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## March 16, 2010

### Philosophy in the EMS Newsletter

#### Posted by David Corfield

The March 2010 edition of the Newsletter of the European Mathematical Society is out, and it includes a short piece by me on realism in mathematics. I was invited to contribute to a series on Platonism by Urs Persson, a series which includes Brian Davies’ ‘Let Platonism Die’ (June 2007), Reuben Hersh’s ‘On Platonism’ and Barry Mazur’s ‘Mathematical Platonism and its Opposites’ (June 2008), and David Mumford’s ‘Why I am a Platonist’ and Philip Davis’ ‘Why I Am A (Moderate) Social Constructivist’ (December 2008). Ulf has himself a riposte to Davies – Let Platonism Live!. I decided to drop the term ‘Platonism’ since I prefer to focus on the issue of the nature of the constraints acting on mathematicians, and don’t wish to tie this wholly to the question of the existence of abstract entities.

Posted at March 16, 2010 12:16 PM UTC

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### Re: Philosophy in the EMS Newsletter

Thanks! My ideas on the background of “platonism” in mathematics, with some art and poetry. Barry Mazur remarked on that: “somehow I would have balanced that list (of “platonic” ideas in math, TR) with “things sought that turned out not to exist” (e.g., the squaring of the circle, trisection of a general angle by Euclidean construction methods) as further hints that a platonic mentality seems to be very energetically at work—in the minds of many mathematicians.” Maybe one could wonder about “platonic mirages” like Hilberts proof theory program.

Posted by: Thomas on March 16, 2010 5:29 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Nice piece, David! Anyone who wants to go straight to it should click here.

Ulf needs a more intelligent spell-checker. He speaks of “rightest political views”, and the “a priory”.

Posted by: John Baez on March 16, 2010 11:41 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

I was poking around old issues at the EMS website and discovered an interview with Alain Connes. I remember Connes because John Baez described his “Noncommutative Geometry” book in such glowing terms that I bought it, much to my later chagrin. I found a Platonic article by Mazur which seemed very fair. Connes holds an adamant view in favor of mathematical realism. He had an interesting discussion with Jean-Pierre Changeux in the earlier book “Conversations on Mind, Matter, and Mathematics”.

March 2008, Interview with Alain Connes (2) –C. Goldstein and G. Skandalis page 29

“whether mathematical reality is something created or something pre-existing is much easier to discuss if one uses the distinction which appears in Godel’s theorem between
“truth” and “provability” of a mathematical statement. I discussed this in details in my book “Triangle of thoughts”…

Posted by: Stephen Harris on March 23, 2010 4:55 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

S Y M M E T RY A Journey into the Patterns of Nature by Marcus du Sautoy
“The universe is built on a plan the profound symmetry of which is somehow present in the inner structure of our intellect.” by Paul Valery

Why, though, is symmetry so pervasive in nature? It is not just a matter of aesthetics. Just as it is for me and mathematics, symmetry in nature is about language. It provides a way for animals and plants to convey a multitude of messages, from genetic superiority to nutritional information. Symmetry is often a sign of meaning, and can therefore be interpreted as a very basic, almost primeval form of communication.

plato.stanford.edu/entries/pythagoras/
“Some of these earlier sources were heavily contaminated by the Neopythagorean view of Pythagoras as the source of all true philosophy, whose ideas Plato, Aristotle and all later Greek philosophers plagiarized. Nicomachus (ca. 50 – ca. 150 CE) assigns Pythagoras a metaphysics that is patently Platonic and Aristotelian and that employs distinctive Platonic and Aristotelian terminology … In the Pythagorean Memoirs, Pythagoras is said to have adopted the Monad and the Indefinite Dyad as incorporeal
principles, from which arise first the numbers, then plane and solid figures and finally the bodies of the sensible world”

“The tendency of people to see purpose in symmetry suggests at least one reason why symmetries are often an integral part of the symbols of world religions.”

SH: The ancient Greeks perceived patterns and attributed them to a Divine provenance, as also in the case of the divine proportion. Symmetry in nature and mathematics accounts for patterns ranging from the hexagonal honeybee cells to the way in which “Moonshine forms a way of explaining the mysterious connection between the monster finite group and modular functions from classical number theory.”

Can you hear the sound of Platonism’s drumbeat vanishing just as the mythological gods have perished when scientific knowledge has replaced primitive belief?!

Posted by: Stephen Harris on March 20, 2010 3:11 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Thomas wrote: “The sketch of the “platonic” mindset seems to fit esp. that of number theorists. An interesting suggestion I received is about Wilfried Owen’s poem “The naming of the parts” as illustration of emotions related to “naming” and the short stories of Tim OBrien “The things they carried” on that and a complex concept of “reality” similar to those of the ancient greeks.”

The idea of the Muses has been hear since at least the days of the ancient Greeks. The eye receives something like two million bits of information per second. Only about two thousand bits of information are passed onto consciousness as a basis for decisions. Perhaps the unconscious can process other conscious ideas and provide unusual creativity. Some people report creativity being passed on in dreams. I think this Muse idea (unconscious) makes a lot more sense than creative input received from an eternal realm of ideas existing beyond space and time. Muses don’t have to be actual personas, just functions of the unconscious. So perhaps everyone has that potential. But I don’t think one has to buy into the notion that everyone is networked into a Jungian Collective Unconscious. If there are archetypes, it is because they are part of the normal evolved brain information that babies are born with that allows them to learn without needing to be taught to learn, early, not language. And because someone writes a historical perspective doesn’t mean that they don’t reveal their own belief inclinations within that primarily historical account. No matter whether they had the intention to do so or not.

Posted by: Stephen Harris on March 22, 2010 5:53 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

I am more interested in e.g. how “platonic” ideas work even when they seem not to work. There I wonder if the story of p-adic Birch Swinnerton-Dyer, where the p-adification of a classical theory run into unexpected troubles, were an interesting example. An other interesting case may be K. Kato’s use of nonexistent “fundamental functions” a guide in ch. III, §1 here. There could be a lot of interesting stuff aside the wellknown cases like F_1, primes”=”knots, numbers”=”function fields,”mysterious functor”,Fontaine-Mazur conjecture, etc.

A history-of-ideas of “platonism” makes sense simply because much nonsense is often associated with that term. There is a funny story by an austrian physicist, “Einsteins Erben”, sketching a possible similar degeneration of the idea of science-mentality, as obviously happened there. That one has to distinguish things shows e.g. Gromov’s essay: On the one hand it fits perfectly into the chain of “platonistic” theorising and he stresses the “platonic” mindset among his colleagues, on the other hand, his questioning of speaking of really huge numbers, or his use of Zellig’s semantics do not. His examples of how hidden symmetries in very simple situations turn into interesting theories and how “good” definitions differ from “unproductive” and “boring” ones are very interesting, independent from ontological beliefs, and it would be great to see if and how other cases of that happen.

Posted by: Thomas on March 22, 2010 1:17 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Thomas wrote: “I am more interested in e.g. how “platonic” ideas work even when they seem not to work.”

I cannot tell what you mean by “platonic”. I think some definition of platonic is necessary because how can you match the concept of platonic with some situation? How do you identify or associate programs in mathematics or number theory that are governed or controlled or fall under the jurisdiction of a platonic event, unless one knows what the word platonic means. Do you mean when some area of mathematics unexpectedly, or it comes as a subjective surprise, is related to another apparently distant area of mathematics.

I don’t think you can give me an example of a mathematical result which required any concept called platonism or any mathematician who required such ideas. David avoided going into the source of a platonic idea, and just focused on whether the idea was considered: discovered or invented.

But if a person wants to say an idea had a platonic inspiration, that means there has to be some (abstract perhaps) place where the idea was discovered. Some place that was available to hold the ideas so that some mathematician could discover them.

So there is a religious/mythological set of beliefs which Pythagoras dreamed up. And Plato adopted many of these ideas. The eternal platonic realm held all mathematical truths. [See Tom Silverman’s post.]
Another idea was Jung’s Archetypes. Jung, Eliade, and Joseph Campbell were all mythologists.

Take the idea of the unicorn which can dip its horn in contaminated waters and purify them. The unicorn is imaginary, with no physical presence unless one can answer,
Where it comes from, what it is, and what it applies to and how it came into existence.

The same criteria apply to platonic inspirations for the mathematician. If one can’t provide concrete answers then platonism is just another religious/mythological/imaginary belief.

A reply like, well so and so believes in Platonism is no evidence. It reminds me of asking a sadhu if he is enlightened.
The sadhu says no, but that his guru is enlightened. Question: How do you know that your guru is enlightened if you are not?
Answer: Because, look at all the other unenlightened aspirants who belong to my order, they all think that the guru is enlightened.

So please provide an example of what your term “platonic insights” means, something you can say for sure is a result of a mathematician receiving platonic inspiration. So where does a “platonic insight” come from if not from a “platonic reality” Manin’s term?
Speaking of computers, Manin said:
“So if he doesn’t have the imagination to distinguish some features of this Platonic reality, he can experiment.” [he = mathematician]

I think it is logical to define an instance where “platonic ideas work” before attempting to figure out where platonic ideas don’t work, or can’t work.

Posted by: Stephen Harris on March 22, 2010 5:13 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

“what you mean by “platonic””: I think the examples make that clear. I refer to what among mathematicians is usually regarded as such, e.g. the idea behind class field theory and Kronecker’s “Jugendtraum”. Gromov’s “h-principle” may be one, but I know too little about that. I don’t think that there is a need to define it. Aside that, to repeat, what could be called “platonic” is a midset and mentality, not some ontological proposition (which however can result in entertaining pieces of literature, like some Borges’ stories). People can believe a lot of things, change their beliefs, or do better things than norishing beliefs with their time. But it is interesting to see which kinds of theories come out of it and how that develops.

Posted by: Thomas on March 22, 2010 6:57 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Here’s what Alain Connes says about mathematical realism. I find it quite congenial to my own way of thinking.

I have no doubt that mathematical reality is something which exists, that it exists independently of my own brain trying to see it, and has exactly the same properties of resistance as external reality. When you want to prove something, or when you examine if a proof is correct or not, you feel the same anguish, the same external resistance as you do with external reality. Some people will tell you that this reality does not exist because it is not “localized” somewhere in space and time. I just find this absurd and I adopt a diametrically opposed point of view: for me even a human being is better described by an abstract scheme than by a material collection of cells – which in any case are totally renewed and replaced over a relatively short period of time and hence possess less meaning or permanence than the scheme itself, which might eventually be reproduced in several identical copies……

If one wants to reduce everything to “matter localized somewhere” one soon meets a wall which comes from quantum mechanics and one finds that this reduction of the outer reality to matter is an illusion that only makes sense at intermediate scales but by no means at a fundamental level. Thus I have no doubt on the subtleness and existence of a reality which can be neither reduced to “matter” nor “localized”.

Posted by: John Baez on March 25, 2010 4:44 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Good, but I always get a little nervous when people reach straight for proof when they want to illustrate mathematical resistance. This lends itself to being interpreted as merely logical resistance. I know Connes elsewhere (see bottom of TWF218) explains how reality is revealed by one’s repeated encountering of the same entities: elliptic functions, modular forms, zeta functions.

Looking to Lautman’s distinction between entities, facts, theories, and ideas, we can see mathematical resistance in all quarters. What’s intriguing me at the moment is the possibility of mathematical accounts of mathematical resistance.

For example, you can run a debate on the inevitability of the quaternions appearing once mathematicians had happily accepted the complex numbers.

The contingentists tell you about Hamilton’s idiosyncratic use of the Coleridgean metaphysical understanding of space and time in his quest to move beyond the 2 dimensions of $\mathbb{C}$.

The inevitabilists point to the classification of normed devision algebras, allowing a 4-dimensional but not 3-dimensional analogue of $\mathbb{C}$.

The contingentists say that the notion of a normed division algebra was invented to make sense of the quaternions.

The inevitabilists point to how the triple $\langle \mathbb{R}, \mathbb{C}, \mathbb{H} \rangle$, or quadruple $\langle \mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O} \rangle$, manifests itself over a range of fields.

And on we go. The inevitabilist will tend to resort to mathematical explanations, which opens them to the charge that that mathematics itself is the result of certain contingent choices. But I’m wondering whether there can be an overwhelming force of evidence from the snugness of fit especially into areas where the concepts had developed independently. Then again this claimed independence is open to challenge.

Posted by: David Corfield on March 25, 2010 10:27 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Alain Connes:

…you feel the same anguish, the same external resistance as you do with external reality.

These seem to be the keywords that describe why mathematics feels “real”.

I suppose everyone knows about Popper’s three worlds?

Alain Connes:

…for me even a human being is better described by an abstract scheme than by a material collection of cells…

To even speak this words requires the mental picture aka abstract scheme of a human body as an interacting cell complex (pun intended), which is neither universal nor very old. I’m sure that the accompanying paradigm of modern medicine, namely “all diseases are diseases of cells”, will change somewhen in the future, just like the comparison of a human mind with a telephone switchboard, customary a hundred years ago, got replaced by the comparison with a computer (which will seem as inept in a hundred years as the switchboard analogy seems to me now).

If one wants to reduce everything to “matter localized somewhere” one soon meets a wall which comes from quantum mechanics…

I don’t understand the rôle that quantum mechanics plays in the argument.

Posted by: Tim van Beek on March 25, 2010 10:45 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

“the rôle that quantum mechanics plays in the argument”: As far as I remember the trialogue book, it is just a plausibility reason that simple concepts of “reality” don’t work anyway, independent from “platonism”. An example without advanced physics gives this nice text of an archeologist on how to present data by drawings. Aside that, Connes takes noncommutativity as fundamental concept, and seems to have the idea of a “quantum platonism”, i.e. that individual minds act as projectors on “platonic ideas”. I’d be curious what composers he thinks at: “Some composers reached, by an hallucinatory work of precision, a level of perfection close to that of some of Riemann’s work.”?

Posted by: Thomas on March 25, 2010 2:40 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Connes wrote:

If one wants to reduce everything to “matter localized somewhere” one soon meets a wall which comes from quantum mechanics…

Tim wrote:

I don’t understand the rôle that quantum mechanics plays in the argument.

Lacking any secret knowledge of what Connes meant, I’d take what he said quite literally: if we try to pursue a simple-minded materialism where we try to reduce the world to ‘objects located somewhere’, we run into a wall. Namely, quantum mechanics says the world isn’t like that. At best, it’s approximately like that under certain circumstances.

Of course, one could switch to a more sophisticated version of materialism which takes quantum mechanics into account…

Posted by: John Baez on March 27, 2010 11:41 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Here is a very nice article by Laurent Lafforgue sketching the “platonic mindset”. Philosopher’s ideas differ (that link explains Connes’ harsh remarks on that).

Posted by: Thomas on March 29, 2010 10:18 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Thanks for pointing this out to me. You can find more of Lafforgue’s philosophical writings here.

Posted by: David Corfield on March 29, 2010 2:19 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

His description of the relation between mathematics and language is very nice too and complements the other text. His quote of S. Weil in one of the texts (“popularisations of science are in part responsible for a general loss of sense of reality”) makes me wonder how Edge etc. work - do they really increase the complexity of their reader’s way of thinking?

Posted by: Thomas on March 30, 2010 1:52 PM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

In hope not to restart silly disputes, here a nice new article of the New Scientist on language shaping perception etc.

Posted by: Thomas on September 6, 2010 11:03 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Here is a talk by Yves André comparing Poincaré’s and Grothendieck’s writings on their “methods”.

Posted by: Thomas on April 29, 2010 11:51 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

News on the “Plato code”

Posted by: Thomas on June 29, 2010 9:52 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

The press release is full of pretty wild claims!

The hidden codes show that Plato anticipated the Scientific Revolution 2,000 years before Isaac Newton, discovering its most important idea…

Posted by: John Baez on June 30, 2010 6:49 AM | Permalink | Reply to this

### Re: Philosophy in the EMS Newsletter

Yes - I wait for the movie

Posted by: T. on June 30, 2010 2:27 PM | Permalink | Reply to this

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text:
Dual system

 Decimal numbers 0 to 15 in the dual system Valence: 8 4 2 1 Zero: 0 0 0 0 One: 0 0 0 1 Two: 0 0 1 0 Three: 0 0 1 1 Four: 0 1 0 0 Five: 0 1 0 1 Six: 0 1 1 0 Seven: 0 1 1 1 Eight: 1 0 0 0 Nine: 1 0 0 1 Ten: 1 0 1 0 Eleven: 1 0 1 1 Twelve: 1 1 0 0 Thirteen: 1 1 0 1 Fourteen: 1 1 1 0 Fifteen: 1 1 1 1

The dual system ( lat. Dualis "containing two"), also called the two-part system or binary system , is a number system that uses only two different digits to represent numbers .

In the usual decimal system , the digits 0 to 9 are used. In the dual system, however, numbers are only represented with the digits of the value zero and one . The symbols 0 and 1 are often used for these digits . Numbers zero through fifteen are listed on the right.

The dual system is the place value system with base 2, so it provides the dyadic (2-adic) representation of numbers ( dyadic ) ( gr. Δύο = two ).

Due to its importance in digital technology , it is the most important number system alongside the decimal system.

The number representations in the dual system are also called dual numbers or binary numbers . The latter is the more general name, as it can simply stand for binary-coded numbers . The term binary number does not specify the way a number is represented, it only states that two different digits are used.

Definition and representation

In the case of numbers in the dual system, the digits are written one after the other without separators, as in the commonly used decimal system, but their place value corresponds to the power of two and not the power of ten. ${\ displaystyle z_ {i}}$

The most significant digit with the value is written on the far left and the lower significant digits with the values up to the right of it in descending order. To represent rational or real numbers , a separating comma is followed by the digits to , which represent the fraction of the number. If this display stops, it looks like this: ${\ displaystyle z_ {m}}$${\ displaystyle z_ {m-1}}$${\ displaystyle z_ {0}}$${\ displaystyle z _ {- 1}}$${\ displaystyle z _ {- n}}$

${\ displaystyle z_ {m} z_ {m-1} \ ldots z_ {0} \ operatorname {,} z _ {- 1} z _ {- 2} \ ldots z _ {- n} \ qquad \ left (m, n \ in \ mathbb {N} \ quad z_ {i} \ in \ {0,1 \} \ right)}$

The value of the binary number results from adding these digits, which are multiplied by their place value beforehand: ${\ displaystyle Z}$${\ displaystyle 2 ^ {i}}$

${\ displaystyle Z = \ sum _ {i = -n} ^ {m} z_ {i} \ cdot 2 ^ {i}}$.

The symbols 0 and 1 are usually used to represent the two digits, analogous to other number systems.

In older literature related to electronic computing , the symbols Low (L) and High (H) are sometimes used in place of 0 and 1. Low then usually stands for the value zero and high for the value one. This assignment is called positive logic , with negative logic the values are assigned the other way round. In computer science, the "digits" true (w) and false (f) or the English translations true (t) and false (f) are also used for binary coded values , whereby false = 0 and true = 1 are usually set.

The symbols L for the value one and 0 for the value zero are (rarely) used.

Negative numbers are written in the binary system as in the decimal system with a preceding minus (-).

Examples

The sequence of digits 1101, for example, does not represent the thousand one hundred and one (as in the decimal system), but the thirteen, because in the dual system the value is calculated through

${\ displaystyle [1101] _ {2} = 1 \ cdot 2 ^ {3} +1 \ cdot 2 ^ {2} +0 \ cdot 2 ^ {1} +1 \ cdot 2 ^ {0} = [13] _ {10}}$

and not through as in the decimal system

${\ displaystyle 1 \ cdot 10 ^ {3} +1 \ cdot 10 ^ {2} +0 \ cdot 10 ^ {1} +1 \ cdot 10 ^ {0} = [1101] _ {10}}$.

For further techniques and examples for converting binary numbers into decimal numbers and vice versa: see section Converting binary numbers into other place value systems .

The bracketing of the results with the subscript 2 or the 10 indicates the basis of the value system used . This makes it easy to see whether the number is displayed in the dual or decimal system. In the literature, the square brackets are often left out and the subscript is sometimes put in round brackets. It is also possible to identify it by adding the capital letter B (for binary ).

Different ways of writing the number twenty-three in the dual system:

• [10111] 2
• 10111 2
• 10111 (2)
• 10111b (so-called Intel Convention)
• 10111B
• 0b10111
• % 10111 (so-called Motorola convention, but also e.g. with DR-DOS DEBUG )
• HLHHH
• L0LLL

history

Development of the dual system

The ancient Indian mathematician Pingala provided the first known description of a number system consisting of two characters in the 3rd century BC. BC before. However, this number system did not have a zero.

A series of eight trigrams and 64 hexagrams are known from the ancient Chinese and Daoist text I Ching . In the 11th century, the Chinese scholar and philosopher Shao Yong developed a systematic arrangement of hexagrams, representing the sequence from 1 to 64, and a method for producing the same. However, there is no evidence that Shao knew how to make calculations in the dual system or that he recognized the concept of value.

Joachim Bouvet transmitted the sixty-four hexagrams from China to Leibniz , 1701

Centuries before the dual system was developed in Europe, Polynesians used binary summaries of numbers to simplify calculations.

At the end of the 17th century, Gottfried Wilhelm Leibniz already felt the dyadic (dyo, Greek = two), i.e. the representation of numbers in the dual system, which he developed, was very important. He saw it as such a convincing symbol of the Christian faith that he wanted to convince the Chinese Emperor Kangxi with it. He wrote to the French Jesuit priest Joachim Bouvet (1656–1730):

“At the beginning of the first day it was 1, which means God. At the beginning of the second day the 2, because heaven and earth were created during the first. Finally at the beginning of the seventh day everything was already there; therefore the last day is the most perfect and the Sabbath , because on it everything is created and fulfilled, and therefore the 7 111 is written, i.e. without a zero. And only if you just write the numbers with 0 and 1, you can see the perfection of the seventh day, which is considered sacred, and of which it is also noteworthy that its characters are related to the Trinity . "
The binary number system in a first draft by Gottfried Wilhelm Leibniz, 1697

On the other hand, his description in a letter to Duke Rudolf von Braunschweig-Wolfenbüttel on January 2, 1697 was somewhat more secular:

“... For one of the main points of the Christian faith ... is the creation of all things out of nothing by the omnipotence of God. Now one can say that nothing in the world represents this better than the origin of the numbers, as it is presented here, by expressing them only with one and zero (or nothing). It will be difficult to find a better example of this mystery in nature and philosophy ... This is all the more fitting because the empty depths and desolate darkness belong to zero and nothing, but the spirit of God with its light belongs to the almighty one. Because of the words of the symbol I thought for a while and finally found this verse to be good: To develop everything from nothing is enough for one (Omnibus ex nihilo ducendis sufficit unum). "

Probably because the precision mechanical skills of the time were insufficient, Leibniz resorted to the decimal system when building his calculating machines .

Page from “Explication de l'Arithmétique Binaire” , 1703

The dual system was fully documented by Leibniz at the beginning of the 18th century in his article Explication de l'Arithmétique Binaire (Histoire de l'Academie Royale des Sciences 1703, published in Paris 1705). Leibniz also confirms the view of Joachim Bouvet , a missionary at the Chinese imperial court, who interpreted the tri- and hexagrams of Fu Hsi (see figure : "Signs of Fu Hsi") as numerals in a certain reading direction. He saw it as an archaic binary system that has been forgotten. This interpretation is now considered very unlikely.

Leibniz also had predecessors in Europe. An earlier treatment of the dual system and other position systems by Thomas Harriot was not published by Thomas Harriot , but was only found in the estate. The first publication of the dual system in Europe is probably in Mathesis Biceps vetus et nova 1670 by the later Spanish bishop Juan Caramuel y Lobkowitz (1606–1682), who also deals with numbers for other bases. Also Blaise Pascal noticed already in De numeris multiplicibus (1654, 1665) that the base 10 is not a necessity.

In 1854, the British mathematician George Boole published a seminal paper detailing a system of logic that became known as Boolean algebra . His logical system paved the way for the realization of electronic circuits that implement arithmetic in the dual system.

application

The dual system became very important in the development of electronic calculating machines, because in digital technology, numbers are represented by electrical states. Two complementary states such as current on / current off or voltage / ground are preferably used, since in this way very error-resistant and simple circuits can be implemented (see binary code ). These two states can then be used as digits. The dual system is the easiest way to calculate with numbers that are represented by these two digits.

Binary numbers are used in electronic data processing to represent fixed-point numbers or whole numbers . Negative numbers are mainly represented as two's complement , which only corresponds to the dual number representation in the positive area. The one's complement , which corresponds to the inverted representation of binary numbers with a preceding one, is used less often. Representing negative numbers in one's complement has the disadvantage that there are two representations for zero, one in the positive and one in the negative. The excess code based on a value range shift offers a further alternative .

In order to approximately represent rational or even real numbers with non-terminating binary number representation in electronic data processing, floating point representations are preferably used in which the number is normalized and divided into mantissa and exponent . These two values are then stored in the form of binary numbers.

Calculation of required places

A binary number with digits can have the maximum value . A four-digit binary number can therefore have at most the value , i.e. 16 - 1 = 15. ${\ displaystyle n}$${\ displaystyle 2 ^ {n} -1}$${\ displaystyle 2 ^ {4} -1}$

Consequently, in the dual system, you can count up to 1023 with your 10 fingers . ${\ displaystyle 2 ^ {10} -1}$

In digital technology, it should be noted that when a binary number is saved, its sign must also be saved. In most cases, the most significant bit in the memory area reserved for the number is used for this purpose. If this memory area is bit large, the maximum value of the positive numbers and the minimum value of the negative numbers are (when the negative numbers are represented in two's complement ) . This is one of the positive numbers and that is the "first negative number". Overall, the number of numbers that can be displayed remains the same . ${\ displaystyle n}$${\ displaystyle 2 ^ {(n-1)} - 1}$${\ displaystyle - (2 ^ {(n-1)})}$${\ displaystyle 0}$${\ displaystyle -1}$${\ displaystyle 2 ^ {n}}$

The number of digits required in the binary system for a given number in the decimal system is ${\ displaystyle n}$

${\ displaystyle \ lfloor \ operatorname {lb} n \ rfloor +1}$.

In this case, referred to the rounding function and the base-2 logarithm of the number . Alternatively, the number of decimal places can be multiplied by 3.322 (+ rounding up), which results in an upper limit, because (one decimal place, actually lb ( ) becomes a maximum of binary places). ${\ displaystyle \ lfloor \ cdot \ rfloor}$${\ displaystyle \ operatorname {lb} n}$${\ displaystyle n}$${\ displaystyle lb (10) \ approx 3 {,} 322}$${\ displaystyle 9 {,} {\ bar {9}}}$${\ displaystyle \ approx 3 {,} 322}$

Basic arithmetic in the dual system

Analogous to the numbers in the decimal system , the common arithmetic basic operations of addition , subtraction , multiplication and division can be carried out with binary numbers . In fact, the required algorithms are even simpler and can be implemented efficiently electronically using logic circuits. The introduction of binary numbers in computer technology therefore brought many advantages.

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0, carry 1

${\ displaystyle {{\ begin {matrix} & 1011_ {2} \\\ operatorname {+} & \ \ \ 11_ {2} \\\ end {matrix}} \ over {\ begin {matrix} & \ \ 1110_ { 2} \\\ end {matrix}}}}$
subtraction example

0 - 0 = 0
0 - 1 = 1, borrow 1
1 - 0 = 1
1 - 1 = 0

${\ displaystyle {\ begin {matrix} & 1011_ {2} \\ {- \} & \ \ 111_ {2} \\\ end {matrix}} \ over {\ begin {matrix} & \ \ \ \ \ 100_ { 2} \ end {matrix}}}$
multiplication example

0 0 = 0 0 1 = 0 1 0 = 0 1 1 = 1 ${\ displaystyle \ cdot}$
${\ displaystyle \ cdot}$
${\ displaystyle \ cdot}$
${\ displaystyle \ cdot}$

${\ displaystyle 1010_ {2} {\ cdot} 11_ {2} = 11110_ {2}}$
division example

0/0 = n.def.
0/1 = 0
1/0 = n.def.
1/1 = 1

${\ displaystyle 1010_ {2} {/} 10_ {2} = 101_ {2}}$

 A. B. M 1 M 2 E. 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1

Binary addition is a fundamental basic operation in the computer world. If you want to add two non-negative binary numbers and add them, you can do that as in the decimal system. You only have to note that in the result, no "two" is noted in the respective position, but a zero and a carryover to the next position. This is done in the same way as the decimal addition, if the addition of a digit results in a ten: ${\ displaystyle A}$${\ displaystyle B}$

The numbers are written on top of each other. Now all binary digits (= bits) from and are processed simultaneously from right to left and a result bit and a flag bit (also called carry ) are generated in each intermediate step . The bits are added up according to the table on the right. The bits of the numbers to be added can be found in columns A and B. The flag bit of the previous intermediate step is in the column . This results in a result bit (E) and a new flag bit ( ) (according to this table, which corresponds to a full adder ). All result bits, strung together from right to left, represent the result. If a flag bit is generated in the last intermediate step, the result is given an additional 1 on the left. ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle M_ {1}}$${\ displaystyle M_ {2}}$

This is best seen with an example. Here the numbers A and B are added together. In each step a marker bit is noted for the next digit.

       A = 10011010 (154)
B = 00110110 (54)
Merker = 01111100
————————
Ergebnis = 11010000 (208)
‗‗‗‗‗‗‗‗

Written subtraction

The subtraction is analogous to the addition.

${\ displaystyle 0-0 = 0}$
${\ displaystyle 0-1 = 1}$, Borrow 1
${\ displaystyle 1-0 = 1}$
${\ displaystyle 1-1 = 0}$

Two numbers in the dual system can be subtracted from each other as shown in the following example:

${\ displaystyle {\ begin {matrix} & 1 & 1 & 0 & 1 & 1 & 1 & 0 \\ - &&& 1 & 0 & 1 & 1 & 1 \\ && {} _ {1} && {} _ {1} & {} _ {1} & {} _ {1} & \ end {matrix} } \ over {\ begin {matrix} = & 1 & 0 & 1 & 0 & 1 & 1 & 1 \ end {matrix}}}$

Here the subtraction 110−23 = 87 is carried out. The small ones in the third row show the carryover. The procedure is the same as that taught in the school for the decimal system. Case 0−1 looks a bit unusual. To clarify the example 2−9 in the decimal system: Imagine a tens place before the two, which results in the subtraction 12−9. The imaginary tens is then passed on as a carry over to the next digit. The same thing happens in the dual system: 0−1 becomes 10−1. As a result, a 1 can be written down; The one imagined in front of the 0 must then be written as a carry over to the next position and also deducted from it.

The procedure does not work (as in the decimal system) if the minuend (1st number) is smaller than the subtrahend (2nd number). If this is the case, a number is subtracted by adding the two's complement of this number. Subtracting a positive number gives the same result as adding the corresponding negative number with the same amount:

${\ displaystyle {\ begin {array} {crcccccccc | l | cr} && 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & {\ text {Minuend}} && 118_ {10} \\ - && 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & {\ text {Subtrahend}} & - & 0 & 1 & 1 & 1 & 1\ & 0 & 0 & 0_ & 1 & 1 & 1 & 0 {\ text {Minuend}} && 118_ {10} \\ + & (-) & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & {\ text {Subtrahend (two's complement)}} & + & (-) 153_ {10} \\ & {\ color {blue} {} _ {0}} & {} _ {1} & {} _ {1} &&& {} _ {1} & {} _ {1} &&& \\\ hline = && 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & {\ text {result (two's complement, negative) }} & = & - 35_ {10} \\\ hline \ hline && 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & {\ text {amount of the result}} && 35_ {10} \\\ end {array}}}$

If the carry (marked in blue) were 1, the two's complement of the result would no longer have to be formed, since the unsigned representation of the positive numbers in the two's complement is the same (see table there ). The carry is added to the leading ones (not shown) of the two's complement, which results in only leading zeros. The above calculation 110−23 = 87 serves as an example:

${\ displaystyle {\ begin {array} {crccccccc | l} && 1 & 1 & 0 & 1 & 1 & 1 & 0 & {\ text {Minuend}} \\ + & (-) & 1 & 1 & 0 & 1 & 0 & 0 & 1 & {\ text {Subtrahend (two's complement)}} \\ & {\ color {blue} { } _ {1}} & {} _ {1} && {} _ {1} &&&&& \\\ hline = && 1 & 0 & 1 & 0 & 1 & 1 & 1 & \ end {array}}}$

Written multiplication

The multiplication is carried out in the same way in the dual system as in the decimal system. Because only 0 and 1 appear as digits, the written multiplication is even easier. The following example, in which the numbers 1100 (12) and 1101 (13) are multiplied, shows the procedure.

First write the task in one line and draw a line below it to make things easier.

1100 · 1101
———————————

The first digit of the second factor is a one and therefore the first factor is right-justified under this one.

1100 · 1101
———————————
1100

For all other ones of the second factor, write the first factor right-justified underneath.

1100 · 1101
———————————
1100
1100
0000
1100

The numbers obtained in this way are then added together to form the result of the multiplication.

1100 · 1101
———————————
1100
+ 1100
+  0000
+   1100
———————————
10011100 (156)

A particularly simple case is the multiplication of a positive binary number by the number 10 (2). In this case, only a 0 has to be appended to the positive binary number:

${\ displaystyle 1101 \ cdot 10 = 11010}$
${\ displaystyle 11010 \ cdot 10 = 110100}$

etc.

Simple commands exist in digital technology for this arithmetic operation.

The Booth algorithm is used when multiplying two two 's complement dual numbers .

Written division

The following algorithms are used when dividing two binary numbers.

Using the example of the division of 1000010/11 (corresponds to 66: 3 in the decimal system)

   1000010 ÷ 11 = 010110 Rest 0 (= 22 im Dezimalsystem) somit mod
− 011
—————
00100
− 011
————
0011
− 011
—————
0

Applying the modulo function with the divisor 10 (2) to positive binary numbers always results in 1 if the last digit of the dividend is 1 and 0 if the last digit of the dividend is 0:

${\ displaystyle 1101 \ mod 10 = 1}$
${\ displaystyle 1100 \ mod 10 = 0}$

For this arithmetic operation, which corresponds to an AND operation with 1, there are simple commands in digital technology.

A particularly simple case is division with the remainder of a positive binary number by the number 10 (2). In this case only the last digit of the dividend has to be deleted. If the last digit of the dividend is 1, this remainder disappears.If with this procedure the number of divisions by 2 corresponds to the number of digits of the dividend, the end result is always 0:

${\ displaystyle 1101 \ div 10 = 110}$
${\ displaystyle 110 \ div 10 = 11}$
${\ displaystyle 11 \ div 10 = 1}$
${\ displaystyle 1 \ div 10 = 0}$

Simple commands exist in digital technology for this arithmetic operation.

Conversion of binary numbers into other place value systems

The small base has the disadvantage that numbers are relatively long and difficult to keep track of in relation to decimal numbers (see table below). This has led to the spread of the hexadecimal system, which the base 16 has. Since 16 is a power of 2, it is particularly easy to convert binary numbers into hexadecimal numbers. For this purpose, four digits of the binary number are replaced by a hexadecimal digit, which also reduces the length of the numbers displayed by a factor of four. The hexadecimal digits with the value 0–15 are usually represented by the number symbols 0–9 and the capital letters A – F (for the values 10–15). As a result, they are relatively easy to read, for example it is easy to determine that EDA5 (16) is larger than ED7A (16), whereas the corresponding binary numbers 1110110110100101 (2) and 1110110101111010 (2) cannot be overlooked so quickly.

 Dual system Decimal system Octal system Hexadecimal system 0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 0 1 2 3 4th 5 6th 7th 8th 9 10 11 12 13 14th 15th 0 1 2 3 4th 5 6th 7th 10 11 12 13 14th 15th 16 17th 0 1 2 3 4th 5 6th 7th 8th 9 A. B. C. D. E. F.

From the dual system to the decimal system

In order to convert a binary number into the corresponding decimal number, all digits are multiplied by their place value (corresponding power of two) and then added.

Example:

${\ displaystyle 1010 _ {(2)} = 1 \ cdot 2 ^ {3} +0 \ cdot 2 ^ {2} +1 \ cdot 2 ^ {1} +0 \ cdot 2 ^ {0} = 1 \ cdot 2 ^ {3} +1 \ cdot 2 ^ {1} = 8 + 2 = 10 _ {(10)}}$

If the binary number ends with a 1, the decimal number is an odd number. If the last digit of the binary number is 0, the decimal number is even.

Example:

${\ displaystyle 101001 _ {(2)} = 1 \ cdot 2 ^ {0} +0 \ cdot 2 ^ {1} +0 \ cdot 2 ^ {2} +1 \ cdot 2 ^ {3} +0 \ cdot 2 ^ {4} +1 \ cdot 2 ^ {5} = 1 \ cdot 2 ^ {0} +1 \ cdot 2 ^ {3} +1 \ cdot 2 ^ {5} = 1 + 8 + 32 = 41 _ {( 10)}}$
${\ displaystyle 101000 _ {(2)} = 0 \ cdot 2 ^ {0} +0 \ cdot 2 ^ {1} +0 \ cdot 2 ^ {2} +1 \ cdot 2 ^ {3} +0 \ cdot 2 ^ {4} +1 \ cdot 2 ^ {5} = 1 \ cdot 2 ^ {3} +1 \ cdot 2 ^ {5} = 8 + 32 = 40 _ {(10)}}$

This procedure can also be written down in the form of a table. To do this, write down the individual digits of a binary number in columns, which are overwritten with the respective place value of the digit. In the following table, the priority is highlighted in orange. In each of the three lines of the white part there is a binary number:

 Status 32 16 8th 4th 2 1 Binary number 0 0 0 1 0 1 5 decimal number 1 0 0 0 1 1 35 0 0 1 0 1 0 10

You now add up all the place values that are above the ones of the binary number and you get the corresponding decimal number with a green background. For example, to calculate the decimal value of the third binary number, the place values 8 and 2 are added. The result is 10.

This table method is also possible for place value systems for other bases; The peculiarity of the dual system is that the respective field entry ('0' or '1') does not have to be multiplied by the value of the position, but is used directly as a selection flag ('no' / 'yes') of this value for addition can be.

From the decimal system to the dual system

There are several ways of converting to the dual system. The division method (also called the modulo method ) is described below using example 41 (10) :

${\ displaystyle \ left. {\ begin {matrix} 41 &: 2 & = & 20 & \ mathrm {remainder} \ \ \ mathbf {1} \\ 20 &: 2 & = & 10 & \ mathrm {remainder} \ \ \ mathbf {0} \\ 10 &: 2 & = & 5 & \ mathrm {remainder} \ \ \ mathbf {0} \\ 5 &: 2 & = & 2 & \ mathrm {remainder} \ \ \ mathbf {1} \\ 2 &: 2 & = & 1 & \ mathrm {remainder} \ \ \ mathbf {0} \\ 1 &: 2 & = & 0 & \ mathrm {remainder} \ \ \ mathbf {1} \ end {matrix}} \ \ right \ uparrow}$

The corresponding binary number results from the notation of the calculated remainders from bottom to top: 101001 (2) .

Another method is the subtraction method. In this case, the greatest possible power of two is subtracted from the decimal number to be converted. If the next largest power of two is greater than the difference of the previous subtraction, the value of the next binary digit is 0. Otherwise, the next binary digit is 1 and the power of two is subtracted. To illustrate this method, let's use the number 41 as an example:

${\ displaystyle \ left. {\ begin {matrix} 41 & -2 ^ {5} & = & 9 & \ mathrm {value} \ \ \ mathbf {1} \\ 9 & -2 ^ {4} & <& 0 & \ mathrm {value } \ \ \ mathbf {0} \\ 9 & -2 ^ {3} & = & 1 & \ mathrm {value} \ \ \ mathbf {1} \\ 1 & -2 ^ {2} & <& 0 & \ mathrm {value} \ \ \ mathbf {0} \\ 1 & -2 ^ {1} & <& 0 & \ mathrm {weighting} \ \ \ mathbf {0} \\ 1 & -2 ^ {0} & = & 0 & \ mathrm {weighting} \ \ \ mathbf {1} \ end {matrix}} \ \ right \ downarrow}$

properties

Divisible by a power of 2

A number, represented by the base n, can be divided by the base n without a remainder (i-fold, i.e. by n i ) as often as the number has zeros at the end (i pieces). A binary number 100101000 2 can therefore be divided by 2 three times (= 2 3 ), since it ends with 3 zeros.

Divisible by 3

Let be a binary number where . We further define the set of ones in even places and the set of ones in odd places . Then applies to the number with respect to the divisibility by ( stands for the number): ${\ displaystyle n = b_ {k} \ dots b_ {0}}$${\ displaystyle b_ {i} \ in \ {0,1 \}}$${\ displaystyle G_ {1} (n) = \ {i \ mid \ exists j.b_ {2j} (n) = 1 \}}$${\ displaystyle U_ {1} (n) = \ {i \ mid \ exists j.b_ {2j + 1} (n) = 1 \}}$${\ displaystyle n}$${\ displaystyle 3}$${\ displaystyle \ #}$

${\ displaystyle | \ #G_ {1} (n) - \ # U_ {1} (n) | \ mod 3 = 0 \ implies n {\ text {is divisible by}} 3}$

In words, a binary number is divisible by 3 without a remainder if the difference in amount between the number of ones in the even positions and the number of ones in the odd positions is divisible by 3.

Example on the number : ${\ displaystyle n = 744628179621 _ {(10)}}$

The number has the following binary representation . It is and and actually${\ displaystyle n = 1010110101011111010011000101001010100101 _ {(2)}}$${\ displaystyle | \ #G_ {1} (n) - \ # U_ {1} (n) | = | 9-12 | = 3}$${\ displaystyle 3 \ mod 3 = 0}$${\ displaystyle 744628179621: 3 = 248209393207}$

format: plain


text:
Freenode is dead -- Long live Libera.Chat!

Freenode is dead. That’s the unfortunate truth. On May 19, an event regularly described as “hostile takeover”, in which the control over the biggest IRC network for free, open-source software has been forcefully taken away from the volunteer group called Freenode Staff who have run it ever since it was founded about 20 years ago, has taken place.

Therefore, all the communities I am involved in and/or manage on Freenode are moving or have already moved to the new network which most of the former Freenode staff have founded: Libera.Chat. This includes:

• AppImage with the channel #appimage
• NewPipe with the channel #newpipe
• Blue Nebula with the channel #blue-nebula

I do not have time to provide a nice summary on the topic as I have previously, e.g., for the youtube-dl incident. In the past days, I have analyzed the available resources (i.e., blog posts, pasted chat logs, and other publicly available information from both sides), and I support the former Freenode Staff in this matter. Large parts of their claims can be verified, something I cannot say about claims from the “new management”. The way this “new management” has literally grabbed power and then started to defame the old staff clearly shows that they don’t have the best intentions, not for the users, not for Freenode as a project, not for IRC.

I have curated some links to interesting resources such as summaries, news, blog posts, resignation letters etc., so read on if you want to take a look yourself.

Update: youtube-dl reinstantiated thanks to EFF

In October, the popular and famous archivist tool youtube-dl was taken down by GitHub due to a DMCA request by the Recording Industry Association of America. I posted a detailed analysis explaining the relation between international, US, EU and German law, and showing why this request was very different from what most people know as the “classic” DMCA takedown request. I felt this was necessary since there were a lot of rumor and wrong claims out on the Internet (not just in forums or on the (anti)social media, but also on media offerings), which did not help the discussion at all. The repository is back now, thanks to the EFF and its team. They filed a counter-notice on behalf of the youtube-dl maintainers, which allowed GitHub to reinstantiate the repository.


text:
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# A certain company that sells only cars and trucks reported

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A certain company that sells only cars and trucks reported [#permalink]

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13 Oct 2013, 00:08
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A certain company that sells only cars and trucks reported that in 2004, revenues from car sales were 20 % more than the revenues from truck sales. In 2005, if the revenues tram truck sales kept unchanged, were the revenues form car sales still greater than the revenue from truck sales?

(1) In 2005, the revenues from car sales were down less than 20 % from 2004.
(2) In 2005, the revenues from car sales were down more than 18 % from 2004.
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Re: A certain company that sells only cars and trucks reported [#permalink]

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A certain company that sells only cars and trucks reported that in 2004, revenues from car sales were 20 % more than the revenues from truck sales. In 2005, if the revenues tram truck sales kept unchanged, were the revenues form car sales still greater than the revenue from truck sales?

In 2004:
{revenues from truck sales} = x.
{revenues from car sales} = 1.2x

In 2005:
{revenues from truck sales} = x.
{revenues from car sales} = ?.

(1) In 2005, the revenues from car sales were down less than 20% from 2004. If the revenue from car sales were down by 0%, then the answer would be YES but if the revenue from car sales were down by 19%, then the answer would be NO (1.2x*0.81<x). Not sufficient.

(2) In 2005, the revenues from car sales were down more than 18% from 2004. Even if the revenue from car sales were down by only 18%, the revenue from car sales still would be less than x --> 1.2x*0.82<x. Sufficient.

_________________
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Joined: 05 Mar 2015
Posts: 962
Re: A certain company that sells only cars and trucks reported [#permalink]

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17 Jun 2016, 10:53
ICsharp wrote:
A certain company that sells only cars and trucks reported that in 2004, revenues from car sales were 20 % more than the revenues from truck sales. In 2005, if the revenues tram truck sales kept unchanged, were the revenues form car sales still greater than the revenue from truck sales?

(1) In 2005, the revenues from car sales were down less than 20 % from 2004.
(2) In 2005, the revenues from car sales were down more than 18 % from 2004.

take easy values for consideration
Let revenues from truck sales in 2004 =Rs 100
revenues from car sales=Rs 120 .
revenues from truck sales in 2005=Rs 100
revenues from car sales in 2005=?(<or> Rs100)
Statement(1) the revenues from car sales were down less than 20 % from 2004
say it was down 19%...19%less of 120=Rs 97.2 (97.2<100)...NO
or it was down 1%...1%less of 120=Rs 118.8 (118.8>100)...YES.
Not sufff....
(2) the revenues from car sales were down more than 18 % from 2004
even if you take a value 18.1% decreased it will be >Rs 100
120(1-18.1/100)=98.....it will always<100(truck sales revenue)
suff...
Ans B
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Re: A certain company that sells only cars and trucks reported [#permalink]

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17 Jun 2016, 14:13
ICsharp wrote:
A certain company that sells only cars and trucks reported that in 2004, revenues from car sales were 20 % more than the revenues from truck sales. In 2005, if the revenues tram truck sales kept unchanged, were the revenues form car sales still greater than the revenue from truck sales?

(1) In 2005, the revenues from car sales were down less than 20 % from 2004.
(2) In 2005, the revenues from car sales were down more than 18 % from 2004.

Let's say revenue from truck sales in 2004= 100
Revenue from car sales in 2004 = 120

(1) In 2005, the revenues from car sales were down less than 20 % from 2004.

If the revenues were 20% less. Then, the actual revenue would have been 120 - 120*20/100= 120-24

But since the revenues are down less then 20%, revenues can be 19% (<revenues by trucks) or 10% (>revenues by trucks)

Not sufficient.

(2) In 2005, the revenues from car sales were down more than 18 % from 2004.
if it were 18%, then the revenues would have been 120- 120*18/100= 120- 22 (approximately)

But since revenues were down more than 18 %; revenues were always less than 100, and less than revenues from trucks. Sufficient.

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Joined: 06 Oct 2013
Posts: 46
Re: A certain company that sells only cars and trucks reported [#permalink]

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29 Jun 2016, 12:24
ICsharp wrote:
A certain company that sells only cars and trucks reported that in 2004, revenues from car sales were 20 % more than the revenues from truck sales. In 2005, if the revenues tram truck sales kept unchanged, were the revenues form car sales still greater than the revenue from truck sales?

(1) In 2005, the revenues from car sales were down less than 20 % from 2004.
(2) In 2005, the revenues from car sales were down more than 18 % from 2004.
.

SOLUTION:

let t be the revenues from Trucks sales , and let c be the revenues from Cars sales .

In 2004, $$c=0.2*t--->t=0.83*c$$.

In 2005, t is same as in 2004.

Statement 1). In 2005, the revenues from car sales were down less than 20 % from 2004.

In 2005, new revenues from car sales= $$c-(<0.2*c)$$ = $$>0.8*c$$.
so it may or may not be less than $$0.83*c.$$

Hence Insufficient.

Statement 2) In 2005, the revenues from car sales were down more than 18 % from 2004.

In 2005 , new revenues from care sales=$$c-(>0.18*c) = <0.82*c.$$

it is always less than $$0.83*c$$.

Hence Sufficient.

Ans B
Manager
Joined: 03 Jun 2015
Posts: 86
A certain company that sells only cars and trucks reported [#permalink]

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16 Feb 2017, 01:08
2004 ---> C = 1.2 T
2005 ---> T is same and Is C still greater than T? Yes/No

1) C was down less than 20%.
Say C IS down by 20%

In 2004 ---> Suppose T = 100 then C will be 120.
In 2005 ---> C is down by 20% so C = 96 and less than 20% will make C>96.
We know T = 100 (remains unchanged)
So if C=97 and T=100....its NO (C is not greater than T)
if C=101 and T=100.....its YES (C is greater than T)
INSUFF

2) C is down more than 18%.
Again say C IS down by 20%.

From above w.k.t C=96 when its down by 20% (which is clearly more than 18%) and if continues to be more down then C<96.....Always NO
Hence SUFF

Option B
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A certain company that sells only cars and trucks reported   [#permalink] 16 Feb 2017, 01:08
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text:
# Idemtracker between article or inbook and book with cite command in biblatex

I have a problem when I cite an article following a book, it seems that the idemtracker doesn't work for documentclass as article or inbook.

Example : biblioessai.tex :

\documentclass[a4paper,12pt]{book}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{csquotes}
\usepackage[french]{babel}

\begin{document}
\cite{article1}

\cite{article2}

\cite{book}
\end{document}

essai.bib :

@article{article1,
author = "My Author",
journal = "A Journal",
title = "Titletitle",
date = "2005",
pages = "49-62",
volume = "12",
number = "1-2"
}

@book{book,
author = "My Author",
origyear = "1994",
pagetotal = "349",
publisher = "A publisher",
title = "An other title",
year = "2004"
}

@article{article2,
author = "My Author",
journal = "A Journal",
title = "An other article",
date = "2008",
pages = "51-63",
volume = "1",
number = "15"
}

After compilation i have :

My Author. Titletitle. Dans : A Journal 12.1-2 (2005)
My Author. An other article. Dans : A Journal 1.15 (2008)
My Author. An other title. Address : A publisher, 2004

And what I would like :

My Author. Titletitle. Dans : A Journal 12.1-2 (2005)
Idem. An other article. Dans : A Journal 1.15 (2008)
Idem. An other title. Address : A publisher, 2004

Any idea? Thanks

• Are you sure you want that bibstyle? It's a bit inconsistent to have a nearly fullcite but only shorten the author with idem. That's why the verbose-styles don't do that per default. (See my answer.) – musicman Jul 2 '14 at 9:19
• I am writing a thesis in french and unfortunately, the french norms are very different from american norms, maybe that's why this surprise you. But, in our norms, we have to present two cites of the same auther in this way even if the reference was not cited before. We use "idem" when different references of the same author are cited consicutively. – user56436 Jul 2 '14 at 12:02
• I understand, see my updated answer. – musicman Jul 2 '14 at 12:44

The "problem" here is that verbose-trad2 always prints a fullcite if a cite is seen for the very first time. After that idem/ibidem etc. will work.

See the definition of bibmacro cite in verbose-trad2.cbx.

So if you repeat in your code

\cite{article1}
\cite{article2}
\cite{book}

the idem will show up.

EDIT: Since you need the idem to show up fullcites, too, it's maybe best to modify the author macro itself:

\usepackage{xpatch}

This will exchange the \printnames{author} command from authortitle.bbx (which is used by verbose-trad2) with a test if it's a idem or not and use the appropriate macro.
I don't know if it's the best solution, but for your MWE it works. Maybe you have also to exchange the editor macro and others as well.


text:
# what evidence is revealed that Victor intends to destroy the creature ? Chapter22

Chapter22 the book 's everbind


text:
# 7.9.5: Limitations On The Use of Negative Feedback

From the fore-going discussion, you may well think that negative feedback can do just about anything, short of curing a rainy day. Such is not the case. Yes, negative feedback can drastically lower distortion and increase bandwidth. Yes, it can have a very profound effect on input and output impedance. And yes, it certainly does stabilize our gains. What then, is the problem? Like all things, negative feedback has specific limitations. The first thing you should note is that $$S$$ is a function of frequency. This was graphically depicted back in Figure 3.3.2. The amount of change seen in impedances and distortion is a function of $$S$$, it follows that these changes must be a function of frequency. Because $$S$$ drops as the frequency increases, the effects of negative feedback diminish as well. For example, if an SP amplifier has an open-loop $$Z_{in}$$ of 200 k$$\Omega$$ and the low frequency $$S$$ is 500, the resulting $$Z_{in}$$ with feedback is 100 M$$\Omega$$. If we increase the input frequency past the open-loop $$f_2$$, the open-loop gain drops and thus, $$S$$ drops. One decade up, $$S$$ will only be 50, so the $$Z_{in}$$ with feedback will only be 10 M$$\Omega$$. If this amplifier has a lower break frequency ($$f_1$$), $$S$$ will drop as the frequency is reduced (below $$f_1$$). The same sort of thing occurs with distortion, however, the harmonics each see a different $$S$$, so the calculation is a bit more involved. Along with the reduction in gain, there is also a change in phase. If the phase around the feedback loop varies from -180$$^{\circ}$$, incomplete cancellation takes place, and thus, the effects of feedback are lessened. The bottom line is that the effects of negative feedback weaken as we approach the frequency extremes.

The other item that must be kept in mind is the fact that negative feedback does not change specific fundamental characteristics of the amplifier. Negative feedback cannot get a circuit to do something beyond its operational parameters. For example, feedback has no effect on clipping level (saturation point). Further, feedback has no effect on slew rate (the maximum rate of output signal change, and an item that we will examine in a later chapter). Actually, when an amplifier slews, feedback is effectively blocked. An accurate output signal is no longer sent back to the input, but the amplifier can't correct for the errors any faster than it already is. In a similar manner, even though feedback can be used to lower the output impedance of a system, this does not imply that the system can produce more output current.

One possible “problem” with negative feedback is really the fault of the designer. It can be very tempting to sloppily design an amplifier with poor characteristics and then correct for them with large amounts of feedback. No matter how much feedback you use, the result will never be as good as a system that was designed carefully from the start. An example of this effect can be seen with TIMD (Transient Inter-Modulation Distortion). TIMD is a function of non-linearities in the first stages of an amplifier, and the excessive application of negative feedback will not remove it. On the other hand, if the initial stages of the system are properly designed, TIMD is not likely to be a problem.1

## References

1See E.M.Cherry and K.P.Dabke, “Transient Intermodulation Distortion- Part 2: Soft Nonlinearity”, Journal of the Audio Engineering Society, Vol.34 No.1/2 (1986): 19–35

This page titled 7.9.5: Limitations On The Use of Negative Feedback is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by James M. Fiore.


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# An object with a mass of  3 kg is lying still on a surface and is compressing a horizontal spring by 60 c m. If the spring's constant is  2 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Mar 20, 2018

The coefficient of static friction is $= 0.04$

#### Explanation:

The mass is $m = 3 k g$

The compression of the spring is $x = 0.6 m$

The spring constant is $k = 2 k g {s}^{-} 2$

The reaction of the spring is $R = k x = 2 \cdot 0.6 = 1.2 N$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The normal reaction of the object is

$N = m g = 3 \cdot 9.8 = 29.4 N$

The coefficient of static friction is

${\mu}_{s} = \frac{R}{N} = \frac{1.2}{29.4} = 0.04$


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Colloquia

# On the interpretation of the Top Quark Mass from Reconstruction

## by Prof. André Hoang (Vienna University)

Friday, 19 January 2018 from to (Europe/Vienna)
 Description The currently most precise measurements of the top quark mass are obtained from kinematic reconstruction obtained at hadron colliders. The observables designed to make these measurements cannot be calculated using analytic methods in QCD and therefore reply on multipurpose Monte-Carlo event generators. As a consequence, the interpretation of the measured top quark masses is theoretically difficult and still an open issue. In my talk I explain the theoretical issues inherent to this problem. It is currently also not yet fully understood, up to a conceptual upper bound, how large the uncertainty is when one interprets the result as measurements of the pole mass. I then explain one approach to shed some light on the interpretation problem using the concept of MC calibration with top mass sensitive observables which can be computed systematically at hadron level using analytic methods for boosted top quarks.


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# LaTeX is putting a border around my figures

I'm trying to embed a few images in a figure, but LaTeX keeps putting a border around three of the sides. Is there any way for me to keep it from adding this border?

The code that I used to create this figure:

\begin{figure}
\centering
\subfloat[Top View]{\label{fig:pat1a}\includegraphics[width=0.5\textwidth]{pat1a}}
\subfloat[Isometric View]{\label{fig:pat1b}\includegraphics[width=0.5\textwidth]{pat1b}}
\caption{Diagrams from Patent Application 12/475,048}
\label{fig:pat1}
\end{figure}

I have included the following packages in my project:

• lineno
• microtype
• verbatim
• hyperref
• array
• graphicx
• subfig

If there's anything that I have to do in each specific figure to prevent this border from appearing, that's fine. However, if there's one configuration thing that I can do at the beginning of the document, that would be even better.

-
Welcome to TeX.sx! Please add a full but minimal working example (MWE) that illustrates your problem. We need to also know which packages you are using. A tip: If you indent lines by 4 spaces, they'll be marked as a code sample. You can also highlight the code and click the "code" button (with "{}" on it) or press CTRL+K. –  Martin Scharrer Nov 13 '11 at 21:38
Please make sure that all images are uploaded using the official stackexchange interface, i.e. the image icon on top of the text field (shortcut: CTRL+G). This ensures that all images are always accessible and do not expire. –  Martin Scharrer Nov 13 '11 at 21:38
Sorry about those formatting mistakes. I've added a list of all of the packages I am using. I'll get to work on a minimal working example now. –  Serplat Nov 13 '11 at 21:47
What package did you use to produce the images? What format are they in (raster, like BMP/PNG/JPG, or vector, like EPS/PDF)? –  Werner Nov 13 '11 at 21:50
They're just PNG images, but apparently they had a transparent border around them. See my answer below. –  Serplat Nov 13 '11 at 21:54

This could be because your images include the border already, before including them. If this is the case, you can trim and clip some of the edges of the included image using the trim=<lx> <ly> <ux> <uy>,clip option of \includegraphics. This trims <lx> from the left, <ly> from the bottom, <ux> from the right and <uy> from the top in big points bp. Here is a minimal example showing this effect:

\documentclass{article}
\usepackage{graphicx}% http://ctan.org/pkg/graphicx
\begin{document}
\begin{figure}
\includegraphics[width=0.3\linewidth,trim=4 4 4 4,clip]{patent.png}
\caption{Diagram from Patent Application 12/475,048}
\end{figure}
\end{document}

The image on the left is used unadjusted from a capture of your images (showing the original border included), while the image on the right has been trimmed/clipped by 4bp on all 4 sides (removing the added border).

-

I figured it out. I'm posting the solution here in-case anyone else has similar issues.

I had saved the images directly out of PowerPoint using "Save as Picture" in the Right-Click menu. This apparently puts a small 1 pixel transparent border around the images.

I suppose LaTeX doesn't render this border transparently, so it was drawing the gray border instead. Cropping the image to remove this transparent border solved my issues.

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The cropping can also be performed within LaTeX. See my answer on how to do this. –  Werner Nov 13 '11 at 21:57
Awesome, thanks. –  Serplat Nov 13 '11 at 22:01

This also happens when you're using \floatstyle{boxed}. If that's the case the solution is to return to plain style:

\floatstyle{plain}
\restylefloat{figure}

As described in Floatstyle how to come back to the default theme

-

draw a box outside of the figure

\begin{figure}
\centering
\subfloat[Top View]{\label{fig:pat1a}\includegraphics[width=0.5\textwidth]{pat1a}}
\setlength\fboxsep{0pt}% distance from figure
\setlength\fboxrule{0.75pt}% border thickness
\fbox{
\subfloat[Isometric View]{\label{fig:pat1b}\includegraphics[width=0.5\textwidth]{pat1b}}
}
\caption{Diagrams from Patent Application 12/475,048}
\label{fig:pat1}
\end{figure}

-


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# Double Coverings of the Double Torus

I'm trying to count all the double coverings of the double torus. I know that the fundamental group of the double torus is

$$\pi_1(X)=\langle a,b,c,d;[a,b][c,d]\rangle$$

where $[a,b]=aba^{-1}b^{-1}$. I also know from the classification theorem for covering spaces that in fact it suffices to count subgroups of index 2. I'm not sure how to do this however.

Here's an idea I've just had - it suffices to count surjective homomorphisms $\pi_1(X)\rightarrow C_2$. These are precisely the maps which send at least one of $a,b,c,d$ to the generator $r$ of $C_2$. Indeed it's easily checked that any such map is a homomorphism. But there are exactly $2^4 -1=15$ such maps. Hence there are 15 double covers of the double torus. Does this sound plausible? To me $15$ seems a bit large...!

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That is not the fundamental group of the double torus. –  Chris Eagle May 13 '12 at 23:20
Your counting argument still works, though. –  Chris Eagle May 13 '12 at 23:22
Sorry - my mistake, I'll edit the question so the fundamental group is right. –  Edward Hughes May 13 '12 at 23:33
Yes thanks - my brain is clearly not working at this time of night! Thanks for verifying the counting argument though! –  Edward Hughes May 13 '12 at 23:35
@LeeMosher: but the subgroups are index 2; doesn't that mean they're automatically normal, hence stabilised by conjugation? –  Ben Millwood Jun 2 '12 at 18:01


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### Section 9-1 : The Heat Equation

Before we get into actually solving partial differential equations and before we even start discussing the method of separation of variables we want to spend a little bit of time talking about the two main partial differential equations that we’ll be solving later on in the chapter. We’ll look at the first one in this section and the second one in the next section.

The first partial differential equation that we’ll be looking at once we get started with solving will be the heat equation, which governs the temperature distribution in an object. We are going to give several forms of the heat equation for reference purposes, but we will only be really solving one of them.

We will start out by considering the temperature in a 1-D bar of length $$L$$. What this means is that we are going to assume that the bar starts off at $$x = 0$$ and ends when we reach $$x = L$$. We are also going to so assume that at any location, $$x$$ the temperature will be constant at every point in the cross section at that $$x$$. In other words, temperature will only vary in $$x$$ and we can hence consider the bar to be a 1-D bar. Note that with this assumption the actual shape of the cross section (i.e. circular, rectangular, etc.) doesn’t matter.

Note that the 1-D assumption is actually not all that bad of an assumption as it might seem at first glance. If we assume that the lateral surface of the bar is perfectly insulated (i.e. no heat can flow through the lateral surface) then the only way heat can enter or leave the bar as at either end. This means that heat can only flow from left to right or right to left and thus creating a 1-D temperature distribution.

The assumption of the lateral surfaces being perfectly insulated is of course impossible, but it is possible to put enough insulation on the lateral surfaces that there will be very little heat flow through them and so, at least for a time, we can consider the lateral surfaces to be perfectly insulated.

Okay, let’s now get some definitions out of the way before we write down the first form of the heat equation.

\begin{align*}&u\left( {x,t} \right) = {\mbox{Temperature at any point }}x{\mbox{ and any time }}t\\ & c\left( x \right) = {\mbox{Specific Heat}}\\ & \rho \left( x \right) = {\mbox{Mass Density}}\\ & \varphi \left( {x,t} \right) = {\mbox{Heat Flux}}\\ & Q\left( {x,t} \right) = {\mbox{Heat energy generated per unit volume per unit time}}\end{align*}

We should probably make a couple of comments about some of these quantities before proceeding.

The specific heat, $$c\left( x \right) > 0$$, of a material is the amount of heat energy that it takes to raise one unit of mass of the material by one unit of temperature. As indicated we are going to assume, at least initially, that the specific heat may not be uniform throughout the bar. Note as well that in practice the specific heat depends upon the temperature. However, this will generally only be an issue for large temperature differences (which in turn depends on the material the bar is made out of) and so we’re going to assume for the purposes of this discussion that the temperature differences are not large enough to affect our solution.

The mass density, $$\rho \left( x \right)$$, is the mass per unit volume of the material. As with the specific heat we’re going to initially assume that the mass density may not be uniform throughout the bar.

The heat flux, $$\varphi \left( {x,t} \right)$$, is the amount of thermal energy that flows to the right per unit surface area per unit time. The “flows to the right” bit simply tells us that if $$\varphi \left( {x,t} \right) > 0$$ for some $$x$$ and $$t$$ then the heat is flowing to the right at that point and time. Likewise, if $$\varphi \left( {x,t} \right) < 0$$ then the heat will be flowing to the left at that point and time.

The final quantity we defined above is $$Q\left( {x,t} \right)$$ and this is used to represent any external sources or sinks (i.e. heat energy taken out of the system) of heat energy. If $$Q\left( {x,t} \right) > 0$$ then heat energy is being added to the system at that location and time and if $$Q\left( {x,t} \right) < 0$$ then heat energy is being removed from the system at that location and time.

With these quantities the heat equation is,

$$$c\left( x \right)\rho \left( x \right)\frac{{\partial u}}{{\partial t}} = - \frac{{\partial \varphi }}{{\partial x}} + Q\left( {x,t} \right)\label{eq:eq1}$$$

While this is a nice form of the heat equation it is not actually something we can solve. In this form there are two unknown functions, $$u$$ and $$\varphi$$, and so we need to get rid of one of them. With Fourier’s law we can easily remove the heat flux from this equation.

Fourier’s law states that,

$\varphi \left( {x,t} \right) = - {K_0}\left( x \right)\frac{{\partial u}}{{\partial x}}$

where $${K_0}\left( x \right) > 0$$ is the thermal conductivity of the material and measures the ability of a given material to conduct heat. The better a material can conduct heat the larger $${K_0}\left( x \right)$$ will be. As noted the thermal conductivity can vary with the location in the bar. Also, much like the specific heat the thermal conductivity can vary with temperature, but we will assume that the total temperature change is not so great that this will be an issue and so we will assume for the purposes here that the thermal conductivity will not vary with temperature.

Fourier’s law does a very good job of modeling what we know to be true about heat flow. First, we know that if the temperature in a region is constant, i.e.$$\frac{{\partial u}}{{\partial x}} = 0$$, then there is no heat flow.

Next, we know that if there is a temperature difference in a region we know the heat will flow from the hot portion to the cold portion of the region. For example, if it is hotter to the right then we know that the heat should flow to the left. When it is hotter to the right then we also know that $$\frac{{\partial u}}{{\partial x}} > 0$$ (i.e. the temperature increases as we move to the right) and so we’ll have $$\varphi < 0$$ and so the heat will flow to the left as it should. Likewise, if $$\frac{{\partial u}}{{\partial x}} < 0$$ (i.e. it is hotter to the left) then we’ll have $\varphi > 0$ and heat will flow to the right as it should.

Finally, the greater the temperature difference in a region (i.e. the larger $$\frac{{\partial u}}{{\partial x}}$$ is) then the greater the heat flow.

So, if we plug Fourier’s law into $$\eqref{eq:eq1}$$, we get the following form of the heat equation,

$$$c\left( x \right)\rho \left( x \right)\frac{{\partial u}}{{\partial t}} = \frac{\partial }{{\partial x}}\left( {{K_0}\left( x \right)\frac{{\partial u}}{{\partial x}}} \right) + Q\left( {x,t} \right)\label{eq:eq2}$$$

Note that we factored the minus sign out of the derivative to cancel against the minus sign that was already there. We cannot however, factor the thermal conductivity out of the derivative since it is a function of $$x$$ and the derivative is with respect to $$x$$.

Solving $$\eqref{eq:eq2}$$ is quite difficult due to the non uniform nature of the thermal properties and the mass density. So, let’s now assume that these properties are all constant, i.e.,

$c\left( x \right) = c\hspace{0.25in}\rho \left( x \right) = \rho \hspace{0.25in}{K_0}\left( x \right) = {K_0}$

where $$c$$, $$\rho$$ and $${K_0}$$ are now all fixed quantities. In this case we generally say that the material in the bar is uniform. Under these assumptions the heat equation becomes,

$$$c\rho \frac{{\partial u}}{{\partial t}} = {K_0}\frac{{{\partial ^2}u}}{{\partial {x^2}}} + Q\left( {x,t} \right)\label{eq:eq3}$$$

For a final simplification to the heat equation let’s divide both sides by $$c\rho$$ and define the thermal diffusivity to be,

$k = \frac{{{K_0}}}{{c\rho }}$

The heat equation is then,

$$$\frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{Q\left( {x,t} \right)}}{{c\rho }}\label{eq:eq4}$$$

To most people this is what they mean when they talk about the heat equation and in fact it will be the equation that we’ll be solving. Well, actually we’ll be solving $$\eqref{eq:eq4}$$ with no external sources, i.e.$$Q\left( {x,t} \right) = 0$$, but we’ll be considering this form when we start discussing separation of variables in a couple of sections. We’ll only drop the sources term when we actually start solving the heat equation.

Now that we’ve got the 1-D heat equation taken care of we need to move into the initial and boundary conditions we’ll also need in order to solve the problem. If you go back to any of our solutions of ordinary differential equations that we’ve done in previous sections you can see that the number of conditions required always matched the highest order of the derivative in the equation.

In partial differential equations the same idea holds except now we have to pay attention to the variable we’re differentiating with respect to as well. So, for the heat equation we’ve got a first order time derivative and so we’ll need one initial condition and a second order spatial derivative and so we’ll need two boundary conditions.

The initial condition that we’ll use here is,

$u\left( {x,0} \right) = f\left( x \right)$

and we don’t really need to say much about it here other than to note that this just tells us what the initial temperature distribution in the bar is.

The boundary conditions will tell us something about what the temperature and/or heat flow is doing at the boundaries of the bar. There are four of them that are fairly common boundary conditions.

The first type of boundary conditions that we can have would be the prescribed temperature boundary conditions, also called Dirichlet conditions. The prescribed temperature boundary conditions are,

$u\left( {0,t} \right) = {g_1}\left( t \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}u\left( {L,t} \right) = {g_2}\left( t \right)$

The next type of boundary conditions are prescribed heat flux, also called Neumann conditions. Using Fourier’s law these can be written as,

$- {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = {\varphi _1}\left( t \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = {\varphi _2}\left( t \right)$

If either of the boundaries are perfectly insulated, i.e. there is no heat flow out of them then these boundary conditions reduce to,

$\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = 0$

and note that we will often just call these particular boundary conditions insulated boundaries and drop the “perfectly” part.

The third type of boundary conditions use Newton’s law of cooling and are sometimes called Robins conditions. These are usually used when the bar is in a moving fluid and note we can consider air to be a fluid for this purpose.

Here are the equations for this kind of boundary condition.

$- {K_0}\left( 0 \right)\frac{{\partial u}}{{\partial x}}\left( {0,t} \right) = - H\left[ {u\left( {0,t} \right) - {g_1}\left( t \right)} \right]\hspace{0.25in} - {K_0}\left( L \right)\frac{{\partial u}}{{\partial x}}\left( {L,t} \right) = H\left[ {u\left( {L,t} \right) - {g_2}\left( t \right)} \right]$

where $$H$$ is a positive quantity that is experimentally determined and $${g_1}\left( t \right)$$ and $${g_2}\left( t \right)$$ give the temperature of the surrounding fluid at the respective boundaries.

Note that the two conditions do vary slightly depending on which boundary we are at. At $$x = 0$$ we have a minus sign on the right side while we don’t at $$x = L$$. To see why this is let’s first assume that at $$x = 0$$ we have $$u\left( {0,t} \right) > {g_1}\left( t \right)$$. In other words, the bar is hotter than the surrounding fluid and so at $$x = 0$$ the heat flow (as given by the left side of the equation) must be to the left, or negative since the heat will flow from the hotter bar into the cooler surrounding liquid. If the heat flow is negative then we need to have a minus sign on the right side of the equation to make sure that it has the proper sign.

If the bar is cooler than the surrounding fluid at $$x = 0$$, i.e. $$u\left( {0,t} \right) < {g_1}\left( t \right)$$ we can make a similar argument to justify the minus sign. We’ll leave it to you to verify this.

If we now look at the other end, $$x = L$$, and again assume that the bar is hotter than the surrounding fluid or, $$u\left( {L,t} \right) > {g_2}\left( t \right)$$. In this case the heat flow must be to the right, or be positive, and so in this case we can’t have a minus sign. Finally, we’ll again leave it to you to verify that we can’t have the minus sign at $$x = L$$ is the bar is cooler than the surrounding fluid as well.

Note that we are not actually going to be looking at any of these kinds of boundary conditions here. These types of boundary conditions tend to lead to boundary value problems such as Example 5 in the Eigenvalues and Eigenfunctions section of the previous chapter. As we saw in that example it is often very difficult to get our hands on the eigenvalues and as we’ll eventually see we will need them.

It is important to note at this point that we can also mix and match these boundary conditions so to speak. There is nothing wrong with having a prescribed temperature at one boundary and a prescribed flux at the other boundary for example so don’t always expect the same boundary condition to show up at both ends. This warning is more important that it might seem at this point because once we get into solving the heat equation we are going to have the same kind of condition on each end to simplify the problem somewhat.

The final type of boundary conditions that we’ll need here are periodic boundary conditions. Periodic boundary conditions are,

$u\left( { - L,t} \right) = u\left( {L,t} \right)\hspace{0.25in}\hspace{0.25in}\frac{{\partial u}}{{\partial x}}\left( { - L,t} \right) = \frac{{\partial u}}{{\partial x}}\left( {L,t} \right)$

Note that for these kinds of boundary conditions the left boundary tends to be $$x = - L$$ instead of $$x = 0$$ as we were using in the previous types of boundary conditions. The periodic boundary conditions will arise very naturally from a couple of particular geometries that we’ll be looking at down the road.

We will now close out this section with a quick look at the 2-D and 3-D version of the heat equation. However, before we jump into that we need to introduce a little bit of notation first.

The del operator is defined to be,

$\nabla = \frac{\partial }{{\partial x}}\vec i + \frac{\partial }{{\partial y}}\vec j\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\nabla = \frac{\partial }{{\partial x}}\vec i + \frac{\partial }{{\partial y}}\vec j + \frac{\partial }{{\partial z}}\vec k$

depending on whether we are in 2 or 3 dimensions. Think of the del operator as a function that takes functions as arguments (instead of numbers as we’re used to). Whatever function we “plug” into the operator gets put into the partial derivatives.

So, for example in 3-D we would have,

$\nabla f = \frac{{\partial f}}{{\partial x}}\vec i + \frac{{\partial f}}{{\partial y}}\vec j + \frac{{\partial f}}{{\partial z}}\vec k$

This of course is also the gradient of the function $$f\left( {x,y,z} \right)$$.

The del operator also allows us to quickly write down the divergence of a function. So, again using 3-D as an example the divergence of $$f\left( {x,y,z} \right)$$ can be written as the dot product of the del operator and the function. Or,

$\nabla \centerdot f = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial z}}$

Finally, we will also see the following show up in our work,

$\nabla \centerdot \left( {\nabla f} \right) = \frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + \frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial y}}} \right) + \frac{\partial }{{\partial z}}\left( {\frac{{\partial f}}{{\partial k}}} \right) = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$

This is usually denoted as,

${\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$

and is called the Laplacian. The 2-D version of course simply doesn’t have the third term.

Okay, we can now look into the 2-D and 3-D version of the heat equation and where ever the del operator and or Laplacian appears assume that it is the appropriate dimensional version.

The higher dimensional version of $$\eqref{eq:eq1}$$ is,

$$$c\rho \frac{{\partial u}}{{\partial t}} = - \nabla \centerdot \,\varphi + Q\label{eq:eq5}$$$

and note that the specific heat, $$c$$, and mass density, $$\rho$$, are may not be uniform and so may be functions of the spatial variables. Likewise, the external sources term, $$Q$$, may also be a function of both the spatial variables and time.

Next, the higher dimensional version of Fourier’s law is,

$\varphi = - {K_0}\nabla u$

where the thermal conductivity, $${K_0}$$, is again assumed to be a function of the spatial variables.

If we plug this into $$\eqref{eq:eq5}$$ we get the heat equation for a non uniform bar (i.e. the thermal properties may be functions of the spatial variables) with external sources/sinks,

$c\rho \frac{{\partial u}}{{\partial t}} = \nabla \centerdot \,\left( {{K_0}\nabla u} \right) + Q$

If we now assume that the specific heat, mass density and thermal conductivity are constant (i.e. the bar is uniform) the heat equation becomes,

$$$\frac{{\partial u}}{{\partial t}} = k{\nabla ^2}u + \frac{Q}{{cp}}\label{eq:eq6}$$$

where we divided both sides by $$c\rho$$ to get the thermal diffusivity, $$k$$ in front of the Laplacian.

The initial condition for the 2-D or 3-D heat equation is,

$u\left( {x,y,t} \right) = f\left( {x,y} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\hspace{0.25in}u\left( {x,y,z,t} \right) = f\left( {x,y,z} \right)$

depending upon the dimension we’re in.

The prescribed temperature boundary condition becomes,

$u\left( {x,y,t} \right) = T\left( {x,y,t} \right)\hspace{0.25in}\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\hspace{0.25in}u\left( {x,y,z,t} \right) = T\left( {x,y,z,t} \right)$

where $$\left( {x,y} \right)$$ or $$\left( {x,y,z} \right)$$, depending upon the dimension we’re in, will range over the portion of the boundary in which we are prescribing the temperature.

The prescribed heat flux condition becomes,

$- {K_0}\nabla u\,\centerdot \,\vec n = \varphi \left( t \right)$

where the left side is only being evaluated at points along the boundary and $$\vec n$$ is the outward unit normal on the surface.

Newton’s law of cooling will become,

$- {K_0}\nabla u\,\centerdot \,\vec n = H\left( {u - {u_B}} \right)$

where $$H$$ is a positive quantity that is experimentally determine, $${u_B}$$ is the temperature of the fluid at the boundary and again it is assumed that this is only being evaluated at points along the boundary.

We don’t have periodic boundary conditions here as they will only arise from specific 1‑D geometries.

We should probably also acknowledge at this point that we’ll not actually be solving $$\eqref{eq:eq6}$$ at any point, but we will be solving a special case of it in the Laplace’s Equation section.


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# Maxwell Distribution of speeds

1. Jun 8, 2015

### Qwurty2.0

1. The problem statement, all variables and given/known data
A gas consisting of 15,200 molecules, each of mass 2.00 x 10-26 kg, has the following distribution of speeds, which crudely mimics the Maxwell distribution:

Number of Molecules - Speed (m/s)
1600 - 220
4100 - 440
4700 - 660
3100 - 880
1300 - 1100
400 - 1320

(a) Determine vrms for this distribution of speeds.

2. Relevant equations
vrms = √(2 * Ek/m)
Ek = (1/2) * n * M * v2

3. The attempt at a solution
Weighted Average
((1600 * 220) + (4100 * 440) + (4700 * 660) + (3100 * 880) + (1300 * 1100) + (400 * 1320)) / 15200
= (9944000 m/s) / 15200
= 654.32 m/s

vrms = √(2 * Ek/m)
Ek = (1/2) * n * M * v2
n = number of moles
M = molar mass
m = mass of single molecule

n = 15200 molecules / 6.02x1023 molecules/mole
= 2.525x10-20 mole

M = 2.00x10-26 kg / molecule * 6.02x1023 molecules / mole
= 0.01204 kg / mol

Ek = (1/2)(2.525x10-20moles)(0.01204 kg / mole)(654.21 m/s)2
= 6.506x10-17kg⋅m/s

vrms = √(2 * (6.506x10-17 kg⋅m/s) / 2.00x10-26 kg)
= 80659.78 m/s ...

The correct answer is 710 m/s so obviously I am either grossly overcomplicating this, or I am using the wrong equation.

2. Jun 8, 2015

### vela

Staff Emeritus
RMS means "root mean square". It's the square root of the mean of the square, so take the velocities and square them, find the weighted average of the squares, and then take the square root of the average.

3. Jun 8, 2015

### Qwurty2.0

I did what you said: I squared the velocities, then I calculated the weighted average (same way as I did above), and then took the square root of the weighted average. I ended up getting 82914.03 m/s, which is close to the answer I originally got but not the correct answer.

What am I missing?

4. Jun 8, 2015

### vela

Staff Emeritus
I got 707 m/s. I gather you're still messing around with calculating the kinetic energy, etc. That's unnecessary. The rms speed depends only on the velocity distribution of the molecules.

Your mistake seems to be that you're calculating the total internal energy of the gas, but $E_k$ is supposed to be the average kinetic energy of a single molecule.

5. Jun 9, 2015

### Qwurty2.0

My bad, I calculated it wrong. I tried again and got the right answer.

Thank you!


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Suppose $f(x) = \begin{cases} 0, & \text{if x is rational},\\ x^2, & \text{if x is irrational} \end{cases}$. Prove that $\lim_{x\to 0} f(x) = 0$

Suppose $f(x) = \begin{cases} 0, & \text{if x is rational},\\ x^2, & \text{if x is irrational} \end{cases}$

Prove that $\lim_{x\to 0} f(x) = 0$

Edit:

My attempt below, though I'm not sure if this approach is correct from the on-set.

For irrational $x$:

$\lim_{\delta x\to 0} f(x) = \frac{d}{dx} x^2 = 2x$

Hence, at $x=0$,

$\lim_{x\to 0} f(x) = \lim_{\delta x\to 0} f(0) = 2(0) = 0$ (proved?)

What about for rational $x$? o.O

My background: I don't have a degree in math, just an undergrad training in engineering. Not sure where the question was referenced from exactly too, but I would peg it at around high school AP/freshmen level. I'm asking this out of an interest in the topic.

• You may had written it in better form: Show that $f$ is continuous and differentiable only at $0$ with limit and derivative $0$ – Groups Oct 2 '15 at 5:48

This proof is pretty easy in my opinion, and can be explained verbally (this is effectively the definition of the limit). Let's say we choose a number very close to $0$. Either the number is rational (and yields $0$), or the number is irrational (and yields a number that is very small). Thus, if we keep choosing irrational numbers closer to $0$ we approach $0$, and if we choose rational numbers close to $0$ we get $0$ exactly. Combining these, as x approaches $0$ the function approaches $0$.
Simply: compare $f(x)$ to $g(x) = 0$ and $h(x) = x^2$ and use the Squeeze Theorem.
Let $\varepsilon > 0$; then $f(x) \leq x^{2} < \varepsilon$ if $|x| < \sqrt{\varepsilon}$, so taking $\delta := \sqrt{\varepsilon}$ suffices.


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# Is it possible to make a grammar LL($1$) which recognizes palindroms?

Is it possible to make an algebraic grammar LL($$1$$) which recognizes palindroms for an alphabet $$\{a,b\}$$?

The language of even-length palindromes $$L_{epal}=\{ww^R \mid w\in\Sigmạ^*\}$$ is a classic example of a non-deterministic context-free language, and you can find the outline of a proof in Hopcroft & Ullman (and other standard texts). The restriction to even-length palindromes simplifies the proof, but it can be extended to the language of all palindromes $$L_{pal}=\{w \mid w=w^R, w\in\Sigmạ^* \}$$. Without doing more work, though, it's evident that $$L_{epal}=L_{pal}\cap (\Sigma\Sigma)^*$$, and deterministic CFLs are closed against intersection with regular languages.


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# Continuity and differentiability of elementary functions

Given a single-variable elementary function (not piecewise), I was wondering if it is continuous and differentiable in all of its maximum domain? (Not considering the trivial examples involving absolute value function)

Also by piecewise I mean functions that can only be expressed in a piecewise manner. E.g. $\cos x$ is not considered piecewise, although it can be expressed in a piecewise manner.

I think the definition of elementary function that I’m going for is the one found on wikipedia: https://en.m.wikipedia.org/wiki/Elementary_function I think it is also required for the function’s domain to at least contain some interval, but I’m not sure.

Actually I know that if an elementary function is differentiable at some point, its derivative at that point is also given by an elementary function, but I am trying to find out if every elementary function is smooth given that it is defined at some point.

I haven’t been able to find anything useful online, so a rigorous proof would be greatly appreciated. Also please try to keep it as simple as possible as my math is bad.

• $f(x)=\sqrt x$ at $x=0$
• $f(x)=\sqrt[3] x$ at $x=0$
• $f(x)=\arcsin x$ at $x=1$


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# 13. + -1 points Find functions f(x) and g(x) so the given function can be expressed as h(x) = f(g(x)). (Use non-trivial functions for f(x) and g(x).) tx), g))

Question

Find function f(x) and g(x)


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Math Help - function of a derivative

1. function of a derivative

find a function whose derivative is 20x^3 plus 6x^2 plus 5

2. Originally Posted by eduk8tedidiot
find a function whose derivative is 20x^3 plus 6x^2 plus 5

What you want to find here is the 'anti-derivative' of the function. The integral of a derivative gives you the antiderivative.

Hence, if we are seeking function $f(x)$, then

$f(x) = \int 20x^3 + 6x^2 + 5 \, dx$

3. Originally Posted by eduk8tedidiot
find a function whose derivative is 20x^3 plus 6x^2 plus 5

Have you heard of an integral? Well anyway, we're looking for the anti-derivative of $20x^3+6x+62+5$

In order to get an $x^3$ the power rule tells us we need an $x^4$ but the derivative of $x^4=4x^3$ so we'd have a 4 out in front when we want a 20, so let's put a 5 out in front of our original guess

Now the derivative of $5x^4=20x^3$ so the first term is done....

Doing similar steps will lead you to an answer of $20x^4+2x^3+5x+c$

Note that when we take the derivative of that, the +c has a derivative of zero, since the question is to find an antiderivative, you can choose any value for c as the answer

4. $f'(x) = 20x^3 + 6x^2 + 5$

$f(x) = 5x^4 + 2x^3 + 5x + C$

$C = konst$

$f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$

$f'(x) = a_1 + 2a_2x + 3a_3x^2 + ...+ na_nx^{n-1}$

$f(x) = C$ C = const. $\rightarrow f'(x) = 0$

$f(x) = kx^m \rightarrow f'(x) = m*k*x^{m-1}$

5. Originally Posted by josipive
$f'(x) = 20x^3 + 6x^2 + 5$

$f(x) = 5x^4 + 2x^3 + 5x + C$

$C = konst$

$f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$

$f'(x) = a_1 + 2a_2x + 3a_3x^2 + ...+ na_nx^{n-1}$

$f(x) = C$ C = const. $\rightarrow f'(x) = 0$

$f(x) = kx^m \rightarrow f'(x) = m*k*x^{m-1}$