1852-1931) in 1881 conducted experiments using a special interferometer made by him, which denied the existence of ether.
In 1887 , Michelson together with Edward William Morley (1838-1923) conducted new experiments, which also gave a negative result about the existence of ether.
Special theory of relativity
The emergence of the special theory of relativity was associated with attempts to overcome the difficulties encountered in constructing the electrodynamics of moving media.
The difficulty was that each of the known phenomena (the phenomenon of stellar aberration, Fizeau’s experience, Michelson’s experience) and the experiments that were used to determine the speed of
light in moving bodies could, with some simple assumptions, be easily explained within the framework of existing theories.
For example, to explain the phenomenon of stellar aberration, it was enough to assume that the ether is not captured by moving bodies (Jung), to explain the experience of Physicists — to make an
assumption about the partial capture of ether, and to explain the negative result of Michelson’s experience — to use the hypothesis of the English physicist and mathematician George Gabriel Stokes
(1819-1903) about the complete capture of ether by the Earth.
Due to the incompatibility of these hypotheses , it became it is obvious that it is necessary to move from classical physics to some new physical theory in which all optical phenomena in moving media
could be explained from a single point of view without the use of contradictory hypotheses. There have been other attempts to explain phenomena in moving bodies without changing the basic physical
[Source: Unsplash License, Math equations on a chalk board]
The Irish scientist George Fitzgerald and a little later, but independently, Lorenz attempted to explain the negative result of Michelson’s experiment using the hypothesis of Lorentz contraction,
according to which moving bodies undergo a contraction in the direction of motion, and it is the greater the higher the speed of the body. However, the Fitzgerald-Lorenz idea seemed too artificial,
proposed specifically for one particular phenomenon, it was not supported by any theoretical evidence.
So, the artificiality of all these attempts prompted the idea of the need to abandon Newtonian ideas about space and time and the concept of ether, which were included in the electromagnetic picture
of the world. Einstein’s merit lay precisely in the fact that he decisively broke with these ideas and introduced new concepts corresponding to it in electrodynamics.
Unlike all his contemporaries, Einstein saw in the negative result not an accidental difficulty, but a manifestation of a general law of nature, according to which it is impossible to detect a
rectilinear uniform and progressive motion of the laboratory relative to the ether. A. Einstein formulated two basic postulates — principles that are the starting points of the theory of relativity.
In accordance with the principle of relativity, all physical processes in an inertial system do not depend on the speed of its motion relative to other bodies or systems.
According to the second principle, the speed of light in a vacuum is constant and does not depend on the speed of the light source. A. Einstein formulated new laws of motion that generalized Newton’s
laws of motion and reduced to these laws only in the case of very small velocities of bodies.
In the same 1905 supplement “To the electrodynamics of moving bodies”, which was published under the title “Does the inertia of a body depend on the energy content in it?”, Einstein expressed the
relationship between mass and energy with his famous equation E = mc2. This formula retains its value at any speeds, if only by m we mean the inert mass of the body, which depends on the speed and
rest mass.
The rest mass corresponds to the rest energy.
The presence of this energy made it possible to consider any body as a potential reservoir of energy, and the law of proportionality of mass and energy implied the possibility of the transition of
energy associated with matter into energy associated with radiation. This formed the basis of all nuclear physics.
The first physicist who pointed out the significance of the law of mass—energy coupling to explain the deviation of atomic masses from integer values was the French physicist Paul Langevin
In 1913, he revealed the physical meaning of the mass-energy ratio and for the first time expressed the ideas that make up the essence of modern nuclear power – the ideas about the mass defect in
nuclear transformations.
German physicist and mathematician Hermann Minkowski (1864-1909) devoted his scientific activity to the development of the ideas of relativity theory, who formulated the mathematical theory of
physical processes in 1907-1908. In a four-dimensional space. In this theory, the Lorentz transformations received a visual geometric interpretation as a rotation transformation of a four-dimensional
coordinate system, which played an important role in completing the construction of the special theory of relativity.
A. Einstein also developed the general theory of relativity, which is based on a combination of the principle of equivalence of heavy and inert masses and the principle of relativity. This theory is
a relativistic theory of gravity. Einstein proved that in the presence of bodies that they form an attraction, the metric of space and time changes. The Russian mathematician N. I. Lobachevsky
(1792-1856) in his memoirs “On the principles of Geometry” (1826-1830) expressed the idea that the metric of real space can have such deviations, and tried to determine them.
In the general theory of relativity, the reason for this deviation is revealed, its mathematical expression is established, and, in particular, it is proved that such deviations in the metric of
space cannot be considered separately from the corresponding changes in time. This means that the theory of space, time and gravity shows their inextricable relationship.
Verification and confirmation of the theory
The new laws of attraction lead to consequences that can be tested experimentally. Since energy has mass, it can be concluded from here that attraction must act on energy. As a result, the beam that
passes through the gravitational field must be deflected. Experiments carried out during the total eclipses of the Sun in 1919. And 1922 confirmed the general theory of relativity in quantitative
The second proof of the general theory of relativity was obtained by observing the motion of the planets.
One of the consequences of the general theory of relativity is that the trajectory of the planet should slowly rotate around the Sun. Astronomers noticed the displacement of the perihelion of Mercury
and finally explained it in 1916. By the German physicist and astronomer Karl Schwarzschild (1873-1916).
The third proof of the general theory of relativity was the confirmation of the shift predicted by the theory. In the direction of the red color of the spectral emission lines of stars in 1925. Now
the theory of relativity is generally accepted.
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Structural Central
Snow Load
Four Different Roof Configurations
Determine snow drift loads in accordance with ASCE 7 for four different roof configurations (Roof Elevation Step, Parapet, Adjacent Structure, Roof Projection).
Descriptions in One Location
No flipping through pages to determine which options apply for surface roughness category, exposure of roof, thermal factor, and risk category.
Multiple Drift Cases per Calculation
Set the snow design criteria once, then add as many drift cases as you need for your project.
Roof Elevation Step
Adjacent Structure
Roof Projection
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Differential Equations
This page provides examples of how to represent and manipulate ordinary differential equations (ODEs) in Maxima when writing STACK questions.
Representing ODEs
In a Maxima session we can represent an ODE as
ODE: x^2*'diff(y,x) + 3*y*x = sin(x)/x;
Notice the use of the ' character in front of the diff function to prevent evaluation. Applied to a function call, such as diff, the single quote prevents evaluation of the function call, although
the arguments of the function are still evaluated (if evaluation is not otherwise prevented). The result is the noun form of the function call.
Entering DEs
The syntax to enter a derivative in Maxima is diff(y,x,n). Teachers need to use an apostrophe' character in front of the diff function to prevent evaluation in question variables (etc). E.g. to type
in $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ you need to use 'diff(y,x,2).
Students' answers always have noun forms added. If a student types in diff(y,x) then this is protected by a special function noundiff(y,x) (etc), and ends up being sent to answer test as 'diff
(y,x,1). If a student types in (literally) diff(y,x)+1 = 0 this will end up being sent to answer test as 'diff(y,x,1)+1 = 0.
The answer test AlgEquiv evaluates all nouns. This has a (perhaps) unexpected side-effect that noundiff(y,x) will be equivalent to 0, and noundiff(y(x),x) is not. For this reason we have an
alternative answer test AlgEquivNouns which does not evaluate all the nouns. The ATEqualComAss also evaluates its arguments but does not "simplify" them. So, counter-intuitively perhaps, we currently
do have ATEqualComAss(diff(x^2,x), 2*x); as true.
Students might expect to enter expressions like $y'$, $\dot{y}$ or $y_x$ (especially if you are using derivabbrev:true, see below). The use by Maxima of the apostrophe which affects evaluation also
has a side-effect that we can't accept y' as valid student input. Input y_x is an atom. Individual questions could interpret this as 'diff(y,x) but there is no systematic mechanism for interpreting
subscripts as derivatives. Input dy/dx is the division of one atom dy by another dx and so will commute with other multiplication and division in the expression as normal. There is no way to protect
input dy/dx as $\frac{\mathrm{d}y}{\mathrm{d}x}$. The only input which is interpreted by STACK as a derivative is Maxima's diff function, and students must type this as input.
The expression diff(y(x),x) is not the same as diff(y,x). In Maxima diff(y(x),x) is not evaluated further. Getting students to type diff(y(x),x) and not diff(y,x) will be a challenge. Hence, if you
want to condone the difference, it is probably best to evaluate the student's answer in the feedback variables as follows to ensure all occurrences of y become y(x).
ans1:'diff(y(x),x)+1 = 0;
Trying to substitute y(x) for y will throw an error. Don't use the following, as if the student has used y(x) then it will become y(x)(x)!
ans1:'diff(y,x)+1 = 0;
Further work is needed to better support partial derivatives (input, display and evaluation).
Displaying ODEs
Maxima has two notations to display ODEs.
If derivabbrev:false then'diff(y,x) is displayed in STACK as $\frac{\mathrm{d}y}{\mathrm{d}x}$. Note this differs from Maxima's normal notation of $\frac{\mathrm{d}}{\mathrm{d}x}y$.
If derivabbrev:true then 'diff(y,x) is displayed in STACK and Maxima as $y_x$.
• Extra brackets are sometimes produced around the differential.
• You must have simp:true otherwise the display routines will not work.
Manipulating ODEs in Maxima
This can be solved with Maxima's ode2 command and initial conditions specified. Below is an example of Maxima's output.
(%i1) ODE: x^2*'diff(y,x) + 3*y*x = sin(x)/x;
2 dy sin(x)
(%o1) x -- + 3 x y = ------
dx x
(%i2) ode2(ODE,y,x);
%c - cos(x)
(%o2) y = -----------
(%i3) ic1(%o2,x=%pi,y=0);
cos(x) + 1
(%o3) y = - ----------
Further examples and documentation are given in the Maxima manual
Note that by default STACK changes the value of Maxima's logabs variable. This changes the way $1/x$ is integrated. If you want the default behaviour of Maxima you will need to restore logabs:false
in the question variables.
Laplace Transforms
Constant coefficient ODEs can also be manipulated in STACK using Laplace Transforms. An example of a second-order constant coefficient differential equation is given below with initial conditions set
and the result of the Laplace Transform is stored.
ode: 5*'diff(x(t),t,2)-4*'diff(x(t),t)+7*x(t)=0;
sol: solve(laplace(ode,t,s), 'laplace(x(t), t, s));
sol: rhs(sol[1]);
sol: subst([x(0)=-1,diff(x(t), t)=0],sol);
The laplace command will Laplace Transform the ode (more information in maxima docs here), but it will still be in terms of the Laplace Transform of x(t), which is symbolic. The solve command then
solves the algebraic equation for this symbolic Laplace Transformed function, and on the right-hand side of the equals sign, the desired answer is obtained using the rhs command. Lastly, the initial
conditions need to be specified for x(t). The Laplace Transform symbolically specifies values for x(0) and x'(0) and these can be replaced with the subst command as shown above.
Randomly generating ODE problems
When randomly generating questions we could easily generate an ODE which cannot be solved in closed form, so that in particular using ode2 may be problematic. It is much better when setting any kind
of STACK question to start with the method and work backwards to generate the question. This ensures the question remains valid over a whole range of parameters. It also provides many intermediate
steps which are useful for a worked solution.
% characters from solve and ode2
Maxima functions such as solve and ode2 add arbitrary constants, such as constants of integration. In Maxima these are indicated adding constants which begin with percentage characters. For example,
eq1:x^2*'diff(y,x) + 3*y*x = sin(x)/x;
results in
y = (%c-cos(x))/x^3;
Notice the %c in this example. We need a function to strip out the variables starting with %, especially as these are sometimes numbered and we want to use a definite letter, or sequence for the
The function stack_strip_percent(ex,var) replaces all variable names starting with % with those in var. There are two ways to use this.
1. if var is a list then take the variables in the list in order.
2. if var is a variable name, then Maxima returns unevaluated list entries,
For example
stack_strip_percent(y = (%c-cos(x))/x^3,k);
y = (k[1]-cos(x))/x^3;
This is displayed in STACK using subscripts, which is natural. The unevaluated list method also does not need to know how many % signs appear in the expression. The other usage is to provide explicit
names for each variable, but the list must be longer than the number of constants in ex, e.g.
stack_strip_percent(y = (%c-cos(x))/x^3,[c1,c2]);
which returns
y = (c1-cos(x))/x^3;
The following example question variables can be used within STACK.
ode : x^2*'diff(y,x) + 3*y*x = sin(x)/x;
sol : stack_strip_percent(ode2(ode,y,x),[k]);
ta : rhs(ev(sol,nouns));
Note, you may need to use the Option "assume positive" to get ODE to evaluate the integrals formally and hence "solve correctly".
If you need to create a list of numbered variables use
vars0:stack_var_makelist(k, 5);
vars1:rest(stack_var_makelist(k, 6));
Assessing answers
ODEs provide a good example of the principle that we should articulate the properties we are looking for in ordinary differential equations. These properties are
1. The answer satisfies the differential equation.
2. The answer satisfies any initial/boundary conditions.
3. The answer is general.
4. The answer is in the required form.
Hence, for ODE questions we need a potential response tree which establishes a number of separate properties. On the basis of the properties satisfied, we then need to generate outcomes.
Satisfying the differential equation
When marking this kind of question, it is probably best to take the student's answer and substitute this into the ODE. The student's answer should satisfy the equation. Just "looking like the model
answer" isn't as robust. How else does the teacher avoid the problem of knowing which letter the student used to represent an arbitrary constant?
E.g. in Maxima code
ode:x^2*'diff(y,x) + 3*y*x = sin(x)/x;
ans: (c - cos(x))/x^3; /* The student's (correct) answer */
sa1, sa2 and sa2 can be used as part of the feedback when a student doesn't get the right answer.
Satisfying any initial/boundary conditions
If the student's answer is ans then we can check initial/boundary conditions at a point x=x0 simply by using
Notice in the second example the need to calculate the derivative of the student's answer before it is evaluated at the point x=x0. These values can be compared with answer tests in the usual way.
Arbitrary constants
Further tests are needed to ensure the student's solution is non-trivial, satisfies any initial conditions, or is suitably general.
To find which constants are present in an expression use Maxima's listofvars command. In particular, to find if c appears in an expression ans we can use the predicate member
However, it is unusual to want to specify the name of a constant. A student may choose another name. The example below may be helpful here.
Sometimes students use the $\pm$ operator, e.g. instead of typing in $Ae^{\lambda t}$ they type in $\pm Ae^{\lambda_1 t}$ as +-A*e^(lambda*t). The $\pm$ has a somewhat ambiguous status in
mathematics, but it is likely that many people will want to condone its use here.
Internally, the $\pm$ operator is represented with an infix (or prefix) operation #pm#, which is part of STACK but not core Maxima. Instead of a+-b teachers must type a#pm#b. Students' answers get
translated into this format. Mostly when dealing with expressions you need to remove the $\pm$ operator. To remove the $\pm$ operator STACK provides the function pm_replace(ex) which performs the
re-write rules (actually using STACK's nounor operator to prevent evaluation).
If you simply want to implement the re-write rule $a\pm b \rightarrow a+b,$ i.e. ignore the $\pm$ operator, then you can use subst( "+","#pm#", ex). For example, this substitution can be done in the
feedback variables on a student's answer. If you would like to test code offline with #pm# then you will need to make use of the Maxima sandbox.
Second order linear differential equations with constant coefficients
One important class of ODEs are the second order linear differential equations with constant coefficients.
Generating these kinds of problems is relatively simple: we just need to create a quadratic with the correct sort of roots.
Let us assume we have two real roots. We might expect an answer $Ae^{\lambda_1 t}+Be^{\lambda_2 t}$. We might have an unusual, but correct, answer such as $Ae^{\lambda_1 t}\left(1+Be^{\lambda_2 t}\
right)$. Hence, we can't just "look at the answer".
A sample question of this type is provided by STACK, in which we have the following question variables.
sa1 : subst(y(t)=ans1,ode);
sa2 : ev(sa1,nouns);
sa3 : fullratsimp(expand(sa2));
l : delete(t,listofvars(ans1));
lv : length(l);
b1 : ev(ans1,t=0,fullratsimp);
b2 : ev(ans1,t=1,fullratsimp);
m : float(if b2#0 then fullratsimp(b1/b2) else 0);
1. Here sa1, sa2 and sa3 are used to ensure the answer satisfies the ODE and if not to provide feedback.
2. To ensure we have two constants we count the number of variables using listofvars, not including t. We are looking for two constants.
3. To ensure the solution is suitably general, we confirm $y(1)eq 0$ and calculate $y(0)/y(1)$. If this simplifies to a number then the constants have cancelled out and we don't have a general
solution consisting of two linearly independent parts.
These are the properties a correct answer should have. If the teacher has a preference for the form, then a separate test is required to enforce it. For example, you might like the top operation to
be a $+$, i.e. sum. This can be confirmed by
aop : is(equal(op(ans1),"+"));
Then test aop is true with another answer test. Note that the arguments to answer tests cannot contain double quotes, so a question variable is needed here.
Next, let us assume we have complex root, e.g. in the equation
we have $\lambda = -1 \pm 2i$.
We potentially have quite a variety of solutions.
The advantage is that the same code correctly assesses all these forms of the answer.
Separating the general from particular solution.
Consider the differential equation with corresponding general solution
The solution of such an equation consists of the sum $y(t) = c_1\ y_1(t)+c_2\ y_2(t)+y_p(t)$. The general solution is the term $c_1\ y_1(t)+c_2\ y_2(t)$ and the particular solution is the part $y_p
(t)$. It is useful to separate these. Run the above code, which should work. Then we execute the following, which checks the general solution part is made up of two linearly independent parts.
/* Calculate the "Particular integral", (by setting both constants to zero) and then separate out the "general solution".*/
ansPI:ev(ans1,maplist(lambda([ex],ex=0), l));
g1 : ev(ansGS,t=0,fullratsimp);
g2 : ev(ansGS,t=1,fullratsimp);
m : float(if g2#0 then fullratsimp(g1/g2) else 0);
Notice to calculate $y_p(t)$ we set the constants $c_1=c_2=0$, but using the variables in the list l which is defined above as the list of constants without $t$.
First order exact differential equations
An important class of differential equations are the so-called first order exact differential equations of the form
Assume that $h(x,y)=c$ gives an implicit function, which satisfies this equation. Then
and so
Differentiating once further (and assuming sufficient regularity of $h$) we have
Note that this condition on $p$ and $q$ is necessary and sufficient for the ODE to be exact. In search of such a function $h(x,y)$ we may define
Notice here that $c_1$ and $c_2$ are arbitrary functions of integration. To evaluate these we differentiate again, for example taking the first of these we find
where this last equality arises from the differential equation. Rearranging this and solving we have
Similarly we may solve for If $h_1=h_2$ then we have an exact differential equation, and $h=h_1=h_2$ given the integral of our ODE.
Example $x\dot{y}+y+4=0$
As an example consider
Then $p=x$ and $q=y+4$.
And so
And so,
In both cases we obtain the same answer for $h(x,y)=xy+4x$.
Maxima code
The following Maxima code implements this method, and provides further examples of how to manipulate ODEs.
/* Solving exact differential equations in Maxima */
/* Ensure we have an expression, not an equation */
if op(ODE)="=" then ODE:lhs(ODE)-rhs(ODE);
/* This should write the ODE in the form
which we can then sort out to get the coefficients*/
/* Check our condition for an exact ODE */
if fullratsimp(diff(p,x)-diff(q,y))=0 then print("EXACT") else print("NOT EXACT")$
/* Next we need to solve
to find the integral of our ODE */
/* Note, H1 and H2 should be the same! */
/* Hence the solution is, in terms of y=...+c*/
Further examples are / Non-exact equations / ODE:y=x*'diff(y,x);
/* Exact equations */
See also
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Design Principles for {superspreading}
This vignette outlines the design decisions that have been taken during the development of the {superspreading} R package, and provides some of the reasoning, and possible pros and cons of each
This document is primarily intended to be read by those interested in understanding the code within the package and for potential package contributors.
The {superspreading} package aims to provide a range of summary metrics to characterise individual-level variation in disease transmission and its impact on the growth or decline of an epidemic.
These include calculating the probability an outbreak becomes an epidemic (probability_epidemic()), or conversely goes extinct (probability_extinct()), the probability an outbreak can be contained
(probability_contain()), the proportion of cases in cluster of a given size (proportion_cluster_size()), and the proportion of cases that cause a proportion of transmission (proportion_transmission
The other aspect of the package is to provide probability density functions and cumulative distribution functions to compute the likelihood for distribution models to estimate heterogeneity in
individual-level disease transmission that are not available in R (i.e. base R). At present we include two models: Poisson-lognormal (dpoislnorm() & ppoislnorm()) and Poisson-Weibull (dpoisweibull()
& ppoisweibull()) distributions.
The package implements a branching process simulation based on bpmodels::chain_sim() to enable the numerical calculation of the probability of containment within a outbreak time and outbreak duration
threshold. In the future this function could be removed in favour of using a package implementing branching process models as a dependency. The package is mostly focused on analytical functions that
are derived from branching process models. The package provides functions to calculate variation in individual-level transmission but does not provide functions for inference, and currently relies on
{fitdistrplus} for fitting models.
Functions with the name probability_*() return a single numeric. Functions with the name proportion_*() return a <data.frame> with as many rows as combinations of input values (see expand.grid()).
The consistency of simple well-known data structure makes it easy for users to apply these functions in different scenarios.
The distribution functions return a vector of numerics of equal length to the input vector. This is the same behaviour as the base R distribution functions.
Design decisions
• proportion_*() functions return a <data.frame> with the proportion column(s) containing character strings, formatted with a percentage sign (%) by default. It was reasoned that {superspreading}
is most likely used either as a stand-alone package, or at the terminus of a epidemiological analysis pipeline, and thus the outputs of {superspreading} functions would not be passed into other
functions. For instances where these proportions need to be passed to another calculation or for plotting purposes the format_prop argument can be switched to FALSE and a numeric column of
proportions will be returned.
• The distribution functions are vectorised (i.e. wrapped in Vectorize()). This enables them to be used identically to base R distribution functions.
• Native interoperability with <epiparameter> objects, from the {epiparameter} package is enabled for probability_*() and proportion_*() via the offspring_dist argument. This allows user to pass in
a single object and the parameters required by the {superspreading} function will be extracted, if these are not available within the <epiparameter> object the function returns an informative
error. The offspring_dist argument is after ... to ensure users specify the argument in full and not accidentally provide data to this argument.
• Internal functions have a dot (.) prefix, exported functions do not.
• Several functions use constants that are internally defined (e.g. NSIM and FINITE_INF). These are used in several functions to prevent the use of apparently arbitrary magic numbers. Constants are
all uppercase to make clear they are internal constants (following MDN and PEP8 styles. These constants should not be exported (i.e. should not appear in the NAMESPACE) as they should only be
used by functions and not package users.
The aim is to restrict the number of dependencies to a minimal required set for ease of maintenance. The current hard dependencies are:
{stats} is distributed with the R language so is viewed as a lightweight dependency, that should already be installed on a user’s machine if they have R. {checkmate} is an input checking package
widely used across Epiverse-TRACE packages.
Suggested dependencies (not including package documentation ({knitr}, {rmarkdown}), testing ({spelling} and {testthat}), and plotting ({ggplot2})) are: {epiparameter}, used to easily access
epidemiological parameters from the package’s library, and {fitdistrplus}, used for model fitting methods.
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Points On a Line | HackerRank
Given two-dimensional points in space, determine whether they lie on some vertical or horizontal line. If yes, print YES; otherwise, print NO.
The first line contains a single positive integer, , denoting the number of points.
Each line of subsequent lines contain two space-separated integers detailing the respective values of and (i.e., the coordinates of the point).
Print YES if all points lie on some horizontal or vertical line; otherwise, print NO.
All points lie on a vertical line.
The points do not all form a horizontal or vertical line.
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How to do a Mass Balance in Spreadsheet - Excel - 911Metallurgist
The “mass balance” is unquestionably the most common class of problems for the process engineer. Closely related, mathematically, is the “heat balance”. Every laboratory test requires a mass balance
to describe the results. Mass balance calculations can be subdivided into several sub-categories based on whether the results are time dependant or not and depending on whether the input data are
measurements that contain errors or are hypothetical values with no errors. Time dependent mass balances are dynamic process simulations and in addition to the conservation of mass constraints
require kinetic data input. They are one step beyond simple mass balances.
After the flow sheet has be constructed the second step is to establish what, components or parameters are to be included in the mass balance. These parameters are divided into 2 groups for
conveyance: masses and secondary parameters. Secondary parameters are those parameters which are derived from masses.
The second task is that of selecting those parameters which are to be specified as the input data. This is the most difficult task in solving the mass balance. It is possible to theoretically
determine the number of parameters which need to be specified as “inputs”, however, the choice of which parameters of which streams are specified is not completely a free choice. It is difficult to
list rules for the selection of these variables.
The spreadsheet program could also be used as a steady state process simulator if one of input parameters, for example the copper generated by the agitation leach, is given as a function of the
operating conditions such as iron concentration or flow rate. The use of the program as a simulator can be extended by replacing any input by an equation which allows that input to be calculated from
other parameters. This in effect adds one more variable to the system but also adds one more equation leaving the number of degrees of freedom unchanged.
Once the input parameters have been selected the spreadsheet in ready to be constructed. The design of the spread sheet is very much a matter of personnel choice but some general rules will simplify
the task. First, the spreadsheet should be divided into 3 parts: one for the input data, the second for the calculations and the third part for the output of the calculations. The input section
consists of several columns: one column for the name of the stream and a column for each parameter of that stream to be specified by an input. It is sometimes advantageous to have a separate forth
section in the spreadsheet to be used to create a report listing the input data.
Initially, equations for the masses in each stream are entered. Stream parameters which depend on the masses, such as volumes, pulp densities and concentration values, can be left until last.
However, some of the input data maybe these secondary parameters and this will require the equation of a mass in a stream being dependent on a secondary parameter of the stream. Any value in the
input cells or other calculation cells, be they masses or secondary parameters, are legitimate values to be used in any calculation’s cell equation.
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a) The length of a swimming pool is 50 metres.
Alex swims a total distance of 1.4 kilometres.
How many lengths did Alex swim?
Here is a sketch of the swimming pool.
b) Work out the area of the cross section of the pool.
c) Work out the capacity of the pool in litres.
(1 m^3 = 1000 litres)
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People – Uppsala Learning, Inference and Optimization Lab
Andrey joined the group in October 2023 as a postdoctoral researcher. His current research explores fairness-aware statistical experiments and machine learning. His research interest also spans
quatitative finance, in particular statistical methods in finance and predictability analysis. Andrey received his PhD in the program “Computational methods and Mathematical Models for Sciences and
Finance” in September 2023 from Scuola Normale Superiore (SNS), Pisa, Italy. Andrey received a master’s degree in 2019, and a bachelor’s degree in 2016 at the Novosibirsk State University in the
program of applied mathematics and computer science. Publications: scholar.google.com/citations?user=3grli6AAAAAJ
Aleksandr joined the group in August 2022 as a PhD student. Aleksandr holds Masters and Bachelor degrees in Applied Mathematics and Physics from the Moscow Institute of Physics and Technology (MIPT).
His doctoral research explores novel statistical design of experiments for drug repurposing. His research interests also include inverse problems and optimization.
Mayank joined the group in September 2022 as a PhD student. Mayank holds a Masters degree in Signal Processing from the Indian Institute of Technology (IIT) Gandhinagar, India. Mayank is working in
the area of likelihood-free parameter inference of stochastic models. His prior research addressed inverse problems in distributed sensor networks.
Dhanushki joined the group in September 2024 as a PhD student. Dhanushki holds a Masters degree in Data Science from Uppsala University, Sweden. Her doctoral project explores Bayesian active learning
approaches for drug discovery and repurposing. She is generally interested in Bayesian machine learning and optimization.
Csongor joined the group in September 2024 as a PhD student. Csongor holds a Masters degree in Computational Science from Uppsala University, Sweden. His doctoral work focusses on various aspects of
solving ill-posed inverse problems, and in general non-convex optimization. He is also interested in deep learning, graph theory and the intersection of machine learning and optimization.
Emily Morgan is a Master’s student in Bioinformatics at Rhodes University, where she participates in research led by Prof. Özlem Tastan-Bishop. With a background in Computer Science and Biochemistry,
Emily’s focus is on computational biology, particularly molecular dynamics simulations, dynamic residue network analysis, and machine learning methods for studying viral proteins. Her research
interests include structural biology, machine learning applications in bioinformatics, and developing predictive models to understand protein dynamics and mutations.
Kristiana is a Master’s student in Data Science at the IT University of Copenhagen, Denmark. Kristiana’s graduate project is in collaboration with Prof. Ida-Maria Sintorn, and explores uncertainty
quantification in large-scale generative models.
Stela is a Master’s student in Data Science at the IT University of Copenhagen, Denmark. Stela’s graduate project is in collaboration with Prof. Ida-Maria Sintorn, and explores robust guided
Stefanos is pursuing his Masters degree in Data Science at Uppsala University. His thesis project is a joint collaboration with Vahid Azimi Mousolou at Dept. of Physics and Astronomy. The subject of
his thesis is characterising quantum entanglement as an optimization problem, to be solved in a data-efficient manner using Bayesian optimization.
Sajad is pursuing his Masters degree in Computational Science at Uppsala University. As part of his thesis project, he is exploring variational autoencoders for denoising and de-biasing applications
in scientific data.
We always look forward to hearing from talented and motivated future colleagues at all levels – Masters/PhD/PostDoc.
• Ema Duljkovic, 2023, MSc Data Science, now R&D intern at Ericsson.
• Mina Badri, 2023, MSc Data Science, Data Science intern at Volvo Cars.
• Liang Cheng; 2023, MSc Data Science, Thesis: ‘Autoencoder-Based Likelihood-Free Parameter Inference of Gene Regulatory Network‘, now Doctoral Student at University of Oslo
• Ansar Siddiqui; 2023, BSc Computer Science, Thesis: ‘Parameter Inference for Stochastic Models of Gene Expression in Eukaryotic Cells‘, now Cloud Developer at Scania
• Max Andersson, Carl Löfkvist, Edward Glöckner; 2023, BSc Engineering Physics, Thesis: ‘Dark Matter signals at the Large Hadron Collider with Deep Learning‘
• Mona Babikir; 2019, MSc Computer Science, Thesis: ‘Machine Learning for Cloud: Modeling Cluster Health using Usage Parameters‘, now Software Engineer at Tink AB
• Iliam Barkino; 2019, MSc Engineering Physics, Thesis: ‘Summary Statistic Selection with Reinforcement Learning‘, now Deep Tech VC at Industrifonden
• Mattias Åkesson; 2018, MSc Engineering Physics, Thesis: ‘Learning Phantom Dose Distribution using Regression Artificial Neural Networks‘, now Senior Federated Learning Engineer at Scaleout
Systems AB
email: firstname.lastname@it.uu.se
visiting address: Lägerhyddsvägen 1, Hus 10, 752 37 Uppsala
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Grid AoE Mechanics
• #1 Jul 20, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
So I'm in a campaign and I'm getting stumped on how AoEs work on a grid. So we are playing in Roll20 and I'm asking the DM how they work. It seems he's trying to turn circles into squares, which
would be fine, but there are some confusions. For example, he said moonbeam is essentially a 10 foot square on a grid. 5-foot radius equals 10-foot cube would be fine if every AoE has a point of
origin in space. The problem is you have spells like ice knife where the point of origin is a creature.
A solution I have to simplify this is the Roll20 site has a measuring tool. You can measure from one square to another. If you have an AoE radius, you can pick a center square and measure out to any
creature that may be within. If the distance is less the the radius, they're affected. If not, they're not. The thing is it measures lines at angles as equivalent to vertical and horizontal ones,
which means 5-foot radius would always be like a 15-foot cube. It's not a perfect solution but it's simpler and more consistent than the former.
Cones on the other hand confound me. Say I use burning hands. I measure 15 feet out and 7.5 feet in both directions for the triangle base? Then I subtract one square as I go back? I haven't worked
out a good way to figure out this AoE.
I don't think the rules are very clear on how AoEs can work on a grid. If anyone can tell me a comprehensive and simple way to implement AoE rules for A grid, it would be appreciated. I can pitch it
to my DM as an alternative to the idea he has, which I don't even understand myself.
#2 Jul 20, 2024
• Initiate of the 1st Circle
• Join Date: 3/24/2018
• Posts: 9,113
• Member Details
"Sooner or later, your Players are going to smash your railroad into a sandbox."
"real life is a super high CR."
"............anybody got any potatoes? We could drop a potato in each hole an' see which ones get viciously mauled by horrible monsters?"
-Ilyara Thundertale
#3 Jul 20, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
Any chance there's something similar for square grids?
#4 Jul 20, 2024
• Initiate of the 1st Circle
• Join Date: 3/24/2018
• Posts: 9,113
• Member Details
It is nowhere near as elegant. Which is why hexes are better.
"Sooner or later, your Players are going to smash your railroad into a sandbox."
"real life is a super high CR."
"............anybody got any potatoes? We could drop a potato in each hole an' see which ones get viciously mauled by horrible monsters?"
-Ilyara Thundertale
#5 Jul 20, 2024
• Adventurer
• Join Date: 9/19/2023
• Posts: 789
• Member Details
This is how we do it at my table. Note that we don't use Roll20, so I don't know how this is implemented there, sorry :(
If the point of origin is a creature, that creature occupies the entire square, and the range is calculated from the edge of its tile outwards. Here are some examples:
- Spirit Guardians (15-ft radius):
- Burning Hands (15-ft cone):
- Thunderwave (15-foot cube):
- Earth tremor (10 ft):
- Ice Knife (taking into acount the last erratum, 5 ft):
If the point of origin is a point in space, then the drawing from Wysperra is how we do it. So, for example:
- Moonbean (5-ft radius):
- Web (20-ft cube):
About the number of affected targets, we follow the rules in XGtE (p. 86, Areas of Effect on a Grid):
If any part of a square is under the template, that square is included in the area of effect. If a creature's miniature is in an affected square, that creature is in the area. Being adjacent to
the edge of the template isn't enough for a square to be included in the area of effect; the square must be entirely or partly covered by the template.
EDIT: added more examples.
#6 Jul 20, 2024
Sillvva Moderator
• Sensei
• Location: Astral Plane
• Join Date: 3/20/2017
• Posts: 12,915
• Member Details
There is also the Area of Effects on a Grid section in Xanathar's Guide to Everything, which describes the template method and the token method.
How I'm posting based on text formatting: Mod Hat On - Mod Hat Off
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#7 Jul 20, 2024
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• Join Date: 2/25/2023
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The DMG provides some guidance for this on Page 251 in the sections called "Areas of Effect" and "Cover" (and some diagrams on nearby pages). Perhaps not intuitively, the rule from the DMG says that
when using a grid and miniatures you actually choose an intersection of grid squares as your point of origin:
AREAS OF EFFECT
The area of effect of a spell, monster ability, or other feature must be translated onto squares or hexes to determine which potential targets are in the area and which aren't.
Choose an intersection of squares or hexes as the point of origin of an area of effect, then follow its rules as normal. If an area of effect is circular and covers at least half a square, it
affects that square.
There aren't any exceptions given for when a creature is the point of origin, so you end up with the point of origin on one of the corners of the square that the creature occupies. This creates
seemingly asymmetrical AoE shapes with respect to the creature's location within the square, but those are the rules as written.
So, for Moonbeam, your DM has it correct.
In the case of Ice Knife, I see two options:
First, you can use the above rule to create an AoE with a radius of 5 feet, as indicated by the "Range/Area" parameter on the D&D Beyond spells database. In cases where the point of origin is a
creature, this creates a situation where a creature on one side will be affected but a nearby creature on the other side wouldn't be. This sort of flies in the face of the text of the spell which
reads "each creature within 5 feet . . .".
The second way is to consider this notation in the D&D Beyond spells database to be an error, since that notation doesn't exist in the hardcopy of the book anyway, and just stick with the text of the
spell, which is that this AoE affects "each creature within 5 feet . . .". However, in terms of trying to place an AoE like this onto a grid, you would have to break the rule for doing so that I've
quoted above, or you would have to significantly enlarge the radius which could have other unintended consequences.
For a spell such as Burning Hands, you are definitely meant to choose a corner of a square as the origin point. Because of the rule for cones that says that the point of origin is not included in the
AoE, the spellcaster of Burning Hands is explicitly unaffected -- so, you could choose one of the back corners of your square if it's advantageous to do so. The Xanathar's Guide to Everything book
has two options for positioning shapes like this onto the grid once the point of origin is chosen -- the Template Method and the Token Method. My guess is that the Template Method should work pretty
well when using a tool such as Roll20 as you could then theoretically position the shape at any angle instead of getting locked into only vertical or diagonal options. Note that the example given in
Xanathar's for using the Template Method for a Cone clearly shows the point of origin at a grid intersection.
Also note that the rules given for the Template Method might make shapes such as Cones a lot more powerful than intended if they are angled just right given that the rule says:
". . . take note of which squares it covers or partially covers. If any part of a square is under the template, that square is included in the area of effect. If a creature’s miniature is in an
affected square, that creature is in the area."
This rule is in conflict with the original DMG rule for circular AoEs as quoted above, so a DM would have to make a ruling on that.
So, this is what the rules actually say on the matter. If you are using an online tool such as Roll20 you might be further constrained by the capabilities of that tool.
#8 Jul 20, 2024
• Adventurer
• Join Date: 9/19/2023
• Posts: 789
• Member Details
@alexdohm9191 If you or your DM are interested in the Dev's opinion on interactions between ranges, grids, and points of origin, you might consider the following tweets:
@enormousturtle Do spells with a range of "Self" have a target? Is the caster the target? Trying to figure out Find Steed's spell sharing.
@JeremyECrawford A range of self means the caster is the target, as in shield, or the point of origin, as in thunderwave (PH, 202).
@samiam8910 How do spells with range Self (X-foot radius) work for bigger creatures? Is the radius from center or creature?
@JeremyECrawford When you create an area of effect with a range of self, your space is the point of origin, whatever your size.
@samiam8910 So an Ancient Dragon with Destructive Wave has the potential to wreck more than a Medium cleric doing the same? (Awesome)
@Jeremy Crawford That's correct.
@ThinkingDM Visualizing one of @JeremyECrawford's reasons why Centaur and Minotaur should not be Large PCs: a 5' aura is 50% bigger and a 10' aura is 33% bigger.
@JeremyECrawford That's exactly right.
@ThinkingDM Is it safe to assume we shouldn’t expect Large PC races?
@JeremyECrawford Yep.
@mrlong78 Thunderwave Spell:Where is the cube? Is the caster at the center of the cube, or on one side of the cube? Caster's choice?
@JeremyECrawford The point of origin of a cubic area of effect, including thunderwave's, is on a face of the cube (PH, pg 204), not inside it.
@RaywkLam Which is correct representing the area of paladin aura (10 ft)?
@JeremyECrawford Is each square 5 ft., and which aura are you talking about?
@RaywkLam Yes, each square is 5 ft. I meant Aura of Protection and Aura of Courage. Thanks.
@JeremyECrawford Aura of Protection/Courage extends 10 ft. all around. (None are correct, unless your circles are polygons.)
@JeremyECrawford A note about D&D spells with a range of "Self (XYZ)": the parenthetical—which says "5-foot radius," "15-foot cone," or something else—means you are the spell's point of
origin, but you aren't necessarily its target. You're creating an effect that originates in your space.
Some of the tweets are related to the following rules from the PHB:
Variant: Playing on a Grid (p. 192)
Ranges. To determine the range on a grid between two things—whether creatures or objects—start counting squares from a square adjacent to one of them and stop counting in the space of the
other one. Count by the shortest route.
Range (Player's Handbook p.202, emphasis mine):
Spells that create cones or lines of effect that originate from you also have a range of self, indicating that the origin point of the spell's effect must be you.
Point of origin (Player's Handbook p. 204, emphasis mine):
Typically, a point of origin is a point in space, but some spells have an area whose origin is a creature or an object.
Note that this last rule does not state that some spells have points of origin 'in' or 'on' a creature or object. Instead, it specifies that the creatures or objects themselves are the origins of the
spells. In these cases, the entire creature or object follows the rules regarding points of origin.
#9 Jul 20, 2024
• Thaumaturgist
• Join Date: 2/22/2020
• Posts: 3,213
• Member Details
So we are playing in Roll20
Roll20 has a way to handle AoEs of all shapes built into it already. You don't need to "figure out" how it would work -- you just pick 'Cone' from the Shapes menu instead of 'Line' in the measuring
tool and measure out to 15 feet, and any token touched by the displayed AoE gets hit by your burning hands
Last edited by AntonSirius: Jul 20, 2024
Active characters:
Green Hill Sunrise, jaded tabaxi mercenary trapped in the Dark Domains (Battle Master fighter)
Mardan Ferres, elven private investigator obsessed with that one unsolved murder (rogue)
Xhekhetiel, halfling survivor of a Betrayer Gods cult (Runechild sorcerer/fighter)
#10 Jul 20, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
So if I understand correctly, you're suggesting that if the point of origin is a point in space (such as moonbeam) that x-foot radius is resection 2x-foot cube (for moonbeam, 5-foot radius means
10-foot cube). But in cases where the point of origin is a creature (such as ice knife) it's effectively a 15-foot cube because it extended outward 5 feet from a square rather than a point in space?
#11 Jul 20, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
So we are playing in Roll20
Roll20 has a way to handle AoEs of all shapes built into it already. You don't need to "figure out" how it would work -- you just pick 'Cone' from the Shapes menu instead of 'Line' in the
measuring tool and measure out to 15 feet, and any token touched by the displayed AoE gets hit by your burning hands
Is that available for players? Or the DM?
#12 Jul 20, 2024
• Thaumaturgist
• Join Date: 2/22/2020
• Posts: 3,213
• Member Details
So we are playing in Roll20
Roll20 has a way to handle AoEs of all shapes built into it already. You don't need to "figure out" how it would work -- you just pick 'Cone' from the Shapes menu instead of 'Line' in the
measuring tool and measure out to 15 feet, and any token touched by the displayed AoE gets hit by your burning hands
Is that available for players? Or the DM?
It should be available for everyone, unless there's some way for a DM to switch it off I'm not aware of
Do you see that Shapes menu on the measuring tool?
Active characters:
Green Hill Sunrise, jaded tabaxi mercenary trapped in the Dark Domains (Battle Master fighter)
Mardan Ferres, elven private investigator obsessed with that one unsolved murder (rogue)
Xhekhetiel, halfling survivor of a Betrayer Gods cult (Runechild sorcerer/fighter)
#13 Jul 20, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
So we are playing in Roll20
Roll20 has a way to handle AoEs of all shapes built into it already. You don't need to "figure out" how it would work -- you just pick 'Cone' from the Shapes menu instead of 'Line' in the
measuring tool and measure out to 15 feet, and any token touched by the displayed AoE gets hit by your burning hands
Is that available for players? Or the DM?
It should be available for everyone, unless there's some way for a DM to switch it off I'm not aware of
Do you see that Shapes menu on the measuring tool?
I knew I could measure lines, I didn't know I could measure shapes. I'll have to look through that when I get the chance.
#14 Jul 20, 2024
• (Perfect)
• Join Date: 11/30/2017
• Posts: 6,985
• Member Details
So if I understand correctly, you're suggesting that if the point of origin is a point in space (such as moonbeam) that x-foot radius is resection 2x-foot cube (for moonbeam, 5-foot radius means
10-foot cube). But in cases where the point of origin is a creature (such as ice knife) it's effectively a 15-foot cube because it extended outward 5 feet from a square rather than a point in
I’ve always operated that when you choose a “point” and you’re playing on a grid, the point is always at an intersection of 2 grid lines. So, yes, a 5’ radius means basically a 10’
square. Yes, this makes circles into squares. But I find it much easier than trying to decide what to do if a given square is only fractionally covered by the radius. So circles are squares and
spheres are cubes.
But even then, cones are just plain wonky. For them, we just kind do the best we can do.
#15 Jul 21, 2024
• (Perfect)
• Join Date: 11/23/2023
• Posts: 137
• Member Details
It should be available for everyone, unless there's some way for a DM to switch it off I'm not aware of
Do you see that Shapes menu on the measuring tool?
I found it. I can do all the shapes. I'll let the DM know about this feature. Should simplify it. Thanks.
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Electrical Power a Relation of Current, Volts, Joule and Coulomb
In the realm of physics, understanding the fundamental concepts of electricity Resistance, Voltage, Current, Joule and Coulomb. One such concept that plays a crucial role in electrical systems is
resistance. Resistance, measured in ohms, impedes the flow of electric current through a conductor. This article delves into the intricate relationship between current, volt, joule and resistance as
power. we will unravel the physics behind these interconnected phenomena.
Understanding Volts and Joules:
Before delving into the relationship between volts to joules, and resistance, it’s imperative to grasp the individual concepts. Volts represent electrical potential difference or voltage, symbolized
by the letter “V.” In simple terms, volts measure the force that drives electric current through a circuit. On the other hand, joules, denoted by “J,” quantify energy. A joule is equivalent to the
work done when a force of one volt moves one coulomb of charge. In essence, volts represent the potential, while joules represent the energy derived from that potential.
The Role of Resistance:
Resistance, denoted by the symbol “R” and measured in ohms (Ω), opposes the flow of electric current in a circuit. This opposition arises due to the collisions between moving electrons and the atoms
of the conductor material. The higher the resistance, the more difficult it becomes for electrons to flow, resulting in decreased current flow. Resistance calculator.
Ohm’s Law:
At the heart of understanding the relationship between volts, joules, and resistance lies Ohm’s law. Ohm’s law states that the current passing through a conductor between two points is directly
proportional to the voltage across the two points and inversely proportional to the resistance. Mathematically, Ohm’s law is expressed as:
V = IR
V is the voltage (in volts)
I is the current (in amperes)
R is the resistance (in ohms)
This equation illustrates that the voltage across a conductor is directly proportional to the current passing through it and inversely proportional to the resistance of the conductor.
Relationship Between Current, Volt, Heat, and Resistance:
The relationship between volts, joules, and resistance becomes apparent when considering the energy dissipated in a resistor. When an electric current flows through a resistor, energy is transformed
and dissipated in the form of heat. This energy dissipation can be calculated using the formula. Here total heat dissipated over time t is denoted by H. Formula of heat produced by current.
H = I²Rt
H = (V²/R)t
Relation of Joules and Coulombs:
V = J/C
I = Q/t
Calculation of electrical Power,
P = VI = (J/C)*(Q/t)
If we put charge to one coulomb i.e. Q = 1C we have,
P = J/t
From above expression we can say one watt is one joule per second.
Practical Implications:
Understanding the relationship between volts, joules, and resistance has practical implications in various fields, including electronics, engineering, and physics research. Engineers utilize this
knowledge when designing circuits to ensure efficient energy transfer and component functionality. By manipulating resistance values, they can control the flow of current and regulate voltage levels
within a circuit.
When the current flow through a conductor it carries electrons which are having negative charge. The negatively charged electron during their flow through conducting wire collide with atoms and
dissipate their energy in the form of heat.
Moreover, in physics research, this understanding aids in the analysis of electrical systems and the determination of energy consumption. Researchers can quantify the energy dissipated in resistors
and use this information to optimize system performance and minimize power losses.
In conclusion, the relationship between volts to joules conversion lies at the core of understanding electrical phenomena. By comprehending Ohm’s law and the principles of energy dissipation in
resistors, we gain insight into the behavior of electrical circuits and systems. This knowledge not only facilitates the design and optimization of electronic devices but also contributes to
advancements in various fields reliant on electrical technologies. As we continue to unravel the mysteries of electricity, the interplay between resistance volts, Joule and Coulomb remains a
cornerstone of modern physics and engineering.
One Reply to “Electrical Power a Relation of Current, Volts, Joule and Coulomb”
1. Thanks for great guide! These fundamental principles are crucial for understanding and designing electrical systems.
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An instance s of the parametrized data type Multiset is a multi-set of elements of type Type, represented as a red-black tree (see [[3] Chapter 13 for an excellent introduction to red-black trees).
The main difference between Multiset and the STL std::multiset is that the latter uses a less-than functor with a Boolean return type, while our Multiset class is parameterized by a comparison
functor Compare that returns the three-valued Comparison_result (namely it returns either SMALLER, EQUAL, or LARGER). It is thus possible to maintain the underlying red-black tree with less
invocations of the comparison functor. This leads to a speedup of about 5% even if we maintain a set of integers. When each comparison of two elements of type Type is an expensive operation (for
example, when they are geometric entities represented using exact arithmetic), the usage of a three-valued comparison functor can lead to considerable decrease in the running times.
Moreover, Multiset allows the insertion of an element into the set given its exact position, and not just using an insertion hint, as done by std::multiset. This can further reduce the running times,
as additional comparison operations can be avoided.
In addition, the Multiset guarantees that the order of elements sent to the comparison functor is fixed. For example, if we insert a new element x into the set (or erase an element from the set),
then we always invoke Compare() (x, y) (and never Compare() (y, x)), where y is an element already stored in the set. This behavior, not supported by std::multiset, is sometimes crucial for designing
more efficient comparison predicates.
Multiset also allows for look-up of keys whose type may differ from Type, as long as users supply a comparison functor CompareKey, where CompareKey() (key, y) returns the three-valued
Comparison_result (key is the look-up key and y is an element of type Type). Indeed, it is very convenient to look-up equivalent objects in the set given just by their key. We note however that it is
also possible to use a key of type Type and to employ the default Compare functor for the look-up, as done when using the std::multiset class.
Finally, Multiset introduces the catenate() and split() functions. The first function operates on s and accepts a second set s2, such that the maximum element in s is not greater than the minimal
element in s2, and concatenates s2 to s. The second function splits s into two sets, one containing all the elements that are less than a given key, and the other contains all elements greater
than (or equal to) this key.
Template Parameters
Type the type of the stored elements.
the comparison-functor type. This type should provide the following operator for comparing two Type elements, namely:
Compare Comparison_result operator() (const Type& t1, const Type& t2) const;
The CGAL::Compare<Type> functor is used by default. In this case, Type must support an equality operator (operator==) and a less-than operator (operator<).
Allocator the allocator type. CGAL_ALLOCATOR is used by default.
The assertion and precondition flags for the Multiset class use MULTISET in their names (i.e., CGAL_MULTISET_NO_ASSERTIONS and CGAL_MULTISET_NO_PRECONDITIONS).
Multiset uses a proprietary implementation of a red-black tree data-structure. The red-black tree invariants guarantee that the height of a tree containing \( n\) elements is \( O(\log{n})\) (more
precisely, it is bounded by \( 2 \log_{2}{n}\)). As a consequence, all methods that accept an element and need to locate it in the tree (namely insert(x), erase(x), find(x), count(x), lower_bound(x)
, upper_bound(x), find_lower(x) and equal_range(x)) take \( O(\log{n})\) time and perform \( O(\log{n})\) comparison operations.
On the other hand, the set operations that accept a position iterator (namely insert_before(pos, x), insert_after(pos, x) and erase(pos)) are much more efficient as they can be performed at a
constant amortized cost (see [4] and [6] for more details). More important, these set operations require no comparison operations. Therefore, it is highly recommended to maintain the set via
iterators to the stored elements, whenever possible. The function insert(pos, x) is safer to use, but it takes amortized \( O(\min\{d,\log{n}\})\) time, where \( d\) is the distance between the given
position and the true position of x. In addition, it always performs at least two comparison operations.
The catenate() and split() functions are also very efficient, and can be performed in \( O(\log{n})\) time, where \( n\) is the total number of elements in the sets, and without performing any
comparison operations (see [6] for the details). Note however that the size of two sets resulting from a split operation is initially unknown, as it is impossible to compute it in less than linear
time. Thus, the first invocation of size() on such a set takes linear time, and not constant time.
The design is derived from the STL multiset class-template (see, e.g, [5]), where the main differences between the two classes are highlighted in the class definition above.
All methods listed in this section can also accept a Type element as a look-up key.
In this case, it is not necessary to supply a CompareKey functor, as the Compare functor will be used by default.
template<class Key , class CompareKey >
iterator find (const Key &key, const CompareKey &comp_key)
searches for the an element equivalent to key in the set. More...
template<class Key , class CompareKey >
size_t count (const Key &key, const CompareKey &comp_key) const
returns the number of elements equivalent to key in the set.
template<class Key , class CompareKey >
iterator lower_bound (const Key &key, const CompareKey &comp_key)
returns an iterator pointing to the first element in the set that is not less than key. More...
template<class Key , class CompareKey >
iterator upper_bound (const Key &key, const CompareKey &comp_key)
returns an iterator pointing to the first element in the set that is greater than key. More...
template<class Key , class CompareKey >
std::pair< iterator, iterator
> equal_range (const Key &key, const CompareKey &comp_key)
returns the range of set elements equivalent to the given key, namely (lower_bound(key), upper_bound(key)) (a const version is also available).
template<class Key , class CompareKey >
std::pair< iterator, bool > find_lower (const Key &key, const CompareKey &comp_key)
returns a pair comprised of lower_bound(key) and a Boolean flag indicating whether this iterator points to an element equivalent to the given key (a const version is
also available).
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An object moving in the xy-plane is acted on by a conservative force described by the potential-energy function U(x,y) where 'a' is a positive constant. Derive an expression for the force f⃗ expressed in terms of the unit vectors i^ and j^. - The Story of Mathematics - A History of Mathematical Thought from Ancient Times to the Modern Day
\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]
This question aims to find an expression for the Force f which is expressed in terms of the unit vectors i^ and j^.
The concepts needed for this question include potential energy function, conservative forces, and unit vectors. Potential Energy Function is a function that is defined as the position of the object
only for the conservative forces like gravity. Conservative forces are those forces that do not depend on the path but only on the initial and final positions of the object.
Expert Answer
The given potential energy function is given as:
\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]
The conservative force of motion in two dimensions is the negative partial derivative of its potential energy function multiplied by its respective unit vector. The formula for conservative force in
terms of its potential energy function is given as:
\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat{i} + \dfrac { dU }{ dy } \hat{j} \Big) \]
Substituting the value of U in the above equation to get the expression for Force f.
\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{i} + \dfrac { d }{ dy } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{j} \Big) \]
\[ \overrightarrow{F} = – \Big( a \dfrac { d }{ dx } \Big( \dfrac{1} {x^2} \Big) \hat{i} + a \dfrac { d }{ dy } \Big( \dfrac{1} {y^2} \Big) \hat{j} \Big) \]
\[ \overrightarrow{F} = 2a \dfrac{ 1 }{ x^3 } \hat{i} + 2a \dfrac{ 1 }{ y^3 } \hat{j} \]
\[ \overrightarrow{F} = 2a \Big( \dfrac{ 1 }{ x^3 } \hat{i} + \dfrac{ 1 }{ y^3 } \hat{j} \Big) \]
Numerical Result
The expression for the force $\overrightarrow {f}$ is expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}$ is calculated to be:
\[ \overrightarrow{F} = \Big( \dfrac{ 2a }{ x^3 } \hat{i} + \dfrac{ 2a }{ y^3 } \hat{j} \Big) \]
Potential energy function is given for an object moving in XY-plane. Derive an expression for the force f expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}.
\[ U(x, y) = \big( 3x^2 + y^2 \big) \]
We can derive an expression for force by taking the negative of the partial derivative of the potential energy function and multiplying it by respective unit vectors. The formula is given as:
\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat {i} + \dfrac { dU }{ dy } \hat {j} \Big) \]
\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } \big( 3x^2 + y^2 \big) \hat {i} + \dfrac { d }{ dy } \big( 3x^2 + y^2 \big) \hat {j} \Big) \]
\[ \overrightarrow{F} = – \big( 6x \hat {i} + 2y \hat {j} \big) \]
\[ \overrightarrow{F} = – 6x \hat {i}\ -\ 2y \hat {j} \]
The expression of force f is calculated to be $- 6x \hat {i}\ -\ 2y \hat {j}$
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The logic AND gate
7. The logic AND gate¶
The AND logic gate has two or more inputs and one output. Its symbol is the following:
The output has a logic high value (1) if all its inputs have a logic high value (1). That is, if input A and input B are high, the output will be high. Hence the name AND in English.
The logical AND function is represented by multiplication, so that the output of the gate will be the logical multiplication of the inputs:
If both inputs are worth one, the output will be worth one, but if either input is worth zero, the output will be worth zero.
The truth table of the logical AND function is as follows:
The NAND logic gate¶
The NAND logic gate has two or more inputs and one output. Its symbol is the following:
The output will be the same as that of an AND gate, but inverted. That is to say that the output will only be worth zero when all the inputs are worth one.
The NAND logic function is represented by negated multiplication, so that the output of the gate will be the logical multiplication of the inputs that is finally inverted:
The truth table of the NAND logic function is as follows:
In the following simulation we can see the operation of the AND logic gate and, below, the operation of the NAND logic gate.
1. Draw the AND logic gate symbol, its logic function, and its truth table.
2. Draw the NAND logic gate symbol, its logic function, and its truth table.
3. Check with the simulator that the truth table of the AND function and that the truth table of the NAND function are correct.
4. In the simulator, add an inverting gate to the output of the AND gate and check that its response is the same as that of the NAND gate.
5. Draw a three-input AND logic gate and its truth table.
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1533 -- Mystery
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 528 Accepted: 64
The Egyptian archeologists have found an old mysterious document that shows the secret key that should be used to open the main room of the largest pyramid. When the linguists translated the old
documen t they found that they still have a problem to open the door, because the document itself is a puzzle whose solution will be the secret key. The old puzzle says that the key is consisted of
some known integers and it shows the number of each integer in the key. It also mentions that there must be some pairs of letters in the key and the lexical order of the key must be as small as
possible. A pair (x, y) of integers is considered to be in a key if there is a y coming after an x in the key. Your task is to help the archeologists to find the proper key.
The input consists of several test cases. In the first line of each test case there are two integers m < 100 and d <= 1000, number of different integers in the key and number of given pairs, then in
the next line there are m integers. The U+1th integer is the number of U's in the key . In the next d lines, in each line there is two integers x and y indicating that the pair ( x,y) is in the key.
The test case with d=0 indicates the end o f file. If there is no solution for an input case you must report "Impossible!" as the output for this case.
In the output, for each test case except the one with d=0 write the key in a separate line. Each two consecutive integers should be separated by exactly a single space.
Sample Input
Sample Output
[Submit] [Go Back] [Status] [Discuss]
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identity property of multiplication calculator
RecommendedScientific Notation QuizGraphing Slope QuizAdding and Subtracting Matrices Quiz Factoring Trinomials Quiz Solving Absolute Value Equations Quiz Order of Operations QuizTypes of angles
quiz. Matrix Inverse Calculator - 4x4 Matrix . Example Which property of multiplication does this equation show? With the Commutative Property of Multiplication, when only multiplication is involved,
numbers can move ("commute") to anywhere in the expression. Property: a + b = b + a 2. Have students use the graphing calculator to review commutative, associative and identity properties as
explained in the TI-73 Equivalency attachment. ... Commutative Property of Multiplication Identity Property of Multiplication All right, and so on. Answers: 3. Which equation illustrates the identity
property of multiplication? (ii) Associative Property : For any three matrices A, B and C, we have (AB)C = A(BC) whenever both sides of the equality are defined. Google Classroom Facebook Twitter. So
multiplying by 1 is the identity property of multiplication. Multiplication property of equality states that if one side of an equation is multiplied with the denominator then multiply the other side
of the equation with the … Analyze the diagram below and complete the instructions that follow. Your email is safe with us. Answer. Solution The numbers are moved around, there is no regrouping. Try
the given examples, or type in your own problem and check your answer with the step-by-step explanations. Examples of identity property of multiplication math 5 distributive. a)(a + bi) × c = (ac +
bci) b)(a + bi) × 0 = 0 c)(a + bi) × (c + di) = (c + di) × (a + bi) d)(a + bi) × 1 = (a + bi) 6. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright
© 2008-2019. Explore the commutative, associative, and identity properties of multiplication. Coolmath privacy policy. The product of any number and 0 is 0. 1 Answers. The Additive Identity
Property. Identity property. A × 0 = 0, where A is any integer. Does the identity property of multiplication apply to division? AI = IA = A. where I is the unit matrix of order n. Hence, I is known
as the identity matrix under multiplication. An interactive math lesson about the commutative, associative, distributive and multiplicative identity properties of multiplication … For any number a,
the following is always true: For example, 45×1 = 45. If A is a non-singular square matrix, there is an existence of n x n matrix A-1, which is called the inverse of a matrix A such that it satisfies
the property:. This property is usually applied when an unknown is a part of addition, and it enables us to single the unknowns out. This property states that when any number is multiplied by 1, it
does not change its identity. Identity Property - The result of multiplying any real number by 1 is the number itself. The symbolism of mathematics takes some time to be completely understood … (iv)
Existence of multiplicative identity : For any square matrix A of order n, we have . You can accept or reject cookies on our website by clicking one of the buttons below. The fraction is called the
multiplicative inverse of (or reciprocal) and vice versa. Fun Facts. Inverse Property of Addition responds that any number added to its opposite will equal zero. Real Life Math SkillsLearn about
investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Let's look at the number 8 . The identity property means that whatever you put in
is whatever comes out. Distributive property: The sum of two numbers times a third number is equal to the sum of each addend times the third number. In this section, we will learn about the
properties of matrix to matrix multiplication. If I is a right identity matrix for A, then the matrix product A.I = A. Display the chart in front of your student and have them look at the products in
the first row or column. Basic-mathematics.com. a × b = b × a. Mathematics. Heather works as . Matrix Subtraction Calculator - 3x3 Matrix. Identity Property of Multiplication. Any number divided by
1, gives the same quotient as the number itself. You will also watch examples that show you how to calculate the additive inverse and multiplicative inverse of a number. The identity property of
multiplication, also called the multiplication property of one says that a number does not change when that number is multiplied by 1. 0. In fact, 1 and -1 are the only two numbers that can be their
own multiplicative inverse. By using this website, you agree to our Cookie Policy. And, when something always works in math, we make it a property: If you believe that your own copyrighted content is
on our Site without your permission, please follow this Copyright Infringement Notice procedure. The "Distributive Law" is the BEST one of all, but needs careful attention. Try the free Mathway
calculator and problem solver below to practice various math topics. Any matrix typically has two different identity matrices: a left identity matrix and a right identity matrix. Calculator;
Subjects. The calculation we get is 3×1=3 [3 lots of 1s] Example 2- Let us consider anyone number and multiply it by 1. So in the figure above, the 2×2 identity could be referred to as I2 and the 3×3
identity could be referred to as I3. You can see this property readily with a printable multiplication chart. So when you multiply any number by 1, the product is that number. Example: 5 + 2 = 2 + 5
=103) Associative Property … Everything you need to prepare for an important exam! We use first party cookies on our website to enhance your browsing experience, and third party cookies to provide
advertising that may be of interest to you. Find an answer to your question “Name the property the equation illustrates-4• (-1/4) = 1 A - Multiplication property of - 1 B - identity property of
division C-Inverse ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. This
property does not take effect on division or subtraction, it applies only on addition and subtraction. Commutative Property for Fraction Addition and Multiplication .. Associative Property for
Fraction Addition and Multiplication . (iii) Matrix multiplication is distributive over addition : For any three matrices A, B and C, we have Multiplication also has its own Identity Property.
Multiplicative Identity Property Calculator: Multiplicative Identity Property Calculator. Simplifying expressions calculator | wyzant resources. Menu. These matrices are said to be square since there
is always the same number of rows and columns. … 8. Example: 3 + 9 = 12 where 12 (the sum of 3 and 9) is a real number.2) Commutative Property of Addition 1. Multiplying numbers by 1 is fun like a
game. • Inverse Operations and Inverse Properties allow us to simplify equations. 7. You will notice that the commutative property fails for matrix to matrix multiplication. By the same token, the
multiplicative inverse of -1 is -1 since -1 Ã -1 is 1. Free Online Scientific Notation Calculator. Also, multiplying by 1 does not change the Identity of a number. 1 = r Commutative property of
Addition Identity of property of multiplication commutative property of multiplication identity of ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer,
then try to use the smart search and find answers to the similar questions. Identity element for addition is 0 and for multiplication is 1. Associative property. Omkar Phatak Aug 5, 2020 . The
multiplication of any number and the identity value gives the same number as the result. The product is the same for both sides, so it shows the commutative property… If you can solve these problems
with no help, you must be a genius! Notice that the multiplicative inverse of 1 is 1 since 1 Ã 1 is 1 . Identity element for addition is 0 and for multiplication is 1. The Identity element or
neutral element is an element which leaves other elements unchanged when combined with them. The Commutative Property of Multiplication, The Associative Property of Multiplication. . Example 1:
Verify the associative property of matrix multiplication for the following matrices. Commutative Property - Numbers or variables are moved around but the answer stays the same. On the RHS we have:
and On the LHS we have: … This is called the identity property of multiplication. Explore the commutative, associative, and identity properties of multiplication. Identity property of multiplication.
For example 4 * 2 = 2 * 4 Lastly, you will also learn that multiplying a matrix with another … Commutative property. Posts. So, the 3× can be "distributed" across the 2+4, into 3×2 and 3×4. if so
give examples. As per identity property of multiplication, if we multiply any integer by 1, then its value remains unchanged, such that; A × 1 = A. When you multiply any number by 1, the product is
that number. This is what it lets us do: 3 lots of (2+4) is the same as 3 lots of 2 plus 3 lots of 4. Solution: Here we need to calculate both R.H.S (right-hand-side) and L.H.S (left-hand-side) of A
(BC) = (AB) C using (associative) property. Email. With the Associative Property of Multiplication, any numbers that are being multiplied together can "associate" with each other. Zero Property Of
Multiplication. Play a game with your child. AA-1 = A-1 A = I, where I is the Identity matrix. Identity Property - The result of multiplying any real number by 1 is the number itself. Identity
property; Division; Numbers; GUEST 2015-12-22 04:53:38. The properties of multiplication of integers are: Closure property; Commutative property; Multiplication by zero; Multiplicative identity;
Associative property; Distributive property; Closure Property. Try a smart search to find answers to similar questions. While we say “the identity matrix”, we are often talking about “an” identity
matrix. They are the commutative, associative, multiplicative identity and distributive properties. 5(1)=5 O a. 1×123 = 123. For example 5 + 0 = 5. Easy! So the identity matrix has the property that
it has ones along the diagonals. The identity property of multiplication states that the product of one and any number is that number. What's the answer to this? ----Have Instagram? Top-notch
introduction to physics. “ ” is the multiplicative identity of a number. For any whole number n, there is a corresponding n×nidentity matrix. Matrix Determinant Calculator - 3x3 & 2x2 Matrix. In
other words, any number multiplied by 1 stays the same. Notice that if we multiply 1 with any number, we get that number back. Verbal Description: If you add two real numbers, the sum is also a real
number. Find an answer to your question “What is the definition of indentity property of multiplication ...” in Mathematics if the answers seem to be not correct or there’s no answer. Matrix
multiplication: if A is a matrix of size m n and B is a matrix of size n p, then the product AB is a matrix of size m p. Vectors: a vector of length n can be treated as a matrix of size n 1, and the
operations of vector addition, multiplication by scalars, and multiplying a matrix by a vector agree with the corresponding matrix operations. Inverse element Multiplicative Identity: For every
square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Inverse Property for Fraction Multiplication where a and b are nonzero. How many “learnhub” numbers are there?
We will only use it to inform you about new math lessons. Email: donsevcik@gmail.com Tel: 800-234-2933 ; Membership Exams CPC Podcast Homework … Properties for Fractions. Example 1: Verify the
associative property of matrix multiplication … Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Educreations is a community where anyone can teach what they know and learn what they
don't. For example 4 * (6 + 3) = 4*6 + 4*3 One stop resource to a deep understanding of important concepts in physics, Area of irregular shapesMath problem solver. An identity matrix is a matrix
whose product with another matrix A equals the same matrix A. For example: 65,148 × 1 = 65,148. The identity property of multiplication simply states that a number equals itself when multiplied by 1.
Verbal Description: If you add two real numbers in any order, the sum will always be the same or equal. Multiplicative Identity Property Calculator. Charles and George learned how to calculate the
area of a rectangle in Math The commutative property of multiplication states that two numbers can be multiplied in either order. This means that you can multiply 1 to any number... and it keeps its
identity! Go ahead and try it with any number you can thing of... It always works! Commutative property: When two numbers are multiplied together, the product is the same regardless of the order of
the multiplicands. You can see this property readily with a printable multiplication chart. The number stays the same! Properties of Multiplication In the last section, we learned how to add long
strings of numbers using the properties of addition. This is the currently selected item. 9. To prevent confusion, a subscript is often used. Multiplication and addition have some similar properties.
Identity element. To understand more about how we and our advertising partners use cookies or to change your preference and browser settings, please see our Global Privacy Policy. As before, a
multiplicative identity only applies to a group of numbers as a whole. Distributive Law. December 29, 2020 Leave a comment Leave a comment a x 1 = a . Here's the 2 x 2 identity matrix, here's the 3 x
3 identity matrix, here's the 4 x 4 matrix. The identity property of 1 says that any number multiplied by 1 keeps its identity. Also, under matrix multiplication unit matrix commutes with any square
matrix of same order. Properties of numbers. If I is a left identity matrix for a given matrix A, then the matrix product I.A = A. So, that's it for the properties of matrix multiplication and
special matrices like the identity matrix I want to tell you about. This is called the identity property of multiplication. 11 "Trigonometry Identity" questions ; Interpolate VBA ; ... solving
quadratic equations using the square root property ; adding bases calculator ; simultaneous equation calculator of three equation ; ... rationalizing the denominator calculator online ;
multiplication solver ; double radical equations ; Flopsy 21 April, 18:19. But when B is the identity matrix, this does hold true, that A times the identity matrix does indeed equal to identity times
A is just that you know this is not true for other matrices B in general. (a + bi) × c = (ac + bci) (a + bi) × 0 = 0 (a + bi) × (c + di) = (c + di) × (a + bi) (a + bi) × 1 = (a + bi) Answers: 1 Show
answers. DM me your math problems! What is the definition of indentity property of multiplication … For instance, let's consider this below mentioned example: Question: Prove that (4 + 3) + 2 = 4 +
(3 + 2) … Identity Property Of Addition & Multiplication. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity property of
multiplication, also called the multiplication property of one says that a number does not change when that number is multiplied by 1. 2 x 8 = 8 x 2. commutative property of multiplication
calculator. Anyway we try to multiply 1 to it, the 8 just keeps coming back as the answer. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. Try
the given examples, or type in your own problem and check your answer with the step-by-step explanations. Try the free Mathway calculator and problem solver below to practice various math topics. The
Multiplicative Identity Property. 25 * (1/25) = 25/25 =1 (identity element of multiplication) x variable 1/x multiplicative inverse (x ≠ 0) x * (1/x) = x/x =1 (identity element of multiplication) 3)
Why learn about the Inverse Operation and Inverse Property? Rejecting cookies may impair some of our websiteâ s functionality. Whole numbers & integers. For instance, Example 1- Let us consider
anyone number and multiply it by 1. The identity property of multiplication states that the product of one and any number is that number. There are four properties involving multiplication that will
help make problems easier to solve. The Additive Inverse Property. The identity property for multiplication may be written as: a … Similarly, it is sometimes necessary to multiply long strings of
numbers without a calculator; this task is made easier by learning some of the properties of multiplication. The most simple feature of multiplication as a mathematical operation is the identity
property. Identity Property for Fraction Addition and Multiplication. Everything you need to prepare for an important exam!K-12 tests, GED math test, basic math tests, geometry tests, algebra tests.
Properties of addition. This matrix is often written simply as \(I\), and is special in that it acts like 1 in matrix multiplication. For example: 874 × 0 = 0. Mathematics, 20.06.2019 18:04.
3.84*21+4*8+3.7*3+3.3*3; John, peter and mary … Responds that any number divided by 1 does not take effect on division or subtraction, it applies on. Solving Absolute value equations Quiz order of
Operations QuizTypes of angles Quiz teach... Says that any number equals itself when multiplied by 1, it does not change the property... Connects the idea of the identity element or neutral element
is an element which leaves other elements unchanged combined! Illustrates the identity matrix has the property that it acts like 1 in matrix …! Explore the commutative, associative, multiplicative
identity property - the result number added to its opposite equal..., 1 and -1 are the only two numbers times a third number 8 = 8 x 2 find answer. Problems with no help, you agree to our Cookie
Policy multiplication identity property.. Calculation we get that number ; numbers ; GUEST 2015-12-22 04:53:38 multiplication chart ; our Story Hire... Important exam the free Mathway calculator and
problem solver below to practice various math topics × 8 = 8 2. Number equals itself when multiplied by 1 be their own multiplicative inverse of number., zero and identity properties as explained in
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regardless of the order of Operations QuizTypes angles. Any connected device it with any number multiplied by 0 is 0 identity property of multiplication calculator for multiplication is 1!! Number
and the identity property of matrix multiplication and special matrices like the property. To single the unknowns out 3.84 * 21+4 * 8+3.7 * 3+3.3 * ;. Are moved around but the answer will always be
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whole teacher is to multiplication. Awards:: Awards:: Pinterest pins, Copyright © 2008-2019 into 3×2 and.! Is explained with examples, or type in your own problem and check your with... To simplify
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polynomials with step-by-step math problem stop resource to a deep understanding of concepts.: DonateFacebook page:: Pinterest pins, Copyright © 2008-2019 … the! Is usually applied when an unknown
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surface area and volume formulas
Calculator online for a the surface area of a capsule, cone, conical frustum, cube, cylinder, hemisphere, square pyramid, rectangular prism, triangular prism, sphere, or spherical cap. Read atleast
one chapter of CBSE Grade 10 Surface Area and Volume book daily to obtain better marks in the tests. We complete the conical part OCD. Browse more Topics under Surface Areas And Volumes … Volume of
three rivers = 3 {(Surface area of a river) × Depth} Q.5. Curved surface area = πrl sq.units . Area is a quantity that expresses the extent of a two – dimensional surface or shape in the plane.
Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. Free math lessons and math homework help from basic math to
algebra, geometry and beyond. Volume And Surface Area Notes For Formulas Download PDF ☞ Class 12 Solved Question paper 2020 ☞ Class 10 Solved Question paper 2020. Part of Math Word Problems For
Dummies Cheat Sheet . Donate Login Sign up. A more detailed explanation (in text and video) of each surface area formula. Surface Area of a Hemisphere. If you faced any problem to find a solution of
Surface Areas … 1. Each solid line segment is edge, and each point at which the edges meet is vertex. • Since some formulas will be … Important Formula For Cylinder. So, curved surface area (C.S.A)
or lateral surface area = 2πr * height Important Formula For Cylinder . i.e Cap Area or Curved surface area of the hemisphere = 1/2 ( 4 π r 2 ) = 2 π r 2 If you're seeing this message, it means we're
having trouble loading external resources on our website. Each rectangular surface is a face. Given a Base edge and Height of the Hexagonal prism, the task is to find the Surface Area and the Volume
of hexagonal Prism. Questions on surface area and volume of solid … • The definitions of a parallelogram and a rhombus. Look at the syllabus for Class 10 and download the latest 2021 book for the
topics which you have studied today. Make sure you always get your answers right in Surface Area and Volume. Volume = 2 3 πr 3 cubic units. Courses. Doing this will support in understanding of each
topic. Leth be the height, l be the slant height and r 1 and r 2 be the radii of the bases (r 1 >r 2) of the frustum of a cone. 011-47340170 . If the total height is 22 cm, diameter of the
cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.). C. S. A of cylinder = ( Perimeter of base) *
Height = 2πrh. A right angle is made up of 90 degrees.A straight line is made up of 180 degrees.If two lines intersect, the sum of the resulting four angles equals 360. These Share this page to
Google Classroom. Let the radius of the sphere be r. Then, 1. Solution: Question 4: In a cylinder, radius is doubled and height is halved, then curved surface area will be (a) halved (b) doubled (c)
same (d) four times Solution: (e) Let the radius … Edges mean sides and vertices mean corners. For more information and examples of these calculations see our pages: Calculating Area,
Three-Dimensional Shapes and Calculating Volume. Total surface area (including both circular ends) = 2πrh + 2πr 2 = 27πr(h + r) 4. Sol. 2. The diameter of a road roller, 120 cm … A table of volume
formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere
and hemisphere. Cuboid: A cuboid is a three-dimensional figure formed by six rectangular surfaces, as shown below. 1. An online geometry formulas in pdf format. By Mary Jane Sterling . Sol. •
Familiarity with the basic properties of parallel lines. Download CBSE Class 10 Surface Areas and Volumes Important Formulas and concepts for exams pdf, Surface Area and Volume revision notes, mind
maps, formulas, examination notes, sure shot questions, CBSE Class 10 Surface Areas and Volumes Important Formulas and concepts for exams. Volume = 4 3 πr 3 cubic units. C. S. A of cylinder = (
Perimeter of base) * Height = 2πrh. There is curved face (Cap Area) and flat face ( base area). Share this page to Google Classroom. 2. Hello friends Today we are going to explain how to find Formula
of Surface Area and Volume #studytask. If its volume is 2002 cm 3, then find its height and total surface area. Then cap area of hemisphere is half surface area of the sphere. Geometric … Get latest
Class 10 Surface Area and Volume topic and chapter wise PDF ebooks and use them for daily reading. Also click … Using the symbols as explained. Get instant scores and step-by-step solutions on
submission. AREA, VOLUME AND SURFACE AREA {4} A guide for teachers ASSUMED KNOWLEDGE • Knowledge of the areas of rectangles, triangles, circles and composite figures. Area and Volume Formula for
geometrical figures - square, rectangle, triangle, polygon, circle, ellipse, trapezoid, cube, sphere, cylinder and cone. Volume and Surface Area Questions and Answers Q.1 A copper wire having 0.20 cm
as the radius of its circular section is one-meter long. I would like to say that after remembering the Surface Areas and Volume formulas you can start the questions and answers solution of the
Surface Areas and Volume chapter. A rectangular solid has six faces, twelve edges, and eight vertices. The frustum of the right circular cone can be viewed as the difference of the two right circular
cones OAB and OCD. Study this list of the … A cylinder's volume is π r² h, and its surface area is 2π r h + 2π r². Where r is the radius of the small circle and R is the radius of bigger circle and
Pi is constant Pi=3.14159. Related Pages Volume Formulas Explained Explanations For … Formulas for Perimeter, Area, Surface Area, and Volume; Formulas for Perimeter, Area, Surface Area, and Volume.
Derive the formula for the curved surface area and total surface area of the frustum of a cone. Solution: Let the radius of spheres = r. Total volume of spheres = Volume of resultant cylinder
Students, teachers, parents, and everyone can find solutions to their math problems instantly. V. SPHERE . Contents: Surface Area and Volume. 3. Area of each end of cylinder = 2πr 2. (3 mark) 6. If
you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Examples: Input : … Volume of cylinder — πr 2 h = [(Area of base) * height] Hollow
Cylinder’s formulae e.g., (Rubber tubes pipes, etc.) Surface area of the hemisphere having two faces. Area Volume Perimeter Surface Area Formulas PDF + Printable. Formulas For Perimeter Area Volume
July 9, 2020 A perimeter is the path that surrounds or encompasses a two-dimensional shape.While Volume is the quantity of a three-dimensional space enclosed by a closed surface. You’ll find
geometric figures showing up frequently in word problems. If the radius of a sphere is doubled, then find the ratio of their volumes. In mathematics, a hexagonal prism is a three-dimensional solid
shape which have 8 faces, 18 edges, and 12 vertices. Surface area and Volume Formulas. Learn how to use these formulas to solve an example problem. Let the radius of a hemisphere be r. Then, 1.
Sample PDF Worksheet. Start New Test ... Customize . Surface area of a Sphere with radius ( r ) = 4 π r 2. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a
frustum of a cone. Note: By clicking the link, only first post will appear. Other questions are in the arrowed list at the bottom. The Surface Area of a Sphere is Surface Area = 4πr² Surface Area = 4
* PI * radius * radius; Surface Area = 4 * 3.14 * 5 * 5 Surface Area = 314. 2. volume and surface area of cone, A cone can also be taken as the triangle that is rotated along one of its vertices. 2.
Take unlimited online tests on Surface Area and Volume. Diameter = … There are a predefined set of basic cone formulas that are used to calculate its curved area, surface area, the volume of a cone,
total surface area etc. This page helps you to get the list of area and volume formulas in geometry. Area Perimeter Volume and Surface Area Formulas. The surface-area-to-volume ratio, also called the
surface-to-volume ratio and variously denoted sa/vol or SA:V, is the amount of surface area per unit volume of an object or collection of objects. Total surface area (including both circular ends) =
2πrh + 2πr 2 = 27πr(h + r) 4. 1. (2 mark) 5. Firstly, determine the volume of a cone and volume of sphere using formula, volume of sphere = 4/3 πr 3 and volume of cone = 1/3 πr 2 h; Further, equate
volume of a cone and volume of sphere; so that we get the radius of sphere. In chemical reactions involving a solid material, the surface area to volume ratio is an important factor for the
reactivity, that is, the rate at which the chemical reaction will proceed.. For a given volume, the object … (3 mark) 8. Prerequisites (Related Formulas) Area Calculations. If the circumference of
the base of a right circular cylinder is 110 cm, then find its base area. 1 square kilometer = 1,000,000 square meters, 1 … VI. We have, for the cylindrical part. Volume The volume of a cone is given
by the formula – Volume = 2 × Pi^2 × R × r^2. Learn More. Related Pages Surface Area of Cuboids Surface Area of Prisms … A-1, Acharya Nikatan, Mayur Vihar, Phase-1, Central Market, New Delhi-110091.
info@entrancei.com The formulas for surface area and volume may look complicated, but they're actually quite simple when you work through them step-by-step. Please refer to the examination notes
which you can use for preparing and revising for exams. The two faces at either ends are hexagons, and the rest of the faces of the hexagonal prism are rectangular. ⇒ π r 2 (14) = 176 ⇒ r 2 = 4 ⇒ r =
2. Create customized worksheets and tests to suit your needs. Example-2 : 15 number of identical spheres are melted and converted into cylinder shape of 10 cm radius and 5.4 cm height is made. You
can also check the list of all Surface Area and Volume Formulas . where a is the base length and h is the height … The surface area of a Torus is given by the formula – Surface Area = 4 × Pi^2 × R ×
r. Where r is the radius of the small circle and R is the radius of bigger circle and Pi is constant Pi=3.14159. Search for courses, skills, and videos. When we calculate the surface area and volume
of composite shapes, we break the shape into its constituting shapes. 3. Total surface area = πrl + πr 2 sq.units . (2 mark) 7. If the total surface area of a cube is 216 cm 2, then find its volume.
A table of surface area formulas and volume formulas used to calculate the surface area and volume of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone,
pyramid, sphere and hemisphere. Download Surface Area and Volume Cheatsheet Below. Volume of cylinder — πr 2 h = [(Area of base) * height] Hollow Cylinder’s formulae e.g., (Rubber tubes pipes, etc.)
Area of each end of cylinder = 2πr 2. The curved surface area of a cone is the multiplication of pi, slant height, and the radius. The radius of a cylinder is 7 cm. This process of calculation is
exactly similar to breaking a bigger problem into a smaller problem, to reach an accurate solution. Geometric figures have names, classifications, and characteristics and are measured in two or more
ways. Search. We have listed top important formulas for Surface Areas and Volume for class 10 chapter 13 which help support to solve questions related to the chapter Surface Areas and Volume. Click
on an exercise link, or a topic link below to start the chapter. Opposite faces are parallel … In this C program to find Volume and Surface Area of Sphere example, We have entered the Radius of a
Sphere = 5. A more detailed explanation (examples and solutions) of each volume formula. HEMISPHERE . 4. • Basic knowledge of congruence and similarity. Surface area and volume (Solid shapes) Chapter
13 Class X Interactive 3d shape, formulas volume and surface area, surface area and volume formulas class 9th and 10th , 3d shape formulas volume and surface area for class 9th, Skip to main content
Search This Blog CBSE Mathematics Beautiful blogs on basic concepts and formulas of mathematics, maths assignments for board classes, maths study … Then find the radius of sphere. So, curved surface
area (C.S.A) or lateral surface area = 2πr * height. Surface area = 4 πr 2 sq.units . Main content. Online calculators and formulas for a surface area and other geometry problems. Math High school …
Review the formulas for the volume of prisms, cylinders, pyramids, cones, and spheres. Angles. Serial order wise Ex 13.1 Ex 13.2 Ex 13.3 Ex 13.4 Ex 13.5 Ex 13.6 Ex 13.7 Ex 13.8 Examples Ex 13.9
(Optional) Concept wise Area Of Cube/ Cuboid Volume Of … Chapter 13 Volume and Surface notes download in pdf which has explanation of Area volume of cylinder, calculation of volume of cylinder,
cylinder volume unit, calculation of volume of a cylinder, calculation volume of cylinder, calculation for volume of a cylinder, calculate … Answer. … Volume of … Area can be understood as the amount
of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. The graphic on this page, is
designed to be a quick reference for calculating the area, surface area and volume of common shapes. • Familiarity with the volume of a rectangular prism. As well as you should learn how to use
formulas of surface area and volume with examples with differential equations in competitive exams. With radius ( r ) 4 made of tin Sheet consists of a river ) × }... Edge, and 12 vertices the shape
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resources on our website 2πrh + 2! Six faces, 18 edges, and characteristics and are measured in two or ways... How to use these Formulas to solve an example problem formed by six rectangular,. Class
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to their math problems instantly to use these Formulas solve. Either ends are hexagons, and 12 vertices of all surface area including. The list of the frustum of a sphere is doubled, then find its
base area ) and flat (... Cone is the radius of its circular section is one-meter long we break the shape into constituting... A more detailed explanation ( in text and video ) of each end cylinder!
School … Take unlimited online tests on surface area = 2πr * height = 2πrh + 2πr.... 12 vertices the unknown defining side lengths, circumferences, volumes or of. R² h, and each point at which the
edges meet is vertex find ratio! A river ) × Depth } Q.5 Formulas PDF + Printable Answers right in area. A three-dimensional figure formed by six rectangular surfaces, as shown below the frustum of
a.... Small circle and Pi is constant Pi=3.14159 of math Word problems for Dummies Cheat Sheet faces the. Circular cone can be viewed as the radius of a right circular cone can be viewed as
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volume formula two.
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From users who are members of Transition to university :
Christian Lawson-Perfect said Ready to use 7 years, 3 months ago
Hannah Aldous said Needs to be tested 7 years, 4 months ago
Elliott Fletcher said Has some problems 7 years, 4 months ago
Stanislav Duris said Has some problems 7 years, 4 months ago
Gave some feedback: Ready to use
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This is fine. I had to add \left and \right around brackets in a couple of places so they stretch around the fractions inside.
This is a really good question, i just have a couple of comments
Main Parts
a) and b) i would use a \displaystyle before the fractions in the questions to make the fractions look bigger.
a) in the very last calculation when you are factorising the quadratic, you write the constant term as a power of a different number like 4^3, i would just leave this as it's simplified number,
x^2-30x-4^3 = 0 could be x^2-30x-64 = 0
b) There is a random ' after the ln on the first line here
I think you are just missing a few full stops after some equations. Specifically, the one before "We then just need to rearrange our equation" and the very last line where you give the final
value of x.
Gave some feedback: Has some problems
Gave some feedback: Has some problems
Gave some feedback: Doesn't work
Gave some feedback: Has some problems
Stanislav Duris commented 7 years, 4 months ago
I feel like some parts in advice need a bit more explanation:
- In part a.iv) I think you should add a bit more explanation why the answer is always 0. Something along the lines that as $b^0 = 1$ then $\log_b(1) = 0$.
- In part b) at the end, "the only value for x is..." I feel should be something along the lines of "the only possible value for x is ...".
- In part c) "Laws for logarithms can also be applied to $\ln$." maybe you could add "as $\ln(x) = \log_e(x)$."
- Also, the last two lines of the advice is only right when p = 2. The right answer in advice always takes a square root rather than the appropriate root. Fix this.
There are some small mistakes in punctuation:
- Some parts are missing a blank space between the = and the gap.
- In advice, I feel like many of the commas you used are unnecesseary, for example "We need to use the rule," is followed by that rule so I would remove the comma so it flows better. You've done
this correctly in part a.iii). Maybe if you want to keep the comma there, you should put comma after the rule/equation on the separate line as well, to make it stand out from the sentence.
Similarly, "Subsituting in our values for x and y gives," I would remove these commas too for the same reason.
- You're missing a few full stops when your sentences finish with an equation in advice. For example, in part a) after the last line in each part i/ii/iii or last line of the whole advice, I
think there should be full stops.
- In advice part b), near the end - "to write our equation as," is followed by two lines of equations. You put full stop after the first line by mistake, but it should be at the end of the second
line. Just move the \text{.} to the end of that expression.
Finally, part c) asks for the answer to be put in the form $\frac{e^{a}}{b}$, but accepts answers in different forms such as ($(\frac{e^{m}}{q})^{1/p}$ or ${e^{a}}*{b^{-1}}$. I think this could
be easily fixed by putting some required/forbidden strings in the string restriction for the question. I would advise trying "(e" and ")^" and "*" as forbidden strings and add "e" as a required
string just to be sure. Hopefully that will fix this tiny problem.
There are 81 other versions that do you not have access to.
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A Symbolic Analysis of Relay and Switching Circuits is the title of a master's thesis written by computer science pioneer Claude E. Shannon while attending the Massachusetts Institute of Technology
(MIT) in 1937,^[1]^[2] and then published in 1938. In his thesis, Shannon, a dual degree graduate of the University of Michigan, proved that Boolean algebra^[3] could be used to simplify the
arrangement of the relays that were the building blocks of the electromechanical automatic telephone exchanges of the day. He went on to prove that it should also be possible to use arrangements of
relays to solve Boolean algebra problems.
The utilization of the binary properties of electrical switches to perform logic functions is the basic concept that underlies all electronic digital computer designs. Shannon's thesis became the
foundation of practical digital circuit design when it became widely known among the electrical engineering community during and after World War II. At the time, the methods employed to design logic
circuits (for example, contemporary Konrad Zuse's Z1) were ad hoc in nature and lacked the theoretical discipline that Shannon's paper supplied to later projects.
Shannon's work also differered significantly in its approach and theoretical framework compared to the work of Akira Nakashima. Whereas Shannon's approach and framework was abstract and based on
mathematics, Nakashima tried to extend the existent circuit theory of the time to deal with relay circuits, and was reluctant to accept the mathematical and abstract model, favoring a grounded
approach.^[4] Shannon's ideas broke new ground, with his abstract and modern approach dominating modern-day electrical engineering.^[4]
The paper is commonly regarded as the most important master's thesis ever due to its insights and influence.^[5]^[6]^[7]^[8] Pioneering computer scientist Herman Goldstine described Shannon's thesis
as "surely ... one of the most important master's theses ever written ... It helped to change digital circuit design from an art to a science."^[9] In 1985, psychologist Howard Gardner called his
thesis "possibly the most important, and also the most famous, master's thesis of the century".^[10] The paper won the 1939 Alfred Noble Prize.
A version of the paper was published in the 1938 issue of the Transactions of the American Institute of Electrical Engineers.^[11]
1. ^ Agarwal, Ravi P; Sen, Syamal K (2014). Creators of Mathematical and Computational Sciences. Cham: Springer International Publishing. p. 425. doi:10.1007/978-3-319-10870-4. ISBN
2. ^ Fox, Charles (2024). Computer Architecture: From the Stone Age to the Quantum Age. San Francisco: No Starch Press. p. 114. ISBN 978-1-7185-0286-4.
3. ^ Caldwell, Samuel H. (1965) [1958]. Switching Circuits and Logical Design, Sixth Printing. New York: John Wiley & Sons. p. 34. ISBN 978-0471129691. “[Shannon] constructed a calculus based on a
set of postulates which described basic switching ideas; e.g., an open circuit in series with an open circuit is an open circuit. Then he showed that his calculus was equivalent to certain
elementary parts of the calculus of propositions, which in turn was derived from the algebra of logic developed by George Boole.”
4. ^ ^a ^b Kawanishi, Toma (2019). "Prehistory of Switching Theory in Japan: Akira Nakashima and His Relay-circuit Theory". Historia Scientiarum. Second Series. 29 (1): 136–162. doi:10.34336/
5. ^ Norman, Jeremy M. (2005). From Gutenberg to the Internet: A Sourcebook on the History of Information Technology. Novato, Calif: Historyofscience.com. p. 749. ISBN 978-0-930405-87-8. OCLC
6. ^ Aleksander, Igor; Morton, Helen (2012). Aristotle's Laptop: The Discovery of our Informational Mind. Series on Machine Consciousness. Vol. 1. World Scientific Publishing. p. 22. doi:10.1142/
8113. ISBN 978-981-4343-49-7.
7. ^ Matthews, Suzanne J.; Newhall, Tia; Webb, Kevin C. (2022). Dive Into Systems: A Gentle Introduction to Computer Systems. San Francisco: No Starch Press. p. 234. ISBN 978-1-7185-0136-2.
8. ^ Domingos, Pedro [@pmddomingos] (2023-12-20). "The most important master's thesis of all time" (Tweet). Retrieved 2024-11-05 – via Twitter.
9. ^ Goldstine, Herman A. (1972). The Computer: From Pascal to von Neumann. p. 119-20.
10. ^ Smith, Nancy Duvergne (2011-08-15). "Claude Shannon: Digital Pioneer's Work Still Reverberates". alum.mit.edu. Retrieved 2024-01-11.
11. ^ Shannon, C. E. (1938). "A Symbolic Analysis of Relay and Switching Circuits" (PDF). Trans. AIEE. 57 (12): 713–723. doi:10.1109/T-AIEE.1938.5057767. hdl:1721.1/11173. S2CID 51638483.
External links
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Eighth Grade
Number and Operations (NCTM)
Understand numbers, ways of representing numbers, relationships among numbers, and number systems.
Understand and use ratios and proportions to represent quantitative relationships.
Compute fluently and make reasonable estimates.
Develop, analyze, and explain methods for solving problems involving proportions, such as scaling and finding equivalent ratios.
Geometry (NCTM)
Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.
Understand relationships among the angles, side lengths, perimeters, areas, and volumes of similar objects.
Create and critique inductive and deductive arguments concerning geometric ideas and relationships, such as congruence, similarity, and the Pythagorean relationship.
Apply transformations and use symmetry to analyze mathematical situations.
Describe sizes, positions, and orientations of shapes under informal transformations such as flips, turns, slides, and scaling.
Measurement (NCTM)
Apply appropriate techniques, tools, and formulas to determine measurements.
Solve problems involving scale factors, using ratio and proportion.
Grade 8 Curriculum Focal Points (NCTM)
Geometry and Measurement: Analyzing two- and three-dimensional space and figures by using distance and angle
Students use fundamental facts about distance and angles to describe and analyze figures and situations in two- and three-dimensional space and to solve problems, including those with multiple steps.
They prove that particular configurations of lines give rise to similar triangles because of the congruent angles created when a transversal cuts parallel lines. Students apply this reasoning about
similar triangles to solve a variety of problems, including those that ask them to find heights and distances. They use facts about the angles that are created when a transversal cuts parallel lines
to explain why the sum of the measures of the angles in a triangle is 180 degrees, and they apply this fact about triangles to find unknown measures of angles. Students explain why the Pythagorean
Theorem is valid by using a variety of methods - for example, by decomposing a square in two different ways. They apply the Pythagorean Theorem to find distances between points in the Cartesian
coordinate plane to measure lengths and analyze polygons and polyhedra.
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How do I determine why my model is infeasible?
When solving optimization models, certain scenarios may arise where the defined constraints cannot be fulfilled, leading to what is known as an infeasible model. Such situations can occur either
during the initial formulation of the model or when there are updates in the data, modifications to limits, or the introduction of new constraints. In such cases, it becomes necessary to either
identify and rectify the underlying cause of the infeasibility or, alternatively, identify a subset of constraints that can be relaxed to achieve a feasible model. In this article, we discuss how to
identify, diagnose, and cope with infeasibility, then we offer some tips and best practices.
How to tell if your model is infeasible?
You can determine the status of a solved Gurobi model by querying the Optimization Status Code or by reviewing the log file. If the model is proven to be infeasible, the status code will be
INFEASIBLE and the log file message will be "Infeasible model". If the model is proven to be either infeasible or unbounded, then status code will be INF_OR_UNBD and the log file message will be
"Model is infeasible or unbounded". For more information on this status, please see How do I resolve the error "Model is infeasible or unbounded"?
Diagnosing and coping with infeasibility
To determine the set of model constraints that cause infeasibility, there are two questions that you can ask the solver:
1. What subset of model constraints is responsible for making the model infeasible?
2. What changes do I need to make to the model to recover feasibility?
To answer the first question, you can compute an Irreducible Infeasible Subsystem (IIS). This is a minimal subset of constraints and variable bounds that is still infeasible if isolated from the rest
of the model. However, if any single constraint or bound from this subsystem is removed, the resulting subsystem is feasible. In Python, you can compute an IIS using the Model.computeIIS() method.
Note that an infeasible model may have multiple IISs. The one returned by Gurobi is not necessarily the one with minimum cardinality; others with fewer constraints or bounds may exist.
To answer the second question, you can compute the smallest (with respect to some specified metric) perturbation that would need to be made to the model in order to recover feasibility. You can do
this in Python with the Model.feasRelax() and Model.feasRelaxS() methods. The latter is a simplified version, hence the "S" suffix.
For instructions and examples on how to use these two features,
For more details, see the Diagnose and cope with infeasibility section in the documentation.
Tips and Best Practices
Prescreen constraints and limits that are more likely to be causing the issue
To reduce the problem size and improve the interpretability of the results, it is helpful to define a list of potentially problematic constraints and limits before running either compute IIS routine
or a feasibility relaxation. When expanding the list of potentially problematic constraints and limits, there are several criteria to take into account:
• constraints and limits that are influenced by changing data,
• constraints and limits that are defined based on user inputs,
• constraints and limits that have been recently modified during model development.
Once you have this list of potentially problematic constraints and limits, you can specify that
• for 'compute IIS', constraints and limits that are not on the list are not eligible for inclusion in the IIS using the ISConstrForce, IISLBForce, IISUBForce, IISSOSForce, IISQConstrForce, and
IISGenConstrForce attributes,
• for feasibility relaxation, only the constraints and limits on the list are eligible for relaxation when using Model.feasRelax() by using the vars and constrs arguments. These two arguments
control which variables whose bounds are allowed to be violated and which linear constraints are allowed to be violated, respectively. For feasibility relaxation, you can also use the IIS results
to get the list of constraints and limits that are potentially problematic.
Compute IIS can be computationally expensive for larger models
In terms of computational complexity, calculating an IIS is relatively inexpensive for linear programming (LP) problems, but can be much more computationally expensive for mixed-integer programming
(MIP) problems. Additionally, the resulting IIS may be large and hard to interpret.
For larger models, particularly MIP models, we recommend using a feasibility relaxation with the objective of minimizing the number of violations. To do this, you need to set the first argument of
Model.feasRelaxS() and Model.feasRelax() to relaxobjtype=2. This is less computationally expensive and, depending on the end goal, can be viewed as an alternative to 'compute IIS'.
Caveat: check that the lower bounds on variables are what you intended
Gurobi automatically adds a lower bound of 0 to variables. This can cause an infeasibility, if this is not what was intended.
Further information
0 comments
Article is closed for comments.
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Big Ideas in Applied Math: Galerkin Approximation
My first experience with the numerical solution of partial differential equations (PDEs) was with finite difference methods. I found finite difference methods to be somewhat fiddly: it is quite an
exercise in patience to, for example, work out the appropriate fifth-order finite difference approximation to a second order differential operator on an irregularly spaced grid and even more of a
pain to prove that the scheme is convergent. I found that I liked the finite element method a lot better^1Finite element methods certainly have their own fiddly-nesses (as anyone who has worked with
a serious finite element code can no doubt attest to). as there was a unifying underlying functional analytic theory, Galerkin approximation, which showed how, in a sense, the finite element method
computed the best possible approximate solution to the PDE among a family of potential solutions. However, I came to feel later that Galerkin approximation was, in a sense, the more fundamental
concept, with the finite element method being one particular instantiation (with spectral methods, boundary element methods, and the conjugate gradient method being others). In this post, I hope to
give a general introduction to Galerkin approximation as computing the best possible approximate solution to a problem within a certain finite-dimensional space of possibilities.
Systems of Linear Equations
Let us begin with a linear algebraic example, which is unburdened by some of the technicalities of partial differential equations. Suppose we want to solve a very large system of linear equations
symmetric and positive definite (SPD). Suppose that
One solution is to consider only solutions subspace
The next step is to convert the system of linear equations
Since we are seeking an approximate solution from the subspace
One can relatively easily show this problem possesses a unique solution ^2Here is a linear algebraic proof. As we shall see below, the same conclusion will also follow from the general Lax-Milgram
theorem. Let In what sense is ^3Note that All of the properties satisfied by the familiar Euclidean inner product and norm carry over to the new Pythagorean theorem). Indeed, for those familiar, one
can show inner product space.
We shall now show that the error
The fact that the error
Thus, the Galerkin approximation , ^4In the sense that then
Variational Formulations of Differential Equations
As I hope I’ve conveyed in the previous section, Galerkin approximation is not a technique that only works for finite element methods or even just PDEs. However, differential and integral equations
are one of the most important applications of Galerkin approximation since the space of all possible solution to a differential or integral equation is infinite-dimensional: approximation in a
finite-dimensional space is absolutely critical. In this section, I want to give a brief introduction to how one can develop variational formulations of differential equations amenable to Galerkin
approximation. For simplicity of presentation, I shall focus on a one-dimensional problem which is described by an ordinary differential equation (ODE) boundary value problem. All of this generalized
wholesale to partial differential equations in multiple dimensions, though there are some additional technical and notational difficulties (some of which I will address in footnotes). Variational
formulation of differential equations is a topic with important technical subtleties which I will end up brushing past. Rigorous references are Chapters 5 and 6 from Evans’ Partial Differential
Equations or Chapters 0-2 from Brenner and Scott’s The Mathematical Theory of Finite Element Methods.
As our model problem for which we seek a variational formulation, we will focus on the one-dimensional Poisson equation, which appears in the study of electrostatics, gravitation, diffusion, heat
flow, and fluid mechanics. The unknown ^5In higher dimensions, one can consider an arbitrary domain Lipschitz boundary. We assume Dirichlet boundary conditions that ^6In higher dimensions, one has
Poisson’s equations then reads^7Laplacian operator.
We wish to develop a variational formulation of this differential equation, similar to how we develop a variational formulation of the linear system of equations in the previous section. To develop
our variational formulation, we take inspiration from physics. If
This integral expression is some kind of variational formulation of our differential equation, as it is an equation involving the solution to our differential equation every averaging function every
will be forthcoming.) It will benefit us greatly to make this expression more “symmetric” with respect to integrate by parts:^8Integrating by parts is harder in higher dimensions. My personal advice
for integrating by parts in higher dimensions is to remember that integration by parts is ultimately a result of the product rule. As such, to integrate by parts, we first write an expression
involving our integrand using the product rule of some differential operator and then integrate by both sides. In this case, notice that divergence theorem to the last term to get nice functions
In particular, if
Compare the variational formulation of the Poisson equation Eq. (8) to the variational formulation of the system of linear equations
Before this unifying theory, we must address the question of which functions continuously differentiable function on square-integrable (^9More specifically, we only insist that weak derivative.
Sobolev space ^10The class of functions satisfying (A) and (B) but not necessarily (C) is the Sobolev space trace operator. Chapter 5 of Evan’s Partial Differential Equations is a good introduction
to Sobolev spaces. The Sobolev spaces
Now this is where things get really strange. Note that it is possible for a function classical solution to weak solution to the differential equation Eq. (5) because, as we have argued, a weak
solution to Eq. (8) need not always solve Eq. (5).^11One can show that any classical solution to Eq. (5) solves Eq. (8). Given certain conditions on regularity theory.
The Lax-Milgram Theorem
Let us now build up an abstract language which allows us to use Galerkin approximation both for linear systems of equations and PDEs (as well as other contexts). If one compares the expressions
bilinear forms: they depend on two arguments (linear transformation in each argument independently if the other one is fixed. For example, if one defines
Implicitly swimming in the background is some space of vectors or function which this bilinear form ^12The connection between vectors and functions is even more natural if one considers a function
Call this space Hilbert space, an inner product space (with inner product Cauchy sequence converges to an element in ^13Note that every inner product space has a unique completion to a Hilbert space.
For example, if one considers the space functions which are zero away from the boundary of The Cauchy sequence convergence property, also known as metric completeness, is important because we shall
often deal with a sequence of entries
With these formalities, an abstract variational problem takes the form
where linear form on Lax-Milgram theorem which establishes existence and uniqueness of solutions to a problem like Eq. (9).
Theorem (Lax-Milgram): Let
1. (Boundedness of
2. (Coercivity) There exists a positive constant
3. (Boundedness of
Then the variational problem Eq. (9) possesses a unique solution.
For our cases, , defined as ^14That is, one has that the equivalent in the sense that
Let us now see how the Lax-Milgram theorem can apply to our two examples. For a reader who wants a more “big picture” perspective, they can comfortably skip to the next section. For those who want to
see Lax-Milgram in action, see the discussion below.
General Galerkin Approximation
With our general theory set up, Galerkin approximation for general variational problem is the same as it was for a system of linear equations. First, we pick an approximation space
Provided the error . To see the
where we use the variational equation Eq. (9) for ^22Compare Eq. (4).
Put simply, ^23Using the fact the norms Céa’s Lemma.
Galerkin approximation is powerful because it allows us to approximate an infinite-dimensional problem by a finite-dimensional one. If we let basis for the space ^24For an arbitrary Thus, plugging in
If we define
We’ve covered a lot of ground so let’s summarize. Galerkin approximation is a technique which allows us to approximately solve a large- or infinite-dimensional problem by searching for an approximate
solution in a smaller finite-dimensional space
Design of a Galerkin approximation scheme for a variational problem thus boils down to choosing the approximation space finite element method. Picking Fourier spectral method. One can use a space
spanned by wavelets as well. The Galerkin framework is extremely general: give it a subspace
Two design considerations factor into the choice of space Smooth functions are very well-approximated by short truncated Fourier expansions, so, if the solution most pairs of basis functions disjoint
supports for most sparse and thus usually much easier to solve. Traditional spectral methods usually result in a harder-to-solve dense linear systems of equations.^25There are clever ways of making
spectral methods which lead to sparse matrices. Conversely, if one uses high-order piecewise polynomials in a finite element approximation, one can get convergence properties similar to a spectral
method. These are called spectral element methods. It should be noted that both spectral and finite element methods lead to ill-conditioned matrices integral equation-based approaches preferable if
one needs high-accuracy.^26For example, only one researcher using a finite-element method was able to meet Trefethen’s challenge to solve the Poisson equation to eight digits of accuracy on an
L-shaped domain (see Section 6 of this paper). Getting that solution required using a finite-element method of order 15! Integral equations, themselves, are often solved using Galerkin approximation,
leading to so-called boundary element methods.
Upshot: Galerkin approximation is a powerful and extremely flexible methodology for approximately solving large- or infinite-dimensional problems by finding the best approximate solution in a smaller
finite-dimensional subspace. To use a Galerkin approximation, one must convert their problem to a variational formulation and pick a basis for the approximation space. After doing this, computing the
Galerkin approximation reduces down to solving a system of linear equations with dimension equal to the dimension of the approximation space.
6 thoughts on “Big Ideas in Applied Math: Galerkin Approximation”
1. The “big idea” described here, which is indeed big, really goes back to Rayleigh. Ritz extended it to what we would recognize today as the (Bubnov) Galerkin method. Finite element interpolation
on triangles dates to Courant (a student of Hilbert’s) in the 1920s, or perhaps teens.
1. There is an excellent article by Martin Gander about the history, https://www.unige.ch/~gander/Preprints/Ritz.pdf, which shows that Ritz created this method in his thesis.
2. Nice writeup! Seems like this might be a typo?
a(x,x) = x^\top Ax \ge \lambda_{\min} = x^\top x
1. The typo should be corrected now. Thanks for catching this!
3. Pingback: more mscharrer's favorites | Hacker News
4. The Golomb-Weinberger principle gives another (seemingly less well-known) perspective on “best approximation”. A related approach is the Nystrom method. Comparing these might be instructive.
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Logic at vismath
Brainstring Advanced
Can you untangle this? The colorful strings of the Brainstring Advanced can be moved using the buttons on the outside. Move them from hole-to-hole to place either one, two, or three buttons of the
same color on each surface. But be careful: Make sure there is no knot inside!
Gear Cube
Are you ready to shift into the next gear? This Gear Cube is the next generation of the Rubik's Cube.
The Gear Cube takes the basic idea of a Rubik's Cube and uses this classic game to build a new concept. This Cube contains 3D-Elements which twist and turn as you shift the sides of the Cube around.
Ivan's Hinge
Manipulate the hinged pieces to create striking patterns: “Ivan's Hinge” consists of colorful pieces hinged together. Now try to fold, twist, and bend it to recreate specific patterns.
For Kids aged 8+
Film: Hotel Hilbert
It seems like a normal day for Fiona Knight when she is on the look out for a hotel room for tonight. But when she enters “Infinit Hotel” her night gets anything but normal. First off, the hotel
manager tells her that every room is taken – just to declare that there will be enough free rooms for every new guest in the same breath...
Film: Julia Robinson and Hilbert's Tenth Problem
A one-hour biographical documentary, Julia Robinson and Hilbert's Tenth Problem tells the story of an important American mathematician against a background of mathematical ideas.
Film: Reason for Math Volume 1: The Language of Math
Reason for Mathis not your ordinary movie about Mathematics. Yes, it teaches you about mathematics but the main goal is to facilitate the understanding and general concepts of this science. Maths is
a way of thinking and has a language of its own. So learning the vocabulary of this foreign language is key to understanding mathematics.
Volume 1 of „Reason for Math“ deals with the basic patterns in addition, subtraction, multiplication and division.
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Math Adventures/Peanut Butter Power - Wikiversity
Answer these questions to get a better understanding of the power of peanut butter.
1. How many food calories are in one pound of peanut butter?
2. How many thermochemical calories are in one pound of peanut butter?
3. How many kWh of energy is stored in one pound of peanut butter?
4. How many kWh of energy is stored in the battery of a Tesla Model 3 (long-range battery) automobile?
5. How many pounds of peanut butter is this equivalent to?
6. How much does the battery in a Tesla Model 3 weigh?
7. How can we use peanut butter to power more devices?
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Template class specifying a continuous subsequence (slice) of a sequence.
The class is used to specify a row or a column span in a matrix ( Mat ) and for many other purposes. Range(a,b) is basically the same as a:b in Matlab or a..b in Python. As in Python, start is an
inclusive left boundary of the range and end is an exclusive right boundary of the range. Such a half-opened interval is usually denoted as \([start,end)\) .
The static method Range::all() returns a special variable that means "the whole sequence" or "the whole range", just like " : " in Matlab or " ... " in Python. All the methods and functions in OpenCV
that take Range support this special Range::all() value. But, of course, in case of your own custom processing, you will probably have to check and handle it explicitly:
const Range
& r, ....)
// process all the data
else {
// process [r.start, r.end)
Template class specifying a continuous subsequence (slice) of a sequence.
Definition types.hpp:630
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Isaac Newton
Newton was born in the village of Woolsthorpe, England, in the year 1642, the year of Galileo's death. Newton's father was a farmer who was uneducated, but owned his own farm; he died before Newton
was born. Newton's mother remarried two years later at which time Newton was sent to live in the care of his grandmother. Newton attended the Free Grammar School in Grantham boarding with a local
family. While he was a good student, he showed none of the genius that he was to demonstrate later. An uncle urged him to attend university at Trinity College, Cambridge, and his mother relented,
even though she originally wanted him to manage her affairs (she apparently had reasonably large properties by this time).
It was at Cambridge that Newton first learned mathematics. Most of the books (many containing handwritten marginal notes) in Newton's library have been reassembled; from these and other sources, it
is known that Newton backtracked from contemporary mathematical works he was trying to read, all the way back to Euclid. He then mastered each work in the forward direction. Isaac Barrow, Newton's
teacher at Cambridge, was a competent and creative mathematician, who must have helped to raise Newton's interests. It is known that Barrow made contributions to the Fundamental Theorem of Calculus.
Newton completed his undergraduate studies in 1665, the same year as the onset of the great plague in London; consequently, he returned to Woolsthorpe. The next two years were the most productive two
years in the history of science and mathematics, a true anni mirabiles. During these two years Newton built on his last year of work at Cambridge to make the revolutionary discoveries whose
elaboration was to occupy him for the rest of his academic life. He discovered the `method of fluxions' (the differential calculus) together with its applications to mechanics, the fact that
integration is the inverse procedure to differentiation (hence the Fundamental Theorem of Calculus), the compostion of white light, and started many of his other investigations.
Throughout his life, Newton was always very reluctant to publish. Many of his papers were only published years after they were written; his full account of the calculus was only to be published after
his death and his Opticks didn't appear until 1704. Some of his manusripts containing results on the calculus were circulated to other mathematicians as early as 1669, but Newton requested that they
be returned. There are many reasons for Newton's tardiness in publication, including distrust of others. Late in his life Newton was involved in a bitter priority dispute with Leibniz over the
invention of the calculus. Part of the conflict over priority for the calculus is concerned with how much knowledge Leibniz had of Newton's unpublished manuscripts and how much of Leibniz's
motivation came from them. There is no doubt that Newton's manuscripts predate Leibniz's work, nor is there doubt that Leibniz published first.
The dispute between Newton and Leibniz and their respective followers was a long one with both Newton and Leibniz showing their worst attributes during the conflict. By strongly following Newton,
avoiding Leibniz's notation, and isolating themselves from continental mathematicians, the next few generations of British mathematicians severely hindered their own progress.
Newton returned to Cambridge in 1667 when it reopened after the plague. He received his Master's Degree in 1668 and was elected to a fellowship. In 1669 Barrow resigned the Lucasian chair to turn to
religion, and he recommended that Newton be appointed in his place. Hence Newton was the second to hold the Lucasian chair (the current holder is Stephen Hawking). In 1687 Newton, with Halley's
encouragement and financial support, published his masterpiece Philosophiae Naturalis Principia Mathematica (The Mathematical Principles of Natural Philosophy),} and Newton soon became as famous as a
scientist-mathematician can be, both among his peers and the general public; he was elected to the Royal Society and to serve as president from 1703 until his death, and he was even subject to
Hogarth's jabs in one of his prints. An appreciation of Newton's discoveries and his role in the scientific community of the time cannot be given here. An excellent source is the biography [W] by
Westfall. The description of Newton's life and work at Cambridge from 1669- 1696 makes fascinating reading.
Since the Principia contains no direct application of the calculus, and since many of its arguments, while geometric, seem motivated by the calculus, many have suspected that Newton obtained much of
what is in the {\it Principia} using calculus, and then rewrote it giving only geometric arguments, possibly with a motive of keeping details of the calculus private. In letters Newton acknowledged
that he did sometimes, but not universally, obtain results in this manner. The geometric recasting, however, was for the purposes of rigor; the calculus was not put on a rigorous foundation for
another two centuries, and Newton did not want to unnecessarily open his work to criticism.
In 1696 Newton left Cambridge to live in London and serve as Warden of the Royal Mint, and he became Master of the Mint in 1699. He served capably in both positions, even becoming wealthy, but he
ceased any creative scientific or mathematical work when he left Cambridge. Newton was knighted by Queen Anne and buried in Westminister Abbey.
Here are two, possibly true, stories about Newton. Newton was elected to parliament to represent Cambridge. It is reported that the only time Newton spoke in parliament was to give the speech, "The
window needs closing." It is often reported that Newton had little sense of humor; his assistant, Humphrey Newton (no relation) said that the only time he saw Newton laugh was when he was asked, "Of
what practical use is geometry?"
Here are four stamps issued by Great Britain in honor of Newton.
Isaac Newton (1642-1727) Isaac Newton (1642-1727)
Great Britain (1987), No. 1172 Great Britain (1987), No. 1173
Isaac Newton (1642-1727) Isaac Newton (1642-11727)
Great Britain (1987), No. 1174 Great Britain (1987), No. 1175
Many countries have issued stamps in honor of Newton, more than any other mathematician except
Copernicus. Here is one issued by Hungary.
Isaac Newton (1642-1727)
Hungary (1977), No. 2485
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Paula Gregory
• Finished
Miele, L. (PI), PHILLIPS, C. (CoPI), Zabaleta, J. (CoPI) & Gregory, P. (CoPI)
21/09/17 → 31/08/21
Project: Research
• Gregory, P. (PI), GREGORY, P. E. (PI), Miele, L. (CoPI), Miele, L. (PI), PHILLIPS, C. (PI), Zabaleta, J. (PI), Miele, L. (CoPI) & Miele, L. (PI)
21/09/17 → 31/08/21
Project: Research
• Gregory, P. (PI) & GREGORY, P. E. (PI)
21/09/17 → 31/08/21
Project: Research
• Gregory, P. (CoPI), ALAM, J. (PI), ALAM, J. (PI), GREGORY, P. E. (PI), ALAM, J. (PI) & GREGORY, P. E. (PI)
17/07/12 → 30/06/18
Project: Research
• ALAM, J. (PI) & Gregory, P. (CoPI)
17/07/12 → 30/06/15
Project: Research
• Gregory, P. (PI), MOLINA, P. (CoPI), GREGORY, P. E. (PI), GREGORY, P. E. (PI), Molina, P. E. (PI), GREGORY, P. E. (PI), Molina, P. E. (PI) & Molina, P. E. (PI)
National Institute on Alcohol Abuse and Alcoholism
5/04/12 → 31/03/18
Project: Research
• MOLINA, P. (CoPI) & Gregory, P. (PI)
5/04/12 → 31/03/14
Project: Research
• MOLINA, P. (CoPI), MOLINA, P. (PI) & Gregory, P. (CoPI)
1/04/12 → 31/05/23
Project: Research
• Gregory, P. (PI), MOLINA, P. (CoPI), GREGORY, P. E. (PI), MOLINA, P. (CoPI), GREGORY, P. E. (PI), Molina, P. E. (PI), GREGORY, P. E. (PI), Molina, P. E. (PI), MOLINA, P. (PI), Molina, P. E. (PI)
& MOLINA, P. (PI)
National Institute on Alcohol Abuse and Alcoholism
1/04/12 → 31/05/22
Project: Research
• Gregory, P. (CoPI), BRANTLEY, P. (PI), GREGORY, P. E. (CoPI), BRANTLEY, P. (PI), BRANTLEY, P. J. (PI), GREGORY, P. E. (PI), BRANTLEY, P. J. (PI), GREGORY, P. E. (PI), BRANTLEY, P. J. (PI) &
GREGORY, P. E. (PI)
National Institute of Diabetes and Digestive and Kidney Diseases
1/04/12 → 31/03/18
Project: Research
• BRANTLEY, P. (CoPI) & Gregory, P. (PI)
1/04/12 → 31/03/16
Project: Research
• MOLINA, P. (PI), SIMON, L. (CoPI), AMEDEE, A. (CoPI), BAGBY, G. J. (CoPI), GUIDO, W. (CoPI), KANTROW, S. (CoPI), KOLLS, J. (CoPI), NELSON, S. (CoPI), SHELLITO, J. (CoPI), WINSAUER, P. (CoPI),
AMEDEE, A. (CoPI), BAGBY, G. J. (CoPI), EDWARDS, S. (CoPI), FERGUSON, T. (CoPI), GILPIN, N. (CoPI), KOLLS, J. (CoPI), NELSON, S. (CoPI), PARSONS, C. (CoPI), SIMON PETER, L. (CoPI), THEALL, K.
(CoPI), Veazey, R. (CoPI), WELSH, D. (CoPI), ZEA, A. (CoPI), Zhang, P. (CoPI), MOLINA, P. (CoPI), SHELLITO, J. (CoPI), WINSAUER, P. (CoPI), KANTROW, S. (CoPI) & Gregory, P. (CoPI)
National Institute on Alcohol Abuse and Alcoholism
5/01/04 → 30/11/23
Project: Research
• Gregory, P. (PI) & GREGORY, P. E. (PI)
1/08/00 → 31/07/07
Project: Research
• GREGORY, P. E. (PI), Gregory, P. (CoPI) & GREGORY, P. E. (PI)
1/08/00 → 31/07/04
Project: Research
• Gregory, P. (PI) & GREGORY, P. E. (PI)
1/09/99 → 31/07/06
Project: Research
• Gregory, P. (PI), MOLINA, P. (PI), MOLINA, P. (PI), SIMON, L. (PI), EDWARDS, S. (PI), FERGUSON, T. (PI), WELSH, D. (PI), SIMON, L. (CoPI), SPITZER, J. (PI), NELSON, S. (PI), GREENBERG, S. (PI),
MASON, C. (PI), MC DONOUGH, K. (PI), SHELLITO, J. (PI), JERRELLS, T. (PI), MC DONOUGH, K. (PI), BAGBY, G. (PI), SPITZER, J. (PI), NELSON, S. (PI), GREENBERG, S. (PI), MASON, C. (PI), MCDONOUGH,
K. (PI), SHELLITO, J. (PI), JERRELLS, T. (PI), BAGBY, G. J. (PI), NELSON, S. (PI), BAUTISTA, A. (PI), PRAKASH, O. (PI), LANCASTER, J. (PI), MOLINA, P. (PI), BROWN, J. (PI), BAGBY, G. J. (PI),
KOLLS, J. (PI), GREENBERG, S. (PI), MASON, C. (PI), MCDONOUGH, K. (PI), SHELLITO, J. (PI), JERRELLS, T. (PI), NELSON, S. (PI), KANTROW, S. (PI), WINSAUER, P. (PI), SHELLITO, J. (PI), AMEDEE, A.
(PI), GUIDO, W. (PI), NELSON, S. (PI), KANTROW, S. (PI), WINSAUER, P. (PI), SHELLITO, J. (PI), BAGBY, G. J. (PI), KOLLS, J. (PI), NELSON, S. (PI), Veazey, R. (PI), AMEDEE, A. (PI), KOLLS, J.
(PI), Zhang, P. (PI), BAGBY, G. J. (PI), THEALL, K. (PI), GILPIN, N. (PI), SIMON PETER, L. (PI), ZEA, A. (PI) & PARSONS, C. (PI)
National Institute on Alcohol Abuse and Alcoholism
1/01/01 → 30/11/21
Project: Research
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October 2021 – Derek Harlan
Excel Error and Error handling Roundup
Technology used: Office 365, Excel 2021
You just typed the closing parentheses of that powerful lookup formula that’s going to save you hours next month on the report. You slowly depress the Enter key while feeling empowered as if the
Excel gods have chosen you! Feeling the click your stomach turns as you watch the fire bestowed upon you turn into a ball of steam that wisps away. Excel is taunting you. Instead of the sales
figures, you see #N/A but what could be wrong?
Excel’s formula errors are unpleasant to see but they’re trying to help us improve our spreadsheet game. In this article, we’ll learn about the types of formula errors that Excel will display, a few
formulas to help us avoid them and some thoughts on good use of error trapping.
What are formula errors
Formula errors are the result that Excel returns when it is unable to evaluate and provide a proper answer to the formula in the cell. The errors always begin with a number sign (i.e. #) and then a
word to give you a clue as to what is causing the error. Below is a table listing common errors and some of their causes
Error (As shown in the cell) Description
#N/A Commonly found in lookup formulas (e.g. Vlookup, Hlookup, Index, Match, etc) if the formula is unable to find what the formula is looking for
#VALUE! Excel’s general error. Simply put, something is wrong with the formula.
#REF! A cell reference is invalid, typical if you delete a row/column that was being referenced by the formula
#DIV/0! Division by zero. The result of which is undefined mathematically.
#NUM! Mathematical operations are being applied to non-numeric cells. You can force text to be treated as numbers using the VALUE() function.
#NAME? Excel doesn’t recognize the formula name, usually the result of a typo (e.g. =SMU(A1:A5) instead of =SUM(A1:A5)
#NULL! The range is not constructed correctly in the formula’s parameters (e.g A5 B7 instead of A5:B7)
#SPILL! Excel is unable to return the results because there isn’t enough room. For example a FILTER function is returning 10 results but there are only 9 blank cells available
Microsoft Excel’s Formula Errors
What formulas can be used to stop errors?
Excel gives us two formulas to handle or trap errors. Using these formulas can make your worksheet more robust and user friendly. They can also allow your workbook to function despite Excel being
unable to calculate a cell.
The IFError function allows you to provide an alternate value or formula if Excel encounters any of these errors: #N/A, #VALUE!, #REF!, #DIV/0!, #NUM!, #NAME?, or #NULL!.
Its usage is:
=IFError(cell reference or formula, value if error)
The cell reference is any cell or formula that you are testing to see if it has or results in an error.
The value if error is what will result if there is an error. Here you can be really creative. A common usage would be a helpful error message consider something like:
=IfError(1/0, “Division by zero is undefined”)
In this case the formula 1/0 (e.g. One divided by 0) would result in #DIV/0!, we can replace that with the more friendly error message explaining that we can’t divide a number by zero.
But you can also use a substitute formula. For example:
=IFError(1/0, A1/B1)
This formula will result in the value of cell A1 divided by B1. Coincidently if the values of A1 divided by B1 would result in an error, then that error will be displayed.
Using the IFERROR formula gives us a lot of flexibility in trapping and resolving errors.
IFNA works very similar to IFERROR except it only works with the #N/A error. Otherwise it’s usage is identical. Recall from the table above that the #N/A error is only encountered with lookup
formulas thus this formula will only trap cases where the lookup value can’t be found in the table.
Its usage is:
=IFNA(Cell Reference or Formula, value if error)
For instance:
=IFNA(Vlookup(A1,A2:B20,2,FALSE),”Value was not found”)
Would display Value was not found instead of #N/A if the lookup value couldn’t be found.
Similar to IFERROR you can use a formula or value in place of the value if error parameter, this can help keep the logic of your workbook functioning.
Consider this example:
As you can see, since New York is not in the list an #N/A is triggered, the first version of the formula returns a 0 which we could then use in our workbook if we need to.
Trap every error?
As you’re building your workbook consider the effects of trapping the error versus knowing there is an error. Sometimes we want to see the error, to know that the workbook is not working as we
intended. For instance in the example above, we are returning 0 for cities not in the list, however since our worksheet is to tell us total sales $0 is also an acceptable answer. We have no easy way
to tell if there were no sales in a city or that the city is not in the list, which might indicate a data issue.
This is a problem known in Computer Science as the semi predicate problem where an error can be confused as a valid value. To prevent this care must be used when selecting what errors we want to trap
and which we want to allow to break the workbook.
Excel has a variety of error codes to help the user understand why their formula isn’t working. Excel also has useful formulas for trapping and resolving potential errors. Careful and considerate use
of these can lead to well constructed and robust workbooks.
Do you want to learn how to automate your financial workpapers? Do you want to get started with VBA?
My book, Beginning Microsoft Excel VBA Programming for Accountants has many examples like this to teach you to use Excel to maximize your productivity! It’s available on Amazon, Apple iBooks and
other eBook retailers!
Learn More:
Function pages on Microsoft Support
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Mathematical universe or mathematical minds?
It is not actually a question, but a subject for debate related to the title of the post: If somehow evolution has equipped us with mathematical minds, it is fair to hypothesize that the "book of
nature is written in the language of mathematics" just because we see it that way. Of course, this does not account for answers to other philosophical problems related to applicability of mathematics
(such as Wigner's puzzle). Still, the (philosophical) investigations to such problems tended to ignored those biological-anthropocentric factors. If I have to draw a question, it would be this: Can
theoretical philosophy be more multi- or inter-disciplinary at least for this topic? Isn't it that philosophy generally avoid science in collaborating on philosophical topics?
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Create HTML Table /w Text Maths - Siri Shortcuts
Where this Shortcut is innovative is its use of text as maths in the calculate function.
For example, if we use a list of objects to fill a table (without headers) is R (total rows) x C (total columns) long. To extract from a list at each Rr (current row) and Cc (current column) we can
use the following formula:
i = [(Rr -1) x C] + Cc
This can be parsed cleverly in Shortcuts using:
Repeat Index – 1
[ANSWER] x count[Headers]cols
[ANSWER] + Repeat Index 2
The benefit is that changing the math written in text above is all that is required to change the logic of the Shortcut’s actual calculations. See Shortcut below.
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Speed Control of Induction Motor: An Overview
7 Speed Control Methods of Induction Motor
This article describes various methods of Speed Control Methods of Induction Motor. An induction motor is a type of electric motor that runs on an alternating current supply. In general, an induction
motor is considered a constant-speed motor, and controlling its speed is a challenging task. Also, the variation in the speed of the induction motor affects its efficiency and performance. Although,
we can employ several methods that allow us to change the speed of an induction motor.
Speed of Induction Motor
In the case of induction motors, two speeds are relevant, namely, synchronous speed and rotor speed.
The synchronous speed is the speed of rotating magnetic flux in the air gap. It is denoted by N[S] and is given by,
Where f is the frequency of AC supply, and P is the number of magnetic poles in the stator winding.
The speed of the rotor in an induction motor is always less than the synchronous speed and is denoted by N[R]. The rotor speed of an induction motor is given by,
Where “s” is the slip. The slip in an induction motor is the difference between synchronous speed and rotor speed.
The torque developed by the induction motor also plays a vital role in speed control. The following equation gives the torque developed by an induction motor.
Where E[2] is the rotor emf, R[2] is the rotor circuit resistance, and X[2] is the rotor reactance.
Speed Control Methods of Induction Motor
In an induction motor, we can control the speed of the motor through Speed Control Methods of Induction Motor. They both the stator side and the rotor side.
Some commonly used methods to control the speed of an induction motor are given below.
From the stator side, the Speed Control Method of 3 Phase Induction Motor:
• Voltage control method
• Frequency control method
• Stator pole change method
• V/f control method
From the rotor side, the speed control methods of the induction motor:
• Rotor resistance control method
• Slip Frequency EMF Injection Method
• Cascading connection method
Let us now discuss each of these methods in detail.
Induction Motor Speed Control Methods From Stator Side
(1). Voltage Control Method:
In the Induction Motor Speed Control Methods the voltage control method is the best, the supply voltage is changed with the help of a starter autotransformer.
According to the torque equation, we get
Under running conditions, the slip in the induction motor is very small. Thus, the term (sX[2])^2 can be neglected.
Also, we know that the rotor emf E[2] is directly proportional to the stator voltage or supply voltage. Thus,
Hence, if we decrease the supply voltage, the torque will also decrease. For a given load, the slip will increase with the decrease in the supply voltage. To maintain the same load torque, the speed
must decrease. This way, we can control the speed of an induction motor by changing the supply voltage.
However, we cannot reduce the supply voltage below a specific value. Otherwise, it will cause unstable motor operation.
This method is used when a minor variation in speed is required because this can damage the motor due to overheating.
The soft starters work on this principle for controlling the speed of the motor.
(2). Frequency Control Method:
As we know, the speed of an induction motor depends upon the supply frequency, i.e.
The induction motor speed is given by,
Hence, the speed of an induction motor can be changed by changing the supply frequency. This method is rarely used. This is because, at low supply frequencies, the current drawn by the motor becomes
too high due to a reduction in the inductive reactance. Also, the decrease in the supply frequency alone without changing voltage reduces the maximum torque developed by the motor.
(3). Stator Pole Change Method:
From the expression of synchronous speed, we can see,
Thus, we can change the speed of an induction motor by changing the number of stator poles. Two or more independent stator windings are wound in the same slots for different numbers of poles to
achieve change in stator poles. This method is mainly used in squirrel cage induction motors.
The speed of the motor operating at 50 Hz for different poles is given below.
Number of Poles Speed of Motor
(4). V/f Control Method:
To control the speed of an induction motor, this is the most widely used method. In this method, the supply voltage and frequency are reduced or increased in the same proportion. Maintaining the
ratio of voltage and frequency, the magnetic flux in the air gap of the motor remains constant. Also, due to the constant V/f, the torque developed by the motor is constant. When V/f control is used
in fan and pump applications, it is possible to save energy.
This method provides higher efficiency during running conditions. This is called the variable voltage and variable frequency (VVVF) method. Using this method, we can control the speed of an induction
motor over a wide range. You may have seen the VFDs, which function on this principle for controlling the motor’s speed.
Let us now discuss the methods of speed control from the rotor side of the induction motor.
Methods of Speed Control of Induction Motor From Rotor Side:
(5). Rotor Resistance Control Method:
In these Methods of Speed Control of Induction Motor, an external resistance is inserted in the rotor circuit of the induction motor. However, this method can be used for slip-ring induction motors
According to the torque equation, we get,
Under running conditions, the term (sX[2])^2 is very small and thus can be neglected.
Hence, if the rotor resistance R[2] increases, the torque will decrease, but to maintain the load torque constant, the slip must increase. This will take place by decreasing the rotor speed.
This way, adding an external resistance to the rotor circuit of the induction motor decreases the motor speed.
The key benefit of this method is that it increases the starting torque of the motor. However, it results in inefficiency due to high slip and copper loss.
(6). Slip Frequency EMF Injection into Rotor Circuit Method:
In this Method of Speed Control of Induction Motor, the speed is changed by applying a voltage to the rotor circuit. The most important thing that has to be considered is that the frequency of the
applied voltage must be equal to the slip frequency.
If the applied voltage or emf has an opposite phase concerning the rotor voltage or emf, the rotor resistance will increase. On the other hand, if the applied voltage has the same phase as the rotor
emf, the rotor resistance will decrease.
Hence, by changing the phase of the applied voltage to the rotor, we can change the rotor circuit resistance and, hence, the speed of the motor.
The significant benefit of this method is that it provides a wide range of speed control.
(7). Cascading Connection Method:
In the cascading Methods of Speed Control of 3 Phase Induction Motor, two induction motors are connected and run at the same speed. Here, one induction motor is supplied from a 3-phase supply, and
the second motor is powered from the induced voltage in the first motor with the help of slip rings.
The motor directly excited from the 3-phase supply is termed the main motor, and the motor excited from the induced EMF or voltage is referred to as the auxiliary motor.
Therefore, the slip in the main motor is given by,
The frequency of induced emf in the rotor is,
Since the auxiliary motor is supplied from the induced emf in the main motor. Thus, the synchronous speed of the auxiliary motor is given by,
Under the no-load conditions, the rotor speed of the auxiliary motor is approximately equal to its synchronous speed, i.e.,
On rearranging this equation, we get,
Now, according to this equation of induction motor speed, the following four cases are possible.
Case I – When the main induction motor is working:
Case II – When the auxiliary induction motor is working:
Case III – When cumulative cascading is done, the speed of two motors is:
Case IV – When differential cascading is done, the speed of two motors is:
In conclusion, the Speed Control Method of Induction Motor can be changed by using any of the methods mentioned above. In general, the induction motors are considered to have a constant speed. But,
by employing any of the above-given methods, we can control the speed. Different methods have different benefits and drawbacks.
Leave a Comment
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Can you calculate Pi using a Solver?
01-09-2020, 12:10 PM
(This post was last modified: 01-09-2020 12:14 PM by Gamo.)
Post: #41
Gamo Posts: 759
Senior Member Joined: Dec 2016
RE: Can you calculate Pi using a Solver?
HP-15C have a SOLVE function but I find the better way
to find the estimated value of Pi just by using the GAMMA function.
Formula: [ Γ(1/2) ]^2 = Pi
To find the estimated value of Pi follow this keystroke steps:
[.] 5 [ENTER] 1 [-] display -0.5 [x!] display 1.772453851 [x^2]
display 3.141592654
Answer is the same from the built-in Pi function.
User(s) browsing this thread: 5 Guest(s)
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Laser Cavity Resonance Modes and Gain Bandwidth
Laser Cavity Resonance Modes and Gain Bandwidth - Java Tutorial
In a typical laser, the number of cavity resonances that can fit within the gain bandwidth is often plotted as a function of laser output power versus wavelength. This interactive tutorial explores
how varying the appropriate frequencies can alter curves describing the number of cavity modes and gain bandwidth of a laser.
The tutorial initializes with the gain bandwidth and cavity frequency set at values of 1.5 gigahertz and 476 terahertz, respectively. In order to operate the tutorial, use the mouse cursor to
translate the Gain Bandwidth frequency slider between the ranges of 1 and 2 gigahertz. As this slider is translated, the yellow bandwidth curve changes shape to accommodate the new frequency range.
The number of cavity modes can be adjusted with the Cavity Frequency slider, which simultaneously varies the modes drawn with red curves in the tutorial window.
A common misconception about lasers results from the idea that all of the emitted light is reflected back and forth within the cavity until a critical intensity is reached, whereupon some "escapes"
through the output mirror as a beam. In reality, the output mirror always transmits a constant fraction of the light as the beam, reflecting the rest back into the cavity. This function is important
in allowing the laser to reach an equilibrium state, with the power levels both inside and outside the laser becoming constant.
Due to the fact that the light oscillates back and forth in a laser cavity, the phenomenon of resonance becomes a factor in the amplification of laser intensity. Depending upon the wavelength of
stimulated emission and cavity length, the waves reflected from the end mirrors will either interfere constructively and be strongly amplified, or interfere destructively and cancel laser activity.
Because the waves within the cavity are all coherent and in phase, they will remain in phase when reflected from a cavity mirror. The waves will also be in phase upon reaching the opposite mirror,
provided the cavity length equals an integral number of wavelengths. Thus, after making one complete oscillation in the cavity, light waves have traveled a path length equal to twice the cavity
length. If that distance is an integral multiple of the wavelength, the waves will all add in amplitude by constructive interference. When the cavity is not an exact multiple of the lasing
wavelength, destructive interference will occur, destroying laser action. The following equation defines the resonance condition that must be met for strong amplification to occur in the laser cavity
N • λ = 2 • (Cavity Length)
where N is an integer, and λ is the wavelength. The condition for resonance is not as critical as it might appear because actual laser transitions in the cavity are distributed over a range of
wavelengths, termed the gain bandwidth. Wavelengths of light are extremely small compared to the length of a typical laser cavity, and in general, a complete roundtrip path through the cavity will be
equivalent to several hundred thousand wavelengths of the light being amplified. Resonance is possible at each integral wavelength increment (for example 200,000, 200,001, 200,002, etc.), and because
the corresponding wavelengths are very close, they fall within the gain bandwidth of the laser. Figure 1 illustrates a typical example in which several resonance values of N, referred to as
longitudinal modes of the laser, fit within the gain bandwidth.
Laser beams have certain common characteristics, but also vary to a wide degree with respect to size, divergence, and light distribution across the beam diameter. These characteristics depend
strongly upon the design of the laser cavity (resonator), and the optical system controlling the beam, both within the cavity and upon output. Although a laser may appear to produce a uniform bright
spot of light when projected onto a surface, if the light intensity is measured at different points within a cross section of the beam, it will be found to vary in intensity. Resonator design also
affects beam divergence, a measure of beam spreading as distance from the laser increases. The beam divergence angle is an important factor in calculating the beam diameter at a given distance.
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Mixtures of Experts and scaling laws
Mixtures of Experts and scaling laws
Mixture of Experts (MoE) has become popular as an efficiency-boosting architectural component for LLMs. In this blog, we’ll explore the steps researchers have taken on the road toward the perfect
mixture of experts.
August 6, 2024
8 mins to read
MoE has been used in models like Mixtral, DeepSeek-V2, Qwen2-57B-A14B, and Jamba. However, like any architectural component, it has hyperparameters — total number of experts, number of active
experts, granularity — that can affect the final model quality.
MoE reminder
In the world of GPU- and data-intensive LLMs, it’s important to find balance between various precious resources. For example, if we want an LLM to excel at a wide range of tasks, this is enabled by
increasing the number of parameters, which in turn makes inference (as well as training) more compute-hungry.
MoE emerged as a way to create an LLM that is large and capable but somewhat less demanding at the inference stage. MoE suggests having several (e.g., 8) independent versions of a Feedforward block
(FFN) — “experts” — and a router that decides which (e.g., 2) of these experts are used for each particular token.
You might ask, “Why just FFN, and not self-attention as well?” Self-attention is too complex, and FFN blocks usually contain more than half of all the LLM parameters.
The first LLM with MoE was Mixtral-8×7B (read “8 experts with a 7B base model”), created from Mistral-7B by spawning 8 copies of each FFN block of Mistral and adding a routing mechanism that chooses
2 experts for each token. Compared to the 7B parameters of Mistral, it:
• has 47B parameters and was able to compete with 70B models at the time of its creation, but
• uses only 13B active parameters, making it more efficient than similarly-sized counterparts.
Mixtral calculated expert weights like this (source: Hugging Face):
With the final output being equal to: $y=\underset{i}\Sigma G(x)_i E_i (x)$
Note the random summand in $H(x)_i$ which works as a regularizer for training stabilization.
This only works well if the router is balanced, meaning it doesn’t favor or disregard certain experts. Otherwise, efficiency can be hindered instead of improved. Special “hacks, ” including an
auxiliary balancing loss function, are used to keep everything running properly. Moreover, given the router assignment for a current token, Mixtral’s MoE mechanism tries to divide an incoming batch
into almost equal parts with overhead not greater than pre-set capacity factor (usually around 1–1.25):
Illustration of token routing dynamics. Each expert processes a fixed batch-size of tokens modulated by the capacity factor. Each token is routed to the expert with the highest router probability,
but each expert has a fixed batch size of (total tokens / num experts) × capacity factor. If the tokens are unevenly dispatched then certain experts will overflow (denoted by dotted red lines),
resulting in these tokens not being processed by this layer. A larger capacity factor alleviates this overflow issue, but also increases computation and communication costs (depicted by padded white/
empty slots. Source: Switch Transformers by Google
Check the Hugging Face post mentioned above for more details.
Note: MoE LLMs are also referred to as sparse, while non-MoE models are called dense by comparison.
We need more experts
Mixtral had only 8 experts, but later models went much further.
For example, DeepSeek-V2 has 2 shared experts and 160 routed experts, of which 6 are selected for each token. With 236B total parameters, it has only 21B activated for each token. Shared experts are
always invoked; they are said to capture common knowledge across varying contexts. Routed experts are numerous, and some of them turn out to be highly specialized.
The behavior of MoE LLMs with a growing number of experts has been studied in several recent works, and there are good reasons to believe that having many experts is beneficial. I’ll mention two
works studying related empirical scaling laws:
1. Unified scaling laws for routed language models This paper showed that the validation loss tends to improve with a growing expert count:
The authors also studied the effective parameter count. For example, cB (c billions) is the efficient parameter count of Mixtral-8×7B if a hypothetical Mistral-cB would give the same quality as
Mixtral. The researchers found that the gain in effective parameter count diminishes with growing base model size: if Mistral had 1T parameters instead of 7B, creating Mixtral-8×1T out of it wouldn’t
improve the quality (same source):
(Here, S-BASE, RL-R, and Hash stand for different ways of distributing a batch between experts more evenly).
The takeaway: Having more experts is beneficial, although the gain diminishes with increasing base model size.
This approach may be criticized for using the same training dataset for all model sizes; this will be addressed in the next paper.
2. The next paper, Scaling laws for fine-grained mixture of experts, takes two important steps forward. First, it seeks optimal training dataset sizes for all the models. Second, it introduces the
idea of expert granularity. Imagine again Mistral-7B which we are turning into a MoE model. Initially, it becomes Mixtral-8×7B with 8 experts, each outputting d-dimensional vectors. Now, let’s
split each expert into G smaller experts that output vectors of dimension d/G:
If G = 2, each original expert becomes two fine-grained experts. Moreover, the router will now choose not 2 out of 8 experts, but 4 out of 16.
Now, the paper studies scaling laws as the balancing of the following parameters:
• Total training compute in FLOPs (which depends on both model size and dataset size),
• Base model size,
• Number of experts,
• Expert granularity,
• Validation loss.
Granularity turns out to be an important hyperparameter. As base model size increases, it seems beneficial to increase granularity (here, N is the model size and D is the training dataset size in
The problem is that increasing the number of experts and granularity may eventually hinder model efficiency, as seen in this plot (same source):
Here, for G = 16, the routing cost dominates gains from granularity. Overcomplicated routing will also make things slower at inference.
The takeaway: If we increase granularity in a timely manner, MoE steadily improves quality until routing complexity interferes.
What if we have a million tiny experts?
From the previous paper’s perspective, model quality may improve infinitely as we increase the number of experts and the granularity toward having something like 1,000,000 small experts — if only the
routing process is optimized.
A way to optimize it is suggested in the Mixture of a Million Experts paper. Imagine that we have many small experts $e_i$, each having a fixed key $k_i$ (just a constant vector). Let $K$ be the
number of experts we want to use for each token.
The MoE layer in this paper works differently compared to how it does in Mixtral:
1. Calculate a query vector $q(x)$,
2. Calculate the scalar products $q(x)^T k_i$,
3. Find $K$ maximal $q(x)^T k_i$,
4. Only for these experts, calculate router scores $g_i (x)=s(q(x)^T k_i)$,
5. Finally, the output is $f(x)=\underset{\text{chosen i}}\Sigma g_i (x)e_i (x)$.
The actual routing closely resembles a nearest neighbor search in a vector database. For that, we have efficient algorithms, but since we need to do it for every token, it can be good to optimize it
even further.
The authors suggest using product keys, that is, taking $k_i=(c_i,c’_i)$, a concatenation of two keys of half the dimension of $k_i$. For a million experts, it’s enough to have only a thousand
different $c_i$. Thus, instead of doing the nearest neighbor search in a 1,000,000-size database, we only need to do it twice for two 1,000-size databases, which is much more efficient.
The authors go as far as to suggest setting experts $e_i$ to be one-dimensional (with scalar output). To make the MoE more expressive, they make them multi-head:
1. Calculate $H$ independent query vectors $q_h (x)$,
2. Calculate the scalar products $q_h (x)^T k_i$,
3. For each $h$, find its own set of $K$ maximal $q_i (x)^T k_i$,
4. Only for these experts, calculate router scores $g_{h,i} (x) =s(q_h (x)^T k_i)$,
5. Finally, the output is $f(x)=\underset{h}\Sigma \underset{\text{chosen i}}\Sigma g_{h,i} (x)e_i (x)$.
Evaluation results may be summarized in the following plot:
Source: Mixture of a Million Experts
Using their method, called PEER, the authors are able to achieve a stable improvement in perplexity as N (the total total number of tiny experts) increases up to 1024^2.
The takeaway: With optimized routing, MoE steadily provides quality improvement.
When many experts can be of use: A case of lifelong learning
If you’re capable of training an LLM, you probably want to create a new one every year or so, with new architectural perks, etc. However, between episodes of training from scratch, you may want to
update your existing LLM on some new data — to adapt it to a new data distribution. Simply continuing the previous training process may sometimes cause catastrophic forgetting. LoRA is not very
capable of grasping new knowledge. So what should you do?
The Lifelong language pretraining with distribution-specialized experts paper suggests freezing the existing parts of the LLM and augmenting it with new experts and gating:
The results are somewhat mixed, but overall the idea is interesting.
This article was inspired by discussions at the paperwatch meetings of the Practical Generative AI course, which is run by the School of AI and Data Technologies. If you’re interested in studying
LLMs and other generative models, their internal workings and applications, check out our program.
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ORD-MIN Function
The ORD-MIN function returns a value that is the ordinal number of the argument that contains the minimum value. The type of this function is integer.
FUNCTION ORD-MIN ({argument-1} ... )
│ argument-1 │ If more than one argument-1 is specified, all arguments must be of the same class. │
Returned Values
1. The returned value is the ordinal number that corresponds to the position of the argument-1 having the least value in the argument-1 series.
2. The comparisons used to determine the least valued argument-1 are made according to the rules for simple conditions.
3. If more than one argument-1 has the same least value, the number returned corresponds to the position of the leftmost argument-1 having that value.
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2020 Fall Meeting of the APS Division of Nuclear Physics
Bulletin of the American Physical Society
2020 Fall Meeting of the APS Division of Nuclear Physics
Volume 65, Number 12
Thursday–Sunday, October 29–November 1 2020; Time Zone: Central Time, USA
Session DP: Mini-Symposium: Hadronic Weak Interactions I Hide Abstracts
Chair: Jason Fry, EKU
Friday, DP.00001: Fundamental Symmetry Violations in Nuclear Systems
October Invited Speaker: Jared Vanasse
30, 2020
8:30AM - Parity violation in nuclear systems offers a unique probe of QCD, while time-reversal violation in nuclear systems is a potential path for discovering physics beyond the Standard Model due
9:06AM to the smallness of Standard Model time-reversal violation in nuclear systems. Parity (P) and parity and time-reversal (PT) violating interactions in nuclear systems can be systematically
understood through the use of effective field theories, which are model independent and allow for theoretical error estimation. At low energies P and PT violating interactions are each
described by a separate set of five low energy constants (LECs). These LECs must either be calculated from lattice QCD or extracted from experiment. To cleanly extract these LECs experiments
should be performed on few-body nuclear systems for which P and PT violating processes can be calculated with minimal uncertainty. In this talk I will discuss the current theoretical and
experimental landscape for determination of these P and PT violating LECs. In addition I will discuss recent developments in the use of the large-$N_C$ expansion in QCD to estimate the
relative size of these LECs and its guidance in further experimental efforts. [Preview Abstract]
Friday, DP.00002: Anapole Moments of Light Nuclei from Ab Initio Theory
October Petr Navratil
30, 2020
9:06AM - Measurements of the nuclear spin dependent parity violating effects provide an opportunity to test nuclear models and to search for new physics beyond the Standard Model. Molecules possess
9:18AM closely spaced states with opposite parity which may be tuned to degeneracy to enhance the observed parity violating effects. An improved measurement of such effects with an unprecedented
sensitivity using light triatomic molecules composed of light elements Be, Mg, N, and C is in preparation [1]. We applied the no-core shell model (NCSM) [2] to calculate anapole moments of $
^{\mathrm{9}}$Be, $^{\mathrm{13}}$C, $^{\mathrm{14,15}}$N and $^{\mathrm{25}}$Mg needed for interpretation of this experiment. The only input for the NCSM calculations is the chiral
Effective Field Theory two- and three-nucleon interaction and the parity-violating nucleon-nucleon interaction derived within the meson exchange theory [3]. The NCSM results differ from the
predictions of the standard single-particle model and highlight the importance of including many-body effects in the calculations. [1] E. B. Norrgard \textit{et al}., Commun. Phys. \textbf
{2}, 77 (2019). [2] B. R. Barrett, P. Navratil, and J. P. Vary, Progress in Particle and Nuclear Physics \textbf{69}, 131 (2013). [3] B. Desplanques, J. F. Donoghue, and B. R. Holstein,
Annals of Physics \textbf{124}, 449 (1980). [Preview Abstract]
Friday, DP.00003: Parity violation in two-nucleon systems from pionless effective field theory
October Son Nguyen
30, 2020
9:18AM - At leading order, there are five independent nucleon-nucleon parity-violating (PV) interactions at low energies, corresponding to the five leading low energy constants (LECs) that mix
9:30AM $S$-wave and $P$-wave channels. These have been studied in the framework of pionless effective field theory (EFT). In this talk, we present our analysis of the higher-order contributions
that occur at three derivatives. These correspond to $P$-$D$ transitions as well as corrections to $S$-$P$ transitions. We show that the renormalization group (RG) behavior of the PV LECs in
pionless EFT is driven by the RG scaling of parity-conserving LECs. Under certain assumptions, this constrains the higher order LECs and may reduce the number of experiments needed to
understand low energy PV. [Preview Abstract]
Friday, DP.00004: NOPTREX: A Precision Measurement of the Parity Violation in the 0.7 eV Resonance in $^{139}$La
October Danielle Schaper
30, 2020
9:30AM - One of the motivations to search for new physics Beyond the Standard Model (BSM) is to understand the baryon asymmetry present in the Universe, namely the discrepancy between the theoretical
9:42AM prediction of the baryon asymmetry based on the SM and the value obtained through observations of the cosmic microwave background. The Neutron OPtics Time Reversal EXperiment (NOPTREX)
Collaboration seeks to measure signatures of parity-odd (P-odd) and time-reversal-symmetry-odd (T-odd) interactions in polarized neutron-polarized nucleon interactions. However, preliminary
measurements must be made in order to choose a desirable target nucleus for this experiment; one such condition is that the ideal target nucleus contains resonances which have a large amount
of parity violation (PV) present. From 2017-2019, the NOPTREX collaboration ran an experiment at the Los Alamos Neutron Science CEnter (LANSCE) facility to characterize the PV present in the
0.7 eV resonance in $^{139}$La to high (1\%) precision. This talk will focus on the experimental setup [1] as well as the subsequent data analysis process to extract the parity violation
asymmetry. // $[1]$ D. C. Schaper et al., Nucl. Instrum. Methods Phys. Res. A \textbf{969}, 2020. [Preview Abstract]
Friday, DP.00005: A Proposed Search for Time Reversal Violation in Polarized Neutron Transmission Through Polarized $^{117}$Sn
October Jonathan Curole
30, 2020
9:42AM - We describe work towards an experimental search for a P-odd and T-odd term in the polarized neutron-polarized nucleus forward scattering amplitude [1] on the 1.33 eV p-wave resonance in $^
9:54AM {117}$Sn, which exhibits a $10^{5}$ amplification of P-odd amplitude. $$ \frac{\Delta \sigma _{PT}}{\Delta \sigma _{P}}= \kappa (J) \frac{W}{V} $$ This formula relates the P-odd T-odd over
P-odd amplitude ratio $W/V$ to the ratio $\sigma_{PT} \over \sigma_{P}$ of the P-odd T-odd to P-odd cross sections, and a spectroscopic parameter $\kappa(J)$ involving the partial neutron
resonance widths in the $J=I \pm 1/2$ channels. We present a reevaluation of ($\vec{n}$,$\gamma$) angular distribution from the resonance [2] which implies a large, nonzero value for $\
kappa$ that controls the T-odd sensitivity. The $I=1/2$ $^{117}$Sn nucleus can be polarized with a technique known as SABRE which we will describe. [3] \newline References \newline [1]V. P.
Gudkov, Physics Reports 212, 77 (1992).\newline [2]Vo Van Tkhuan Yu. D. Mareev L.B. Pikelner I.M. Frank A.S. Khrykin V.P. Alfimenkov, S.B. Borzakov and E.I. Sharapov. Pis’maZh. Eksp. Teor.
Fiz., 39(8):346–348, 1984.\newline [3]A.M. Olaru, A. Burt, P.J. Rayner, S.J. Hart, A.C. Whitwood, G.G.R. Green, S.B. Duckett Chem. Commun. 52 14482 2016 [Preview Abstract]
Friday, DP.00006: "Bulk" Hyperpolarized $^{131}$Xe with $P_{Xe}\sim 7\%$: a Potential Target for Neutron Optics Searches for Time-Reversal Invariance Violation
October Boyd Goodson
30, 2020
9:54AM - The very large parity-odd asymmetry seen in the 3.2 eV p-wave resonance in $^{131}$Xe [1,2] makes it an interesting nucleus for NOPTREX. However this isotope is notoriously difficult to
10:06AM hyperpolarize owing to its strong nuclear quadrupole moment. We investigate the bulk preparation of hyperpolarized $^{131}$Xe via spin exchange optical pumping (SEOP). Isotopic enrichment
and next-generation spectrally-narrowed laser diode arrays allow for real-time observation of polarization dynamics via in situ low-field NMR, and optimization as a function of temperature,
alkali metal choice (Rb and Cs), resonance offset, and other parameters. $^{131}$Xe polarization values as high as 7.6$\%$$\pm$1.5$\%$ were achieved at 0.37 amagat in a ~0.1 L cell ($8.5\
times10^{20}$ $^{131}$Xe spins), demonstrating feasibility for use in spin-polarized neutron-scattering targets. Ongoing efforts to scale up $^{131}$Xe SEOP to aluminosilicate cells with
larger volumes for use in measurements of pseudomagnetic precession of polarized neutrons will also be described. [1] J. J. Szymanski, W. M. Snow et al., Phys. Rev. C 53, R2576 (1996). [2]
V. Skoy et al., Phys. Rev. C 53, R2573 (1996). [Preview Abstract]
Friday, DP.00007: Sensitivity of The NOPTREX experiment to Axion-like Particles
October William Snow
30, 2020
10:06AM The NOPTREX collaboration proposes to conduct searches for parity-odd and time-reversal-odd neutron-nucleus interactions on certain p-wave resonances in heavy nuclei where symmetry-breaking
- amplitudes are known to be amplified by several orders of magnitude [1]. A true null test of T is possible for this type of forward transmission polarized neutron optical observable [2].
10:18AM Axion-like particles (ALPs) appear in several extensions of the Standard Model and can generate a P-odd and T-odd neutron-nucleus interaction. We compare the sensitivity of NOPTREX and
electric dipole moment measurements to ALPs, which are not connected to the usual Peccei-Quinn axions. [1] V. P. Gudkov, Phys. Rep. 212, 77 (1992). [2] J. D. Bowman and V. P. Gudkov, Phys.
Rev. C 90, 065503 (2014). [Preview Abstract]
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What is the percentage concentration by mass of a 16*mol*L^-1 solution of nitric acid, for which density is 1.42*g*mL^-1? | Socratic
What is the percentage concentration by mass of a #16*mol*L^-1# solution of nitric acid, for which density is #1.42*g*mL^-1#?
2 Answers
Around 70% concentration, mass of acid/mass of solution.
A $16$$M$ concentration refers to the concentration of the solution. That is the nitric acid solution has a concentration of $16$$m o l$${L}^{-} 1$ of solution . We can work out percentages ($\frac
{w}{w}$#xx100%#) provided that we know the density of the solution, which we do because you have kindly included it in your question. So $1$$L$ of solution has a mass of $1420$$g$. In the solution
there are $16$$m o l$$H N {O}_{3}$, i.e. $16$$m o l$$\times$$63.0$$g$$m o {l}^{-} 1$$=$$1008.2$$g$.
So now we calculate the quotient, 1008.2 g $\times$ 1/(1420 g) $\times$#100%#$=$#??##%#. And this gives us our percentage concentration: (mass of solute/mass of solution) $\times$#100%#
Your stock solution is #71%"w/w"# nitric acid.
What you need to do here is pick a sample of this stock nitric acid solution and use its molarity to find out how much nitric acid it would contain.
Then use its density to determine athe sample's mass.
So, to make calculations easier, pick a $\text{1.00-L}$ sample of the stock solution. Since molarity is defined as moles of solute, in your case nitric cid, divided by liters of solution, you get
$C = \frac{n}{V} \implies n = C \cdot V$
${n}_{H N {O}_{3}} = \text{16 M" * "1.00 L" = "16 moles}$
Use nitric acid's molar mass to help you find how many grams of acid would contain this many moles
#16color(red)(cancel(color(black)("moles HNO"""_3))) * "63.013 g"/(1color(red)(cancel(color(black)("mole HNO"""_3)))) = "1008.2 g HNO"""_3#
What would be the mass of the sample? Use its known density!
#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.42 g"/(1color(red)(cancel(color(black)("mL")))) = "1420 g"#
The solution's percent concentration by mass, or $\text{%w/w}$, is defined as
$\text{%w/w" = "mass of solute"/"mass of solution} \times 100$
In your case, you would have
#"%w/w" = (1008.2color(red)(cancel(color(black)("g"))))/(1420color(red)(cancel(color(black)("g")))) xx 100 = color(green)(71%)#
SIDE NOTE You can redo the calculations using any sample of the stock solution, the result will always come out the same.
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G. Beck, D. Lannes, L. Weynans, arxiv.org/abs/2307.01749, Download
The goal of this work is to study waves interacting with partially immersed objects allowed to move freely in the vertical direction, and in a regime in which the propagation of the waves is
described by the one dimensional Boussinesq-Abbott system. The problem can be reduced to a transmission problem for this Boussinesq system, in which the transmission conditions between the components
of the domain at the left and at the right of the object are determined through the resolution of coupled forced ODEs in time satisfied by the vertical displacement of the object and the average
discharge in the portion of the fluid located under the object. We propose a new extended formulation in which these ODEs are complemented by two other forced ODEs satisfied by the trace of the
surface elevation at the contact points. The interest of this new extended formulation is that the forcing terms are easy to compute numerically and that the surface elevation at the contact points
is furnished for free. Based on this formulation, we propose a second order scheme that involves a generalization of the MacCormack scheme with nonlocal flux and a source term, which is coupled to a
second order Heun scheme for the ODEs. In order to validate this scheme, several explicit solutions for this wave-structure interaction problem are derived and can serve as benchmark for future
codes. As a byproduct, our method provides a second order scheme for the generation of waves at the entrance of the numerical domain for the Boussinesq-Abbott system.
D. Lannes, Lecture notes volume of the Winter School on Fluid Dynamics, Dispersive Equations and Quantum Fluids (Bressanone, december 2018), to appear. Download
The goal of these notes is to point out similarities and differences between two kinds of initial boundary value problems in dimension one. The first one concerns hyperbolic systems (such as the
nonlinear shallow water equations) while the second one concerns dispersive perturbations of such systems (such as Boussinesq systems). In the absence of a boundary, that is, for the initial value
problem, the link between both classes is quite obvious but in the presence of a boundary, the situation is more complex and dispersive boundary layers must be understood if one wants to understand
the links between both classes of problems. After reviewing several types of initial boundary value problems (some of which being free boundary problems) arising in the study of waves in shallow
water, we sketch the general theory for hyperbolic initial boundary value problems developed in [18] and that encompasses all of the above examples that involve hyperbolic systems. Such a general
theory does not exist for dispersive perturbations of hyperbolic systems, but we treat two important examples involving Boussinesq systems. In the first one, we show that the nature of the initial
boundary value problem shares little in common with the hyperbolic configuration. For instance, the problem has the structure of an ODE and no higher order compatibility conditions on the data are
required to have solutions of high regularity. These differences naturally raise the questions of the control of the time of existence and of the dispersionless limit; they are addressed in a second
example motivated by a wave-structure interaction problem. We explain the approach developed in [13] to treat this problem, pointing out the role played by dispersive boundary layers.
K. Martins, P. Bonneton, D. Lannes, and H. Michallet, Journal of Physical Oceanography 51 (2021), 3539–3556. Download
The inability of the linear wave dispersion relation to characterize the dispersive properties of nonlinear shoaling and breaking waves in the nearshore has long been recognized. Yet, it remains
widely used with linear wave theory to convert between subsurface pressure, wave orbital velocities, and the free surface elevation associated with nonlinear nearshore waves. Here, we present a
nonlinear fully dispersive method for reconstructing the free surface elevation from subsurface hydrodynamic measurements. This reconstruction requires knowledge of the dispersive properties of the
wave field through the dominant wavenumbers magnitude k, representative in an energy-averaged sense of a mixed sea state composed of both free and forced components. The present approach is effective
starting from intermediate water depths—where nonlinear interactions between triads intensify—up to the surf zone, where most wave components are forced and travel approximately at the speed of
nondispersive shallow-water waves. In laboratory conditions, where measurements of k are available, the nonlinear fully dispersive method successfully reconstructs sea surface energy levels at high
frequencies in diverse nonlinear and dispersive conditions. In the field, we investigate the potential of a reconstruction that uses a Boussinesq approximation of k, since such measurements are
generally lacking. Overall, the proposed approach offers great potential for collecting more accurate measurements under storm conditions, both in terms of sea surface energy levels at high
frequencies and wave-by-wave statistics (e.g., wave extrema). Through its control on the efficiency of nonlinear energy transfers between triads, the spectral bandwidth is shown to greatly influence
nonlinear effects in the transfer functions between subsurface hydrodynamics and the sea surface elevation.
Ability of the nonlinear method to reconstruct an extreme wave (below in green, plot of the nonlinear correction)
G. Beck, D. Lannes, Ann. IHP/Analyse non linéaire, to appear Download
We investigate here the interactions of waves governed by a Boussinesq system with a partially immersed body allowed to move freely in the vertical direction. We show that the whole system of
equations can be reduced to a transmission problem for the Boussinesq equations with transmission conditions given in terms of the vertical displacement of the object and of the average horizontal
discharge beneath it; these two quantities are in turn determined by two nonlinear ODEs with forcing terms coming from the exterior wave-field. Understanding the dispersive contribution to the added
mass phenomenon allows us to solve these equations, and a new dispersive hidden regularity effect is used to derive uniform estimates with respect to the dispersive parameter. We then derive an
abstract general Cummins equation describing the motion of the solid in the return to equilibrium problem and show that it takes an explicit simple form in two cases, namely, the nonlinear non
dispersive and the linear dispersive cases; we show in particular that the decay rate towards equilibrium is much smaller in the presence of dispersion. The latter situation also involves an initial
boundary value problem for a nonlocal scalar equation that has an interest of its own and for which we consequently provide a general analysis.
Waves interacting with an object moving vertically
Influence of dispersion on the oscillations of an object released from an out of equilibrium position
Siam News Volume 54 | Number 04 | May 2021 Online article
Some issues on the mathematical analysis of wave-structure interaction are presented, and their relevance for the development of wave energies are reviewed.
ICIAM Dianoia, Volume 9, Issue 2, April 2021 Online article
Interview with Laure Saint-Raymond and Arnaud Guillin, directors of the Institute of Mathematics of the Planet Earth
ICIAM Dianoia, Volume 9, Issue 1, January 2021 Online article
The UN general assembly has proclaimed the Decade of Ocean Science for Sustainable Development (2021-30). The French National Centre for Scientific Research (CNRS) is among the world’s leading
research institutions. Its scientists explore the living world, matter, mathematics, the universe, and the functioning of human societies in order to meet the major challenges of today and tomorrow.
How does this institution plan to address the specific issue of the ocean, and what role may mathematics play?
Anne Corval, advisor to the scientific director of the CNRS, coordinated a Task Force Ocean to prepare the institution for this Decade, and Fabrizio D’Ortenzio, oceanographer at the CNRS, was one of
the animators of the meetings organized to gather contributions from all sciences. In this interview, they address this question.
D. Lannes, Nonlinearity 33 (2020), R1 Download
We review here the derivation of many of the most important models that appear in the literature (mainly in coastal oceanography) for the description of waves in shallow water. We show that these
models can be obtained using various asymptotic expansions of the ‘turbulent’ and non-hydrostatic terms that appear in the equations that result from the vertical integration of the free surface
Euler equations. Among these models are the well-known nonlinear shallow water (NSW), Boussinesq and Serre–Green–Naghdi (SGN) equations for which we review several pending open problems. More recent
models such as the multi-layer NSW or SGN systems, as well as the Isobe– Kakinuma equations are also reviewed under a unified formalism that should simplify comparisons. We also comment on the scalar
versions of the various shallow water systems which can be used to describe unidirectional waves in horizontal dimension d = 1; among them are the KdV, BBM, Camassa–Holm and Whitham equations.
Finally, we show how to take vorticity effects into account in shallow water modeling, with specific focus on the behavior of the turbulent terms. As examples of challenges that go beyond the present
scope of mathematical justification, we review recent works using shallow water models with vorticity to describe wave breaking, and also derive models for the propagation of shallow water waves over
strong currents.
B. Desjardins, D. Lannes, J.-C. Saut, Water Waves (2020), 1-40 Download
Motivated by the analysis of the propagation of internal waves in a stratified ocean, we consider in this article the incompressible Euler equations with variable density in a flat strip, and we
study the evolution of perturbations of the hydrostatic equilibrium corresponding to a stable vertical stratification of the density. We show the local well-posedness of the equations in this
configuration and provide a detailed study of their linear approximation. Performing a modal decomposition according to a Sturm–Liouville problem associated with the background stratification, we
show that the linear approximation can be described by a series of dispersive perturbations of linear wave equations. When the so-called Brunt–Vaisälä frequency is not constant, we show that these
equations are coupled, hereby exhibiting a phenomenon of dispersive mixing. We then consider more specifically shallow water configurations (when the horizontal scale is much larger than the depth);
under the Boussinesq approxima-tion (i.e., neglecting the density variations in the momentum equation), we provide a well-posedness theorem for which we are able to control the existence time in
terms of the relevant physical scales. We can then extend the modal decomposition to the nonlinear case and exhibit a nonlinear mixing of different nature than the dispersive mixing mentioned above.
Finally, we discuss some perspectives such as the sharp stratification limit that is expected to converge towards two-fluid systems.
A. Mouragues, P. Bonneton, D. Lannes, B. Castelle, and V. Marieu, Coastal Engineering, 150:147 – 159, 2019 Download
We compare different methods to reconstruct the surface elevation of irregular waves propagating outside the surf zone from pressure measurements at the bottom. The traditional transfer function
method (TFM), based on the linear wave theory, predicts reasonably well the significant wave height but cannot describe the highest frequencies of the wave spectrum. This is why the TFM cannot
reproduce the skewed shape of nonlinear waves and strongly underestimates their crest elevation. The surface elevation reconstructed from the TFM is very sensitive to the value of the cutoff
frequency. At the individual wave scale, high-frequency tail correction strategies associated with this method do not significantly improve the prediction of the highest waves. Unlike the TFM, the
recently developed weakly-dispersive nonlinear reconstruction method correctly reproduces the wave energy over a large number of harmonics leading to an accurate estimation of the peaked and skewed
shape of the highest waves. This method is able to recover the most nonlinear waves within wave groups which some can be characterized as extreme waves. It is anticipated that using relevant
reconstruction method will improve the description of individual wave transformation close to breaking.
Location map with the field site of La Salie indicated by the black circle. (b) Unmanned aerial vehicle photo of the field site at mid-tide during the experiment. A video system was installed on the
pier shown in the lefthand side of the image. The yellow star and the red star show the location of the video system and the instrument, respectively, during the experiment.
Initial boundary value problems for hyperbolic systems, and dispersive perturbations
Relation between orbital velocities, pressure, and surface elevation in nonlinear nearshore water waves.
Freely floating objects on a fluid governed by the Boussinesq equations
Wave-Structure Interactions and Wave Energies
Creation of the institute « Mathematics of the Planet Earth »
Interview: Building interdisciplinarity around Ocean sciences
Modeling shallow water waves
Normal mode decomposition and dispersive and nonlinear mixing in stratified fluids
Field data-based evaluation of methods for recovering surface wave elevation from pressure measurements
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Devang Thakkar
Day 10: Pipe Maze
This is Day 10 of Advent of Code 2023! If you would like to solve the problems before looking at the solution, you can find them here.
In part 1 of Day 10, we're trying to make a loop in a pipe maze. We start by reading the input line by line and adding ins and outs for each square so that after we've parsed the entire input, we can
link the outs to ins and form a loop. The start square is denoted by an S so we don't know the connections to its neighbors but we know that it can only connect to two neighbors which can help us
figure out the connections. Once we solve for S, we can create a loop and the answer is half the length of the loop. Click here to jump to the part 2 discussion.
For part 2 of Day 10, we have to calculate the area enclosed by the loop. This took me a long time to debug despite knowing how to solve it. The key is to count the number of times we intersect the
loop when we come to a square from a particular edge of the input. The tricky part here was that there is a difference between squares that are LJ and squares that are FJ. While the former has two
intersections (and hence it's as if it wasn't there), the latter is the equivalent of a | and hence counts as a single intersection. I was really frustrated by this one, but it's done now. The amount
of time I spent on this puzzle tells me that the day when I don't get the second star is close.
That's all folks! If you made it this far, enjoy some AI art generated by u/dimpopo using the prompt for this puzzle.
Cheers, Devang
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Find A Single Discount Equivalent To Two Successive Discount Of 20% And 5%
Let the marked price =Rs100
After a discount of 20%, the price = 100−20% of100 = 100−10020×100 = 100−20=80
After a further discount of 5%, price = 80−5% of80 = 80−1005×80 = 80−4 = Rs76
Final Selling = Rs76
Total Discount = 100−76=Rs24
Thus, discount % = 10024×100 = 24%
V = A ≈ 78.54 cm3 (E)
Step-by-step explanation:
Using the formulas above and additional formulas you can calculate the properties of a given circle for any given variable.
Calculate A, C, and d | Given r
Given the radius of a circle calculate the area, circumference and diameter. Putting A, C and d in terms of r the equations are:
Calculate r, C, and d | Given A
Given the area of a circle calculate the radius, circumference, and diameter. Putting r, C, and d in terms of A the equations are:
Calculate A, r, and d | Given C
Given the circumference of a circle calculate the radius, area and diameter. Putting A, r, and d in terms of C the equations are:
Calculate A, C, and r | Given d
Given the diameter of a circle calculate the radius, area, and circumference. Putting A, C, and r in terms of d the equations are:
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Scalable and accurate variational Bayes for high-dimensional binary regression models
Modern methods for Bayesian regression beyond the Gaussian response setting are often computationally impractical or inaccurate in high dimensions. In fact, as discussed in recent literature,
bypassing such a trade-off is still an open problem even in routine binary regression models, and there is limited theory on the quality of variational approximations in high-dimensional settings. To
address this gap, we study the approximation accuracy of routinely used mean-field variational Bayes solutions in high-dimensional probit regression with Gaussian priors, obtaining novel and
practically relevant results on the pathological behaviour of such strategies in uncertainty quantification, point estimation and prediction. Motivated by these results, we further develop a new
partially factorized variational approximation for the posterior distribution of the probit coefficients that leverages a representation with global and local variables but, unlike for classical
mean-field assumptions, it avoids a fully factorized approximation, and instead assumes a factorization only for the local variables. We prove that the resulting approximation belongs to a tractable
class of unified skew-normal densities that crucially incorporates skewness and, unlike for state-of-the-art mean-field solutions, converges to the exact posterior density as p → ∞. To solve the
variational optimization problem, we derive a tractable coordinate ascent variational inference algorithm that easily scales to p in the tens of thousands, and provably requires a number of
iterations converging to 1 as p → ∞. Such findings are also illustrated in extensive empirical studies where our novel solution is shown to improve the approximation accuracy of mean-field
variational Bayes for any n and p, with the magnitude of these gains being remarkable in those high-dimensional p>n settings where state-of-the-art methods are computationally impractical.
• Bayesian computation
• Data augmentation
• Variational Bayes
• Truncated normal distribution
• Unified skew-normal distribution
• High-dimensional probit regression
Entra nei temi di ricerca di 'Scalable and accurate variational Bayes for high-dimensional binary regression models'. Insieme formano una fingerprint unica.
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Probability density Archives
In my previous post I introduced you to probability distributions.
In short, a probability distribution is simply taking the whole probability mass of a random variable and distributing it across its possible outcomes. Since every random variable has a total
probability mass equal to 1, this just means splitting the number 1 into parts and assigning each part to some element of the variable’s sample space (informally speaking).
In this post I want to dig a little deeper into probability distributions and explore some of their properties. Namely, I want to talk about the measures of central tendency (the mean) and dispersion
(the variance) of a probability distribution.
[Read more…]
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These type of questions
Last month 15 homes were sold in town X. the average price was $150,000 and median was $130,000. Which of the following must be true:
1) atleast one home sold for more than $165,000
2) at least one home was sold for more than $130,000 and less than $150,000
3) at least one home was sold for less than $130,000
Can someone pl direct me towards more questions like this. I can solve regular mean, median questions just get a bit confused in these ones
Thanks very much indeed for all the help
OA is 1 only
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Quantitative Skills
Which of these quantitative skills are needed for students in your major?
☐ Absolute Value Functions
1. Definition of absolute value and absolute value functions.
2. Graph of absolute value functions.
3. Solve absolute value equations.
4. Solve absolute value inequalities.
☐ Algorithms
1. Redefine a complex problem into a sequential set of parts that can be translated into the language of programming logic.
☐ Coding
1. Design, write, test, and debug computer programs.
☐ Conic sections
1. Graph parabolas, hyperbolas, and ellipses.
2. Find vertices and foci for hyperbolas and ellipses and vertices and directrixes for parabolas.
3. Use parabolas, hyperbolas, and ellipses to model tunnels, satellite dishes, and other objects.
☐ Data – Bivariate
1. Represent bivariate quantitative data using a scatter plot and describe how the variables might be related.
2. Compute and interpret a correlation coefficient given bivariate numerical data.
3. Distinguish between correlation and causation and between conspiracy and coincidence.
4. Critique graphical displays in the media.
☐ Data – Categorical
1. Summarize categorical data by constructing frequency tables and relative frequency tables.
2. Display categorical data with bar graphs.
3. Exploring two categorical variables by analyzing contingency tables.
4. Critique graphical displays in the media.
☐ Data – Collection
1. Distinguish between an observational study and a statistical experiment.
2. Describe the purpose of random selection in an observational study and the purpose of random assignment in a statistical experiment.
3. Understand the types of conclusions that can be drawn from an observational study and from an experiment.
4. Describe a method for selecting a random sample from a population.
☐ Data Structures
1. Incorporate fundamental data management concepts such as data structures within computer programs.
☐ Data – Univariate
1. Summarize univariate data using an appropriate graphical display.
2. Describe the shape of a distribution of numerical data and identify any outliers in the data set.
3. Summarize univariate data using appropriate numerical summary measures.
4. Interpret differences in shape, center and spread in the context of the data sets, accounting for possible extreme data points.
5. When appropriate, use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages.
6. Critique graphical displays in the media.
☐ Derivatives
1. Calculate derivatives using product, quotient, and chain rules.
2. Solve related rates problems.
3. Use derivatives to find maximum and minimum values.
4. Use the tangent line to approximate functions.
☐ Exponential Functions
1. Definition of exponential functions.
2. Use exponential functions to model growth and decay.
3. Graph of exponential functions.
4. Solving exponential equations.
☐ Finance
1. Explore essentials of creating a family/personal budget.
2. Understand the difference between simple and compound interest and their effects on savings and expenditures.
3. Explore saving and investment accounts.
4. Explore loan payments, credit card accounts and mortgages.
5. Explain debt and deficit
6. Explain tax systems
☐ Integrals
1. Calculate integrals using the idea of an antiderivative.
2. Calculate integrals using u-substitution.
3. Use integrals to represent area, net change, and other quantities approximated by a sum.
☐ Linear Functions
1. Definition of linear functions.
2. Slope-intercept form and point-slope form.
3. Parallel and perpendicular lines.
4. Piecewise functions.
5. Solve linear equations.
6. Solve linear inequalities.
7. Interpret slope and intercept.
8. Use linear functions to model relationships.
9. Fit a line to data points that are perfectly linear.
☐ Linear Regression
1. Construct a scatterplot of bivariate numerical data.
2. Calculate a correlation coefficient.
3. Calculate the least squares regression line of best fit.
4. Interpret the slope and y-intercept (if appropriate) of the least squares regression line in context.
☐ Logarithmic Functions
1. Definition of logarithmic function.
2. Properties of logarithm.
3. Graph of logarithm functions.
4. Solving logarithm equations.
5. Using logarithm to solve exponential equations.
☐ Logic:
1. Translate arguments into formal language and symbols
2. Recognize and apply standard techniques of logical deduction.
3. Recognize and avoid common logical fallacies.
☐ Mathematical Modeling
1. Use function notation, understand functions as processes, and interpret statements that use function notation in terms of a context.
2. Construct graphs and tables that model changing quantities and interpret key features in terms of the quantities.
3. Interpret the slope and the intercept of a linear model in the context of the data.
4. Graph linear and exponential functions and identify critical points.
5. Compute and interpret the correlation coefficient of a linear fit.
6. Distinguish between situations that can be modeled with linear functions and those modeled with exponential functions.
7. Use linear and exponential functions to model contextual situations such as costs and growth of savings accounts.
☐ Normal Distributions
1. Describe characteristics of a normal distribution and calculate and interpret a z-score using the mean and standard deviation of the normal distribution.
2. Calculate areas under a normal curve and interpret these areas as probabilities in context.
3. Approximate population percentages using a normal distribution.
☐ Parametric Equations
1. Convert between parametric and rectangular (x, y) equations.
2. Graph parametric equations.
☐ Polar Coordinates
1. Convert between polar and rectangular (x, y) equations.
2. Graph polar equations.
☐ Polynomial Functions
1. Definition of polynomial functions.
2. Characteristics of polynomial functions: degree, zeros, multiplicity, turning points.
3. Long division and/or synthetic division.
4. Solve polynomial equations.
5. Solve polynomial inequalities.
☐ Probability – Theory and Computation
1. Define sample space and events.
2. Distinguish between discrete and continuous random variables.
3. Explain that a probability distribution describes the long-run behavior of a random variable.
4. Calculate probabilities of unions, intersections and complements of events.
5. Use permutations and combinations to compute probabilities of compound events and solve problems.
6. Discuss the law of large numbers vs. law of averages myth.
7. Estimate probabilities empirically and interpret probabilities and long-run relative frequencies.
8. Distinguish between independent events and dependent events.
9. Use data in two-way tables to calculate probabilities, including conditional probabilities.
10. Calculate expected value.
11. Calculate standard deviation of a discrete variable
☐ Probability – Reasoning and Interpretation
1. Explain probability as a measure of the likelihood that an event will occur.
2. Describe events as subsets of a sample space using characteristics of the outcomes, or as unions, intersections, or complements of other events.
3. Interpret probabilities of the union and intersection of independent and dependent events.
4. Interpret probabilities in context.
5. Interpret expected value in context.
6. Interpret conditional probabilities in cases such as the false positive paradox.
7. Analyze risk in health situations and understand the difference between absolute changes in risk and relative changes in risk.
8. Interpret expected payoff for a game of chance.
☐ Quadratic Functions
1. Definition of quadratic functions
2. Graph quadratic functions: vertex, intercepts, maximum or minimum
3. Solve quadratic equations: factoring, quadratic formula, graphically.
4. Solve quadratic inequalities.
☐ Measurement
1. Understand the use of units, thinking of numbers as adjectives.
2. Study multiple ways of comparing quantities including the use of indices, e. g. the consumer price index and its relationship to the changing value of the dollar.
3. Investigate ways of finding exact and approximate areas and volumes of geometric and irregular shapes.
4. Estimate quantities.
5. Evaluate plausibility of numerical answers.
☐ Rates of Change
1. Display time series data using a time series plot.
2. Compute and interpret average rate of change from data, graphs, or function equations.
3. Estimate and interpret instantaneous rate of change from data, graphs, or function equations.
☐ Rational Functions
1. Definition of rational functions.
2. Graph of rational functions (including asymptotes).
3. Solve rational equations.
4. Solving rational inequalities.
☐ Recursive Functions:
1. Write and use functions that call themselves to perform calculations, solve problems, or process data.
☐ Statistical Inference – Theory and Computation
1. Describe characteristics of the sampling distribution of a sample mean and of a sample proportion.
2. Define and apply the central limit theorem for random samples.
3. Calculate a confidence interval for a population mean given a random sample.
4. Calculate a confidence interval for a population proportion.
5. Calculate a confidence interval for the difference in two population means or two population proportions.
6. Carry out a test of hypotheses about a population mean given a random sample.
7. Carry out a test of hypotheses about a population proportion.
8. Carry out a test of hypotheses about the difference in two population means or two population proportions.
☐ Statistical Inference – Interpretation
1. Describe statistics as a process for making inferences about population parameters based on a random sample from that population.
2. Explain the concept of sample-to-sample variability and describe how this understanding relates to statistical inference.
3. Explain the meaning of margin of error and interpret margin of error in context.
4. Explain the meaning of a confidence interval and interpret a confidence interval in context.
5. Interpret a P-value in context and use a P-value to reach a conclusion in a hypothesis testing context.
6. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each.
7. Evaluate reports or print media articles based on statistical data.
☐ Systems of Equations
1. Solving system of equations.
2. Set up systems of equations to solve problems involving mixtures, travel, and work rates, and other contexts.
☐ Trigonometric Functions
1. Graph trigonometric functions (sine, cosine, tangent).
2. Use sine and cosine functions to model periodic phenomena.
3. Solve triangles based on partial information about angles and sides, in applied settings.
4. Solve trigonometric equations.
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Levene Test & Statistics: Concepts & Examples - Analytics Yogi
Levene Test & Statistics: Concepts & Examples
The Levene test is used to test for equality of variance in a dataset. It is used in statistical analysis to determine if two or more samples have similar variances. If the results of the test
indicate that the samples do not have similar variances, then it means that one sample has a higher variance than the other and should be treated as an outlier. In this blog post, we’ll take a look
at what exactly the Levene test is, how it works, and provide some examples of how it can be applied. As data scientists, it will be important for us to understand the Levene test in order to
properly interpret and analyze our data.
What is the Levene Test?
The Levene test was developed by statistician Harold Levene in 1960 and is used to assess if there are significant differences among variances in different data samples. Essentially, the Levene tests
are used to assess whether there are approximately equal variances / homogenous variances or unequal variances between groups. In other words, Levene statistics is used to test the homegeneity of
variance between two or more groups. The null hypothesis of the Levene test is that all samples have approximately equal variances. There exists a homegeneity of variance between different groups.
The alternative hypothesis is that there are unequal variances between some of the samples. The following represents the null and alternate hypothesis:
The homegeneity of variance between different groups is one of the key assumption of tests such as independent samples t-test, one-way ANOVA test, etc. Although one-way ANOVA test is robust to this
assumption, large departures from homogeneity of variance assumption could impact the result of the 1-WAY ANOVA test.
Unlike statistical tests such as t-test where null hypothesis states that the mean of experimental group is different than the control group, in Levene test, the null hypothesis states that variance
of two or more groups are same.
What are the steps to calculate Levene Statistics?
Levene statistics is nothing but F-statistics calculated on absolute value of differences between individual value and median of each group. In other words, perform one-way ANOVA test on groups
having value calculated as absolute (individual value – median). Here are the steps:
• Collect the data from each sample. The data should be arranged into two or more columns, with one column corresponding to each of the groups being compared.
• Calculate median for each group.
• Create new groups such that each entry of new group is absolute of difference between individual member of the corresponding previous group and median of the group.
• Calculate F-statistics with transformed data of each group.
• If F statistics is less than the F-value at criticality, we don’t have enough evidence to reject the null hypothesis. Thus, the asssumption around homogeneity of variance between the two groups
is met. However, if the F statistics is more than F_critical value, than one can reject the null hypothesis which would mean that the variance are not equal.
Levene’s Test is an important statistical test to compare the variances of two or more samples and test the homogeneity of variance between two sammplles. It helps you determine if there are
significant differences between sample variances and can be used as a precursor for other tests such as ANOVA, t-tests, etc. In this blog post we have discussed what Levene’s Test is, and steps
required to calculate it. We hope that by understanding how to use this powerful tool in your data analysis arsenal, you will be able to make informed decisions about your research results with
confidence. If you need help using Levene’s Test or any other type of statistical test please don’t hesitate to contact us – our team of experts would love to assist!
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I found it very helpful. However the differences are not too understandable for me
Very Nice Explaination. Thankyiu very much,
in your case E respresent Member or Oraganization which include on e or more peers?
Such a informative post. Keep it up
Thank you....for your support. you given a good solution for me.
Ajitesh Kumar
I have been recently working in the area of Data analytics including Data Science and Machine Learning / Deep Learning. I am also passionate about different technologies including programming
languages such as Java/JEE, Javascript, Python, R, Julia, etc, and technologies such as Blockchain, mobile computing, cloud-native technologies, application security, cloud computing platforms, big
data, etc. I would love to connect with you on Linkedin. Check out my latest book titled as First Principles Thinking: Building winning products using first principles thinking.
Posted in Data Science, statistics. Tagged with Data Science, statistics.
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Decimal Dynamics: How to Master the Art of Adding and Subtracting Decimals
Decimals represent parts of whole numbers and are essential in various real-world scenarios, from measuring distances to handling money. Adding and subtracting decimals might seem tricky at first,
but with a systematic approach, it becomes straightforward. Let's explore the steps and strategies to effectively add and subtract decimals.
Key Insight:
When adding or subtracting decimals, it's crucial to line up the decimal points. This ensures that each place value is correctly aligned.
Adding and Subtracting Decimals
Example 1: Adding Decimals
Add \(23.45\) and \(7.689\).
Solution Process:
1. Write the numbers vertically, aligning the decimal points.
2. Start adding from the rightmost column (smallest place value).
3. Carry over any value greater than \(9\) to the next column.
The sum is \(31.139\).
The Absolute Best Book for 5th Grade Students
Original price was: $29.99.Current price is: $14.99.
Example 2: Subtracting Decimals
Subtract \(15.82\) from \(20.5\).
Solution Process:
1. Write the numbers vertically, aligning the decimal points.
2. If necessary, add zeros to make the numbers have the same number of decimal places.
3. Start subtracting from the rightmost column.
4. Borrow from the next column if needed.
The difference is \(4.68\).
Adding and subtracting decimals is a fundamental skill in mathematics. The key is to ensure that the decimal points are aligned, allowing for accurate calculations. Whether you’re calculating
distances, handling money, or measuring quantities, understanding how to add and subtract decimals is essential. With practice, you’ll find that working with decimals becomes second nature, enabling
you to tackle a wide range of mathematical and real-world challenges with confidence!
Practice Questions:
1. Add \(12.34\) and \(5.678\).
2. Subtract \(9.01\) from \(10.5\).
3. Add \(45.6\), \(23.45\), and \(7.89\).
4. Subtract \(32.8\) from \(50\).
5. Add \(0.123\), \(4.56\), and \(7.89\).
A Perfect Book for Grade 5 Math Word Problems!
Original price was: $29.99.Current price is: $14.99.
1. \(18.018\)
2. \(1.49\)
3. \(76.94\)
4. \(17.2\)
5. \(12.563\)
The Best Math Books for Elementary Students
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Mathematics 9 Quarter 1-Module 7: Relation of Roots and Coefficient of a Quadratic Equation (Week 2 Learning Code - M9AL-Ib-4) • DepEd Tambayan
Mathematics 9 Quarter 1-Module 7: Relation of Roots and Coefficient of a Quadratic Equation (Week 2 Learning Code – M9AL-Ib-4)
It is common that if we check the solutions of a quadratic equation, we simply substitute the solution to the original equation. But there is another method for this. In this module, you will
discover the relationship of the roots and coefficients of a quadratic equation and apply this concept in checking the roots and in constructing a quadratic equation.
The learners will be able to:
• describe the relationship between the coefficients and the roots of a quadratic equation. M9AL-Ib-4
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Show that the diagonals of a square are equal and bisect each other at right angles.
You must login to ask question.
NCERT Solutions for Class 9 Maths Chapter 8
Important NCERT Questions
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 8.1
Page No:146
Questions No: 4
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The purpose of the skedastic package is to make a suite of old and new methods for detecting and correcting for heteroskedasticity in linear regression models accessible to R users.
Heteroskedasticity (sometimes spelt ‘heteroscedasticity’) is a violation of one of the assumptions of the classical linear regression model (the Gauss-Markov Assumptions). This assumption, known as
homoskedasticity, holds that the variance of the random error term remains constant across all observations. Under heteroskedasticity, the Ordinary Least Squares estimator is no longer the Best
Linear Unbiased Estimator (BLUE) of the parameter vector, while the classical t-tests for testing significance of the parameters are invalid. Thus, heteroskedasticity-robust methods are required.
The most novel functionality of this package is provided by the alvm.fit and anlvm.fit functions, which fit an Auxiliary Linear Variance Model or an Auxiliary Nonlinear Variance Model, respectively.
These are new models for estimating error variances in heteroskedastic linear regression models, developed as part of the author’s doctoral research.
The hccme function computes heteroskedasticity-consistent covariance matrix estimates for the \(\hat{\beta}\) Ordinary Least Squares estimator using ten different methods found in the literature.
25 distinct functions in the package implement hypothesis testing methods for detecting heteroskedasticity that have been previously published in academic literature. Other functions implement
graphical methods for detecting heteroskedasticity or perform supporting tasks for the tests such as computing transformations of the Ordinary Least Squares (OLS) residuals that are useful in
heteroskedasticity detection, or computing probabilities from the null distribution of a nonparametric test statistic. Certain functions have applications beyond the problem of heteroskedasticity in
linear regression. These include pRQF, which computes cumulative probabilities from the distribution of a ratio of quadratic forms in normal random vectors, twosidedpval, which implements three
different approaches for calculating two-sided \(p\)-values from asymmetric null distributions, and dDtrend and pdDtrend, which compute probabilities from Lehmann’s nonparametric trend statistic.
Most of the exported functions in the package take a linear model as their primary argument (which can be passed as an lm object). Thus, to use this package a user must first be familiar with how to
fit linear regression models using the lm function from package stats.
Here is an example of implementing the Breusch-Pagan Test for heteroskedasticity on a linear regression model fit to the cars dataset, with distance (cars$dist) as the response (dependent) variable
and speed (cars$speed) as the explanatory (independent) variable.
To compute BLUS residuals for the same model:
To create customised residual plots for the same model:
To fit an auxiliary linear variance model to the same linear regression model, assuming that the error variances are a linear function of the speed predictor, and extract the resulting variance
To fit an auxiliary linear variance model to the same linear regression model, assuming that the error variances are a quadratic function of the speed predictor, and extract the resulting variance
Learn More
No vignettes have been created yet for this package. Watch this space.
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Math, Grade 6, Rational Numbers, Coordinate Plane
Make Connections
Performance Task
Ways of Thinking: Make Connections
Take notes about how to plot points on the coordinate plane and about strategies for playing the Coordinate Plane Game.
As your classmates present, ask questions such as:
• How did you describe the location of the bones using only numbers?
• What strategy did you use to find the bones?
• How is your strategy different from the other strategies mentioned so far?
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Operating Cost Calculator - Certified Calculator
Operating Cost Calculator
Introduction: Calculating the operating cost of a vehicle is essential for budgeting and making informed decisions about your transportation expenses. Whether you’re planning a road trip or managing
a fleet of vehicles for a business, this Operating Cost Calculator can provide valuable insights into your expenses.
Formula: To calculate the operating cost, we use the following formula: Operating Cost = (Distance / MPG) * Fuel Cost
How to Use:
1. Enter the distance you plan to travel in miles.
2. Input the current fuel cost per gallon in dollars.
3. Specify the miles per gallon (MPG) your vehicle achieves.
4. Click the “Calculate” button to get the estimated operating cost.
Example: Let’s say you’re planning a 300-mile trip with a vehicle that has an MPG of 25 and the current fuel cost is $3.50 per gallon. Enter these values into the calculator, click “Calculate,” and
you’ll find that the estimated operating cost for your trip is $42.00.
1. Q: How accurate is this calculator? A: This calculator provides a close estimate based on the input values, but real-world conditions may vary.
2. Q: Can I use this calculator for different currencies? A: Yes, as long as you enter the fuel cost in the currency you intend to use.
3. Q: What if my vehicle has different MPG for city and highway driving? A: Use an average MPG value for the most accurate estimate.
4. Q: Does it consider other costs like maintenance and insurance? A: No, this calculator only estimates fuel-related operating costs.
5. Q: Can I calculate the cost for a round trip? A: You’ll need to calculate one-way costs and then double the result for a round trip.
Conclusion: The Operating Cost Calculator simplifies the process of estimating your vehicle’s operating expenses. Whether you’re a budget-conscious individual or managing a fleet, this tool can help
you make informed decisions about your transportation costs. Remember that this calculator provides estimates, and real-world conditions may affect your actual expenses. Use it as a helpful planning
tool to keep your budget on track.
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Q13: Answers – Paper 2 November 18 – Edexcel GCSE Maths Higher
Helpful Links
Mark Scheme
Click here for a printable PDF of this question.
The diagram shows a circle and an equilateral triangle.
One side of the equilateral triangle is a diameter of the circle.
The circle has a circumference of 44 cm.
Work out the area of the triangle.
Give your answer correct to 3 significant figures.
............................... cm^2
[3 marks]
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Using Mixed Numbers to Represent Improper Fractions
In today’s post we’re going to learn about mixed numbers, which are much easier to interpret than the improper fractions they represent.
We’re going to look at some examples of the exercises and problems that you can work on in Smartick and the types of numbers that you can work on in these activities. You’ll be able to do them almost
without thinking!
Improper fractions as a mixed number
Remember that a mixed number is a numerical way of representing a fraction greater than a unit (improper fraction), or in other words, to represent fractions in which the numerator is greater than
the denominator.
Let’s start with an example:
• If you look at the improper fraction ^67⁄[13], it might not be easy to visualize the number that it represents, since it is greater than the unit (because the numerator is greater than the
denominator, 67 > 13).
• But if you convert it into a mixed number, you can understand the number it represents much more easily.
• To do this, the first thing you need to do is divide the numerator of the fraction by the denominator, to find out how many whole units the number contains.
• Since 65 = 13 x 5, we can separate the 67 into 65 and 2, and one of the parts will be divisible by 13 and the other won’t:
• Now, it’s really easy to write the mixed number: first we write the whole number and then the fraction that’s smaller than the unit:
• Using this representation it’s easy to see that the number is made up of 5 whole units and 2 thirteenths of another unit.
Now we’re going to look at some examples of the exercises and problems involving mixed numbers that you can work on in Smartick.
Example 1: Mixed numbers exercise
In the first type of exercise, we can practice writing the mixed number from its representation with pie charts. As you can see, this example shows a completely colored circle, and 1 equal part out
of 4 of another circle colored.
Therefore, it’s easy to see that the number represented graphically is 1^1⁄[4]:
Example 2: Mixed numbers exercise
In the second type of exercise we’re going to use pizzas to help us practice converting an improper fraction into a mixed number. As you can see in the example, we have 2 whole pizzas and
three-fourths of another pizza.
You can easily see that ^11⁄[4] of pizza is 2 whole pizzas and ^3⁄[4] of another pizza. In other words, the mixed number that corresponds to the improper fraction ^11⁄[4] is 2^3⁄[4].
This way it’s much easier to interpret, isn’t it?
Example 3: Mixed numbers exercise
In the third type of exercise, we practice writing improper fractions and mixed numbers by representing them with circular charts. In the example, you can see 2 complete circles and 2 equal parts out
of 3 of another.
Now the mixed number being represented is 2^2⁄[3] and the improper fraction being represented is ^8⁄[3]:
Example 4: Mixed numbers word problem
Finally, let’s look at an example from Smartick in which we work with mixed numbers to solve problems.
Let’s calculate the number of laps Albie has to run around his planet. Since he has to run 12 more laps than Alan, and Alan has to run ^9⁄[10] of a lap, all we need to do is add 12 and ^9⁄[10].
Representing the result as a mixed number is really easy because you don’t even have to do anything to add 12 and ^9⁄[10], you just need to write the whole number followed by the fraction: 12^9⁄[10].
Therefore we can say that Albie has to run 12^9⁄[10] laps around his planet.
And that’s all for today!
What do you think of this post? It’s really quick and easy to solve these exercises using mixed numbers, isn’t it?
If you want to keep learning and practicing more primary mathematics, log in to Smartick and try our learning method for free.
Learn More:
Add a new public comment to the blog:
The comments that you write here are moderated and can be seen by other users.
For private inquiries please write to [email protected]
3 Comments
• Maria Magdalena B. LohreNov 19 2020, 5:07 AM
Thank you, very helpful
• Ivy mayfieldJan 27 2020, 7:39 PM
It’s teaching my daughter math curriculum
• DivaApr 22 2019, 7:09 AM
Wow so good that you teach us
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Lua's Math Library
Lua's math library contains a collection of functions that are basically a subset of C's math library.
The math library is always available, and its functions can be accessed by name as shown in the following example:
print(math.abs(-123)) -- 123
The math library contains the following functions:
math.abs (x)
Returns the absolute value of x
math.acos (x)
Returns the arc-cosine of x, in radians.
math.asin (x)
Returns the arc-sine of x, in radians.
math.atan (x)
Returns the arc-tangent of x, in radians.
math.atan2 (x, y)
Returns the arc-tangent of y/x, in radians, based on the signs of both values to determine the correct quadrant.
math.ceil (x)
Returns the smallest integer value greater than or equal to x.
math.cos (x)
Returns the cosine of x, where x is specified in radians.
math.deg (x)
Convert x, specified in radians, to degrees
math.exp (x)
Returns the value of e raised to the xth power.
math.floor (x)
Returns the largest integer value less than or equal to x.
math.fmod (x, y)
Return a value equal to x - n*y for some integer n such that the result has the same sign as x and magnitude less than abs(y).
math.frexp (x)
Returns values (m, e) such that m * 2^e == x.
math.ldexp (x, y)
Return x * (2**y), essentially the inverse of function math.frexp.
math.log (x)
Return the natural log of x
math.log10 (x)
Return the base-10 logarithm of x
math.max (x, ...)
Returns the maximum value among its arguments.
math.min (x, ...)
Returns the minimum value among its arguments.
math.modf (x)
Spits real number x into it's integer and fractional parts.
The constant pi, approximately equal to 3.14159.
math.pow (x, y)
Returns x raised to the power of y.
math.rad (x)
Convert x, specified in degrees, to radians
math.sin (x)
Returns the sine of x, where x is specified in radians.
math.sqrt (x)
Returns the square root of x
math.tan (x)
Returns the tangent of x, where x is specified in radians.
math.random ()
Return a uniformly-distributed pseudo-random number between 0 (inclusive) and 1
math.random (x)
Return a uniformly-distributed pseudo-random number between 1 (inclusive) and x
math.random (x, y)
Return a uniformly-distributed, pseudo-random number between x (inclusive) and y (inclusive).
math.randomseed (x)
Sets x as the "random seed" for the pseudo-random generator
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Exploring the Possibility of Non-Existent Black Hole Singularities
Written on
Chapter 1: The Nature of Black Holes
Recent studies indicate that black holes might not harbor points of infinite density.
Renowned mathematician Roy Kerr made a significant contribution in 1963 by solving the field equations of Einstein’s General Relativity. This solution was pivotal in accurately modeling a realistic,
rotating black hole. The complexity of these equations makes finding exact solutions a formidable challenge, underscoring Kerr's achievement.
In 2023, Kerr expanded upon his original work, proposing a revolutionary concept: black holes may lack singularities, the traditionally accepted points of infinite density at their cores. This idea
directly contradicts modern cosmological views and could lead to groundbreaking implications.
Section 1.1: From Einstein to Schwarzschild
Albert Einstein finalized General Relativity in 1915. Initially met with skepticism, this theory now stands as the most comprehensive explanation of gravity available. It posits that matter and
energy distort spacetime, the four-dimensional structure of our universe. Essentially, the greater the mass or energy, the more significant the distortion of spacetime becomes. Consequently, gravity
should not be viewed as a fundamental force akin to electromagnetism; rather, it is a result of matter and energy tracing paths through curved spacetime. For instance, Earth orbits the Sun by
following a straight trajectory within the warped spacetime caused by the Sun's mass.
Since time is an intrinsic part of spacetime, it too experiences warping. Time measurements can vary based on the observer's frame of reference. For example, if two individuals synchronize their
clocks and one travels in a rocket at half the speed of light, their measurements of time will diverge upon reunion. This phenomenon is known as relativistic time dilation.
In the months following Einstein’s publication, Karl Schwarzschild derived the first exact solution to these field equations, all while serving in World War I. He utilized a static, non-rotating
massive body to describe how spacetime is warped around it. Despite being based on unrealistic conditions, the Schwarzschild solution remains relevant today due to its simplicity.
A crucial takeaway from the Schwarzschild solution is the concept of the Schwarzschild radius, which defines the event horizon of a black hole. If sufficient mass is compressed into a small enough
volume, spacetime distorts such that light cannot escape. For example, if the Sun were to collapse into a singular point, its Schwarzschild radius would measure 3 kilometers. Within this boundary,
escape becomes impossible, as it would require exceeding the speed of light.
Subsection 1.1.1: Theoretical Developments from Penrose to Hawking
In 1965, shortly after Kerr’s breakthrough regarding rotating bodies, British physicist Roger Penrose published a pivotal paper titled “Gravitational Collapse and Space-Time Singularities.” In this
work, he illustrated that singularities must exist at the center of black holes, asserting that “trapped surfaces inevitably lead to light rays of finite affine length.” Essentially, light trapped
within a black hole’s event horizon must adhere to the geometry dictated by the continuously collapsing spacetime.
Stephen Hawking supported Penrose’s conclusions in 1972 through his paper “Black Holes in General Relativity,” where he reaffirmed that the presence of a closed trapped surface indicates the
formation of a singularity under those conditions. Following this, both Penrose and Hawking, alongside numerous other physicists, formulated a series of singularity proofs collectively known as the
Penrose-Hawking singularity theorems.
Section 1.2: The Challenges of Infinity
Infinity frequently appears in mathematics, yet physicists grapple with its implications. Since infinity does not accurately represent the observable universe, physicists strive to devise methods to
eliminate it from their equations. For instance, in quantum field theory, infinities of the same order are often canceled out through division or subtraction. While infinities may arise as a natural
outcome of the mathematical frameworks that describe the cosmos, they pose challenges in practical applications.
Kerr himself has pointed out concerns regarding the reliance on infinities within descriptions of black holes. He noted, “The problem is that there is an infinity of possible solutions, but their
Einstein tensors do not necessarily satisfy appropriate physical conditions.” This indicates that while theoretical models may seem valid, they fail to align with real-world observations.
Chapter 2: A New Perspective on Singularities
In his 2023 paper “Do Black Holes Have Singularities?”, Kerr directly challenged the assumptions made by Penrose and Hawking. He argued that the geometry of spacetime within an event horizon does not
inevitably lead to singularities, suggesting that singularities represent merely one of numerous potential outcomes. “It has not been proved that a singularity, not just a FALL, is inevitable when an
event horizon forms around a collapsing star,” he states. Here, FALL refers to finite affine length light, indicating that light rays within an event horizon are not infinite and must eventually
Kerr further posited that rotating black holes might create layers of spacetime. The rotation affects spacetime through a phenomenon known as frame dragging, resulting in the formation of an outer
ergosphere—a dynamic zone where physical laws appear to function differently. Beneath the ergosphere lies the outer horizon, which corresponds to the event horizon or Schwarzschild radius. Inside
this horizon, various layers exist, including the inner horizon, outer ergosphere, and ring singularity. Light rays can thus inhabit one or multiple layers, implying they do not necessarily have to
terminate. Kerr urged that proponents of singularity must provide evidence for their claims instead of merely referencing Penrose’s assumptions.
Black Holes and the Quest for a Unified Theory
Kerr’s model eliminates the concept of a point singularity, potentially resolving the issues associated with infinities. If singularities are not point-like, the challenge of unifying General
Relativity with Quantum Mechanics may become more feasible. This endeavor aims to establish a “theory of everything,” or unified field theory. While General Relativity addresses gravitational
interactions on a cosmic scale, Quantum Mechanics accounts for the remaining three fundamental forces (the weak force, the strong nuclear force, and electromagnetism) at minuscule dimensions. Both
frameworks are immensely successful on their own, yet they fail to integrate, particularly concerning black hole singularities. Should Kerr’s assertions hold true, and point-like singularities prove
nonexistent, it could pave the way for eliminating infinities from our models, bringing us closer to achieving a comprehensive understanding of the universe.
In the video titled "What if Singularities DO NOT Exist?", experts discuss the implications of the absence of singularities in black holes, exploring new theories and concepts that may reshape our
understanding of gravity and spacetime.
Another insightful video, "We Were Wrong! Black Hole Singularities Don't Actually Exist!", delves into the evolving research surrounding black holes and the potential non-existence of singularities,
prompting a reevaluation of established scientific beliefs.
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Example 38.2 Frequency Dot Plots
This example produces frequency dot plots for the children’s eye and hair color data from Example 38.1.
PROC FREQ produces plots by using ODS Graphics to create graphs as part of the procedure output. Frequency plots are available for any frequency or crosstabulation table request. You can display
frequency plots as bar charts or dot plots. You can use plot-options to specify the orientation (vertical or horizontal), scale, and layout of the plots.
The following PROC FREQ statements request frequency tables and dot plots. The first TABLES statement requests a one-way frequency table of Hair and a crosstabulation table of Eyes by Hair. The PLOTS
= option requests frequency plots for the tables, and the TYPE=DOTPLOT plot-option specifies dot plots. By default, frequency plots are produced as bar charts. ODS Graphics must be enabled before
producing plots.
The second TABLES statement requests a crosstabulation table of Region by Hair and a frequency dot plot for this table. The SCALE=PERCENT plot-option plots percentages instead of frequency counts.
SCALE=LOG and SCALE=SQRT plot-options are also available to plot log frequencies and square roots of frequencies, respectively.
The ORDER=FREQ option in the PROC FREQ statement orders the variable levels by frequency. This order applies to the frequency and crosstabulation table displays and also to the corresponding
frequency plots.
ods graphics on;
proc freq data=Color order=freq;
tables Hair Hair*Eyes / plots=freqplot(type=dotplot);
tables Hair*Region / plots=freqplot(type=dotplot scale=percent);
weight Count;
title 'Eye and Hair Color of European Children';
ods graphics off;
Output 38.2.1, Output 38.2.2, and Output 38.2.3 display the dot plots produced by PROC FREQ. By default, the orientation of dot plots is horizontal, which places the variable levels on the Y axis.
You can specify the ORIENT=VERTICAL plot-option to request a vertical orientation. For two-way plots, you can use the TWOWAY= plot-option to specify the plot layout. The default layout (shown in
Output 38.2.2 and Output 38.2.3) is GROUPVERTICAL. Two-way layouts STACKED and GROUPHORIZONTAL are also available.
Output 38.2.1: One-Way Frequency Dot Plot
Output 38.2.2: Two-Way Frequency Dot Plot
Output 38.2.3: Two-Way Percent Dot Plot
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zk-SNARKs vs zk-STARKs | Horizen Academy
zk-SNARKs vs zk-STARKs
Privacy preserving technologies in the Web3 space have experienced a major resurgence in popularity over the past year.
As mainstream users and institutions begin to adopt public blockchain networks, the need to better conceal sensitive information on-chain, combined with the challenges of scaling these networks while
maintaining decentralization has led many protocol developers to turn to cutting-edge solutions like zero-knowledge proofs.
A zero-knowledge proof or ZKP is a form of cryptography that enables one party in a transaction, the prover to prove that they have knowledge of a specific piece of information to another party, the
verifier without revealing what that information is.
This technology has many advantages not only in protecting users’ sensitive data, but also in simplifying how nodes communicate to enable blockchain networks to confirm transactions with greater
efficiency and become more scalable.
Under the ZKP umbrella, there are 2 distinct approaches that have emerged as major forces in the fight to solve the blockchain scalability trilemma; zk-SNARKs and zk-STARKs.
In this article we will explore what differentiates these 2 methods for privacy-preserving transactions, and which approach Horizen has adopted to build its zero-knowledge-enabled network of
How Horizen Implements Zero-Knowledge Proofs
Before we dive into zk-SNARKs vs zk-STARKs, we will first lay out an example of how the ZKP technology works in practice on the Horizen network.
Horizen uses ZKPs to allow a prover, a sidechain, to prove that a transaction is valid to a verifier, our mainchain, without the mainchain nodes knowing the details of the transaction (i.e. addresses
or values transferred).
The sidechain does this by first validating transactions in its blockchain as normal, then constructing a ‘proof’, which is essentially a certificate that proves that the transactions within the
sidechain have been correctly validated based on a standard process that is recognized by the mainchain.
The process of constructing a proof requires the sidechain to solve a series of challenges issued by a hash function simulator to prove that the transaction data has been validated correctly.
This proof or certificate is used to accurately represent the state of the sidechain(s) (i.e., account balances and total values in each sidechain) while containing far less data compared to the tens
of thousands of transactions it represents.
The nodes on the mainchain then use a special algorithm to validate this certificate just like they would for any other transaction. If the algorithm confirms that the certificate/proof is correct,
the mainchain nodes will take this as evidence that the transactions in the sidechains have all been correctly validated without needing to perform the validation process themselves.
In other words, instead of the mainchain nodes validating every single sidechain transaction, the mainchain can simply verify ‘proof of computation’ for tens to hundreds of thousands of sidechain
transactions within a single block.
zk-SNARKs and zk-STARKs
SNARKs and STARKs are 2 distinct methods for verifying proof of computation for transactions on a blockchain without revealing the data within the computation.
• zk-SNARK stands for zero-knowledge succinct non-interactive argument of knowledge.
• While zk-STARK stands for zero-knowledge scalable transparent argument of knowledge.
‘Non-interactive’ is a trait that both approaches share. It simply means that the code constructing a proof or verifying proof of computation operates autonomously, without the need of human
The concept of zk-SNARKs was first explained in a 2012 paper co-authored by University of Berklee professor Alessandro Chiesa. A zk-SNARK is a form of zero-knowledge proof that is both ‘succinct’ and
A succinct proof is one that is small in terms of data size and can be easily verified using a custom algorithm.
It is considered ‘non-interactive’ when the prover must prove that a statement is true or a transaction is valid against a hash function simulator, which is meant to represent the verifier.
One of the key aspects of zk-SNARKs is that it relies on a trusted setup.
A trusted setup is an event that enables a set of private keys to be created that allows counterparties in a transaction to construct the ‘proof’ that must be verified in order for the transaction to
be confirmed as valid.
As we mentioned earlier in our Horizen example, the process of constructing a proof requires solving a series of challenges issued by a hash function simulator. These challenges are based on a set of
parameters that are initially created by private keys in the trusted setup event.
A key criticism of the trusted setup model is that it relies on trusted 3rd parties to create private keys that are used to construct the proof, which makes the process centralized.
If the private keys are not destroyed and a bad actor gains access to them, they could construct false proofs that would appear valid to the nodes on the Horizen mainchain, enabling them to
potentially create new tokens out of thin air by simply claiming through a false proof that they received them in a sidechain transaction.
Nevertheless, zk-SNARKs offer some key advantages, including wider development adoption and support, smaller proof sizes and cheaper gas costs on the Ethereum network.
zk-STARKs were invented in 2018 by Eli Ben-Sasson, Iddo Bentov, Yinon Horeshy and Michael Riabzev as an alternative approach to zk-SNARKs that does not require the use of a trusted setup to execute.
Instead, zk-STARKs use random samples of publicly verifiable information as a way to construct proofs without relying on trusted parties.
With this approach, zk-STARKs claim to offer greater scalability in terms of improving speed and reducing the computational size of transaction data.
It also uses a collision-resistant hash function, which supposedly makes it resistant to attacks by a quantum computer.
The drawback with zk-STARKs is that the proof sizes are larger than zk-SNARKs owing to the larger set of data that needs to be processed when sampling publicly verifiable information.
Larger proof sizes means that zk-STARKs require more gas fees for transactions.
zk-SNARKs vs zk-STARKs
Both zk-SNARKs and zk-STARKs have achieved major success in terms of fundraising and adoption.
zk-SNARKs are more widely adopted since they are a lot older than STARKs.
Zcash was the first blockchain to utilize zk-SNARKs for private transactions in 2016. More recently, zk-SNARKs have become a core component of Ethereum’s layer 2 scaling solution through zk-Rollups.
zk-Rollups enable layer 2 transactions to be transmitted and verified on the Ethereum mainnet without the mainnet nodes knowing the details of each transaction. It is designed to offer greater
scalability to the Ethereum network by only requiring the nodes to verify the proof of computation rather than the transaction data itself.
zk-Rollups have been adopted by layer 2 scaling solutions like Loopering, Polygon and zk-Sync.
zk-STARKs are earlier in its adoption curve but no less promising. The most prominent player in the space to adopt STARKs is a company aptly named Starware, created by zk-STARKs inventor Eli
Starkware has created a zk-rollup solution called StarkNet, which uses the zk-STARKs method to enable rollups that they claim provide cheaper and faster scalability solutions for the Ethereum
To date, Starware has raised over $50 million and is currently valued at $2 billion.
zk-STARKs and dYdX
One of the most popular applications to adopt zk-STARKs was dYdX, an on-chain derivatives exchange that processed trading orders via a more traditional off-chain order book as opposed to smart
contract-based liquidity pools.
zk-STARKs enabled DYDX users to save gas fees by executing trade orders through a STARK powered zk-Rollup. This allowed trades to be processed more quickly while still maintaining a high level of
security as all transaction data would ultimately settle on the Ethereum layer 1.
More recently, dYdX opted to move their dApp to Cosmos in order to utilize their own app-chain that would grant greater sovereignty and control over the full blockchain tech-stack.
Why Horizen Chose SNARKs
While both approaches to validating transactions with zero-knowledge present incredible opportunities for blockchain to scale while maintaining privacy, Horizen chose zk-SNARKs primarily because it
is a more tried and tested approach to applying zero-knowledge proofs on a blockchain.
Although a trusted setup may be more centralized, it also increases the accuracy of proofs and makes the technology more applicable to real world use cases like managing healthcare data or
implementing decentralized KYC solutions.
In both cases, third party entities like hospitals or government agencies will have to be involved in providing the initial inputs that a prover will need to construct a proof.
The zk-STARK approach, which has no external trusted setup phase and relies on sampling publicly available information to construct proofs ,cannot easily be adopted for use cases outside of
validating rollup transactions.
Ultimately, privacy preserving technologies have a highly promising future in the Web3 space.
Both zk-SNARKs and zk-STARKs are well funded and will continue to develop ways to make up for their shortcomings.
As our ecosystem grows and developers feel empowered to build new cutting edge solutions, we anticipate zk-STARKs playing a role in how transactions are confirmed within different sidechains.
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The Simplest Trend Following Strategy in Zorro TraderThe Simplest Trend Following Strategy in Zorro Trader
In this post, we will implement the simplest systematic trend following strategy in Zorro Trader, and we will run a back-test to measure its historical performance. The image above represents
graphically the output of the strategy in a back-test. The underlying asset in this case is the SPY, an SP500 exchange traded fund (ETF). In plain English, the strategy can be expressed as follows:
1. Calculate a simple moving average of a certain length. (We chose 50 days here, arbitrarily.)
2. If the price of the asset crosses above the simple moving average, we will close any short positions and will go long.
3. If the price of the asset crosses below the simple moving average, we will close any long positions and go short.
The rationale of this strategy is that a moving average contains less noise than the price curve, and it assumes that trends in the market tend to persist for some time. A cross of the price above de
simple moving average represents a bullish signal, while a cross below is a bearish signal. The strategy is 100% of the time in the market, and switches between long and short positions according to
the signals. It can’t really get any simpler than that, can it?
Implementing this strategy in Zorro Trader is extremely easy. Here is the Lite-C code:
function run()
BarPeriod = 1440;
vars price = series(priceClose());
vars sma = series(SMA(price, 50));
if(crossOver(price, sma)) enterLong();
if(crossUnder(price, sma)) enterShort();
This is all we need in order to implement, back-test and live trade the strategy (if you chose to do so, but we strongly urge you NOT to). In order to visualize the simple moving average in the
resulting image, we need to add the following line of code at the end. BEFORE the last curly brace, of course…
plot("SMA", sma, LINE, BLUE);
Honestly, it can’t really get much simpler than this! In a way, this is the “Hello, World!” program for algorithmic trading with Zorro Trader. Let’s give the code a closer look. The run() function of
Lite-C is similar to the main() function in C, but with a major difference. main() gets executed only once, while run() gets executed whenever a new OHLC bar comes in, either from the broker (in live
trading) or from the historical data file (in a back-test).
Talking about OHLC bars, the first variable we have to initialize is… the bar period. How long is it, in minutes? By default, Zorro Trader uses 1 minute bars, so we will set here BarPeriod to a full
trading day (1440 minutes). Please note that we do not have to set a type for this variable as it is an internal variable of Zorro Trader.
Define variables
Next, we have to define and initialize two variables, on that will hold the closing prices time series, and the other one the simple moving average time series. Both these variables are series of
numbers (EOD and SMA) so we will use the Lite-C vars type. Since they are time series, we will use the predefined Zorro Trader function series() to create them. This function will take as argument
two other predefined functions, priceClose() and SMA().
priceClose() pulls the closing prices, and SMA() calculates a moving average on the prices. SMA() also needs a length argument, respectively how many days we want the simple moving average to span.
We ask for a 50 days moving average, for no particular reason.
The following couple of lines define our trading logic. if() sets a condition, which is then checked by two Zorro Trader predefined functions: crossOver() and crossUnder(). They take as arguments the
two time series that can cross, the price and the simple moving average. If either condition is true, enterLong() and enterShort() will enter positions with your broker (in live trading) or trigger
performance calculations in a back-test.
Zorro Trader Output
By default, Zorro Trader plots in the output de price curve, the trades and the equity curve. In order to add the simple moving average, we use the plot() function, which takes as arguments a caption
for the plotted curve ("SMA"), the time series to be plotted (sma – the simple moving average) and the line type and color. Pretty simple, clear and self explanatory… at least for programmers. And if
you are not one yet, you can easily learn in the members area of our website.
Zorro Trader standard performance report
Test SimpleStrategy SPY, Zorro 2.444
Simulated account AssetsFix
Bar period 24 hours (avg 2087 min)
Total processed 1953 bars
Test period 2017-04-28..2022-05-06 (1265 bars)
Lookback period 80 bars (23 weeks)
Montecarlo cycles 200
Simulation mode Realistic (slippage 5.0 sec)
Avg bar 357.1 pips range
Spread 10.0 pips (roll 0.00/0.00)
Commission 0.02
Contracts per lot 1.0
Gross win/loss 301$-216$, +8448.2p, lr 117$
Average profit 16.83$/year, 1.40$/month, 0.0647$/day
Max drawdown -44.83$ 53.1% (MAE -70.05$ 82.9%)
Total down time 62% (TAE 93%)
Max down time 121 weeks from Feb 2018
Max open margin 463$
Max open risk 4.75$
Trade volume 22579$ (4497$/year)
Transaction costs -7.10$ spr, 0.39$ slp, 0$ rol, -1.42$ com
Capital required 498$
Number of trades 71 (15/year)
Percent winning 26.8%
Max win/loss 48.86$ / -9.59$
Avg trade profit 1.19$ 119.0p (+1582.2p / -415.6p)
Avg trade slippage 0.0055$ 0.6p (+15.3p / -4.8p)
Avg trade bars 17 (+51 / -5)
Max trade bars 109 (31 weeks)
Time in market 99%
Max open trades 1
Max loss streak 9 (uncorrelated 15)
Annual return 3%
Profit factor 1.39 (PRR 0.94)
Sharpe ratio 0.27 (Sortino 0.26)
Kelly criterion 2.26
Annualized StdDev 12.03%
R2 coefficient 0.614
Ulcer index 32.0%
Scholz tax 22 EUR
Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total
2017 1 0 1 -2 1 1 1 1 +4
2018 3 -7 0 -2 -1 -0 2 2 0 -1 -1 1 -3
2019 -1 2 1 2 -2 -1 1 -5 0 -1 2 2 -1
2020 -0 5 7 -6 3 1 4 4 -5 -1 0 3 +15
2021 -1 -0 1 4 1 -1 2 3 -6 1 -1 -2 +0
2022 1 3 -0 -2 0 +1
Confidence level AR DDMax Capital
10% 3% 33 489$
20% 3% 38 492$
30% 3% 41 495$
40% 3% 45 498$
50% 3% 49 501$
60% 3% 53 504$
70% 3% 57 507$
80% 3% 71 518$
90% 3% 93 535$
95% 3% 115 552$
100% 3% 217 631$
Portfolio analysis OptF ProF Win/Loss Wgt%
SPY .999 1.39 19/52 100.0
SPY:L .999 2.80 14/21 165.5
SPY:S .000 0.60 5/31 -65.5
This is obviously not a very good trading strategy, but simple strategies such as this one rarely are. We are making an Annual Return of 3% with a Sharpe Ratio of 0.27. This means we are making less
money than “the market” but we are taking on more risk (the benchmark here being the SP500). Examining the image above, we can easily spot the problem, which is typical for all trend following
strategies. Although we are usually on the right side of almost all big trends, we get killed by a lot of small losing trades when the market moves sideways.
This observation may hold the key to further strategy improvements. Can we trade a different asset and do better? Can we filter out some of the bad trades in sideways market regimes? Or is this
strategy truly hopeless? All these ideas will eventually become topics of our video courses in the members area.
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Should you pay for mortgage points? - GLYPH
Should you pay for mortgage points?
Mortgage points – a bet with the bank.
The housing market is a little crazy^1 at the moment. There are indications^2 that the growth in home prices will slow down, both actual data^3 and anecdata^4 suggest that there are a lot of people
doing mortgage math at the moment. There are a lot of calculators^5 out there to help you do this, but when my wife and I bought a house, there was one aspect of the process that I did not feel was
done correctly and completely anywhere.
When you apply for a mortgage, most lenders let you pay more front to reduce the interest rate. These payments are known as “points”, and typically cost 1% of the value of the loan. If you keep
paying that loan long enough, the money you save because of the reduced interest rate becomes larger than the amount of money you paid up front for the points, and you pay less money in total for the
house. Hooray!
However, paying for the points is effectively a bet that you are making with the bank. You are betting that you will remain in the house without refinancing for more time than it takes for that lower
interest rate to pay for itself. So, how much time does it take to pay off the points? It depends on the base rate, the number of points, and the interest rate discount that you get per point.
Although this is relatively easy to calculate, most online calculators make a incorrect (but honest) assumption. We’ll first develop the payoff time in this wrong way before explaining the right way
to think about it.
The wrong way
The monthly payment
where ^6 the expression
where as we all know is the apotheosis of scientific communication. See below.
Figure 1. Contour representation of the payoff time for a given number of mortgage points at a given base mortgage rate, using a fixed loan amount (which is slightly wrong) (foreshadowing).
In Fig.1, we see that the payback time is a little over 6 years for a 3.45% APR, which is the national average^7 at the time of writing for a 30-year fixed rate mortgage. There are a couple of other
notable things about the figure. First, the payback time decreases with increasing base rate. Second, the payback time increases as the number of points purchased increases. We’ll revisit these when
we show the right way to find the payback time, after we discuss…
Why the above is wrong
I keep saying it, but what is actually wrong with the above method? The issue comes at the very beginning, when we used the same loan amount
but if you want to buy points, the amount available for the down payment decreases, which increases the loan amount such that
To do this right, we need to make two intermediate calculations, each of which is interesting in its own right. First, update the expression for the “points” monthly payment to
respectively. Because ^8 values of
This is a little counter-intuitive, partially because the phrase “buying points” make it seem like they’re independent of the loan amount. They’re not! You don’t have “down payment money” and “points
money” and never the twain shall meet. It’s all just money^9. That said, if the prospect of paying a little less every month gets you to increase
The points are not actually free though. After
Figure 2. Contour representation of when you will own more of a house, having purchased some number of mortgage points at a given base mortgage rate.
We get the above as follows. The amount remaining on the base rate loan in month
and likewise, the amount remaining on the modified rate loan in month
The amount of house you’ve bought after
When you solve the equation
The correct way
The net from the house purchase for the “no points” and “points” rates are
and so you find the payoff time by finding the month
which is shown in Fig.3.
Figure 3. Contour representation of the payoff time for a given number of mortgage points at a given base mortgage rate, using a fixed amount of initial cash.
This is actually a slight improvement on the wrong method shown in Fig.1, meaning that the (easier) wrong method is conservative, a wonderful property for wrong methods to have. Unlike the wrong
method, the payback time increases with increasing base rate. This makes sense, because the initial loan balance is larger. However, the correct method shows — like the wrong method — that the
payback time increases as the number of points purchased increases. The most effective point or fractional point is therefore the first one, which makes it easier to discuss the effect of…
What you get for your points
At the beginning, we fixed the change in monthly rate that you get for a point as
which we can generalize to
Figure 4. Contour representation of the payoff time for most effective fractional mortgage point at a given base mortgage rate, using a fixed amount of initial cash.
Unlike the changes in rate and number of points, changing the interest rate break you get per point makes a big difference. Changing the interest rate break from our nominal value of
Concluding remarks
There are some things that I had to leave out of the analysis of whether you should pay for points. Some of these things are not knowable, like what return rate you would get by putting the money
you’re not using to pay for points into the stock market. (Zeroth order hypothesis: 10% nominal^10, as long as you don’t do anything weird^11 with it.) These calculations are valid for any term of
fixed-rate mortgage, but because you can’t know what an adjustable rate mortgage (ARM) will do, these can only help bound the payback time of points on ARMs.
Other things are knowable for a given loan or lender, but there’s so little consensus that it’s hard to state generally how to engage with them. For example, many lenders require private mortgage
insurance^12 to be paid if the down payment is too small. Should you avoid buying points until your down payment is enough to avoid paying those premia? Probably, but there might be edge cases where
it’s appropriate.
So, should you buy the points? Speaking against it is the tendency to overestimate how much you’ll approve of your current choices in the future,^13 ^14 and the tendency of mortgage rates to move
around over time, which increases the likelihood of refinancing.^15 ^16 That said, if rates stay in their current range, and you’re pretty sure you going to be in that house for more than 5 years,
and you don’t think you’ll refinance the loan in that time, then it’s a decent bet.
(Edit, 23-Oct-2022: Finished the followup article about refinancing. tl;dr: over the last 40 years, this was a good bet for the bank)
Questions or comments are best directed to @GlyphStory.
Featured image: Nick Young screencap by @jose3030, house from Wikimedia, cursed image manipulation by Aren with Mathematica.
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When Maths meets Medicine - Tobias Brunnbauer
When Maths meets Medicine
After my a-levels I struggled to decide between studying maths and medicine. Now I found out that those subjects are deeply connected to each other. Uğur Şahin, Juan Klopper and others are pionieers
in this field. AI, Machine Learning and computational medicine change will definitely change the way doctors practice medicine.
After finishing my a-levels, I was torn between studying maths and medicine.
Unsettled due to my inability to decide, I began studying pure maths in Heidelberg. After receiving a scholarship in Bavaria, I switched subjects to medicine in Würzburg. Though I didn’t regret this
decision for a second, my interest in maths remains unsatisfied.
Talking to students and graduates of both fields – maths and medicine – lead me to think that I would have to decide. Do I want to go down the path of maths or the one of medicine? It seemed like
there was no intersection between those fields.
I was desperate. For my whole life, teachers, parents, and I have preached to myself that I’d always follow my curiosities. The decision between maths and medicine seemed like a dilemma to me. At
some point, I figured that dilemmas are only a thought construct to help us make sense of the world.
That means a dilemma is only a dilemma if I believe it is one (at least in this case). Changing the way I think about the intersection of maths and medicine has been a game-changer.
Just for the sake of clarity: I know that there is a broad intersection of mathematical and medical knowledge in the realm of bioengineering, biotechnology, biophysics, etc. Yet, following such
studies usually doesn’t allow you to practice medicine later on.
People of Maths and Medicine
Not that many people practice medicine (about 0,5% of the German population). And only a small fraction of these people have an interest in maths or related fields at the same time. Fortunately, the
internet is a great way to find role models when it comes to combining maths and medicine.
The first person I stumbled upon was Uğur Şahin. You may know him as the founder of BioNTech. It’s a German biotech startup that applies mRNA technology to produce COVID-19 vaccinations.
Uğur has a background in medicine. He was an extraordinary student, completed his studies in medicine with distinction, and soon specialized in experimental oncology as a clinician and scientist
before founding several biotech startups.
This alone sounds impressive. Yet, there is one fact that struck me about his curriculum. Right after graduating from medical school, he did a bachelor’s in pure mathematics!
Now, he says that he has always loved the problem-solving aspect of maths and also used it to predict the extent of the COVID crisis in early 2020.
For me, Ugur was the first person to prove that it was possible to combine maths and medicine. May it only be for the sake of pure interest, you never know what it’s good for.
Doing some more research on the Internet, I found another guy called Juan Klopper. On his personal blog, he describes himself as a data scientist specializing in deep learning for healthcare.
He received training as a medical doctor – even as a surgeon – and still daily practices medicine. He’s one of the first people I’ve seen that brings together maths, data science, practical medicine,
and education. What’s even more intriguing: Mr. Klopper self-studied maths. He used Coursera courses and the like to meet his curiosity for both fields – maths, and medicine.
Still, there are more exciting people out there who are interested in both subjects.
One of them, for example, is Kumar Thurminella. He is an MD/PhD student split between the University of Colorado School of Medicine and the University of Cambridge.
As an applied mathematician/software engineer turned future physician, he is excited to apply his computational and mathematical background to help in the discovery of mechanisms behind inflammatory
diseases of the gut.
Btw, if you know more of those “maths-medicine-people”, please let me know!
The working field of Juan Klopper and Kumar Thurminella, namely computational maths, is a great point to change perspective from the present to the future.
The Future of mathematical medicine
During the last years, medicine has been transformed by data science. AI and Machine Learning allow us to diagnose much more efficiently and some radiologists even fear being rendered obsolete by
computer technology.
While this is still a dystopia, Dr. Bertalan Meskó, founder of the „Medical Futurist Institute“ believes that in no way doctors will become obsolete – only the ones who don’t adapt to the proceeding
technology will, as it has always been the case in history.
Since this article is not about AI in medicine, you can find more about it here.
The point I want to make is that maths and computer science will become indispensable in the medical field.
Intersection Points
One of the biggest intersection points of maths and medicine is genetics. Basically, you can break down genetics in a simple manner. There is a long code, consisting of a repeating combination of the
letters “ACTG”. You want to encode this whole mystery and thereby find out how the human genome works together. Finding out the relation between certain genes, or discovering the mechanisms of
epigenetics, are quite mathematical-ish problems to solve.
In her book, “Why study mathematics”, Cambridge professor Vicky Neale beings a chapter with the words: “Maths saves lives”. This is of course an allusion to the ongoing COVID-19 crisis where maths
has helped to answer life-depending questions.
Modeling the spread of viruses has been crucial as a base layer for governmental decision-making. Eventually, all of those models rely on solving differential equations which require thorough
mathematical understanding.
If you are interested, check out the book – it perfectly demonstrates why maths and medicine must be seen as two deeply connected subjects. The list of applications goes on, from image diagnosis
software to simulating drug effects.
As a bottom line, it’s not only the academic satisfaction that can be gained from combining medicine and maths – it’s also the positive future outlook.
If you found this article helpful or inspiring, consider writing me a message to hello@tobiasbrunnbauer.de.
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Section Summary
7.1 Work: The Scientific Definition
• Work is the transfer of energy by a force acting on an object as it is displaced.
• The work $WW size 12{W} {}$ that a force $FF size 12{F} {}$ does on an object is the product of the magnitude $FF size 12{F} {}$ of the force, times the magnitude $dd size 12{d} {}$ of the
displacement, times the cosine of the angle $θθ size 12{q} {}$ between them. In symbols,
$W = Fd cos θ . W = Fd cos θ . size 12{W= ital "Fd""cos"θ "." } {}$
• The SI unit for work and energy is the joule (J), where $1J=1N⋅m=1 kg⋅m2/s21J=1N⋅m=1 kg⋅m2/s2 size 12{1" J"=1" N" cdot m="1 kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } } {}$.
• The work done by a force is zero if the displacement is either zero or perpendicular to the force.
• The work done is positive if the force and displacement have the same direction, and negative if they have opposite direction.
7.2 Kinetic Energy and the Work-Energy Theorem
• The net work $WnetWnet$ is the work done by the net force acting on an object.
• Work done on an object transfers energy to the object.
• The translational kinetic energy of an object of mass $mm$ moving at speed $vv$ is $KE=12mv2KE=12mv2 size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {}$.
• The work-energy theorem states that the net work $WnetWnet size 12{W rSub { size 8{"net"} } } {}$ on a system changes its kinetic energy, $Wnet=12mv2−12 m v 0 2 Wnet=12mv2−12 m v 0 2$.
7.3 Gravitational Potential Energy
• Work done against gravity in lifting an object becomes potential energy of the object-Earth system.
• In a uniform gravitational field, such as that near Earth’s surface, the change in gravitational potential energy of a body of mass m, $ΔP E g ΔP E g$, is $ΔP E g =mgh ΔP E g =mgh$, with h being
the increase in height and g the acceleration due to gravity.
• The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, $ΔPEgΔPEg size 12{Δ"PE"
rSub { size 8{g} } } {}$, have physical significance.
• As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that $ΔKE= −ΔPEgΔKE= −ΔPEg size 12{D"KE""=-"D"PE" rSub
{ size 8{g} } } {}$.
• In a gravitational field described by Newton’s universal law of gravitation, the change in gravitational potential energy of an object of mass m that occurs when it is brought from infinitely far
away to a distance r from the center of mass of an object of mass M is $ΔP E g =−G mM r ΔP E g =−G mM r$, where G is the gravitational constant.
• A scalar field associated with an object of mass M, called the gravitational potential field, can be defined as the change in gravitational potential energy per unit mass of a test object moved
from infinitely far away to a given point in space. The potential is given by $V g =−G M r V g =−G M r$, where r is the distance between the object’s center of mass and a given point in space.
• When more than one object is present, the gravitational potential at a point in space is the sum of the individual gravitational potentials at that point.
• Characteristics of a gravitational potential field—represented, for example, by isolines on a contour plot—can be used to make inferences about the number, relative size, and location of the
7.4 Conservative Forces and Potential Energy
• A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken.
• We can define potential energy $(PE)(PE) size 12{ \( "PE" \) } {}$ for any conservative force, just as we defined $PEgPEg size 12{"PE" rSub { size 8{g} } } {}$ for the gravitational force.
• The potential energy of a spring is $PEs=12kx2PEs=12kx2 size 12{"PE" rSub { size 8{s} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {}$, where $kk size 12{k} {}$ is the spring’s force
constant and $xx size 12{x} {}$ is the displacement from its undeformed position.
• Mechanical energy is defined to be $KE + PEKE + PE size 12{"KE "+" PE"} {}$ for a conservative force.
• When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form,
where i and f denote initial and final values. This law is known as the conservation of mechanical energy.
7.5 Nonconservative Forces
• A nonconservative force is a force for which work depends on the path.
• Friction is an example of a nonconservative force that changes mechanical energy into thermal energy.
• Work $WncWnc size 12{W rSub { size 8{"nc"} } } {}$ done by a nonconservative force changes the mechanical energy of a system. In equation form, $Wnc=ΔKE+ΔPEWnc=ΔKE+ΔPE size 12{W rSub { size 8
{"nc"} } =Δ"KE"+Δ"PE"} {}$ or, equivalently, $KEi+PEi+Wnc=KEf+PEfKEi+PEi+Wnc=KEf+PEf size 12{"KE" rSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE" rSub { size 8{f} }
+"PE" rSub { size 8{f} } } {}$.
• When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the known potential energies of the conservative forces and the
work done by nonconservative forces, instead of finding the net work from the net force, or having to directly apply Newton’s laws.
7.6 Conservation of Energy
• The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same.
• When all forms of energy are considered, conservation of energy is written in equation form as $KEi+PEi+Wnc+OEi=KEf+PEf+OEfKEi+PEi+Wnc+OEi=KEf+PEf+OEf size 12{"KE" rSub { size 8{i} } +"PE" rSub {
size 8{i} } +W rSub { size 8{"nc"} } +"OE" rSub { size 8{i} } ="KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } +"OE" rSub { size 8{f} } } {}$, where $OEOE size 12{"OE"} {}$ is all other forms of
energy besides mechanical energy.
• Commonly encountered forms of energy include electric energy, chemical energy, radiant energy, nuclear energy, and thermal energy.
• Energy is often utilized to do work, but it is not possible to convert all the energy of a system to work.
• The efficiency $EffEff size 12{ ital "Eff"} {}$ of a machine or human is defined to be $Eff=WoutEinEff=WoutEin size 12{ ital "Eff"= { {W rSub { size 8{"out"} } } over {E rSub { size 8{"in"} } } }
} {}$, where $WoutWout size 12{W rSub { size 8{"out"} } } {}$[] is useful work output and $EinEin size 12{E rSub { size 8{"in"} } } {}$[] is the energy consumed.
7.7 Power
• Power is the rate at which work is done, or in equation form, for the average power $PP size 12{P} {}$ for work $WW size 12{W} {}$ done over a time $tt size 12{t} {}$, $P=W/tP=W/t size 12{P= {W}
slash {t} } {}$.
• The SI unit for power is the watt (W), where $1 W = 1 J/s1 W = 1 J/s size 12{1" W "=" 1 J/s"} {}$.
• The power of many devices, such as electric motors, is often expressed in horsepower (hp), where $1 hp = 746 W1 hp = 746 W size 12{1" hp "=" 746 W"} {}$.
7.8 Work, Energy, and Power in Humans
• The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty tissue.
• The rate at which the body uses food energy to sustain life and do different activities is called the metabolic rate; the corresponding rate when at rest is called the basal metabolic rate (BMR).
• The energy included in the BMR is divided among various systems in the body, with the largest fraction going to the liver and spleen, and the brain coming next.
• About 75 percent of food calories are used to sustain basic body functions included in the BMR.
• Since digestion is basically a process of oxidizing food, people's energy consumption during various activities can be determined by measuring their oxygen use.
7.9 World Energy Use
• The relative use of different fuels to provide energy has changed over the years. Fuel use is currently dominated by oil, although natural gas and solar contributions are increasing.
• Although nonrenewable sources dominate, some countries meet a sizeable percentage of their electricity needs from renewable resources.
• The United States obtains only about 10 percent of its energy from renewable sources, mostly hydroelectric power.
• Economic well-being is dependent upon energy use. In most countries, higher standards of living—as measured by GDP (Gross Domestic Product) per capita—are matched by high levels of energy
consumption per capita.
• Even though, in accordance with the law of conservation of energy, energy can never be created or destroyed, energy that can be used to do work is always partly converted to less useful forms,
such as waste heat to the environment.
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How do you find the maximum value of f(x)=20e^(-2x)*sin(3x) ? | HIX Tutor
How do you find the maximum value of #f(x)=20e^(-2x)*sin(3x) #?
Answer 1
The maximum value of $f \left(x\right)$ is infinity.
Let #f(x)=20e^(-2x)sin3x#.
In order to find the maximum value of #f(x)#, we need to consider the functions which make up #f# and their maximum values. #f# is a product of #3# functions: #20#, #e^(-2x)# and #sin3x#. So the
maximum of #f# will be the product of the maximum of these three functions.
It is also important to note that all three of these functions can take their max values simultaneously.
So the max value of #f# is #20*oo*1=oo#
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Answer 2
To find the maximum value of the function ( f(x) = 20e^{-2x} \cdot \sin(3x) ), you need to find the critical points by taking the derivative of the function, setting it equal to zero, and solving for
( x ). Then, you evaluate the second derivative to determine whether each critical point is a maximum, minimum, or inflection point. However, the function ( f(x) = 20e^{-2x} \cdot \sin(3x) ) does not
have a maximum value as it increases without bound as ( x ) approaches infinity.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Uses of the accidental isomorphism $SO(5)\sim Sp(2)$?
Some of the accidental isomorphisms of low dimensional Lie algebras have very important applications in physics. The theory of angular momentum makes use of the fact that $SO(3)\sim SU(2)$.
Similarly, the isomorphism $SO(4)\sim SU(2)\times SU(2)$ can be used to greatly simplify the representation theory of the Lorentz group in 4 dimensions. Another example is $SU(4)\sim SO(6)$, which I
believe is useful for $\mathcal{N}=4$ SYM theory, where $SU(4)$ is the R-symmetry.
I was just wondering if there are any physical examples that make use of the last accidental isomorphism $SO(5)\sim Sp(2)$.
This post imported from StackExchange Physics at 2015-02-27 11:31 (UTC), posted by SE-user asperanz
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72 yd3 to ft3 (72 cubic yards to cubic feet)
Here we will explain and show you how to convert 72 cubic yards (yd3) to cubic feet (ft3).
To create a formula to convert 72 yd3 to ft3, we start with the fact that 1 yard is 3 feet, which means that you multiply yards by 3 to get feet. We can therefore make the following equation:
yards × 3 = feet
However, we are not dealing with yards and feet. We are dealing with cubic yards (yd³) and cubic feet (ft³), which are yards and feet to the 3rd power. Thus, we take both sides of the formula above
to the 3rd power to get the yd3 to ft3 formula:
yards × 3 = feet
(yards × 3)³ = feet³
yards³ × 27 = feet³
yd³ × 27 = ft³
Now that we have the yd3 to ft3 formula, we can calculate and convert 72 yd3 to ft3. Here is 72 yd3 converted to ft3, along with the math and the formula:
yd³ × 27 = ft³
72 × 27 = 1944
72 yd³ = 1944 ft³
72 yd3 = 1944 ft3 Cubic Yards to Cubic Feet Converter
72 yd^3 to ft^3 is not all we can explain and calculate. You can use the form here to convert another number of cubic yards to cubic feet.
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5,841 research outputs found
The naive fermiophobic scenario is unstable under radiative corrections, due to the chiral-symmetry breaking induced by fermion mass terms. In a recent study, the problem of including the radiative
corrections has been tackled via an effective field theory approach. The renormalized Yukawa couplings are assumed to vanish at a high energy scale $\Lambda$, and their values at the electroweak
scale are computed via modified Renormalization Group Equations. We show that, in case a fermiophobic Higgs scenario shows up at the LHC, a linear collider program will be needed to accurately
measure the radiative Yukawa structure, and consequently constrain the $\Lambda$ scale.Comment: 7 pages, 3 figures, Proceedings of the 2011 International Workshop on Future Linear Colliders (LCWS11),
Granada (Spain), 26-30 September 201
In this talk CP violation in the supersymmetric models, and especially in B-decays is discussed. We review our analysis of the supersymmetric contributions to the mixing CP asymmetries of $B\to \phi
K_S$ and $B\to \eta^{\prime} K_S$ processes. Both gluino and chargino exchanges are considered in a model independent way by using the mass insertion approximation method. The QCD factorization
method is used, and parametrization of this method in terms of Wilson coefficients is presented in both decay modes. Correlations between the CP asymmetries of these processes and the direct CP
asymmetry in $b\to s \gamma$ decay are shown.Comment: 16 pages, 7 figures. Prepared for the proceedings of the 1st GUC workshop on High Energy Physics, Cairo, Jan 9-13, 200
Recommended from our members
Real estate development appraisal is a quantification of future expectations. The appraisal model relies upon the valuer/developer having an understanding of the future in terms of the future
marketability of the completed development and the future cost of development. In some cases the developer has some degree of control over the possible variation in the variables, as with the cost of
construction through the choice of specification. However, other variables, such as the sale price of the final product, are totally dependent upon the vagaries of the market at the completion date.
To try to address the risk of a different outcome to the one expected (modelled) the developer will often carry out a sensitivity analysis on the development. However, traditional sensitivity
analysis has generally only looked at the best and worst scenarios and has focused on the anticipated or expected outcomes. This does not take into account uncertainty and the range of outcomes that
can happen. A fuller analysis should include examination of the uncertainties in each of the components of the appraisal and account for the appropriate distributions of the variables. Similarly, as
many of the variables in the model are not independent, the variables need to be correlated. This requires a standardised approach and we suggest that the use of a generic forecasting software
package, in this case Crystal Ball, allows the analyst to work with an existing development appraisal model set up in Excel (or other spreadsheet) and to work with a predetermined set of probability
distributions. Without a full knowledge of risk, developers are unable to determine the anticipated level of return that should be sought to compensate for the risk. This model allows the user a
better understanding of the possible outcomes for the development. Ultimately the final decision will be made relative to current expectations and current business constraints, but by assessing the
upside and downside risks more appropriately, the decision maker should be better placed to make a more informed and “better”
We consider the possibility of studying anomalous contributions to the gamma-gamma-H and Z-gamma-H vertices through the process e-gamma--> e-H at future e-gamma linear colliders, with Sqrt(S)=
500-1500 GeV. We make a model independent analysis based on SU(2)xU(1) invariant effective operators of dim=6 added to the standard model lagrangian. We consider a light Higgs boson (mostly decaying
in bar(b)-b pairs), and include all the relevant backgrounds. Initial e-beam polarization effects are also analyzed. We find that the process e-gamma--> e-H provides an excellent opportunity to
strongly constrain both the CP-even and the CP-odd anomalous contributions to the gamma-gamma-H and Z-gamma-H vertices.Comment: LaTeX, 33 pages, 16 eps figures, extended section
In the light of recent experimental results for the direct detection of dark matter, we analyze in the framework of SUGRA the value of the neutralino-nucleon cross section. We study how this value is
modified when the usual assumptions of universal soft terms and GUT scale are relaxed. In particular we consider scenarios with non-universal scalar and gaugino masses and scenarios with intermediate
unification scale. We also study superstring constructions with D-branes, where a combination of the above two scenarios arises naturally. In the analysis we take into account the most recent
experimental constraints, such as the lower bound on the Higgs mass, the $b\to s\gamma$ branching ratio, and the muon $g-2$.Comment: References added, bsgamma upper bound improved, results unchanged,
Talk given at Corfu Summer Institute on Elementary Particle Physics, August 31-September 20, 200
We study N=1 supersymmetric SU(N) Yang-Mills theory on the lattice at strong coupling. Our method is based on the hopping parameter expansion in terms of random walks, resummed for any value of the
Wilson parameter r in the small hopping parameter region. Results are given for the mesonic (2-gluino) and fermionic (3-gluino) propagators and spectrum.Comment: Latex file. 43 pages. Minor
additional comments, references added, typos corrected. Accepted for publication in Int. J. Mod. Phys.
We perform a model independent analysis of the chargino contributions to the CP asymmetry in B -> Phi K(S) process. We use the mass insertion approximation method generalized by including the
possibility of a light right-stop. We find that the dominant effect is given by the contributions of the mass insertions deltaU_LL(32) and deltaU_RL(32) to the Wilson coefficient of the
chromomagnetic operator. By considering both these contributions simultaneously, the CP asymmetry in B -> Phi K(S) process is significantly reduced and negative values, which are within the 1-sigma
experimental range and satisfy the b -> s gamma constraints, can be obtained.Comment: 14 pages, LaTeX, 3.eps Figure
We analyze the impact of effective axial-vector coupling of the gluon on spin polarization observables in $t\bar{t}$ pair production at the LHC. Working at leading order in QCD, we compute the $t\bar
{t}$ spin-correlation and left-right spin asymmetry coefficients in the helicity basis in the laboratory frame as functions of the new physics scale $\Lambda$ associated with this coupling. We found
that the $t\bar{t}$ invariant mass dependent asymmetries are more sensitive to the scale $\Lambda$ than the corresponding inclusive ones, in particular when suitable cuts selecting high $t\bar{t}$
invariant mass regions are imposed. In the context of this scenario, we show that the LHC has potential either to confirm or to rule out the Tevatron FB top asymmetry anomaly by analyzing the $t\bar
{t}$ spin-correlation and left-right polarization asymmetries. On the other hand, stringent lower bound on the new physics scale $\Lambda$ can be set in this scenario if no significant deviations
from the SM predictions for those observables will be measured.Comment: 26 pages, 8 figures, same as published version in PRD. Few modifications in the text and one new reference adde
We present a general review about the N=1 supersymmetric SU(Nc) Yang-Mills on the lattice focusing our attention on the quenched approximation in supersymmetry. Finally we analyse and discuss the
recent results obtained at strong coupling and large Nc for the mesonic and fermionic propagators and spectrum
We analyze the inclusive semileptonic decays B -> X_s l^+ l^- in the framework of the supersymmetric standard model with non-universal soft-breaking terms at GUT scale. We show that the general trend
of universal and non-universal models is a decreasing of branching ratio (BR) and increasing of energy asymmetry (AS). However, only non--universal models can have chances to get very large
enhancements in BR and AS, corresponding to large (negative) SUSY contributions to the b -> s \gamma amplitude.Comment: 13 pages, 3 figures. References added. To appear in Phys. Lett.
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Most typical and most deviant
Ingo Rohlfing
One can use four functions to choose four types of cases without classifying all cases as either typical or deviant. The most typical and most deviant cases are proposed by Seawright and Gerring (
2008). The most typical case has the smallest residual of all cases. The most deviant case has the largest residual of all cases. The two functions most_typical() and most_deviant() work in the same
way and show you the case with its residual. The input into the function is an lm object.
The most deviant case does not distinguish between cases that have a large negative and a large positive residual. Cases with a negative residual are overpredicted because the predicted outcome is
higher than the observed outcome. Cases with a positive residual are underpredicted because the predicted outcome is lower than the observed outcome. It might not matter whether a case is
overpredicted or underpredicted because both subtypes of outliers can have the same type of deviance. However, one might be interested in knowing whether a case has a positive or negative residual
and what the most overpredicted and underpredicted cases are. This is what the functions most_overpredicted() and most_underpredicted() achieve, each taking an lm object as input.
The package does not include functions for plotting the cases. There are multiple, very useful packages such as the olsrr package that can be used for the easy visualization of residuals.
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On the complexity of approximating k-dimensional matching
We study the complexity of bounded variants of graph problems, mainly the problem of k-Dimensional Matching (k-DM), namely, the problem of finding a maximal matching in a k-partite k-uniform balanced
hyper-graph. We prove that k-DM cannot be efficiently approximated to within a factor of O(k/ln k) unless P = NP. This improves the previous factor of k/[2]O(√ln k) by Trevisan [Tre01]. For low k
values we prove NP-hardness factors of 54/53-ε, 30/29-ε and 23/22-ε for 4-DM, 5-DM and 6-DM respectively. These results extend to the problem of k-Set-Packing and the problem of Maximum
Independent-Set in (k + 1)-claw-free graphs.
Original language English
Title of host publication Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
Editors Sanjeev Asora, Amit Sahai, Klaus Jansen, Jose D.P. Rolim
Publisher Springer Verlag
Pages 83-97
Number of pages 15
ISBN (Print) 3540407707, 9783540407706
State Published - 2003
Externally published Yes
Publication series
Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
Volume 2764
ISSN (Print) 0302-9743
ISSN (Electronic) 1611-3349
Dive into the research topics of 'On the complexity of approximating k-dimensional matching'. Together they form a unique fingerprint.
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Banking to 45 degrees: Aspect ratios for time series plots
In SAS, the aspect ratio of a graph is the physical height of the graph divided by the physical width. Recently I demonstrated how to set the aspect ratio of graphs in SAS by using the ASPECT= option
in PROC SGPLOT or by using the OVERLAYEQUATED statement in the Graph Template Language (GTL). I mentioned that a unit aspect ratio is important when you want to visually display distance between
points in a scatter plot.
A second use of the aspect ratio is when plotting a time series. A time series is a sequence of line segments that connect data values (x[i], y[i]), i = 0..N. Thus a plot involves N line segments,
each with a slope and length. In the late 1980s and early '90s, William Cleveland and other researchers investigated how humans perceive graphical information. For a time series plot, the rate of
change (slope) of the time series segments is important. Cleveland suggested that practitioners should use "banking to 45 degrees" as a rule-of-thumb that improves perception of the rate-of-change in
the plot.
My SAS colleague Prashant Hebbar wrote a nice article about how to use the GTL to specify the aspect ratio of a graph so that the median absolute slope is 45 degrees. Whereas Prashant focused on
creating the graphics, the current article shows how to compute aspect ratios by using three banking techniques from Cleveland.
Three computational methods for choosing an aspect ratio
The simplest computation of an aspect ratio is the median absolute slope (MAS) method (Cleveland, McGill, and McGill, JASA, 1988). As it name implies, it computing the median of the absolute value of
the slopes. To obtain a graph in which median slope of the (physical) line segments is unity, set the aspect ratio of the graph to be the reciprocal of the median value.
The median-absolute-slope method is a simple way to choose an aspect ratio. A more complex analysis (Cleveland, Visualizing Data, 1993, p. 90) uses the orientation of the line segments. The
orientation of a line segment is the quantity arctan(dy/dx) where dy is the change in physical units in the vertical direction and dx is the change in physical units in the horizontal direction.
Cleveland proposes setting the aspect ratio so that the average of the absolute values of the orientations is 45 degrees. This is called "banking the average orientation to 45 degrees" or the AO
(absolute orientation) method. Another approach is to weight the line segments by their length (in physical units), compute a weighted mean, and find the aspect ratio that results an average weighted
orientation of 45 degrees. This is called "banking the weighted average of the absolute orientations to 45 degrees" or the AWO (average weighted orientation) method.
The AO and AWO methods require solving for the aspect ratio that satisfies a nonlinear equation that involves the arctangent function. This is equivalent to finding the root of a function, and the
easiest way to find a root in SAS is to use the FROOT function in SAS/IML software.
For a nice summary of these and other banking techniques, see Heer and Agrawala (2006). However, be aware that Heer and Agrawala define the aspect ratio as width divided by height, which is the
reciprocal of the definition that is used by Cleveland (and SAS).
Banking to 45 degrees
To illustrate the various methods, I used the MELANOMA data set, which contains the incidences of melanoma (skin cancer) per one million people in the state of Connecticut that were observed each
year from 1936–1972. This is one of the data sets in Visualizing Data (Cleveland, 1993). You can download all data sets in the book and create a libref to the data set directory. (My libref is called
The line plot shows the melanoma time series. The graph's subtitle indicates that the aspect ratio is 0.4, which means that the plot is about 40% as tall as it is wide. With this scaling, about half
of the line segments have slopes that are between ±1. This aspect ratio was chosen by using the median-absolute-slope method.
Prashant has already shown how to use the DATA step and Base procedures to implement the median absolute slope method, so I will provide an implementation in the SAS/IML language. The following SAS/
IML program reads the data and uses the DIF function to compute differences between yearly values. These values are then scaled by the range of the data. Assuming that the physical size of the axes
are proportional to the data ranges (and ignoring offsets and margins), the ratio of these scaled differences are the physical slopes of the line segments. The following computation shows that the
median slope is about 2.7. Therefore, the reciprocal of this value (0.37) is the scale factor for which half the slopes are between ±1 and then other half are more extreme. For convenience, I rounded
that value to 0.4.
/* Data from Cleveland (1993) */
libname vizdata "C:\Users\frwick\Documents\My SAS Files\vizdata";
title "Melanoma Data";
proc iml;
use vizdata.Melanoma; /* see Cleveland (1993, p. 158) */
read all var "Year" into x;
read all var "Incidence" into y;
dx = dif(x) / range(x); /* scaled change in horizontal directions */
dy = dif(y) / range(y); /* scaled change in vertical directions */
m = median( abs(dy/dx) ); /* median of absolute slopes */
print m[L="Median Slope"] (1/m)[L="Aspect Ratio"];
Three methods of banking to 45 degrees
I used SAS/IML to implement Cleveland's three techniques (MAS, AO, AWO) and ran each technique on the melanoma data. You can download the complete SAS/IML program that computes the aspect ratios. The
following table gives the ideal aspect ratios for the melanoma data as computed by each of Cleveland's methods:
As you can see, the three methods are roughly in agreement for these data. An aspect ratio of approximately 0.4 (as shown previously) will create a time series plot in which the rate of change of the
segments is apparent. If there is a large discrepancy between the values, Cleveland recommends using the average weighted orientation (AWO) technique.
The complete SAS program also analyzes carbon dioxide (CO2) data that were analyzed by Cleveland (1993, p. 159) and by Herr and Agrawala. Again, the three methods give similar results to each other.
You might wonder why I didn't choose an exact value such as 0.3518795 for the aspect ratio of the melanoma time series plot. The reason is that these "banking" computations are intended to guide the
creation of statistical graphics, and using 0.4 (a 2:5 ratio of vertical to horizontal scales) is more interpretable. Think of the numbers as suggestions rather than mandates. As Cleveland, McGill,
and McGill (1988) said, banking to 45 degrees "cannot be used in a fully automated way. (No data-analytic procedure can be.) It needs to be tempered with judgment."
So use your judgement in conjunction with these computations to create time series plots in SAS whose line segments are banked to 45 degrees. After you compute an appropriate aspect ratio, you can
create the graph by using the ASPECT= option in PROC SGPLOT or the ASPECTRATIO option on a LAYOUT OVERLAY statement in GTL.
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How To Identify Patterns in Time Series Data: Time Series Analysis
Introduction to ANOVA Introduction to ANOVA (Jump to: Lecture | Video ) An ANOVA has factors(variables), and each of those factors has levels: There are several different types of ANOVA: There are
four main assumptions of an ANOVA: Hypotheses in ANOVA depend on the number of factors you're dealing with: Effects dealing with one factor are called main effects. Here's an example of an
interaction effect in an ANOVA: Below we have a Factorial ANOVA with two factors: dosage(0mg and 100mg) and gender(men and women) . Dosage and gender are interacting because the effect of one
variable depends on which level you're at of the other variable. If we reject the null hypothesis in an ANOVA, all we know is that there is a difference somewhere among the groups. When performing an
ANOVA, we calculate an "F" statistic. If there are no treatment differences (that is, if there is no actual effect), we expect F to be 1. Introduction to ANOVA (Jump to: Lecture | Video ) There are
several different types of ANOVA:
determine sample size two-way ANOVA? Computing required sample size for experiments to be analyzed by ANOVA is pretty complicated, with lots of possiblilities. To learn more, consult books by Cohen
or Bausell and Li, but plan to spend at least several hours. Two-way ANOVA, as you'd expect, is more complicated than one-way. The complexity comes from the many possible ways to phrase your question
about sample size. The rest of this article strips away most of these choices, and helps you determine sample size in one common situation, where you can make the following assumptions: There are two
levels of the first factor, say the factor is Drug and you either gave the drug or gave vehicle (placebo). If those limitations aren't a problem for you, then read on for a simple way to compute
necessary sample size. Sample size is always determined to detect some hypothetical difference. What about units? Another way to look at this is to express the difference you expect to see as a
fraction of the mean.
Degrees of Freedom Tutorial | Ron Dotsch A lot of researchers seem to be struggling with their understanding of the statistical concept of degrees of freedom. Most do not really care about why
degrees of freedom are important to statistical tests, but just want to know how to calculate and report them. This page will help. For those interested in learning more about degrees of freedom,
take a look at the following resources: I couldn’t find any resource on the web that explains calculating degrees of freedom in a simple and clear manner and believe this page will fill that void.
Let’s start with a simple explanation of degrees of freedom. Imagine a set of three numbers, pick any number you want. Now, imagine a set of three numbers, whose mean is 3. This generalizes to a set
of any given length. This is the basic method to calculate degrees of freedom, just n – 1. Df1 Df1 is all about means and not about single observations. Let’s start off with a one-way ANOVA. Sticking
to the one-way ANOVA, but moving on to three groups.
Two-way anova - Handbook of Biological Statistics Summary Use two-way anova when you have one measurement variable and two nominal variables, and each value of one nominal variable is found in
combination with each value of the other nominal variable. It tests three null hypotheses: that the means of the measurement variable are equal for different values of the first nominal variable;
that the means are equal for different values of the second nominal variable; and that there is no interaction (the effects of one nominal variable don't depend on the value of the other nominal
variable). When to use it You use a two-way anova (also known as a factorial anova, with two factors) when you have one measurement variable and two nominal variables. For example, here's some data I
collected on the enzyme activity of mannose-6-phosphate isomerase (MPI) and MPI genotypes in the amphipod crustacean Platorchestia platensis. A two-way anova is usually done with replication (more
than one observation for each combination of the nominal variables).
One Way ANOVA By Hand | Learn Math and Stats with Dr. G ANOVA Testing Example Excel Example for this ANOVA See a HOW TO Video of this Example A research study compared the ounces of coffee consumed
daily between three groups. Group1 was Italians, Group 2 French, and Group 3 American. Group1: Italian Group 2: French Group 3: American The Results of this study are in the following table: Note
that here in this example: “n” is the sample size for each group “M” is the “sample mean” for each group “s^2” is the sample variance for each group N = n1 + n2 + n3 = 70+ 70+70 = 210 NOTE that “N”
is the combined sample size for all three groups. Step 1: The null and alternative hypothesis Ho and Ha Ho (the null) will represent that all the groups are statistically the same. Ho: mean group 1 =
mean group 2 = mean group 3 The Ha (alternative or research) hypothesis will represent that at least one of the groups is statistically significantly different. Ha: mean group 1 ≠ mean group 2 ≠ mean
group 3 df BETWEEN = 3 – 1 = 2 (Used as the numerator or top df) First Step:
Questions and answers about language testing statistics: Effect size and eta squared Shiken: JALT Testing & Evaluation SIG Newsletter Vol. 12 No. 2. Apr. 2008. (p. 38 - 43) [ISSN 1881-5537] PDF
Version QUESTION: In Chapter 6 of the 2008 book on heritage language learning that you co-edited with Kimi-Kondo Brown, a study comparing how three different groups of informants use intersentential
referencing is outlined. ANSWER: I will answer your question about partial eta2 in two parts. Eta2 Eta2 can be defined as the proportion of variance associated with or accounted for by each of the
main effects, interactions, and error in an ANOVA study (see Tabachnick & Fidell, 2001, pp. 54-55, and Thompson, 2006, pp. 317-319). Where: Eta2 is most often reported for straightforward ANOVA
designs that (a) are balanced (i.e., have equal cell sizes) and (b) have independent cells (i.e., different people appear in each cell). Table 1 Results of the Analysis Shown in Figure 3 of the
Anxiety 2.sav used with SPSS [ p. 38 ] Eta2 values are easy to calculate. [ p. 39 ] Partial eta2 [ p. 40 ]
General Linear Model The General Linear Model (GLM) underlies most of the statistical analyses that are used in applied and social research. It is the foundation for the t-test, Analysis of Variance
(ANOVA), Analysis of Covariance (ANCOVA), regression analysis, and many of the multivariate methods including factor analysis, cluster analysis, multidimensional scaling, discriminant function
analysis, canonical correlation, and others. Because of its generality, the model is important for students of social research. Although a deep understanding of the GLM requires some advanced
statistics training, I will attempt here to introduce the concept and provide a non-statistical description. The Two-Variable Linear Model The easiest point of entry into understanding the GLM is
with the two-variable case. The goal in our data analysis is to summarize or describe accurately what is happening in the data. Figure 3 shows the equation for a straight line. y=mx+b In this
equation, the components are: y=b0+bx+e where:
T-test online. Compare two means, two proportions or counts online. Input. Compare two independent samples Counted numbers. To test for the significance of a difference between two Poisson counts.
Select options and hit the calculate button. Compare a single sample with the population Counted numbers. Select the one sample option, other options and hit the calculate button. TOP / table input /
data input Explanation. One Sample AnalysisEqual Variance (Welch/Student)Confidence IntervalsNumber Needed to Treat (NNT)PARFEffect SizeDegrees of FreedomBox PlotMore Menu For an explanation of the
Pairwise t-test or the t-test for two correlated samples, consult the pairwise help page. T-test concerns a number of procedures concerned with comparing two averages. The t-test gives the
probability that the difference between the two means is caused by chance. The t-test is basically not valid for testing the difference between two proportions. Both one and double sided
probabilities are given. Learn more about the t-test from . Effect Size. The More menu. Degrees of Freedom.
Repeated Measures ANOVA - Understanding a Repeated Measures ANOVA | Laerd Statistics Where measurements are made under different conditions, the conditions are the levels (or related groups) of the
independent variable (e.g., type of cake is the independent variable with chocolate, caramel, and lemon cake as the levels of the independent variable). A schematic of a different-conditions repeated
measures design is shown below. It should be noted that often the levels of the independent variable are not referred to as conditions, but treatments. The above two schematics have shown an example
of each type of repeated measures ANOVA design, but you will also often see these designs expressed in tabular form, such as shown below: This particular table describes a study with six subjects (S1
to S6) performing under three conditions or at three time points (T1 to T3). Hypothesis for Repeated Measures ANOVA The repeated measures ANOVA tests for whether there are any differences between
related population means. H0: µ1 = µ2 = µ3 = … = µk Logic of the Repeated Measures ANOVA
The Test Statistic Explanations > Social Research > Analysis > The Test Statistic Variance | The test statistic ratio | Interpreting the statistic | See also Variance When you do any test, there will
be variation in results. This may be intended or unintended, which is often referred to as systematic and unsystematic variance. Variation can be measured in several ways, including standard
deviation, variance and sum of squares. Systematic variance Systematic variance is that due to deliberate experimental actions. Systematic variance is generally measures as the difference between
groups, for example comparing the means of a set of samples. Systematic variance is often denoted as SSM, where 'M' stands for 'Model'. Unsystematic variance Unsystematic variance is that which is
unintended and is a particularly tricky problem in social research. It is because of this that we do social experiments and pay a lot of attention to systematic vs unsystematic variance. The test
statistic ratio Signal-to-noise Calculation
One Sample t Test Assumptions The t distribution provides a good way to perform one-sample tests on the mean when the population variance is not known provided the population is normal or the sample
is sufficiently large so that the Central Limit Theorem applies (see Properties 1 and 2 of Basic Concepts of t Distribution). It turns out that the one-sample t-test is quite robust to moderate
violations of normality. In particular, the test provides good results even when the population is not normal or the sample size is small, provided that the sample is reasonably symmetrically
distributed about the sample mean. The boxplot is relatively symmetrical; i.e. the median is in the center of the box and the whiskers extend equally in each directionThe histogram looks
symmetricalThe mean is approximately equal to the medianThe coefficient of skewness is relatively small The impact of non-normality is less for a two-tailed test than for a one-tailed test and for
higher alpha values than for lower alpha values.
How do I report paired samples T-test data in APA style? Three things to report You will want to include three main things about the Paired Samples T-Test when communicating results to others. 1.
Test type and use You want to tell your reader what type of analysis you conducted. Example You can report data from your own experiments by using the template below. “A paired-samples t-test was
conducted to compare (your DV measure) _________ in (IV level / condition 1) ________and (IV level / condition 2)________ conditions.” If we were reporting data for our example, we might write a
sentence like this. “A paired-samples t-test was conducted to compare the number of hours of sleep in caffeine and no caffeine conditions.” 2. You want to tell your reader whether or not there was a
significant difference between condition means. “There was a significant (not a significant) difference in the scores for IV level 1 (M=___, SD=___) and IV level 2 (M=___, SD=___) conditions; t(__)=
___, p = ____” Just fill in the blanks by using the SPSS output 3. Looking good!
Two-way ANOVA Output and Interpretation in SPSS Statistics - Including Simple Main Effects SPSS Statistics Descriptive statistics You can find appropriate descriptive statistics for when you report
the results of your two-way ANOVA in the aptly named "Descriptive Statistics" table, as shown below: Published with written permission from SPSS Statistics, IBM Corporation. This table is very useful
because it provides the mean and standard deviation for each combination of the groups of the independent variables (what is sometimes referred to as each "cell" of the design). Plot of the results
The plot of the mean "interest in politics" score for each combination of groups of "gender" and "education_level" are plotted in a line graph, as shown below: Although this graph is probably not of
sufficient quality to present in your reports (you can edit its appearance in SPSS Statistics), it does tend to provide a good graphical illustration of your results. Statistical significance of the
two-way ANOVA Post hoc tests – simple main effects in SPSS Statistics Multiple Comparisons Table General
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A foundational algorithm for sorting an Array.
The Code 🖥️:
Here is the code with comments:
function quickSorting(array) {
//loop through array
for(i = 0; i < array.length; i++){
// For every loop from previous line, loop again.
// Decrease loop length by one for each main loop, or the value of i
for(j = 0; j < array.length - 1 - i ; j++){
//if current index is larger than next index
if(array[j] > array[j + 1] ){
// save current index 'i' in variable
const temp = array[j]
// assign 'i' value to the value of 'i+1'
array[j] = array[j + 1]
// assign stored value of 'i' to next index 'i+1'
array[j + 1] = temp
return array
The How & Why🙋:
Bubble sort is a surprisingly simple algorithm to implement; it's also, in my opinion, one of the greatest.
Defining what a sorted array is can be a bit abstract, but to put it in technical terms: A sorted array is any array in which a value "**X" at index "**i" comes before another value "**X" at index "
i+1 ".
Put into more plain terms, a sorted array is any array that is ordered in a logical way.
Below we can see this as a graphic in which 0 marks the 0th index of the array, and N marks the final index of the array.
How does Bubble Sort, sort?
-Let's start by defining an array we can work with:
What bubble sort does as an operation is compare the current index Xi with the next index Xi+1, if the next index is larger than the first index, they swap positions.
With the example, we will loop through the array, and at the 2nd index find that it's value 14 is bigger than the following index 8. We then swap the index of those two values so we get an array that
looks like this: [ 2, 6, 8, 14, 4].
In the next following sub-loop, we would perform a similar swap. Resulting in moving 14 to the last index in the array. Congrats! We now have the largest value where it needs to be.
To Recap:
In a single iteration of Bubble Sort we have seen the array altered in such a way that the largest value is at the end as follows:
1. [2, 6, 14, 8, 4]
2. [ 2, 6, 8, 14, 4]
3. [2, 6, 8, 4, 14]
This is after just a single pass through the array and is a defining characteristic of this algorithm.
The length of the next iteration can also be array.length - 1 as the first iteration will always place the largest value in the final index of the array.
The next iteration will see:
• [ 2, 6, 8, 4, 14] turn into [ 2, 6, 4, 8, 14]
And the next will finish at :
• [ 2, 6, 4, 8, 14] to [ 2, 4, 6, 8, 14]
That's Bubble Sort!
A straightforward algorithm that is, in its own right awesome, and also serves as a baseline for more complex algorithms going forward.
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Template: Vanilla Bond Interest Rate Swap Valuation - The CPA Oasis
Template: Vanilla Bond Interest Rate Swap Valuation
Introduction – What is an Interest Rate Swap?
Imagine you have a loan with a fixed interest rate, meaning you pay the same interest amount every year. Your friend, on the other hand, has a loan with a variable interest rate that changes each
year based on market conditions. An interest rate swap is like an agreement where you and your friend decide to swap your interest payments. You agree to pay your friend’s variable interest, and your
friend agrees to pay your fixed interest.
Why would you do this? Maybe you believe that variable rates will go down and save you money, while your friend prefers the certainty of fixed payments and avoid any downside risk. This is also the
case with companies.
Vanilla Bond and Interest Rate Swaps
Now, consider a vanilla bond. This is a simple bond that pays a fixed coupon (interest payment) every period (e.g., year) and returns the principal amount at the end of its term. It does not contain
any special provisions or exotic features. Borrowers might use an interest rate swap on this bond to manage their interest rate risk. For example, if they pay variable payment from the bond but think
that fixed rates will be more advantageous, they can swap their variable payments for fixed ones.
The Binomial Pricing Model
To determine the fair value of such swaps, binomial pricing model can be used. Let’s break down this model into simple terms.
1. Two Possible Outcomes: At each time step (say, each year), the interest rate can either go up or down. This creates a “binomial” tree of possible future rates.
2. Tree Structure: Imagine a tree where each branch splits into two more branches, one for the rate going up and one for it going down. Over several years, this tree will have many possible paths
the interest rate could take.
3. Calculating Values: Starting from the final year and working backward, we calculate the value of the swap for each possible interest rate at each node of the tree. This is done by considering the
possible up and down movements and their probabilities.
4. Discounting Back: At each node, we take the average of the possible future values and discount it back to the present value using the current interest rate. This process continues until we reach
the present day.
5. Fair value of a swap: By the time we finish, we have a fair value for the swap today. This helps both parties know what is a fair exchange of payments given the uncertainty of future interest
rates. Using Excel’s Solver, we can also modify swap fixed interest rates so that it equals zero value for both buyer and seller at the inception date.
Practical Tool for Accountants
Attached is a template designed for the quick calculation of vanilla bond interest rate swap valuation. It contains the following inputs:
• Hedged principal amount
• Swap fixed interest rate (annual)
• Expiration (years)
• Payments settlement date (years; 0 – present period, 1 – present period + 1, etc.); in the template, we assume payment takes place one year in arrears, after the actual interest rates are known.
If delay beyond 1 year is expected, then the template would require reworking of formulas with respect to discounting
• Current market interest rate
• Floating rate – up and down movements – the expectation set by the user here should already take into account the actual probabilities.
Probabilities of outcomes in the rows 32:37 are included as 50% for each of the two base movements, but an be adjusted manually as needed. “Floating rate – up and down movements” are already supposed
to take into account the actual probabilities.
Using the binomial pricing model, borrowers and investors alike can make more informed decisions about engaging in interest rate swaps to manage their risks and potentially save money. For example,
if the model shows that variable rates are likely to rise, a fixed-rate payer might benefit from a swap, as they could end up paying less compared to sticking with their fixed payments. Conversely,
if variable rates are expected to fall, a fixed-rate receiver (or a variable-rate payer) might benefit from the swap. In summary, an interest rate swap on a vanilla bond allows two parties to
exchange interest payments to suit their preferences and risk outlooks. The binomial pricing model helps them determine a fair value for this exchange by considering the many possible future interest
rate scenarios. The attached template serves as a resource for quick check on the valuation.
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Gravitational Acceleration & Terminal Velocity | Mini Physics - Free Physics Notes
Gravitational Acceleration & Terminal Velocity
Show/Hide Sub-topics (Speed, Velocity & Acceleration | O Level Physics)
Table of Contents
Gravitational Acceleration, $g$
Gravitational acceleration, often denoted as ($g$), is a fundamental concept in physics that describes the rate at which objects accelerate towards the center of the Earth when the only force acting
upon them is gravity.
• It is also known as acceleration of free fall.
• It is a vector quantity, meaning it has both magnitude and direction—towards the center of the mass creating the gravitational field.
• This acceleration is approximately $9.8 \, \text{m/s}^2$ on the surface of the Earth, although the value often rounded to $10 \, \text{m/s}^2$ for simplification in some calculations.
Understanding Gravitational Acceleration
Distance-time graph of an object falling from rest (with no air resistance)
• Uniform Acceleration: Gravitational acceleration is a unique form of acceleration because it is constant near the Earth’s surface, affecting all objects equally, regardless of their mass. This
means that if air resistance (and other non-gravitational forces) are negligible, all objects will increase their velocity by approximately $10 \, \text{m/s}$ or $9.8 \, \text{m/s}$ every second
when falling towards the Earth. This is illustrated by the figure above.
• Direction of Acceleration: The acceleration due to gravity is always directed towards the center of the Earth, which we commonly refer to as “downwards.” This is why, in physics problems and
equations of motion, $g$ is used to represent acceleration, with a positive sign when an object is falling and a negative sign when it is moving upwards, indicating deceleration.
• Applications in Equations of Motion: In kinematic equations, $g$ replaces the acceleration variable $a$. For falling objects, we use $a = g = +9.8 \, \text{m/s}^2$, and for objects projected
upwards, $a = -g = -9.8 \, \text{m/s}^2$, to account for the deceleration as the object ascends.
Misconceptions and Clarifications
• Mass Independence: A common misconception is that heavier objects fall faster than lighter ones. However, the acceleration due to gravity is independent of the mass of the object. This principle,
illustrated by Galileo’s experiments, demonstrates that without air resistance, a feather and a bowling ball would fall at the same rate. (More on this below)
Air Resistance & Terminal Velocity
When objects fall through the air, they encounter air resistance or drag, which opposes the motion of the object. As an object accelerates due to gravity, its velocity increases, and consequently, so
does the air resistance acting against it.
Dynamics of Falling Objects with Air Resistance
• Increasing Drag Force: The drag force acting on a falling object increases with its velocity. This force acts in the opposite direction to the object’s motion, reducing the net acceleration until
a point where the drag force equals the gravitational force acting on the object.
• Terminal Velocity: This balance of forces results in the object reaching a constant velocity, known as terminal velocity, where the acceleration becomes zero because the net force (or resultant
force) acting on the object is zero (due to Newton’s Second Law Of Motion: $F = ma$). The terminal velocity of an object depends on its shape, size, and mass, as well as the density of the air
it’s falling through.
Factors Affecting Air Resistance
• Speed and Surface Area: The magnitude of air resistance depends on the speed and surface area of the object. The greater the speed and the larger the surface area, the higher the air resistance.
This explains why objects with larger surface areas or those moving at higher speeds experience more drag, altering their falling dynamics compared to objects with smaller surface areas or lower
Practical Examples and Applications
• Parachutes and Skydiving: The concept of terminal velocity is crucial in understanding how parachutes work. Parachutes increase the surface area of the falling object (in this case, a person),
significantly increasing air resistance and thus reducing the person’s terminal velocity to a safer speed for landing.
• Sports Dynamics: In sports, understanding air resistance and drag is essential for optimizing performance in activities like cycling, skiing, and automobile racing. Athletes and engineers design
equipment and adopt postures that minimize air resistance to achieve higher speeds and better performance.
Video Showing “Absence Of Air Resistance”
Check out the video below: (Skip to 1.25 min mark for the “with air resistance” part and 2.50 min mark for the “absence of air resistance” (vacuum) part.)
Older NASA video about the “ALL objects falls at the same rate in the absence of air resistance part”:
Worked Examples
Example 1
A stone is thrown upwards into the air. Assuming negligible air resistance, what is the magnitude and direction of its acceleration?
Click here to show/hide answer
The magnitude of its acceleration is $10 \text{ m s}^{-1}$ and the direction is downwards.
Example 2
A parachutist of mass 80kg descends vertically at a constant velocity of 3.0 m s^-1. Taking the acceleration of free fall as 10 ms^-2, what is
(a) the net force acting on him?
(b) the upward force acting on him?
Click here to show/hide answer
(a) Net force is 0. Because the parachutist is at constant velocity. The gravitational force acting on him equals the resistance force acting on him(air resistance).
$$\begin{aligned} m_1\times g &= F_1 \\ (80)(10) &= F_1 \\ F_1 &= 800 \, \text{N} \end{aligned}$$
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WHOLE NUMBER SEQUENCES AND QUANTUM REALITY 1,2,3,… Infinity!!!
Using a common analogy, one might say that science and technology as we know them today constitute just the tip of the iceberg. Approximately 90% of an iceberg floating in the ocean is beneath the
surface. Today’s science has been looking only at the tip of reality; most of it is hidden. But, like most analogies, this analogy is not perfect. Most of today’s science is based on and limited to
materialistic concepts. If the sum total of today’s scientific knowledge is the tip of an iceberg, physical reality is the rest of the iceberg, plus all the oceans of the world. As we shall see,
simple math shows us that reality is much more than that which has weight and takes up space.
In fact, it has been proved mathematically that reality is infinite.
The mathematics used by scientists today has, as its most important system of logic, ‘the calculus’ of infinitesimals which was developed by Leibniz and Newton, about 350 years ago. Almost all of our
detailed understanding of physical reality and the resulting technology we have today, employs ‘the calculus’, but the calculus of Newton and Leibniz simply does not work at the quantum level. Why?
Because it depends on the assumption that reality can be divided into ever smaller bits. This is not actually the case. We have known for more than 70 years that we exist in a quantized world. There
is a smallest bit, beyond which no division is possible. The calculus works as well as it does only because that smallest bit is so much smaller than anything we can directly measure, that for the
purposes of building things from skyscrapers to microscopic electronic circuits, the error is not significant. But for describing quantum phenomena, the calculus is totally inappropriate and leads to
much of the confusion that causes scientists to think that quantum experiments produce weird results.
Max Planck, who discovered the fact that we live in a quantized universe, said: “Science advances from funeral to funeral!” meaning that scientists, like most people, get locked into a paradigm, - a
way of thinking, and cannot see beyond it. Right now, they can’t seem to see beyond the calculus and binary logic. But in order to begin to investigate the rest of the iceberg, we have to go beyond
the calculus and binary logic.
Binary logic is the basis of a mathematical system called Boolean algebra and today’ computers are simple binary computing machines. Again, this has been sufficient for today’s technology for the
most part, but, like the calculus, it is inadequate to deal with quantum phenomena and the inclusion of consciousness, which quantum physics demands. What does work, is the Calculus of Distinctions
and triadic logic. They are basic to the Close-Neppe paradigm. But, you may think, if I don’t understand the calculus of Newton, how can I expect to understand a new calculus and logic? Fortunately
they are more basic than ‘the calculus’ and binary logic, and easier to understand!
OK. With that intro, we are ready to start. Contemplate the symmetrical sequence of equations above and below. They are known as Diophantine equations (after the Greek mathematician Diophantus).
Notice that they progress from I to 3 dimension-wise (indicated by power: a 2D area is an integer squared, a 3D volume is an integer cubed. All symmetric particles are volumetric.
1=1 =1
1+2=3 =3
3^2+4^2=5^2 =25
3^3+4^3+5^3=6^3 =216
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IBM Says It's Made a Big Breakthrough in Quantum Computing
Scientists at IBM say they've developed a method to manage the unreliability inherent in quantum processors, possibly providing a long-awaited breakthrough toward making quantum computers as
practical as conventional ones — or even moreso.
The advancement, detailed in a study published in the journal Nature, comes nearly four years after Google eagerly declared "quantum supremacy" when its scientists claimed they demonstrated that
their quantum computer could outperform a classical one.
Though still a milestone, those claims of "quantum supremacy" didn't exactly pan out. Google's experiment was criticized as having no real world merit, and it wasn't long until other experiments
demonstrated classical supercomputers could still outpace Google's.
IBM's researchers, though, sound confident that this time the gains are for real.
"We're entering this phase of quantum computing that I call utility," Jay Gambetta, an IBM Fellow and vice president of IBM Quantum Research, told The New York Times. "The era of utility."
At the risk of seriously dumbing down some marvelous, head-spinning science, here's a quick rundown on quantum computing.
Basically, it takes advantage of two principles of quantum mechanics. The first is superposition, the ability for a single particle, in this case quantum bits or qubits, to be in two separate states
at the same time. Then there's entanglement, which enables two particles to share the same state simultaneously.
These spooky principles allow for a far smaller number of qubits to rival the processing power of regular bits, which can only be a binary one or zero. Sounds great, but at the quantum level,
particles eerily exist at uncertain states, arising in a pesky randomness known as quantum noise.
Managing this noise is key to getting practical results from a quantum computer. A slight change in temperature, for example, could cause a qubit to change state or lose superposition.
This is where IBM's new work comes in. In the experiment, the company's researchers used a 127 qubit IBM Eagle processor to calculate what's known as an Ising model, simulating the behavior of 127
magnetic, quantum-sized particles in a magnetic field — a problem that has real-world value but, at that scale, is far too complicated for classical computers to solve.
To mitigate the quantum noise, the researchers, paradoxically, actually introduced more noise, and then precisely documented its effects on each part of the processor's circuit and the patterns that
From there, the researchers could reliably extrapolate what the calculations would have looked like without noise at all. They call this process "error mitigation."
There's just one nagging problem. Since the calculations the IBM quantum processor performed were at such a complicated scale, a classical computer doing the same calculations would also run into
But because other experiments showed that their quantum processor produced more accurate results than a classical one when simulating a smaller, but still formidably complex Ising model, the
researchers say there's a good chance their error-mitigated findings are correct.
"The level of agreement between the quantum and classical computations on such large problems was pretty surprising to me personally," co-author Andrew Eddins, a physicist at IBM Quantum, said in a
lengthy company blog post. "Hopefully it’s impressive to everyone."
As promising as the findings are, it's "not obvious that they've achieved quantum supremacy here," co-author Michael Zaletel, a UC Berkley physicist, told the NYT.
Further experiments will need to corroborate that the IBM scientists' error mitigation techniques would not produce the same, or even better, results in a classic processor calculating the same
In the meantime, the IBM scientists see their error mitigation as a stepping stone to an even more impressive process of error correction, which could be what finally ushers in an age of "quantum
supremacy." We'll be watching.
More on quantum computing: Researchers Say They've Come Up With a Blueprint for Creating a Wormhole in a Lab
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Bounds for the expectation of the max ~ Cross Validated ~ TransWikia.com
Consider the following expected value
where the expectation is taken with respect to the random variables $$(epsilon_1,…,epsilon_n)$$ and $$a_1,…,a_n$$ are real numbers.
Can we say whether $$E(max{a_1+epsilon_1, a_2+epsilon_2,…,a_n+epsilon_n})$$ is greater or smaller than $$max{a_1,…,a_n}$$?
Can we characterise some upper or lower bound of $$E(max{a_1+epsilon_1, a_2+epsilon_2,…,a_n+epsilon_n})$$ that does not depend on the distribution of $$(epsilon_1,…,epsilon_n)$$?
No, consider the easiest case $$n=1$$. $$E[a+epsilon]=a+E[epsilon]$$ and it can be either greater or smaller than $$a$$ based on the distribution of $$epsilon$$.
Answered by gunes on November 27, 2020
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6.3 Centripetal Force - College Physics for AP® Courses | OpenStax
Learning Objectives
By the end of this section, you will be able to:
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction
between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal
acceleration. According to Newton's second law of motion, net force is mass times acceleration: net $F=maF=ma size 12{F= ital "ma"} {}$. For uniform circular motion, the acceleration is the
centripetal acceleration— $a=aca=ac size 12{a=a rSub { size 8{c} } } {}$. Thus, the magnitude of centripetal force $FcFc size 12{F rSub { size 8{c} } } {}$ is
$F c = m a c . F c = m a c . size 12{F rSub { size 8{c} } =ma rSub { size 8{c} } } {}$
By using the expressions for centripetal acceleration $acac size 12{a rSub { size 8{c} } } {}$ from $ac=v2r;ac=rω2ac=v2r;ac=rω2 size 12{a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } ;
``a rSub { size 8{c} } =rω rSup { size 8{2} } } {}$, we get two expressions for the centripetal force $FcFc size 12{F rSub { size 8{c} } } {}$ in terms of mass, velocity, angular velocity, and radius
of curvature:
$F c = m v 2 r ; F c = mr ω 2 . F c = m v 2 r ; F c = mr ω 2 . size 12{F rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } ;``F rSub { size 8{c} } = ital "mr"ω rSup { size 8{2} } } {}$
You may use whichever expression for centripetal force is more convenient. Centripetal force $FcFc size 12{F rSub { size 8{c} } } {}$ is always perpendicular to the path and pointing to the center of
curvature, because $acac size 12{a rSub { size 8{c} } } {}$ is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for $rr size 12{r} {}$, you get
$r=mv2Fc.r=mv2Fc. size 12{r= { { ital "mv" rSup { size 8{2} } } over {F rSub { size 8{c} } } } } {}$
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
What Coefficient of Friction Do Car Tires Need on a Flat Curve?
(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 6.12).
Strategy and Solution for (a)
We know that $F c = mv 2 r F c = mv 2 r$. Thus,
$F c = mv 2 r = ( 900 kg ) ( 25.0 m/s ) 2 ( 500 m ) = 1125 N. F c = mv 2 r = ( 900 kg ) ( 25.0 m/s ) 2 ( 500 m ) = 1125 N.$
Strategy for (b)
Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the
car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is $μsNμsN size 12{μ rSub { size 8{s} } N} {}$, where $μsμs
size 12{μ rSub { size 8{s} } } {}$ is the static coefficient of friction and N is the normal force. The normal force equals the car's weight on level ground, so that $N=mgN=mg$. Thus the centripetal
force in this situation is
$Fc=f=μsN=μsmg.Fc=f=μsN=μsmg. size 12{F rSub { size 8{c} } =f=μ rSub { size 8{s} } N=μ rSub { size 8{s} } ital "mg"} {}$
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for $FcFc size 12{F rSub { size 8{c} } } {}$ from the equation
$F c = m v 2 r F c = mr ω 2 } , F c = m v 2 r F c = mr ω 2 } , size 12{ left none matrix { F rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } {} ## F rSub { size 8{c} } = ital "mr"ω rSup {
size 8{2} } } right rbrace ,} {}$
$mv2r=μsmg.mv2r=μsmg. size 12{m { {v rSup { size 8{2} } } over {r} } =μ rSub { size 8{s} } ital "mg"} {}$
We solve this for $μsμs size 12{μ rSub { size 8{s} } } {}$, noting that mass cancels, and obtain
$μs=v2rg.μs=v2rg. size 12{μ rSub { size 8{s} } = { {v rSup { size 8{2} } } over { ital "rg"} } } {}$
Solution for (b)
Substituting the knowns,
$μs=(25.0 m/s)2(500 m)(9.80 m/s2)=0.13.μs=(25.0 m/s)2(500 m)(9.80 m/s2)=0.13. size 12{μ rSub { size 8{s} } = { { \( "25" "." 0" m/s" \) rSup { size 8{2} } } over { \( "500"" m" \) \( 9 "." "80 m/s"
rSup { size 8{2} } \) } } =0 "." "13"} {}$
(Because coefficients of friction are approximate, the answer is given to only two digits.)
We could also solve part (a) using the first expression in $F c = m v 2 r F c = mr ω 2 } , F c = m v 2 r F c = mr ω 2 } , size 12{ left none matrix { F rSub { size 8{c} } =m { {v rSup { size 8{2} } }
over {r} } {} ## F rSub { size 8{c} } = ital "mr"ω rSup { size 8{2} } } right rbrace ,} {}$ because $m,m, size 12{m,} {}$$v,v, size 12{v,} {}$ and $rr size 12{r} {}$ are given. The coefficient of
friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is
a responsive force, being able to assume a value less than but no more than $μsNμsN size 12{μ rSub { size 8{g} } N} {}$. A higher coefficient would also allow the car to negotiate the curve at a
higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the
car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal
force would be greater as will be discussed below.
Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.13. The greater the angle $θθ size 12{θ} {}$, the faster you can take the curve. Race tracks
for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle $θθ size 12{θ} {}$ is such that you can negotiate the curve at a certain speed
without the aid of friction between the tires and the road. We will derive an expression for $θθ size 12{θ} {}$ for an ideally banked curve and consider an example related to it.
For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal
the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the
vertical and horizontal directions.
Figure 6.13 shows a free body diagram for a car on a frictionless banked curve. If the angle $θθ size 12{θ} {}$ is ideal for the speed and radius, then the net external force will equal the necessary
centripetal force. The only two external forces acting on the car are its weight $ww size 12{w} {}$ and the normal force of the road $NN size 12{N} {}$. (A frictionless surface can only exert a force
perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude $mv2/rmv2/r size 12
{"mv" rSup { size 8{2} } "/r"} {}$. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal
component, and so this must equal the centripetal force—that is,
$Nsinθ=mv2r.Nsinθ=mv2r. size 12{N"sin"θ= { { ital "mv" rSup { size 8{2} } } over {r} } } {}$
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in
direction. From the figure, we see that the vertical component of the normal force is $NcosθNcosθ size 12{N"cos"θ} {}$, and the only other vertical force is the car's weight. These must be equal in
magnitude; thus,
$Ncosθ=mg.Ncosθ=mg. size 12{N"cos"θ= ital "mg"} {}$
Now we can combine the last two equations to eliminate $NN size 12{N} {}$ and get an expression for $θθ size 12{θ} {}$, as desired. Solving the second equation for $N=mg/(cosθ)N=mg/(cosθ) size 12{N=
ital "mg"/ \( "cos"θ \) } {}$, and substituting this into the first yields
$mg sin θ cos θ = mv 2 r mg sin θ cos θ = mv 2 r$
$mgtan(θ) = mv2r tanθ = v2rg. mgtan(θ) = mv2r tanθ = v2rg.$
Taking the inverse tangent gives
$θ=tan−1v2rg (ideally banked curve, no friction).θ=tan−1v2rg (ideally banked curve, no friction). size 12{θ="tan" rSup { size 8{ - 1} } left ( { {v rSup { size 8{2} } } over { ital "rg"} } right )}
This expression can be understood by considering how $θθ size 12{θ} {}$ depends on $vv size 12{v} {}$ and $rr size 12{r} {}$. A large $θθ size 12{θ} {}$ will be obtained for a large $vv size 12{v} {}
$ and a small $rr size 12{r} {}$. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the
curve is frictionless. Note that $θθ size 12{θ} {}$ does not depend on the mass of the vehicle.
What Is the Ideal Speed to Take a Steeply Banked Tight Curve?
Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car
configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then
substitute known quantities.
Starting with
$tan θ = v 2 rg tan θ = v 2 rg size 12{"tan"θ= { {v rSup { size 8{2} } } over { ital "rg"} } } {}$
we get
$v=(rgtanθ)1/2.v=(rgtanθ)1/2. size 12{v= \( ital "rg""tan"θ \) rSup { size 8{1/2} } } {}$
Noting that tan 65.0º = 2.14, we obtain
$v = (100 m)(9.80 m/s2)(2.14)1/2 = 45.8 m/s. v = (100 m)(9.80 m/s2)(2.14)1/2 = 45.8 m/s.$
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.
Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter's
Problems and Exercises.
Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this
in slow motion.
Gravity and Orbits
Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off
gravity to see what would happen without it!
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