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## définition - Probability interpretations
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# Probability interpretations
The word probability has been used in a variety of ways since it was first applied to the mathematical study of games of chance. Does probability measure the real, physical tendency of something to occur or is it a measure of how strongly one believes it will occur, or does it draw on both upon both these elements? In answering such questions, we interpret the probability values of probability theory.
There are two broad categories of probability interpretations which can be called "physical" and "evidential" probabilities. Physical probabilities, which are also called objective or frequency probabilities, are associated with random physical systems such as roulette wheels, rolling dice and radioactive atoms. In such systems, a given type of event (such as the dice yielding a six) tends to occur at a persistent rate, or "relative frequency", in a long run of trials. Physical probabilities either explain, or are invoked to explain, these stable frequencies. Thus talk about physical probability makes sense only when dealing with well defined random experiments.[citation needed] The two main kinds of theory of physical probability are frequentist accounts (such as those of Venn, Reichenbach and von Mises)[citation needed] and propensity accounts (such as those of Popper, Miller, Giere and Fetzer).[citation needed]
Evidential probability, also called Bayesian probability (or subjectivist probability), can be assigned to any statement whatsoever, even when no random process is involved, as a way to represent its subjective plausibility, or the degree to which the statement is supported by the available evidence. On most accounts, evidential probabilities are considered to be degrees of belief, defined in terms of dispositions to gamble at certain odds. The four main evidential interpretations are the classical (e.g. Laplace's) interpretation, the subjective interpretation (de Finetti and Savage), the epistemic or inductive interpretation (Ramsey, Cox) and the logical interpretation (Keynes and Carnap).[citation needed]
Some interpretations of probability are associated with approaches to statistical inference, including theories of estimation and hypothesis testing. The physical interpretation, for example, is taken by followers of "frequentist" statistical methods, such as R. A. Fisher, Jerzy Neyman and Egon Pearson.[citation needed] Statisticians of the opposing Bayesian school typically accept the existence and importance of physical probabilities, but also consider the calculation of evidential probabilities to be both valid and necessary in statistics. This article, however, focuses on the interpretations of probability rather than theories of statistical inference.
The terminology of this topic is rather confusing, in part because probabilities are studied within so many different academic fields. The word "frequentist" is especially tricky. To philosophers it refers to a particular theory of physical probability, one that has more or less been abandoned. To scientists, on the other hand, "frequentist probability" is just what philosophers call physical (or objective) probability. Those who promote Bayesian inference view "frequentist statistics" as an approach to statistical inference that recognises only physical probabilities. Also the word "objective", as applied to probability, sometimes means exactly what "physical" means here, but is also used of evidential probabilities that are fixed by rational constraints, such as logical and epistemic probabilities.
“ It is unanimously agreed that statistics depends somehow on probability. But, as to what probability is and how it is connected with statistics, there has seldom been such complete disagreement and breakdown of communication since the Tower of Babel. Doubtless, much of the disagreement is merely terminological and would disappear under sufficiently sharp analysis. ” — Savage (1954), page 2
## Philosophy
The philosophy of probability presents problems chiefly in matters of epistemology and the uneasy interface between mathematical concepts and ordinary language as it is used by non-mathematicians. Probability theory is an established field of study in mathematics. It has its origins in correspondence discussing the mathematics of games of chance between Blaise Pascal and Pierre de Fermat in the seventeenth century, and was formalized and rendered axiomatic as a distinct branch of mathematics by Andrey Kolmogorov in the twentieth century. In its axiomatic form, mathematical statements about probability theory carry the same sort of epistemological confidence shared by other mathematical statements in the philosophy of mathematics.[1][2]
The mathematical analysis originated in observations of the behaviour of game equipment such as playing cards and dice, which are designed specifically to introduce random and equalized elements; in mathematical terms, they are subjects of indifference. This is not the only way probabilistic statements are used in ordinary human language: when people say that "it will probably rain", they typically do not mean that the outcome of rain versus not-rain is a random factor that the odds currently favor; instead, such statements are perhaps better understood as qualifying their expectation of rain with a degree of confidence. Likewise, when it is written that "the most probable explanation" of the name of Ludlow, Massachusetts "is that it was named after Roger Ludlow", what is meant here is not that Roger Ludlow is favored by a random factor, but rather that this is the most plausible explanation of the evidence, which admits other, less likely explanations.
Thomas Bayes attempted to provide a logic that could handle varying degrees of confidence; as such, Bayesian probability is an attempt to recast the representation of probabilistic statements as an expression of the degree of confidence by which the beliefs they express are held.
Though probability initially may have had lowly motivations, its modern influence and use is widespread ranging from medicine, through practical pursuits, all the way to the higher-order and the sublime.
## Classical definition
The first attempt at mathematical rigour in the field of probability, championed by Pierre-Simon Laplace, is now known as the classical definition. Developed from studies of games of chance (such as rolling dice) it states that probability is shared equally between all the possible outcomes, provided these outcomes can be deemed equally likely.[3]
The theory of chance consists in reducing all the events of the same kind to a certain number of cases equally possible, that is to say, to such as we may be equally undecided about in regard to their existence, and in determining the number of cases favorable to the event whose probability is sought. The ratio of this number to that of all the cases possible is the measure of this probability, which is thus simply a fraction whose numerator is the number of favorable cases and whose denominator is the number of all the cases possible.
Pierre-Simon Laplace, A Philosophical Essay on Probabilities[4]
The classical definition of probability works well for situations with only a finite number of equally-likely outcomes.
This can be represented mathematically as follows: If a random experiment can result in N mutually exclusive and equally likely outcomes and if NA of these outcomes result in the occurrence of the event A, the probability of A is defined by
$P(A) = {N_A \over N}$.
There are two clear limitations to the classical definition.[5] Firstly, it is applicable only to situations in which there is only a 'finite' number of possible outcomes. But some important random experiments, such as tossing a coin until it rises heads, give rise to an infinite set of outcomes. And secondly, you need to determine in advance that all the possible outcomes are equally likely without relying on the notion of probability to avoid circularity—for instance, by symmetry considerations.
## Frequentism
For frequentists, the probability of the ball landing in any pocket can be determined only by repeated trials in which the observed result converges to the underlying probability in the long run.
Frequentists posit that the probability of an event is its relative frequency over time,[3] i.e., its relative frequency of occurrence after repeating a process a large number of times under similar conditions. This is also known as aleatory probability. The events are assumed to be governed by some random physical phenomena, which are either phenomena that are predictable, in principle, with sufficient information (see Determinism); or phenomena which are essentially unpredictable. Examples of the first kind include tossing dice or spinning a roulette wheel; an example of the second kind is radioactive decay. In the case of tossing a fair coin, frequentists say that the probability of getting a heads is 1/2, not because there are two equally likely outcomes but because repeated series of large numbers of trials demonstrate that the empirical frequency converges to the limit 1/2 as the number of trials goes to infinity.
If we denote by $\textstyle n_a$ the number of occurrences of an event $\mathcal{A}$ in $\textstyle n$ trials, then if $\lim_{n \to \infty}{n_a \over n}=p$ we say that $\textstyle P(\mathcal{A})=p$
The frequentist view has its own problems. It is of course impossible to actually perform an infinity of repetitions of a random experiment to determine the probability of an event. But if only a finite number of repetitions of the process are performed, different relative frequencies will appear in different series of trials. If these relative frequencies are to define the probability, the probability will be slightly different every time it is measured. But the real probability should be the same every time. If we acknowledge the fact that we only can measure a probability with some error of measurement attached, we still get into problems as the error of measurement can only be expressed as a probability, the very concept we are trying to define. This renders even the frequency definition circular.[citation needed]
## Logical, epistemic, and inductive probability
It is widely recognised that the term "probability" is sometimes used in contexts where it has nothing to do with physical randomness.[citation needed] Consider, for example, the claim that the extinction of the dinosaurs was probably caused by a large meteorite hitting the earth. Statements such as "Hypothesis H is probably true" have been interpreted to mean that the (presently available) empirical evidence (E, say) supports H to a high degree. This degree of support of H by E has been called the logical probability of H given E, or the epistemic probability of H given E, or the inductive probability of H given E.
The differences between these interpretations are rather small, and may seem inconsequential. One of the main points of disagreement lies in the relation between probability and belief. Logical probabilities are conceived (for example in Keynes' Treatise on Probability) to be objective, logical relations between propositions (or sentences), and hence not to depend in any way upon belief. They are degrees of (partial) entailment, or degrees of logical consequence, not degrees of belief. (They do, nevertheless, dictate proper degrees of belief, as is discussed below.) Frank P. Ramsey, on the other hand, was skeptical about the existence of such objective logical relations and argued that (evidential) probability is "the logic of partial belief" ("Truth and Probability", 1926, p. 157). In other words, Ramsey held that epistemic probabilities simply are degrees of rational belief, rather than being logical relations that merely constrain degrees of rational belief.
Another point of disagreement concerns the uniqueness of evidential probability, relative to a given state of knowledge. Rudolf Carnap held, for example, that logical principles always determine a unique logical probability for any statement, relative to any body of evidence.[citation needed] Ramsey, by contrast, thought that while degrees of belief are subject to some rational constraints (such as, but not limited to, the axioms of probability) these constraints usually do not determine a unique value.[citation needed] Rational people, in other words, may differ somewhat in their degrees of belief, even if they all have the same information.
## Propensity
Propensity theorists think of probability as a physical propensity, or disposition, or tendency of a given type of physical situation to yield an outcome of a certain kind or to yield a long run relative frequency of such an outcome.[citation needed] This kind of objective probability is sometimes called 'chance'.
Propensities, or chances, are not relative frequencies, but purported causes of the observed stable relative frequencies. Propensities are invoked to explain why repeating a certain kind of experiment will generate a given outcome type at a persistent rate,[citation needed] which are known as propensities or chances. Frequentists are unable to take this approach,[citation needed] since relative frequencies do not exist for single tosses of a coin, but only for large ensembles or collectives. In contrast, a propensitist is able to use the law of large numbers to explain the behaviour of long-run frequencies. This law, which is a consequence of the axioms of probability, says that if (for example) a coin is tossed repeatedly many times, in such a way that its probability of landing heads is the same on each toss, and the outcomes are probabilistically independent, then the relative frequency of heads will (with high probability) be close to the probability of heads on each single toss. This law allows that stable long-run frequencies are a manifestation of invariant single-case probabilities. In addition to explaining the emergence of stable relative frequencies, the idea of propensity is motivated[citation needed] by the desire to make sense of single-case probability attributions in quantum mechanics, such as the probability of decay of a particular atom at a particular time.
The main challenge facing propensity theories is to say exactly what propensity means. (And then, of course, to show that propensity thus defined has the required properties.) At present, unfortunately, none of the well-recognised accounts of propensity comes close to meeting this challenge.
A propensity theory of probability was given by Charles Sanders Peirce.[6][7][8][9] A later propensity theory was proposed by philosopher Karl Popper, who had only slight acquaintance with the writings of C. S. Peirce, however.[6][7] Popper noted that the outcome of a physical experiment is produced by a certain set of "generating conditions". When we repeat an experiment, as the saying goes, we really perform another experiment with a (more or less) similar set of generating conditions. To say that a set of generating conditions has propensity p of producing the outcome E means that those exact conditions, if repeated indefinitely, would produce an outcome sequence in which E occurred with limiting relative frequency p. For Popper then, a deterministic experiment would have propensity 0 or 1 for each outcome, since those generating conditions would have same outcome on each trial.[citation needed] In other words, non-trivial propensities (those that differ from 0 and 1) only exist for genuinely indeterministic experiments.
A number of other philosophers, including David Miller and Donald A. Gillies, have proposed propensity theories somewhat similar to Popper's.[citation needed]
Other propensity theorists (e.g. Ronald Giere[citation needed]) do not explicitly define propensities at all, but rather see propensity as defined by the theoretical role it plays in science. They argue[citation needed], for example, that physical magnitudes such as electrical charge cannot be explicitly defined either, in terms of more basic things, but only in terms of what they do (such as attracting and repelling other electrical charges). In a similar way, propensity is whatever fills the various roles that physical probability plays in science.
What roles does physical probability play in science? What are its properties? One central property of chance is that, when known, it constrains rational belief to take the same numerical value. David Lewis called this the Principal Principle,[citation needed] a term that philosophers have mostly adopted.[citation needed] For example, suppose you are certain that a particular biased coin has propensity 0.32 to land heads every time it is tossed. What is then the correct price for a gamble that pays \$1 if the coin lands heads, and nothing otherwise? According to the Principal Principle, the fair price is 32 cents.
## Subjectivism
Gambling odds reflect the average bettor's 'degree of belief' in the outcome.
Subjectivists, also known as Bayesians or followers of epistemic probability, give the notion of probability a subjective status by regarding it as a measure of the 'degree of belief' of the individual assessing the uncertainty of a particular situation. Epistemic or subjective probability is sometimes called credence, as opposed to the term chance for a propensity probability.
Some examples of epistemic probability are to assign a probability to the proposition that a proposed law of physics is true, and to determine how probable it is that a suspect committed a crime, based on the evidence presented[citation needed].
Gambling odds don't reflect the bookies' belief in a likely winner, so much as the other bettors' belief, because the bettors are actually betting against one another. The odds are set based on how many people have bet on a possible winner, so that even if the high odds players always win, the bookies will always make their percentages anyway.[citation needed]
The use of Bayesian probability raises the philosophical debate as to whether it can contribute valid justifications of belief.
Bayesians point to the work of Ramsey[citation needed] and de Finetti[citation needed] as proving that subjective beliefs must follow the laws of probability if they are to be coherent.[10]
The use of Bayesian probability involves specifying a prior probability. This may be obtained from consideration of whether the required prior probability is greater or lesser than a reference probability[clarification needed] associated with an urn model or a thought experiment. The issue is that for a given problem, multiple thought experiments could apply, and choosing one is a matter of judgement: different people may assign different prior probabilities, known as the reference class problem. The "sunrise problem" provides an example.
## Practical controversy
This difference in point of view has also many implications both for the methods by which statistics is practiced, and for the way in which conclusions are expressed. When comparing two hypotheses and using some information, frequency methods would typically result in the rejection or non-rejection of the original hypothesis at a particular significance level, and frequentists would all agree that the hypothesis should be rejected or not at that level of significance.[dubious ] However, there is no normative methodology to choose levels of significance.[citation needed] Bayesian methods would suggest that one hypothesis was more probable than the other, but individual Bayesians might differ about which was the more probable and by how much, by virtue of having used different priors; but that's the same thing as disagreeing on significance levels,[dubious ] except significance levels are just an ad hoc device which are not really a probability, while priors are not only justified by the rules of probability,[dubious ] but there is definitely a normative methodology to define beliefs; so even if a Bayesian wanted to express complete ignorance (as a frequentist claims to do[dubious ] but does it wrong[dubious ]), they could do it with the maximum entropy principle. The most important distinction between the frequentist and Bayesian paradigms,[citation needed] is that frequentist makes strong distinctions between probability, statistics, and decision-making, whereas Bayesians unify decision-making, statistics and probability under a single philosophically and mathematically consistent framework, unlike the frequentist paradigm which has been proven to be inconsistent[dubious ], especially for real-world situations where experiments (or "random events") can not be repeated more than once.[citation needed] Bayesians would argue that this is right and proper[citation needed] — if the issue is such that reasonable people can put forward different, but plausible, priors and the data are such that the likelihood does not swamp the prior, then the issue is not resolved unambiguously at the present stage of knowledge and Bayesian statistics highlights this fact. They would argue[citation needed] that any approach that purports to produce a single, definitive answer to the question at hand in these circumstances is obscuring the truth. But "frequentists" do not claim to produce "a single, definitive answer to the question at hand".
An alternative solution, is the eclectic view, which accepts both interpretations: depending on the situation, one selects one of the two interpretations for pragmatic, or principled, reasons.[11]
## Prediction
An alternative account of probability emphasizes the role of prediction – predicting future observations on the basis of past observations, not on unobservable parameters. In its modern form, it is mainly in the Bayesian vein.[citation needed] This was the main function of probability before the 20th century,[12] but fell out of favor compared to the parametric approach, which modeled phenomena as a physical system that was observed with error, such as in celestial mechanics.[citation needed] The modern predictive approach was pioneered by Bruno de Finetti, with the central idea of exchangeability – that future observations should behave like past observations.[12] This view came to the attention of the Anglophone world with the 1974 translation of de Finetti's book,[12] and has since been propounded by such statisticians as Seymour Geisser.
## Axiomatic probability
The mathematics of probability can be developed on an entirely axiomatic basis that is independent of any interpretation: see the articles on probability theory and probability axioms for a detailed treatment.
## References
1. ^ Laszlo E. Szabo, A Physicalist Interpretation of Probability (Talk presented on the Philosophy of Science Seminar, Eötvös, Budapest, 8 October 2001.)
2. ^ Laszlo E. Szabo, Objective probability-like things with and without objective indeterminism, Studies in History and Philosophy of Modern Physics 38 (2007) 626–634 (Preprint)
3. ^ a b 'Interpretations of Probability', Stanford Encyclopedia of Philosophy [1], accessed 23 December 2006
4. ^ Laplace, P. S., 1814, English edition 1951, A Philosophical Essay on Probabilities, New York: Dover Publications Inc.
5. ^ Spanos, A. Statistical foundations of economic modelling
6. ^ a b Miller, Richard W. (1975). "Propensity: Popper or Peirce?". British Journal for the Philosophy of Science 26 (2): 123–132. DOI:10.1093/bjps/26.2.123.
7. ^ a b Haack, Susan; Kolenda, Konstantin, Konstantin; Kolenda (1977). "Two Fallibilists in Search of the Truth". Proceedings of the Aristotelian Society 51 (Supplementary Volumes): 63–104. JSTOR 4106816.
8. ^ Burks, Arthur W. (1978). Chance, Cause and Reason: An Inquiry into the Nature of Scientific Evidence. University of Chicago Press. pp. 694 pages. ISBN 0-226-08087-0.
9. ^ Peirce, Charles Sanders and Burks, Arthur W., ed. (1958), the Collected Papers of Charles Sanders Peirce Volumes 7 and 8, Harvard University Press, Cambridge, MA, also Belnap Press (of Harvard University Press) edition, vols. 7-8 bound together, 798 pages, online via InteLex, reprinted in 1998 Thoemmes Continuum.
10. ^ Probability Theory; The Logic of Science, by E T Jaynes.
11. ^ Cox, D.R. (2006) Principles of Statistical Inference. Cambridge University Press. ISBN 0-521-68567-2
12. ^ a b c
• Laurence Jonathan Cohen (1989) An Introduction to the Philosophy of Induction and Probability. Oxford Univ. Press.
• Donald A. Gillies (2000) Philosophical Theories of Probability. London: Routledge. (a comprehensive monograph covering the four principal current interpretations: logical, subjective, frequency, propensity)
• Antony Eagle, ed. (2011) "Philosophy of Probability: Contemporary Readings". London: Routledge.
• Ian Hacking (1975) Emergence of Probability.
• Paul Humphreys, ed. (1994) Patrick Suppes: Scientific Philosopher, Synthese Library, Springer-Verlag.
• Vol. 1: Probability and Probabilistic Causality.
• Vol. 2: Philosophy of Physics, Theory Structure and Measurement, and Action Theory.
• Jackson, Frank, and Robert Pargetter (1982) "Physical Probability as a Propensity," Noûs 16(4): 567–583.
• Andrei Khrennikov (2009) Interpretations of probability (2nd ed.) Walter de Gruyter (covers mostly non-Kolmogorov probability models, particularly with respect to quantum physics)
• David Lewis (1986) Philosophical Papers, Vol. II. Oxford Univ. Press.
• Brian Skyrms (2000) Choice and Chance, 4th ed. Wadsworth.
• Jan Von Plato (1994) Creating Modern Probability. Cambridge University Press.
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https://forum.knime.com/t/can-knime-de-pivot-work-with-a-pivot-table/75908 | 1,713,827,220,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00255.warc.gz | 240,787,220 | 7,227 | # Can KNIME de-pivot / work with a pivot table?
Hi there,
I am new to using KNIME and am wondering if KNIME can work with a pivot table? I generated a pivot table with Excel, so the rows with a value > 1 need to be duplicated in order to get a flat table. Can KNIME do this?
I am sorry if there is already an answer but I really couldn’t find one.
Best regards,
Marie
Hi, @MTAB01
Welcome to the KNIME Community Forum.
If you give an example with data, we will be able to help you better. It is also important to clarify what you want.
The more specific you are and the more context you give, the more help you will get.
Br
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I have a table like this:
Attribute 1 Attribute 2 Value
Apple red 1
Banana yellow 2
So there is the information that there is one red apple and two yellow bananas in two rows. I think it is neccessary for KNIME to have a flat table with three rows: one for the red apple and two rows with each a yellow banana. Can KNIME work with this and duplicate the yellow banana so that there are two banana-rows or do I have to do this myself?
Or can KNIME work with the above table and “understand” that I have to datasets for banana?
Hi @MTAB01 , welcome to the KNIME community.
I think I understand where you are coming from but if the value 2 is originally derived from the sum of two rows each containing 1, then KNIME can’t know this. The source data at the lowest level of granularity would be required unless you can subsequently write some rules to “build” the data. There are ways to build data based on duplicating existing data such as the “one row to many” node(which may be of assistance here but i can’t tell from the brief example) but there isn’t really a generic way of achieving this.
If you only have summarised data then effectively you’ve lost information. Suppose the “2” were actually 100 and the original source data were 25, 25 and 50… How could KNIME infer this?
Edit… Having reread your original question alongside the example maybe the One Row to Many node is a possible solution if the “2” simply represents the number of rows required
If you have a pivot table in Excel, then normally you’d have the raw data from which that pivot was derived. It would generally be better to supply that raw data to KNIME and then work from there. Excel pivot tables and KNIME pivots are very different things.
[btw. I’ve moved your post to the “KNIME Analytics Platform” section as the “Node Development” section of the forum is more about the technical details of the creation of new nodes, rather than general KNIME usage,
so you’ll likely get more responses in the more general area].
Thank you very much, I get your point. I don’t have direct access to the raw data in this special case, but I will try the One Row to Many node, the “2” does represent the number of rows required. Thanks again
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Mayoress writes:
<<Somewhere I learned a math concept called RULE 72 where you take your annual return times something and divide by 72 and that gives you the number of years it would take to double your investment.
I forget how it works. Please explain. >>
The Rule of 72 provides a very rough measure of how long it will take in years for your money to double when it's invested at a given rate of return. You divide your rate of return into the number 72 to get the years it takes for doubling the investment.
Say you get a 9% rate of return. Divide that into 72 and you get 8. That means in roughly 8 years you will have doubled your money if you get that 9% return each year on average.
Regards..Pixy
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Groebner basis to solve linear system of equations
I am trying to solve a linear system of equation modulo some prime $p$. I have a matrix which gives me the coefficients of the polynomials (i.e., first row would be ($a, c, b, d$) for $ax+by+cz+d =0$). I would usually use solve_right(), writing my system as $Ax=b$ mod $p$, but I am intrigued by the use of Groebner basis. I have read about their implementation but I am confused at the notion of ideal to define before solving the system. I would like to solve the system in the ring $Z/pZ$. Also, can I feed a matrix to the function or should I first convert the matrix in a bunch of equations ? Any clarification would be welcome !
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Groebner bases are somehow overkill: they are used for polynomial systems (involving equations like $ax^3y^4+by^5z+cxz+d=0$).
Here your system is linear, so linear algebra (matrix computations) is definitely the appropriate tool. It is much faster, can handle larger system, etc.
If you want to work modulo some prime p, you should define your matrices over the finite field GF(p):
sage: matrix(GF(5), [[1, 2, 4, 2], [2, 4, 4, 0], [0, 4, 4, 1], [2, 0, 0, 2]])
If you really want to use ideals, you can easily define your polynomials over GF(p), here is an example:
sage: R.<x,y,z> = PolynomialRing(GF(5)) ; R
Multivariate Polynomial Ring in x, y, z over Finite Field of size 5
sage: P = 2*x+3*y+5*z+2
sage: Q = 2*x+y+5*z+2
sage: S = x+y+z
sage: I = R.ideal([P,Q,S]) ; I
Ideal (2*x - 2*y + 2, 2*x + y + 2, x + y + z) of Multivariate Polynomial Ring in x, y, z over Finite Field of size 5
sage: I.variety()
[{y: 0, z: 1, x: 4}]
Note that when the variety is not zero dimensional, you might encounter issues.
sage: I = R.ideal([P,Q])
sage: I.variety()
ValueError: The dimension of the ideal is 1, but it should be 0
In linear words, this case corresponds to to having a non-trivial kernel, a case that is very well handled with matrices, see the following list of examples:
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CFDguy September 4, 2008 12:23
Heating of a metal piece inside a furnace
How to go about simulating the process of heating a metal inside a furnace? Are all 3 modes of heat transfer to be considered? There is no combustion taking place. It will be a transient problem where the metal gets heated inside the furnace and I want to monitor the core temperature of the metal.
Thanks, CFDguy
Ant September 5, 2008 05:05
Re: Heating of a metal piece inside a furnace
Hi,
Whenever the temperature goes beyond 1000 C raditaion heat transfer becomes significant. What will be important is to specify the emmisivity of the inner furnace walls and metal surface.
You do not need to model the thickness of furnace walls. You can specify the same in the wall boundary conditions. Get proper values of conductivity of the furnace walls. Specify external temperature and heat transfer coefficient/ heat loss/constant temperature for furnace walls.
Model the metal thickness only if you want to get the temperature variation along the thickness. What is the mode of heating?
Regards, Ant
CFDguy September 5, 2008 08:41
Re: Heating of a metal piece inside a furnace
Thanks Ant. The heat transfer into the metal piece takes place by conduction, convection and radiation. Since the wall temperature of this metal piece will increase with time, I guess UDF needs to be used for this transient BC?(for proper modeling of conduction) The same would apply for convection as well - transient HTC? How significant would convection be considering this is a small furnace with negligible air flow? Mode of heating is electric, I hope I understood your question.
Thanks again, CFDguy
All times are GMT -4. The time now is 11:12. | 459 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-44 | latest | en | 0.874392 |
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wd:Q4658263
A Million Random Digits with 100,000 Normal Deviates is a random number book by the RAND Corporation, originally published in 1955. The book, consisting primarily of a random number table, was an important 20th century work in the field of statistics and random numbers. It was produced starting in 1947 by an electronic simulation of a roulette wheel attached to a computer, the results of which were then carefully filtered and tested before being used to generate the table. The RAND table was an important breakthrough in delivering random numbers, because such a large and carefully prepared table had never before been available. In addition to being available in book form, one could also order the digits on a series of punched cards. The table is formatted as 400 pages, each containing 50 lines of 50 digits. Columns and lines are grouped in fives, and the lines are numbered 00000 through 19999. The standard normal deviates are another 200 pages (10 per line, lines 0000 through 9999), with each deviate given to three decimal places. There are 28 additional pages of front matter. The main use of the tables was in statistics and the experimental design of scientific experiments, especially those that used the Monte Carlo method; in cryptography, they have also been used as nothing up my sleeve numbers, for example in the design of the Khafre cipher. The book was one of the last of a series of random number tables produced from the mid-1920s to the 1950s, after which the development of high-speed computers allowed faster operation through the generation of pseudorandom numbers rather than reading them from tables. The book was reissued in 2001 (ISBN 0-8330-3047-7) with a new foreword by RAND Executive Vice President Michael D. Rich. It has generated many humorous user reviews on Amazon.com.The digits (sequence A002205 in the OEIS) begin: 10097 32533 76520 13586 34673 54876 80959 09117 39292 74945
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original title: A Million Random Digits with 100,000 Normal Deviates
date of publication: 1955
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https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_5&direction=prev&oldid=145559 | 1,627,457,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00049.warc.gz | 123,894,995 | 10,289 | # 2021 AMC 12B Problems/Problem 5
## Problem
The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90\deg$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$
## Solution
$\boxed{\textbf{(D)} ~7}$
~ pi_is_3.14 | 167 | 442 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-31 | latest | en | 0.58586 |
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## Clarification and intuition request for rationally equivalent algebraic cycles
I am having some difficulty lining up the definition and my intuition for rational equivalence of cycles. My intuition is based off of the idea that two cycles being rationally equivalent is analogous to the two cycles being homotopic.
If it is requested of me to state the definition of rational equivalence, I will; for the moment I will refrain and simply link to the wikipedia page for the chow ring and adequate equivalence relation.
Here are two examples that I have thought of that are giving me trouble.
1. In $\mathbb{k}[x,y]$, consider the line given by $x = 0$ and the unit circle. By taking the rational function $\frac{x^2+y^2 - 1}{x}$ on $\mathbb{k}[x,y]$, we get that these two are rationally equivalent. These two being homotopic is only semi-okay with me. I can imagine making the circle larger and larger, making it look more and more like a line. But it seems to me that the final step of taking it to the line is not allowed by homotopy.
2. Consider the 2-torus $\mathbb{T}$, thought of as sitting in $\mathbb{A}^3$ with center at the origin. Let $S_1$ be (one of the) circles obtained by intersecting $\mathbb{T}$ with the $xy$-plane, and let $a$ be the regular function defining it. Let $S_2$ be one of the circles obtained by intersecting with the $xz$-plane, and $b$ the regular function defining it. Then the rational function $\frac{a}{b}$ gives a rational equivalence of $S_1$ and $S_2$. This goes very much against my intuition.
Hopefully somebody will tell me where I am going wrong in my thinking.
-
If you formally replace the interval by the affine line in your definition of homotopy and force the result to be an equivalence relation, then rational equivalence is what you get. But perhaps you knew that. – Donu Arapura Oct 3 at 13:42
Any codimension 1 cycle that is defined by a regular function is already zero in the Chow group. So your question should not be "Why are the line $x=0$ and the unit circle rationally equivalent to each other?" but rather "Why is the line $x=0$ rationally equivalent to zero?" (and ditto for the unit circle).
The answer is that either of these divisors can be "moved to infinity" and hence out of the affine plane altogether. (E.g. you can think of the graph of the regular function $x$ as a path that takes the line at $x=0$ to the (missing) line $x=\infty$. | 633 | 2,572 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2013-20 | latest | en | 0.92937 |
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# Please look at the maths below:
£0.00 @
They say only people with an IQ with 120 and over are able to figure this out. If: 2 + 3 = 10 7 + 2 = 63 6 + 5 = 66 8 + 4 = 96 Then: 9 + 7 = ? Read More
6y, 11m agoPosted 6 years, 11 months ago
They say only people with an IQ with 120 and over are able to figure this out.
If:
2 + 3 = 10
7 + 2 = 63
6 + 5 = 66
8 + 4 = 96
Then:
9 + 7 = ?
6y, 11m agoPosted 6 years, 11 months ago
Options
(20)
banned#1
9 + 7 = 16
just because the rest of the maths was bad, no reason not to be able to get one right
#2
16 x 9 = 144
#3
Are you x'ing the 1st number by the sum of both numbers?
eg. 2+3 = 10 is actually 2(2+3)=10.
#4
n1ck09
16 x 9 = 144
Yes, I agree.
#5
I reckon the answer is 144?
Doh- just saw the answer was already there. take off 50 iq points!
#6
yep 144
#7
jubbyme
9 + 7 = 16
just because the rest of the maths was bad, no reason not to be able to get one right
oooo we got the same answer. We must be right. The others are hopeless aint they?
#8
I dont know the answer, my IQ is 4, just need to know how to solve it
banned#9
2(2+3) = 10
7(7+2) = 63
6(6+5) = 66
8(8+4) = 96
So...for 9+7
9(9+7) = 144
#10
n1ck09
16 x 9 = 144
got it now, cheers
#11
9+7=144
calc
1st +2nd number x 1st number
banned#12
2 + 3 = 5 --------- this answer × first No.
2 × 5 = 10
7 + 2 = 9
7 × 9 = 63
6 + 5 = 11
6 × 11 = 66
8 + 4 = 12
8 × 12 = 96
THEN; 9 + 7= 16
9 × 16 = 144
suspended#13
144
Got there in the end :)
6 + 5 = 66 - this made it easy
#14
#15
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609
banned#16
Fairly basic. I doubt the claim that it shows you have an IQ above 120 is true too.
#17
numptyj
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609
easy as pie:whistling:
#18
9 + 7 = 16
9 * 16 = 144
#19
So easy! 144! Even my little boy can do that (well, almost).
#20
jubbyme
9 + 7 = 16
just because the rest of the maths was bad, no reason not to be able to get one right
Prize for the funniest answer. :thumbsup: | 1,117 | 2,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-22 | latest | en | 0.880118 |
http://amuconference.org/metallica/intense/8751729850e89c7cae2a8120f143a- | 1,679,615,018,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00560.warc.gz | 3,579,393 | 12,868 | Use this statistics calculators for frequency distribution, Minimum & Maximum, median, mode and The number of local extremums depends on the function, but for a polynomial of degree \$ n \$, there are, at most, \$ n-1 \$ local extremums. To lift safely, a one rep max calculator is essential for determining what you can handle on the squat rack, bench, deadlift and more. Find the absolute minimum and absolute maximum of f (x,y) = 192x3 +y2 4xy2 f ( x, y) = 192 x 3 + y 2 4 x y 2 on the triangle with vertices (0,0) ( 0, 0), (4,2) ( 4, 2) and (2,2) ( We say that the function f(x) has a global maximum at x=x 0 on the interval I, The immense value becomes the absolute maximum from the results you get, while the smallest LBW = Lean body weight in kg. Since the endpoints are not included, they can't be the global extrema, and this interval has no global minimum or maximum. - An electrical diversity calculator based on appendix A of the On-Site Guide. I can draw a graph and look at the intervals in between and on the sides of the critical points to The best part is that you dont need any prior experience or knowledge. Hence, although f (x) has several local maxima, f (x) does not have a global maximum. Choose either maximum or minimum (for this question you will do both) Saddle Points are used in the study of calculus. Simple. It is a global maximum or a local Search: Gmax Catch Rate Calculator. This is called your borrowing power. Examples. The input value of \(x_1\) for which Enter the values into the function f (x). example. Find min/max values. Trim Comparisons. Calculate the global minimum and global maximum of the function on the interval from MATH 110 at Pennsylvania State University, World Campus 158, Saskatchewan for \$469,000 CAD Land, 0 bathrooms. Not all functions have a (local) minimum/maximum. Find min/max values. Search: Gmax Catch Rate Calculator. This home affordability calculator provides a simple answer to the question, How much house can I afford? Your one-rep max is the max weight you can lift for a single repetition for a given exercise. The above calculator is an online tool which shows output for the given input. Tap for more steps - (3x + 1)(x - Enter the endpoints, a and b, into the function f (x). This calculator helps you work out the most you could borrow from the bank to buy your new home. Determining global maxima and minima. Observed Hb = the patients current hemoglobin in g/dl.
With its sloping roofline, sporty styling, and fascinating variable-compression engine, the Infiniti QX55 stands out as a luxurious and well-equipped midsize 2-row SUV. Local maximum is the greatest element in a subset or a given range of a function. Absolute minima & maxima (closed intervals) AP.CALC: FUN4 (EU), FUN4.A (LO), FUN4.A.3 (EK) Google Classroom Facebook Twitter. Extrema are points on a function where a maximum or minimum exists. Settings # popular sets (click to use):decimal,octal,hexadecimal: Show detailed settings: Enter your numbers here # Data summary # How many numbers detected: 3: Detected numbers: 2, 10, 3.3: Numbers base: 10 (decimal) Weekly Subscription \$2.49 USD per week until cancelled. Purdue University Global ("Institution") and/or its licensors may change these Terms from time to time. The limits of functions can be considered both at points and at infinity. CALVERT FORMULA FOR CARBOPLATIN DOSING: Total Dose (mg) = (target AUC) x (GFR + 25) Calvert AH, Newell DR, Gumbrell LA, et al. for a constant function the minimum and the maximum are identical. Finding absolute extrema on a closed interval. RAID Calculator FS Series Ultra-high performance, all-flash storage solutions for latency-sensitive workloads. You now have a much clearer picture of your local extrema. Spring Promotion To calculate the maximum demand we can Then using the plot of the function, you can determine whether the points you find were a local minimum or a local maximum. The calculator was developed by Henkel and the Wuppertal Institute (Germany). Discontinuous. First-type data elements (separated by spaces or commas, etc. A global minimum, also known as an absolute minimum, is the smallest overall value of a set, function, etc., over its entire range. Extremum is called maximum or minimum point of the function. Global - English. Alternatively, you can go straight to Email. Consider the set {a | i N}. Practice: Absolute minima & maxima (closed intervals) On the last page you learned how to find local extrema; one is often more interested in finding global extrema: . Consider the set {a | i N}. | 1,086 | 4,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-14 | latest | en | 0.857775 |
https://boolean.h.uib.no/mediawiki/index.php?title=Commutative_Presemifields_and_Semifields&direction=prev&oldid=334&printable=yes | 1,656,749,466,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00151.warc.gz | 191,798,860 | 10,870 | # Background
For a prime ${\displaystyle p}$ and a positive integer ${\displaystyle n}$ let ${\displaystyle \mathbb {F} _{p^{n}}}$ be the finite field with ${\displaystyle p^{n}}$ elements. Let ${\displaystyle F}$ be a map from the finite field to itself. Such function admits a unique representation as a polynomial of degree at most ${\displaystyle p^{n}-1}$, i.e.
${\displaystyle F(x)=\sum _{j=0}^{p^{n}-1}a_{j}x^{j},a_{j}\in \mathbb {F} _{p^{n}}}$.
The function ${\displaystyle F}$ is
• linear if ${\displaystyle F(x)=\sum _{j=0}^{n-1}a_{j}x^{p^{j}}}$,
• affine if it is the sum of a linear function and a constant,
• DO (Dembowski-Ostrim) polynomial if ${\displaystyle F(x)=\sum _{0\leq i\leq j,
• quadratic if it is the sum of a DO polynomial and an affine function.
For ${\displaystyle \delta }$ a positive integer, the function ${\displaystyle F}$ is called differentially ${\displaystyle \delta }$-uniform if for any pairs ${\displaystyle a,b\in \mathbb {F} _{p^{n}}}$, with ${\displaystyle a\neq 0}$, the equation ${\displaystyle F(x+a)-F(x)=b}$ admits at most ${\displaystyle \delta }$ solutions.
A function ${\displaystyle F}$ is called planar or perfect nonlinear (PN) if ${\displaystyle \delta _{F}=1}$. Obviously such functions exist only for ${\displaystyle p}$ an odd prime. In the even case the smallest possible case for ${\displaystyle \delta }$ is two (APN function).
For planar function we have that the all the nonzero derivatives, ${\displaystyle D_{a}F(x)=F(x+a)-F(x)}$, are permutations.
## Equivalence Relations
Two functions ${\displaystyle F}$ and ${\displaystyle F'}$ from ${\displaystyle \mathbb {F} _{p^{n}}}$ to itself are called:
• affine equivalent if ${\displaystyle F'=A_{1}\circ F\circ A_{2}}$, where ${\displaystyle A_{1},A_{2}}$ are affine permutations;
• EA-equivalent (extended-affine) if ${\displaystyle F'=F''+A}$, where ${\displaystyle A}$ is affine and ${\displaystyle F''}$ is afffine equivalent to ${\displaystyle F}$;
• CCZ-equivalent if there exists an affine permutation ${\displaystyle {\mathcal {L}}}$ of ${\displaystyle \mathbb {F} _{p^{n}}\times \mathbb {F} _{p^{n}}}$ such that ${\displaystyle {\mathcal {L}}(G_{F})=G_{F'}}$, where ${\displaystyle G_{F}=\lbrace (x,F(x)):x\in \mathbb {F} _{p^{n}}\rbrace }$.
CCZ-equivalence is the most general known equivalence relation for functions which preserves differential uniformity. Affine and EA-equivalence are its particular cases. For the case of quadratic planar functions the isotopic equivalence is more general than CCZ-equivalence, where two maps are isotopic equivalent if the corresponding presemifields are isotopic.
# On Presemifields and Semifields
A presemifield is a ring with left and right distributivity and with no zero divisor. A presemifield with a multiplicative identity is called a semifield. Any finite presemifield can be represented by ${\displaystyle \mathbb {S} =(\mathbb {F} _{p^{n}},+,\star )}$, for ${\displaystyle p}$ a prime, ${\displaystyle n}$ a positive integer, ${\displaystyle \mathbb {S} =(\mathbb {F} _{p^{n}},+)}$ additive group and ${\displaystyle x\star y}$ multiplication linear in each variable.
Two presemifields ${\displaystyle \mathbb {S} _{1}=(\mathbb {F} _{p^{n}},+,\star )}$ and ${\displaystyle \mathbb {S} _{2}=(\mathbb {F} _{p^{n}},+,\circ )}$ are called isotopic if there exist three linear permutations ${\displaystyle T,M,N}$ of ${\displaystyle \mathbb {F} _{p^{n}}}$ such that ${\displaystyle T(x\star y)=M(x)\circ N(y)}$, for any ${\displaystyle x,y\in \mathbb {F} _{p^{n}}}$. If ${\displaystyle M=N}$ then they are called strongly isotopic. Each commutative presemifields of odd order defines a planar DO polynomial and viceversa:
• given ${\displaystyle \mathbb {S} =(\mathbb {F} _{p^{n}},+,\star )}$ let ${\displaystyle F_{\mathbb {S} }(x)={\frac {1}{2}}(x\star x)}$;
• given ${\displaystyle F}$ let ${\displaystyle \mathbb {S} _{F}=(\mathbb {F} _{p^{n}},+,\star )}$ defined by ${\displaystyle x\star y=F(x+y)-F(x)-F(y)}$.
Hence two quadratic planar functions ${\displaystyle F,F'}$ are isotopic equivalent if their corresponding presemifields are isotopic. Moreover, we have:
• ${\displaystyle F,F'}$ are CCZ-equivalent if and only if ${\displaystyle \mathbb {S} _{F},\mathbb {S} _{F'}}$ are strongly isotopic;
• for ${\displaystyle n}$ odd, isotopic coincides with strongly isotopic;
• if ${\displaystyle F,F'}$ are isotopic equivalent, then there exists a linear map ${\displaystyle L}$ such that ${\displaystyle F'}$ is EA-equivalent to ${\displaystyle F(x+L(x))-F(x)-F(L(x))}$. | 1,387 | 4,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 60, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.68004 |
https://www.physicsforums.com/threads/prove-that-for-n-1-integral.270128/ | 1,539,908,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512161.28/warc/CC-MAIN-20181018235424-20181019020924-00165.warc.gz | 1,068,316,584 | 15,316 | # Homework Help: Prove that for n> 1 integral
1. Nov 7, 2008
### Daggy
1. The problem statement, all variables and given/known data
Prove that for n> 1
$$\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \frac{n}{n^2 - 1}$$
2. Relevant equations
3. The attempt at a solution
Tried substitute x = cosh theta, then
$$\frac{\mathrm{dx}}{\mathrm{d}\theta} = \sinh \theta$$
$$\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{0}^{\infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n }$$
I'm getting in the right direction here? I'm really stuck..
Last edited: Nov 7, 2008
2. Nov 7, 2008
### HallsofIvy
Re: Integral
Isn't something missing here?
3. Nov 7, 2008
Re: Integral
n > 1
4. Nov 7, 2008
### gabbagabbahey
Re: Integral
I think he meant your attempt at a solution
5. Nov 7, 2008
### Daggy
Re: Integral
well I believe that mentioning n > 1 also is quite important... I've tried to use sinh-substitution, but can't really say I'm getting any wiser about it.
6. Nov 7, 2008
### Dick
Re: Integral
Use cosh(x)=(e^x+e^(-x))/2 and sinh(x)=(e^x-e^(-x))/2. Now try to integrate it. It's really pretty easy.
7. Nov 7, 2008
### Daggy
Re: Integral
$$\int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{x = 0}^{x = \infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n } \mathrm{d}\theta$$
$$= \int\limits_{x = 0}^{x = \infty} \frac{e^{\theta} - e^{-\theta}}{(e^\theta)^n} \mathrm{d}\theta$$
Don't know what to do here.
Last edited: Nov 8, 2008
8. Nov 7, 2008
### Dick
Re: Integral
Split it into two integrals. Use rules of exponents. And you are missing a factor of 1/2, I think.
9. Nov 8, 2008
### Daggy
Re: Integral
$$\frac{1}{2} \int\limits_{x = 0}^{x = \infty} {e^{\theta - n\theta}} \mathrm{d}\theta - \frac{1}{2} \int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta} \mathrm{d}\theta =$$
Then integrating and substitute back, don't really seem to get rid of those exponentials..
10. Nov 8, 2008
### Dick
Re: Integral
Just continue. Write down the antiderivatives. n>1. The contribution from infinity vanishes. It's the theta=0 part that counts. And e^0=1. See, no exponential?
11. Nov 8, 2008
### HallsofIvy
Re: Integral
Your original substitution x= cosh(theta) has a fundamental problem: the lower limit of integration is x= 0 and cosh(theta) is never 0. I would suggest x= tan(y) instead.
12. Nov 8, 2008
### Daggy
Re: Integral
Is that a problem? when integrating and substituting back the cosh dissappears does'nt it?
13. Nov 8, 2008
### Dick
Re: Integral
Yeah, it's a problem. If you use x=cosh(theta), 1+cosh(theta)^2 isn't equal to sinh(theta)^2. It's the other way around. Use x=sinh(theta). (You had said you were using sinh-substitution, and I didn't notice you had it backwards.) It looks almost the same as x=cosh(theta) except for a sign difference.
Last edited: Nov 8, 2008
14. Nov 8, 2008
### Daggy
Re: Integral
So
$$\int\limits_{0}^{\intfty}{\frac{\cosh{\theta}}{(\sinh{\theta} + \cosh{\theta})^n}}\mathrm{d}\theta = \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{\frac{e^{x} + e^{-x}}{(e^x)^n}}{\mathrm{d}\theta}$$
and
$$\frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{\theta - n\theta}}\mathrm{d}\theta + \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta}}\mathrm{d}\theta$$
and then factorize e^{1 - n} out?
$$\frac{e^{1 - n}}{2} \int\limits_{x = 0}^{x = \infty}{cosh \theta}~ \mathrm{d}\theta$$
15. Nov 8, 2008
### Dick
Re: Integral
Noooo. And you can change your limits to theta=0 etc, ok? Your first integral is e^((1-n)*theta). That's EASY to integrate.
16. Nov 8, 2008
### Daggy
Re: Integral
ahh. figured out, thanks | 1,395 | 3,728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-43 | latest | en | 0.747408 |
https://ohbug.com/uva/10726/ | 1,725,856,651,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00735.warc.gz | 405,552,046 | 2,478 | # Coco Monkey
This is the story of a time long ago. There were these sailors stranded on a beautiful island full off coconut trees. And what do our sailors do with the coconut trees? They collect all the coconuts and divide those among themselves and the monkeys that they brought with them! History tells us that there were S sailors on the island, and they collected C coconuts in total. At nightfall they decided to leave the coconuts in a secret place and divide it equally in the morning. But as the night grew darker, the sailors began to get impatient. At midnight, one of the sailors got up. He dug up the coconuts and saw that after dividing the coconuts equally into S baskets, there are exactly M coconuts left. These, as you must have guessed, goes to the M monkeys that they have brought with them. Now, our greedy sailor took one of those baskets for himself and left the remaining S − 1 ones for his commodores. History repeats. So we have another sailor succumbing to the temptation of juicy coconuts. He again, divided the remaining coconuts equally in S baskets only to find M coconuts extra. One basket for himself, M coconuts for the M monkeys and the remaining S − 1 baskets for his fellows – that’s how it goes. Before the night was over each and every sailor played his part exactly once. At dawn when they started to divide the remaining coconuts, they found that there’s no extra coconut left for the monkeys. But the monkeys would not complain, nor would any of the sailors. They had their share alright. Historians, as one would expect, can be trusted for facts not for the figures. So whatever they would say about the values of S, C and M cannot be trusted entirely. But given the values of S, M and an interval for C one can identify possible values for C. As a programmer you only need to count the number of valid C values on a given range [low, high] inclusive. Input The first line of the input gives you the number of test cases, T (1 ≤ T ≤ 15). Then T test cases follow. Each of the test cases consists of four integers and is presented one in each line. The first integer gives you the number of sailors S (1 < S ≤ 100), then the next integer gives you the number of monkeys M (0 ≤ M < S). The next two integers will give you the value of low and high (0 ≤ low ≤ high ≤ 108). Output For each of the test cases, you need to print one line of output. The output for each test case starts with the test case number, followed by the number of valid C values in the range [low, high]. For the exact format of output, please consult the sample input/output section. Illustration: The first test case gives you 3 sailors, 2 monkeys and a range of [49, 50] to choose from. If you’d try to do the math in hand, you’d see that only 50 would satisfy the scenario. So the answer for the first test case has to be 1. Here’s the brief illustration: Sailor # Total Coconuts His share Monkeys share 1 50 16 2 2 32 10 2 3 20 6 2 Sample Input 3 3 2 49 50 4 2 5 10000 Remaining Coconuts 32 20 12
2/2 6 4 1 100000000 Sample Output Case 1: 1 Case 2: 10 Case 3: 357 | 762 | 3,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.971805 |
https://www.kaysonseducation.co.in/questions/p-span-sty_6128 | 1,674,944,464,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00129.warc.gz | 832,588,963 | 11,936 | Obtain the equation of the sphere where the points (1, 0, 1) and (5, 4, 5) are the extremities of a diameter. Deduce the equation in Cartesian form. Also find the radius and the centre of the sphere. : Kaysons Education
Obtain The Equation Of The Sphere Where The Points (1, 0, 1) And (5, 4, 5) Are The Extremities Of A Diameter. Deduce The Equation In Cartesian Form. Also Find The Radius And The Centre Of The Sphere.
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Question
Solution
Correct option is
Let be the position vectors of the extremities (1, 0, 1) and (5, 4, 5) of the diameter.
Let be the position vector of a general point (x, y, z) on the sphere.
The equation of the required sphere is
…(1)
…(2)
(2)
SIMILAR QUESTIONS
Q1
Find the equation of the sphere which passes through the points (1,–3, 4), (1, –5, 2), (1, –3, 0) and has its centre on the plane x + y + z = 0.
Q2
Chord AB is a diameter of the sphere with coordinates of A as (3, 2, –2). Find the coordinates of B.
Q3
A plane passes through a fixed point (a, b, c) and cuts axes in A, B, C. Find the locus of the centre of the sphere OABC
Q4
Obtain the equation of the sphere described on the join of the points (2, –3, 4) and B(–5, 6, –7) as a diameter. | 504 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-06 | latest | en | 0.788852 |
https://math.stackexchange.com/questions/2338319/two-dimensional-recurrence-relation | 1,718,553,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00701.warc.gz | 353,055,777 | 36,730 | # two-dimensional recurrence relation
How to solve the following 2D recurrence relation?
$$F(n_1, n_2)=a\bigl[F(n_1-1, n_2)+F(n_1+1, n_2)+F(n_1, n_2-1)+F(n_1, n_2+1)\big]$$ with $$1\leq n_1\leq L_1,\ 1\leq n_2\leq L_2$$. a is a positive real number. Boundary condition: $$F(m_1, m_2)=1$$.
*$$n_1$$ and $$n_2$$ are the coordinate indices and ($$m_1$$, $$m_2$$) is a point in the lattice.
I checked similar questions on this topic and I tried to use the Generating function method as explained Solving two-dimensional recurrence relation $a_{i,\ j}\ =\ a_{i,\ j-1}\ +\ a_{i-1,\ j-1}$ The difficulty is the boundary condition.
The problem I'm considering is the first passage time of a particle searching for a target in 2-D lattice of size $$L_1 \times L_2$$. Here $$(n_1, n_2)$$ is the position of a particle at time t=0 and $$(m_1, m_2)$$ is the position of the target. $$F(n_1, n_2)=F(n_1, n_2,s)$$ is the Laplace transform (s is the Laplace variable) of the first passage time probability.
• Please use latex conventions. For example $n$ indexed by $1$ is rendered by : dollar sign followed by letter a followed by underscore symbol followed by digit 1 followed by dollar sign. Commented Jun 27, 2017 at 18:06
• I am afraid that there is no other solution than the zero function. Commented Jun 27, 2017 at 18:19
• boundary condition is corrected. Commented Jun 27, 2017 at 22:58
• Hint: find when $\lambda_1^{n_1}\lambda_2^{n_2}$ satisfies the recurrence relation. Linear combinations give the general solution.
– J.G.
Commented Jun 27, 2017 at 22:59
• @J.G. Thanks for your advice. The problem is while there is only one recursion relation, there are two undetermined variables $\lambda_1$ and $\lambda_2$. Commented Jun 28, 2017 at 0:03
Following a suggestion in my comment, we attempt $F=\lambda_1^{n_1}\lambda_2^{n_2}$ so $\frac{1}{a}=\frac{1}{\lambda_1}+\lambda_1+\frac{1}{\lambda_2}+\lambda_2$. Define $\mu_i:=\lambda_i+\frac{1}{\lambda_i}$ so $\lambda_i=\frac{\mu_i\pm\sqrt{\mu_i^2-4}}{2}$ and $\mu_2=\frac{1}{a}-\mu_1$. The most general solution is a linear combination over choices of $\mu_1$, an integral rather than a sum. The result is$$\int_Sd\mu_1\left(a_1^+\left(\mu_1\right)\left(\frac{\mu_1+\sqrt{\mu_1^2-4}}{2}\right)^n+a_1^-\left(\mu_1\right)\left(\frac{\mu_1-\sqrt{\mu_1^2-4}}{2}\right)^n+a_2^+\left(\mu_1\right)\left(\frac{\frac{1}{a}-\mu_1+\sqrt{\left(\frac{1}{a}-\mu_1\right)^2-4}}{2}\right)^n+a_2^-\left(\mu_1\right)\left(\frac{\frac{1}{a}-\mu_1-\sqrt{\left(\frac{1}{a}-\mu_1\right)^2-4}}{2}\right)^n\right)$$for arbitrary functions $a_i^\pm$. The integration range $S$ is an exercise for the reader, and so is finding a solution satisfying the desired boundary condition. It's constrained by the fact that constraints on the $\lambda_i$ constrain the $\mu_i$ (and may in some cases cause one of the values for a $\lambda_i$ consistent with a given $\mu_i$ consistent with that range to be outside it, meaning a coefficient has to be $0$). | 1,008 | 2,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.799751 |
https://community.particle.io/t/mics-2614-library/22026 | 1,604,067,126,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00074.warc.gz | 253,963,736 | 9,837 | # MICS-2614 library
Hello…
I am working on MICS-2614 Ozone sensor. I tried to find out library for MICS-2614, but i can’t. I don’t know how to make code for it. I also refered datasheet, but i found only schematic, not library or any equation. Can anyone suggest me how to make library of MICS-2614 or suggest me links to get tips for coding?
I have not used this chip, but I read the data sheet and I believe this should work. Keep in mind that I have not tested this!
Connect pin F (heater+) to 3V3.
Connect pin C (heater-) to a 160 ohm resistor, which is then connected to GND. Note that the data sheet says 82 ohm, but that’s for a 5V supply.
Connect pin G to 3V3. The data sheet indicates 5V, but the analog inputs can only be used up to 3.3V and you should be able to use 3.3V for the measuring circuit.
Connect pin D to an analog input pin (A0-A6). Also connect pin D to GND using a resistor. The minimum is 820 ohms, but I’d use 10K.
The sensor is a variable resistance between pin G and D, depending on the level of ozone. The resistance varies from 3K to 60K ohms.
By adding the external 10K resistor, you have a voltage divider circuit, so the voltage at the analog input will vary between 0.471V and 2.538V. This corresponds to an analog input value that you can read using analogRead() of 584 to 3150. Note that the numbers are backwards, 584 is the maximum concentration of 1000 ppb and 3150 is the minimum concentration of 10 ppb.
Good luck!
Can you suggest me any equation for ppb calculation?
This took some thinking:
``````double value = (double) analogRead(A0);
double ppb = Math.pow(10, (2566-(value-584) / 1283) + 1)
``````
You might need to play around with the constants a little to get it to work out.
2566 = maximum value you get, which is actually the minimum amount of ozone
584 = minimum value you get, which is the maximum amount of ozone, which hopefully you’ll never see
1283 = a factor used to get the full range to work out to two orders of magnitude on a log10 scale
Thank you so much, Now i will try with this equations.
Thank you once again.
This Equation is not working. It gives me infinite value.
Oops. Sorry, I think I missed parentheses. It should be:
``````double ppb = Math.pow(10, ((2566-(value-584)) / 1283) + 1)
``````
Thank you. Now it is seeming that it is working. Can you tell me these values are right or wrong?
Can i ask how you get this equation?
I have no way of knowing for sure, but the values look plausible.
The values from analogRead should range from 584 to 3150, but are reversed where 3150 is the minimum (10 ppb) and 584 is the maximum (1000 ppb).
The first thing I did was adjust to 0 to 2566 by subtracting 584 = (value - 584).
Then make it so 0 is the smallest ozone concentration instead of the largest
2566 - (value above) = 2566 - (value - 584)
According to the data sheet, the scale is logarithmic, so the lowest value is 10, the largest value = 1000 and the middle value is 100 ppb. The middle value is 2566 / 2 = 1283.
The next step is to convert the scale from 0 to 2566 to 1 to 3, that’s ((2566-(value-584)) / 1283) + 1.
Then the Math.pow(10, ((2566-(value-584)) / 1283) + 1) converts the 1 to 3 to the range of 10 (10^1) to 1000 (10^3) using a log10 scale.
1 Like
Can you tell me readings are right or wrong?
How could @rickkas7 possibly know how much ozone is near your sensor right now?
As he said above, the values are “plausible” and that means in the right range. Do you have source of ozone you can test with to see how the values go up or down?
2 Likes
I ordered this sensor. Right now i am taking readings from proteus simulation as you can see in above image. But i think my reading are wrong, because according to data sheet range of sensor is 10 to 1000 ppb.
How can i make equation for ppb calculation from above graph? I made equation, but it gives me wrong reading.
A few things to note about that graph:
1. The X and Y on that graph are “backwards” from the equation you want–you need to measure resistance and calculate O3 ppb.
2. The graph has been calibrated to a normalized resistance value for 100ppb. You need to think about how you would come up with this number since the datasheet only says it is between 3 and 60 kohm.
3. The graph is for 25 degrees C and 50% relative-humidity and might not be valid at other temperatures and humidity levels.
If you have MATLAB or similar, you can do a polynomial fit to some points from the graph (just remember that the Y axis is the input and the X axis is the output. Finding the order of the polynomial to fit requires some experience or guess work. If you have not studied linear algebra, it will be hard to explain how this polynomial fit is obtained, but I am sure if you search many references will come up.
When I fit a polynomial to a few points from that graph and assume a third-order equation, I get the following equation which seems to fit the points well. If you had actual calibration data from the sensor you are using, you could do better.
``````Rnorm = Rmeasured / R100ppb; /* You must find R100ppb to calibrate */
x = Rnorm;
O3ppb = 1.15354664573491*x^3 + 0*x^2 + 95.0232835796925*x + 3.82316977457257;
``````
1 Like
@bko, just a note - the second order term is zero
I left it in just to show that there is not a mistake in the order of the equation.
1 Like
@bko
Thank you so much.
I saw Proteus Simulation with this Equation. I think it is ruuning.
I ordered Sensor. When it will come, I will test and give updates.
Bko,
Could you please elaborate how to create “reverse” polynomial equation?
I am using excel, and i understand how to get the equation taking X or ppm as input, but can not figure out how to reproduce your suggestion.
Thanks
Hi Wally,
Reverse here just means that when you read the graph, pretend the Y axis is the X axis and vice-versa. No big deal. It is like you printed the graph and rotated it 90 degrees.
I have never tried this in Excel, but it has a function called LINEST that looks like it might work. It does a linear least-squares fit to the points you give it. You want something like `=LINEST(Y0:Y10, X0:X10^{1,2,3}` for a third-order fit to 11 points.
There are many ways to fit equations to data and the way I did it was to sample several points off of the graph in [x,y] = [R, ppm] format, like [0.065, 10], [1, 100], [6.75, 1000], etc. I then formed the Vandermonde matrix with the x-coordinates and computed the inverse matrix. I then multiplied this inverse by the y-coordinate vector to get the polynomial coefficient vector. | 1,751 | 6,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-45 | latest | en | 0.908135 |
https://getacho.com/what-is-5601-in-scientific-notation/ | 1,679,882,141,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946584.94/warc/CC-MAIN-20230326235016-20230327025016-00461.warc.gz | 330,489,068 | 12,000 | # What is 5601 in Scientific Notation?
Scientific notation is a way of representing a number in a more compact form. In scientific notation, numbers are written as a product of two parts: a coefficient and a base. The coefficient is a number between 1 and 10, and the base is a power of 10. So, 5601 in scientific notation is written as 5.601 x 10³.
## How to Calculate 5601 in Scientific Notation?
In order to calculate 5601 in scientific notation, you need to first determine the coefficient and the base. To do this, you need to move the decimal point so that the coefficient is between 1 and 10. In this case, you move the decimal point three places to the left, so that the coefficient is 5.601. The base is 10 to the power of the number of places you moved the decimal point, which in this case is three. Therefore, 5601 in scientific notation is 5.601 x 10³.
## Uses of Scientific Notation
Scientific notation is often used to represent very large or very small numbers. It is especially useful in scientific calculations, as it allows for more precise and efficient calculations. For example, it is much easier to calculate the product of two numbers written in scientific notation than it is to calculate the product of two numbers written in standard form.
## Limitations of Scientific Notation
Although scientific notation is useful in some cases, it can be confusing and difficult to read. This is why it is not used in everyday calculations. Instead, it is used mainly for scientific and technical applications. In addition, some calculators cannot handle numbers written in scientific notation, so care must be taken when using them.
##
In conclusion, 5601 in scientific notation is 5.601 x 10³. Scientific notation is a useful way of representing very large or very small numbers, and is often used for scientific calculations. However, it can be difficult to understand and some calculators cannot handle numbers written in scientific notation, so care must be taken when using them. | 436 | 2,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-14 | latest | en | 0.948781 |
https://quizizz.com/en-in/area-of-rectangles-and-parallelograms-worksheets | 1,712,979,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00135.warc.gz | 432,818,366 | 33,787 | Area Review
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http://icopiedyou.com/im-a-generalist-that-likes-dipping-down-to-where-my-brain-hurts-and-then-coming-back-up-again-category-theory-really-hits-all-the-right-notes-with-me-while-a-lot-of-it-is-over-my-head-it-_does_-d/ | 1,610,838,395,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00031.warc.gz | 47,285,399 | 23,035 | # I’m a generalist that likes dipping down to where my brain hurts and then coming back up again. Category theory really hits all the right notes with me. While a lot of it is over my head, it _does_ do the magic thing of loosely tying lots of diverse things together using a consistent language. So even if I don’t understand their peculiar way of writing, simply seeing two different examples of a category together in the same place tells me that if I know one, I know the other. This is what happened with mathematical morphology. I saw these features in this graphics program that’s used for medical imaging. [I’m not a doctor but I love playing with images and trying new things]. – ImageJ. I wondered, “What is “erode”? What is “dialate?” What is “watershed function?” “what is Laplacian of Gaussian?” and discovered mathematical morphology. Things like Infimum and supremum fascinated me. It actually made logical sense and I could experiment visually. I found it it was a complete lattice. Not knowing what that was, I looked into that. Then I saw other examples of complete lattices on nLab (the Category theory site). — and it’s as if a whole bunch of math I didn’t know before, I could now at least partially wrap my brain around because it should work similarly to what I’m learning – and it does. I’m lucky in a way: I don’t worry about Gödel’s incompleteness and never did. I started with computing, which “sidestepped” Gödel by introducing steps and time and a different way to think of functions. Church, Turing, Von Neumann, with Lambda Calculus in the 1930s and 1940s, then Claude Shannon in 1948 with Information Theory. — this is the kind of family I was raised with. I didn’t know about set theory or Gödel’s incompleteness until my 20s, long after I’d already worked with computing-type things. My own mathematics abilities stopped at Trig. I got a D in Calculus because I didn’t care about slicing circles and could not understand why I couldn’t simply use rough estimates under the curves as I was used to pixels and grids not smooth things. So, it never became a dilemma for me. But the discontinuity between ways-to-think-of-math always hummed in the background of my mind and it was only a few years ago that I really paid attention to what was bothering me.
I’m a generalist that likes dipping down to where my brain hurts and then coming back up again. Category theory really hits all the right notes with me.
While a lot of it is over my head, it _does_ do the magic thing of loosely tying lots of diverse things together using a consistent language.
So even if I don’t understand their peculiar way of writing, simply seeing two different examples of a category together in the same place tells me that if I know one, I know the other.
This is what happened with mathematical morphology. I saw these features in this graphics program that’s used for medical imaging. [I’m not a doctor but I love playing with images and trying new things]. – ImageJ.
I wondered, “What is “erode”? What is “dialate?” What is “watershed function?” “what is Laplacian of Gaussian?” and discovered mathematical morphology.
Things like Infimum and supremum fascinated me. It actually made logical sense and I could experiment visually.
I found it it was a complete lattice. Not knowing what that was, I looked into that. Then I saw other examples of complete lattices on nLab (the Category theory site). — and it’s as if a whole bunch of math I didn’t know before, I could now at least partially wrap my brain around because it should work similarly to what I’m learning – and it does.
I’m lucky in a way: I don’t worry about Gödel’s incompleteness and never did. I started with computing, which “sidestepped” Gödel by introducing steps and time and a different way to think of functions.
Church, Turing, Von Neumann, with Lambda Calculus in the 1930s and 1940s, then Claude Shannon in 1948 with Information Theory. — this is the kind of family I was raised with. I didn’t know about set theory or Gödel’s incompleteness until my 20s, long after I’d already worked with computing-type things.
My own mathematics abilities stopped at Trig. I got a D in Calculus because I didn’t care about slicing circles and could not understand why I couldn’t simply use rough estimates under the curves as I was used to pixels and grids not smooth things.
So, it never became a dilemma for me. But the discontinuity between ways-to-think-of-math always hummed in the background of my mind and it was only a few years ago that I really paid attention to what was bothering me.
I still love the notion of infinitesimals and infinities but I also kind of think they’re artifacts of our brain’s compression faculties smoothing lines and shapes out for us.
I usually stop at “I grasp the concept”. I don’t do much actual mathematics. I’d say I probably don’t do any. But I like knowing what’s happening when there’s math happening. So, if i’m working with images, or video or music, I like knowing what the various computer operations are doing and to know that I need to understand the basic algorithms at play, and to understand the basic algorithms at play, I need to know what mathematical concept these algorithms are enacting on my computer and to know what that mathematical concept is about, I need to know if it’s modeling any physics that I can imagine. Sometimes it isn’t modeling any physics I can imagine and that’s when I need to find other sources for analogy. That’s where category theory is nice for me.
—- | 1,229 | 5,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-04 | latest | en | 0.958846 |
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# If 45 is 120% of a number, what is 80% of the same number?
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If 45 is 120% of a number, what is 80% of the same number? [#permalink]
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09 Aug 2019, 01:10
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If 45 is 120% of a number, what is 80% of the same number?
A. 30
B. 32
C. 36
D. 38
E. 41
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Re: If 45 is 120% of a number, what is 80% of the same number? [#permalink]
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09 Aug 2019, 01:27
45 = 1.2x
x = 45/12*10
4/5x = 4/5*45/12*10 = 30.
Re: If 45 is 120% of a number, what is 80% of the same number? [#permalink] 09 Aug 2019, 01:27 | 500 | 1,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-24 | latest | en | 0.836369 |
https://articlereview.biz/case-study/how-to-write-introduction-for-math-ia | 1,701,339,370,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00370.warc.gz | 127,592,619 | 36,349 | ## How to Structure & Format Your Maths IA
• March 23, 2023
• 1 Introduction:
• 2.1 Introduction:
• 2.3 Good Introduction:
• 2.4 Mathematical Background:
• 2.5 Exploration:
• 2.6 Conclusion:
• 2.7 Good Conclusion:
• 2.8 References:
## Introduction:
Mathematical exploration is an important part of the International Baccalaureate (IB) program, and writing a math Internal Assessment (IA) is a key requirement for students seeking to earn an IB diploma. The IA provides an opportunity for students to delve deeper into a mathematical concept of their choice, and to demonstrate their understanding and application of mathematical concepts and techniques . Math IA examples can serve as valuable references when embarking on your IA journey.
Mathematics Analysis and Approaches (Math AA) is one of the two courses offered by the IB for Mathematics. Math AA focuses on developing students’ mathematical knowledge and skills in pure mathematics, including algebra, calculus, geometry, and trigonometry. Mathematics Applications and Interpretation (Math AI) is the other course that focuses on applying mathematics to real-world contexts, including statistics, probability, and modeling.
Both Math AA and Math AI require students to complete an IA by the end of the course and the IA is an extremely important part of a student’s math grade. However, writing a successful math IA requires more than just a solid grasp of mathematical concepts ; it also requires careful planning and structure. In this blog post, we’ll explore the key elements that make up a strong math IA structure, including the introduction, mathematical background, exploration, and conclusion. We’ll discuss the purpose and content of each section, and provide tips and strategies for crafting a clear and effective IA. Whether you’re just getting started on your math IA, or are looking to refine and improve your existing work, this blog post will provide valuable insights and guidance to help you succeed.
## Structure of the Math IA:
The IA for both Math AA and AI follows a similar structure. Students are required to select a topic from one of the four areas of study: Algebra, Functions and Equations, Circular Functions and Trigonometry, and Calculus. The IA should be approximately 12-20 pages long, excluding appendices, and must contain the following components:
## Introduction :
In this section, you should introduce the topic you have chosen and explain why it is important or interesting. You should also include a clear statement of the aim and objectives of your IA . A well-crafted introduction, like in the following maths IA examples , can engage your readers and set the stage for your exploration:
In this math IA, I will explore the topic of calculus. Calculus is a branch of mathematics that deals with rates of change and slopes of curves. I will look at some calculus concepts such as derivatives and integrals and try to explain them in simple terms. This will help readers understand the basics of calculus.
This introduction is too broad and does not provide a clear focus for the IA. It is also not engaging and doesn’t grab the reader’s attention.
## Good Introduction:
Have you ever wondered how the trajectory of a soccer ball is calculated after it’s been kicked? What about the math behind predicting the weather or designing roller coasters? All of these scenarios involve the application of calculus, a branch of mathematics that allows us to understand the behavior of curves and rates of change. In this IA, I will explore the concepts of calculus through real-world examples, demonstrating how it can be used to solve complex problems and enhance our understanding of the world around us.
Why it’s good:
This introduction sets up the IA by engaging the reader with real-world scenarios and providing a clear focus on exploring calculus. It also hints at the importance and relevance of the topic, making the reader interested to learn more.
## Mathematical Background :
This section should include a brief overview of the mathematical concepts and techniques that you will be using in your IA. It is important to show a deep understanding of mathematical theory and to provide clear explanations of any equations or formulas you will be using. You can find that in most Math IA examples or sample IAs often begin with a solid mathematical foundation:
• The background is too general and does not provide specific information on the mathematical concepts that will be used in the IA.
• It does not indicate the level of mathematical knowledge that is required to understand the IA.
• It does not provide any context or motivation for why these mathematical concepts are important.
## Good Mathematical Background:
• The background provides specific information on the mathematical concepts that will be used in the IA.
• It indicates the level of mathematical knowledge that is required to understand the IA.
• It provides context and motivation for why these mathematical concepts are important.
• It is written in clear and concise language that is easy to understand.
• It may include references or additional resources that readers can use to refresh their knowledge of the relevant mathematical concepts.
## Exploration :
This is the most substantial section of your IA and should demonstrate your ability to use mathematical tools to solve a real-world problem. You should explain the methodology used to investigate the problem and provide clear evidence of your calculations and reasoning. It is important to show your work clearly, including any graphs or diagrams you create look at some of t hese math ia exploration examples
• The exploration is too simplistic and does not demonstrate a deep understanding of the mathematical concepts being discussed.
• The methodology used to explore the mathematical concepts is not clearly described, making it difficult for readers to follow the logic of the argument.
• There is no data or evidence provided to support the conclusions drawn in the exploration.
• The exploration does not connect the mathematical concepts to any real-world applications or examples.
## Good Exploration Section:
• The exploration demonstrates a deep understanding of the mathematical concepts being discussed, and provides a clear explanation of how these concepts are being applied.
• The methodology used to explore the mathematical concepts is clearly described, making it easy for readers to follow the logic of the argument.
• The exploration is supported by data or evidence, which helps to strengthen the argument being made.
• The exploration connects the mathematical concepts to real-world applications or examples, which helps to provide context and motivation for why these concepts are important.
• The exploration may include diagrams, graphs, or other visual aids to help illustrate the mathematical concepts being discussed.
• The exploration may include examples of how the mathematical concepts have been used in previous research, or how they can be applied to solve real-world problems.
## Conclusion :
In this section, you should summarize your findings and explain how they relate to the original aim and objectives of your IA. You should also reflect on any limitations of your investigation and suggest potential areas for further research . A poorly-crafted conclusion, like in the following math IA conclusion examples , can leave a bad impression:
In conclusion, we have explored the concepts of calculus and shown how they can be applied to solve problems. We have seen that calculus is a powerful tool for understanding rates of change and curves, and can be used to model a wide range of phenomena. We hope that this IA has given you a better understanding of calculus and its applications.
This conclusion is too general and does not summarize the specific findings of the IA. It also does not provide any insights or recommendations for further research or applications.
## Good Conclusion:
In conclusion, we have demonstrated how linear regression can be used to model real-world data and make predictions. Our analysis has shown that this technique can be used to effectively model and predict various phenomena, from the price of stocks to the spread of infectious diseases. However, there are still many areas where linear regression can be improved, such as accounting for nonlinear relationships or dealing with outliers. Future research in this area could explore these issues and develop new techniques for improving the accuracy and reliability of linear regression models.
This conclusion provides a clear summary of the main findings of the IA, and suggests potential avenues for further research and development. It also highlights the relevance and importance of the topic, and indicates that there is still much to be learned and discovered in this field.
## References :
It is important to include a list of all sources that you have used in your IA. This includes any textbooks, articles, websites, or other resources that you have consulted. You should also provide clear citations within your IA to show where you have used information from these sources.
Here are a few pointers on how to format your IA well:
• Title Page: The title page should include the title of your IA, your name, your candidate number, the date, and the word count.
• Page Numbers: All pages of your IA should be numbered, including the title page and appendices.
• Font and Size: Use a clear, legible font such as Times New Roman or Arial in size 12. Use a larger font for headings and subheadings.
• Line Spacing: Use double line spacing throughout your IA, except in tables, equations, and diagrams.
• Appendices: Any additional material, such as raw data, should be included in appendices at the end of your IA. Make sure to refer to these appendices in the main body of your IA.
• Graphs and Diagrams: All graphs and diagrams should be labeled clearly and should have appropriate titles and axes. Use colors and shapes to distinguish between different data points or lines. Make sure that the labels and titles are legible and that the scales are appropriate for the data being presented.
• Equations: All equations should be presented clearly, with variables and constants clearly labeled. Use appropriate notation and make sure that equations are properly formatted, with fractions, exponents, and other mathematical symbols clearly presented.
• Tables: All tables should be labeled clearly and should have appropriate column headings. Make sure that the tables are presented in a logical order and that any units of measurement are clearly indicated.
• Language: Use clear and concise language throughout your IA. Avoid using jargon or technical terms unless they are necessary. Make sure that your writing is grammatically correct and that you use appropriate punctuation.
• Proofreading: Finally, make sure that you proofread your IA carefully before submitting it. Check for spelling and grammar errors, and make sure that all of your equations, graphs, and tables are properly labeled and formatted. It can be helpful to have someone else read through your IA to check for errors or inconsistencies.
Here are a few additional tips to help you write a successful IB Math AA and AI IA:
• Choose a topic that you are interested in: You will be spending a lot of time on your IA, so it is important to choose a topic that you find engaging and challenging. This will help you stay motivated and focused throughout the process.
• Use a variety of sources: In order to demonstrate a deep understanding of the mathematical concepts and techniques used in your IA, it is important to use a variety of sources and to cite them properly. This demonstrates that you have researched your topic thoroughly and are able to apply the knowledge you have gained from a range of different sources. Textbooks and academic journals can provide a strong foundation for your research, as they often present complex mathematical concepts in a clear and concise manner. Reputable websites, such as those associated with educational institutions or professional organizations, can also provide useful information and insights.
• Use real-world examples: The IA is an opportunity to apply your mathematical knowledge and skills to real-world problems. Whenever possible, use examples that are relevant to your own life or that demonstrate the practical applications of the mathematical concepts you are studying. By selecting examples that are relevant and meaningful to you, you can not only create a more engaging project but also deepen your understanding of the mathematical concepts you are studying. The IA provides an opportunity to demonstrate the practical applications of mathematics and showcase the importance of this field in solving real-world problems.
In conclusion, the structure of your IB Math AI and AA IA is crucial to the success of your project. Your IA should be well-organized, clearly written, and contain all necessary components, such as an introduction, background information, data collection and analysis, and a conclusion. By following the guidelines provided by the IB, selecting an appropriate topic, and using effective mathematical tools, you can create a compelling IA that showcases your skills and knowledge in mathematics. Remember to plan ahead, manage your time effectively, and seek help if needed. With these tips in mind, you can excel in your IB Math IA and AA IA and achieve your academic goals.
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## How to Structure Your Math IA Format
Are you getting ready to present your Math IA? In this article, we’re breaking down the best way to format yours in a way that promotes independent thinking and personal engagement. By following this guide, you’ll be able to score highly in your mathematics course.
## Understanding What Math IA is All About
Students are required to investigate an area of mathematics and then present it as a written body of work.
Based on this, it is clear that the student needs to know the appropriate mathematical language and relevant key terms to engage in the topic.
While there are different perspectives when it comes to formatting your maths IA, we recommend the steps outlined below.
By following this tried and tested structure for your math IA, each student will be able to present their findings consistently and optimally
Personal connection and engagement with the subject matter are key. Familiarise yourself with the steps below, and then dive into your chosen topic. Utilising technology can be helpful in conducting a thorough exploration of the topic.
The use of mathematics to prove and explain concepts is applicable to various contexts. This assessment is a great way for students to expand their knowledge and learn valuable skills.
Your assessment needs to define the concept and aim of the work. This will help to keep your data analysis focused. These ideas need to be conveyed in your writing. This is an area of maths that is not strictly bound to the number!
The examiner will be looking at the quality of the idea and the body of work in relation to the assessment criteria.
## How to Structure and Format Maths IA
According to student reviews, IB Maths is a struggle for many students working towards getting their diplomas. Apart from the dreaded mathematics exam , you’re also expected to write up the exploration of a topic.
You can think of the IA as a written mathematical presentation that will impact your final grade.
Many IB students find it hard to study mathematical concepts. This process can help you reflect on new relevant logic that you may not have known previously.
## Math IA Structure: What is the Internal Assessment?
The internal assessment is an individual evaluation that focuses on subject-related work.
Alongside the criteria, samples of the student’s work (oral performances, portfolios, lab reports, and essays) are also submitted to the IB for the final grade.
It should show personal engagement with the topic at hand. In this case, it is about the investigation of and correlation between mathematical concepts.
## Breakdown of the Math IA Assessment : Appropriate Mathematical Language
Let’s take a look at the criteria for your Math IA. Knowing how you’re being graded will make it much easier to make sure you’re ticking all the boxes.
## Presentation (4 marks)
The first criterion is about the presentation, with the aim of assessing the general organization and coherence of your IA.
Although students tend to focus on the complexity of math that their exploration demonstrates, a full 4 points are rewarded for the clarity of your explanations and structure.
In order to score in the top range here, make sure your IA is clearly structured. We’ve shared the optimal format in the next section.
## Mathematical Communication (4 marks)
The second criterion looks largely at the mathematical language you have used, such as:
• Terminology
Ensure that these three components are accurate and consistent throughout your IA.
Terms like “plug in” or “put in” should be replaced with mathematically sophisticated words like “substitute.”
## Personal Engagement (3 marks)
To achieve the top marks for personal engagement, your engagement must be truly authentic and drive the exploration forward.
It needs to be independent and unique. It should display a degree of creativity in that you present mathematical ideas in your own way and explore the topic from various different perspectives.
This involves making predictions about things you may be interested in, and then finding ways to manipulate the problem, formula, or question to encompass those areas.
## Reflection (3 marks)
The IB needs us to do more than just show what we’ve done.
During the reflective stage, connect the results with the initial aims. By doing this, you can determine findings throughout the process.
It is about evaluating the research to pick up on all evidence that goes beyond what a typical mathematical test would.
The IB is all about learning, so be sure to show the marker your growth throughout the IA.
## Use of mathematics (6 marks)
This section looks at the quality of the maths and how relevant it is to the exploration.
The IB is measuring relevancy by checking that you only included maths that is directly intended to answer the research question.
It is also worth noting that the maths produced should be at a similar level to the math you cover in your syllabus. This doesn’t mean that you’re confined to only looking into topics that are covered on your syllabus, but it should be of the same rigour!
## Introduction:
Like almost all of your internal assessments, your Maths IA has to begin with a super clear introduction that sets the context and aim of the whole exploration.
It is a great place to show your ‘personal engagement’ with the topic you’ve chosen for your IA.
Be sure to account for your interest in the top, its relevance in your life, your prior knowledge about it, what you wish to achieve, and how you’ll arrive at an answer.
You may also include any personalised problem statements and explain how you aim to achieve a solid investigation on the topic.
The body of your IA exploration should focus on the particular topic you have chosen to investigate and the relevant mathematical material that will address the intended aim of the work.
A pivotal point to consider is the level and clarity of the mathematics you use – the IB rewards a lot of marks for the use and communication of Mathematics, so keep this in mind when you start writing your Maths IA up!
For more info on how to write the exploration, check out our complete Math IA Guide .
## Math IA investigation:
As with all assessments, you also need to include a solid conclusion that summarises the research and work you’ve done.
What conclusions did you reach, and did you succeed in exploring the aim that you set out at the beginning of the Math IA!
Importantly, make sure to also discuss some of the challenges in your IA and what you would/could explore with more time and more words.
Finally, zoom out and think about the further implications of your study.
Did your learning affect your life in any way, or how might it affect the lives of others? How has your involvement allowed you to reflect on different mathematics topics?
There is no specific word count for your Math IA, but the IB advises that the exploration should be around 12-20 pages long.
## Fonts and Spacing for Your Math IA:
There are no specific requirements on which font you should use, but going with Arial or Times New Roman is generally recommended, with double line spacing and font size of 12.
You may present your work in a word processing software (like Microsoft Word or Pages), or it can be handwritten.
## Diagrams and Graphs:
You should include relevant graphs, tables, and diagrams.
Do not simply place these simply as appendices at the end of the essay – they should be fully and clearly labelled to ensure that the examiner knows what you’ve included and why.
It forms an essential part of your research and shows that you fully understand the examples you have included in your analysis.
## Bibliography and Citations:
Your report should include a full bibliography with all sources at the end of the report.
In addition to a bibliography at the end, you must acknowledge all direct quotes that you use throughout your essay.
## Topic Ideas for SL and HL mathematics
There are many ideas you can explore for the assessment, including graph theory, surface area, geometry, calculus equations, statistics, linear regression, modelling statistics, the SIR model, etc.
In your development, you may investigate the correlation between different topics or ideas within Mathematics SL and HL or your AI SL.
Whatever you choose, remember that you will create a new exploration of complex ideas. It’s wise to form your own opinions based on substantive evidence.
## Mathematical Concepts Conclusion: The IB Structure
So there we have it, a well organized exploration of the idea and IB layout of your Mathematics IA !
We hope this will give you the push you need to realise your potential and understand complex and unfamiliar mathematics. As with all mathematics, personal involvement by means of practising is best if you want to score highly in your HL syllabus.
## Lanterna Resources and Opportunities for Math IA
If you need a bit more of a boost, we’ve got free resources for IB students that you might find helpful! For more personal engagement, feel free to reach out to your instructor or tutor.
Lanterna also offers the following to support students:
• Revision Courses : These offer a helping hand when it matters the most to boost your grades! Look out for these during the Winter and Easter breaks before the final exam.
• Summer Courses : When you are about to start your first or final year of the IB, the summer presents the perfect opportunity to get ahead.
• Online Private Tuition : One-on-one support from the comfort of your own home, whenever and wherever you need it.
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## How To Write Your Mathematics Internal Assessment
Mathematics Internal Assessment (IA) is a crucial component of the IB Diploma Programme. It is a student-led exploration that enables the learner to delve deeper into a mathematical concept or problem of their interest. The Math IA is a 12-20 page report that showcases the student’s ability to conduct independent research, analyze data, apply mathematical concepts, and present their findings in a clear and coherent manner.
These are the reasons why the IA is invaluable:
1.Holistic assessment: The IA gives an opportunity to students to showcase their research skills, and is a more holistic assessment of their mathematical abilities.
2.Developement of critical skills: Problem-solving, logical reasoning, data analysis, and communication are crucial for academic and professional settings.
3.Application of Mathematics in real-world contexts: This allows students to have a more relatable experience with the abstract Mathematical knowledge they have gleaned in their coursework.
4.Development of research and analytical skills that can be used in the university and research environments.
5.A solid IA serves as a good indication to universities that a student is well-prepared to undertake independent study or research. Writing a Math IA can be daunting, but with proper planning and execution, it can be a rewarding experience that helps you understand the beauty and power of mathematics.
In this guide, we will discuss how to organize your Mathematics IA.
## IA Cover Page
It should have the following elements: A)Title: The title of the Math IA should be clear, concise, and reflective of the exploration’s focus. It should give the reader an idea of what the IA is about.
B) IB Candidate Number: This is your unique identification number provided by the International Baccalaureate organization. It typically consists of three letters followed by three digits (e.g., ABC123). The candidate number helps identify your work during assessment and ensures anonymity.
C) Session: Indicate the examination session during which the IA was submitted. For example, it could be “May 2023” or “November 2023,” depending on the IB exam session.
D) Page Count: 12-20 Pages in length with double spacing. The page length per subsection is not set, but one can imagine it should correspond to the marking rubric.
E) Course and Level: Specify whether the Math IA is for Standard Level (SL) or Higher Level (HL) Mathematics. This is crucial as the level of sophistication and depth of mathematical exploration may vary between SL and HL.
F) Date of Submission: Include the date when you submitted the Math IA. This helps track the timeline of your work.
G) School Name and Code: Mention the name of your school and the school code. This information helps identify the institution where the IA was completed.
H) Candidate Name: Include your full name on the cover page. This ensures that your IA is properly identified and credited to you.
I) Abstract (Optional): Some schools may require an abstract to be included on the cover page. An abstract is a concise summary of the IA, typically consisting of 100-150 words. It should provide an overview of the research question, the methods used, and the main findings.
J) Declaration of Authenticity (If Required): Some schools or examiners may ask you to sign a declaration of authenticity on the cover page, stating that the work is entirely your own and has not been plagiarized or copied from other sources.
## IA Important Rubric Requirements
(A) Mathematical Presentation (0-4) In order to get the full 4 marks for this criterion, there must be a coherent structure, accompanied with clear explanations. A general guideline for this is to break the IA into:
## Section 1: Introduction
Introduction – Why, what, then how.
Why? Your IA introduction should include a rationale for why you have chosen your topic for your Mathematical Exploration (the name of this IA). You should find some personal way to engage with your chosen topic to satisfy this requirement. Choose a topic you’re genuinely interested in, state said interest explicitly and use your own personal examples where possible.
What? This is where you define the objective for your IA, and state what you want to achieve with this aim. Picking a topic – specifically an aim – should be considered carefully and in conjunction with your tutor/teacher to ensure there is sufficient depth to your topic (as this depends on whether you’re taking SL or HL Math). Make your aim explicitly – this is important.
Some examples of previous IA topics are listed below (these are basic topics and not finalised research questions) and in the appendix. However, remember that there ought to be some personal engagement within the topic-choosing process: “Why planes travel a curved route and not a seemingly direct route” “Does the stock market’s returns warrant its variance?” “Projectile motion” “L’Hôptal’s rule and evaluating limits” “Image rotations using rotational matrices”
How? You must outline how your exploration topic relates to your specific curriculum, how you’ve completed the exploration, and provide any necessary background information – your classmates should be able to understand your IA if they were to read it.
Then how: This means your plan of action. The level of difficulty must be appropriate for the your level (SL or HL).
## Section 2 (Body): Theory & Calculation
Theory Provide only the relevant theory needed to reach a conclusion/understanding of your aim. If there is a particular method (in mathematics, there are often numerous ways to reach the same answer) that you’ve used you should explain the method and why you’ve used this method.
Calculation For this section you must include all formulae and assumptions (i.e., the actual numbers) used to make your calculations and the mathematical steps that you took to reach your aim. Note assumptions’ pertinence if someone wants to repeat your exploration. After going through your mathematical work you must explain how they relate to your exploration topic. Depending on the type of exploration in which you are partaking you should use appropriate graphs, tables, x-y-z planes, or other methods of presenting your results. See below.
As can be seen from the figures above, figures are labelled appropriately. Calculations come with brief explanations and connect the earlier theory with the specific scenario in your exploration.
Note that the theory section should include not only the explanation of relevant theories but also the related background information. Likewise, ensure that the calculation section is comprehensive and includes all necessary formulas and mathematical steps.
## Section 3: Reflection, Conclusion, and Bibliography of Mathematical Exploration
Conclusion Your conclusion is a continuation of section 1 and 2. You are answering your aim from your introduction (section 1) with the theory and calculations in from section 2. This should be done in a clear, concise, and coherent manner. Not only should you explain the results and implications of your calculations, but you ought to relate this to the aim raised in your introduction. You may also include much of the reflection in your conclusion if you prefer a more integrated approach. Note the IB says the following regarding where the reflection should be placed: “Substantial evidence means that the critical reflection is present throughout the exploration. If it appears at the end of the exploration it must be of high quality and demonstrate how it developed the exploration in order to achieve a level 3.” This implies a preference for integration but it does not mean you are excluding yourself from a level 3/3 grade for the reflection rubric.
Reflection Your reflection should occur throughout your IA; however, you may also include a separate section depending on the layout of your IA. Here’s what you should do:
• Consider limitations and extensions of your conclusion.
• Similarly consider strengths and weaknesses.
• Relate the mathematics within the exploration to your personal knowledge (or personal engagement).
• Raise future research questions.
• The IB states your reflection must be “crucial, deciding or deeply insightful. It will often develop the exploration by addressing the mathematical results and their impact on the student’s understanding of the topic.”
Bibliography A detailed bibliography is required so you must keep all sources which you utilise throughout your IA process. You should include a thorough bibliography to support your introduction, background, theory, and perhaps calculations. Types of relevant sources include online databases, your school textbook, or specific theories found both online and physically.
Mathematical communication
• This rubric assess the student’s ability to organize his research and findings in a coherent manner. The fluency of the mathematical language used is assessed via the following ways:Notation
• Symbols (choose universally accepted ones)
• Terminology (All key terms and variables must be defined appropriately when first introduced)
Personal Engagement
A unique part of this IA is the personal engagement. This rubric assess a student’s ability to make the Mathematics his/her own – Infusing creativity in the exploration of ideas from multiple angles. The student must develop his or her own independent understanding of the topic he/she is exploring. For example, you could write in the first person why you have decided to pursue your interest in Fourier series. You could link it with your interest in understanding heat conduction. You discuss the challenges you encountered during your research, discuss how you felt, and emphasize the mathematical concepts you have learned during the investigative process.
Use of Mathematics
This rubric assesses how well a student uses the Mathematics in the exploration. Simplicity is encouraged, and the level of difficulty should be pegged at the level of the course. The Mathematics must be precise, and exhibiting a clear logical structure. Note the difference between receiving a 6/6 for the use of mathematics rubric for HL/SL according to the IB:
SL – “Relevant mathematics commensurate with the level of the course is used. The mathematics explored is correct. Thorough knowledge and understanding are demonstrated.”
HL – “Relevant mathematics commensurate with the level of the course is used. The mathematics explored is precise and demonstrates sophistication and rigour. Thorough knowledge and understanding are demonstrated.”
In both cases you should use mathematics of a similar level to what you are studying in your respective studies. However, in HL, the mathematics that is explored must be precise and shows sophistication and rigour. Students must show the mastery and use of complex concepts, be able to see the Mathematical problem from various viewpoints, and see the ways to link different areas of Mathematics.
E.g., use of mathematics carries a weighting of up to 6/20 while reflection carries a weighting of up to 3/20. Hence, you should expect to spend more pages on calculations than your reflection.
Topics to spur interest (note these topics are not developed but are included to stimulate your own more developed aims):
In conclusion, a well-structured layout is crucial for a successful Math IA, including an introduction, main body, analysis, and conclusion. Meeting the rubric requirements is also essential for a high grade, which includes criteria such as presentation, communication, and reflection. By following these guidelines, you can write an engaging and informative Math IA.
If in doubt, reach out to experienced tutors at Quintessential Education for extra help and guidance. Start your journey towards academic success today!
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## Maths Internal Assessment: Advice & Tips
Choosing the right subjects can set up a student for life. make sure your child starts the ib on the right foot by speaking with an eib consultant to pick the perfect subjects..
For your Mathematics Internal Assessment, you have to complete a Mathematical Exploration on a topic of your choice. Whilst sometimes feared by students, this is a chance for you to explore an area of mathematics that interests you! It is a piece of written work that is graded out of 20 marks in total and contributes to 20% of your total grade. Here we look at the structure of the assessment, how to choose a topic, how to lay out your assignment and what to include.
Structure and how to choose a topic
The Maths IA forms 20% of your overall grade for Maths studies, SL and HL. The IBO recommend that it is 6-12 pages in length and includes maths that is commensurate with the level of difficulty of the course. This can be partly formed by maths studied within the IB maths syllabus, but it is also recommended that you look elsewhere for inspiration.
You could look at the optional modules (available at HL) that you do not study for potential starting points and build from here. Alternatively, think about an area which interests you where maths could be applied. Personally, I chose to investigate the bottle flip for my Maths IA as this became well known whilst we were preparing our IA ideas. Choosing something personal will allow you to score highly on the personal engagement criteria, which we will talk more about below.
A good way to structure your internal assignment would be the following:
• Introduction – Consider the background to your topic and where there is an opportunity for mathematical exploration
• Rationale – Why did you chose your topic (linking this to the introduction)
• Aim – What you would like to achieve from your exploration. Make sure these are clear and focused aims and show evidence supporting your personal engagement
• Main body – Where the maths comes in. Look at different perspectives on the problem and discuss their respective merits and outcomes
• Conclusion – What did you learn, what were your limitations and what areas are there for further exploration and application of your research?
Whilst this is a good way of approaching the assessment, the IBO state that this is definitely not the only way. If you think there is a better way do use it as this may allow you to show personal engagement and score highly.
Marking criteria
The IA marking criteria are split into 5 categories, which we break down below:
Communication (4 marks)
The way you communicate in the Internal Assessment needs to be coherent, well organised, concise and complete. Conciseness has been emphasised as the IBO have recognised that some IAs have been too long and this has detracted from the quality of students’ work. To reiterate, 6-12 pages of concise and rigorous critical thinking is a good guide regarding how much to write.
In addition, the IBO have recommended that all graphs, tables and diagrams should be embedded in your main body (not as appendices). This will help the examiner read through your work clearly and so will improve your chances of scoring a better mark. Similarly, citing when and where other people’s ideas are used throughout your essay is key, as well as having a bibliography at the end. This will ensure you steer clear of plagiarism by highlighting to the examiner which work is your own – which will also improve your chances of scoring highlight in personal engagement.
Mathematical presentation (3 marks)
Firstly, the IBO recommend that you define key terms where required. Definitions, which can be found in textbooks or journal articles (please remember to reference!), will help the examiner follow your exploration and shows that you understand the topic. Similarly, ensure that you are using appropriate mathematical language such as notation, symbols and terminology. Try not to be too complex in the language you use as, if used wrongly, this may show a lack of understanding and drag both your mathematical presentation and communication marks down. However, please do use mathematical language where necessary. Define it, and explain clearly what it is referring to in your exploration.
Furthermore, try to make your work look visually pleasing by using multiple methods of presentation. These may be formulae, diagrams, tables, charts, graphs or models, and should be used to demonstrate your understanding of the topic. Please type out your own equations and draw your own graphs (for example, use Desmos and print screen, limiting any branding that is visible) and ensure all graphs have axes labels and headings. Finally, and most importantly, ensure all graphs have a purpose! This will demonstrate an ability to critically highlight the key elements of your exploration in a concise and coherent manner.
Personal engagement (4 marks)
Often the trickiest of the 5 criteria to score highly in, personal engagement requires you to demonstrate your connection with the topic. There are a number of things you can do that will improve your mark in this criterion:
• Speak in first person – This will show that you are the one who undertook and researched this project and that you value it
• Present the exploration in your own style – Differentiating yourself from other students in this way will show you have not just followed a template approach
• Talk about the challenges you faced – Doing so will improve your marks in the personal engagement and reflection criteria
• Discuss how you felt (surprised, frustrated) as your investigation progressed – Again, this shows your personal connection with your exploration
• Highlight the mathematical concepts you have learned – How has this exploration developed your mathematical ability?
Why Pi matters
Reflection (3 marks)
Proving yourself wrong or getting a result you were not expecting is not necessarily a bad thing as it will offer you a chance to reflect and score highly on this criterion. Try to recognise what caused the result that you were not expecting and, as importantly, how this could be fixed. Also reflect on the successes of your exploration – Did you succeed in proving something, obtaining an optimum result or modelling a difficult concept? Discussing both the strengths and the weaknesses of your work here will help you to score highly.
Moreover, try to discuss different approaches that you have taken or could have taken and how these may have impacted your results. For example, did you use a simple model to begin with (as I did in my IA), which allowed you to isolate the key equations, and build a more complex structure from here? Evaluating potential approaches from different perspectives will show strong critical analysis.
Furthermore, if you have space you could discuss the implications of your work or areas for further exploration. Do the results you have obtained imply any policy changes or develop our understanding of a topic area? Alternatively, could you apply your research to other fields, or explore further areas within the topic? If so, please try to include these to score highly for reflection.
Use of mathematics (6 marks)
Unlike many people think, you do not have to use mathematics that is way beyond the course level to score highly on this criterion. The IBO recommend that you use maths that is “commensurate with the level of your course”. This may take a syllabus topic, such as calculus, and develop it to go beyond what is covered in the IB course. Alternatively, it may apply this knowledge to a new scenario, as I did using differentiation when modelling the bottle flip. Ensure that, whichever level you choose, your exploration remains at a level where you can include “creativity or (a) personalised problem” to ensure you score highly on the personal engagement criterion as well.
One useful source of topics could be the optional modules that you do not cover as part of your course. If you are a Standard Level mathematician you have free reign to choose one of the Higher Level optional modules for inspiration. For example, a student interested in computer science may choose to explore the use of Dijkstra’s and Kruskal’s algorithms in graph theory. It is important to remember that the IB is more concerned about how you show your knowledge and understanding of the mathematical topic you choose. Sophisticated mathematics may include understanding and use of challenging mathematical concepts, looking at a problem from different perspectives or seeing underlying structures to link different areas of mathematics. The maths need not be complicated, just well explored.
Once you have written your exploration, it is important to proof read it and ensure that there are no grammatical errors and all references use the same structure. Moreover, please make sure that all of the 5 areas listed in the mark scheme have been addressed explicitly. Here is a link to a very useful proof-reading checklist to ensure you have everything included.
Whichever topic you choose to do your Maths IA on, there are 3 important things to remember:
• Be engaged! Show your personal enthusiasm for the topic through writing in first person, using your own style and describing the challenges you faced and how they made you feel.
• Aim to be clear, consistent and concise! Be clear with your use of mathematical terminology by defining it where necessary; be consistent with your layout and referencing style; and be concise when writing your work – longer does not always mean better.
• Have fun! This is your chance to explore an area of maths that is likely to not be covered by the syllabus and interests you. Make the project your own and enjoy the experience as much as you can.
Best of luck!!
If you need any help with your Maths IA or any of your other IB subjects, we offer tuition, talks and consultations about everything IB. Please get in touch if you would like to find out more, please see the relevant links at the top of the page, or check out a similar blog post from the great team over at Lanterna Education on how to structure and format your Maths IA here .
## Revision/Exam Tips: IB History
Tok tips: the essay, what level of maths should you choose in the ib, why igcse is a good preparation for the ib, third culture kids and the ib, using past papers to revise, which igcse subjects should you choose for university, why ib by joe thomas.
## IB Math IA Guide - Math IA Shenanigans That No One Will Tell Yeh!
Ace Your IB Math IA with our ultimate guide for 2023! Get top marks and ace your IA with ease. Discover proven tips, tricks and strategies to nail your Math IA today!
## Table of content
Criterion a - mathematical presentation: (levels- 0, 1, 2, 3, 4), criterion b - mathematical communication: (levels- 0, 1, 2, 3, 4), criterion c - personal engagement: (levels - 0, 1, 2, 3), criterion d - reflection: (levels - 0, 1, 2, 3), criterion e - use of mathematics: (levels - 0, 1, 2, 3, 4, 5, 6), introduction, body of your exploration.
IB Math students will tell you how they’re always on the edge of their seats for some help, but IB Math IA takes that anxiety to an entirely different level. The reality is far from frightening; nonetheless, IB Math IA can be handled well with a unique IB Math IA topic in hand and lots of coffee!
But does that guarantee a dependable 7?
It takes more than just a perfect IB Math IA topic to ace.
From researching several IB Math IA examples to planning the mathematical working of your exploration, your IB Math IA structure will get you into trouble if you don’t give it the time it demands. With all the varied content available in bulk online, the process is bound to become anything but easy.
But worry not!
You are at the right place - The Ultimate Guide to IB Math IA!
You should also know about the updated course structure of IB Mathematics. Students are allowed to opt for any one of the following four courses in Math:
• Analysis and Approaches ( AA HL ) - Higher Level
• Analysis and Approaches (AA SL) - Standard Level
• Applications and Interpretation (AI HL) - Higher Level
• Applications and Interpretation (AI SL) - Standard Level
Also, here’s a great surprise for all students!
Patrick Jones , the creator of PatrickJMT Math Videos, acknowledged as the best Math teacher globally with over 1.2 million subscribers on YouTube, has gotten on board with our team at Nail IB! How great is that! He is already working on creating an entire Nail IB video course, and it will prove to be a wholesome guide for you as you tread on your IB Math journey! You should check out his excellent, world-renowned content here !
Before moving any further, we insist you check out our Free IB Resources for IB Mathematics SL and IB Mathematics HL. These are specially assembled for your benefit and will surely assist you on your IB Math journey!
For an absolute hold on IB Math, check out our premium notes designed and curated specially for you, be it IB Mathematics SL or IB Mathematics HL . These bundles are not just limited to messages but offer past year papers and How-to Guides for Extended Essays, Internal Assessments, and more; examples included! You’re in for a smooth ride with these by your side;)
You can also stream our webinar on How to Write an IB Math Internal Assessment in under 30 minutes and hear directly from a recent IB graduate to understand the fundamental pointers and some fantastic hacks to lay the foundation of your IB Math IA. Getting the proper guidance ensures you a 7 in the subject you have feared for too long. Click here to watch it now!
First things first, let’s understand the criteria. Unless we acknowledge the requirements against which our exploration is scored, it’ll be equivalent to a shot in the dark. The conditions, irrespective of whether you opt for SL, HL (AA or AI), are as follows:
Assessment is done on the conciseness, brevity, and clarity/coherence of your investigation. The proper structure must be given to your IA. As per IB guidelines,
a coherent exploration is,
• Logically developed
• Meets the Aim.
Also, a well-organized exploration,
• Includes Introduction
• Describes the Aim of the investigation and
• Has a Conclusion
Assessment is done on the appropriateness of the mathematical terminology, notation, and symbols used to progress the exploration. Marks and notes should be correctly used as are used in IB textbooks. For example, x2 should not be written as x^2.
If used, different mathematical representation tools such as tables, graphs, and diagrams must be relevant to the working and be commented on/explained well. Avoid inconsistent use of Mathematical terminology. Applying ICT Tools(for example- GeoGebra and Desmos ) should be made wisely. For Calculations, Graphic Display Calculators can also be used, but that doesn't outdo math formulas' importance.
Assessment is done on the personal involvement shown. The sure shot way to ensure Engagement is, first and foremost, by going ahead with a topic that interests you (something unique or that affects real-life situations). Personal Engagement is seen throughout the exploration by:
• Independent thinking and creativity showed by the student
• Making the Math Idea your own
• Investigating the idea from varied perspectives
• Exploring different possibilities
Avoid portraying superficial interest. Opportunities for demonstration of personal Engagement should be noticed.
Assessment is done on the evaluation and analysis of the investigation. Mentioning the significance of your exploration results, discussing possible limitations, and justifying why you chose the procedure you did can portray a fair reflection of the IA. Merely explaining your results will get you only a score of 1 out of 3. According to IB guidelines, a review should be meaningful and critical.
A meaningful reflection includes
• Considering limitations in the work
• Comparing different Mathematical Approaches
• Commenting on the Learning
A critical reflection entails,
• Considering What Next
• Discussing the implications of the results
• Discussing the strengths and weaknesses of the approaches
• Considering different perspectives
Reflection is an analysis of the student's work, seen throughout the exploration, not just the Conclusion.
Assessment is done on the implementation of Mathematics in the IA. It is essential to understand that the Math used should be on par with the course, nothing too simple and nothing you need help understanding. Also, the Mathematics used should be fully understood and engrained by you. Unfamiliar Mathematics, if used, should be explained well by giving personal examples. As per IB guidelines, students are expected to produce work that is,
• Commensurate with the level of the course(should either be part of the syllabus at a similar level or slightly beyond)
• Relevant Mathematics used means Math which supports the development of the exploration towards the completion of its Aim
To score higher levels in Criterion E, it is crucial to understand the meaning of the following terms:
• Precise Mathematics is error-free and uses an appropriate level of accuracy at all times
• Sophistication means the Math should be commensurate with the HL syllabus
• Rigour involves clarity of logic and language when making Mathematical arguments and calculations.
Here is a fully annotated sample IB Math IA, going through which you can gain a lot of insight (Read the annotations properly).
Another example, the Breaking the Code investigation(fully annotated)- is given here for IB Mathematics HL. The IA explores encryption and decryption in the context of Mathematics.
• Going through the report, we see that the document needs more structure in the beginning since the Introduction does not mention the Rationale or Aim. You don't want to be committing such a blunder.
• Moving further, we see that the body encompasses Math, which is well explained, thereby excelling on criterion E when graded on the SL scale. To excel in HL, the math should have been more rigorous than descriptive.
• Toward the end, the report needs to include a Reflection on the results obtained, which doesn't fare well for the IA. Potential implications of the topic need to be included, and the Conclusion seems bland.
Another interesting annotated sample- for IB Mathematics SL- Regularisation of Irregular Verbs: When can I use the words swimming and know correctly? is given for reference here . Understand which key points have been missed and which have been taken care of. The more you go through sample IAs, the better your chances of preparing an investigation that'll be scored well.
Like any other exploration, your IB Math IA expects you to give it a fair shot of effort and interest. If you see it as a burden, it will undoubtedly become one. Equally important is to draw an analogy between the topic of your choice and the math involved. Taking care of these points in general, let's understand all that goes into making one's IB Math IA (a brief outline):
• It can have a personal story attached; mention it in your Introduction. If not, explain how the topic underhand impacts real-world situations and motivates you to land on it. Your passion for the IA idea shows in your work, and you don't want to be doing your investigation just for the sake of it.
• f you're curious about Fibonacci numbers, the Golden ratio , and nature alike, try looking for relationships among them on the Internet. This will entail going through many research papers, publications, and journals and finally settling on mathematical findings and proofs that will help you investigate that particular something you wish to explore. For example, if you want to study how Traffic Jams have math running in the background , research it in detail since Traffic snarls are an imminent pain for us all. Only you will have to pick up parts you think are relevant and understandable to you.
• The cycle of Inquiry, Action, and Reflection in learning is vital. Learning the implication of Plagiarism, Collusion, and Duplication of Work is essential to keep one's IA transparent and impressive.
• Besides citing references in the bibliography section, ensure you include it in the body as a footnote or in the exploration itself. Citing credible sources shows how transparent your work is and helps examiners cross-check for correctness. Acknowledging the author's work is essential to the IA-making process.
With this, let's discuss the Structure/Layout of the Investigation. There are numerous guidelines available all over the Internet. Regarding the IA length, though you should keep 6 -12 pages as the prescribed length, your focus should be on including all that pertains to your idea and ruling out everything that's not. So don't set out with a mindset to refrain from exceeding six pages; set out to include everything you know needs to be. Similarly, it is advised to make sure the Math used is suitable for SL and HL levels.
Without any further adieu, let's highlight what the layout of the IA should look like:
• Sets the background of your exploration and gives an argument for your topic choice. Your Rationale tells why you chose so and so topic.
• This is where you define your investigation's objective and tell what you wish to achieve with this idea.
• Let's say, for instance, you have opted for Math HL; for a simple mathematical investigation that scores six on the Math SL grade scale, you might end up with a mere four on the HL. The difficulty of your opted Math subject should reflect in your Internal Assessment. It would help if you also outlined the areas of mathematics you will cover in your investigation.
• Elaborate on the method you used for the exploration and justify why you chose to proceed with that particular method.
• Use relevant mathematical tools like labelled graphs, charts, etc., for your mathematical work and explain them in the IA context.
• State your results relating them to the Aim of your Internal Assessment. The significance and impact of these results should be highlighted, as well. In addition, briefly tell how the exploration was helpful to you and all you have gained from it. Possibilities of extension should be mentioned. The bibliography is for you to cite the sources used by you in the making of the IA. We suggest you use the Citation Machine for additional guidance in the bibliography.
Now that we're comfortable with the IB Math IA structure let's look at some interesting IB Math IA topics that will get your creative juices flowing and help kickstart your Math IA journey today!
• Simulating models to study and forecast weather patterns. (You could come up with a personal account that led you to land on something like this)
• Exploring the different probabilities associated with a game of your choice; for example, Solitaire(if you're a game buff).
• Investigating the Math associated with the Global Positioning System(GPS) and the intricacies of the technology involved.
• Exploring Fermat's little theorem or Goldbach's conjecture (one of the most significant unsolved problems in Mathematics).
• Finding the volume and surface area of an egg, apple, mango or any other real-world object using Calculus's power (Simulation could be used). A good IA on modelling manages to score an easy 15-16 marks out of 20.
• Investigating the structural designs of bridges that prevent collapse under loading.
• Studying complex roots graphically.
• Exploring how guitar frets are arranged in Pythagoras Ratios.
• Comparing which will prove beneficial: lump-sum payment of a lottery prize or fee done in instalments?
• Understanding how ISBN codes and Credit Card Codes can be cracked.
And that's a wrap!
Just like any other IA, IB Mathematics IA needs to be started early so that you don't end up compiling just anything at the last minute!
Give it the time it needs, and it will surely pay off. It might seem heavy, but once you decide to pursue an idea of your liking, there will be no turning back! Keep in mind the essential pointers and win the battle courageously!
Want some A-quality guidance? Look no further; at Nail IB, we have assembled premium content for you to ace your IBs, and you should check out our resources for a smooth IB experience. Click here for top-notch IB resources or to assess how your prep is going! Our exclusive Nail IB course, created by Patrick Jones, will be out soon too, so stay tuned, as there is no way you would want to miss the holy grail every Math IB student wants!!
## IB Resources you will love!
Nan + free ib flashcards, -1 + free ia samples, nan + ib videos by experts, -1 + ib sample practice questions, ib resources for nan + subjects.
• Maths Studies
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## Writing A Good IA Introduction
By shotaway , June 9, 2008 in Maths Studies
• Start new topic
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## __inthemaking 64
Is this for SL, math studies or HL? I did math SL and we didn't need an intro.
[quote name='__inthemaking' post='18112' date='Jun 10 2008, 11:09 AM']Is this for SL, math studies or HL? I did math SL and we didn't need an intro.[/quote] Maths SL. From my understanding it's not something big, but an introduction is required.
Just introduce what u plan on accomplishing in your IA and how u plan on doing it =P
Yeah, it's not something very extensive. I remember I had that IA and I introduced it in a couple of lines and still got a 7. I'd copy and paste what I had but I'm still waiting on my results. Don't stress too much over it.
[quote name='shotaway' post='18118' date='Jun 9 2008, 11:40 PM']Maths SL. From my understanding it's not something big, but an introduction is required.[/quote] Really?! I never did an intro for my maths SL IAs. Is it really a requirement then, because I ended up with a 18/20 overall on my IA and a 7 in math (I took the May 2007 exam) with no intro. All I did was type up all the given data and questions exactly and answer the questions.
Actually ... You don't have to make an intro for a math IA. However, putting the information nicely with a good introduction that explains the planning for the IA will be better, and helps you to get a better score.
I just wrote my internal assessment on matrices. a pretty easy way to introduce your paper is like talking to a 7 year old boy. some kind of; in this paper i will try to find the relationship of a pair of matrices elevated to a serie of numbers ......
• 11 months later...
Best way to do it is to connect the first step to your conclusion at the end so that its clear to the marker that you've made the connection. My teacher said that your IA should read like an essay rather than answering questions, you shouldn't even put the numbers or anything, so it should just flow.
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## IB Math SL Internal Assessment: Directions
• Statistical Analysis
Length: 12-20 pages - no word count; logic, precision and clarity count more than length.
Method:
• An introduction that states your rationale/purpose for the topic
• Cite any references or direct quotes using MLA format
• State all definitions and explanations of concepts
• Use proper notation, applicable graphs, and mathematical computations
• Personal engagement
• A conclusion that ties up the major ideas, including whether the results are reasonable. What went wrong? Why? Think TOK
• Use the rubric to guide you
## Introduction
Can an examiner clearly identify your topic?
How did you collect your data? What questions did you ask? Did you do a survey? Who answered it? Did you give it personally or over social media?
Do you have well defined processes and well defined parameters?
Step by step explanations.
What do you need to find to complete your process?
What else do you need to know?
All processes should be in the order that you will complete them. Correlation must be completed before Linear Regression, so it should be described in the introduction before linear regression.
Make sure your pages are numbered and that your work is double spaced. IB prefers Arial font.
## Data Analysis
1. If creating a graph, make sure everything is labeled
2. Show ALL work, every step
3. Include all equations - use the format from class, do not Google the equations
4. After each process, give a brief explanation to the results of the process and how they connect to your topic
5. Create your model, then use technology to compare
6. Analyze for extrapolating and interpolating data
1. Data should be collected through survey, observation, or research
2. Data should be relevant
3. Data should be sufficient in both quantity and quality: 30-50 for a good model
4. Data should be organized in a form that is appropriate for analysis
5. If you are using a survey, you must include the survey - it can be in the paper or added as an appendix
6. Data taken from another source must be cited and must be raw, unanalyzed data; do not use percentages for Chi Squared
1. Create a meaningful conclusion, bringing back together all of your processes and summarizing the results
2. Explain how each process connects to bring you to the conclusion of either proving or disproving your hypothesis
3. Indicate validity - did the processes you used properly help you get to the conclusion you wanted to achieve? Is there any other process that could have been added or anything you would have done differently?
4. Discuss reliability of your model if using statistics
5. Think of real world applications, do outside factors affect the outcome?
6. Could this project lead you into more research or analysis? How?
7. BE REFLECTIVE!
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## IB Math Studies Internal Assessments
IA Introduction Rough Draft
Title- Must be at the top of the page
Font Arial Size 12
1inch Margins
Single spacing or greater
Portrait Orientation
Make sure to email a copy to yourself
Introduction must include
1) Title
(Leave a space)
2) Explanation of why research is important to you
(leave a space)
3) STATEMENT OF TASK- (write the words) statement of what you will be doing in the project and how you will be doing it. This could be 1 or 2 sentences only.
4) Detailed Plan
a. What data will be collected, population, description of the sampling process, if you are conducting the survey, write about it, or if you are conducting an experiment describe the process.
b. How will this data be organized?
c. Write about the 3 math processes that will be used (2 simple, 1 further) Write about the relevancy of each math process. Explain why that math process is critical to your paper.
d. Write a preliminary conclusion. Write about what you think is going to happen. You can refer back to this statement when you do the section on validity.
#### IMAGES
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When writing an introduction for a guest speaker, begin by welcoming the audience to the event or speech, note that it is an honor to be able to introduce the speaker, provide an overview of the speaker’s significance to the community or bu...
3. How Do You Write an Introduction to a Project?
An introduction to a project, paper or verbal presentation engages an audience and provides a concise preview that includes the background of the project, clarifies the points examined and explains the conclusions. An introduction sometimes...
4. How to Structure & Format Your Maths IA
In this section, you should introduce the topic you have chosen and explain why it is important or interesting. You should also include a clear
5. How to Structure and Format Your Math IA
Like almost all of your internal assessments, your Maths IA has to begin with a super clear introduction that sets the context and aim of the whole exploration.
6. How To Write Your Mathematics Internal Assessment
Why? Your IA introduction should include a rationale for why you have chosen your topic for your Mathematical Exploration (the name of this IA).
7. Maths Internal Assessment: Advice & Tips
Aim to be clear, consistent and concise! Be clear with your use of mathematical terminology by defining it where necessary; be consistent with your layout and
8. IB Math IA (Ultimate Guide For 2023)
It can have a personal story attached; mention it in your Introduction. If not, explain how the topic underhand impacts real-world situations
9. IB Math IA Complete Guide Part 1: An Introduction
Access all videos at https://mrflynnib.com. This video introduces the IB math IA for both Analysis and Approaches, and Applications and
10. Writing A Good IA Introduction
Best way to do it is to connect the first step to your conclusion at the end so that its clear to the marker that you've made the connection. My
11. IB Math SL Internal Assessment: Directions
Introduction · Can an examiner clearly identify your topic? · How did you collect your data? What questions did you ask? · Do you have well defined
12. Jeannette Alvarez
Write about the 3 math processes that will be used (2 simple, 1 further)
13. Math SL IA Introduction
Expressing ideas clearly · Identifying a clear aim for the exploration · Focusing on the aim and avoiding irrelevance · Structuring ideas in a logical manner
14. Writing & Math IB Internal Assessment
Math Studies students can approach the IA Project as they would a science lab. | 14,794 | 73,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.895387 |
https://edurev.in/course/quiz/attempt/18524_Test-Introduction-to-Communication-System/5e843f9b-d38a-4d9d-87a9-102e8d43b3a9 | 1,695,528,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00253.warc.gz | 257,693,493 | 45,819 | Test: Introduction to Communication System
Test: Introduction to Communication System - Electronics and Communication Engineering (ECE)
Test Description
10 Questions MCQ Test Topicwise Question Bank for Electronics Engineering - Test: Introduction to Communication System
Test: Introduction to Communication System for Electronics and Communication Engineering (ECE) 2023 is part of Topicwise Question Bank for Electronics Engineering preparation. The Test: Introduction to Communication System questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Introduction to Communication System MCQs are made for Electronics and Communication Engineering (ECE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Introduction to Communication System below.
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Test: Introduction to Communication System - Question 1
The very high frequency (VHF) range extends from
Detailed Solution for Test: Introduction to Communication System - Question 1
Very high frequency (VHF) has a frequency range of 30-300 MHz and wavelength range of 10 to 1 m.
Test: Introduction to Communication System - Question 2
Match List-I (Frequencies spectrum) with List-II (Frequency range) and select the correct answer using the codes given-below the lists: List-I A. Extremely low frequency B. Low frequencies C. Voice frequencies D. Very low frequencies List-II 1. 30-300 kHz 2. 300-3000 Hz 3. 3-30 kHz 4. 30-300 Hz
Test: Introduction to Communication System - Question 3
Match List-I with List-li and select the correct answer using the codes given below the lists: List-I A. Channel B. Carrier C. Modulation D. Distortion List-II 1. The process by which some characteristics of a carrier is varied by ah information signal. 2. Any undesirable change in an information signal. 3. A signal that can be modulated by an information signal. 4. A path for transmission of signal.
Detailed Solution for Test: Introduction to Communication System - Question 3
Explanation : A-4, B-3, C-1, D-2
a) Transmission channel, the path between two nodes of a network that a data communication follows. The term can refer to the physical cabling that connects the nodes on a network, the signal that is communicated over the pathway or a subchannel in a carrier frequency.
b) In telecommunications, a carrier wave, carrier signal, or just carrier, is a waveform that is modulated (modified) with an information bearing signal for the purpose of conveying information.
c) Modulation is a process by which a carrier signal is altered according to information in a message signal.
d) Distortion is any undesired change in the amplitude or phase of any component of an information signal that causes a change in the overall waveform of the signal.
Test: Introduction to Communication System - Question 4
The communication medium causes the signal to be
Detailed Solution for Test: Introduction to Communication System - Question 4
The communication medium is the channel through which the message travels from the transmitter to the receiver. During the process of transmission and reception the signal gets distorted due to noise introduced in the system.
Test: Introduction to Communication System - Question 5
The function of the input transducer in a communication system is
Detailed Solution for Test: Introduction to Communication System - Question 5
The message from the information source may or may not be electrical in nature. In a case when the message produced by the information source is not electrical in nature, an input transducer is used to convert it into a time-varying electrical signal.
Test: Introduction to Communication System - Question 6
Two key barriers to human communication are
Test: Introduction to Communication System - Question 7
Communication is the process of
Detailed Solution for Test: Introduction to Communication System - Question 7
Communication is simply the basic process of exchanging information.
Test: Introduction to Communication System - Question 8
Test: Introduction to Communication System - Question 9
Which of the following is not a major communication medium?
Detailed Solution for Test: Introduction to Communication System - Question 9
Explanation : In the communication process, a medium is a channel or system of communication—the means by which information (the message) is transmitted between a speaker or writer (the sender) and an audience (the receiver).
Wire is not a major communication medium.
In a wired network, data is transmitted over a physical medium.
Nowadays, wireless networks are being used. So wired connection is not a major communication medium.
Test: Introduction to Communication System - Question 10
The process of transmitting two or more information signals simultaneously over the same channel is called
Detailed Solution for Test: Introduction to Communication System - Question 10
Multiplexing may be defined as the process of combining several message signals together and send them over the same communication channel, i.e. FDM, TDM and CDM.
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https://people.maths.bris.ac.uk/~matyd/GroupNames/128/C2.(C8s7D4).html | 1,604,142,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00141.warc.gz | 460,454,268 | 6,330 | Copied to
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## G = C2.(C8⋊7D4) order 128 = 27
### 3rd central extension by C2 of C8⋊7D4
p-group, metabelian, nilpotent (class 3), monomial
Series: Derived Chief Lower central Upper central Jennings
Derived series C1 — C2×C4 — C2.(C8⋊7D4)
Chief series C1 — C2 — C22 — C2×C4 — C22×C4 — C22×C8 — C2×C4×C8 — C2.(C8⋊7D4)
Lower central C1 — C2 — C2×C4 — C2.(C8⋊7D4)
Upper central C1 — C23 — C2×C42 — C2.(C8⋊7D4)
Jennings C1 — C2 — C2 — C22×C4 — C2.(C8⋊7D4)
Generators and relations for C2.(C87D4)
G = < a,b,c,d | a2=b4=c8=d2=1, dbd=ab=ba, ac=ca, ad=da, bc=cb, dcd=b2c-1 >
Subgroups: 340 in 146 conjugacy classes, 60 normal (44 characteristic)
C1, C2 [×7], C2 [×2], C4 [×4], C4 [×8], C22 [×7], C22 [×10], C8 [×4], C2×C4 [×6], C2×C4 [×4], C2×C4 [×14], D4 [×6], C23, C23 [×8], C42 [×2], C22⋊C4 [×4], C4⋊C4 [×2], C4⋊C4 [×7], C2×C8 [×4], C2×C8 [×4], C22×C4 [×3], C22×C4 [×3], C2×D4 [×2], C2×D4 [×5], C24, C2.C42, C4×C8 [×2], D4⋊C4 [×4], D4⋊C4 [×2], C2×C42, C2×C22⋊C4 [×2], C2×C4⋊C4 [×3], C2×C4⋊C4, C22×C8 [×2], C22×D4, C22.4Q16 [×2], C23.65C23, C24.3C22, C2×C4×C8, C2×D4⋊C4 [×2], C2.(C87D4)
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C2×C4 [×6], D4 [×4], C23, D8 [×2], SD16 [×2], C22×C4, C2×D4 [×2], C4○D4 [×4], C42⋊C2, C4×D4 [×2], C4⋊D4, C22.D4, C4.4D4, C422C2, C2×D8, C2×SD16, C4○D8 [×2], C24.C22, C4×D8, C4×SD16, C88D4, C87D4, C4.4D8, C42.78C22, C2.(C87D4)
Smallest permutation representation of C2.(C87D4)
On 64 points
Generators in S64
```(1 57)(2 58)(3 59)(4 60)(5 61)(6 62)(7 63)(8 64)(9 17)(10 18)(11 19)(12 20)(13 21)(14 22)(15 23)(16 24)(25 35)(26 36)(27 37)(28 38)(29 39)(30 40)(31 33)(32 34)(41 49)(42 50)(43 51)(44 52)(45 53)(46 54)(47 55)(48 56)
(1 27 55 22)(2 28 56 23)(3 29 49 24)(4 30 50 17)(5 31 51 18)(6 32 52 19)(7 25 53 20)(8 26 54 21)(9 60 40 42)(10 61 33 43)(11 62 34 44)(12 63 35 45)(13 64 36 46)(14 57 37 47)(15 58 38 48)(16 59 39 41)
(1 2 3 4 5 6 7 8)(9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24)(25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56)(57 58 59 60 61 62 63 64)
(1 55)(2 8)(3 53)(4 6)(5 51)(7 49)(9 19)(10 31)(11 17)(12 29)(13 23)(14 27)(15 21)(16 25)(18 33)(20 39)(22 37)(24 35)(26 38)(28 36)(30 34)(32 40)(41 63)(42 44)(43 61)(45 59)(46 48)(47 57)(50 52)(54 56)(58 64)(60 62)```
`G:=sub<Sym(64)| (1,57)(2,58)(3,59)(4,60)(5,61)(6,62)(7,63)(8,64)(9,17)(10,18)(11,19)(12,20)(13,21)(14,22)(15,23)(16,24)(25,35)(26,36)(27,37)(28,38)(29,39)(30,40)(31,33)(32,34)(41,49)(42,50)(43,51)(44,52)(45,53)(46,54)(47,55)(48,56), (1,27,55,22)(2,28,56,23)(3,29,49,24)(4,30,50,17)(5,31,51,18)(6,32,52,19)(7,25,53,20)(8,26,54,21)(9,60,40,42)(10,61,33,43)(11,62,34,44)(12,63,35,45)(13,64,36,46)(14,57,37,47)(15,58,38,48)(16,59,39,41), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,55)(2,8)(3,53)(4,6)(5,51)(7,49)(9,19)(10,31)(11,17)(12,29)(13,23)(14,27)(15,21)(16,25)(18,33)(20,39)(22,37)(24,35)(26,38)(28,36)(30,34)(32,40)(41,63)(42,44)(43,61)(45,59)(46,48)(47,57)(50,52)(54,56)(58,64)(60,62)>;`
`G:=Group( (1,57)(2,58)(3,59)(4,60)(5,61)(6,62)(7,63)(8,64)(9,17)(10,18)(11,19)(12,20)(13,21)(14,22)(15,23)(16,24)(25,35)(26,36)(27,37)(28,38)(29,39)(30,40)(31,33)(32,34)(41,49)(42,50)(43,51)(44,52)(45,53)(46,54)(47,55)(48,56), (1,27,55,22)(2,28,56,23)(3,29,49,24)(4,30,50,17)(5,31,51,18)(6,32,52,19)(7,25,53,20)(8,26,54,21)(9,60,40,42)(10,61,33,43)(11,62,34,44)(12,63,35,45)(13,64,36,46)(14,57,37,47)(15,58,38,48)(16,59,39,41), (1,2,3,4,5,6,7,8)(9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24)(25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56)(57,58,59,60,61,62,63,64), (1,55)(2,8)(3,53)(4,6)(5,51)(7,49)(9,19)(10,31)(11,17)(12,29)(13,23)(14,27)(15,21)(16,25)(18,33)(20,39)(22,37)(24,35)(26,38)(28,36)(30,34)(32,40)(41,63)(42,44)(43,61)(45,59)(46,48)(47,57)(50,52)(54,56)(58,64)(60,62) );`
`G=PermutationGroup([(1,57),(2,58),(3,59),(4,60),(5,61),(6,62),(7,63),(8,64),(9,17),(10,18),(11,19),(12,20),(13,21),(14,22),(15,23),(16,24),(25,35),(26,36),(27,37),(28,38),(29,39),(30,40),(31,33),(32,34),(41,49),(42,50),(43,51),(44,52),(45,53),(46,54),(47,55),(48,56)], [(1,27,55,22),(2,28,56,23),(3,29,49,24),(4,30,50,17),(5,31,51,18),(6,32,52,19),(7,25,53,20),(8,26,54,21),(9,60,40,42),(10,61,33,43),(11,62,34,44),(12,63,35,45),(13,64,36,46),(14,57,37,47),(15,58,38,48),(16,59,39,41)], [(1,2,3,4,5,6,7,8),(9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24),(25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56),(57,58,59,60,61,62,63,64)], [(1,55),(2,8),(3,53),(4,6),(5,51),(7,49),(9,19),(10,31),(11,17),(12,29),(13,23),(14,27),(15,21),(16,25),(18,33),(20,39),(22,37),(24,35),(26,38),(28,36),(30,34),(32,40),(41,63),(42,44),(43,61),(45,59),(46,48),(47,57),(50,52),(54,56),(58,64),(60,62)])`
44 conjugacy classes
class 1 2A ··· 2G 2H 2I 4A ··· 4L 4M ··· 4R 8A ··· 8P order 1 2 ··· 2 2 2 4 ··· 4 4 ··· 4 8 ··· 8 size 1 1 ··· 1 8 8 2 ··· 2 8 ··· 8 2 ··· 2
44 irreducible representations
dim 1 1 1 1 1 1 1 2 2 2 2 2 2 type + + + + + + + + + image C1 C2 C2 C2 C2 C2 C4 D4 D4 D8 SD16 C4○D4 C4○D8 kernel C2.(C8⋊7D4) C22.4Q16 C23.65C23 C24.3C22 C2×C4×C8 C2×D4⋊C4 D4⋊C4 C2×C8 C22×C4 C2×C4 C2×C4 C2×C4 C22 # reps 1 2 1 1 1 2 8 2 2 4 4 8 8
Matrix representation of C2.(C87D4) in GL5(𝔽17)
1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 16 0 0 0 0 0 16
,
4 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 16 0 0 0 1 0
,
13 0 0 0 0 0 14 3 0 0 0 14 14 0 0 0 0 0 12 12 0 0 0 5 12
,
1 0 0 0 0 0 1 0 0 0 0 0 16 0 0 0 0 0 16 0 0 0 0 0 1
`G:=sub<GL(5,GF(17))| [1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,16,0,0,0,0,0,16],[4,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,16,0],[13,0,0,0,0,0,14,14,0,0,0,3,14,0,0,0,0,0,12,5,0,0,0,12,12],[1,0,0,0,0,0,1,0,0,0,0,0,16,0,0,0,0,0,16,0,0,0,0,0,1] >;`
C2.(C87D4) in GAP, Magma, Sage, TeX
`C_2.(C_8\rtimes_7D_4)`
`% in TeX`
`G:=Group("C2.(C8:7D4)");`
`// GroupNames label`
`G:=SmallGroup(128,666);`
`// by ID`
`G=gap.SmallGroup(128,666);`
`# by ID`
`G:=PCGroup([7,-2,2,2,-2,2,2,-2,224,141,736,422,58,2019,248,2804,172]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^2=b^4=c^8=d^2=1,d*b*d=a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,d*c*d=b^2*c^-1>;`
`// generators/relations`
×
𝔽 | 3,946 | 6,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-45 | longest | en | 0.361203 |
http://research.stlouisfed.org/fred2/series/OPENRPNGA156NUPN?rid=258 | 1,397,785,955,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00600-ip-10-147-4-33.ec2.internal.warc.gz | 180,092,034 | 17,296 | # Openness at constant prices for Nigeria
2010: 55.35245 Percent (+ see more)
Annual, Not Seasonally Adjusted, OPENRPNGA156NUPN, Updated: 2012-09-17 11:45 AM CDT
Click and drag in the plot area or select dates: 1yr | 5yr | 10yr | Max to
Exports plus Imports divided by GDP is the total trade as a percentage of GDP. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: openk
Source: University of Pennsylvania
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(a) Openness at constant prices for Nigeria, Percent, Not Seasonally Adjusted (OPENRPNGA156NUPN)
Exports plus Imports divided by GDP is the total trade as a percentage of GDP. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: openk
Openness at constant prices for Nigeria
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Ask a Physicist: The Shapiro Delay
Monday, February 01, 2016
Adhersh from India, wants to know:
What will be the behavior of light when it passes between two highly massive bodies?
Let’s elaborate a bit on your question, to make it into that most powerful of theoretical physics tools: a thought experiment!
Imagine we’re out in deep space—in spacesuits, of course—far from any gravitational influence, and illuminated only by the distant starlight. We’re far away from each other, but much, much farther from anything else.
At a certain exact time that we’ve agreed upon beforehand, I shine a laser pointer toward you, aiming precisely at a light-sensitive panel (like a solar panel, which converts light into an electrical signal) installed on your suit. When the laser light strikes this solar panel, a computer in the suit notes that it’s received a signal and the time that it received it—finding that it was a precise seven seconds after the time we agreed I would send the signal.
Since the light signal took seven seconds to get from me to you, we can deduce that you and I are some 2.1 billion meters—or seven “light-seconds”—apart.
Light-seconds is a measure of distance, like light-years: since the speed of light isconstant, it's often convenient to talk about astronomical distances in terms of howlong it would take a ray of light to make the journey.
Now, imagine the same scenario, but instead of us being alone there are two massive objects between us—equidistant from the path the laser will take, as shown below:
If just one of those objects were there, it would deflect the laser beam to one side in a process called gravitational lensing, and the signal would never reach you at all, landing somewhere among the distant stars! With both bodies the same mass and same distance from the laser's trajectory, however, it takes the same path it originally would have, and reaches you just the same—or does it?
When the agreed-upon time comes, I send my laser signal shooting through the gap between these two planets, across a few billion meters of space, to the photoreceptor panel in your suit. When the light hits it, a computer in your suit records that, just as before, it's received a signal. Unlike before, however, it notes that it was slightly MORE than seven seconds between when the signal was sent and when it was received! This strange time-warping phenomenon is called the Shapiro Delay
How can this be? There are two ways to interpret it: the most intuitively obvious is that a change in gravitational potential—going from "high" to "low" to "high" again, the way it would as it approaches and then leaves the vicinity of those massive bodies—slows down light. Light moves slower than c all the time—whenever it's traveling through anything other than the vacuum of space. However, one of the fundamental assumptions of modern physics is that the speed of light in a vacuum is constant, regardless of what it's around. To get around this when trying to work out the math of how prominent this effect should be, it's instead assumed that time itself is slowed down by this change.
(Note: this is different from ordinary gravitational time dilation, where clocks tick faster the closer they are to a massive object—since our signal is starting and ending at the same gravitational potential, we shouldn't see any unusual redshift or blueshift, as we'd get from gravitational time dilation.)
But time, as we've discussed before, is just a construct—a way of ordering events. To those who can stretch their imaginations into the fourth dimension and embrace "timeless physics", this scenario can be seen another way—as a warping of space.
You've probably seen the "elastic sheet" model of gravity, demonstrating the way mass warps spacetime. While the analogy isn't perfect, it provides an interesting insight here. In the first image above, where there's no appreciable mass, the light ray takes a straight path from me to you, and takes seven seconds to make the trip. In the second image, you can imagine the fabric of spacetime being bent "down" into the page by the gravity of the massive bodies. Suddenly, the straightest path from me to you bends into the third dimension, meaning a photon would have to travel "downhill" and back "up" to make it to its destination. Even if it were moving at a constant speed, it would still take longer to make the trip, because it's actually traveling a greater distance.
The really neat thing about all this is that you can think of the situation either way—as a stretching of either space or time. Since your question happens in three-dimensional space rather than the 2D example above, it might be hard to imagine a fourth spatial dimension to "stretch" things into, and this is where time comes in very handy—you can simply imagine treat time as the fourth dimension and imagine that it's being stretched by the gravity of these planets. Equivalently, if your mind's eye is up to the task, you can imagine that there's a literal fourth spatial direction—perpendicular to the three in which we seem to exist—and that space is being stretched into it, giving the light a longer path to travel.
This equivalence is one of the more mind-bending properties of relativity theory, and it's part of why physicists use the term "spacetime" when they're talking about the fabric of the universe—space and time, when you get to the nitty-gritty of it, behave in such a way that it doesn't really make sense to talk about one without the other.
This is an extremely insightful question, Adhersh—it cuts to the heart of one of the fundamental tests of Einstein’s theory of general relativity, and helps get at the deeper meaning of one of modern physics’ most crucial assumptions. The fact that you’re wondering about things like this bodes well for your future in physics!
Keep thinking!
Positron
Posted by Unknown
1 Comment:
Anonymous said...
so when we find any new object's distance Far in universe by its light reaching to us that means Its not correct distance cause time delay because lots of gravitational object obstruct it.
is it true?
Monday, February 1, 2016 at 9:21 PM | 1,319 | 6,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-24 | latest | en | 0.962274 |
https://web2.0calc.com/questions/my-accelerated-math-question | 1,540,182,991,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00493.warc.gz | 863,753,393 | 5,724 | +0
# My Accelerated Math Question
0
273
2
Mr.Hayes is spraying grass seed across a lawn. The lawn is in a circular shape with a diameter of 28 yards. What is the area of the lawn to the nearest square yard? Use 3.14 for pi.
Guest May 23, 2017
#1
+7324
+1
area of circle = π * (radius)2
radius = 28 yd / 2
area of circle = π * (14 yd)2
area of circle = 196π yd2
area of circle ≈ 196(3.14) yd2 ≈ 615 yd2
hectictar May 23, 2017
#1
+7324
+1
area of circle = π * (radius)2
radius = 28 yd / 2
area of circle = π * (14 yd)2
area of circle = 196π yd2
area of circle ≈ 196(3.14) yd2 ≈ 615 yd2
hectictar May 23, 2017
#2
+473
0
area of circle = π * (radius)2
radius = 28 yd / 2 | 303 | 705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-43 | latest | en | 0.736191 |
https://bookofproofs.github.io/branches/geometry/euclidean-geometry/elements-euclid/book--1-plane-geometry/angles-and-sides-in-a-triangle-iv.html | 1,702,287,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00782.warc.gz | 162,543,575 | 3,972 | # Proposition: 1.25: Angles and Sides in a Triangle IV
### (Proposition 25 from Book 1 of Euclid's “Elements”)
If two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight lines greater than the (corresponding) angle (in the latter). * Let $ABC$ and $DEF$ be two triangles having the two sides $AB$ and $AC$ equal to the two sides $DE$ and $DF$, respectively (That is), $AB$ (equal) to $DE$, and $AC$ to $DF$. * And let the base $BC$ be greater than the base $EF$. * I say that angle $BAC$ is also greater than $EDF$.
### Modern Formulation
If two triangles ($$\triangle{ABC}$$, $$\triangle{DEF}$$) have two sides of one triangle ($$\overline{AB}$$, $$\overline{AC}$$) respectively equal to two sides of the other ($$\overline{DE}$$, $$\overline{DF}$$), where the base of one $$(\overline{BC})$$ is greater than the base of the other $$(\overline{EF})$$, then the angle $$(\angle{BAC})$$ contained by the sides of the triangle with the longer base $$(\triangle{ABC})$$ is greater in measure than the angle $$(\angle{EDF})$$ contained by the sides of the other triangle $$(\triangle{DEF})$$.
In shorter words, if $$\overline{AB}=\overline{DE}$$, $$\overline{AC}=\overline{DF}$$, and $$\overline{BC} > \overline{EF}$$, then $$\angle{BAC} > \angle{EDF}$$.1
Proofs: 1 Corollaries: 1
Proofs: 1 2 3
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Calahan
@Casey
@Fitzpatrick
### References
#### Adapted from CC BY-SA 3.0 Sources:
1. Callahan, Daniel: "Euclid’s 'Elements' Redux" 2014 | 520 | 1,663 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-50 | latest | en | 0.779612 |
https://metricsystemconversion.info/03-the-perfect-gas-law/ | 1,575,984,113,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00559.warc.gz | 448,664,472 | 23,447 | RONALD SMITH: Last time, we talked about
how planets retain their atmospheres. And I wonder if there are
any questions about that discussion we had last time? Just to review, we defined the
escape velocity, we defined the molecular speed, and then
we talked about the relationship between those
two in regards to how an atmosphere can be retained
by a planet. Anything on that? OK, let’s get started then
with a new subject. I promised I would spend a few
minutes beginning today to talk about the system of units
we’re going to be using. So if you don’t object, I’m
going to spend about 10 minutes or so on that. I’m sure it’s a review for most
the SI system of units. We’ll be using that throughout
the course primarily, although there are some traditional
units that come up in meteorology and oceanography
that don’t necessarily fit into this SI system. SI is– well, it’s French. It’s Systeme International. We just say the International
System of Units. It might also be called
the metric system. And you’ll see why
in just a minute. So the three basic– or the foundation blocks for
the SI system of units are mass, for which we use the unit
kilograms, length, for which we use the unit meter, and
time, for which we use the unit seconds. Now, there are some other
fundamentals that involve electric field strength,
magnetic field strength. We’re not going to
be using those. So I’m going to just work with
these three and then see what we can build up based
on this foundation. So with these as a foundation,
we’ve got a bunch of really simple things we
can write down. For example, speed, the rate at
which something is moving, is going to be meters per
second, just taking the length of meters and the
time of seconds. By the way, in current
scientific convention, it’s more appropriate to write this
m, seconds to the minus one. Sometimes I’ll use this. Sometimes I’ll use that. Today, this is the
more preferred. In the scientific literature,
you’ll normally find this inverse operator used to
indicate that it’s per second. Acceleration, of course, is
going to be meters per second per second, so we could write
that as meters, second to the minus two. It’s how fast is the
speed changing. A few other easy ones, area
would be meters squared. Volume would be meters cubed. And then we’ll move into
the ones that are a little bit trickier. Let’s start with force. Now, the SI unit of force is
the Newton, named after Sir Isaac, of course. But it’s not a fundamental
unit. We can derive it from these. And the way I do that is just
to remember Newton’s Law, F equals ma. So the unit of force, which is
a Newton, can also be written as the product of– well, I can write it this way. It’s going to be kilograms,
meters, per second squared. So that’s another way
to write a Newton. We could do a pressure. Pressure is going to
be the subject of most of today’s lecture. A pressure is a force per unit
area, a force per unit area. The SI name for it is the
Pascal, named after the French scientist who made a fundamental
breakthrough in understanding pressure. And of course, we can write it
in terms of the three building blocks by realizing, if that’s
a force per unit area, then a pascal is going to
be a kilogram, meter per second squared. And then per area, so it’s
meters to the minus two. Let’s simplify that. So it’s kilograms, meters to
the minus one, seconds to the minus two. That is the pressure unit
called the Pascal. Energy, well, the way I think
of energy is that it’s the amount of work done as I
push something along. It’s usually the product of the
force times the distance. Force pushing over
some distance is an amount of energy. Well, that makes it easy if you
can remember that, because then I can take the
unit of force– oh, by the way, the unit of
energy in the SI system is going to be the Joule,
J-O-U-L-E. And of course, that is going to be this times an
extra distance, so it’s going to be kilograms, meters
squared, seconds to the minus two. That will be a Joule. Power, well, power is the rate
of expending energy. How fast are you using
or creating energy? The SI system unit for power
is the familiar watt. You’ll find it stamped
on your light bulbs. It’s a power unit. And we know immediately what
that’s going to be, because we have energy here. But this is per unit time, so
that’s going to be kilograms, meters squared, second
to the minus three. I’ve changed that two to a three
to get it into a per unit time system. Any questions on this yet? Well, one we’ll be using today
also is the mass density. It’s how much mass of a
fluid or an object is there per unit volume. It’s a mass per unit volume. So this is a really
trivia one. There’s no name for it, but it’s
going to be kilograms per meter cubed. Or kilograms, meter to
the minus three. So that’s not bad once you
get the hang of it. And what you’re going to be
doing in this course is to be using various formulas,
computing quantities. And to improve your odds of
getting it right, I recommend that you check the units on
every calculation that you do. If the units don’t work out
wrong as well. So let me give you an
example of that. The one we’re going to be
working on today is the perfect gas law. One way to write it is
P equals rho RT. P is the pressure, rho is the
density, the mass density, R is the gas constant, and
T is the temperature. Now, let me write out
the units for this. Pressure, we already know– where did I put pressure? There it is. Pressure is kilograms, meters
to the minus one, seconds to the minus two. Now, that should be equal to
the product of the units of all these other things. The units of mass density
we’ve already said are kilograms per meter cubed. The units of the gas constant
I’ll give you. It’s joules per kilogram
per degree Kelvin. And the temperature will
be in Kelvins as well. So is that going
to cancel out? Well, it’s not completely clear
yet, because we haven’t taken apart this joule yet to
see what’s inside that. But it looks like we’re going
to get rid of the kilograms OK, and it looks like
the Kelvins are going to cancel out. But what is in that joule? Well, that joule is here. It is a kilogram, meter squared
per second squared. And it looks like that’s going
to work, because we’ve got a meters cubed downstairs. That’s going to take that meters
squared and make it into a meters to the minus one,
and then that’s going to exactly balance with
the left-hand side. You see how that works? So this is a calculation that
should be going on in the background whenever you are
working on a numerical problem to be sure you’ve got
the units right. No questions on that? Well, the focus today is talking
about pressure and the perfect gas law. How many of you have seen
the perfect gas law before in courses? Most of you. Well, let’s first imagine a box
full of gas molecules, but they’re moving around. We know what the typical
molecular speed is. They’re colliding off each
other, but they’re also occasionally bouncing off the
wall of that chamber. And every time they do that,
they impart a little bit of force to the wall that they
bounce off of, and that’s called pressure. It’s the repeated bouncing of
molecules off the side of a box that gives rise to this
quantity we call pressure. It’s going to depend, of course,
on the number of molecules, their speed, and
their mass, in principle. At least, it might depend
on these things. Now, if you’ve taken a course
in chemistry, you probably have seen the perfect gas law
written this way: pressure times volume equals mRT. That’s probably the most
familiar way to write the perfect gas law in a
chemistry course. Here, P is the pressure, V is
the volume of the container that you have it in, m is
the number of moles– let me write this out,
number of moles. This is the gas constant. That’s the temperature,
of course, in Kelvins. That’s the volume of
the container. And that’s the pressure. Do you remember what
a mole is? A mole is a certain number
of molecules. Avogadro’s number– which is, if I remember, it’s
6.02 times 10 to the 23– is a number of molecules
of any gas in a mole. So this would be the number of
formula is that it seems to be independent of the
mass of the molecule. While I speculated that this
might depend on the mass, it turns out that it doesn’t
depend on the mass. You would think that a molecule
that has a heavier mass would impart more force
as it bounces off the wall. But you may remember from last
time that at a given temperature, heavier molecules
move more slowly. So in fact, those two factors
cancel out in the perfect gas law. So the pressure you get depends
only on the number of molecules that you
have, not on the mass of those molecules. That’s a bit of a surprise,
so be aware of that. What gets a little bit
confusing is that in atmospheric science, we
don’t use the perfect gas law in this form. So I’m going to give you the
form in which we will be using it in this class. Stop me if you have questions. We’re going to write the perfect
gas law as P equals rho RT, where this is the
pressure again, this is the mass density, that is
the gas constant for the gas in question– we call it the specific gas
constant, not the universal gas constant– and that again is the
temperature in degrees Kelvin. What we’ve done here– I think you can see it if
you compare the two– basically we’ve said that the
air density, the mass density is going to be– it’s going to be the number of
moles per unit volume times the molecular weight. So the more molecules you have
and the heavier each molecule is in a given volume,
that’s going to determine the mass density. So I’ve used this formula, if
you like, to rewrite that so that it looks like this
form that we use in atmospheric science. Question? STUDENT: I have a question. For Avogadro’s number, is it
10 to the negative 3 or negative 23? PROFESSOR: 10– what
did I write there? Oh, thank you. 10 to the 23. 10 to the plus 23. That’s the number of
molecules in a– thanks very much. Yeah, that’s your job
out there, to keep me honest on this. So what is this specific
gas constant, then? If you follow the math through,
you can see that we’ve defined the specific
gas constant as being the universal gas constant divided
by the molecular weight. So when you’re using air,
that’ll be one number. When you’re using hydrogen,
that’ll be a different number, and so on. So what’s the advantage
of this? We seem like we’ve made things
more complicated, because we no longer have a universal
gas constant. It’s because we want to get
at this mass density. That’s important in atmospheric
science. We want to know, how
dense is the air? And that’s why we want it. We don’t want to work in terms
of number of molecules. We want to work in terms
of the mass of the air. So let’s do an example. First of all, for air,
then, let me put it subscript air there. The average molecular weight
for air is 29. This is 8,314. And so that turns out
to be roughly 287. And the units on that
are joules per kilogram per Kelvin. So that’s the specific gas
constant for air, which is the gas we have most abundantly, of
course, in our atmosphere. Let’s work out a quick
example of that. Let’s say– and I’ll try to make it somewhat
similar to this room– let’s say the temperature
is 15 degrees Celsius and the air density,
I somehow know that’s 1.2 kilograms per cubic meter. What will be the pressure? Well, first of all, we have to
convert this to Kelvins, so it’s going to be
15 plus 273.1. That’s going to be
it into the formula that’s going to be 1.2 times
287 times 288.1. And that comes out to
be 99,221.7 pascals. The unit on that is going
to be pascals. 1.2, which is the air density,
the specific gas constant, 287, and the temperature
expressed in Kelvins, 288.1. Questions on that? Now, what good is this? This is a very useful formula,
but it’s not as useful as one might think in every
application. First of all, for air, we
can take that as known. But in general, as you move
around the atmosphere, the other three things
will be changing. And so if I know one of these,
like temperature, well, that formula’s pretty useless,
because I don’t know either of the other two. So this formula is best used
when you know two of those quantities and need
to get the third. For example, if you knew density
and temperature, that would give you pressure. If you knew pressure and
density, you could solve that for temperature, and so on. So it’s useful, but it’s not
everything we would like. Question? STUDENT: Just going back to the
pascals, do you want us to express it in pascals or
kilograms per meter cubed? PROFESSOR: For pressure, you
should express it in pascals. Or what is sometimes a more
frequent unit in meteorology is a millibar. Millibar, which is sometimes
written as a hectopascal, which is one one-hundredth of
a pascal [correction: one hundred Pascals]. So this would be 992.217
hectopascals. In the meteorological
literature, you’ll often find hectopascals used instead
of pascals. It’s easy to do the
conversion. Just divide by 100 if you’re
going that way, or multiply by 100 if you’re going that way. Now, there is an application, a
direct application, for the perfect gas law that I’m going
to show you now that is really of fundamental importance for
how the atmosphere works. And so I’m going to go through
this a little carefully, because it is something we’ll be
meeting over and over again in the course. And there’s a very simple idea
that I’m sure you are aware of, and that is warm air rises,
and cold air sinks. I’d like to actually
prove that to you. It seems like a trivial
thing, but I’d like to prove that to you. And to do that, I’m going to
have to define something called the buoyancy force. The buoyancy force is a pressure
force on an object immersed in a liquid or a fluid
in a gravity field. Now, this is very easy to
imagine, because if you’ve ever taken a basketball or a
beach ball into a pool and tried to push it down in the
water, you know there’s a rather large force resisting
that trying to make that ball quickly lift back
up to the top. That’s the buoyancy force. And for example, here’s
or your beach ball. You’re trying to hold it down
there with your hand. There’s something very
strong pushing it up. What’s pushing that ball up? What’s the physics of that? Anybody? What’s pushing that ball up? Yeah. STUDENT: The displaced water? PROFESSOR: Yes. But how does it work? In the back? STUDENT: Is it because the
ball is less dense than the water is? PROFESSOR: The ball is less
dense than the water, but I’m looking for a more detailed
mechanism. Yes. STUDENT: The same amount of
force is the weight of the baseball bat? PROFESSOR: Yes. But actually how does it act? What’s the physics? How is acting on that ball? So you’re right. It depends on the
water displaced. That’s going to be
Archimedes’ Law. I’m going to put that on the
board in just a moment. Yes. STUDENT: [INAUDIBLE] outside pressure pushing
down on the water? PROFESSOR: Well, it’ll have
a bit to do with that, but that’s not going to be
having to do with the pressure coming down here. It’s going to have to do with
variations in pressure within that liquid. Anybody else? Yes. STUDENT: [INAUDIBLE] PROFESSOR: Yes. So as you go down in this fluid,
the pressure’s getting greater and greater. That means the pressure acting
up on the bottom is greater than the pressure acting
down on the top. So that’s why I said it’s got to
be a liquid with some mass in a gravity field. Because only in a gravity
field will there be that increase in pressure
as you go down. So when you push that beach ball
down there, realize the pressure at the bottom of the
ball is greater than the pressure at the top
of the ball. And that is what’s causing
this buoyancy force. So that’s step one. By the way, let’s quantify that
using the comment that was made earlier. What is Archimedes’ Law? Archimedes’ Law said that that
buoyancy force is equal to the weight of the water displaced,
or the weight of the– let’s call it water– the weight
of the fluid displaced. So in order to compute that
force, we just have to know how much water would be there if
the object were not there. In this case, it would be
the volume of the object multiplied times the density
of the fluid. But it’s the weight, not the
mass, so this has to be multiplied by little g, the
acceleration of gravity. If you have something of mass
m, its weight is the product of mass and the acceleration
of gravity. So that’s Archimedes’ Law. That’s the buoyancy force. Now, it’s acting in the
atmosphere all the time whenever you have a little
parcel of air that’s at a different temperature than
its surroundings. And that’s what I want to work
out, and that’s where the perfect gas law is going to be
a very nice thing to have. So I’ve worked out an example. I’ve imagined a little
piece of air– maybe it’s about this big– that’s got a certain pressure. I’m going to use the subscript
p, because I’m calling this a parcel, a little
parcel of air. It’s got a density, and it’s
got a temperature. And then surrounding it
is the environment. That’ll be the pressure in the
environment, the density of the environment, and the
temperature of the environment. And my goal is to find out
the buoyancy force acting on that parcel. I want to know, if it’s warm, is
it going to rise, or is it going to sink? And so on and so forth. Now, we’re going to have to
make some assumptions, but they’re going to be very
good assumptions. The first assumption is we’re
going to assume that the pressure in the parcel is equal
to the pressure of the environment. So the pressure here is equal
to the pressure there. Why would that be? If you had air that was at
different pressure than its environment, let’s say at
greater pressure, it would immediately expand until
the pressure matched. If you don’t believe that, blow
up a balloon so you got the pressure in that thing a
little bit higher than the environment, and then pop it. Well, the instant you pop it,
now the rubber is gone. You’ve got that high-pressure
air, and what does it do? It immediately expands
in order to equalize the pressure. So this idea of equalizing
pressure happens very, very rapidly, and that’s why I can
assume that these two pressures are equal. Let me put in some
numbers to this. Let’s say that these pressures
are equal to 80,000 pascals. The temperature of
the environment let’s say is 275 Kelvin. The temperature of the parcel
let’s say is 277 Kelvin, so just a two-degree difference
between the two. And I’m going to compute
the density for both. Rho for the environment is
going to be P for the environment over R and TE. So it’ll be 80,000 divided by
287 divided by 275, and that’s going to be 1.0136. The units will be kilograms
per cubic meter. That’s the density of air
in the environment. The density of air in the parcel
is going to be the same pressure, 80,000, the same gas
constant, 287, but the temperature’s a little
bit different. It’s 277. So that’s going to be 1.0063
kilograms per cubic meter. So what I’ve shown you here
is that the density of the environment is a little bit
greater than the density of the parcel itself. Now, what does that mean
in terms of buoyancy? I think I’ll move
back over here. Here’s my parcel. The gravity force pulling down
on that is going to be the mass of the parcel
it’s going to be the volume times the density of the parcel,
rho sub p times g. The buoyancy force acting up
is going to be the volume– well, we’re using Archimedes’
Law now, so it’s going to be this quantity here. It’s going to be the volume
again times the density of the environment– that’s the fluid that’s
been displaced– times g. Well, now you can see
immediately what’s going to happen here. The V’s are the same for both,
g is the same for both, but the densities appear
differently. The down force is related to
the density of the parcel. The up force is related to the
density of the environment. In our case, the density of the
environment is less, so– is that right? Greater. So this one is going to
be a little bit less. That’s going to be a
little bit greater. And the net buoyancy
force is up. So what I’ve proven here is that
a parcel of air, if it’s equilibrated its pressure with
the environment, is going to be less dense. Therefore, it’s going to have a
buoyancy force that’s going to make it rise. Well, this is probably the basic
physics of what happens in the atmosphere to generate
all the wind circulations, to generate clouds, sea breezes. Almost everything you can think
of in the atmosphere, any air motion, probably can be
tracked back to this simple little idea, that temperature
differences, if the pressure is equilibrated, will generate
buoyancy forces, either up or down. Now, if I had chosen a cooler
temperature for the parcel, let’s say 273, of course, then
everything would be reversed. The parcel would be denser than
air, this vector would be larger than that one, and the
parcel of air would sink. So it works both ways. Now, this is a tricky argument,
a number of steps. So I’d be pleased to stop for
a minute or two and take questions on this. Yes. STUDENT: So it equalizes
pressure, but at the expense of equalizing temperature? PROFESSOR: That’s right. So that’s a very
good question. The question is, why does
it equalize pressure? Why doesn’t it equalize
temperature or density? Well, they’re different
quantities. Pressure is a force per unit
area, and that’s the thing that wants to equalize. There’s no quick process– temperature might equalize over
an hour or two, because they’re in contact with each
other, but not that instantaneous equilibration
like you get with a balloon popping. That’s pressure equilibration,
and it’s fast and it’s physical. And the other two either are
slower, or just there’s no tendency for that at all. But that’s right. That’s the key part of the
argument, isn’t it, that of these three quantities we’re
talking about, the pressure wants to equilibrate, but
the other two do not. And that’s what leads rise to
the whole concept of buoyancy, warm air rising, cold
air sinking. It all comes from the way the
pressure equilibrates. That’s key. Other questions on this? Anything? Well, that went through
pretty quickly. I wanted to– oh, wait. Let’s do another example. I want to do another
example, because– let’s say that I’ve got my
parcel, and everything’s defined as before. But now I’ve got– let’s say– what did I use? I’ve got helium in
here, helium. And I’ve got air out here. The pressures are equal. In this case, I’m going
to say the temperature are equal as well. But are the densities equal? Do you think the densities
are going to be equal in these two cases? No. And let’s see how that’s
going to work out. The density of the environment
is going to be the pressure of the environment with the gas
constant for air, 287, and then the temperature
of the environment. The density of the parcel is
going to be pressure of the parcel over– now, let’s see. What’s going to be the gas
constant for helium? 8,314 divided by the molecular
weight of helium, which you recall is four. The gas constant for helium’s
about 2,079. So look, even if the pressures
are the same and the temperatures are the same,
because they’re different gases, the densities are going
to be very, very different. And so the helium balloon is
going to have a smaller mass, smaller density than the air
that it’s displaced. It’s going to rise. It’s going to have a buoyancy
force that rises. And in this case, it comes in
through the different gas constant, which in turn arises
because of the different molecular weights. So later on in the course, in
the lab, we’ll be launching helium-filled balloons, and
you can think back at that moment and realize, ah, that’s
what’s going on. That’s why there’s
a buoyancy force. That’s why that balloon wants
to rise is because it has a different gas constant, because
it has a different molecular weight. It’s a lighter gas. Each molecule has a smaller
mass than the air molecules do. Any questions there? I can’t leave this
subject without mentioning mixtures of gases. So we imagine this same box, and
it’s got some A molecules, and it’s got some B molecules. And they’re all bouncing
around off the walls and so on. There’s a mixture of gas A and
gas B. What is the deal there? When you mix two gasses
together, what relationship do they have to each other? I can tell you pretty clearly
what’s going to happen. The temperatures are going
to quickly equilibrate. Even if the masses are
different, because they’re bouncing into each other
frequently, thousands of times per second, the temperature of
the A and B molecules will quickly come to the
same value. The pressure is additive. In other words, we can define
the pressure that the A molecules are making, we can
call that PA, and the pressure that the B molecules are
making, that’s PB. And the total pressure,
P Total, is just the sum of the two. So we use this term
partial pressure– partial pressure of A, partial
pressure of B– and they add up to give the total pressure. So for example, if the pressure
in this room– let’s call it P Total
for the moment– is about 1,013 millibars– or that is to say 1,013
with two more decimal places pascals– part of that is due to the
nitrogen molecules. That’s the partial pressure
of the nitrogen. Part of it’s due to the
oxygen molecules. Part of it’s due to the argon. There’s also some water
vapor in this room. Water vapor is contributing
something to that total pressure. So when you’re measuring
pressure in a gas, you’re measuring the sum of all the
pressures of the components within that gas. Usually, we don’t need to know
that, but occasionally, that’s the way we keep track of how
much of these other gases you have. Someone might say, well,
the partial pressure of water vapor is three millibars today
or something like that. That’s the contribution water
vapor is making to the total pressure on this
particular day. So it’s a useful quantity. Let me remind you what the
atmospheric composition is for our atmosphere. For air on Earth,
it’s primarily nitrogen, oxygen, and argon. I’m going to give you the number
two ways: by volume, which is what the
chemists say– I prefer to remember that that
is by molecule, by the number of molecules– and I’m going to also give
it to you by mass. For nitrogen, it’s 78.1% by
volume and 75.5% by mass. In other words, 78% of the
molecules in this room are nitrogen, but 75.5% of the mass
of the gas in this room is the nitrogen. Oxygen, 21.0% and 23.2%. Argon, 0.9% and 1.3%. Just remember, there’s this
difference because the molecules have different
masses. Some are heavier. Some are lighter. So whether you’ve counted
up the molecules and are representing the fraction that
way, or whether you’re counting up the masses, you’re
going to get slightly different numbers for the two. Now I’ve chosen, and the
convention is to define that part as being the air, because
these proportions are constant everywhere you go in
the atmosphere. If I go to the North Pole, the
Equator, the South Pole, if I go high in the atmosphere,
winter or summer, these proportions are unchanging. So we call that air. But then there are other
gases as well. And sometimes they’re
called trace gases. Sometimes they’re just called
variable gases. They’re found in varying
proportions depending where you are. Let me give you an example. Probably the most important
one is water, water vapor, H2O. And it’s found anywhere from– well, from let’s say one part
per 100, 10 to the minus two, to really as small as you want
to go, maybe 10 to the minus five, by volume. CO2, another very important gas,
is more thoroughly mixed, but not perfectly mixed. A typical value these days might
be about 395 parts per million by volume, ppmv. So
we’re using this method, we’re counting molecules. I could write that as
395 times 10 to the minus six by volume. That varies only up and down by
about 5%, plus or minus 5%. So that’s nearly well mixed,
but not quite as thoroughly mixed as these gases within
the atmosphere– within the air. Some other molecules I mentioned
last time on the slide you should be aware
of are methane, nitrous oxide–N2O– and ozone. And just be aware that those and
a few other gases will pop up from time to time in this
course, and we’ll be wanting to know what their partial
pressure is, what their mass ratio is, what their ratio
by molecules is. We can convert back and forth
between these different measures using the formulas
that I’ve given you today. Any questions here? We’ve actually covered a lot of
I think somewhat confusing material, so I want to
be sure we take a few more minutes for questions. Yes. STUDENT: For the water vapor,
the numbers there, it’s 10 to the negative 2 – there is
10 to the negative 5. PROFESSOR: Negative five. Thank you. For example, I don’t have an
instrument with me to measure this– we’ll be doing it in
lab– but yesterday and today have been rather humid days. So this means that this number
is going to be a little larger than it would have been last
week, when we had a drier atmosphere. So that’s an example of how
that number fluctuates. This fraction isn’t changed
between last week and this week, but this one has. So these are variable ones,
and these are constant proportions. Yes. STUDENT: When it’s 100%
humidity, about where is that range? PROFESSOR: Well, so that depends
on the temperature. So the relative humidity–
measure of how much water vapor you have to the maximum
that can be held in the vapor state. And because that second
number is so strongly temperature-dependent,
I can’t give you a fixed number for this. It’ll depend on the
temperature. But we’ll talk about that later
on, because that’s so important for how clouds
form and so on. I think I may have time to do
one other thing before we quit today, and that’s to talk about
how density and pressure change with altitude. First of all, just some
background information. The typical sea-level density,
of course it varies from place to place. But if you want to work out a
problem and you’re not given enough information, you should
know, for example, in this room, the density is about 1.2
kilograms per cubic meter. And a typical sea-level pressure
is about 1,013 millibars or 1,013 two
more zeros pascals. So let’s take that as just
basic climatological information. But now I’m interested
in how those numbers change with height. Typically, if I plot pressure on
the x-axis and height using the letter z on the y-axis,
it looks like this. It decreases rapidly at first,
then less rapidly, and so on, asymptotically, but never
actually reaching zero. And if I plot air density, it
looks very much the same. It’s such a simple relationship
that we’d like to be able to have a
formula for it. And there is a nice one,
but we have to make an approximation now. We have to assume that the
temperature is approximately constant with height, which is
not a very good approximation, but we’re looking here just
to get a rough– maybe an estimate at the 10% level or
the 20% level, something in that range. But if this approximation is
used, then I can write down a formula for the pressure as
a function of height. It’s the pressure at sea level
times e to the minus z over H sub S. And the density follows exactly
the same formula. The density at sea level,
rho sub SL– I’m using Greek letter
rho for density– e to the minus z over H sub S.
Now, if you’re familiar with this exponential function,
you would have already recognized it here. This is the behavior of the
exponential function. It drops rapidly at first and
then more slowly as you go on. This thing is called the density
for the scale height, the scale height for
the atmosphere. It is a measure of how fast
the pressure and density decrease as you go up. And there’s a very simple
formula for it. It’s RT over g, the gas constant
times the temperature divided by the surface
gravity. Let’s work it out for Earth. Air has a gas constant of 287. Let’s take 288 for the
temperature of the air and 9.81 for the acceleration
of gravity. That turns out to be
approximately 8,400 meters. Every time you go up 8,400,
meters, you tick off another fractional decrease
in atmospheric pressure and density. We just have a minute
left, so I can do a quick example of this. Let’s say we’ve got an aircraft flying at 37,000 feet. That’s typically what an
airliner would fly at. And you’d like to know what is
the pressure and density of the air just outside
the cabin? Of course, the cabin itself is
pressurized so you can breathe and maintain consciousness,
but what is the air temperature and pressure
just outside? Well, first of all, we have
to convert this to meters. That’s going to be
11,278 meters. And then I’m going to put
it into this formula. So the pressure at that height
z is going to be the pressure at sea level, 101,300,
times e to the minus 11,278 divided by 8,400. I hope you know how to do that
in your scientific calculator with the minus sign in there. Practice that. That comes out to be– let’s see– 26,524 pascals. Well, that’s about a quarter of
the pressure at sea level. And density would be something
very similar. It’ll be 1.2 times e to the
minus 11,278 over 8,400. And that’s going to come
out to be 0.31. Units are kilograms
per cubic meter. So that also is about
a quarter of what you started with. So that’s not much. In other words, at typical
airliner flight level, the density and pressure that you’re
flying through is only about one quarter that you have
here at sea level, and that’s why the cabin has
to be pressurized. We’re really out of time now,
so we’ll move on to some new material on Wednesday. | 8,448 | 33,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-51 | latest | en | 0.941746 |
https://tbc-python.fossee.in/convert-notebook/Electronic_Devices_and_Circuit_Theory_by_R_L_Boylestad_and_Louis_Nashlesky/Chapter1.ipynb | 1,590,544,089,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00498.warc.gz | 569,330,336 | 37,488 | # Chapter 1: Semiconductor Diodes¶
## Example 1.1 Page Number-13¶
In [8]:
k=1.38*(10**(-23)) #boltzmann's constant
t=273+27 #converting given temperature to Kelvin
q=1.6*(10**(-19)) #charge on an electron
# V=(k*t)/q
V=(k*t)/q
V=V*1000 #converting result in millivolts
print "Thermal Voltage=",V,"mV"
Thermal Voltage= 25.875 mV
## Example 1.2 (a) Page Number-18¶
In [10]:
Id= 1 #in mA, current across diodes
#from the standard graph for Ge,Si, and GaAs diodes
Vge=0.2
Vsi=0.6
Vgaas=1.1
print "Voltage across Germanium diode=",Vge,"V"
print "Voltage across Silicon diode =",Vsi,"V"
print "Voltage across GaAs diode =",Vgaas,"V"
Voltage across Germanium diode= 0.2 V
Voltage across Silicon diode = 0.6 V
Voltage across GaAs diode = 1.1 V
## Example 1.2 (b) Page Number-18¶
In [11]:
Id= 4 #in mA, current across diodes
#from the standard graph for Ge,Si, and GaAs diodes
Vge=0.3
Vsi=0.7
Vgaas=1.2
print "Voltage across Germanium diode=",Vge,"V"
print "Voltage across Silicon diode =",Vsi,"V"
print "Voltage across GaAs diode =",Vgaas,"V"
Voltage across Germanium diode= 0.3 V
Voltage across Silicon diode = 0.7 V
Voltage across GaAs diode = 1.2 V
## Example 1.2 (c) Page Number-18¶
In [12]:
Id=30 #in mA, current across diodes
#from the standard graph for Ge,Si, and GaAs diodes
Vge=0.42
Vsi=0.82
Vgaas=1.33
print "Voltage across Germanium diode=",Vge,"V"
print "Voltage across Silicon diode =",Vsi,"V"
print "Voltage across GaAs diode =",Vgaas,"V"
Voltage across Germanium diode= 0.42 V
Voltage across Silicon diode = 0.82 V
Voltage across GaAs diode = 1.33 V
## Example 1.2 (d) Page Number-18¶
In [18]:
#Average value for Germanium
Vg=(0.2+0.3+0.42)/3
#Average value for Silicon
Vs=(0.6+0.7+0.82)/3
#Average value for GaAs
Vgs=(1.1+1.2+1.33)/3
print "Average Volatge value for Germanium Diode=",round(Vg,3),"V"
print "Average Volatge value for Silicon Diode=",round(Vs,3),"V"
print "Average Volatge value for GaAs Diode=",round(Vgs,3),"V"
Average Volatge value for Germanium Diode= 0.307 V
Average Volatge value for Silicon Diode= 0.707 V
Average Volatge value for GaAs Diode= 1.21 V
## Example 1.2 (e) Page Number-18¶
In [22]:
#comparing average values in d with the standard knee voltages
#Average value for Germanium
Vg=(0.2+0.3+0.42)/3
#Average value for Silicon
Vs=(0.6+0.7+0.82)/3
#Average value for GaAs
Vgs=(1.1+1.2+1.33)/3
kge=0.3
ksi=0.7
kgaas=1.2
print "Very close correspondence between knee voltage and average voltage"
print "Germanium",kge,"V vs",round(Vg,3),"V"
print "Silicon",ksi,"V vs",round(Vs,3),"V"
print "GaAs",kgaas,"V vs",round(Vgs,3),"V"
Very close correspondence between knee voltage and average voltage
Germanium 0.3 V vs 0.307 V
Silicon 0.7 V vs 0.707 V
GaAs 1.2 V vs 1.21 V
## Example 1.2(a) Page Number-22¶
In [4]:
Id=2*(10**(-3)) #in ampere
Vd=0.5 #in volts
rd=Vd/Id
print "dc resistance=",rd,"ohms"
dc resistance= 250.0 ohms
## Example 1.2(b) Page Number-22¶
In [3]:
Id=20*(10**(-3)) #in ampere
Vd=0.8 #in volts
rd=Vd/Id
print "dc resistance=",rd,"ohms"
dc resistance 40.0 ohms
## Example 1.2(c) Page Number-22¶
In [7]:
#Id=-Is
Id=1*(10**(-6)) #in ampere
Vd=-10 #in volts
rd=abs(Vd)/Id
rd=rd/(10**(6))
print "dc resistance=",rd,"Mohms"
dc resistance= 10.0 Mohms
## Example 1.3(a) Page Number-24¶
In [2]:
# drawing tangent at Id=2mA and choosing any random points n the tangent to gwt two set of values of Id and Vd
Id1=4*(10**(-3)) #IN ampere
Id2=0 #IN ampere
Vd1=0.76 #IN VOLTS
Vd2=0.65 #IN VOLTS
X=Id1-Id2
Y=Vd1-Vd2
rd=Y/X
print "ac resistance=",rd,"ohms"
ac resistance= 27.5 ohms
## Example 1.3(b) Page Number-24¶
In [3]:
# drawing tangent at Id=2mA and choosing any random points n the tangent to gwt two set of values of Id and Vd
Id1=30*(10**(-3)) #IN ampere
Id2=20*(10**(-3)) #IN ampere
Vd1=0.80 #IN VOLTS
Vd2=0.78 #IN VOLTS
X=Id1-Id2
Y=Vd1-Vd2
rd=Y/X
print "ac resistance=",rd,"ohms"
ac resistance= 2.0 ohms
## Example 1.3(c) Page Number-24¶
In [8]:
#calculating Dc resistance
#Case-1
Id1=2*(10**(-3)) #in ampere
Vd1=0.7 #in volts
Rd=Vd1/Id1
rd=27.5 #ac resistance in ohms
if Rd>rd:
print "Dc resistance=",Rd,"ohms exceeds ac resistance=",rd,"ohms"
else:
print "Dc resistance=",Rd,"ohms didnot exceeds ac resistance=",rd,"ohms"
#Case-2
Id1=25*(10**(-3)) #in ampere
Vd1=0.79 #in volts
Rd=Vd1/Id1
rd=2 #ac resistance in ohms
if Rd>rd:
print "Dc resistance=",Rd,"ohms exceeds ac resistance=",rd,"ohms"
else:
print "Dc resistance=",Rd,"ohms didnot exceeds ac resistance=",rd,"ohms"
Dc resistance= 350.0 ohms exceeds ac resistance= 27.5 ohms
Dc resistance= 31.6 ohms exceeds ac resistance= 2 ohms
## Example 1.4 Page Number-40¶
In [9]:
#Equation- change in Cvz=(Tc*Vz*(t1-t0))/100%
Tc=0.072 #unit %/celsius
t1=100 #in celsius
t0=25 #in celsius
Vz=10 #in volts
Cvz=(Tc*Vz*(t1-t0))/100
nVz=Vz+Cvz #new Vz
print "New potential across zener diode=",nVz,"V"
New potential across zener diode= 10.54 V
## Example 1.5 Page Number-43¶
In [15]:
#Equation wavelength(x)=c/f,where c=speed of light and f=frequency of the light
c=3*(10**(8))*(10**(9)) #in nm/s
x1=(c/(400*(10**12))) #in nm
x2=c/(750*(10**12)) #in nm
print "The range of Wavelength for the frequency of Visible lightis",x1,"nm to",x2,"nm"
The range of Wavelength for the frequency of Visible lightis 750 nm to 400 nm | 2,109 | 5,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.576096 |
https://discourse.julialang.org/t/in-place-map-of-vector-functions-across-slices-of-multidimensional-arrays/78684 | 1,660,197,433,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00028.warc.gz | 217,200,650 | 7,182 | In-place map of vector functions across slices of multidimensional arrays
Hi,
I am trying to in-place map some vector functions across slices of multidimensional arrays
For instance, given three matrices:
`A, B, C = rand(4, 3), rand(4, 3), rand(4, 3)`
I wish to efficiently compute the cross product across the rows of `B` and `C`, and store the results in `A`.
So I was wondering, what is the correct/best practice for doing it?
(Sorry if this is a duplicated question, since the operation need is elementary, but I couldn’t find a way for it…)
The best I could get was something like:
``````# I didn't find a way to cast a 2-D Array into nested Array{Array{Float64}},
# so the line below is just for initializing a storage
tmp = [rand(3) for _ in range(1, 4)]
# 155.633 ns (6 allocations: 768 bytes)
# 154.115 ns (6 allocations: 768 bytes)
@btime map!((b, c) -> cross(b, c), tmp, collect(eachrow(\$B)), collect(eachrow(\$C)))
``````
If you are working with 3-component arrays representing position vectors in space (so that you have lots of operations on small arrays of length 3), then I would strongly consider using StaticArrays.
In your example you would just have 1d arrays of `SVectors` and could use ordinary broadcasting:
``````julia> using StaticArrays, LinearAlgebra
julia> A, B, C = rand(SVector{3,Float64}, 4), rand(SVector{3,Float64}, 4), rand(SVector{3,Float64}, 4);
julia> A .= B .× C
4-element Vector{SVector{3, Float64}}:
[0.10928103138317774, -0.01855287736466356, -0.023819251068657343]
[0.16095133726726152, 0.34506454774262896, -0.2178141205786981]
[-0.453417349873811, 0.34516082042036367, 0.2587571993077106]
[0.5297917349791741, -0.2985033529890534, -0.3202434018984674]
``````
In general, StaticArrays (or similar) are by far the most efficient and convenient way to work with physical coordinate vectors.
2 Likes
Thanks a lot, this is super fast!
Would you recommend also using `StaticArrays` for broadcasting vector functions other than the cross product?
Yes.
The point is that if you know ahead of time (at compile time) that you have lots of vectors of length 3 (or any other small fixed size), then `StaticArrays` lets you tell this information to the compiler to get efficient storage (e.g. inlined into arrays, stored in registers, …) and fast operations (e.g. unrolled loops, specialized implementations for particular sizes…).
Note also if you are coming from Matlab or Python or R and have been trained to rearrange all computations into “vectorized” form, one at a time (e.g. first a vectorized cross product, then a vectorized dot product, …), you might want to re-think your habits. Writing your own loop is perfectly fine in Julia too, and can avoid lots of inefficient temporary arrays or suboptimal repeated passes over arrays doing cheap operations (like cross products). Broadcasted operations can still be convenient, but then you should take advantage of Julia’s fusion for nested “dot calls.” See also this blog post.
4 Likes
awesome!
A second thought here.
I guess this is going to limit the capability for applications like sequentially broadcasting multiple vector functions along different dimensions, is that right?
If so, I guess we still don’t have a general solution…?
(Your solution solves my application, very much appreciated!)
Just saw your edit on encouraging implementing for-loop instead of trying to vectorize.
That makes a lot of sense, thanks!
Also note that, in the event that you are using matrices instead of vectors of StaticArrays, you should try to work along columns, not across rows, since Julia’s `Array` type is column major, like Matlab, and unlike numpy.
sure!
If you are doing more general operations along different dimensions of multidimensional arrays, I think looping is a lot more powerful than trying to shoehorn it into broadcasting — especially using Julia’s `CartesianIndex` machinery to easily express loops over `N`-dimensional arrays in code that is generic for any `N`. For example, see this code for wave propagation in `N` dimensions.
3 Likes | 1,016 | 4,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.856572 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/quantirisk/how-to-estimate-value-at-risk-given-skewness-and-kurtosis-in-excel-1n5o | 1,656,979,570,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104506762.79/warc/CC-MAIN-20220704232527-20220705022527-00750.warc.gz | 510,500,069 | 27,981 | ## DEV Community is a community of 871,688 amazing developers
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Wynn Tee for Quantitative Risk Solutions PLT
Posted on • Originally published at addin.qrstoolbox.com
# How to estimate value-at-risk given skewness and kurtosis in Excel
Risk practitioners often use value-at-risk (VaR) to quantify the potential future loss of an investment. As a risk metric, VaR answers questions like "What is the worst daily/monthly/yearly return going to be 90%/95%/99% of the time?"
## Terminology
In statistical jargon, VaR is simply the percentile of the return distribution at a given confidence level. The above diagram illustrates VaR at the 95% confidence level.
Sometimes, VaR is defined in terms of a loss (instead of return) distribution. In that case, the above diagram would flip horizontally and VaR would be in the right tail.
## Mean and variance
If the mean and variance of the return distribution are given, VaR can be estimated by assuming returns follow the normal distribution and calculating the percentile of the normal distribution that fits the given mean and variance.
To estimate VaR at the 95% confidence level in Excel when the mean and variance of the return distribution are 10% and 0.01 respectively, enter the formula `=NORM.INV(1-0.95, 10%, SQRT(0.01))` in a cell.
To estimate VaR when the previous mean and variance are for the loss (instead of return) distribution, change the formula to `=NORM.INV(0.95, 10%, SQRT(0.01))`.
## Skewness and kurtosis
In reality, returns are more likely to follow a non-normal distribution with negative skewness and large kurtosis. According to journalist Martin Wolf, many investment strategies have a high probability of small gains and a low probability of huge losses.
If the skewness and kurtosis of the return distribution are given in addition to the mean and variance, VaR can be estimated by assuming returns follow a non-normal parametric distribution and calculating the percentile of the distribution that fits the given moments.
There are many non-normal parametric distributions, but the Pearson and Johnson distributions are the only ones that fit all possible combinations of skewness and kurtosis. No other parametric distribution has such a feature.
Unfortunately, Excel does not come with native functions for using the Pearson and Johnson distributions.
## QRS.VAR functions
Fortunately, QRS Toolbox for Excel includes functions for estimating VaR from not only the normal distribution, but also the Pearson and Johnson distributions. The names of the functions all begin with QRS.VAR.
QRS.VAR.NORMAL returns VaR for the normal distribution. It accepts the confidence level, mean, standard deviation, and choice of tail as parameters. It is simpler to use compared to the two previous formulas involving NORM.INV.
QRS.VAR.PEARSON and QRS.VAR.JOHNSON return VaR for the Pearson and Johnson distributions respectively. They accept the same parameters as QRS.VAR.NORMAL in addition to skewness and kurtosis.
## Pearson Type IV example
There are 5 main types of Pearson distributions, namely Types I, III, IV, V, and VI. To the best of our knowledge, QRS.VAR.PEARSON is the only publicly available Excel function for estimating VaR from all 5 types of Pearson distributions.
The Pearson Type IV distribution deserves special mention. It is a natural alternative to the normal distribution in risk applications, because it is not only unimodal and unbounded, but also skewed (squared skewness < 32) and fat-tailed (kurtosis > 3).
To try QRS.VAR.PEARSON yourself, add QRS Toolbox to your instance of Excel and start your free trial of QRS.VAR.PEARSON. Then, open the example workbook.
In the workbook:
• Cell B1 contains the confidence level, 99%.
• Cell B2 contains the mean, 10%.
• Cell B3 contains the variance, 0.01.
• Cell B4 contains the skewness, -2.
• Cell B5 contains the kurtosis, 16.
``````=QRS.PEARSON.TYPE(B2, B3, B4, B5)
``````
To determine the type of Pearson distribution that fits the given moments, enter the formula `=QRS.PEARSON.TYPE(B2, B3, B4, B5)` in cell B6. The result is IV.
``````=QRS.VAR.PEARSON(B1, B2, B3, B4, B5)
``````
To estimate VaR for the corresponding Pearson Type IV distribution, enter the formula `=QRS.VAR.PEARSON(B1, B2, B3, B4, B5)` in cell B7. The result is -23.7%, which means the investment is estimated to lose at most 23.7% in 99 out of 100 periods.
## Johnson SU example
There are 3 types of Johnson distributions, namely SU, SL, and SB. To the best of our knowledge, QRS.VAR.JOHNSON is the only publicly available Excel function for estimating VaR from all 3 types of Johnson distributions by moment matching.
Software that support fitting the Johnson distributions by moments typically use a method by Hill, Hill, and Holder, but that method does not always return a result. QRS.VAR.JOHNSON uses a proprietary method that is more reliable.
To try QRS.VAR.JOHNSON yourself, start your free trial of QRS.VAR.JOHNSON and reopen the workbook from the previous example.
``````=QRS.JOHNSON.TYPE(B2, B3, B4, B5)
``````
To determine the type of Johnson distribution that fits the given moments, enter the formula `=QRS.JOHNSON.TYPE(B2, B3, B4, B5)` in cell B6. The result is SU.
``````=QRS.VAR.JOHNSON(B1, B2, B3, B4, B5)
``````
To estimate VaR for the corresponding Johnson SU distribution, enter the formula `=QRS.VAR.JOHNSON(B1, B2, B3, B4, B5)` in cell B7. The result is -24.3%, which is in the same ballpark as the result from the previous example.
## Final remarks
If you would like to use QRS.VAR.PEARSON and QRS.VAR.JOHNSON beyond your free trial periods, you may purchase the right to use them indefinitely for as little as USD 19.00. If you find the QRS.VAR functions useful, please share this page with other potential users. | 1,452 | 5,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | longest | en | 0.877994 |
https://www.mathpax.com/convolution-as-seen-in-virtual-work-nov-2023/ | 1,726,619,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00626.warc.gz | 804,080,702 | 20,952 | # Convolution as seen in Virtual Work Nov 2023
Convolution as seen in a Virtual Work (Structural Analysis) Problem
Linear or nonlinear?
Recently, I read a post featuring an example virtual work problem in structural analysis.
On the first page, there was the usual integration equation in which the strain energy of a virtual moment was multiplied by the strain energy of a single point load. Nothing unusual at all about the problem, or the approach, or the equation.
My surprise was in seeing the shortcut calculation. The diagram is at the bottom of Page 1 (the diagram with two triangles). The integration equation and the shortcut are near the top of Page 2 (the third equation). Here’s the link to the PDF.
Indeed the shortcut is somewhat explained in the linked PDF below. Refer to Page 10.
I had never seen the shortcut used, nor the table which explained it.
I decided to explore this on my own.
My exploration resulted in a few charts as follows. All of these are live interactive (built using Plotly).
Hover over chart elements (on a mobile device, touch) to see data values.
I used Python, Sympy, and Jupyter to program my solutions.
My first objective was to verify the coefficient, , used in the equation on Page 2 (*).
After working that out, I proceeded to the general case of the convolution of two triangles as follows
• Triangle 1:
• left side at: x = 0
• right side at: x = 1
• apex at: x = a
• Triangle 2:
• left side at: x = 0
• right side at: x = 1
• apex at: x = b
There are 3 parts to the integration, labeled , , amd . I add , , and to obtain , the result of the integration.
```# In this example, a <= b
x, a, b = symbols('x, a, b')
Q = integrate((x/a)*(x/b), (x, 0, a))
R = integrate((1-x)/(1-a)*(x/b), (x, a, b))
S = integrate((1-x)/(1-a)*(1-x)/(1-b), (x, b, 1))
T = Q + R + S
T = simplify(T)
```
I obtained the following general equation.
As I say at the top of this post: “Linear or nonlinear?”
I had fully expected all results, for all and to be nonlinear. Turns out, that is generally true with the exception where or .
If , then the general equation reduces to
$\dpi{110}&space;f(0,b)=\cfrac{0^2+b^2-2b}{6b(0-1)}$
and then to
$\dpi{110}&space;f(0,b)=\cfrac{b^2-2b}{-6b}$
then
$\dpi{110}&space;f(0,b)=\cfrac{b-2}{-6}$
then
$\dpi{110}&space;f(0,b)=\cfrac{2-b}{6}$
which is linear!
If , then the general equation reduces to
$\dpi{110}&space;f(0,b)=\cfrac{a^2+1-2}{6(a-1)}$
and then to
$\dpi{110}&space;f(0,b)=\cfrac{a^2-1}{6(a-1)}$
then
$\dpi{110}&space;f(0,b)=\cfrac{a+1}{6}$
which is linear also!
(*) To close the loop, when and , then
$\dpi{110}&space;f(a,b) = f(0,0.5) = \cfrac{1}{4}$
And when and , then
$\dpi{110}&space;f(a,b) = f(0.5,1) = \cfrac{1}{4}$
I derived another special case as shown below. This pattern has two linear elements and the integration is for the pattern squared.
```# In this example, a <= b
x, a, b, c, d, L1, L2 = symbols('x, a, b, c, d, L1, L2')
Q = b + x * (c-b)/(L1)
R = c + (x-L1) * (d-c)/(L2-L1)
Q2 = integrate(Q**2, (x, 0, L1))
R2 = integrate(R**2, (x, L1, L2))
T = Q2 + R2
T = simplify(T)
```
$\dpi{110}&space;F_{special\;1}(a,b,c,L_1,L_2)=\cfrac{1}{3}\,[L_1(a^2+ab)+L_2b^2+(L_2-L_1)(bc+c^2)\]$
I note that the most general convolution of two independent linear segments is:
```# In this example, a <= b
x, a, b, c, d, L1, L2 = symbols('x, a, b, c, d, L1, L2')
R1 = a + (x-L1) * (b-a)/(L2-L1)
R2 = c + (x-L1) * (d-c)/(L2-L1)
T = integrate(R1*R2, (x, L1, L2))
T = simplify(T)
```
$\dpi{110}&space;F_{special\;2}(a,b,c,d,L_1,L_2)=\cfrac{1}{6}\,(L_2-L_1)[2(ac+bd)+(ad+bc)\]$
If and , then this equation reduces to
$\dpi{110}&space;F_{special\;2}(a,b,a,b,L_1,L_2)=\cfrac{1}{3}\,(L_2-L_1)[a^2+b^2+ab\]$
or better yet
$\dpi{110}&space;F_{special\;2}(a,b,a,b,L_1,L_2)=\cfrac{1}{3}\,(L_2-L_1)[(a+b)^2-ab\]$ | 1,345 | 3,829 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.909138 |
https://nm.mathforcollege.com/chapter-08-03-runge-kutta-2nd-order-method/ | 1,721,367,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00568.warc.gz | 375,586,155 | 10,308 | # Chapter 08.03 Runge-Kutta 2nd Order Method
###### Prerequisites & Objectives
Prerequisites for Runge-Kutta 2nd Order Method [PDF] [DOC]
Objectives of Runge-Kutta 2nd Order Method[PDF] [DOC]
###### Textbook Chapter
A Textbook Chapter on Runge-Kutta 2nd Order Method[PDF] [DOC]
###### Digital Audiovisual Lectures
Runge-Kutta 2nd Order Method: Background [YOUTUBE9:46] [
Runge-Kutta 2nd Order Method: Formulas [YOUTUBE 10:57] [
Runge-Kutta 2nd Order Method: Heun’s Method [YOUTUBE 9:27] [
Runge-Kutta 2nd Order Method: Midpoint Method [YOUTUBE 10:45] [
Runge-Kutta 2nd Order Method: Ralston’s Method Part 1 of 2 [YOUTUBE 7:09] [
Runge-Kutta 2nd Order Method: Ralston’s Method Part 2 of 2 [YOUTUBE 5:00] [
Runge-Kutta 2nd Order Method: Derivation Part 1 of 2 [YOUTUBE 7:08] [
Runge-Kutta 2nd Order Method: Derivation Part 2 of 2 [YOUTUBE 5:31] [
###### Multiple Choice Test
Test Your Knowledge of Runge-Kutta 2nd Order Method [HTML] [PDF] [DOC]
###### Presentations
PowerPoint Presentation of Runge-Kutta 2nd Order Method [PDF] [PPT]
###### Worksheets
Worksheet of Runge-Kutta 2nd Order Method [MAPLE] [MATHCAD] [MATHEMATICA] [MATLAB]
Convergence Worksheet of Runge-Kutta 2nd Order Method [MAPLE] [MATHCAD] [MATHEMATICA] [MATLAB]
###### Blog Entries
Comparing Runge-Kutta 2nd Order Methods[BLOG]
Runge-Kutta 2nd Order Method Equations Derived [BLOG]
A MATLAB Program for Comparing Runge-Kutta 2nd Order Methods [BLOG]
###### Wolfram Demonstration
Global and Local Errors in Runge-Kutta Methods [NBP]
###### Examples from Other Majors
Chemical Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY]
Civil Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY]
Computer Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY]
Electrical Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY]
Industrial Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY]
Mechanical Engineering Example of Runge-Kutta 2nd Order Method [PDF] [DOC] [PHY] | 630 | 2,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.766731 |
http://shoplist.dedasys.com/CS146SecLab/4/ | 1,624,450,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00257.warc.gz | 41,847,029 | 3,979 | ## Lab 4: Encryption with lists and procedures.
1. Write a program that has one button and one label.
• The label should display the contents of a variable. (use `-textvariable`)
• The variable contains a number (initialized to 0).
• When you click the button, it invokes a procedure that adds 1 to the variable. Don't forget the scoping rules for local/global scope.
The results should resemble this (after clicking the Incr button a few times.
2. The fibonacci series is generated by
1. Initialize a list as {1 1}
2. Create each new element by adding the end and end-1 elements of the list
ie: {1 1 (1+1=2)} then {1 1 2 (1+2=3)}, etc.
3. This can be implemented with three variables. Two to hold the previous two elements of the list, and one to hold the new sum.
Modify the previous program to display the next fibonacci number in a series each time you click the button.
All you need to change is the contents (and perhaps name) of the procedure.
3. One more modification to the previous program. Rewrite the procedure so that it adds a random integer between 0 and 9 to the previous value.
Run it a couple of times. Do you get the same set of numbers each time?
Most computer applications use a pseudo-random number generator that uses some fancy math to calculate a random number.
The number isn't actually random, it's just a random distribution. If you can force the random number generator to start at a known value, you'll always get the same values after that.
This is useful when you are debugging an application that uses random numbers and you want it to behave consistently.
The `srand()` function sets the random number generator start point. The Tcl interpreter calls this with the current time-in-seconds when it starts. This ensures that you always get a different pattern of random numbers when you run an application.
A line like this will force the random number generator to start from 0.
``````
expr srand(0)
``````
Add this to the program and check that you get the same random numbers each time you run it.
4. Here is the complete code from the lecture example.
The variable `alpha` is defined within the `encrypt` procedure. That makes it local to the `encrypt` procedure.
``````
################################################################
# proc encrypt {}--
# encrypt the contents of .plain and insert into .encrypt
# Arguments
# NONE
#
# Results
# .encrypt window is modified.
#
proc encrypt {txt} {
# Fill the lookup table
set alpha \
{a b c d e f g h i j k l m n o p q r s t u v w x y z}
# Iterate through characters
foreach char [split \$txt {}] {
set pos [lsearch \$alpha \$char]
# If character is in list, encrypt it, else keep the character
if {\$pos >= 0} {
set pos2 [expr {(\$pos + 13) %26}]
set char2 [lindex \$alpha \$pos2]
} else {
set char2 \$char
}
append encrypt \$char2
}
return \$encrypt
}
################################################################
# proc encryptWin {inWin outWin}--
# Read characters from inWin and insert encrypted characters
# into outWin
# Arguments
# inWin: Name of a text widget to read input from
# outWin: Name of a text widget to write encrypted text to
# Results
# outWin widget contents are modified.
#
proc encryptWin {inWin outWin} {
set txt [\$inWin get 0.0 end]
\$outWin delete 0.0 end
\$outWin insert 0.0 [encrypt \$txt]
}
set w [text .plain]
grid \$w -columnspan 3
set w [text .encrypt]
grid \$w -columnspan 3
set w1 [button .b_encrypt -text "Encrypt" -command encryptWin .plain .encrypt ]
set w2 [button .b_exit -text "Exit" -command exit]
grid \$w1 \$w2
``````
This version of the `encrypt` procedure defines the `alpha` variable in the global scope. That way it only needs to be defined once, which makes the `encrypt` procedure more efficient. (The assignment is done once, instead of every time the `encrypt` procedure is invoked.
Replace the previous encrypt procedure with this one and test it.
This code won't work - fix it.
``````
# Fill the lookup table
set alpha {a b c d e f g h i j k l m n o p q r s t u v w x y z}
################################################################
# proc encrypt {}--
# encrypt the contents of .plain and insert into .encrypt
# Arguments
# NONE
#
# Results
# .encrypt window is modified.
#
proc encrypt {txt} {
# Iterate through characters
foreach char [split \$txt {}] {
set pos [lsearch \$alpha \$char]
# If character is in list, encrypt it, else keep the character
if {\$pos >= 0} {
set pos2 [expr {(\$pos + 13) %26}]
set char2 [lindex \$alpha \$pos2]
} else {
set char2 \$char
}
append encrypt \$char2
}
return \$encrypt
}
``````
5. The previous `encrypt` procedure uses a fixed offset (Caesar) cipher.
Modern ciphers like DES use a random number and a secret seed to change the offset for each character.
Rework the `encrypt` procedure to use `expr srand(\$seed)` and `rand()` to generate a better encryption.
Write a decrypt procedure to convert back to plaintext to test this.
6. The previous encryption procedures used a single list and calculated offsets. We can write an encryption procedure that uses two lists - one for the plain text, and one for the cipher text:
``````
set plainChars {a b c d e f g h i j k l m n o p q r s t u v w x y z}
set encryptChars {z y x w v u t s r q p o n m l k j i h g f e d c b a}
proc encrypt {txt} {
global plainChars
global encryptChars
foreach char [split \$txt {}] {
set pos [lsearch \$plainChars \$char]
# If character is in list, encrypt it, else keep the character
if {\$pos >= 0} {
set char2 [lindex \$encryptChars \$pos}
} else {
set char2 \$char
}
append encrypt \$char2
}
return \$encrypt
}
``````
This code has a small typo bug in it.
Cut/Paste it into your program and fix the bug, then write a decrypt procedure to convert cipherText into plainText.
7. The previous example uses a fixed relationship between plainText and cipherText. It always converts an 'a' to a 'z', etc.
The `encryptChars` array can be created on the fly with code like this
``````
################################################################
# proc buildEncryptChars {orig}--
# return a list of reordered characters
# Arguments
# orig A list of list elements to be reordered.
#
# Results
# No side effects.
#
proc randomizeList {orig} {
set origLen [llength \$orig]
for {set i 0} {\$i < \$origLen} {incr i} {
set pos [expr {int(rand()*[llength \$orig])}]
lappend new [lindex \$orig \$pos]
set orig [lreplace \$orig \$pos \$pos]
}
return \$new
}
``````
Modify the previous example by adding this procedure and then adding a wrapper around the `randomizeList` procedure to initialize the `encryptChars` variable.
The new procedure should accept a seed value so that you can decrypt a message later.
8. A useful encrypting/decryption program would allow you to load text from a file and then encrypt or decrypt it.
Add a Load button to this procedure that will call a procedure that reads a set of text and inserts it into the upper text widget so that you can then encrypt or decrypt it.
Add a Save button that will use the `tk_getSaveFile` command to get a filename, open the file in write mode, and save the data from the lower text window.
The results should look like this:
9. An even more useful application would have buttons to Load and Save from each text window, a button to clear the text and scrollbars attached to the windows.
Add an argument to the Load and Save procedures in the previous example to select the window that will have data loaded to or saved from.
Modify the `buildGUI` procedure to create separate Load and Save buttons for each text window to load/save into that window.
Also modify the buildGUI procedure to contain scrollbars attached to each text widget.
The results should look something like this:
10. Instead of splitting the incoming text into characters, it can be kept as words.
In that case, we can use two lists of words to generate a code message instead of a cipher message.
``````
set plain {IBM Microsoft Buy Sell Stock}
set code {{Blue Dog} Gate Bake Bury Cake}
``````
Write the procedures so that the phrase
``````
If IBM is higher than Microsoft, buy their Stock
``````
becomes
``````
If Blue Dog is higher than Gate, Bake their Cake
``````
No online solution for this one. See if you can figure it out. | 2,043 | 8,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.865535 |
https://laurenceanywaysthemovie.com/what-can-161-be-divisible-by/ | 1,701,845,610,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00195.warc.gz | 419,136,177 | 12,931 | # What can 161 be divisible by?
## What can 161 be divisible by?
When we list them out like this it’s easy to see that the numbers which 161 is divisible by are 1, 7, 23, and 161.
## What are the factors of 161?
Factors of 161: 1, 7, 23 and 161.
Is 160 divisible by any number?
When we list them out like this it’s easy to see that the numbers which 160 is divisible by are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, and 160.
### Is 163 divisible by any number?
When we list them out like this it’s easy to see that the numbers which 163 is divisible by are 1 and 163. What is this? You might be interested to know that all of the divisor numbers listed above are also known as the Factors of 163. Not only that, but the numbers can also be called the divisors of 163.
### Why is 163 a prime number?
The number 163 is divisible only by 1 and the number itself. For a number to be classified as a prime number, it should have exactly two factors. Since 163 has exactly two factors, i.e. 1 and 163, it is a prime number.
Is 343 divisible by any number?
As 343 is not even, it is not divisible by any even number. Try the divisibility tests with 3, 5 and 10. It is not divisible by them.
## What is the LCM of 161?
Solution: The factors of 161 are 1, 7, 23, 161 and factors of 81 are 1, 3, 9, 27, 81. Therefore, the Lowest Common Multiple (LCM) of 161 and 81 is 13041 and Greatest Common Factor (GCF) of 161 and 81 is 1.
## What type of number is 161?
161 (number)
← 160 161 162 →
Cardinal one hundred sixty-one
Ordinal 161st (one hundred sixty-first)
Factorization 7 × 23
Divisors 1, 7, 23, 161
Is 160 a prime numbers?
160 is an even composite number. 160 has more than 2 factors, hence it is a composite number. Factors of 160 are all the integers that 160 can be divided into. The factors of 160 are written as 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, and 160.
### What are the factors of 160?
The factors of 160 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 and 160.
### What is the prime factorization of 163?
The number 163 is prime and therefore its factors are only the numbers 1 and 163 itself. Hence, it has only one prime factor that is the number itself, i.e. 163. So, the prime factorization of 163 can be written as 1631 where 163 is prime.
Why is 163 not a prime number?
## What are all the numbers that 161 is divisible by?
The numbers that 161 is divisible by are 1, 7, 23, and 161. You may also be interested to know that all the numbers that 161 is divisible by are also known as the factors of 161. In view of this, is 161 a prime number?
## Is 161 an odd number?
Yes, 161 is a deficient number, that is to say 161 is a natural number that is strictly larger than the sum of its proper divisors, i.e., the divisors of 161 without 161 itself (that is 1 + 7 + 23 = 31 ). 161 is an odd number, because it is not evenly divisible by 2. What is an even number?
What is the prime factorization of 161?
To be 161 a prime number, it would have been required that 161 has only two divisors, i.e., itself and 1. However, 161 is a semiprime (also called biprime or 2 -almost-prime), because it is the product of a two non-necessarily distinct prime numbers. Indeed, 161 = 7 x 23, where 7 and 23 are both prime numbers.
### Is 161 a semiprime?
However, 161 is a semiprime (also called biprime or 2 -almost-prime), because it is the product of a two non-necessarily distinct prime numbers. Indeed, 161 = 7 x 23, where 7 and 23 are both prime numbers. | 1,052 | 3,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-50 | longest | en | 0.94813 |
http://www.chegg.com/homework-help/questions-and-answers/imagine-cart-rests-frictionless-surface-car-100-kg-30-basketballs-1-kg--throw-basketballs--q3300964 | 1,475,058,125,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661349.6/warc/CC-MAIN-20160924173741-00039-ip-10-143-35-109.ec2.internal.warc.gz | 402,400,670 | 13,935 | Imagine you are on a cart that rests on a frictionless surface (you + car = 100 kg) with 30 basketballs (1 kg each). You throw the basketballs off with a speed of 3 m/s relative to the cart every ten seconds. After 5 minutes (when all the basketballs have been thrown), how fast in m/s are you going? You throw the basketballs in the same direction each time. Use conservation of linear momentum to find your velocity after a few throws. You should find a pattern and it will become simpler how to determine the answer without much more tedious work. (The answer is not 1.11) | 135 | 575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2016-40 | latest | en | 0.953806 |
https://math.stackexchange.com/questions/3850467/sequence-of-continuous-functions-on-0-1-pointwise-converging-to-an-unbounded | 1,702,326,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00210.warc.gz | 421,516,332 | 37,044 | # Sequence of continuous functions on $[0,1]$ pointwise converging to an unbounded function
I have spent a few hours trying to find an example of a sequence of continuous functions $$f_n$$ $$[0,1]\rightarrow \mathbb{R}$$ that pointwise converge to a function $$f$$: $$[0,1]\rightarrow \mathbb{R}$$ that is unbounded.
Attempt: I have so far only managed to think of an example where $$f_n:(0,1]\rightarrow \mathbb{R}$$ when we define: $$f_n = \frac{n}{nx+2}$$
This sequence converges to $$f=\frac{1}{x}$$, which is clearly not bounded on the interval.
Edit thanks to Olivier Moschetta
My issue is when $$x=0$$ the sequence does not pointwise converge to a function on $$[0,1]$$.
Can anyone help me fix this example?
• Aren't the $f_n$ continuous at $0$? Their discontinuity point is $x=-2/n$ which is not in $[0,1]$. The problem is more that $f_n(0)=n/2$ is divergent, so $f_n$ do not converge pointwise to a function defined on $[0,1]$. Oct 3, 2020 at 21:21
• @OlivierMoschetta Yes you are correct, I just fixed my wording. Thanks for the help! Oct 3, 2020 at 21:27
It's pretty easy to "fix" your example. Define $$f_n(x)=\frac{nx}{nx^2+2}$$. At $$x=0$$ the sequence converges to $$0$$, at any other point to $$\frac{1}{x}$$.
You can take, for instance, $$f_n\colon[0,1]\longrightarrow\Bbb R$$ defined by$$f_n(x)=\begin{cases}2n^2x&\text{ if }x\leqslant\frac1{2n}\\n&\text{ if }\frac1{2n}\leqslant x\leqslant\frac1n\\\frac1x&\text{ otherwise.}\end{cases}$$Each $$f_n$$ is continuous and the sequence $$(f_n)_{n\in\Bbb N}$$ converges pointwise to$$\begin{array}{ccc}[0,1]&\longrightarrow&\Bbb R\\x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$ | 584 | 1,694 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-50 | latest | en | 0.757411 |
https://learnit.itu.dk/local/coursebase/view.php?s=ft&view=public&ciid=734 | 1,628,044,589,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00377.warc.gz | 351,078,972 | 13,823 | Official course description:
Full info last published 17/05-21
### Introduction to Data Science and Programming
##### Course info
Language:
English
ECTS points:
15
Course code:
BSINDSP1KU
Participants max:
95
Offered to guest students:
no
Offered to exchange students:
no
Offered as a single subject:
no
##### Programme
Level:
Bachelor
Programme:
BSc in Data Science
##### Staff
Course manager
Associate Professor
Teacher
Part-time Lecturer
Teacher
Postdoc
Teaching Assistant
Teaching Assistant (TA)
Teaching Assistant
Teaching Assistant (TA)
Teaching Assistant
Teaching Assistant (TA)
Semester
Efterår 2021
Start
30 August 2021
End
31 December 2021
##### Abstract
This course is an introduction to programming, data science and related foundations.
##### Description
The subjects covered in the course include:
• Computer fundamentals
• Problem analysis, program design and implementation
• Python programming:
• Simple data types, methods, fields, variables, expressions
• Objects, special Python objects (Strings, Lists, Files), classes and class design, object-oriented design (object-oriented concepts: encapsulation, polymorphism, inheritance)
• Basic logic and its relation to Boolean types and operations, decision structures
• Loop structures
• Data collections
• Algorithm design and recursion
• Testing and documentation
• Foundational maths for programming and Data Science:
• Basic formal reasoning (basic first order logic, set theory, sequences and sums)
• Basic probabilities
• Induction and recursion
• Basic graph theory
• Introductory Data Science:
• Basic statistical and visual summaries and basic reporting
• Data curation and preparation for analysis
• Basic Machine Learning
• Basic definitions of graph theory and basic Network Analysis
• Big data and distributed computing
##### Formal prerequisites
The course is mandatory for students on first semester BSc in Data Science.
The course is only open for students enrolled in BSc in Data Science.
##### Intended learning outcomes
After the course, the student should be able to:
• Analyse a problem, with an aim to construct a short program in Python script to solve it.
• Design a program (given an analysis).
• Implement a program (given a design).
• Test the program (including explaining whether it works as desired, and measuring to what degree the testing supports such conclusions).
• Apply basic Python constructions.
• Evaluate and explain whether or how a basic Python construction is appropriate to solve a problem.
• Describe what a data science project is.
• Prepare data for a data science analysis.
• Create and explain basic statistical and visual summaries for datasets.
• Assess basic probabilities for events.
• Create and explain basic data science analyses (basic machine learning and basic network analysis).
• Describe and apply formal definitions, and construct induction proofs.
• reflect on and apply basic data structures for data science
##### Learning activities
There are 14 weeks of teaching activities.
Note that the course also includes weekly homework assignments and up to two mandatory coding tests, which are listed under mandatory activities.
The lectures cover topics in Data Science, Python programming, and any Supplementary Mathematics Foundation (beyond Gymnasium Maths) for this as listed above in the course content.
The exercise sessions are practical sessions associated with the topics covered in the weekly lectures. The exercises are also aimed at additionally equipping the students with the skill sets necessary for the successful completion of homework assignments.
##### Mandatory activities
The course includes:
1) a maximum of 12 mandatory homework assignments, and
2) a minimum of 1 and a maximum of 2 short coding tests on premises, approximately 10 lectures and 20 lectures into the course.
All mandatory activities are pass/fail. Any coding parts of mandatory activities must be 100% correct for a pass. Any analysis parts of mandatory activities must be sound and well justified. The students will be granted a second attempt at all mandatory activity. The second attempt deadline will be set one week after the first deadline.
Homework Assignments
• Each homework assignment is closely tied to the weekly progression of the course, and also closely relates to the topics addressed in the lectures and the corresponding exercise sessions.
• The homework assignments require the students to closely observe the submission deadlines. The students are required to submit correct solutions to the mandatory homework assignments to be considered eligible for the exam. We allow for a maximum of 1 assignment to be failed/not submitted.
Coding Tests
• The coding test mandatory activity is a checkpoint to assess the adoption level for the skills required to participate in the remaining modules of the semester.
• The coding tests are based on the more basic programming exercises reviewed in Exercise Sessions.
The student will receive the grade NA (not approved) at the ordinary exam, if the mandatory activities are not approved and the student will use an exam attempt.
##### Course literature
Distributed course notes, articles and chapters as needed.
##### Student Activity Budget
Estimated distribution of learning activities for the typical student
• Preparation for lectures and exercises: 10%
• Lectures: 20%
• Exercises: 20%
• Assignments: 30%
• Exam with preparation: 20%
##### Ordinary exam
Exam type:
A: Written exam on premises, Internal (7-point scale)
Exam variation:
A33: Written exam on premises on paper with restrictions
Exam duration:
4 hours
Aids allowed for the exam:
Written and printed books and notes
Pen
E-books and/or other electronic devices
• To access books and notes only.
##### Time and date
Ordinary Exam - on premises Thu, 6 Jan 2022, 15:00 - 19:00 | 1,175 | 5,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-31 | latest | en | 0.826483 |
https://www.maasaimarathon.org/category/marathon-km-pace-calculator/ | 1,601,594,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402132335.99/warc/CC-MAIN-20201001210429-20201002000429-00234.warc.gz | 896,596,873 | 130,084 | Categories
## Marathon Km Pace Calculator
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This free pace calculator computes pace, time, and distance, given values for marathon (26. 22 miles / 42. 195 km), 4:41/mile or 2:55/km, 5:10/mile or 3:13/km . Calculate your finish time for popular race distances (5k, 10k, 10 mile, half marathon, marathon) based on your expected pace. calculate the pace needed to hit This free pace calculator computes pace, time, and distance, given values for two of the variables. it can also be used for training purposes through the multipoint pace calculator, convert between units of pace, and estimate a finish time. learn more about heart rate zones and different types of exercise, or explore hundreds of other calculators addressing math, finance, health, fitness, and.
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## Explore The Lewis Clark Trail
small residuary infection following their staging surgery edwards km pneumococcal infections: curative strategies and pitfalls [url= des waldgebiets und gleichnamigen naturparks schönbuch ca 17 km südlich von stuttgart und hat ca 8500 einwohner 12 jul 2020 5k, 5 mile, 10k, 10 mile, half marathon, and marathon times for paces you can plug in your goal time to our training pace calculator tool to see first, to easily determine your pace per mile and/or pace per kilometer from a . | 1,709 | 6,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-40 | latest | en | 0.752845 |
http://en.allexperts.com/q/Aeronautical-Engineering-1809/2013/5/contra-coaxial-rotating-helicopters.htm | 1,493,026,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119225.38/warc/CC-MAIN-20170423031159-00376-ip-10-145-167-34.ec2.internal.warc.gz | 118,457,801 | 7,352 | You are here:
# Aeronautical Engineering/Contra / Coaxial rotating Helicopters.
Question
QUESTION: Dear Paul
http://en.wikipedia.org/wiki/Helicopter_rotor
http://en.wikipedia.org/wiki/Contra-rotating
http://en.wikipedia.org/wiki/Contra-rotating_propellers
http://en.wikipedia.org/wiki/Coaxial_rotors
http://www.maxxprod.com/mpi/mpi-266.html
http://dspace.mit.edu/handle/1721.1/65309
http://dspace.mit.edu/handle/1721.1/61906
I want to know the advantages and disadvantages of implementing
Contra rotating helicopters design over the conventional main rotor system in terms of manufacturing cost, efficiency, power, speed etc
Thanks & Regards,
Prashant S Akerkar
ANSWER: Prashant - The main advantages of contra-rotating rotors is the balanced torque which does not have to be countered by the tail rotor (though you still need some directional control), the added thrust from two rotors, and the fact that blades can advance on the left and right sides simultaneously. The disadvantages include flow interactions and noise from the two rotors passing close to each other (especially from one passing through the wake of the other), weight of the second rotor, and mechanical complexity of the rotor head. Depending on the mission, a co-axial rotor can be a winner, but it is probably a heavy-lift situation where noise and vibration can be tolerated.
Paul
[an error occurred while processing this directive]---------- FOLLOW-UP ----------
QUESTION: Dear Paul
Thank you.
The Design complexity will increase if more than two rotors in opposite directions are to be driven in odd and even configuration.
i.e.
3,5,7 etc Odd
4,6,8 etc Even
Increasing the Number of Rotors (3,4,5,6,7,8) can affect the Helicopter speed ?.
Will 3,5,7 etc Odd configuration also work ?
OR
Only 2,4,6,8 etc Even Configuration work ?
Thanks & Regards,
Prashant S Akerkar
ANSWER: I'm not sure if you are talking about rotors or rotor blades. In any case, I don't know the effects of even or odd number of blades. All I can say is that machines designed for high speed have fewer rotor blades than machines designed for high lift at static conditions.
---------- FOLLOW-UP ----------
QUESTION: Dear Paul
Thank you.
I am referring to the Number of Rotors (2,3,4,5,6,7,8) and not the Rotor Blades.
The Design complexity will increase if more than two rotors in opposite directions are to be driven in odd and even configuration.
i.e.
3,5,7 etc Odd
4,6,8 etc Even
Increasing the Number of Rotors (3,4,5,6,7,8) can affect the Helicopter speed ?.
Will 3,5,7 etc Odd configuration also work ?
OR
Only 2,4,6,8 etc Even Configuration work ?
Thanks & Regards,
Prashant S Akerkar
Prashant - If you want to increase helicopter speed, you have to increase the airspeed through the rotor disc. Adding a third rotor, for example, may add some power at the expense of more drag, more engine power required, more vibration, and more weight in the rotor, head, and engine. The second and third downstream rotors are in a really bad flow field. At some point, the negative factors overcome the positive factors. Although, I do not know how to analyze this, I suspect that N rotors will slow the helicopter relative to N-1 rotors (N > 1). I have no idea about even and odd combinations of rotors.
Paul
p.s. I should have said N > 2
Questioner's Rating
Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Dear Paul Thank you. Thanks & Regards, Prashant S Akerkar
• Ask a Question
Aeronautical Engineering
Volunteer
#### Paul Soderman
##### Expertise
Aeronautics, Aerodynamics, Fluid Mechanics, Aeroacoustics, Noise Control, Muffler Design, Wind Tunnel Research.... I know nothing about India - do not ask about schools, jobs, application requirements, career choices, etc. for India. Please, no text message verbiage; I prefer full words in full sentences. Thanks.
##### Experience
38 years as research engineer at NASA
Publications
AIAA, NASA
Education/Credentials
B.S. and M.S. Aeronautical Engineering - U. of Washington, Graduate work Standford U.
Awards and Honors
AIAA Associate Fellow (American Institute of Aeronautics and Astronautics) | 1,039 | 4,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-17 | latest | en | 0.862038 |
http://www.askiitians.com/iit-jee-atomic-structure/relation-between-kinetic-energy-and-wavelength/ | 1,406,791,589,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510272680.42/warc/CC-MAIN-20140728011752-00345-ip-10-146-231-18.ec2.internal.warc.gz | 366,867,841 | 14,690 | ```Relation between Kinetic Energy and Wavelength:
K.E (E) =1/2 mv2
v = √2E / m ⇒ λ = h/ mv
λ = h / √2E.m
Illustration:
An e-, a proton and an alpha particle have K.E of 16 E, 4 E and E respectively. What’s the qualitative order of their Broglie wavelengths?
(A) λe > λP > λa (B) λe > λp = λa
(C) λP < λe < λa (D) λa < λe = λP
Solution:
λ = h / √2mK.E . Hence (B) is correct.
Illustration:
What is the de-Broglie wavelength of electron having K.E. of 5 eV?
Solution:
K.E. = 1/2 mv2
v =√ 2K.E / m
Now = λ = h / mv
= h / √2K.E.m
= 6.62 x 10-34 / √ 2 x 5 x 1.6 x 10-19 x 9.1 x 10-31
= 5.486 x 10-10 m
```
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PHOTOELECTRIC EFFECT Sir J.J. Thomson, observed... | 776 | 2,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-23 | latest | en | 0.65167 |
https://study.com/academy/topic/ap-physics-1-kinematics-homeschool-curriculum.html | 1,585,777,621,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00217.warc.gz | 708,819,929 | 26,642 | # Ch 2: AP Physics 1 Kinematics: Homeschool Curriculum
The kinematics unit of this AP Physics 1 Homeschool course is designed to help homeschooled students learn about principles of kinematics. Parents can use the short videos to introduce topics, break up lessons and keep students engaged.
## Who's it for?
This unit of our AP Physics 1 Homeschool course will benefit any student who is trying to learn about kinematics concepts. There is no faster or easier way to learn about kinematics. Among those who would benefit are:
• Students who require an efficient, self-paced course of study to learn kinematics equations, projectile motion, kinematics graphs, scalars and vectors.
• Homeschool parents looking to spend less time preparing lessons and more time teaching.
• Homeschool parents who need a physics curriculum that appeals to multiple learning types (visual or auditory).
• Gifted students and students with learning differences.
## How it works:
• Students watch a short, fun video lesson that covers a specific unit topic.
• Students and parents can refer to the video transcripts to reinforce learning.
• Short quizzes and a kinematics unit exam confirm understanding or identify any topics that require review.
## Kinematics Unit Objectives:
• Recognize kinematics as the study of objects' motion.
• Differentiate between scalars and vectors.
• Learn how to determine the position of an object at any moment in time.
• Distinguish between an object's distance and its displacement.
• Discover the difference between an object's speed and its velocity.
• Calculate an object's acceleration using the appropriate equation.
• Understand scientific figures and how to provide answers using scientific notation.
• Solve two-step accelerated motion problems using the big five kinematic equations.
• Learn how to use graphs to represent kinematics.
• Analyze objects' motion using ticker tape diagrams.
• List the uses of vector diagrams.
• Describe motion using position vs. time graphs.
• Find the slope of a line on position vs. time graphs.
• Describe motion based on velocity vs. time graphs.
• Ascertain acceleration using the slope of a velocity vs. time graph.
• Find an object's displacement using a velocity vs. time graph.
• Provide qualitative descriptions of graphs of motion.
• Determine the speed, time and height in free fall problems.
• Use position vs. time and velocity vs. time graphs to represent free fall motion.
• Explore an object's acceleration moving under the influence of gravity.
• Understand what projectile motion is and how to solve projectile motion problems.
• Identify the list of equations used to represent objects' motion.
23 Lessons in Chapter 2: AP Physics 1 Kinematics: Homeschool Curriculum
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 673 | 3,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-16 | longest | en | 0.911037 |
http://www.pearltrees.com/stuartcalimport/modifiable-objects/id3756834 | 1,369,336,435,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703728865/warc/CC-MAIN-20130516112848-00025-ip-10-60-113-184.ec2.internal.warc.gz | 641,100,243 | 11,056 | # Modifiable Objects
Get flash to fully experience Pearltrees
## Tweezers
A variety of tweezers
## Network topology
Network topology is the arrangement of the various elements ( links , nodes , etc.) of a computer [ 1 ] [ 2 ] or biological network . [ 3 ] Essentially, it is the topological [ 4 ] structure of a network, and may be depicted physically or logically. Physical topology refers to the placement of the network's various components, including device location and cable installation, while logical topology shows how data flows within a network, regardless of its physical design. Distances between nodes, physical interconnections, transmission rates, and/or signal types may differ between two networks, yet their topologies may be identical. A good example is a local area network (LAN): Any given node in the LAN has one or more physical links to other devices in the network; graphically mapping these links results in a geometric shape that can be used to describe the physical topology of the network.
A tag cloud with terms related to Web 2.0 A tag cloud ( word cloud , or weighted list in visual design) is a visual representation for text data, typically used to depict keyword metadata (tags) on websites, or to visualize free form text. Tags are usually single words, and the importance of each tag is shown with font size or color. [ 1 ] This format is useful for quickly perceiving the most prominent terms and for locating a term alphabetically to determine its relative prominence. When used as website navigation aids, the terms are hyperlinked to items associated with the tag. [ edit ] History In the language of visual design, a tag cloud (or word cloud) is one kind of "weighted list", as commonly used on geographic maps to represent the relative size of cities in terms of relative typeface size.
## Tag cloud
In linear algebra , an orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors (i.e., orthonormal vectors). Equivalently, a matrix Q is orthogonal if its transpose is equal to its inverse : which entails where I is the identity matrix . An orthogonal matrix Q is necessarily invertible (with inverse Q −1 = Q T ), unitary ( Q −1 = Q * ), and normal ( Q * Q = QQ * ).
## Orthogonal matrix
Illustration of electrophoresis Illustration of electrophoresis retardation Electrophoresis is the motion of dispersed particles relative to a fluid under the influence of a spatially uniform electric field . [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] This electrokinetic phenomenon was observed for the first time in 1807 by Ferdinand Frederic Reuss ( Moscow State University ), [ 7 ] who noticed that the application of a constant electric field caused clay particles dispersed in water to migrate.
## Glossary
A glossary , also known as an vocabulary , or clavis , is an alphabetical list of terms in a particular domain of knowledge with the definitions for those terms. Traditionally, a glossary appears at the end of a book and includes terms within that book that are either newly introduced, uncommon, or specialized. A bilingual glossary is a list of terms in one language defined in a second language or glossed by synonyms (or at least near-synonyms) in another language.
## Shuttle
The original meaning of the word shuttle is the device used in weaving to carry the weft. By reference to the continual to-and-fro motion associated with that, the term was then applied in transportation and then in other spheres. Thus the word may now also refer to:
## Hub
Hub , or Hubs may refer to: [ edit ] Wheels Bicycle hub , the central part of a bicycle wheel Locking hubs , accessory on four-wheel drive vehicles Wheel hub assembly , an automotive part [ edit ] Buildings
A web is a silken structure created by a spider. Web may also refer to: [ edit ] Computing World Wide Web or "the Web", a hypertext system that operates over the Internet Web 2.0 , a perceived transition of the Web from a collection of Web sites to a full-fledged computing platform serving Web applications WorldWideWeb , the first web browser and editor Web (web browser) , the web browser included with GNOME desktop environment (previously known as Epiphany ) Web.com , a public company that offers websites and other services for small businesses and consumers Webs (web hosting) , a website which allows users to create free websites WEB , a computer programming system created by Donald Knuth to implement literate programming
## Web
In biochemistry and structural biology , secondary structure is the general three-dimensional form of local segments of biopolymers such as proteins and nucleic acids (DNA/RNA). It does not, however, describe specific atomic positions in three-dimensional space, which are considered to be tertiary structure . Secondary structure can be formally defined by the hydrogen bonds of the biopolymer, as observed in an atomic-resolution structure.
## Protein secondary structure
In computer science , a lookup table is an array that replaces runtime computation with a simpler array indexing operation. The savings in terms of processing time can be significant, since retrieving a value from memory is often faster than undergoing an 'expensive' computation or input/output operation. [ 1 ] The tables may be precalculated and stored in static program storage, calculated (or "pre-fetched" ) as part of a program's initialization phase ( memoization ), or even stored in hardware in application-specific platforms. Lookup tables are also used extensively to validate input values by matching against a list of valid (or invalid) items in an array and, in some programming languages, may include pointer functions (or offsets to labels) to process the matching input.
## Lookup table
A tumor suppressor gene , or anti-oncogene , is a gene that protects a cell from one step on the path to cancer. When this gene is mutated to cause a loss or reduction in its function, the cell can progress to cancer, usually in combination with other genetic changes. [ edit ] Two-hit hypothesis
## Tumor suppressor gene
Phosphatase and tensin homolog ( PTEN ) is a protein that, in humans, is encoded by the PTEN gene . [ 2 ] Mutations of this gene are a step in the development of many cancers . PTEN acts as a tumor suppressor gene through the action of its phosphatase protein product. This phosphatase is involved in the regulation of the cell cycle , preventing cells from growing and dividing too rapidly. [ 3 ] It is one of the targets of an oncomiR, MIRN21 . This gene was identified as a tumor suppressor that is mutated in a large number of cancers at high frequency.
## PTEN (gene)
The Sun is Earth's primary source of light. About 44% of the sun's electromagnetic radiation that reaches the ground is in the visible light range. Visible light (commonly referred to simply as light ) is electromagnetic radiation that is visible to the human eye , and is responsible for the sense of sight . [ 1 ] Visible light has a wavelength in the range of about 380 nanometres to about 740 nm – between the invisible infrared , with longer wavelengths and the invisible ultraviolet , with shorter wavelengths. Primary properties of visible light are intensity , propagation direction, frequency or wavelength spectrum , and polarisation , while its speed in a vacuum, 299,792,458 meters per second, is one of the fundamental constants of nature. Visible light, as with all types of electromagnetic radiation (EMR), is experimentally found to always move at this speed in vacuum.
## Light
From Wikipedia, the free encyclopedia Rate (mathematics) , a specific kind of ratio, in which two measurements are related to each other Rate of travel, or velocity Naval rating or rate, terms used to designate specialty or seniority of enlisted naval personnel. Rate of a ship , a term indicating a sail ship's firepower in the British Royal Navy Bit rate , number of bits that are conveyed or processed per unit of time Rates , a Portuguese parish and town located in the municipality of Póvoa de Varzim In science and finance | 1,702 | 8,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-20 | longest | en | 0.897076 |
https://de.slideshare.net/Diramar/the-beauty-of-mathematics-presentation | 1,620,583,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00332.warc.gz | 230,018,462 | 30,305 | Diese Präsentation wurde erfolgreich gemeldet.
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# The Beauty Of Mathematics
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### The Beauty Of Mathematics
1. 1. Here is an interesting and lovely way to look at the beauty of mathematics, and of God, the sum of all wonders. The Beauty of Mathematics Wonderful World
2. 2. 1 x 8 + 1 = 9 12 x 8 + 2 = 9 8 123 x 8 + 3 = 9 8 7 1234 x 8 + 4 = 9 8 7 6 12345 x 8 + 5 = 9 8 7 6 5 123456 x 8 + 6 = 9 8 7 6 5 4 1234567 x 8 + 7 = 9 8 7 6 5 4 3 12345678 x 8 + 8 = 9 8 7 6 5 4 3 2 123456789 x 8 + 9 = 9 8 7 6 5 4 3 2 1
3. 3. 1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 +10= 1111111111
4. 4. 9 x 9 + 7 = 88 98 x 9 + 6 = 888 987 x 9 + 5 = 8888 9876 x 9 + 4 = 88888 98765 x 9 + 3 = 888888 987654 x 9 + 2 = 8888888 9876543 x 9 + 1 = 88888888 98765432 x 9 + 0 = 888888888 Brilliant, isn’t it?
5. 5. 1 x 1 = 1 11 x 11 = 1 2 1 111 x 111 = 1 2 3 2 1 1111 x 1111 = 1 2 3 4 3 2 1 11111 x 11111 = 1 2 3 4 5 4 3 2 1 111111 x 111111 = 1 2 3 4 5 6 5 4 3 2 1 1111111 x 1111111 = 1 2 3 4 5 6 7 6 5 4 3 2 1 11111111 x 11111111 = 123456787654321 111111111 x 111111111 = 12345678987654321 And look at this symmetry:
6. 6. 101% From a strictly mathematical viewpoint: What Equals 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been in situations where someone wants you to GIVE OVER 100% How about ACHIEVING 101%? What equals 100% in life? Now, take a look at this…
7. 7. Here’s a little mathematical formula that might help Answer these questions:
8. 8. If: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Is represented as: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
9. 9. If: H-A-R-D-W-O-R- K 8+1+18+4+23+15+18+11 = 98% And: K-N-O-W-L-E-D-G-E 11+14+15+23+12+5+4+7+5 = 96%
10. 10. But: A-T-T-I-T-U-D-E 1+20+20+9+20+21+4+5 = 100% THEN , look how far the love of God will take you: L-O-V-E-O-F-G-O-D 12+15+22+5+15+6+7+15+4 = 101%
11. 11. Therefore, one can conclude with mathematical certainty that: While Hard Work and Knowledge will get you close, and Attitude will Get you there , It’s the Love of God that will put you over the top!
12. 12. Have a great day… and that God bless you! Diramar setembro,2008 www.slideshare.net/Diramar www.pdmfedc.multiply.com (Internet search) | 1,209 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-21 | latest | en | 0.386368 |
http://numbernut.com/numbersandcounting/activities/number_4card_count1-10.html | 1,721,653,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00868.warc.gz | 21,922,866 | 5,441 | Basic Math | Basic-2 Math | Prealgebra | Workbooks | Glossary Glossary | Standards | Site Map | Help
# Pick-a-Card: Counting to Ten
This activity was designed to help you count values from one (1) to ten (10). The top of the activity will ask you to find a number of shapes. The card below will have a variety of shapes. You need to find the card with the right shape and the right number of shapes. For example, you will be asked to find "four pentagons" and you need to click the card with four five-sided figures. All of the questions combine random values.
Good luck and have fun.
## Directions
This is one of the NumberNut four-card activities. Your question will be displayed at the top of the screen. Under that question, you will see four (4) possible answers. Just click on the card with the correct answer. The next screen will show you the correct answer.
You get a happy face for every correct answer and a sad face for every wrong answer. The quiz is over after ten (10) questions. Take the quiz again and again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations.
RELATED LINKS LESSONS: - NumberNut.com: Numbers ACTIVITIES - Pick-a-Card: Counting 1 to 10 - Memory Challenge: Counting 1 to 10 - Memory Challenge: Names of Numbers - More or Less: Numbers 1-10 - Before and After: Numbers 1-10 - Memory Challenge: Numbers 0 to 9 - Memory Challenge: Chinese Numbers - Memory Challenge: Mayan Numbers - Memory Challenge: Roman Numerals - Pick-a-Card: Recognizing Numbers 1-30 - More or Less: Numbers 1-30 - Before and After: Numbers 1-30 - Pick-a-Card: Recognizing Numbers 1-50 - Pick-a-Card: Recognizing Numbers 1-100 - More or Less: Numbers 1-100 - Before and After: Numbers 1-100 - Pick-a-Bar: Recognizing Numbers 1-1,000 - More or Less: Numbers 1-1,000 - Before and After: Numbers 1-1,000 - Pick-a-Bar: Roman Numerals 1-1,000 - Pick-a-Bar: Recognizing Numbers 1-10,000 - More or Less: Numbers 1-10,000
- Overview - Shapes-Colors - Numbers - Addition - Subtraction - Multiplication - Division - Operations - Dates & Times > Activities
* The custom search only looks at Rader's sites.
Go for site help or a list of mathematics topics at the site map! | 577 | 2,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.840941 |
https://lists.cam.ac.uk/pipermail/cl-isabelle-users/2013-February/msg00290.html | 1,638,525,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362619.23/warc/CC-MAIN-20211203091120-20211203121120-00099.warc.gz | 439,935,677 | 2,650 | # Re: [isabelle] Trying to reduce equality proofs from 2 to 1 step
On 27.02.2013 23:21, Gottfried Barrow wrote:
(A = B) ====> (A seO B) --> (A = B),
where "====>" means I have to convert (A=B) into the form (A seO B)
before anything can be proved, and I can't get the simplifier to convert
(A=B) to something like (A seO B), which I suppose is for good reason,
since "=" is used everywhere. (In a manner of speaking, I manually
convert (A=B) to (A seO B) as the first step in the proof.)
Actually, this kind of thing is good use case for introduction rules. Given a rule X: "A se0 B ==> A = B", you could perform the proof with
by (rule X) simp
or, for a structured proof:
proof (rule X)
...
qed
If declared with [intro?], rule (and hence proof) will automatically pickup this rule.
-- Lars
This archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc. | 253 | 882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.930896 |
https://medium.com/@alexanderdouglas/unrestricted-generality-in-medieval-logic-and-divine-omnipotence-fc7a46c1b6d?source=rss-5eb869eee506------2 | 1,553,584,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204857.82/warc/CC-MAIN-20190326054828-20190326080828-00054.warc.gz | 558,574,950 | 28,326 | # Absolute Generality, Syllogistic, and Divine Omnipotence
In an earlier post, I wrote about Spinoza’s rejection of a standard theological argument against his proposition, to the effect that God produces everything it is in his power to produce. The argument, roughly, is that if God produces everything it is in his power to produce, then there must be some limit on his power. Supposing that God produces a race of giants to the extent of his power, take the tallest giant in the race; it must be that God could not have produced a taller one.
The argument seems to involve an implicit rejection of infinite domains — otherwise we could say that in such a case there is no tallest giant produced by God, just as there is no largest prime number. But to think of such a case involves reference to an actual rather than merely potential infinity, and as Poincaré pointed out (before criticising the ideas of Russell and Cantor), such an idea was traditionally alien to mathematical theory:
The notion of infinity had long since been introduced into mathematics, but this infinity was what philosophers call a becoming. Mathematical infinity was only a quantity susceptible of growing beyond all limit; it was a variable quantity of which it could not be said that it had passed, but only that it would pass, all limits.
In this case we could say, of some group of giants, that there exists outside that group some taller giant, but we cannot speak of the group of all the giants and say that for any one in that group there is another, taller giant in the same group. For that group would then have to be an actual infinity — a quantity that ‘has passed all limits’ — for reasons Poincaré explained in detail with reference to paradoxes of non-predicative definitions.
Within the traditional theory of inference — the syllogistic — there is a more interesting restriction on generality. All syllogistic terms must have instances (this is to preserve the traditional Square of Opposition, among other things). Without negative terms there is no way to speak with absolute generality; we can say that all S are P but there is no way to express the thought that everything is P. With negative terms it becomes formally possible to express thoughts about everything. We can have, SaP; S’aP. By the law of excluded middle, everything must be P. But this formal possibility is ruled out if all terms have instances. For if P is a term then so is P’. But, as Neville Keynes showed, ordinary rules of conversion can allow us to draw inferences from SaP; S’aP that are contradictory unless P’ is taken as empty:
SaP = SeP’ = P’eS
S’aP = S’eP’ = P’eS’ = P’aS
If both all and no P’ are S then there must be no P’, but then P’ is an empty term.
The upshot is that syllogistic provides no way to speak of everything — not even within a domain. For if S and P are non-empty terms within any domain comprising a model for inferences — what Lewis Carroll called the ‘universe of discourse’ — then S’ and P’, if they are to be accepted as terms, must also be non-empty within that domain; after all they can combined with the other terms and each other according to the various quantities and qualities. Absolute generality is implicitly rejected in the syllogistic, so long as we respect the existential import of all terms. The universe of discourse must be inexhaustible by the terms of our discourse.
A.N. Prior (Formal Logic, second edition, 129–30) complains that if we take seriously the stricture forbidding non-empty terms then we can also generate the result that SeP and S’eP’ will never be jointly false. Yet, he notes, if we take ‘man’ for S and ‘non-man’ for P we seem to have a clear countermodel to this rule: no men are non-men, but no non-men are non-non-men (i.e. men). This countermodel is, however, perhaps already implicitly rejected by the inexhaustibility condition on the universe of discourse, since men and non-men together seem to exhaust the domain of things under discussion. If we form a term, M, such that SaM and PaM, then M will be a term for everything altogether, whether a man or not, and M’ will then be empty.
It follows from this inexhaustibility condition that, within the constraints of the syllogistic, we can’t reason about what is in God’s power. To do so, we would have to take what is in God’s power as a term. And then its negative, given divine omnipotence, will be empty.
How then could theologians reason about divine omnipotence? I suppose it is the kind of thing that can be shown but not said. For any group of things we conceive or speak of as being in God’s power, there will be other things we are not conceiving or speaking of that are also in God’s power — God’s power is then the ‘universe of discourse’, which, like all such universes within syllogistic, is inexhaustible by our powers of expression, no matter how far we extend them.
Spinoza, following Descartes (or so I argue), probably rejected the syllogistic as a complete model of inference. But elsewhere he seems to appeal to the rejection of absolute generality — namely in his arguments against the divisibility of infinite extension in the Scholium to Proposition 17 of the Ethics. At least, as I read those arguments, they consist of a denial that infinite extension could be composed of any set of finite regions.
Alison Peterman’s excellent paper on extension in Spinoza puts the point in this way. Spinoza argues that finite bodies can’t compose an infinite extension for the same reason that points can’t compose a line. But, Peterman notes, points can’t compose a line because they are zero-dimensional and lines are one-dimensional. Does it follow that infinite extension is four-dimensional?! Peterman suggests that Spinoza’s analogy aims at a more general result: finite bodies cannot compose infinite extension because they are the wrong sort of thing to compose it.
In the background here, I think, is the sort of sceptical attitude towards actual infinity found in Aristotle’s Metaphysics. But it is difficult to see what grounds that rejection if not some more general rejection of absolute generality: why could we not, e.g., speak of all the finite bodies in a group such that for any finite set of bodies in that group there are others besides? Such a group, provided the bodies didn’t overlap, would compose an infinitely extended body.
Spinoza aimed, of course, to demonstrate his philosophy more geometrico. The problem is that there was no formal theory of inference to explain what governs geometrical reasoning (for that, at least for Euclidean geometry, we have to wait for Schwäbhauser, Szmielew, and Tarski). Certainly the syllogistic won’t work; Descartes showed in the Rules for the Direction of the Mind that it doesn’t cover ordinary arithmetical inferences. But in the absence of such a theory we need to answer questions such as that of the permissibility of absolute generality. In the absence of a theory of inference, it is unsurprising that Spinoza should appear to have not made up his mind about absolute generality. | 1,579 | 7,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-13 | latest | en | 0.970659 |
http://www.2classnotes.com/5th-class/lines-and-angle/ | 1,600,437,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00452.warc.gz | 159,563,659 | 12,107 | Friday , September 18 2020
# NCERT 5th Class (CBSE) Mathematics: Lines And Angle
CHECK IN
A protractor is used to measure angles.
This angle is measured on the inner scale as the baseline arm point to 0° on the inner scale. The other arm crosses 50° showing that ∠ AOB = 50°.
Angles are named according to their measures.
## गुरु और चेला 5 NCERT CBSE Hindi Book Rimjhim Chapter 12
गुरु और चेला 5th Class NCERT CBSE Hindi Book Rimjhim Chapter 12 गुरु और चेला – प्रश्न: … | 150 | 477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.751922 |
https://blog.autarkaw.com/2018/02/21/computational-time-for-forward-elimination-steps-of-naive-gaussian-elimination-on-a-square-matrix/ | 1,723,571,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00266.warc.gz | 108,890,821 | 11,568 | # Computational Time for Forward Elimination Steps of Naive Gaussian Elimination on a Square Matrix
Problem Statement
How much computational time does it take to conduct the forward elimination part of the Naïve Gauss Elimination method on a square matrix?
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## 0 thoughts on “Computational Time for Forward Elimination Steps of Naive Gaussian Elimination on a Square Matrix”
1. Nate says:
Nice post, many times you hear the cost is O(n^3) but can easily forget why or what the lower order terms are.
One thing that comes to mind is that this is true as long as the computation is arithmetically intense. That is, the number of floating point operations to memory accesses is large. In the case of Gaussian elimination, this is true and so counting FLOPS is a good measure of cost. However, in a operation such as a sparse matrix-vector product (where you have low arithmetic intensity, 1 fused multiply-add per memory access) it is the memory bandwidth that limits performance.
2. ncollier says:
Nice post, we are often reminded that the cost is O(n^3) but often forget why or what the lower order terms are.
One things that comes to mind is that modeling the FLOPS is a good measure of performance only if the computation has a high arithmetic intensity. That is, the number of floating point operations to memory accesses is high. This is true of Gaussian Elimination. However, in an operation such as a sparse matrix-vector multiply the arithmetic intensity is low (1 fused multiply add per memory access) and the operation becomes bandwidth limited.
1. Yes, so true Nate. This may be very rudimentary way to introduce students to computational time. Many variables enter into the equation for computational time as you pointed out! Watch soon for computational time for Gauss Jordan method of finding inverse! | 413 | 1,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-33 | latest | en | 0.928784 |
http://classroom.synonym.com/calculate-diameter-hexagon-2689.html | 1,513,429,398,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588072.75/warc/CC-MAIN-20171216123525-20171216145525-00155.warc.gz | 56,779,105 | 9,108 | A hexagon is a six-sided, two dimensional figure. Its diameter is measured from a vertex, or intersection of two sides, through the center, to the opposite side to the point where two sides intersect. A common example of hexagons in nature are beehives. To calculate the diameter, you need to measure the length of one of the hexagon's sides.
### Step 1
Measure the length of one side of the hexagon with the ruler and record the length.
### Step 2
Measure the other sides of the hexagon to make sure the hexagon is regular. In a regular hexagon, all six sides will be equal. If the hexagon is irregular, it will not have a diameter.
### Step 3
Multiply the side length by 2 to calculate the diameter of the hexagon. In a hexagon, the radius, or distance from the intersection of two sides to the center of the hexagon, is equal to the side length, and the diameter equals twice the radius. For example, if each side in your hexagon equals 13 inches, your diameter would be 26 inches. | 232 | 990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-51 | latest | en | 0.898543 |
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# NGR MNFD
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1 year ago | 0 | 1,423 | 5,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.913435 |
https://1pdf.net/8th-gradesciencestaarreview-_58668a6fe12e89f5593c7de0 | 1,550,866,638,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247526282.78/warc/CC-MAIN-20190222200334-20190222222334-00245.warc.gz | 472,347,333 | 19,832 | Formula chart! 8th gradeScienceSTAARReview Objective2:Force,Motion,&Energy 8.6.A&demonstrate& and&calculate&how& unbalanced& forces&change&the& …
Objective 2: Force, Motion, & Energy 8.6.A demonstrate and calculate how unbalanced forces change the speed or direction of an object’s motion
Force A Force is a Push or a Pull that can change motion.
How Force is Measured Newton - The SI unit used to measure force. The symbol for Newton is N. Formula chart!
Net force= 3N to the right
Net Force= mass x acceleration
F
=
m x a
I am a roller skater with a mass of 72kg. If I am accelerating toward a wall at 2 3.7m/s , what will be the amount of force at which I hit the wall?
Spring Scale – Measures Force in Newtons (N).
Net Force When more than one force acts on an object, the forces combine to form a Net Force. The combination of all the forces acting on an object is the Net Force.
Net Force = 2 ĺ ADD forces in the same direction Magnitude is the size of a force.
Net Force = 1 ĸ SUBTRACT forces in opposite directions.
8.6.C investigate and describe applications of Newton’s law of inertia, law of force and acceleration, and law of action-‐reaction such as in vehicle restraints, sports activities, amusement park rides, Earth’s tectonic activities, and rocket launches
Newton’s 1st Law Newton’s First Law: An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. This law is often called the Law of Inertia
Examples of Newton’s 1st Law:
x
Car suddenly stops and you strain against the seat belt (vehicle restraints) because our bodies want to keep moving
x x x
When riding a horse, the horse suddenly stops and you fly over its head Ketchup stays in the bottom (at rest) until you bang (outside force) on the end of the bottom Can you think of another example?____________________________________________
Newton’s 2nd Law Newton’s Second Law: Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object). It can be measured by F=MxA This law is often called the Law of Acceleration Calculate ____F____ = 1000 kg x 0.5 m/s/s ____F____ = ___500 N___ Examples of Newton’s 2nd Law: x Hitting a baseball- the harder the hit, the faster the ball goes accelerating x A grocery cart filled with lots of food vs. an empty grocery cart x The positioning of football players – massive players on the line with lighter (faster to accelerate) players in the backfield x Can you think of another example? ______________________________________________
Newton’s 3rd Law Newton’s Third Law: For every action there is an equal and opposite re-action. For every force there is a reaction force that is equal in size, but opposite in direction. This law is often called the Law of Action-Reaction. Examples of Newton’s 3rd Law: x Momentum of the car moving forward and the car comes to a sudden stop, our body pushes against the seat (action) belt and the seat belt pushes back (reaction). x When you lean on the wall to rest, the weight on the wall provides the reaction force and the wall x
x
pushes back on you (reaction force) with the same force. As the gases move downward, the rocket moves in the opposite direction. Can you think of another example?____________________________________________
Use the Arrows to show Action and Reaction in the pictures below.
Sudden stop
Leaning on wall
Forces may move an object
Rocket lifting off
Forces may transfer between objects.
Types of Forces Balanced – Forces that are equal in magnitude but opposite in direction. Balanced forces do not cause a change in the motion of objects.
Unbalanced - Force that cause a change in the motion of an object. One force must be larger than the other.
Speed, Velocity & Acceleration
8.6.B differentiate between speed, velocity, and acceleration
Speed is the rate used to measure the distance traveled over a period of time.
Velocity is a measure of the speed in a given direction.
Question: A green helicopter is moving up at 30 kilometers per hour. A blue helicopter is moving down at 30 kilometers per hour. A.
Are the helicopters’ speeds the same? Explain.
B.
Are the velocities the same? Explain.
x Acceleration is the change of velocity over a period of time. x If speed or direction changes, then you have acceleration.
In your own words, explain the differences between speed, velocity, and acceleration.
7.7.A contrast situations where work is done with different amounts of force to situations where no work is done such as moving a box with a ramp and without a ramp, or standing still
Work Work is the amount of force applied times the distance over which it is applied. In order for work to occur or happen… THE OBJECT MUST MOVE IN THE DIRECTION OF THE FORCE APPLIED.
Work = Force x distance W=f x d Formula chart!
Solve: 1. 2.
A force of 825 N is needed to push a car across a lot. Two student push the car 35 m . How much work is done? You push against the wall for 3min with a force of 10 N. How much work is done? Explain.
Work or No Work?
Leaning
Pushing
Leaning on wall: Work or No Work?
Pushing skateboard forward: Work or No Work?
Standing in the rain: Work or No Work?
Lifting
6.8.A compare and contrast potential and kinetic energy
Standing
Potential Energy Forms of Energy Potential
Description of Energy Energy that is stored in an object.
Example: The rubber band chicken. As the rubber band is stretched and placed in the hold position, the rubber band will store energy.
Kinetic Energy Forms of Energy Kinetic
Description of Energy Energy of motion; Based on the mass and speed of the moving object.
Example: The flying rubber band chicken. As the rubber band is released it becomes energy in motion.
Potential to Kinetic Energy
When the coaster is at its highest point on the track, it has it the greatest potential energy. As the coaster loses height it gains speed: PE is transformed into KE. As the coaster gains height it loses speed: KE is transformed into PE. 6.8.C calculate average speed using distance and time measurements
Average speed = distance time
Formula chart!
s=
Solve: 1. 2.
You arrive in my class 45 seconds after leaving math which is 90 meters away. How fast did you travel? You need to get to class, 200 meters away, and you can only walk in the hallways at about 1.5 m/s. (if you run any faster, you’ll be caught for running). How much time will it take to get to your class?
Graphing Motion Time (sec)
Distance (m)
1 2 3 4 5 6
5 10 15 30 35 40
Speed = distance time
S=d t
Race 50
Distance (meters)
6.8.D measure and graph changes in motion
d/t
40 30
distance
20 10 0 1
2
3
4
Time (sec)
5
6
6.9.C demonstrate energy transformations such as energy in a flashlight battery changes from chemical energy to electrical energy to light energy
Energy Energy is the ability to do work. Forms of Energy: 1. Electrical 2. Chemical 3. Radiant/Solar 4. Nuclear 5. Mechanical
Categories of Energy Potential 1. Chemical 2. Mechanical 3. Nuclear
1. 2. 3. 4. 5.
Kinetic Radiant / Sunlight Thermal / Heat Electrical Sound Mechanical
* Mechanical Energy can be both potential and kinetic.
Electrical Energy Forms of Energy Electrical
Description of Energy Delivered by tiny charged particles called electrons, this form of energy is typically moved through a wire.
Example: Lighting or Electricity
Description of Energy Energy that travels as light Solar Energy – energy from the Sun only Radiant Energy – energy from all other light sources
Nuclear Energy Forms of Energy Nuclear
Description of Energy Energy stored in the nucleus of an atom — the energy that holds the nucleus together.
Example: Nuclear power plants split the nuclei of uranium atoms.
Thermal Energy Forms of Energy Thermal / Heat
Description of Energy The vibration and movement of the atoms and molecules within substances. As an object is heated up, its atoms and molecules move and collide faster.
Example: Geothermal - heat from the earth.
Mechanical Energy Forms of Energy Mechanical
Description of Energy Potential energy stored in objects by tension. Kinetic energy when machine parts are moving.
Example: Gears or compressed spring; moving parts
Sound Energy Forms of Energy Sound
Description of Energy The movement of energy through substances. Sound is produced when a force causes an object or substance to vibrate.
Example: Moving guitar strings
Chemical Energy Forms of Energy Chemical
Description of Energy Energy stored within the bonds of atoms and molecules.
Example: Gasoline, Batteries, or Food
Energy Transformations Energy can change from one form to another. Example: Kinetic Energy can turn into potential energy and back again. Chemical Energy can be used to create Electrical Energy and Electrical Energy can be used to create Heat Energy
Law of Conservation – Energy cannot be created or destroyed but can only change from one form to another. Chemical - Electrical
Batteries made of chemicals – Creates electricity to turn on the light bulb. Sunlight – Photosynthesis produces glucose Nuclear – Electrical
Nuclear Energy - Power Plant changes energy into electricity for homes
Mechanical - Sound
Speaker movement –Vibrations create sound
Energy Transformations Chemical - Mechanical
Thermal – Electrical
Gas – Engine turns blade to cut grass
Heat from the Earth – Power Plant changes it to electricity for homes | 2,364 | 9,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-09 | longest | en | 0.87669 |
https://shakyradunn.com/slide/encm369-assebly-lanuage-review-1gtqxu | 1,606,331,537,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184123.9/warc/CC-MAIN-20201125183823-20201125213823-00719.warc.gz | 477,597,626 | 10,600 | # ENCM369 Assebly Lanuage Review
Assembly Language Review Being able to repeat on the Blackfin the things we were able to do on the MIPS 01/26/20 Review of 50% OF ENCM369 in 50 minutes 1 Assembly code things to review 50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS Being able to ADD and SUBTRACT the contents of two data registers Being able to bitwise AND and bitwise OR the contents of two data registers Being able to place a (small) required value into a data register Being able to place a (large) required value into a data register Being able to write a simple void function (returns nothing) Being able to write a simple int function (returns and int)
Being able to ADD and SUBTRACT the contents of two memory locations IF YOU CAN DO THE SAME THING ON THE BLACKFIN THEN THATS 50% OF THE LABS AND 50% OF EXAMS ACED 01/26/20 2 / 28 Being able to ADD and SUBTRACT the contents of two data registers It makes sense to ADD and SUBTRACT values stored in data registers Blackfin DATA registers R0, R1, R2 and R3 R0 = R1 + R2; // Addition e.g. 4 + 6 10 (Decimal) 0x14 + 0x16 0x2A (Hexadecimal) R3 = R1 R2; // Subtraction e.g. 4 - 6 8 (Decimal) 0x14 - 0x16 0xFFFFFFFE (Hexadecimal) 01/26/20 Review of 50% OF ENCM369 in 50 minutes
3 / 28 Being able to bitwise AND and OR the contents of two data registers It makes sense to perform OR and AND operations on bit- patterns stored in data registers. NEVER perform ADD and SUBTRACT operations on bitpatterns stored in data registers. (Although SOMETIMES get the correct answer code defect) Blackfin DATA registers R0, R1, R2 and R3 R0 = R1 & R2; // Bitwise AND e.g. B11001100 & B01010101 B01000100 R3 = R1 | R2; // Bitwise OR e.g. B11001100 | B01010101 B11011101 01/26/20 Review of 50% OF ENCM369 in 50 minutes 4 / 28
Is it a bit pattern or a value? Hints from C++ If the code developer is consistent when writing the code then Bit patterns are normally stored as unsigned integers e.g. unsigned int bitPattern = 0xFFA2345FF Values are normally stored as signed integers e.g. signed int fooValue = -1; or int fooValue = -1; where the word signed is understood. Understood means its there but not actually written down (which means that it sometimes causes defects in your code) Note that bitPattern = 0xFFFFFFFF and fooValue = -1 are STORED as the SAME bit pattern 0xFFFFFFFFF in the registers and memory of MIPS and Blackfin processor 01/26/20 Review of 50% OF ENCM369 in 50 minutes 5 / 28 Being able to place a required
value into a data register 1 Like the MIPS, the Blackfin uses 32 bit instructions all registers are the same size to ensure maximum speed of the processor (highly pipelined instructions). The 32 bit Blackfin instruction for placing a value into a data register has two parts to have16 bits available for describing the instruction and 16 bits for describing the signed 16 bit value to be put into a signed 32 bit data register. This means that you have to use 2 32-bit instructions to put large values into a data register (SAME AS MIPS). 01/26/20 Review of 50% OF ENCM369 in 50 minutes 6 / 28
Placing a value into a data register Similar to MIPS, different syntax R1 = 0; legal -- 0 = 0x0000 (signed 16 bits); (becomes the signed 32 bit 0x00000000 after auto sign extension of the 16-bit value 0x0000) R0 = 33; legal -- 33 = 0x0021 (signed 16 bits) (becomes the signed 32 bit 0x00000021 after auto sign extension of the 16-bit value 0x0021) R2 = -1; legal -- -1 = 0xFFFF (signed 16 bits) (becomes the signed 32 bit 0xFFFFFFFF after auto sign extension of the 16-bit value 0xFFFF)
R3 = -33; legal -- -33 = 0xFFDE (signed16 bits) (becomes the signed 32 bit 0xFFFFFFDE after auto sign extension of the 16-bit value 0xFFDE) 01/26/20 Review of 50% OF ENCM369 in 50 minutes 7 / 28 Placing a large value into a data register This approach does not work for any large value R1 = 40000; DOES NOT WORK WITH MIPS EITHER illegal -- as 40000 cant be expressed as a signed 16-bit value it is the positive 32 bit value 0x00009C40
If the assembler tried to take the bottom 16 bits of the decimal 40000 and sign extend it then this would happen 16-bit hex value 9C40 (1001 1100 0100 0000) becomes 32-bit hex value after sign extension 0xFFFF9C40 which is a negative value Therefore it is illegal to try to put a 32-bit value directly into a register; just as it would be illegal to try in MIPS. 01/26/20 Review of 50% OF ENCM369 in 50 minutes 8 / 28 Placing a large value into a data register If the assembler tried to take the bottom 16 bits of the decimal 40000 and sign extend it then this would happen 16-bit hex value 9C40 (1001 1100 0100 0000) becomes 32-bit hex value after sign extension 0xFFFF9C40 which is a negative value illegal just as it would be in MIPS // Want to do R1 = 40000
// Instead must do operation in two steps as with MIPS #include R1.L = lo(40000); // Tell assembler to put bottom // 16-bits into low part of R1 register R1.H = hi(40000); // Tell assembler to put top // 16-bits into high part of R1 register 01/26/20 9 / 28 Placing a large value into a data register A common error in the laboratory and exams is getting this two step thing wrong . Forgetting the second step is easy to do just as easy to forget on Blackfin as on MIPS // Want to do R1 = 41235 R1.L = lo(41235); // bottom 16-bits into low part of R1 register R1.H = hi(41325); // top 16-bits into high part of R1 register FORGOTTEN SECOND STEP RECOMMENDED SYNTAX TO AVOID CODE DEFECTS
#define LARGEVALUE 41235 // C++ - like syntax R1.L = lo(LARGEVALUE); R1.H = hi(LARGEVALUE); Yes you CAN put multiple Blackfin assembly language instructions on one line 01/26/20 Review of 50% OF ENCM369 in 50 minutes 10 / 28 A void function returns NO VALUE extern C void SimpleVoidASM(void) #include Things in red were cut-and-pasted using the editor to save Lab. time .section program;
.global _SimpleVoidASM; _SimpleVoidASM: _SimpleVoidASM.END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 11 / 28 A simple int function return a value extern C int SimpleIntASM(void) #include Things in red were cut-and-pasted using the editor .section program; .global _SimpleIntASM; _SimpleIntASM: R0 = 7; // Return 7 _SimpleIntASM.END: RTS;
01/26/20 Review of 50% OF ENCM369 in 50 minutes 12 / 28 Being able to ADD and SUBTRACT the contents of two memory locations Lets set up a practical situation A background thread is putting values into an array. Processor could be MIPS or Blackfin For background thread read interrupt service routine or ISR. ISR work in parallel with the foreground thread that is doing the major work on the microprocessor Write a subroutine (returns int) that adds together the first two values of this shared array 01/26/20
Review of 50% OF ENCM369 in 50 minutes 13 / 28 Start with a copy of the int function extern C int SimpleIntASM(void) #include Things in red were cut-and-pasted using the editor .section program; .global _SimpleIntASM; _SimpleIntASM: R0 = 7; // Return 7 _SimpleIntASM.END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 14 / 28
Modify to be extern C int AddArrayValuesASM(void) #include Things in red were cut-and-pasted using the editor .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: R0 = 7; // Return 7 _AddArrayValuesASM.END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 15 / 28 Add a data array #include .section L1_data; .byte4 _fooArray[2];
Things in red were cut-and-pasted using the editor // Syntax for building an array // of 32-bit values .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : R0 = 7; // Return 7 _AddArrayValuesASM .END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 16 / 28 Plan to return sum, initialize sum to 0 #include .section L1_data;
.byte4 _fooArray[2]; Things in red were cut-and-pasted using the editor .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: #define sum_R0 R0 sum_R0 = 0; // register int sum; // sum = 0; _AddArrayValuesASM .END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 17 / 28
Place the memory address of the start of the array into a pointer register . Other code Things in red were cut-and-pasted using the editor .section L1_data; .byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0; #define pointer_to_array_P1 P1 P1.L = lo(_fooArray); P1 is a POINTER register
(address register) // register int * pointer_to_array P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0]; _AddArrayValuesASM .END: RTS; 01/26/20 Review of 50% OF ENCM369 in 50 minutes 18 / 28 Read the contents of the first array location into register R1 and add to sum_R0; . Other code Things in red were cut-and-pasted using the editor .section L1_data;
.byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM : #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0; #define pointer_to_array_P1 P1 P1L = lo(_fooArray); // register int * pointer_to_array P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0]; R1 = [pointer_to_array_P1]; sum_R0 = sum_R0 + R1; // int temp = fooArray[0]; // sum = sum + temp _AddArrayValuesASM.END: RTS;
01/26/20 Review of 50% OF ENCM369 in 50 minutes 19 / 28 Read the contents of the second array location into register R1 and add to sum_R0; . Other code Things in red were cut-and-pasted using the editor .section L1_data; .byte4 _fooArray[2]; .section program; .global _AddArrayValuesASM; _AddArrayValuesASM: #define sum_R0 R0 // register int sum; sum_R0 = 0; // sum = 0;
#define pointer_to_array_P1 P1 // register int * pointer_to_array P1.L = lo(_fooArray); P1.H = hi(_fooArray); // pointer_to_array = &fooArray[0]; R1 = [pointer_to_array_P1]; // int temp = fooArray[0]; sum_R0 = sum_R0 + R1; // sum = sum + temp R1 = [pointer_to_array_P1 + 4]; sum_R0 = sum_R0 + R1; // temp = fooArray[1]; // sum = sum + temp _AddArrayValuesASM .END: RTS; 01/26/20 20 / 28 Add code to .ASM (assembly) file
01/26/20 Review of 50% OF ENCM369 in 50 minutes 21 / 28 Assignment 1, Q1 Demo answer 01/26/20 Review of 50% OF ENCM369 in 50 minutes 22 / 28 Assembly code things to review 50% of ENCM369 in 50 minutes YOU ALREADY KNOW HOW TO DO THESE THINGS ON THE MIPS Being able to ADD and SUBTRACT the contents of two data registers Being able to bitwise AND and bitwise OR the contents of two data registers Being able to place a (small) required value into a data register
Being able to place a (large) required value into a data register Being able to write a simple void function (returns nothing) Being able to write a simple int function (returns and int) Being able to ADD and SUBTRACT the contents of two memory locations IF YOU CAN DO THE SAME THING ON THE BLACKFIN THEN THATS 50% OF THE LABS AND 50% OF EXAMS ACED 01/26/20 23 / 28
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https://strategywiki.org/wiki/Professor_Layton_and_the_Curious_Village/Puzzles_76-100 | 1,702,159,767,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00456.warc.gz | 605,625,885 | 28,877 | Caution! This guide contains information about each puzzle including all three hints and the correct answer. If you do not want to see the solution to a puzzle and spoil the fun of figuring it out, scroll down carefully. The answers are hidden within the spoiler tags, so don't peek in them unless you're really stuck!
Puzzle 076
Name: A Tile Square
Trigger: Talk to Pauly
Chapter: 5
Picarats: 30
Note: This Puzzle exist only in US-Version.
Description: You have at your disposal a large number of tiles like the one shown below. If you were to take these tiles and try to make a square, what is the fewest number of tiles you'd need?
• Hint 1: Since the tiles have a different length and width, you'll need to find a number that can be divided by both dimensions. Of course, the puzzle doesn't end there. You'll need to do a little creative thinking in order to find the fewest number of tiles that allow you to form a square.
• Hint 2: Don't forget the puzzle also gives you the thickness of the tiles.
• Hint 3: As the sides of the tiles are 10 and 12 inches long, the smallest common multiple of the two works out to 60. Therefore, you'll need to arrange a 5x6 tile square. That's a total of 30 tiles. Simple, right? Too simple, in fact. There's a way to make a square using even fewer tiles.
Spoiler
Puzzle 076
Name: Flower Garden
Trigger: Talk to Pauly
Chapter: 5
Picarats: 30
Note: This Puzzle exist only in EU-Version.
Description: You decide to rent some space in a flower garden. Informing the owner of the garden that you need enough space for 12 plants, he tells you that there are four suitable allotments available, A-D. The rent for each allotment is calculated by area, but on top of the rent for the 12 plant-sized spaces, there is also a separate charge for the fence around the whole allotment. With this in mind, can you tell which allotment has the most expensive rent?
• Hint 1: You can find the answer by adding up the length of fence around all the allotments, but there is another way...
• Hint 2: At first glance, doesn't A look like it has the shortest length of fence? Yes, it does!
• Hint 3: Each individual square has four sides, so an allotment of 12 square will have 48 side in total, with some shared between squares. The number of sides being shared by neighbouring squares will differ greatly depending on the shape of the allotment.
Spoiler
Puzzle 077
Name: Which Job?
Chapter: 5
Picarats: 20
Description: Two corporations have put out Help Wanted ads. Aside from the information below, the two companies' offers are exactly the same. From a purely financial standpoint, which one should you work for? Company A will pay you \$100,000 a year a give you a \$20,000 raise yearly. Company B will pay you \$50,000 every six months and give you a \$5,000 raise every six months.
• Hint 1: If you take the time to add things, this puzzle can be solved in a minute. Go on, try and add up how much money you'd receive from both companies each year for a few years.
• Hint 2: For Company A, you can just look at your yearly salary to get your total income for the year. Company B will give you a raise halfway through the year, so you'll need to calculate two six-month periods to get your yearly pay. Which one offers more money?
• Hint 3: Let's add up how much you'd make in your first year at either company. Company A gives you \$100,000 for your first year of work. Company B gives you \$50,000 for your first half year and then gives you a raise. This means that for the next six-month period, you will make \$55,000. Add those together and you have your yearly pay for year one: \$105,000.
Spoiler
Puzzle 078
Name: Water Pitchers
Trigger: Talk to Crouton
Location: Restaurant
Chapter: 5
Picarats: 60
Description: You have one 16-quart pitcher full of water, one empty nin-quart pitcher, and one empty seven-quart pitcher. Using nothing but these three pitchers, can you divide the water evenly so that the 16-and nine-quart pitchers are each holding exactly eight quarts of water?
• Hint 1: That's right, it's time for another good, old-fashioned pitcher puzzle. This time around, the shortest solution requires 15 moves. As always, don't try shifting the liquid back and forth aimlessly. One tip that can make this process easier is to focus on the discrepancy between how much each pitcher can hold. Other than that, just do what you've been doing all along on these puzzles, and you'll come out on top.
• Hint 2: After you move a larger pitcher filled with water to a smaller pitcher, some water will remain in the larger pitcher. Pay attention to this leftover amount. Think about how you can use this leftover water in your pours to reach an answer.
• Hint 3: 16. 10. 9. 6. Once you've made it this far, you're only four more moves away from the solution.
Spoiler
```Answer: It takes 15 moves to complete this puzzle. Starting out, you have (16, 0, 0) First, pour the 16 pitcher into the 9 (7, 9, 0). Then the 9 pitcher goes into the 7 (7, 2, 7). Next, pour the 7 into the 16 pitcher (14, 2, 0). Now pour the 9 into the 7 again (14, 0, 2). The 16 goes into the 9 pitcher (5, 9, 2). Again, pour the 9 pitcher into the 7 (5, 4, 7). Now pour the 7 pitcher back into the 16 (12, 4, 0). The 9 goes into the 7 (12, 0, 4), followed by the 16 pitcher into the 9 (3, 9, 4). Next, the 9 pitcher is poured into the 7 (3, 6, 7). Now pour the 7 back into the 16 (10, 6, 0), followed by the 9 into the 7 (10, 0, 6). Almost there! Pour the 16 into the 9 (1, 9, 6). Now the 9 goes into the 7 (1, 8, 7). Finally, the 7 goes back into the 16 (8, 8, 0).
```
Puzzle 079
Name: Apples and Oranges
Trigger: Talk to Crouton
Location: Restaurant
Chapter: 5
Picarats: 40
Description: Some careless deliveryman loaded two shipments of fruit into the wrong warehouses. As you can see in the picture below, the oranges are currently in the apple warehouse and vice versa. Can you correct the mistake and put all the fruit in its proper place?
• Hint 1: There are two small areas between the warehouses where you can stash an additional two boxes. Use these to your advantage when shifting boxes back and forth.
• Hint 2: Find a pattern for sending boxes over and repeat it until solved. As long as you're gradually shifting the boxes to the proper warehouses, you will finish the puzzle eventually. Don't be afraid of moving a few boxes back to the wrong warehouse on occasion. Sometimes you need to go backward to go forward.
• Hint 3: Above all, you must find a method for moving things around. This is by no means a difficult problem, but if you just send over boxes to the opposing warehouse, you'll block the entrance and get stuck. Think about how you can avoid obstructing the warehouses' entrances.
Spoiler
```Answer: It takes 26 moves to complete this puzzle. Move 2 Oranges in both niches. Move 3 Apples (from the top row or bottom row) left past the niche. Move both Oranges right in top row (or bottom row). Move 2 Apples back right past the niches. Move another 2 Oranges (from the top row or bottom row) in the niches. Move 5 Apples (not the one in the middle) left past the niches. Move both Oranges in the niches right in bottom row (or top row). Move 2 Apples back right. Move the last 2 Oranges in the niches. Now move all Apples left on the Red Fields. All that remains is to move 2 Oranges right.
```
Puzzle 080
Name: Too Many Queens 1
Trigger: Talk to Flick
Location: Restaurant
Chapter: 5
Picarats: 20
Description: In chess, the queen can move the full length of the board diagonally, vertically, and horizontally. See if you can place four queens on this 4x4 chessboard. There's a catch though! You must arrange the pieces so that no queen blocks another's line of movement. Good luck!
• Hint 1: It's not like you have to solve the puzzle in a limited number of moves, so go ahead and check out all the possibilities. Here's a tip: try arranging the pieces in a way that allows for a line of symmetry between them.
• Hint 2: The four corner spaces on the board should be left unoccupied. The four pieces will form a perfectly symmetrical shape.
• Hint 3: You don't need to place any pieces in the four center squares of the board either. Now that you've eliminated those spaces and the corner spaces, you should have a pretty good idea about where your pieces should go.
Spoiler
Answer: There are several solutions to this problem. Here's one: Place a queen in the 1st column, 3rd row. Another queen goes in the 2nd column, 1st row. A third queen goes in the 3rd column, 4th row. The last queen goes in the 4th column, 2nd row.
Puzzle 081
Name: Too Many Queens 2
Trigger: Talk to Flick
Location: Restaurant
Chapter: 5
Picarats: 40
Description: In chess, the queen can move the full length of the board diagonally, vertically, and horizontally. See if you can place five queens on this 5x5 chessboard. There's a catch though! You must arrange the pieces so that no queen blocks another's line of movement.
• Hint 1: Here's a hint to get you started: place one queen in the dead center of the board.
• Hint 2: Once you place one piece in the board's center, you'll only have four pieces left to place. The remaining our pieces will surround the center in a symmetrical shape.
• Hint 3: Don't put any pieces in the four corner spaces of the board. You can also ignore the eight spaces directly surrounding the center space.
Spoiler
Answer: There are several ways to solve this problem. Here's one: Place a queen in the 1st column, 1st row. Another queen goes in the 2nd column, 3rd row. The 3rd column queen goes in the 5th row. The 4th column queen is placed in the 2nd row. The final queen goes in the 5th column and 4th row.
Puzzle 082
Name: Too Many Queens 3
Trigger: Talk to Flick
Location: Restaurant
Chapter: 5
Picarats: 60
Description: In chess, the queen can move the full length of the board diagonally, vertically, and horizontally. Let's try something a little different this time. See if you can arrange three queens on this 5x5 chessboard so that no more pieces can be placed on the board. Make sure you place the pieces so that no queen blocks another's line of movement.
• Hint 1: This one's a bit of a puzzler, but if you check everything thoroughly, you'll find the answer sooner or later. Don't put anything in the center square. Remember that you have to arrange the pieces so that no queen blocks another's line of movement. Even if you think you've got the answer, if one of your queens turns red, it means that you haven't got the placement just right yet.
• Hint 2: One of your three queens needs to go in a corner space.
• Hint 3: Two of your queens need to be placed within the eight squares that directly surround the center space.
Spoiler
Answer: There are several ways to solve this problem. Here's one: One queen goes in the 2nd column, 3rd row. The second queen goes in the 3rd column, 1st row. The final queen goes in the 5th column, 5th row.
Puzzle 083
Name: Too Many Queens 4
Trigger: Talk to Flick
Location: Restaurant
Chapter: 5
Picarats: 80
Description: In chess, the queen can move the full length of the board diagonally, vertically, and horizontally. You have three queens positioned on an 8x8 chessboard. Place the remaining five queens on the board so that no piece blocks another's line of movement. This is a tough one!
• Hint 1: You're sure you can't think this one out by yourself? Oh, all right. Here's a hint: leave the four corner spaces empty.
• Hint 2: Place one queen in the space third from the top in the far-left column. Place another queen in the far-right column, three spaces from the bottom.
• Hint 3: One queen goes in the space one to the right of the upper-left corner. Place another queen in the bottom row four spaces from the right.
Spoiler
Answer: One queen goes in the 1st column, 3rd row. Another queen goes in the 2nd column, 1st row. The next queen goes in the 3rd column, 7th row. The 5th column has a queen in the 8th row. The last queen goes in the 8th column, 6th row.
Puzzle 084
Name: Which Boxes to Move?
Trigger: Talk to Stachen
Location: Sewer Exit
Chapter: 7
Picarats: 30
Description: In preparation for your big move, you've packed your belongings into 20 boxes and arranged them as shown below. With everything packed, you are now ready to label each box with its contents. In order to do so though, you'll need to move a few boxes around. How many of the boxes can't be labeled without rearranging the stacks?
• Hint 1: Fourteen boxes are visible in the illustration. You have to use what you can see to visualize where the hidden boxes are. You can clearly see all the boxes in the top two levels of the pile, so you only need to worry about the two layers of boxes closest to the ground.
• Hint 2: The boxes are stacked upon each other, so any box not directly on the ground must have another box supporting it from below. Knowing this, you can infer that there are no fewer than six boxes in the layer that's second from the bottom, and no fewer than nine boxes resting directly on the floor.
• Hint 3: Working from the ground up, you can deduce there are nine boxes on the first layer, six boxes on the second layer, four boxes on the third layer, and then one box that rests on the entire stack. That gives you a total of 20 boxes, which is how many the puzzle says you have. With all boxes accounted for, all you need to do is figure out which of these hidden boxes are completely surrounded by other boxes on all sides.
Spoiler
Puzzle 085
Name: Weekend Getaway
Trigger: Tap the Laytonmobile
Location: Outside the Sewer
Chapter: 7
Picarats: 50
Note: This Puzzle exist only in US-Version.
Description: You and your girlfriend went on a road trip over the weekend. On the way to your destination, you drove 180 miles, and your girlfriend drove the rest of the way. Coming home on the exact same roads, your girlfriend drove the first 150 miles and then you got behind the wheel for the last leg of the journey. So what is the difference in miles between the distance you drove and the distance your girlfriend drove?
• Hint 1: Distances both to and from your destination were exactly the same. On the way there, you drove 180 miles, and on the way home, you drove one way minus 150 miles.
• Hint 2: Think about the distance your girlfriend drove. On the way there, she drove one way minus 180 miles. On the way home, she drove 150 miles. The total distance each person drove must include the trip out and the trip back. If you combine the earlier information about each leg of the trip, you'll see that your girlfriend drove a total of "one way minus 30 miles."
• Hint 3: Using the principle in Hint Two and the information from Hint One, you can express the distance you drove as "one way plus 30 miles." with that said, the difference between the distance you drove and the distance your girlfriend drove should be clear as day.
Spoiler
Puzzle 085
Name: Train Speed
Trigger: Tap the Laytonmobile
Location: Outside the Sewer
Chapter: 7
Picarats: 50
Note: This Puzzle exist only in EU-Version.
Description: A train with a length of 100m takes 30 seconds to travel over a 400m bridge. Assuming that the train's speed is constant, how many kilometres per hour is it travelling at?
• Hint 1: You're probably thinking that this is a really bothersome calculation problem. Actually, there's a really easy way to work it out. Just make sure you get all the numbers right before you start.
• Hint 2: A 100m train travelling over a 400m bridge in 30 seconds. So you have to divide 400 by 30, right? Well, the train starts travelling over the bridge the moment the front of the train is on the bridge, and it finishes once the end of the train has moved off the bridge. How far has the train travelled in this time?
• Hint 3: Can you see that the train must travel 500m in order to travel completely over the bridge? If so, the rest is easy. 500m is 0.5km. 30 seconds is 0.5 minutes.
Spoiler
Puzzle 086
Name: Squares and Circles
Trigger: Talk to Sylvain
Chapter: 7
Picarats: 30
Description: Sylvain brought you this diagram to see if you can help him with it. Several circles and squares are pictured in the diagram below. How many times larger is the area of the blue square when compared to that of the red square?
• Hint 1: Do you see the circle that touches the sides of the big blue square? A smaller square sits inside the circle and touches it. Since the square is smaller than the circle, you can rotate it within the circle.
• Hint 2: If you rotate the middle square 45 degrees, its corners will touch the sides of the large blue square. At the same time, notice that the rotation has made it so that the red square's corners now make contact with the middle square. Having trouble visualizing the rotation? Try drawing it on your screen.
• Hint 3: From the rotation described in the second hint, draw two perpendicular lines from the outer circle through the middle to divide the squares into four quadrants. Do this and you'll see that the middle square's area is equal to half of the blue square's. Go and try it for yourself.
Spoiler
Puzzle 087
Name: Ferris Wheel Riddle
Trigger: Tap on the Ferris Wheel
Location: Ferris Wheel
Chapter: 7
Picarats: 50
Description: There are 10 two-seater cars attached to the fair's Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. The wheel began operation at 10 in the morning and shut down 30 minutes later. What's the maximum number of people that could have taken a ride on the wheel in that time period?
• Hint 1: You may think all you have to do is add a few things up, but there's a trick to this problem that's easy to overlook. Think about the way Ferris wheels have to work. If two people catch the first ride of the day, how much time will pass before they get off the ride?
• Hint 2: If the wheel has to stop promptly at 10:30 then the operator won't allow anyone on who can't get off the Ferris wheel by 10:30. So what does that mean? It means that when two passengers step out of their gondola at 10:30, all other gondolas on the Ferris wheel should be empty as well.
• Hint 3: The first pair of people to board the Ferris wheel at 10:00 will get off the ride 10 minutes later. Since the ride itself takes 10 minutes, no one will get off until 10:10. Then, from 10:10 until shutdown at 10:30, people will get off the ride at a rate of two per minute. Calculate the number of people coming off the ride and you'll have your answer, but be careful, because it's easy to make a mistake.
Spoiler
Puzzle 088
Name: In a Hole
Trigger: Talk to Sylvain
Location: Shack Path
Chapter: 7
Picarats: 30
Note: This Puzzle exist only in US-Version.
Description: A tennis ball has rolled its way down into a hole. This particular hole is extremely deep and has a sharp bend in the middle, making the ball impossible to retrieve by hand. To make matters worse, the ground around the hole is made of hard clay, so digging the ball out isn't an option. However, you have something incredibly commonplace on hand that you can use to get the ball out. What do you use to get the ball out? Answer in five letters.
• Hint 1: "You can't reach the bottom of the hole, so how about trying to find some tool that can," you say? No, no, that's not the way. Is there some way you can get the ball to come to the mouth of the hole?
• Hint 2: Think about the particular characteristics of a tennis ball. It bounces, it's light... Well, it has many interesting properties, but what happens when you throw a tennis ball into a lake?
• Hint 3: If a great rainstorm were to come along right now, you could probably retrieve your ball without doing anything at all. Think about why that is. And just what is "rain," anyway?
Spoiler
Puzzle 088
Name: Leaky Tank
Trigger: Talk to Sylvain
Location: Shack Path
Chapter: 7
Picarats: 30
Note: This Puzzle exist only in EU-Version.
Description: A 2.5m deep water tank has water poured into it for 8 hours, starting at 9am. The water level rises by 60cm in this time. However, it seems that water is leaking out again at night, because the next morning the water level has dropped by 20cm. If the water level continues to rise 40cm a day in this way, on what day will the tank first overflow?
• Hint 1: The total rise in water level each day is 40cm, just as the puzzle says. But if you think about it, isn't there an important point missing? Yes, you're on the right track now.
• Hint 2: In 8 hours starting at 9am - in other words, until 5pm - water enters the tank and the water level rises by 60cm. For the next 16 hours, water leaks out of the tank, bringing the water level down by 20cm.
• Hint 3: At 5pm on day 1, the water level is 60cm. At 9am on day 2, it's 40cm. At 5pm on day 2, 1m. At 9am on day 3, 80cm. At 5pm on day 3, 1.4m. At 9am on day 4, 1.2m. Notice anything?
Spoiler
Puzzle 089
Name: Which Way?
Trigger: Tap on the Stone Tablet
Location: Under the Shack
Chapter: 7
Picarats: 30
Description: The path you are on forks to the left and right in front of the sign seen below. Your gut feeling tells you that this sign reveals the direction you need to go. Find an arrow within the picture like the yellow one on the side of the board. When you find it, draw a line around it as neatly as possible. Keep in mind that the arrow you are searching for may not be the same size as the one pictured below.
• Hint 1: There's no way around it. You just have to search the picture long and hard for the answer. The puzzle mentions the directions left and right, so there's a good chance the hidden arrow won't point upward like the example, but sideways.
• Hint 2: The arrow hidden in the picture is a bit larger than the example.
• Hint 3: Don't bother searching the left half of the sign.
Spoiler
Answer: The arrow points left and is a bit bigger than the example. The base of the arrow is right up against the right edge of the sign. On this right side, there are several rectangles. The fourth rectangle from the top makes up the base of the arrow. The top of the arrow is made up by the first vertical line from the right and the two diagonal lines that reach all the way to the right side.
Puzzle 090
Name: Get the Ball Out! 2
Trigger: Tap on the Door
Location: Underground Path
Chapter: 7
Picarats: 50
Description: Can you get the red ball out of the maze? Slide obstructing blocks out of the way to clear a path for the ball. This problem can be solved in as few as 14 moves.
• Hint 1: Sure, the puzzle looks a little daunting at first, but take heart. If you use the five open spaces in the box, you can solve the puzzle. Think about how to move things around so that you create some wiggle room for the bigger blocks.
• Hint 2: The solution requires that you move that big yellow block into the upper-right portion of the screen. Here's what you can do to start making room for that move to take place. Move the blue block at the bottom over to the right and slide the purple block into the space that has a hole. From here, if you move the blue block at the bottom so that it sits directly beneath the yellow block, you'll free up a space on the right that you can work with.
• Hint 3: This hint starts from where Hint Two left off. Move the purple block into the lower-right space and then bring down the green block above it. Next, move the purple block in the upper right as far to the right as you can and move the blue block in the upper left to the far left. Now you should be able to move that big yellow block up and to the right. Whit that block out of the way, the rest should be easy.
Spoiler
```Answer: The shortest solution to this puzzle has 14 moves. First, move the bottom blue block all the way to the right (1). Then, move the bottom-most purple block into the space with the hole (2). Next, slide the bottom blue block all the way to the left (3). The right-most purple block can now slide down and to the right in one move (4). The green block now moves all the way down (5). The top-most purple block moves all the way to the right (6). The top blue block moves to the left (7). Now the yellow block can move all the way up and to the right (8). Move the red ball all the way down and to the right (9), then move the left-most purple block right and up (10). This allows you to now move the left green block up (11) and the bottom blue block to the left (12). Move the purple block up and left (13), and finally move the red ball out (14).
```
Puzzle 091
Name: Pattern Matching
Trigger: Tap on the Picture hanging on the Right Wall
Location: Underground Area
Chapter: 7
Picarats: 40
Description: The large shape below is made up of a pattern. A section of the shape has been removed. Of the options A, B, C, and D, which one should you insert into the large shape to complete the pattern?
• Hint 1: The first thing you need to do is identify the pattern within the large shape. It's a simple pattern made up of squares, Xs, and circles running diagonally down and to the right.
• Hint 2: If you've determined what belongs in the blank spaces, you just need to find the option that matches the missing spaces. The right answer may be rotated, so check each selection from all angles to make sure you have the right one.
• Hint 3: The correct piece has two circles, three squares, and three Xs. Armed with that knowledge, all you have to do is make sure that the piece you select matches the pattern when rotated into position.
Spoiler
Puzzle 092
Name: Wash Up
Trigger: Talk to Sylvain
Chapter: 8
Picarats: 30
Description: You need to wash your face, but all 13 water valves in the sewer piping are shut tight. You have to open the valves to get the water flowing to your sink. So here's your challenge. Direct the water all the way to your sink by opening as few valves as possible.
• Hint 1: It goes without saying that if you open every valve in sight, the water will make it to your faucet. To keep things simple, though, let's start near the source. It's safe to say that you'll have to open one of the three valves directly surrounding the water source. Keep the number of valves you open in addition to this one as low as possible, and you'll have your answer.
• Hint 2: There are three more valves positioned near the faucet. Work backward from each of these and see where they go. Do this right, and you'll eliminate quite a few possible valves.
• Hint 3: You really only need to open two valves. Start from the valves near the water source and trace a route. If you don't find yourself at the faucet after opening a second valve, you made a wrong turn somewhere. This theory also works in reverse. Keep performing test runs like this and you'll find the solution.
Spoiler
Answer: You only need to open two valves. Follow the start of the pipe to the first intersection and turn left, the turn right at the next intersection. The first valve you come to needs to be opened. The other valve is by the end of the pipes. From the end of the pipe, go straight up and turn left. The valve here needs to be opened as well. Hit SUBMIT.
Puzzle 093
Name: Over the River
Trigger: Talk to Ramon
Location: Manor Border
Chapter: 6
Picarats: 30
Description: Help Stachenscarfen move the wolf, sheep, and the cabbage from one side of the river to the other while obeying the following rules.
• In addition to its captain, the raft can only support one animal or item at once.
• When Stachenscarfen isn't near, the wolf will eat the sheep, and you'll have to start over.
• The sheep will eat the cabbage when Stachenscarfen isn't around. If you let the sheep have its way, you'll have to start over.
You can shuttle the raft back and forth as many times as you like, but the shortest solution takes seven moves.
• Hint 1: Think about your first move here. If you take the wolf over first, the sheep will eat that poor, helpless cabbage. Meanwhile, if you take the cabbage over first, the sheep will fall victim to that hungry wolf. So now do you have a better idea about which of the three to move first?
• Hint 2: The key to solving this puzzle is being flexible in the way you go about solving it. Don't forget that you can bring a single creature or item with you when you return to the left bank of the river. When your cabbage or sheep is in danger, you can always bring it back with you.
• Hint 3: The sheep should travel first. When you return to the left bank, you can bring the wolf or the cabbage with you. However, remember that no matter what you bring over next, you'll have to take something back to the left bank to keep your sheep or cabbage from being eaten. Now you should be able to solve this puzzle with ease.
Spoiler
```Answer: This puzzle is solvable in seven moves. First, take the sheep across. Then return to the left bank empty handed. Take the cabbage to the right bank. Exchange the cabbage for the sheep and return the sheep to the left bank. Now take the wolf to the right bank and leave it with the cabbage. Return to the left bank empty handed. Last, take the sheep back over to the right bank.
```
Puzzle 094
Name: Get the Ball Out! 4
Trigger: Tap on the Gate
Location: Tower Floor 1
Chapter: 9
Picarats: 70
Description: This perplexing door has a device on it that contains a small red ball in the upper-left corner. If you guide the ball to the hole in the lower right, it looks like the door might open. Slide blocks out of the way and move the ball to the hole. The shortest possible route involves 21 moves.
• Hint 1: If you're persistent in your efforts, you'll eventually get the ball to the hole, but there are tricks to streamline the process. The large block in the puzzle is relatively limited in range or movement, so you're going to have to go do more work with the smaller and more maneuverable blocks. If you focus on them and plan out a route, things should go smoothly.
• Hint 2: If you move the green block on the right down, you can fill the space you create with a purple block from the right. If you can pull that off, you'll be able to move the blue block in the upper left, which will allow you to move the red ball out of its slot.
• Hint 3: There is no set solution to this problem. Here's one way to start out. Move the lower-left purple blocks, then slide the lower blue block to the left. Next, drop the green block on the right down. The rest is up to you.
Spoiler
```Answer: (1) Move bottom-left purple down, (2) move the other purple left, (3) then move Blue left, (4) Green down, (5) Purple down, (6) Purple right, down, (7) Blue right, (8) Ball down, right, (9) Green up, (10) Yellow left, (11) Purple left, down, (12) Blue down, (13) Ball right, up, (14) Blue up, left, (15) Purple up, left, (16) Ball down, (17) Purple right, up, (18) Purple up, (19) Ball left, (20) Green up, (21) Ball to the hole.
```
Puzzle 095
Name: A Magic Square
Trigger: Tap on Question Mark
Location: Tower Floor 2
Chapter: 9
Picarats: 60
Description: You need to solve this magic square in order to proceed. A magic square is a set of numbers organized in a square so that adding any string of three numbers, be they horizontal, vertical, or diagonal, results in the same total. One and two have already been placed on the square for you. Complete the rest of the square to open the lock.
• Hint 1: If you have the patience to experiment with all seven numbers, you'll run across the right answer eventually. On the other hand, not everyone has the patience to do that, so here's a hint. The number five goes in the center space.
• Hint 2: Still having trouble? All right then, here's a big hint for you. The sum of each horizontal, vertical, and diagonal pillar is 15.
• Hint 3: You know that the number five goes in the center space. You also know that the sum of each string of numbers is 15. Fifteen minus five is 10, so each pair of numbers that surrounds five should add up to 10. Pair one with nine, two with eight, and so forth. Follow this principle as you arrange your numbers, and the answer is yours.
Spoiler
Answer: The top row, from left to right is 2, 9, 4. The middle row from left to right is 7, 5, 3. The bottom row from left to right is 6, 1, 8.
Puzzle 096
Name: Take the Stairs
Trigger: Talk to Pavel
Location: Tower Floor 3
Chapter: 9
Picarats: 30
Note: This Puzzle exist only in US-Version.
Description: You have business on the eighth floor of a 10-story building. It took you 48 seconds to make your way from the first floor to the fourth. If you keep moving at the same speed, how long will it take you to reach the eighth floor from the fourth.
• Hint 1: To solve this puzzle, you must base the time it takes to climb the rest of the way on the amount of time it took to make it to the fourth floor. How many flights of stairs did you climb between the first and fourth floors again?
• Hint 2: If you start on the first floor, you'll travel through the second-, third-, and fourth-floor stairs before you reach the fourth floor. When you continue on from the fourth floor, you'll have to climb the fifth-, sixth-, seventh-, and eighth-floor stairs before you reach the eighth floor.
• Hint 3: There are three flights of stairs between the first and fourth floors. Between the fourth floor and the eighth floor, you have an additional four flights of stairs to climb. If you've got all that, then you just have to do the math.
Spoiler
Puzzle 096
Name: On the Stairs
Trigger: Talk to Pavel
Location: Tower Floor 3
Chapter: 9
Picarats: 30
Note: This Puzzle exist only in EU-Version.
Description: Jim, Mike, Steve and Tom are standing on the stairs. You know the following about them:
1. Jim and Tom are not next to each other.
2. Steve and Mike are one step apart.
3. Tom is on a higher step than Steve.
So, which of A, B, C and D is Jim?
• Hint 1: This puzzle is about conditions. Ask yourself: what if Jim were A? What if he were B? Start with a possible answer and see if it fits the conditions of the question.
• Hint 2: Steve and Mike are next to each other. In other words, they are one of the groups AB, BC or CD.
• Hint 3: Combine the information in the second hint with the fact that Tom is above Steve. You can then add the information that Jim is not next to Tom, so...
Spoiler
Puzzle 097
Name: Princess in a Box 1 (US) / Maiden's Escape (EU)
Trigger: Tap on the Question Mark
Location: Tower Floor 4
Chapter: 9
Picarats: 60
Description: Tired of leading a sheltered life, this princess is trying to escape her castle. Armed guards, however, are blocking her path. Slide the blocks out of the way to move the red one out the exit to the right. Her freedom depends on you. Can you do it?
• Hint 1: You might think you'll get stuck within moments of starting this puzzle, but as long as you aren't repeating the same moves over and over, you'll get that block out eventually.
• Hint 2: This trick should give you some wiggle room. Try to move the blue blocks so that both of them are directly above or below the red block. As a matter of fact, hold off on thinking about how to move the red block to the exit until you finish this step.
• Hint 3: Here's one more trick that should help you move those blocks around. While sliding blocks around, see if you can't move two green blocks to the immediate right of the red block. Then move the purple block to the right of those.
Spoiler
```Answer: The shortest possible route involves 40 moves. Move 2 Green Blocks right and Purple Block right, then 2 Green Blocks left and up. Move Blue Block right and Red Block down. Move 2 of the bottom Green Blocks left in the gap. Move Green Block left, down and another Green Block left. Move Purple Block and then the Blue Block up. Move one Green Block left and another one down. Now move Blue Block and then Red Block right. Move 2 Green Blocks and then the Blue Block down. Next move 2 Green Blocks and Purple Block left. Move a Green Block left, up to move Blue Block up. Move Green Block up, right to move the Red Block right. Move one Green Block right, down, then move Blue Block down. Move one Green Block down, left and move Purple Block left. Move Green Block left, down and another one twice left and move Blue Block up. Move Green Block right three times to move Red Block up. Move the remaining 3 Green Blocks down and left in the gap. All that remains is to move the Red Block right on the Exit Field.
```
Puzzle 098
Name: Card Order
Trigger: Talk to Martha
Location: Tower Floor 5
Chapter: 9
Picarats: 70
Description: You've placed on joker and four aces with different suits facedown on a table. Use the hints below to determine the position for each card.
1. . The club is to the immediate right of the heart.
2. . Neither the diamond nor the joker is next to the spade.
3. . Neither the joker nor the diamond is next to the club.
4. . Neither the diamond nor the spade is next to the heart.
• Hint 1: If the club is to the immediate right of the heart, the heart can't be the rightmost card. Neither the diamond nor the spade are next to the heart. You know that the club is to the immediate right of the heart. So the card to the left of the heart is either the joker, or the heart is the leftmost card itself.
• Hint 2: The heart is to the immediate left of the club, and neither the joker nor the diamond are next to the club, so either the club is the rightmost card, or the spade lies to the right of it. Don't forget that the diamond, heart, and joker can't be next to the spade. Therefore, the spade is the rightmost card, and the club lies to its immediate left.
• Hint 3: From earlier hints, you know that from the right side, the cards appear in this order: spade, club, then heart. As for the two remaining cards, you know that the diamond can't be next to the heart, so the joker must be the fourth card from the right.
Spoiler
Answer: From left to Right, the cards are: Diamond, Joker, Heart, Club, and Spade.
Puzzle 099
Name: 33333!
Trigger: Tap on the Question Mark
Location: Tower Floor 6
Chapter: 9
Picarats: 70
Description: Use each of the numbers one through nine exactly once to fill in the blanks and complete this equation: 00000-0000=33333.
• Hint 1: At its core, this is a simple math problem, so you're just going to have to work it out if you want to solve it. However, there is a way to cut down on the amount of work you need to do. Try thinking about the first and last digits for each number. The leftmost digit in the upper number is "4."
• Hint 2: There are two possible solutions, but some digits are located in the same place for both answers. For example, for both solutions, "7" is the leftmost digit of the bottom number.
• Hint 3: This is the last hint you're going to get. For the top number, going from the left, the first three digits are "4," "1," and "2."
Spoiler
Answer: There are two possible answers to this problem. One has the top row as 41286 and the bottom row as 7953. Thus, 41286 - 7953 = 33333
Puzzle 100
Name: Seven Squares
Trigger: Tap on the Question Mark
Location: Tower Floor 7
Chapter: 9
Picarats: 70
Description: Your task is to draw lines between the pins on this board to form seven squares. The seven squares do not have to be uniform in size, but you can only use each pin once. All righty then, give it a shot.
• Hint 1: Most of the squares you make will be tilted 45 degrees to the side, and their sizes will vary as well. Start by looking for pins you can connect to create squares at a diagonal.
• Hint 2: You want more specifics? All right, here's the location of one of the squares. Connect the four pins in the top-left corner to form a tiny square. Just so you know, this is the only square on the board that isn't tilted.
• Hint 3: The largest square contains the pin that's third from the top on the left column and the bottom pin from the far-right column. You also need to form a small diagonal block using the two pins lined up diagonally on the bottom-left portion of the board. There are two more small squares just like this one on the board.
Spoiler
Answer: First, make a tiny square with the four pins in the top-left corner. Next, draw a line from the third pin in the first row to the fifth pin in the second row. Connect this pin to the third pin in the fourth row. This pin connects to the second pin in the third row. Complete the square.
The next square starts with the fourth pin in the first row. Connect this pin to the last pin in the fourth row. Connect this pin with the second pin in the last row, then connect this pin with the first pin in the third row. Complete the square.
Now draw a line from the fifth pin in the first row to the last pin in the second row. Connect this pin to the fourth pin in the third row, then draw a line from this pin to the fourth pin in the second row. Complete the square.
The next square starts with the third pin in the second row. Draw a line to the last pin in the third row. This pin connects to the last pin in the last row. Now draw a line to the second pin in the fifth row. Complete the square.
Start the next square with the third pin in the third row. This pin connects to the fourth pin in the fourth row, with connects to the fourth pin in the fifth row. Draw a line to the second pin in the fourth row, then complete the square.
The last square starts with the first pin in the fourth row. This pin connects to the third pin in the fifth row. Now draw a line to the first pin in the last row, then connect to the first pin in the fifth row. Complete the square. | 10,302 | 41,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.943837 |
https://developer.aliyun.com/article/1431331 | 1,716,414,632,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00335.warc.gz | 186,651,179 | 25,647 | # Python每日一练(20230412) 队列实现栈、二叉树序列化、交换链表节点
## 1. 用队列实现栈
• void push(int x) 将元素 x 压入栈顶。
• int pop() 移除并返回栈顶元素。
• int top() 返回栈顶元素。
• boolean empty() 如果栈是空的，返回 true ；否则，返回 false
• 你只能使用队列的基本操作 —— 也就是 push to backpeek/pop from frontsizeis empty 这些操作。
• 你所使用的语言也许不支持队列。 你可以使用 list （列表）或者 deque（双端队列）来模拟一个队列 , 只要是标准的队列操作即可。
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
[null, null, null, 2, 2, false]
• 1 <= x <= 9
• 最多调用100pushpoptopempty
• 每次调用 poptop 都保证栈不为空
https://edu.csdn.net/practice/25389378
class MyStack:
def __init__(self):
"""
"""
self.queue = []
self.help = []
def push(self, x):
"""
Push element x onto stack.
"""
while len(self.queue) != 0:
self.help.append(self.queue.pop(0))
self.queue.append(x)
while len(self.help) > 0:
self.queue.append(self.help.pop(0))
def pop(self):
"""
Removes the element on top of the stack and returns that element.
"""
de = self.queue.pop(0)
return de
def top(self):
"""
Get the top element.
"""
de = self.queue[0]
return de
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
if len(self.queue) == 0:
return True
return False
# %%
myStack = MyStack();
myStack.push(1)
myStack.push(2)
print(myStack.top())
print(myStack.pop())
print(myStack.empty())
2
2
False
## 2. 二叉树的序列化与反序列化
• 树中结点数在范围 [0, 104]
• -1000 <= Node.val <= 1000
https://edu.csdn.net/practice/25389379
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Codec:
def serialize(self, root):
"""Encodes a tree to a single string.
:type root: TreeNode
:rtype: str
"""
if root == None:
return "null,"
left_serialize = self.serialize(root.left)
right_serialize = self.serialize(root.right)
return str(root.val) + "," + left_serialize + right_serialize
def deserialize(self, data):
"""Decodes your encoded data to tree.
:type data: str
:rtype: TreeNode
"""
def dfs(queue):
val = queue.popleft()
if val == "null":
return None
node = TreeNode(val)
node.left = dfs(queue)
node.right = dfs(queue)
return node
from collections import deque
queue = deque(data.split(","))
return dfs(queue)
# Your Codec object will be instantiated and called as such:
# codec = Codec()
# codec.deserialize(codec.serialize(root))
## 3. 两两交换链表中的节点
• 链表中节点的数目在范围 [0, 100]
• 0 <= Node.val <= 100
https://edu.csdn.net/practice/25389380
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
def __init__(self):
def initList(self, data):
for i in data[1:]:
node = ListNode(i)
p.next = node
p = p.next
return r
ret = []
return
while node != None:
ret.append(node.val)
node = node.next
return ret
class Solution(object):
while p != None and p.next != None:
q, r = p.next, p.next.next
prev.next = q
q.next = p
p.next = r
prev = p
p = r
# %%
s = Solution()
print(l.convert_list(s.swapPairs(l1)))
[2, 1, 4, 3]
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Python编程实验五：文件的读写操作
Python编程实验五：文件的读写操作
8 0 | 1,273 | 3,340 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.2389 |
https://www.aqua-calc.com/calculate/gravel-volume-to-weight/substance/caribsea-coma-and-blank-freshwater-coma-and-blank-super-blank-naturals-coma-and-blank-kon-blank-tiki | 1,685,501,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00307.warc.gz | 669,685,118 | 8,313 | # Weight of CaribSea, Freshwater, Super Naturals, Kon Tiki
## caribsea, freshwater, super naturals, kon tiki: volume to weight
### Weight, i.e. how many oz, lbs, g or kg in 1 cubic inch of CaribSea, Freshwater, Super Naturals, Kon Tiki
gram 26.25 ounce 0.93 kilogram 0.03 pound 0.06
### Volume
centimeter³ 16.39 inch³ 1 foot³ 0 meter³ 1.64 × 10-5
• About CaribSea, Freshwater, Super Naturals, Kon Tiki
• 1 cubic centimeter of CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1.60185 gram [g]
• 1 cubic meter of CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1 601.85 kilograms [kg]
• 1 cubic inch of CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 0.92593 ounce [oz]
• 1 cubic foot of CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 100.00023 pounds [lbs]
• CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1.6 gram per cubic centimeter or 1 601.85 kilogram per cubic meter, i.e. density of caribSea, Freshwater, Super Naturals, Kon Tiki is equal to 1 601.85 kg/m³. In Imperial or US customary measurement system, the density is equal to 100 pound per cubic foot [lb/ft³], or 0.93 ounce per cubic inch [oz/inch³] .
• CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
• Calculate how many pounds or kilograms of “CaribSea, Freshwater, Super Naturals, Kon Tiki” fills a specified volume in an aquarium or pond. For instance, calculate how many pounds or kilograms of a selected gravel, substrate or sand in a liter or gallon capacity of your aquarium or pond.
#### Foods, Nutrients and Calories
GREEN BEAN SAUTE, UPC: 826766718290 weigh(s) 90 grams per metric cup or 3 ounces per US cup, and contain(s) 29 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
41170 foods that contain Fatty acids, total polyunsaturated. List of these foods starting with the highest contents of Fatty acids, total polyunsaturated and the lowest contents of Fatty acids, total polyunsaturated
#### Gravels, Substances and Oils
CaribSea, Freshwater, African Cichlid Mix, White weighs 1 169.35 kg/m³ (73.00014 lb/ft³) with specific gravity of 1.16935 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Bisodium sulfate [Na2SO4 or Na2O4S] weighs 2 671 kg/m³ (166.74508 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-23, liquid (R23) with temperature in the range of -95.56°C (-140.008°F) to 4.45°C (40.01°F)
#### Weights and Measurements
A kilotesla is a SI-multiple (see prefix kilo) of the magnetic flux density unit tesla and equal to 1 thousand teslas (1000 T)
The pressure forces on any solid surface by liquid or gas, do not apply to any single point on that surface. Instead, these forces are spread equally along the whole surface.
gr/US c to oz/ft³ conversion table, gr/US c to oz/ft³ unit converter or convert between all units of density measurement.
#### Calculators
Calculate gravel and sand coverage in a cylindrical aquarium | 989 | 3,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-23 | longest | en | 0.607992 |
http://aptitude.brainkart.com/aptitude/gate-exam/ce-civil-engineering/structural-engineering/5/ | 1,620,957,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989616.38/warc/CC-MAIN-20210513234920-20210514024920-00439.warc.gz | 3,783,122 | 11,090 | Structural Engineering (Test 5)
Gate Exam : Ce Civil Engineering
| Home | | Gate Exam | | Ce Civil Engineering | | Structural Engineering |
Structural Engineering
| Structural Engineering |
Q.1
A 12 mm thick plate is connected to two 8 mm thick plates, on either side through a 16 mm diameter power driven field rivet as shown in the figure below. Assuming permissible shear stress as 90 MPa and permissible bearing stress as 270 MPa in the rivet, the rivet value of the joint is
A. 56.70 kN
B. 43.29 kN
C. 36.19 kN
D. 21.65 kN
Explaination / Solution:
No Explaination.
Workspace
Report
Q.2
A propped cantilever of span L carries a vertical concentrated load at the mid-span. If the plastic moment capacity of the section is MP, the magnitude of the collapse load is
A. 8Mp/L
B. 6Mp/L
C. 4Mp/L
D. 2Mp/L
Explaination / Solution:
Workspace
Report
Q.3
Consider the following statements :
1. On a principal plane, only normal stress acts.
2. On a principal plane, both normal and shear stresses act.
3. On a principal plane, only shear stress acts
4. Isothermal state of stress is independent of frame of reference.
Which of the above statements is/are correct ?
A. 1 and 4
B. 2 only
C. 2 and 4
D. 2 and 3
Explaination / Solution:
No Explaination.
Workspace
Report
Q.4
A continuous beam is loaded as shown in the figure below. Assuming a plastic moment capacity equal to MP , the minimum load at which the beam would collapse is
A. 4MP/L
B. 6MP/L
C. 8MP/L
D. 10MP/L
Explaination / Solution:
No Explaination.
Workspace
Report
Q.5
In Marshall method of mix design, the coarse aggregate, fine aggregate, fines and bitumen having respective values of specific gravity 2.60, 2.70, 2.65 and 1.01, are mixed in the relative proportions (% by weight) of 55.0, 35.8, 3.7 and 5.5 respectively. The theoretical specific gravity of the mix and the effective specific gravity of the aggregates in the mix respectively are
A. 2.42 and 2.63
B. 2.42 and 2.78
C. 2.42 and 2.93
D. 2.64 and 2.78
Explaination / Solution:
Workspace
Report
Q.6
A column is supported on a footing as shown in the figure below. The water table is at a depth of 10 m below the base of the footing
The safe load (kN) that the footing can carry with a factor of safety 3 is
A. 282
B. 648
C. 945
D. 1269
Explaination / Solution:
No Explaination.
Workspace
Report
Q.7
A pre-tensioned concrete member of section 200mm × 250mm contains tendons of area 500 mm2 at centre of gravity of the section. The prestress in the tendons is 1000 N/mm2. Assuming modular ratio as 10, the stress (N/mm2) is concrete is
A. 11
B. 9
C. 7
D. 5
Explaination / Solution:
No Explaination.
Workspace
Report
Q.8
The maximum tensile stress at the section X -X shown in the figure below is
A. 8P/bd
B. 6P/bd
C. 4P/bd
D. 2P/bd
Explaination / Solution:
No Explaination.
Workspace
Report
Q.9
For the fillet weld of size ‘s ’ shown in the adjoining figure the effective throat thickness is
A. 0.61S
B. 0.65S
C. 0.70S
D. 0.75S
Explaination / Solution:
No Explaination.
Workspace
Report
Q.10
Two people weighing W each are sittingon a plank of length L floating on water at L/4 from either end. Neglecting the weight of he plank, the bending moment at the centre of the plank is
A. WL/8
B. WL/16
C. WL/32
D. zero | 1,002 | 3,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.832781 |
http://www.freedictionary.org/?Query=Plane%20surveying | 1,719,026,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00032.warc.gz | 39,432,343 | 3,743 | Search Result for "plane surveying":
```
The Collaborative International Dictionary of English v.0.48:Plane \Plane\, a. [L. planus: cf. F. plan. See Plan, a.]
Without elevations or depressions; even; level; flat; lying
in, or constituting, a plane; as, a plane surface.
[1913 Webster]
Note: In science, this word (instead of plain) is almost
exclusively used to designate a flat or level surface.
[1913 Webster]
Plane angle, the angle included between two straight lines
in a plane.
Plane chart, Plane curve. See under Chart and Curve.
Plane figure, a figure all points of which lie in the same
plane. If bounded by straight lines it is a rectilinear
plane figure, if by curved lines it is a curvilinear plane
figure.
Plane geometry, that part of geometry which treats of the
relations and properties of plane figures.
Plane problem, a problem which can be solved geometrically
by the aid of the right line and circle only.
Plane sailing (Naut.), the method of computing a ship's
place and course on the supposition that the earth's
surface is a plane.
Plane scale (Naut.), a scale for the use of navigators, on
which are graduated chords, sines, tangents, secants,
rhumbs, geographical miles, etc.
Plane surveying, surveying in which the curvature of the
earth is disregarded; ordinary field and topographical
surveying of tracts of moderate extent.
Plane table, an instrument used for plotting the lines of a
survey on paper in the field.
Plane trigonometry, the branch of trigonometry in which its
principles are applied to plane triangles.
[1913 Webster]
The Collaborative International Dictionary of English v.0.48:Surveying \Sur*vey"ing\, n.
That branch of applied mathematics which teaches the art of
determining the area of any portion of the earth's surface,
the length and directions of the bounding lines, the contour
of the surface, etc., with an accurate delineation of the
whole on paper; the act or occupation of making surveys.
[1913 Webster]
Geodetic surveying, geodesy.
Maritime surveying, or Nautical surveying, that branch of
surveying which determines the forms of coasts and
harbors, the entrances of rivers, with the position of
islands, rocks, and shoals, the depth of water, etc.
Plane surveying. See under Plane, a.
Topographical surveying, that branch of surveying which
involves the process of ascertaining and representing upon
a plane surface the contour, physical features, etc., of
any portion of the surface of the earth.
[1913 Webster]``` | 603 | 2,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.885301 |
https://www.hpmuseum.org/forum/showthread.php?tid=19088&pid=166001&mode=threaded | 1,726,253,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00795.warc.gz | 771,704,752 | 5,621 | Gamma function, SinhIntegral, CoshIntegral
11-07-2022, 02:31 PM
Post: #5
Albert Chan Senior Member Posts: 2,703 Joined: Jul 2018
RE: Gamma function, SinhIntegral, CoshIntegral
(11-06-2022 06:27 PM)robmio Wrote: HP PRIME --> Gamma (4/5, -6) --> 294.845140024
MATHEMATICA --> Gamma [4/5, -6] --> 238.757-172.621*i.
Mathematica Gamma, converted to HP Prime Gamma
Note: it is not Abs[Gamma[4/5,-6]] ≈ 294.624
Gamma[4/5] + Abs[Gamma[4/5] - Gamma[4/5,-6]]
= Gamma[4/5] + (Gamma[4/5] - Gamma[4/5,-6]) / (-1)^(4/5)
= 294.845140024 ...
HP Prime Gamma, back to Mathematica Gamma:
CAS> gamma(a,x) := when(x<0, [Gamma(a),Gamma(a,x)] * [1+(-1)^a,-(-1)^a], Gamma(a,x))
CAS> gamma(4/5,-6.) → 238.757077078-172.62130796*i
Quote:I hadn't really noticed: why does HP PRIME return the absolute value?
It is just a guess, but some integral result is more elegant.
CAS> int(e^x^3) → 1/3*(Gamma(1/3,-x^3) - Gamma(1/3))
CAS> Ans(x=6.) → 5.96393809188e91
Mathematica:
∫(e^x^3) = -(x Γ(1/3, -x^3))/(3 (-x^3)^(1/3))
∫(e^x^3, x=0..6) ≈ (5.964E91 + 0.7733*i) - (-0.4465 + 0.7733*i) ≈ 5.964E91
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Messages In This Thread Gamma function, SinhIntegral, CoshIntegral - robmio - 11-06-2022, 06:27 PM RE: Gamma function, SinhIntegral, CoshIntegral - lrdheat - 11-06-2022, 07:03 PM RE: Gamma function, SinhIntegral, CoshIntegral - robmio - 11-06-2022, 07:24 PM RE: Gamma function, SinhIntegral, CoshIntegral - lrdheat - 11-06-2022, 08:45 PM RE: Gamma function, SinhIntegral, CoshIntegral - Albert Chan - 11-07-2022 02:31 PM RE: Gamma function, SinhIntegral, CoshIntegral - robmio - 11-07-2022, 03:34 PM RE: Gamma function, SinhIntegral, CoshIntegral - robmio - 11-08-2022, 11:49 AM
User(s) browsing this thread: 1 Guest(s) | 696 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.478967 |
https://www.ucl.ac.uk/lifelearning/courses/anova-glms-using-spss | 1,558,826,350,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258453.85/warc/CC-MAIN-20190525224929-20190526010929-00464.warc.gz | 971,460,464 | 9,099 | # ANOVA: General Linear Models (GLMs) with SPSS
• 5 hours
• 1 day
• 02 July 2019
## Overview
This one-day course on analysis of variance (ANOVA) takes a hands-on approach to learning.
You'll learn how to choose, run, interpret and report a variety of ANOVA models available in SPSS within the general linear model (GLM) function.
SPSS software will be used for demonstration and practice throughout.
This course is delivered by UCL's Centre for Applied Statistics Courses (CASC) - part of the UCL Great Ormond Street Institute of Child Health (ICH).
## Course content
ANOVA is a branch of statistical analysis used for comparing numerical data between groups and categories, with the flexibility to increase model size and complexity and control for other numeric variables.
You'll be introduced to the underlying principles of ANOVA and the extensions and applications of this type of statistical analysis are discussed.
The structure, application and interpretation of the following models will be covered:
• one-way ANOVA
• factorial ANOVA
• univariate ANOVA
• multivariate ANOVA
• repeated measures ANOVA
## Use of SPSS
The conceptual basis of this course is applicable to other statistical packages, but practical elements of the course are conducted using SPSS. Example datasets are provided for you to perform analyses on.
The course will take place in a cluster room, with access to a computer and version 22 of SPSS throughout the day.
You're welcome to bring your own laptop if preferred, but please ensure your software is licenced before attending. Where possible, we recommend that you use a recent version of SPSS (e.g. 19-22) for maximum compatibility with the notes provided during the course.
## Learning outcomes
By the end of this course you should be able to:
• understand what ANOVA is and the basics of how it works
• know the data checks and parametric assumptions for ANOVA
• know/understand the steps to carrying out an ANOVA in SPSS
• select the appropriate test for different study designs
• read and interpret SPSS output for ANOVA
You should also have some understanding of similarities between ANOVA and regression.
## Eligibility
Basic knowledge of introductory statistics and of SPSS software are desirable.
## Cost and concessions
The fees are as follows:
• External delegates (non UCL) - £200
• UCL staff, students, alumni - £100*
• ICH/GOSH staff and students - free
* valid UCL email address and/or UCL alumni number required upon registration
Prices include printed course materials, refreshments (and lunch for non-ICH participants).
## Certificates
You can request a certificate of attendance for this course once you've completed it. Please send your request to ich.statscou@ucl.ac.uk
Include the following in your email:
• the name of the completed course for which you'd like a certificate
• how you'd like your name presented on the certificate (if the name/format differs from the details you gave during registration)
## Cancellations
You can cancel your booking up to five working days before the start of the course for a full refund, but please give as much notice as possible. Places cancelled or changed after this point won't be eligible for a refund. Please send all cancellation requests to the course administrator
## Find out about other statistics courses
CASC's stats courses are for anyone requiring an understanding of research methodology and statistical analyses. The courses will allow non-statisticians to interpret published research and/or undertake their own research studies.
Find out more about CASC's full range of statistics courses, and the continuing statistics training scheme (book six one-day courses and get a seventh free.)
Course team
## Eirini Koutoumanou - Course Lead
Eirini joined GOS ICH in 2008 as the first CASC Teaching Fellow and was promoted to Senior Teaching Fellow in 2014. She has a Bachelor’s degree in Statistics from the Athens University of Economics and Business and a Master's degree in Statistics from Lancaster University, and she's currently studying for a PhD at ICH. Eirini started teaching at ICH straight after her student days, putting into practice and further developing her passion for statistics teaching. She's played an instrumental role in the formation of CASC and hopes to see it develop further.
Student review
“I found the course and the learning materials very useful. The learning materials were very detailed giving me the opportunity to refer back to them anytime.”
“It's a very useful course on ANOVA.” | 963 | 4,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.895468 |
http://lists.contesting.com/pipermail/towertalk/2013-November/137934.html | 1,569,072,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574501.78/warc/CC-MAIN-20190921125334-20190921151334-00066.warc.gz | 111,051,749 | 4,803 | [TowerTalk] K9YC vertical dipole test....
Peter FORBES prforbes at bigpond.net.au
Fri Nov 15 07:19:06 EST 2013
```Hi Dan,
Some great animations!
Firstly, I should set the scene as to why I posed the original question.
If you go back to the original question asked by James Setzler, K1SD on
14/11/13, James was talking particularly about the 80mx and 40mx
performance of his HF2V.
The excellent work that Jim K9YC did with his analysis and presentation -
everything I agree with, both from a theoretical and a practical (50 years
DXing) point of view - but remember that this analysis is for 40mx, 20mx and
10mx.
What needs to be considered is that on 80mx (also 160mx) and to a lesser
extent 40mx, the angle of arrival and takeoff for DX signals is actually
very much higher than one would expect.
The ARRL antenna handbook for example, quotes figures like: for 40mx
99% of the time signal below 35 degrees, 50% of time above 25 degrees and
99% of the time above 10 degrees.
These figures have been backed up by countless research papers over the
years
For 80mx and 160mx the signals are at even greater elevation angles. (and
I should add, varies with time as the ionosphere rises/tilts/ Dlayer
appears, sporadic clouds in summer etc. etc.)
******************
Although there are many other factors to consider with ionospheric
propagation, if we take for example Jim's analysis
http://k9yc.com/VerticalHeight.pdf of a 40mx vertical ground plane antenna
at ground level, 33 feet and 40 feet, we find that:
City ground from 33 -> 50 degrees the ground level antenna outperforms the
33 feet antenna
Rocky ground from 25 -> 60 degrees the ground level antenna outperforms
the 33 feet antenna
Average ground from 23 -> 57 degrees the ground level antenna outperforms
the 33 feet antenna
Pastoral ground from 19 -> 57 degrees the ground level antenna outperforms
the 33 feet antenna
Very good ground from 17 -> 60 degrees the ground level antenna
outperforms the 33 feet antenna
As I have said, there are many other factors to consider, but with some 50%
of signals often arriving and departing at above 25 degrees, it is really
only city situations where you can clearly state that the elevated ground
plane will almost always outperform the ground mounted antenna.
>From years of experience using vertical antennas, this high angle effect on
lower frequency signals is definitely something that needs to be considered
and especially on 80mx (and 160mx).
Jim's excellent analysis shows quite clearly what the trend is as you move
from poor to good local ground conditions.
My question related to just how effective measuring vertically polarised
signals at approximately zero elevation angle is with regard to DXing on
80mx and 40mx.
Given the choice, most would opt for an elevated vertical. However on 80mx
and to a lesser extent on 40mx a ground mounted antenna on good ground can
be an effective DX antenna.
Cheers
Peter VK3QI
-----Original Message-----
From: TowerTalk [mailto:towertalk-bounces at contesting.com] On Behalf Of Dan
Maguire
Sent: Friday, 15 November 2013 9:15 PM
To: towertalk at contesting.com
Subject: Re: [TowerTalk] K9YC vertical dipole test....
VK3QI wrote:
>>> Are you really measuring what DXers are really after when comparing
vertical antennas at different heights above ground and ground mounted.?
Putting aside for the moment the distinction between measuring and modeling,
I wanted to investigate what VK3QI said about "DX angles" (in this case 10°)
vs "zero" angle. First thing I did was create an adjustable vertical dipole
model that also has ground conditions that can be set via a variable.
Here's the AutoEZ "Variables" sheet showing (to start) the base (variable B)
at 20 ft, the length (variable D) at 35 ft, and the ground type (variable G)
as "Average".
With those starting conditions I used the Resonate button to reset the
length (variable D). After that the length was not changed.
Then I set up a series of test cases with the base at 0.5 ft and with the
ground characteristics varying through all the non-water choices, from
Extremely Poor to Very Good, per the EZNEC definitions of such.
Here's an animated gif showing the results. The value for "G" (Ground type
index) may be seen in the lower right corner.
As the ground gets better the gain at 10° gets better (pretty much). Then I
ran a similar series of test cases with the base of the dipole at 40 ft.
Again, look in the lower right corner to see the "G" value for any given
frame of the animation.
That was interesting. With the base at 40 ft the gain at 10° gets _worse_
as the ground gets better. More of the energy is going into the second
(higher) lobe. However, note that the outer ring is the same for both these
animations so you can compare magnitudes as well as pattern shapes. For any
given ground type the "DX angle" (10°) gain is always higher for the higher
dipole. For example, here's a comparison for Average ground.
So that seems to address the first point that VK3QI made. Gain at "DX
angles" (in this case 10°) gets better as the dipole is raised and it
doesn't matter what the ground conditions are. Now for his point about
actual measurements:
>>> Measuring a vertical at another location 5 miles away, but at the same
relative height is really measuring the ability of the vertical antenna to
couple to the ground to produce a vertically polarised ground wave.
And that seems to be in regards to page 76 from the K9YC presentation:
K9YC measured an increase in gain of 9.5 dB as the dipole base was raised
from 0.5 ft to 40 ft. If both he and the 3 watt transmitter at W6GJB had
been located on the wheat fields of Kansas that would have been one thing,
but they are both in the Santa Cruz mountains and I'm betting it's not
exactly 5 miles of flat ground (as NEC assumes) between them. Hence I'm not
sure how to use NEC to verify the K9YC measurements since there is no Far
Field at an elevation of 0°. So I fudged a bit.
Here's an animation with the green dot marker at 1° elevation, not exactly
the same as equal heights but close. The ground type is fixed at Very Poor
per K9YC's comments. The dipole base ranges through 0.5, 10, 20, 30, and 40
ft.
As the base height is raised the gain at 1° increases by about 6.6 dB (from
-19.34 dBi to -12.72 dBi). That's not identical to the measured increase of
9.5 dB but it seems to verify that the gain at "ground level" (almost)
should increase as the dipole height is increased, which seems to
corroborate what K9YC measured.
Not sure what any of this proved or didn't prove, just thought I'd share it.
Here's the model file I used if anyone else would like to play with it.
(Note that you can't just click on the link to open the model. Save it to
your computer then open it from within AutoEZ, as explained in Step 3 of the
AutoEZ Quick Start guide.)
This model will work just fine with the free demo version of AutoEZ which
http://ac6la.com/autoez.html
Dan, AC6LA
http://ac6la.com
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``` | 1,916 | 7,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-39 | latest | en | 0.941403 |
https://www.convertunits.com/from/kattha/to/tunnland | 1,660,107,012,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00268.warc.gz | 627,162,389 | 12,643 | ## ››Convert kattha [Nepal] to tunnland
kattha tunnland
How many kattha in 1 tunnland? The answer is 14.604733727811.
We assume you are converting between kattha [Nepal] and tunnland.
You can view more details on each measurement unit:
kattha or tunnland
The SI derived unit for area is the square meter.
1 square meter is equal to 0.0029585798816568 kattha, or 0.00020257677659833 tunnland.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kattha [Nepal] and tunnland.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of kattha to tunnland
1 kattha to tunnland = 0.06847 tunnland
10 kattha to tunnland = 0.68471 tunnland
20 kattha to tunnland = 1.36942 tunnland
30 kattha to tunnland = 2.05413 tunnland
40 kattha to tunnland = 2.73884 tunnland
50 kattha to tunnland = 3.42355 tunnland
100 kattha to tunnland = 6.8471 tunnland
200 kattha to tunnland = 13.69419 tunnland
## ››Want other units?
You can do the reverse unit conversion from tunnland to kattha, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 475 | 1,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.75744 |
https://socratic.org/questions/57e474347c0149023a91fc86 | 1,627,228,971,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00519.warc.gz | 545,825,416 | 6,269 | # To produce a mass of 279*g mass of iron metal, what quantities of aluminum, and ferric oxide are required?
Sep 25, 2016
Approx. $135 \cdot g$ of aluminum metal, and $400 \cdot g$ of iron oxide are required.
#### Explanation:
We need (i) a stoichometrically balanced equation:
$F {e}_{2} {O}_{3} + 2 A l \rightarrow A {l}_{2} {O}_{3} + 2 F e$
And (ii) the molar quantity of iron metal produced:
$=$ $\frac{279 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1}$ $=$ $5.00 \cdot m o l$.
Since a $5.00 \cdot m o l$ of iron metal were produced, the given stoiohiometry requires at least a $2.50 \cdot m o l$ quantity of $\text{ferric oxide}$, and a $5.00 \cdot m o l$ quantity of $\text{aluminum metal}$.
Given these molar quantities, we calculate equivalent masses of $2.50 \cdot m o l \times 159.69 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g of iron oxide.
And $5.00 \cdot m o l \times 26.98 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g of aluminum metal.
This would be an expensive way to produce iron. | 337 | 995 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-31 | latest | en | 0.713234 |
https://math.stackexchange.com/questions/2673795/perform-the-modified-eulers-method-given-a-point-and-a-stepsize | 1,585,778,213,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00152.warc.gz | 606,524,391 | 32,109 | # Perform the modified Euler's Method given a point and a stepsize
Consider the following ODE
$$\dfrac{dy}{dx} = -\dfrac{2x}{y}~~~~\text{where}~~~~y(0)=1$$
Given that $y(0.7)=0.141421$ to $6$ digit precision, use the modified Euler's method to estimate $y(0.8)$ using $h=0.1$ and work to $5$ digit precision.
How do I do this? I can do it from scratch, but given $y(0.7)$ is throwing me off for some reason.
Thanks in advance for any help!
Given the ODE $y'=f(x,y)$ then Euler's method says $y(x_{n+1})=y(x_n)+h\cdot f(x_n,y(x_n))$ where in your case $f(x,y) = -\frac{2x}{y}$, $x_{n+1}=0.8$, $x_n=0.7$ and $h=x_{n+1}−x_n=0.1$. The modified Euler's method says $$y(x_{n+1}) = y(x_n) + \frac{h}{2}\cdot[f(x_n,y(x_n)) + f(x_{n+1},\tilde{y}(x_{n+1})) ]$$ where $ỹ (x_{n+1})=y(x_n)+h\cdot f(x_n,y(x_n))$ is the value you would get using Euler's method. It's just to plug in and compute all the terms here (and do the computations to the given precision).
The analytical solution to the ODE is $y(x) = \sqrt{1-2x^2}$ and $y(0.7)$ is in fact $0.141421$ to $6$ significant digits. It's just provided to allow you to perform the method from $x=0.7$. The reason $x=0.7$ is choosen is probably since the analytical solution does not exist (as a real function) for $x>0.707$ so the values you obtain using Euler's vs modified Euler's will likely be very different (and both wrong in this case).
• @RBadger I get something similar $-0.25928$ when working to 5 digit precision (but I did it very fast so might have made some mistakes) – Winther Mar 3 '18 at 16:48 | 536 | 1,556 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-16 | latest | en | 0.833384 |
https://brilliant.org/discussions/thread/helphint-required/ | 1,516,781,028,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00442.warc.gz | 622,264,100 | 23,312 | ×
# Help/Hint required.
Q1)Find the number of 4 digit numbers with distinct digits chosen from the set {0, 1, 2, 3, 4, 5} in which no two adjacent digits are even.
I'm not sure about the answer. I'm getting 60. There's a high chance I may be wrong.
Q2) Let a, b, c > 0. If $$\dfrac 1 c , \dfrac 1 b \quad and \dfrac 1 a$$ are in arithmetic progression, and if $$a^2+b^2,b^2+c^2,c^2+a^2$$ are in a geometric progression, prove that a=b=c
Somehow, I managed to get a=b=c. but I'm not really confident of my approach.
Q3) Let n be a positive integer such that 2n + 1 and 3n + 1 are both perfect squares. Show that 5n + 3 is a composite number.
I tried this
Let $$2n+1=k^2$$ and $$3n+1=m^2$$
$$5n+3=m^2+k^2+1$$
Then I used modulo to bash it out. am I on the right track? Because this doesn't seem to lead to the answer.
Thanks!
Note by Mehul Arora
2 years, 4 months ago
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Q1) Let the odd digit be $$O$$ and even digit be $$E$$.So by product rule we can work out each case. So hereby I give the possible combinations:
$OEOO = 3\times 3 \times 2 \times 1 = 18 \\ OOEO = 3\times 2 \times 3 \times 1 =18 \\ EOOO = 2\times 3 \times 2 \times 1 = 12 \\ OOOE = 3\times 2 \times 1 \times 3 = 18 \\ EOOE = 2\times 3 \times 2 \times 2 = 24 \\ OEOE = 3\times 3 \times 2 \times 2 = 36 \\ EOEO = 2\times 3 \times 2 \times 2 = 24 \\ \Rightarrow 18+18+12+18+24+36+24 = \boxed{150}$
- 2 years, 4 months ago
Thanks Nihar! :D
- 2 years, 4 months ago
Saale dost ko thanks bolta hain :P
- 2 years, 4 months ago
Ahahahahahh xD
Chal nahi bolta thanks xD
Waapas liya thanks xD
- 2 years, 4 months ago
For the record, here are the numbers:
1 3210,,5210,,1230,,5230,,1250,,3250,,3410,,5410,,1430,,5430,,1450,,3450,,3012,,5012,,3412,,5412,,1032,,5032,,1432,,5432,,1052,,1452,,3052,,3452,,3014,,5014,,3214,,5214,,1034,,5034,,1234,,5234,,1054,,3054,,1254,,3254,,5031,,5231,,5431,,3051,,3251,,3451,,5013,,5213,,5413,,1053,,1253,,1453,,3015,,3215,,3415,,1035,,1235,,1435,,2301,,2501,,2341,,2541,,2103,,2503,,2143,,2543,,2105,,2145,,2305,,2345,,4301,,4501,,4321,,4521,,4103,,4503,,4123,,4523,,4105,,4305,,4125,,4325,,5301,,3501,,5321,,3521,,5341,,3541,,5103,,1503,,5123,,1523,,5143,,1543,,3105,,3125,,3145,,1305,,1325,,1345,,2310,,2510,,2130,,2530,,2150,,2350,,2314,,2514,,2134,,2534,,2154,,2354,,4310,,4510,,4130,,4530,,4150,,4350,,4312,,4512,,4132,,4532,,4152,,4352,,2531,,2351,,2513,,2153,,2315,,2135,,4531,,4351,,4513,,4153,,4315,,4135,,5310,,3510,,5130,,1530,,3150,,1350,,5312,,3512,,5132,,1532,,3152,,1352,,5314,,3514,,5134,,1534,,3154,,1354,
- 2 years, 4 months ago
Comp. Science Gawd is here. _/_
- 2 years, 4 months ago
For the 2nd one, We get that $$b=2ac/a+c$$ Substitute this in the second equation, $$( {b}^2+{c}^2)/({a}^2+{b}^2)=({ c}^2+{a}^2)/({b}^2+{c}^2)$$ By simplifying we get, a=b=c
- 2 years, 4 months ago
Oh, I like question 3. The main way to show that a number is composite, is to factorize it and then show that the factors are not $$\pm1$$
Staff - 2 years, 4 months ago
Wait in Q1 when there asking for no two adjacent digits are even, doesn't that mean none of the digits are even ? O.O
- 2 years, 4 months ago
Not necessarily I guess. The placement of the digits could be OEOO
- 2 years, 4 months ago
Oh, so no two adjacent digits are both even?
- 2 years, 4 months ago
Exactly :P
- 2 years, 4 months ago
In that case I got 150 as an answer :(.
550 doesn't make sense because the maximum number of four digit numbers with distinct digits from the given set is $$5 \cdot 5 \cdot 4 \cdot 3 = 300$$ , which is less than your answer.
- 2 years, 4 months ago
I got 60 as the answer. Could you please elaborate on how you got 150?
- 2 years, 4 months ago
Oh! I guess i mistook the question for a,b,c and d not necessarily being distinct. Lemme work this one out. Till then, please work the other questions :)
- 2 years, 4 months ago
Well I just took it in cases. You need at least one even and at most two evens.
So you just work from there.
- 2 years, 4 months ago
Oh okay, thanks! :D
- 2 years, 4 months ago
- 2 years, 4 months ago
1) 278
- 2 years, 4 months ago | 1,917 | 4,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-05 | longest | en | 0.809695 |
https://llllllll.co/t/teletype-challenge-bartok-palindrome/5188 | 1,670,217,523,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00085.warc.gz | 392,486,450 | 9,500 | # Teletype Challenge : Bartok Palindrome
Sorry, no code yet. Just problems in need of a solution.
I planned to participate in last week’s Disquiet Junto involving a musical palindrome. During research, I was reminded of Bartok’s palindrome in Music for Strings, Percussion, and Celesta (1936), 3rd Movement (Adagio). The xylophone part has an interesting 4 bar rhythmic pattern in a palindrome format.
As I thought through how to make use of this pattern, I thought about using Teletype to script out the pattern. And… problems ensued.
Wondering if anyone has a suggestion for how to approach scripting note divisions? The Bartok pattern has quarter, eighth, sixteenth, triplet-eighths, and triplet-sixteenth note values. I initially thought about placing the values in a TT pattern in millisecond values, but could figure out how to assign those to the M value.
I appreciate TT’s simplicity. Still, without nested/bracketed loop and decision structures, I’m at a loss how to solve this problem.
Any ideas?
EDIT Later:
After a couple days of talk and me mostly scratching my head, @tambouri came up with an appropriate solution. It makes of of TT’s ER operator, added to the mix by @sam Thi sis an ingenious solution to creating a quite complex rhythmic pattern in TT. I hope by placing it up top, it will encourage others to take a look at the ER operator.
Solution (only requires a trigger to jack 1 - a quite fast trigger to really hear the pattern.)
``````INIT :
SCRIPT 2 // sets X and p.n 0 location
TR.TIME A 25 // just a value that worked for me
SCRIPT 1:
P.N 0
Z ER P.HERE 12 X // the Euclidian Rhythm pattern statement
IF Z : TR.PULSE A
X MOD ADD X 1 12
IF EZ MOD X 6 : P.NEXT
SCRIPT 2: // for testing or playful resets
P.N 0
P.I 0
X 0
PATTERN 0 :
1 1 0 0 0 0 1 1 1 1 2 2 3 3 4 6 4 4 6 4 3 3 2 2 1 1 1 1 0 0 0 0``````
4 Likes
Sadly, I have absolutely nothing technical to offer, but I love the connection to Bartok–I adore Bartok.
This reminds me of his fifth string quartet, which is in effect a palindrome, if only in the macro, structural sense. “Arch” form as it is sometimes referred to.
It is a benchmark piece of music for me personally. Among the high points of human achievement. I was grateful to have heard it performed live last year at the LA Phil as part of their chamber series. Astonishing!
5 Likes
Envious!
(more characters)
Not at my teletype, but a few thoughts on alternate starting places
I haven’t investigated the rhythms mode in the just type beta, but per the instructions it seems this could perhaps be a logical application
More things I’ve not investigated, but the euclidean rhythm ops added to the 1.2 firmware might be useful here
If I have some time this eve I’ll give the Metro solution a swing
2 Likes
Thanks for this! I’d forgotten about the doc addendums. Will give this a thorough read thru this evening.
Put the notes durations (in ms) in a pattern and use DEL accessing those values ?
finding the smallest duration unit could be useful as well, since all of te duration would be multiples of it.
That was my idea also.
There are triplets and duplets in the Barton pattern. Hard to get to where needed with integer-based math?
@tambouri I spent some time TT’s eclidian rhythm operator last evening. Really fantastic for generating all sorts contrasting patterns. I will definitely continue to explore, but I was not able to generate triplet patterns.
If I understand correctly, triplets may not be a part of this system.
It depends.
If your smallest triplet value is 10ms, the smallest duplet value can be 15ms and there is no prob. Depends on the tempo you want to use, though.
I broke down the Barton pattern into ms values for each note length, with a base tempo of 60 BPM.
So a quarter note would be 1000ms. Triplet eighth note = 333 and triplet sixteenth = 166. For both triplet types, a bit of math fudging would be necessary to get on solid tempo. For the triplet eighths, we would need to add 1ms to 1 of the triplet parts. For triplet sixteenths, add 1 to 2 parts.
The pattern looks like:
``````1000
1000
500
500
333
334
333
250
250
167
166
167
250
250
250
250
167
166
167
250
250
333
334
333
500
500
1000
1000
``````
Now, how to actually get TT code designed to make this work… that’s the challenge.
3 Likes
Ok. I believe I’ve come up with Euclidean Rhythm operator (ER) statements that could work for all but the sixteenth notes in the Barton Palindrome pattern.
The smallest value is a sixteenth triplet, so I came up with ER patterns of 24 steps.
``````triplet sixteenth - 1, 24, 0
sixteenth - ?
triplet eighth - 2, 24, 0
eighth - 3, 24, 0
quarter - 6, 24, 0
``````
Still left with the sixteenth note being able to fit into the length of a bar comprised of 24 steps.
Edit: hmm… If I double the number of steps to 48, this may solve the problem. I’m not at TT, but will give this a thorough test later.
2 Likes
I didn’t have time to play with this last night, and now I’m back at my desk at work, but this is how I was thinking you could use the euclidean rhythm generator. Per my reading of the ER writeup and no actual testing on the teletype, so not unlikely that I’m missing something, but … hear me out
Using a clock 1/12 the length of the desired 1/4 note in your pattern the following could create triggers when the ER returns a 1. and you could shift through a pattern of fill settings every 1/8th of the piece which would be every 6 beats of the 1/12 clock
i believe ER FILL LENGTH STEP would work as follows
so a script might work something like this.
SCRIPT 1
P.N 0
X MOD ADD X 1 12 x counts 0-11
IF EZ MOD X 6 : P.NEXT move to the next pattern every 1/8note = 1/6 of the 12 count
IF ER P.HERE 12 X : TR.PULSE 1 execute a trigger if ER returns 1
So you would now choose a fill amt for every 1/8 of the piece. (1/8 as this seems to be the smallest scale that the division shifts). so you could put the following into Pattern 0 (split up into bars only to make it easier to see my thought process)
bar 1 - 1 1 0 0 0 0 1 1 1/4 note, 1/4 rest, 1/4 rest, 1/4 note
bar 2 - 1 1 2 2 3 3 4 6 1/4 note, 1/8 notes, 1/8t notes, 1/16 notes, 16t notes…
bar 3 - 4 4 6 4 3 3 2 2
bar 4 - 1 1 1 1 0
Repeat disclaimer, this is off the top of my head and likely has some issues. 1 thing I’m definitely unsure of is if ER 0 12 STEP would work to fill a pattern w/ no active steps to take into account the rests. And this doesn’t solve for duration yet
4 Likes
Very nice, @tambouri. You came directly to the 1 beat version of what it took me multiplepasses to get to.
Still required is some fancy pattern resetting to get the correct number of note values per full pattern, but that seems doable.
I fear for the squirrel running the clocking wheel, tho.
2 Likes
Im pretty certain that ER 0 12 works exactly as expected
1 Like
seeing a problem where the order of the commands I used are likely causing it to skip step 0 and advance before the first beat, maybe that’s cause this?
so maybe have a reset script and change up the order of the commands a little
SCRIPT 1 :
P.N 0
IF ER P.HERE 12 X : TR.PULSE 1
X MOD ADD X 1 12
IF EZ MOD X 6 : P.NEXT
SCRIPT 2 :
X 0
P.N 0
P.I 0
PATTERN 0 :
1 1 0 0 0 0 1 1 1 1 2 2 3 3 4 6 4 4 6 4 3 3 2 2 1 1 1 1 0 0 0 0
This way it should trigger the 1st step before moving to X = 1
1 Like
Nope - throws a ‘not enough parameters’ error
This line is a character too long it seems. Drat!
Solved by storing P.HERE in a var Z.
`````` P.N 0
Z P.HERE
IF ER Z 12 X : TR.PULSE A
....``````
I think @GoneCaving is stating that `ER 0 12 step` should always return a `0`. It does. (Or it should, if it doesn’t it’s a bug.)
The C source that underlies the op is here, you can see at the beginning of the `euclidean` function that it returns a `0` if `fill` is less than 1.
1 Like
Yep, that’s clearly what I meant. Post was from memory and without checking, but meant that a rest is certainly possible.
1 Like | 2,256 | 8,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.871876 |
https://mathematica.stackexchange.com/questions/120474/does-meanmarker-boxwhiskerchart-exclude-outliers | 1,716,310,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00005.warc.gz | 347,253,641 | 40,472 | # Does "MeanMarker" (BoxWhiskerChart) exclude outliers?
When you use the "MeanMarker" function within BoxWhiskerChart, does the mean shown on the box plot exclude the near and/or far outliers identified from the plot?
If not, is there a (relatively simple) way to have the mean marker shown on the plot exclude outliers?
In answer to the first question, no, the mean does not exclude outliers:
SeedRandom[1]
r = RandomVariate[NormalDistribution[0, 1], 100]^2;
fence = Quantile[r, 0.95]
mean1 = Mean[r]
mean2 = Mean @ DeleteCases[r, x_ /; x > fence]
3.33858
1.01835
0.770329
BoxWhiskerChart[r, {{"Outliers"}, {"MeanMarker"}},
GridLines -> {{}, {fence, mean1, mean2}}]
Regarding the second question one could calculate these as I did for mean2 above and plot them with Epilog but there may be a simpler way. I'll see what I can find.
• Have you looked at TrimmedMean[], by any chance? Jul 12, 2016 at 6:20
• @J.M. Thanks for the hint! Despite that being around since version 6 I don't think I have ever used it. Jul 12, 2016 at 6:27
ClearAll[trimmedMeansF]
trimmedMeansF[dir_: Directive[Thick, Opacity[1, Red]]] := BoxWhiskerChart[
ConstantArray[#, 2] & /@ TrimmedMean[Transpose@#, {0, .05}],
{{"Fences", None}, {"Whiskers", Opacity[0]}, {"MedianMarker",
Opacity[0]}, {"MeanMarker", .8, dir}}] &;
Example:
SeedRandom[1]
r = RandomVariate[NormalDistribution[0, 1], 100]^2;
rr = {r, 1 + r, 2 + r};
Legended[Show[BoxWhiskerChart[rr,{{"Outliers"}, {"MeanMarker", Directive[Thick, Yellow]}},
ChartStyle -> 1, GridLines -> {None, Mean@Transpose@rr}],
trimmedMeansF[Directive[Thick, Cyan]]@rr],
LineLegend[Directive[Thick, #] & /@ {Yellow, Cyan}, {"means", "trimmed means"}]]
Alternatively, post-process to modify the position of the mean markers:
means1 = Mean@Transpose@rr;
means2 = TrimmedMean[Transpose@rr, {0, .05}];
bwc = BoxWhiskerChart[rr, {{"Outliers"}, {"MeanMarker", Directive[Thick, Yellow]}},
GridLines -> {None, means1}, PlotLabel -> TableForm[{means1, means2},
TableHeadings -> {{"means", "trimmedmeans"}, {"data1", "data2", "data3"}}],
ChartStyle -> 1, ImageSize -> 400];
Row[{bwc, bwc /. With[{sel = Select[Cases[bwc, _LineBox, Infinity],
Not[FreeQ[#, Alternatives @@ means1]] &]}, | 688 | 2,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.814999 |
https://groups.yahoo.com/neo/groups/pcgen/conversations/topics/77677?var=1&l=1 | 1,492,939,763,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118310.2/warc/CC-MAIN-20170423031158-00454-ip-10-145-167-34.ec2.internal.warc.gz | 781,208,172 | 20,425 | ## Re: [pcgen] Re: [DOCS] JEP formulas
Expand Messages
• Hallelujah! I see the light, I m now a JEP convert! JEP brings some fantastic capabilities too PCGen. My new favorite is the use of Boolean operators. Let take
Message 1 of 35 , Jun 30, 2004
Hallelujah! I see the light, I'm now a JEP convert!
JEP brings some fantastic capabilities too PCGen. My new favorite is the use
of Boolean operators.
Let take for example my ridiculously complicated formula:
((CL+2)/2)+((((CL-3)/2).TRUNC)MIN2)-((((CL-2)/2).TRUNC)MIN3)-((CL/14).TRUNC)
This progression ends at 16 at 20th level so all this math is just to get -4
in at the right levels. With JEP it can be expressed this way:
CL-(CL>=4)-(CL>=8)-(CL>=13)-(CL>=18)
Very simple and easy to read, and it works, just tested it.
The four Boolean expressions read "Class Level is equal to or greater than x
level, if its true it returns a 1 if false it returns a zero. Very useful.
I'm working on updating the docs on this subject now.
--
~ Eddy
~ Doc Chimp, Data Tamarin
~ PCGen BoD Documentation Second
• ... If I remember correctly (no guarantee) but a--e.com was exactly right.
Message 35 of 35 , Jul 7, 2004
Mark Coletti wrote:
> On Fri, 02 Jul 2004 08:45:25 -0700, Kevin Brown
> <kevin_brown@...> wrote:
> > >>http://gmgen.sourceforge.net/wiki/
> > >>
> > >>I just hit that website, and the links are full of hard-core porn
> > >>sites... someone might wanna check into that.
> > >
> > >
> > > Deleted!
> > > That is so not cool.
> >
> > The changes should be able to be rolled back thanks to it being a
> wiki page, heheh.
>
> QuantumGIS had the same problem recently. In fact, I wonder if it was
> the same asshat. Was the offender posting links hosted by a--e.com,
> which is Russian?
>
> QuantumGIS locked down its WIki. It's a PITA, but was necessary.
> Apparently this had occured regularly.
>
>
> MAC
> --
> I'm taking reality in small doses to build immunity.
>
>
> PCGen's release site: http://pcgen.sourceforge.net
> PCGen's alpha build: http://rpg.plambert.net/pcgen
> PCGen's FAQ:
> http://rpg.plambert.net/pcgen/current/_docs/
>
>
> <http://us.ard.yahoo.com/SIG=129aj5q7s/M=295196.4901138.6071305.3001176/D=groups/S=1705016061:HM/EXP=1089240609/A=2128215/R=0/SIG=10se96mf6/*http://companion.yahoo.com>
>
>
>
> ------------------------------------------------------------------------
>
> * To visit your group on the web, go to:
> http://groups.yahoo.com/group/pcgen/
>
> * To unsubscribe from this group, send an email to:
> pcgen-unsubscribe@yahoogroups.com
> <mailto:pcgen-unsubscribe@yahoogroups.com?subject=Unsubscribe>
>
> * Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service <http://docs.yahoo.com/info/terms/>.
>
>
If I remember correctly (no guarantee) but a--e.com was exactly right.
Your message has been successfully submitted and would be delivered to recipients shortly. | 836 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-17 | latest | en | 0.853872 |
https://socratic.org/questions/how-do-you-combine-4c-1-5ab-d | 1,686,372,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00641.warc.gz | 609,158,018 | 5,765 | # How do you combine (4c+1)-(5ab+d)?
$4 c + 1 - 5 a b - d$
$4 c + 1 - 5 a b - d$ | 44 | 81 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-23 | latest | en | 0.326228 |
https://fr.mathworks.com/matlabcentral/cody/solutions/1901595 | 1,571,283,151,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00304.warc.gz | 504,212,344 | 15,453 | Cody
# Problem 1074. Property dispute!
Solution 1901595
Submitted on 15 Aug 2019 by David Hill
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [0 0 5 10]; b = [3 2 6 6]; c = [3 2 2 6]; assert(isequal(disputed_region(a,b),c))
2 Pass
a = [0 1 2 3]; b = [-1 2 4 5]; c = [0 2 2 2]; assert(isequal(disputed_region(a,b),c))
3 Pass
a = [3 4 6 1]; b = [2 1 2 2]; c = []; assert(isequal(disputed_region(a,b),c))
4 Pass
a = [5 18 14 6]; b = [11 23 17 18]; c = [11 23 8 1]; assert(isequal(disputed_region(a,b),c))
5 Pass
a = [100 97 3 2]; b = [99 93 12 12]; c = [100 97 3 2]; assert(isequal(disputed_region(a,b),c))
6 Pass
delta = 0.5; a = [0 0 5 10]+delta; b = [3 2 6 6]+delta; c = [3 2 2 6]+delta; assert(isequal(disputed_region(a,b),c))
7 Pass
a = [0 0 1 1]; b = [2 2 1 1]; c = []; assert(isequal(disputed_region(a,b),c)) | 405 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-43 | latest | en | 0.519249 |
http://betterlesson.com/lesson/resource/2586610/student-work-1 | 1,477,560,256,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721174.97/warc/CC-MAIN-20161020183841-00052-ip-10-171-6-4.ec2.internal.warc.gz | 26,591,897 | 20,005 | Student Work 1
# Navigating Road Blocks in Problem Solving
Unit 5: Multiply or Divide with Word Problems
Lesson 5 of 6
## Big Idea: Students can solve real world problems using their understanding of multiplication and division and recognizing their relationship.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Fluency (Math)
50 minutes
### Sarah Maffei
##### Similar Lessons
###### Multiplication Properties
4th Grade Math » Multiplication and Division Meanings
Big Idea: Multiplication Properties are rules that can be used to help with math computation.
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Memphis, TN
Environment: Urban
###### Naming Arrays
3rd Grade Math » Understanding Multiplication
Big Idea: After partners have worked with each other and have some practice creating arrays, it is time to challenge them to define the concept demonstrated by an array and how it can be used to solve problems.
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Troy, MI
Environment: Suburban | 215 | 982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-44 | longest | en | 0.891478 |
http://nrich.maths.org/public/leg.php?code=-58&cl=2&cldcmpid=1209 | 1,475,051,665,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661327.59/warc/CC-MAIN-20160924173741-00193-ip-10-143-35-109.ec2.internal.warc.gz | 200,318,721 | 6,149 | # Search by Topic
#### Resources tagged with Manipulating algebraic expressions/formulae similar to Nim-like Games:
Filter by: Content type:
Stage:
Challenge level:
### There are 13 results
Broad Topics > Algebra > Manipulating algebraic expressions/formulae
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Sums of Pairs
##### Stage: 3 and 4 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Algebra Match
##### Stage: 3 and 4 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Perimeter Expressions
##### Stage: 3 Challenge Level:
Create some shapes by combining two or more rectangles. What can you say about the areas and perimeters of the shapes you can make?
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### Algebra from Geometry
##### Stage: 3 and 4 Challenge Level:
Account of an investigation which starts from the area of an annulus and leads to the formula for the difference of two squares.
### Matchless
##### Stage: 3 and 4 Challenge Level:
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### Temperature
##### Stage: 3 Challenge Level:
Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same? | 639 | 2,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-40 | longest | en | 0.89836 |
https://nl.mathworks.com/matlabcentral/cody/problems/1818-07-common-functions-and-indexing-3/solutions/947808 | 1,579,392,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00085.warc.gz | 580,474,909 | 15,564 | Cody
# Problem 1818. 07 - Common functions and indexing 3
Solution 947808
Submitted on 31 Aug 2016 by Sofie Overgaard Pedersen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
eMat = [13 -1 5;-22 10 -87]; Ref = [1 1 1;eMat(2,:)]; user = MyFunc(); assert(isequal(Ref,user))
eMat = 13 -1 5 -22 10 -87 eMat2 = 1 1 1 -22 10 -87
2 Pass
Ref = [13 -1 5;-22 10 -87]; [eMean Mat] = MyFunc(); assert(isequal(Ref,Mat))
eMat = 13 -1 5 -22 10 -87 eMat2 = 1 1 1 -22 10 -87 | 223 | 585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.652625 |
https://stats.stackexchange.com/questions/209116/central-limit-theorem-when-the-dimension-size-increases-with-the-sample-size | 1,582,690,632,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00464.warc.gz | 556,930,214 | 31,941 | # Central Limit Theorem when the dimension size increases with the sample size
Let $X_1, X_2,\ldots, X_n \in \mathcal{R}^d$ and be zero-mean, unit variance random variables. Here the dimension ($d$) is a function of the sample size($n$) i.e, $d=f(n)$. For example $d = \sqrt{n}$.
1. When can we apply vectorized central-limit-theorem to the following sum?
$$S_n=\frac{X_1+X_2+\cdots+X_n}{\sqrt{n}}$$
2. Where can one find a comprehensive literature for above?
• To have a guess, MVN means all linear combinations are normal. This could be a way to approach the "finite variance" needed for CLT to work. Basically you're trying to show that $\sqrt {n}t^T\overline {X}_n$ is quadratic in $t$ as $n\to\infty$. For some functions $d=f (n)$ this won't work - my guess is anything which grows faster than $\sqrt {n}$ will have infinite variance (unless you have strong negative correlation between component of $X_i$) – probabilityislogic Apr 25 '16 at 1:02
• I don't see why $d>\sqrt{n}$ should give infinite variance. The normalization from $t$ should take care of it. – Vivek Bagaria Apr 25 '16 at 1:18
• The asymptotic covariance matrix will be $d \times d$, which means it will be $\sqrt{n} \times \sqrt{n}$. But since $n$ is tending to $\infty$ and the CLT is an asymptotic result, I don't see how one can successfully define an infinite dimensional square matrix. It might be a matrix of finite values, but it will have countably infinite entries. – Greenparker Apr 25 '16 at 3:10
• Greenparker: An infinite dimensional rv is gaussian iff every finite subset is a gaussian. This helps us deal with infinite matrices. – Vivek Bagaria Apr 25 '16 at 9:35
• It's not apparent that the increase in $d$ is any problem at all provided you explain how you intend to add vectors of different dimensions. If you fix $k\ge 1$ and simply ignore all components beyond the $k^\text{th}$ one (and presuming $d(n)$ is a nondecreasing function so that a limit can even be defined), you may apply the usual CLT (assuming it holds for your variables) to demonstrate the first $k$ components of $S$ converge in distribution to an MVN. – whuber Apr 25 '16 at 14:31 | 597 | 2,149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-10 | latest | en | 0.858712 |
https://kidsworksheetfun.com/two-digit-addition-color-by-number/ | 1,656,826,190,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00636.warc.gz | 393,202,303 | 21,852 | # Two Digit Addition Color By Number
Two digit addition coloring worksheets categories. This 2 digit addition with regrouping mystery pictures activity is designed to make learning math more fun to use choose and print the right type of problem for your class.
Summer Color By Code Double Digit Addition And Subtraction Addition Code Color Digit Double In 2020 Double Digit Addition Addition And Subtraction Math Coloring
### This 2 digit addition color by number resource includes 5 pages for introducing or reviewing 2 digit addition.
Two digit addition color by number. 2 digit addition coloring. When we talk concerning two digit addition coloring worksheets below we can see some similar pictures to inform you more. Two color by numbers worksheets that students must first solve double digit addition questions to know what color belongs to what number.
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Some of the worksheets for this concept are adding 1 and 2 digit numbers with regrouping color the name fun math game s double digit addition regrouping work addition color by number butter ies name addition practice 1 double digit subtraction regrouping work. Strengthen your second grader s ability to tackle his addition facts by adding up the number strings and. Your kids are sure to learn faster when crayons enter the picture.
Math addition color by number double digit math addition coloring worksheets and double digit multiplication color by number are three of main things we will show you based on the post title. 2493 views downloads. 2 digit addition is an essential skill for second graders who need t.
Practicing three digit addition is fun with these winter themed color by number puzzles. Doubling as a color by number coloring page this 2nd grade math worksheet makes practicing two digit subtraction fun. With more related things like 2 digit subtraction color by number two digit addition with regrouping worksheets and 2 digit addition coloring worksheets.
2 digit addition coloring displaying top 8 worksheets found for this concept. Get three more double. Practice two digit subtraction 10.
Kids love solving mysteries and that s exactly what they ll do that with these coloring worksheets. 2 digit addition is an essential skill for second graders who need t. 1 st 2 nd 3 rd homeschool.
This is a great activity for those students who need ad. Included are two 3 digit addition pages without regrouping two 3 digit addition pages with regrouping and two 3 digit addition pages with mixed practice. One worksheet reveals a seahorse the other shows a sailboat great way for students to practice the skill while getting immediate feedback.
Download two digit addition coloring worksheets hd widescreen wallpaper from the above resolutions from the directory addition posted by worksheetschool on 01 03 2019 if you don t find the exact resolution you are looking for then go for. If your little ones are learning addition these are the perfect little worksheets to help them have fun with math. Activities printables math centers.
Print all of our color by number addition coloring pages today and give them to your kids and their friends and your class. You will love the no prep print and go ease of these 2nd grade go math 2 6 2 digit addition color by number printables.
Easter Color By Code Double Digit Addition And Subtraction With Regrouping Double Digit Addition Addition And Subtraction Subtraction
This Spring Three Digit Addition Color By Number Activity Is The Perfect Spring T Math Coloring Worksheets Addition Coloring Worksheet Addition And Subtraction
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2nd Grade Go Math 4 6 2 Digit Addition Color By Numbers Math Coloring Worksheets Go Math 2nd Grade Math Worksheets
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Color By Number Practice 2 Digit Addition Freebie Free Math Printables Addition Freebie Math Coloring
Halloween Activities 3 Digit Addition Color By Number With Without Regrouping Addition Coloring Worksheet Printable Math Worksheets Math Addition Worksheets
New Year S Three Digit Addition Color By Number Math Coloring Worksheets Fun Math Worksheets Color Worksheets
Sums Of 2 One Digit Numbers Color By Numbers Kindergarten Math Worksheets Free Kindergarten Math Worksheets Addition And Subtraction
Color By Number Spring Mystery Pictures Double Digit Addi Teaching Subtraction Sight Words Kindergarten Color Worksheets
Math Theme Color By Code Two Digit Addition And Subtraction No Regrouping Addition And Subtraction Math Coloring Double Digit Addition
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Two Digit Addition Coloring Worksheets Free Addition Coloring Worksheet Math Coloring Worksheets Color Worksheets
Halloween Activities 3 Digit Addition Color By Number With Without Regrouping Addition Coloring Worksheet Printable Math Worksheets Math Addition Worksheets
Ocean Animal Color By Code Two Digit Addition And Subtraction No Regrouping Addition And Subtraction Subtraction Activities Subtraction | 1,021 | 5,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-27 | latest | en | 0.876058 |
https://www.beatthegmat.com/search.php?author_id=161608&sr=posts&sid=08d3e49bff6b8ef9cd08ae358bcee5cb | 1,618,912,258,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00221.warc.gz | 733,972,090 | 9,736 | ## Search found 90 matches
#### violens
##### violens
A violin constructed to have improved sound would sound different from the best-sounding existing violins. To professional violinists, a violin that sounds different from the best-sounding existing violins sounds less like a violin and therefore worse than the best-sounding existing violins. Profess...
by sandeep_thaparianz
Wed Jul 18, 2012 10:00 am
Forum: Critical Reasoning
Topic: violens
Replies: 2
Views: 815
#### ITR
##### ITR
The people who are least likely to be audited by the Internal Revenue Service this year are those who have been audited since 1985 and who were found to have made no mistakes in filing their returns during that audit. Of the following people, who is MOST likely to be audited by the IRS? (A) A person...
by sandeep_thaparianz
Sun Jul 15, 2012 3:12 am
Forum: Critical Reasoning
Topic: ITR
Replies: 1
Views: 1118
#### question
###### Critical Reasoning
+1 for B
by sandeep_thaparianz
Sun Jul 15, 2012 2:38 am
Forum: Critical Reasoning
Topic: question
Replies: 8
Views: 2260
#### Percent problem..from DS 1000 series
###### Data Sufficiency
has to be E
by sandeep_thaparianz
Sun Jul 08, 2012 4:39 am
Forum: Data Sufficiency
Topic: Percent problem..from DS 1000 series
Replies: 6
Views: 3346
#### Statistics
###### Data Sufficiency
thats because first option says 4 or more that means 4,5,6,7, and so on
and second option says 2 or few that means 0,1,2
so only option left is 3
hope is understood ur question properly
by sandeep_thaparianz
Sun Jul 08, 2012 4:37 am
Forum: Data Sufficiency
Topic: Statistics
Replies: 4
Views: 1005
#### Proposition 13
##### Proposition 13
With Proposition 13, if you bought your house 11 years ago for $75,000, your property tax would be approximately$914 a year (1 percent of $75,000 increased by 2 percent each year for 11 years); and if your neighbor bought an identical house next door to you for$200,000 this year, his tax would be ...
by sandeep_thaparianz
Sat Jul 07, 2012 6:46 am
Forum: Critical Reasoning
Topic: Proposition 13
Replies: 4
Views: 1203
#### IR
###### GMAT Integrated Reasoning
A seems to be the best
by sandeep_thaparianz
Fri Jul 06, 2012 9:43 pm
Forum: GMAT Integrated Reasoning
Topic: IR
Replies: 5
Views: 2141
#### integrated reasoning
###### GMAT Integrated Reasoning
question mentions min years of exp. to become a principal
for analyst it says<3 that means min =0
for associate it says 2 but can be relaxed(last point) so min again =0
for principal it says 4+ so min =4
sum all mins and we will have 4
by sandeep_thaparianz
Fri Jul 06, 2012 9:27 pm
Forum: GMAT Integrated Reasoning
Topic: integrated reasoning
Replies: 5
Views: 2178
#### French-manufactured public toilet
###### Critical Reasoning
What is the question??
You have written the argument and options.
by sandeep_thaparianz
Thu Jul 05, 2012 10:21 am
Forum: Critical Reasoning
Topic: French-manufactured public toilet
Replies: 3
Views: 1387
#### the product of 5 smallest prime numbers
###### Problem Solving
is it first five prime nos??
by sandeep_thaparianz
Wed Jul 04, 2012 9:05 am
Forum: Problem Solving
Topic: the product of 5 smallest prime numbers
Replies: 4
Views: 954
#### accelerated growth
###### Sentence Correction
i chose E..but explanations says opposite to what i was thinking.
by sandeep_thaparianz
Tue Jul 03, 2012 8:05 pm
Forum: Sentence Correction
Topic: accelerated growth
Replies: 4
Views: 1732
#### accelerated growth
##### accelerated growth
The Federal Reserve announcement said that growth had accelerated after slowing in the second quarter and that the policy makers remain concerned about the prospects of inflation, even though there are few signs of higher energy prices driving up the cost of other goods so far. • that growth had a...
by sandeep_thaparianz
Tue Jul 03, 2012 8:04 pm
Forum: Sentence Correction
Topic: accelerated growth
Replies: 4
Views: 1732
#### modifiers
###### Sentence Correction
A is the OA
B is wrong because the author is non essential modifier hence needs to be offset by commas. Likewise the poet modifies Walt Whitman. Again it is acting as a non essential modifier.
If in last option had been is replaced by are then it is grammatically correct.
by sandeep_thaparianz
Tue Jul 03, 2012 7:17 am
Forum: Sentence Correction
Topic: modifiers
Replies: 3
Views: 786
#### Simple SC gone wrong
###### Sentence Correction
by sandeep_thaparianz
Tue Jul 03, 2012 4:12 am
Forum: Sentence Correction
Topic: Simple SC gone wrong
Replies: 3
Views: 864
#### pest control
##### pest control
Pests had destroyed grape, celery, chili pepper crops, sugar beet and walnut in the region, but in the 1880s, more effective pest-control methods saved the citrus industry. • Pests had destroyed grape, celery, chili pepper crops, sugar beet and walnut in the region, but in the 1880s, more effectiv...
by sandeep_thaparianz
Tue Jul 03, 2012 12:42 am
Forum: Sentence Correction
Topic: pest control
Replies: 6
Views: 1434 | 1,416 | 5,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-17 | latest | en | 0.926552 |
https://www.physicsforums.com/threads/quantum-communication.876616/ | 1,519,164,893,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813109.36/warc/CC-MAIN-20180220204917-20180220224917-00017.warc.gz | 904,239,445 | 17,273 | # B Quantum communication
1. Jun 23, 2016
### ed777
is it possible to entangle two particles and then separate them buy a large distance, and then interfere with one of the particles and observe this in the other, like a quantum communication system that is faster than light?
2. Jun 23, 2016
Staff Emeritus
No, it is not.
3. Jun 23, 2016
### entropy1
Suppose you have two stone-throwers sitting back-to-back throwing stones at two oppositely positioned targets. Let's call the targets A and B. Now let's assume that the stone-throwers both repeatedly throw stones simultanuously at the targets, and let's suppose they have a chance of exactly 50 percent to hit the target, and 50 percent to miss it. Suppose behind each target stands an observer, checking for each stone if it hits or misses. Let's call them Alice and Bob.
What we can firstly conclude is that Alice and Bob both observe 50 percent of the stones hitting their observed target, and 50 percent missing it.
Now suppose the stone-throwers simultanuously hit or miss their targets. There would still be a 50/50 chance of hitting or missing each target, but if one was hit, the other would also be hit, and similarly for misses. This would be a measureble effect if Alice and Bob would compare their notes! They would measure a 100 percent match between hits and likewise between misses.
Now suppose that in midflight of the stones, Bob would change something to target B that would influence whether it would be hit or missed while keeping the probability ratio of being hit 50/50. Now Alice and Bob still would observe 50 percent hits and 50 percent misses. However, if they would compare their notes, they would notice that there wouldn't be a 100 percent match between hits on one side and misses on the other. Sometimes there would be a miss where the other got a hit.
So, in the latter case, Alice and Bob would still see a 50/50 ratio hits/misses on their own target. Bob didn't change Alice's ratio. The difference between the targets only became apparent when Alice and Bob compared notes.
Note: To be more precise, all we know, is that Alice's and Bob's observations stay the same, and that the hit/miss correlation between them may vary; whether Bob's change in his target really does not at all affect Alice's results, we cannot know. After all, we can't know what 'would have happened' would Bob have decided otherwise. In any case, any 'influence' Bob would pose on Alice would perfectly blend into Alice's results, since both exhibit a purely random 50/50 ratio.
Last edited: Jun 23, 2016
4. Jun 25, 2016
### ed777
what about spooky action at a distance?
5. Jun 25, 2016
### Staff: Mentor
We could have another long thread about whether spooky action at a distance even exists - we already have several of these - but even if it does it cannot be used for communication.
There are many many threads here explaining why. | 669 | 2,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-09 | longest | en | 0.975719 |
http://excel.tips.net/T011627_Creating_an_Amortization_Schedule.html | 1,474,879,542,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660712.3/warc/CC-MAIN-20160924173740-00163-ip-10-143-35-109.ec2.internal.warc.gz | 94,482,128 | 9,080 | # Creating an Amortization Schedule
Please Note: This article is written for users of the following Microsoft Excel versions: 97, 2000, 2002, and 2003. If you are using a later version (Excel 2007 or later), this tip may not work for you. For a version of this tip written specifically for later versions of Excel, click here: Creating an Amortization Schedule.
Mary would like to use Excel to create an amortization schedule for her home mortgage. Problem is, she doesn't know enough about finance to know which of the financial worksheet functions she should use to do the calculations.
It actually is fairly easy to come up with the right calculations. At its simplest, a mortgage payment consists of two parts: principle and interest. Given some basic information such as how much you are borrowing (your principal), what your interest rate is, and how many monthly payments you need to make, you can then come up with your amortization schedule. Try this out:
1. Open a blank worksheet.
2. In cell B1 put your interest rate.
3. Give cell B1 a name, such as "Rate".
4. In cell B2 put the number of months you need to pay.
5. Give cell B2 a name, such as "Term".
6. In cell B3 put how much you are borrowing.
7. Give cell B3 a name, such as "Principal".
8. In cell A6 put the number 1. This represents your payment number.
9. In cell B6 put this formula: =Principal. The amount you put into cell B3 should now also appear in cell B6.
10. In cell C6 put this formula: =PPMT(Rate/12,\$A6,Term,Principal). This is the amount you will pay toward your principal in this payment. (The PPMT function returns the amount of principle for a given payment.)
11. In cell D6 put this formula: =IPMT(Rate/12,\$A6,Term,Principal). This is the amount you will pay in interest in this payment. (The IPMT function returns the amount of interest for a given payment.)
12. In cell E6 put this formula: =PMT(Rate/12,Term,Principal). This is the total amount of the payment.
13. Copy everything from row 6 to row 7.
14. Change cell A7 to the following formula: =A6+1.
15. Change cell B7 to the following formula: =B6+C6. This cell now contains the new principal balance for your loan.
16. Copy row 7 down as many rows as you need.
17. Add any explanatory labels desired in the ranges A1:A3 and A5:E5. (See Figure 1.)
18. Figure 1. A simple amortization schedule.
Remember that I said that this creates a simple amortization schedule. It doesn't take into account varying interest rates, refinancing, non-monthly payments, additional payments, escrow amounts, or any number of other variables. In such instances you would be better to look for a ready-made amortization template. There are any number of them available online, including these from Microsoft:
```http://office.microsoft.com/en-us/templates/TC001056620.aspx
```
You can also find a very good explanation of amortization schedules at this page:
```http://www.tvmcalcs.com/calculators/apps/excel_loan_amortization
```
ExcelTips is your source for cost-effective Microsoft Excel training. This tip (11627) applies to Microsoft Excel 97, 2000, 2002, and 2003. You can find a version of this tip for the ribbon interface of Excel (Excel 2007 and later) here: Creating an Amortization Schedule.
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David 07 Feb 2014, 18:25
How do I create a spreadsheet with this formula???????
Monthly Pay=[ rate + rate /([1+rate]^months -1)] X principle
Where rate of %6 means 6/1200
Mark Russell 21 Apr 2012, 13:30
The values in cells A1 through A3 are labels. You need to enter the amounts in cells B1 through B3 that correspond to the labels in column A, and then name the cells in column B using the name manager.
Bob Manne 21 Apr 2012, 09:07
I tried this exactly as described, yet all of the formulas in B6:E6 show up as ?NAME?
I tried the values in A1:A3 as
Rate: term: principal:
Rate term principal
"Rate" "Term" "Principal"
and got the same result.
What is missing here?
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http://mathhelpforum.com/calculus/105451-double-check-these-derivatives-print.html | 1,526,908,989,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864186.38/warc/CC-MAIN-20180521122245-20180521142245-00437.warc.gz | 190,210,145 | 3,576 | # Double check these derivatives
Printable View
• Oct 1st 2009, 07:39 AM
CFem
Double check these derivatives
Clearly I'm missing at least one important step.
(1) Find an equation of the line tangent to the curve y1 given below that is parallel to the given line y2.
y
1=6x * sqrt(x)
y2 = 9 + 9x
y1 = 6x(x^.5)
y1 = 6x^(3/2)
y1' = (3/2)(6x^.5)
y1' = 9x^(1/2)
Seems to match the criteria, but it's wrong?
(2) For what values of r does the function y = erx satisfy the equation y'' + 3y' - 40y = 0?
(3) f (x) = 9 cos(x) + 5/7 cot(x)
Seemed simple.
9(-sin(x)) + (5/7)(-csc^2(x))
(4) 6-6tan(x) /sin(x)-cos(x)
limit as x goes to pi/4. Bottom is 0 = ??
(5) Implicit differentiation cos(x) + sqrt(y) = 7
-sin(x) + (1/2)y^(-1/2) * (dy/dx) = 0
(1/2)y^(-1/2) = sin(x)
y^(-1/2) = 2sin(x)
1/sqrt(y) = 2sin(x)
y = (1/2sin(x))^2
y = 1/4sin(x)^2
Again, not sure where I went wrong.
(6) 5sqrt(x) + sqrt(y) = 5
(5/2)x^(-1/2) + (1/2)y^(-1/2) * (dy/dx) = 0
I can't put this answer into a form that doesn't suck, but it was 5/sqrt(x) over 1/sqrt(y)
(7) f(x) = sin(4x) + ln(4x)
f'(x) = cos(4x) + 1/4x
This one seems simple, but I'm guessing the problem is with the Trig part.
• Oct 1st 2009, 10:18 AM
DeMath
Quote:
Originally Posted by CFem
Clearly I'm missing at least one important step.
(1) Find an equation of the line tangent to the curve y1 given below that is parallel to the given line y2.
y
1=6x * sqrt(x)
y2 = 9 + 9x
y1 = 6x(x^.5)
y1 = 6x^(3/2)
y1' = (3/2)(6x^.5)
y1' = 9x^(1/2)
Seems to match the criteria, but it's wrong?
$\displaystyle y = 6x\sqrt x = 6\sqrt {{x^3}} \Rightarrow {y^\prime } = 9\sqrt x .$
The equation of the line tangent:
$\displaystyle y - 6\sqrt {x_0^3} = 9\sqrt {{x_0}} \left( {x - {x_0}} \right)$
Now use the condition of parallelism $\displaystyle k_1=k_2$ of two straight lines $\displaystyle y_1= k_1 x + a$, $\displaystyle {y_2} = {k_2}x + b.$
Then you get: $\displaystyle 9 = 9\sqrt{x_0}\Leftrightarrow x_0 = 1.$
Finally, you have: $\displaystyle y - 6 = 9\left( {x - 1} \right) \Leftrightarrow y = 9x - 3.$
• Oct 1st 2009, 11:27 AM
DeMath
Quote:
Originally Posted by CFem
Clearly I'm missing at least one important step.
(4) 6-6tan(x) /sin(x)-cos(x)
limit as x goes to pi/4. Bottom is 0 = ??
This one seems simple, but I'm guessing the problem is with the Trig part.
$\displaystyle \mathop {\lim }\limits_{x \to \pi /4} \frac{{6 - 6\tan x}}{{\sin x - \cos x}} = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{1 - {{\tan }^2}x}}{{\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =$
$\displaystyle = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{{{\cos }^2}x -{{\sin }^2}x}}{{{{\cos }^2}x\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =$
$\displaystyle = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\cos }^2}x\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =$
$\displaystyle = - 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{\cos x + \sin x}}{{{{\cos }^2}x\left( {1 + \tan x} \right)}} =$
$\displaystyle = - 6 \cdot \frac{{\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}}} {{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}\left( {1 + 1} \right)}} = - 6\sqrt 2 .$ | 1,342 | 3,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2018-22 | latest | en | 0.829326 |
https://im.kendallhunt.com/MS_ACC/students/1/3/18/practice.html | 1,721,252,818,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00262.warc.gz | 286,373,964 | 26,379 | # Lesson 18
Using Long Division
Let’s divide whole numbers.
### Problem 1
Andre and Jada both found $$657 \div 3$$ using the partial quotients method, but they did the calculations differently, as shown here.
1. How is Jada's work the same as Andre’s work? How is it different?
2. Explain why they have the same answer.
### Problem 2
Here is a long-division calculation of $$917 \div 7$$.
1. There is a 7 under the 9 of 917. What does this 7 represent?
2. What does the subtraction of 7 from 9 mean?
3. Why is a 1 written next to the 2 from $$9-7$$?
### Problem 3
Han's calculation of $$972 \div 9$$ is shown here.
1. Find $$180 \boldcdot 9$$.
2. Use your calculation of $$180 \boldcdot 9$$ to explain how you know Han has made a mistake.
3. Identify and correct Han’s mistake.
### Problem 4
Find each quotient.
### Problem 5
The mass of one coin is 16.718 grams. The mass of a second coin is 27.22 grams. How much greater is the mass of the second coin than the first? Show your reasoning.
(From Unit 3, Lesson 15.)
### Problem 6
One micrometer is a millionth of a meter. A certain spider web is 4 micrometers thick. A fiber in a shirt is 1 hundred-thousandth of a meter thick.
1. Which is wider, the spider web or the fiber? Explain your reasoning.
2. How many meters wider?
(From Unit 3, Lesson 15.) | 373 | 1,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.902775 |
https://www.physicsforums.com/threads/calculating-induced-current-in-a-solenoid-loop-circuit.592936/ | 1,720,795,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00530.warc.gz | 694,372,298 | 16,533 | # Calculating Induced Current in a Solenoid-Loop Circuit
• geostudent
In summary, the conversation discussed a problem involving a cylindrical solenoid with a two-turn rectangular loop and a battery. The goal was to determine the magnitude and direction of the current in the loop one microsecond after it was connected to the battery. The conversation included equations for magnetic field, inductance, and current, and the general approach was deemed correct although there may have been some errors in the calculations.
geostudent
## Homework Statement
A cylindrical solenoid 30 cm long with a radius of 8 mm has 400 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 190 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.)
## Homework Equations
B = μ$_{0}$NI /d
L= μ$_{0}$N$^{2}$$\pi$R$^{2}$/d
I = emf/R * [1-e$^{-(R/L)t}$]
emf(induced)= d$\Phi$/dt
## The Attempt at a Solution
I just want to check my reasoning here and get advice on how to approach a problem like this.
1. Since the Current is varying with time, I used I = emf/R * [1-e$^{-(R/L)t}$] to find I at t=1microsecond and got I= .20194 Amperes
2. Used B = μ$_{0}$NI /d to find B=1.76229E-4 Tesla
3. Induced Emf = -d$\Phi$/dt where $\Phi$ = ∫B*dA
I took dA to be the cross-sectional Area of the rectangle
Then induced (EMF*Number of turns in rectangle)/R =I
I keep getting relatively close answers but not correct. I don't think I'm thinking of this n the right way. Where am I at fault?
The general approach looks alright, although I'm not sure what you're doing with the value of the current in the coil or the value of the B field for that particular time. It'll be the rate of change of the B field that you'll need, no?
Even so, the value you're getting for the current looks a bit odd. What values did you calculate for the inductance and the time constant?
## What is induced current in a coil?
Induced current in a coil is the flow of electric charge that is generated in a coil of wire when the magnetic field through the coil changes. This change in magnetic field can be created by moving the coil in a stationary magnetic field, changing the strength of the magnetic field, or changing the orientation of the coil with respect to the magnetic field.
## How is induced current different from direct current?
Induced current is different from direct current in that it is not produced by a battery or other external source of electric charge. Instead, it is created by the changing magnetic field passing through the coil, which induces a voltage that causes the current to flow.
## What is Faraday's law of induction?
Faraday's law of induction states that the induced electromotive force in a closed circuit is proportional to the rate of change of the magnetic flux through the circuit. In other words, the faster the magnetic field changes, the greater the induced current in the coil will be.
## What factors affect the strength of induced current in a coil?
The strength of induced current in a coil is affected by several factors, including the strength of the magnetic field, the number of turns in the coil, the speed at which the coil moves through the magnetic field, and the resistance of the coil.
## What are some practical applications of induced current in a coil?
Induced current in a coil has many practical applications, such as in generators, transformers, and electric motors. It is also used in devices such as metal detectors, induction cooktops, and wireless charging systems.
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1K | 1,064 | 4,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.943681 |
http://forums.wolfram.com/mathgroup/archive/2006/Jun/msg00285.html | 1,726,488,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00680.warc.gz | 10,223,635 | 9,191 | Re: Or in a Select question
• To: mathgroup at smc.vnet.net
• Subject: [mg67108] Re: [mg67066] Or in a Select question
• From: János <janos.lobb at yale.edu>
• Date: Fri, 9 Jun 2006 01:09:14 -0400 (EDT)
• References: <200606080854.EAA12393@smc.vnet.net> <20060608195255.M47050@sarj.ca>
• Sender: owner-wri-mathgroup at wolfram.com
```Hi ggroup,
Some times ago I had to study mathematical logic, but looks like in
my case that knowledge was very much concentrated close to the scalp,
so when the hair started to fall out so did the logic grabbing to it
from underneath :)
I still would like to know why did I get the error message below for
Cases when I replaced the === with the =!=
That was my original reason to ask the list :)
Thanks a lot,
János
On Jun 8, 2006, at 3:52 PM, ggroup at sarj.ca wrote:
> You are missing a basic identity in logic:
>
> Assume A & B are True/False statements, then
>
> !A || !B == !( A && B )
> !A && !B == !( A || B )
>
> As far as I can tell, the logic you've used in the first example is:
> First Pair don't match OR Second Pair don't match
>
> Symbolically,
> A = First Pair match
> B = Second Pair match
> !A || !B
>
> What you want is:
> First Pair don't match AND Second Pair don't match
>
> Symbolically,
> A = First Pair match
> B = Second Pair Match
> !A && !B
>
> Depending on how you like to think about these things, it may be
> more obvious
> to write the equivalent of the last example:
> Neither the first nor the second pair match
>
> Symbolically
> A = First pair match
> B = Second pair match
> !(A||B)
>
> Hope that helps!
>
>
> On Thu, 8 Jun 2006 04:54:07 -0400 (EDT), János wrote
>> Hi,
>>
>> I have a list
>>
>> lst={a,b,c}
>>
>> I make another list from it the following way:
>>
>> In[2]:=
>> pp = Partition[Tuples[lst,
>> 2], 2, 1]
>> Out[2]=
>> {{{a, a}, {a, b}},
>> {{a, b}, {a, c}},
>> {{a, c}, {b, a}},
>> {{b, a}, {b, b}},
>> {{b, b}, {b, c}},
>> {{b, c}, {c, a}},
>> {{c, a}, {c, b}},
>> {{c, b}, {c, c}}}
>>
>> From here I would like to select all the elements whose sublists
>> contain only different elements. So my "logical" selection was:
>>
>> In[54]:=
>> Select[pp,
>> #1[[1,1]] =!= #1[[1,2]] ||
>> #1[[2,1]] =!= #1[[2,
>> 2]] & ]
>> Out[54]=
>> {{{a, a}, {a, b}},
>> {{a, b}, {a, c}},
>> {{a, c}, {b, a}},
>> {{b, a}, {b, b}},
>> {{b, b}, {b, c}},
>> {{b, c}, {c, a}},
>> {{c, a}, {c, b}},
>> {{c, b}, {c, c}}}
>>
>> Well, that did not do any damage to the list. After some time I came
>> up with this one:
>>
>> In[49]:=
>> Complement[pp, Select[pp,
>> Xor[#1[[1,1]] =!=
>> #1[[1,2]],
>> #1[[2,1]] =!= #1[[2,
>> 2]]] & ]]
>> Out[49]=
>> {{{a, b}, {a, c}},
>> {{a, c}, {b, a}},
>> {{b, c}, {c, a}},
>> {{c, a}, {c, b}}}
>>
>> That looks OK, but also looks too complicated. Why my "logical" one
>> does not work here?
>>
>> Interestingly if I just use either the left or right side of the Or,
>> that partial select is working. For example:
>>
>> In[65]:=
>> Select[pp, #1[[1,1]] =!=
>> #1[[1,2]] & ]
>> Out[65]=
>> {{{a, b}, {a, c}},
>> {{a, c}, {b, a}},
>> {{b, a}, {b, b}},
>> {{b, c}, {c, a}},
>> {{c, a}, {c, b}},
>> {{c, b}, {c, c}}}
>>
>> Now if I try with Cases and conditional pattern matching then the
>> selection for sublists with identical elements works:
>>
>> In[98]:=
>> Cases[pp, {u_, v_} /;
>> u[[1]] === u[[2]] ||
>> v[[1]] === v[[2]]]
>> Out[98]=
>> {{{a, a}, {a, b}},
>> {{b, a}, {b, b}},
>> {{b, b}, {b, c}},
>> {{c, b}, {c, c}}}
>>
>> If I change here the === to =!=, then I do not get again that I
>> expect:
>>
>> In[103]:=
>> Cases[pp,
>> (({u_, v_} /; u[[1]]) =!=
>> u[[2]] || v[[1]]) =!=
>> v[[2]]]
>> From In[103]:=
>> \!\(\*
>> RowBox[{\(Part::"partd"\),
>> ":", "\<\"Part specification \\!\\(u \[LeftDoubleBracket] 2 \
>> \[RightDoubleBracket]\\) is longer than depth of object. \
>> \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\
>> \\\", ButtonFrame->None, ButtonData:>\\\"General::partd\\\"]\\)
>> \"\>"}]\) From In[103]:= \!\(\* RowBox[{\(Part::"partd"\), ":",
>> "\<\"Part specification \\!\\(v \[LeftDoubleBracket] 2 \
>> \[RightDoubleBracket]\\) is longer than depth of object. \
>> \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\
>> \\\", ButtonFrame->None, ButtonData:>\\\"General::partd\\\"]\\)
>> \"\>"}]\) Out[103]= {} From In[104]:= Part::"partd":"Part
>> specification \!\(u \[LeftDoubleBracket] 2 \ \[RightDoubleBracket]\)
>> is longer than depth of object. \ \!\(\*ButtonBox[\"More\[Ellipsis]
>> \",
>> ButtonData:>\"General::partd\"]\)" From In[104]:= Part::"partd":"Part
>> specification \!\(v \[LeftDoubleBracket] 2 \ \[RightDoubleBracket]\)
>> is longer than depth of object. \ \!\(\*ButtonBox[\"More\[Ellipsis]
>> \",
>> ButtonData:>\"General::partd\"]\)"
>>
>> if I change =!= to only != then I still do not get that I expect:
>>
>> In[108]:=
>> Cases[pp, {u_, v_} /;
>> u[[1]] != u[[2]] ||
>> v[[1]] != v[[2]]]
>> Out[108]=
>> {}
>>
>> Obviously I am not GETting something here :)
>>
>>
>> János
>> P.S. It is 5.1 on OSX 10.4.6. I know that Or evaluates in a non-
>> traditional way and looked the Appendix - that is how I ended up with
>> Xor.
>>
>> ----------------------------------------------
>> Trying to argue with a politician is like lifting up the head of a
>> corpse.
>> (S. Lem: His Master Voice)
>
>
>
----------------------------------------------
Trying to argue with a politician is like lifting up the head of a
corpse.
(S. Lem: His Master Voice)
```
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• Next by thread: Re: Or in a Select question | 1,998 | 5,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.879624 |
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Re: Simple Logic Problem with Perl
by JamesNC (Chaplain)
on Oct 19, 2005 at 23:19 UTC ( #501472=note: print w/replies, xml ) Need Help??
in reply to Simple Logic Problem with Perl
Here's my stab at it:
```use strict;
use warnings;
use Data::Dumper;
my \$sample1 = ['str1 str2 1',
'str3 str4 2',
'str5 str6 3',
'str7 str8 4',
'str9 str10 5'];
my \$sample2 = ['str1 str2 0',
'str3 str4 4'];
my \$sample3 = ['str1 str2 3',
'str3 str4 4'];
get_column(\$sample1);
sub get_column
{
my \$ar = shift;
my \$all;
foreach my \$str ( @{\$ar} )
{
my @ar = (split " ",\$str);
push @{\$all}, [ @ar ];
}
my @clean;
my \$diff;
foreach my \$i ( 1..\$#{\$all} )
{
\$diff = abs(\$all->[\$i][2]- \$all->[\$i-1][2] );
print "\$i\t\$diff\n";
if ( \$diff == 1 )
{
push @clean, \$all->[\$i-1][0];
}
if ( \$diff > 1 )
{
push @clean, \$all->[\$i-1][0], \$all->[\$i-1][1], \$all->[\$i][
+0], \$all->[\$i][1];
}
if( \$diff == 1 && \$i == \$#{\$all} ){
push @clean, \$all->[\$i][0], \$all->[\$i][1];
}
}
print Dumper \@clean;
return ;
}
JamesNC
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# A university spent \$1.8 million to install solar panels atop a parking garage. These panels will have a capacity of 300 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the
A university spent \$1.8 million to install solar panels atop a parking garage. These panels will have a capacity of 300 kilowatts (kW) and have a life expectancy of 20 years. Suppose that the discount rate is 30%, that electricity can be purchased at \$0.10 per kilowatt-hour (kWh), and that the marginal cost of electricity production using the solar panels is zero. Hint: It may be easier to think of the present value of operating the solar panels for 1 hour per year first. Approximately how many hours per year will the solar panels need to operate to enable this project to break even? 12,667.14 28,953.46 Ο Ο Ο Ο 18,095.91 16,286.32 If the solar panels can operate only for 16,286 hours a year at maximum, the project break even. Continue to assume that the solar panels can operate only for 16,286 hours a year at maximum. In order for the project to be worthwhile (i.e., at least break even), the university would need a grant of at least
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Blank 2 options: \$234,041.05 or \$126,022.11 or \$180,031.58
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A 358-kg boat is sailing 15.9° north of east at a speed of 1.80 m/s. Thirty seconds later, it is sailing 36.6° north of east at a speed of 3.68 m/s. During this time, three forces act on the boat: a 29.7-N force directed 15.9° north of east (due to an auxiliary engine), a 23.9-N force directed 15.9° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east.
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# LectureNotes28 - ECS 120 Lesson 28 – Approximation...
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Unformatted text preview: ECS 120 Lesson 28 – Approximation Algorithms, Randomized Algorithms Oliver Kreylos Wednesday, June 6th, 2001 In our discussion of time complexity, we have seen that there are problems that are solvable in principle, but require so much time and/or space to run, that they cannot in practice be solved for any but the smallest input sizes. But what should one do if facing such a problem in practice? Often, there are ways to get around the prohibitive time requirement by either making the problem slightly easier, or by accepting a solution that might be wrong in very few cases. Today we will address both these approaches. 1 Approximation Algorithms As an early example of an NP-complete problem, we have seen the VERTEX- COVER problem: Given a graph G = ( V,E ), is there a vertex cover, i. e., a subset of nodes, that touches all edges in E and contains not more than k vertices, where k is a given constant? Posed as a language, this problem becomes VERTEX-COVER = h G,k i G has a vertex cover of at most k vertices Typically, this problem is asked in another form: Instead of asking whether some vertex cover exists, the task is to find the smallest possible vertex cover: Given a graph G = ( V,E ), find a vertex cover of G having the smallest possible number of vertices. Typically, optimization problems like this one are even more difficult to decide than the related yes/no problems: VERTEX- COVER is NP-complete, but MIN-VERTEX-COVER is NP-hard, i. e., it is not even in NP itself. 1 But the problem can be solved quite easily if the requirement of finding the optimal solution is relaxed: In practice, it is often sufficient to find a solution that is not much worse than the theoretical best solution. We can formalize this notion by the following definition: Definition 1 ( k-optimal Approximation) If P is a minimization prob- lem, and A is a polynomial-time approximation algorithm for P that always returns a solution that is not more than k times optimal, then A is a k- optimal approximation algorithm for P . The following algorithm is 2-optimal for MIN-VERTEX-COVER: 2-MIN-VERTEX-COVER On input h G i , where G = ( V,E ) is a graph, 1. Find an unmarked edge e ∈ E that does not touch, i. e., does not share a vertex with, any marked edge in E . If there is no such edge, go to step 4. 2. Mark the found edge e . 3. Go to step 1. 4. Output all vertices touching marked edges. This algorithm definitely runs in polynomial time, and it also produces a cor- rect vertex cover: If there were any edge in E that is not covered by a selected vertex, it cannot share any vertices with any marked edge. But then it would have been marked by step 2 of the algorithm, and both vertices touched by it would have been included in the vertex cover. That is a contradiction....
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Previous articleOnline Casino Activities Galore! | 556 | 2,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | longest | en | 0.955768 |
http://en.wikipedia.org/wiki/Balaban_10-cage | 1,433,277,926,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036266.4/warc/CC-MAIN-20150601214356-00109-ip-10-180-206-219.ec2.internal.warc.gz | 66,260,152 | 9,513 | # Balaban 10-cage
Balaban 10-cage
The Balaban 10-cage
Named after A. T. Balaban
Vertices 70
Edges 105
Diameter 6
Girth 10
Automorphisms 80
Chromatic number 2
Chromatic index 3
Properties Cubic
Cage
Hamiltonian
In the mathematical field of graph theory, the Balaban 10-cage or Balaban (3-10)-cage is a 3-regular graph with 70 vertices and 105 edges named after A. T. Balaban.[1] Published in 1972,[2] It was the first (3-10)-cage discovered but is not unique.[3]
The complete list of (3-10)-cage and the proof of minimality was given by O'Keefe and Wong.[4] There exists 3 distinct (3-10)-cages, the other two being the Harries graph and the Harries-Wong graph.[5] Moreover, the Harries-Wong graph and Harries graph are cospectral graphs.
The Balaban 10-cage has chromatic number 2, chromatic index 3, diameter 6, girth 10 and is hamiltonian. It is also a 3-vertex-connected graph and a 3-edge-connected graph.
The characteristic polynomial of the Balaban 10-cage is : $(x-3) (x-2) (x-1)^8 x^2 (x+1)^8 (x+2) (x+3) (x^2-6)^2 (x^2-5)^4 (x^2-2)^2 (x^4-6 x^2+3)^8$. | 373 | 1,065 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-22 | latest | en | 0.835854 |
http://msyangmath.com/math-10/math-10-practice-midterm-2014/ | 1,618,213,046,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00179.warc.gz | 64,752,713 | 8,700 | ## Math 10 Practice Midterm 2014
Multiple Choice:
Part A:
1D 2E 3D 4B 5D 6D 7D 8A 9C
10C 11A 12A 13A 14B 15B
Part B:
1A 2A 3B 4B 5D 6C 7C 8A 9A 10C 11A 12D
13A 14B 15C 16A 17B 18B 19A 20A 21B 22A
Short answers:
Part C:
1) 78 2) 339 3) 45 4) 440 (or 441 depending which conversions you use) 5) 9 | 176 | 307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | latest | en | 0.251708 |
https://se.mathworks.com/matlabcentral/cody/problems/405-back-to-basics-15-classes/solutions/2116622 | 1,590,630,851,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00473.warc.gz | 541,331,836 | 15,619 | Cody
# Problem 405. Back to basics 15 - classes
Solution 2116622
Submitted on 4 Feb 2020 by Takeyoshi Terui
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### Test Suite
Test Status Code Input and Output
1 Pass
x = @sin; y_correct = 'function_handle'; assert(isequal(input_class(x),y_correct))
2 Pass
x = {1}; y_correct = 'cell'; assert(isequal(input_class(x),y_correct))
3 Pass
x = [1]; y_correct = 'double'; assert(isequal(input_class(x),y_correct))
4 Pass
x = uint32(1); y_correct = 'uint32'; assert(isequal(input_class(x),y_correct))
5 Pass
x = 'abcd'; y_correct = 'char'; assert(isequal(input_class(x),y_correct)) | 206 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-24 | latest | en | 0.379765 |
http://anas.pk/page/3/ | 1,513,442,763,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588294.67/warc/CC-MAIN-20171216162441-20171216184441-00166.warc.gz | 12,997,087 | 10,630 | ## C++ Program to Separate Digits of a Positive Integer
Suppose you want to write a program in C++ which will take a positive integer from the user and output its individual digits. For example, if the input provided is 12345 then it will print 1,2,3,4 and 5 separately as integers. In this article we shall look at a few different ways of accomplishing exactly this.
First of all, look at the following program. It is probably the simplest way of separating digits of an integer. It makes use of the fact that when the division operator (/) is used with two integer operands then it gives us only the whole part of the quotient and the remainder operator (%) gives us the remainder when dividend is divided by the divisor. In this particular case, we use 10 with both these operators as divisor. When we use it with the remainder operator we get the last digit of the number entered by the user and when we use it with the division operator then we get the original number without that last digit. In this way, repeating these two steps until the original number becomes zero, we get each individual digit of the original number form right to left i.e. from least significant to most significant digit. Continue reading C++ Program to Separate Digits of a Positive Integer
## How to enable Welcome Screen in Adobe Illustrator CS5
Default setting in Adobe Illustrator CS5 is to hide the Welcome Screen (Well, I don’t know why 🙂 ?). However, some people find it really helpful in their day to day work flow to have it open when no other document is open for editing. So, if you are one of those people then don’t worry. It is extremely easy to enable it so that it will show every time you start illustrator until you don’t open a document or create a new one. Also it will reappear if at sometime you close all of the open documents in your workspace. To enable it, follow the two simple steps given below:
1. In the menu bar click on Help > Welcome Screen, with no document open. The welcome screen will appear in the middle of the window.
2. Click once on the check box besides “Don’t show again” to un check it in the lower left hand corner of the welcome screen to enable it for future. Continue reading How to enable Welcome Screen in Adobe Illustrator CS5
## Does traffic exchange websites help in affiliate marketing?
I signed up on some traffic exchange websites like hit2hit, leadclub etc. a few days ago and now I am getting a good amount of traffic on my personal website which is in turn helping my website in SEO. But I am asking this question because I have noticed that on these traffic exchange platforms, most people click on the link of website, wait it to load while they continue to do their other tasks on their computer. After a while when the minimum time required to get credit passes, they go back to their browser and confirm their visit of that website by clicking on the appropriate image or by following any other confirmation method of that particular traffic exchange website.
In short, I think that traffic obtained from these sources doesn’t give us “readers”. Am I right in my imagination? If yes, then do you think some one from such kind of traffic is going to buy something from a company if I become their affiliate and put my affiliate link in rotation on such a website?
And most importantly if anyone of you has personal experience of affiliate marketing on traffic exchange websites then please share some of your statistics with me. For example, on average, how much visits you had to deliver to make a sale? I mean a rough hits/sale ratio If you don’t mind.
Thank you very much in advance!!
## Recursive C++ Program to Convert a Decimal Integer into Binary Number
Here is a very simple and recursively written C++ program to convert a decimal integer number into its binary equivalent. However it returns the binary equivalent in the form of a string of characters but you can very easily write an other function to convert this string into an actual binary number as well. Continue reading Recursive C++ Program to Convert a Decimal Integer into Binary Number | 846 | 4,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-51 | latest | en | 0.887086 |
http://www.docstoc.com/docs/104587999/Experiments-In-Optics | 1,430,205,127,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246660743.58/warc/CC-MAIN-20150417045740-00134-ip-10-235-10-82.ec2.internal.warc.gz | 450,656,790 | 53,829 | Experiments In Optics by changcheng2
VIEWS: 22 PAGES: 17
• pg 1
``` First Year Laboratory
Experiments
in optics
2009 - 2010
Physics, Imperial College London v. 09/2007 (Damzen)
1
Experiments in Optics
Introduction
Welcome to Experiments in Optics. For the next 2 weeks, you will do a variety of experiments that
illustrate the ray and wave nature of light. Recent years have seen a revolution in optical technology
(e.g. lasers in CD- and DVD-players, telecommunications and medicine), and optics also remains
one of the main diagnostic measurement techniques in a wide range of industrial and research fields.
However, the aims of this part of the lab go beyond learning some optics - in particular, we want you
to appreciate that measurements are never exact and therefore that experiment and "theory" will
never agree precisely. The best we can hope for is that experiment and theory are consistent, bearing
in mind the errors of the experiment and the approximations of the theory.
Everyone knows that measurements have errors but it is easy to forget that theories have their "error"
- i.e. approximations - and this is illustrated in the present topic. In optics, we consider light to be
described by rays or waves or photons, and sometimes by all three in the same experiment!
Physical Model of Light Name given to model
Rays Geometrical Optics
Waves Physical Optics
Photons Quantum Optics
One important thing to learn in any branch of science is what level of approximation is required to
solve a particular problem. For example, quantum optics is meant to provide a complete description
of light but if you tried to use it to design a simple lens you would (probably) never get an answer. It
turns out that an extremely high quality lens can be designed using the concept of rays (geometrical
optics) and some knowledge of the wave model (physical optics). On the other hand, the basic
principle of operation of a laser requires the quantum model - no amount of clever playing with rays
or waves can fully explain the light emission in a laser. In Experiments in Optics, we shall do
experiments that use the models of light that are called Geometrical (Ray) Optics and Physical
(Wave) Optics. Ray and wave pictures of light are fairly straightforward, and sufficiently
comprehensive, to account for most optical phenomena and devices
In Experiments in Optics you will work with a partner. There are 4 sessions (2 per week) and the
following table summarises the schedule of experiments:
Lab Experiment
Session
1 Single Slit Diffraction
2 Two-slit Interference and Gratings
3 Lenses
4 Polarisation
2
Working Practices
Safety
Laser Safety: During the course of this series of experiments, you will use a compact diode laser of
wavelength 670nm (red). The power of this laser is approximately 1 mW. Lasers produce a highly
collimated (parallel) beam of light: the eye could focus this to a very small spot as small as 10 m in
diameter, giving a power density of 1 kW per square cm (but of course only over a tiny area). Retinal
damage may occur before the "blink reflex" closes the eye. Therefore,
NEVER LOOK DIRECTLY INTO THE LASER
NEVER POINT THE LASER AT ANY OTHER PERSON
DO NOT USE MIRRORS TO REFLECT LASER BEAMS AROUND THE LAB
Electrical Safety: In this experiment you will also use various items of mains-powered electrical
equipment: this should not be tampered with in any way. If you are in any doubt regarding the safety
of any piece of equipment or any experimental procedure, you should consult a demonstrator before
proceeding.
Trip Hazards: You are often working in darkened conditions, so it is especially important that bags
and coats are stowed thoughtfully and passageways around benches are kept clear.
Start by writing the day and time, and title of the experiment. As you do each part of the lab, it is
essential to keep a clear written record in your lab notebook. However, do not spend a long time
engrossed in your lab book - never more than 10 minutes at a time - remember this is a practical
laboratory, not an exercise in writing. Every 10 or 15 minutes, record what you have done in your
book. Write clearly, draw lots of clearly labelled sketches, write down any conclusions you have
drawn or decisions you have made. It is vital that you describe or draw what you actually see. If a
trace on a screen has lots of bumps, wiggles and spots then draw them. Don't draw what you think
you might get from a perfect experiment: you might be throwing away important details.
Working Practice
We expect you to work hard in the laboratory. Concentrate on the experiments and work
methodically. If you get stuck, first talk the problem through with your partner; if you cannot resolve
the issue after 5 minutes, ask a demonstrator.
Keep Optical Components Clean!
Do not touch the surface of an optical component with your bare hands. If lenses, gratings, polarisers
etc get covered with fingerprints and dirt, they will scatter light and, when quantitative measurements
are being made, may produce quite spurious effects.
3
INTERFERENCE and DIFFRACTION
When two or more waves are superimposed, the resultant amplitude (and hence intensity) depends on
the amplitudes and phases of the component waves. This process is known either as interference or
diffraction, depending upon the geometry of the particular experiment. When there are a finite
number of waves (e.g. light from two very narrow slits or two distinct optical paths) we talk of
interference but if there is a continuum of waves (e.g. from light passing through a single slit of non-
negligible width) then we talk of diffraction.
The aim of this set of experiments is to observe the interference and diffraction patterns obtained
when a laser illuminates various apertures and to compare them with those predicted according to the
wave theory of 1ight. Analysis of diffraction close to the diffracting aperture, known as near-field
diffraction, can become quite mathematically complex. A much simpler case is the so-called far-
field diffraction regime in which the plane of observation is very far from the aperture (ideally at
infinity) – as seen in Figure 1a. Due to the physical size of optical bench this is not always practical
and an equivalent effect is achieved by making all observations at the focal plane of a lens – as seen
in Figure 1b. From Figure 1b it is noted that rays diffracted at angle are observed on the viewing
screen at a distance x from the lens axis where = x/f (for small angles).
diffracted rays at angle parallel rays meet at a
a) b)
meet at infinity point in focal plane
input x x
light
~
f
L very distant Lens, f observing
diffracting object observing screen
screen =x/f (small angles)
Figure 1: Input light beam (e.g. from laser) is diffracted into rays at new angles. The far-field diffraction
pattern can be observed a) on a very distant screen (ideally at infinity) or b) in the focal plane of a lens.
Experiment 1: Diffraction by a Single Slit & by a Strand of Hair
There are basically two parts to this 3 hour session. First, you will investigate the diffraction pattern
of a single slit. Secondly you will examine the diffraction pattern of a strand of hair and based on
single slit diffraction and Babinet‟s principle determine its diameter.
List of Equipment
1.5 m triangular profile optical bench, with 5 saddles
Diode laser, post-mounted, with stabilised 5 V power supply
bi-convex lens f =500 mm, bi-convex lens f=1000 mm
holder for slides
set of slides (diffracting slits and gratings)
holder for observing screen, post-mounted, with white card
precision translation stage with photodiode (+power supply and portable voltmeter)
metre rule
Single Slit Diffraction Theory: The theoretical derivation of the intensity of light in the “far-field”
diffraction pattern of a single slit is found by summing the amplitudes of light from each elementary
point on the slit. This procedure follows from Huygens' Principle and the amplitude in this case is the
electric field of light and has both magnitude and phase. The intensity pattern is found by squaring
the amplitude pattern. For a slit of width a, the intensity I(x) of the diffraction pattern as a function
of distance x in the observation plane is given, for small angles = x/f, by
4
sin ax
2
I( x ) I 0 ax f
f
(1)
where I0 is the intensity at the centre, is the
wavelength and f is the focal length of the focussing 1.0
lens. Note that the quantities a, x, and f in Eq. (1) all
0.8
have dimensions of length and should all be expressed
Intensity
in the same units, e.g. mm. Figure 2 shows the single 0.6
slit diffraction pattern.
0.4
Figure 2: The single slit diffraction pattern of Eq.(1), with
the scaling constant I0=1. The horizontal axis is in units of
the quantity x .
a 0.2
f
0.0
-10 -5 0 5 10
Distance
From equation (1), the zeros of intensity occur at x-
coordinates xm given by
axm
f m
m= 1, 2, 3,...
f
which can be re-arranged as xm m m 1, 2, 3 .... (2)
a
EXPERIMENT 1a: SINGLE-SLIT DIFFRACTION.
Figure 3 shows the basic set-up for the experiments on interference and diffraction. In these
experiments, the light source is a red diode laser emitting at a mean wavelength of = 670 1 nm.
This laser is powered by a special stable 5 volt power supply. [This is the type of laser device used in
bar-code readers at Supermarket check-out counters and within DVD players. The basic diode laser
inside the housing is an ultra-compact device and it is easy to forget that it may cause permanent
damage if pointed at the eye. Therefore - NEVER LOOK DIRECTLY INTO THE LASER -
NEVER POINT THE LASER AT ANY OTHER PERSON.]
lens, Photodiode/slit on
focal length f translation stage
LASER
f
diffracting object voltmeter
Figure 3: Basic arrangement for measurement of the intensity of interference and diffraction patterns.
The object is placed close to the laser output. Use the f = 0.5 m lens for measurements on the
photodiode assembly. Use the f=1m lens for visual observations on a white screen.
For this experiment, use the slide marked "Single Slit". The exact width of this slit is not known but
it is about 60 m. To measure the intensity a photodiode is used together with a narrow slit (=100
m) in front of it. The photodiode output is a voltage that is linearly proportional to the light
intensity. The slit/photodiode are provided as a single unit that attaches to a precision translation
slide (the slit/photodiode can be rotated by loosening the screw on the mounting bracket of the
translation slide). The translation slide has a total range of 25 mm. Make sure that the photodiode slit
is vertical (loosen the screw and rotate the slit/photodiode if necessary). Use the bi-convex lens with
5
f=500mm and ensure the distance between the centre of the lens and the photodiode slit is accurately
located at the focal length of the lens to observe the far-field diffraction pattern. Looking by eye,
check that the centre of the diffraction pattern is approximately coincident with the photodiode slit
when the translation stage is set at its mid-point (about 12.5 mm) (make minor adjustments by tilting
the laser and re-centring the “single slit” in its holder, if necessary). You should then measure the
intensity over the full range of the micrometer movement.
First, find the approximate centre of the diffraction pattern (the largest intensity) and choose the
scale on the voltmeter (either 2 V FSD or 200 mV FSD).
Second, get a feel for how often you will have to sample the diffraction pattern to get a good
representation of it: every 1 mm, every 0.5 mm, every 0.25mm...? A little preliminary work here
will save you a lot of effort later. [Make sure you know how to read the micrometer: ask a
demonstrator if you have any doubt].
Finally, make your set of measurements. You will have probably already noticed that background
illumination affects your measurements so try to ensure that this remains constant and/or is very
much reduced during the complete run. You may notice that the meter reading fluctuates more-
or-less randomly. This is called "noise": what are the causes of this noise? How might you reduce
the effects of noise physically or mathematically in your results?
Plot your measured intensity profile (by hand, on graph paper) as a function of x for x = 0 to 25 mm.
Find the position of the first zeroes (m = 1) and hence estimate the slit width.
Experiment 1b: DETERMINING THE DIAMETER OF A HAIR
Babinet's Principle. Suppose that the amplitude (electric field) of the light wave at a certain point x
in a diffraction pattern of a certain object is A1(x). Now, replace the object by its complementary
form: the complementary form can be pictured as the (photographic) "negative", i.e., the
complementary object is transmitting where the origina1 is opaque, and is opaque where the original
is transmitting. For example, the complementary object corresponding to a slit is a thin opaque strip
or wire. (i.e. a slit has transmission function T1(x‟)=1 for |x‟|<a/2 and T1=0 otherwise, the
complementary wire has T2(x‟)=0 for |x‟|<a/2 and T2=1 otherwise). Let the complementary screen
produce an amplitude A2(x) at point x in the diffraction pattern. Babinet's principle states that
A1(x) + A2(x) = A(x) (3)
where A(x) is the amplitude of the wave at point x in the diffraction pattern when there is no
diffraction screen present at all. This theorem results from the reasoning that no screen at all (T=1
everywhere) = sum of screen (T1(x‟))+ complement of screen(T2(x‟)) and hence the sum of the two
complementary screens‟ diffraction patterns = the diffraction pattern with no screen. Hence,
according to Babinet, in regions where A(x) = 0, A1(x) = -A2(x). Since intensity is proportional to
the square of the amplitude (i.e. I(x)=|A(x)|2), the intensity diffraction pattern of the object is
(surprisingly!?) the same as the complementary object in these regions.
In the case of the set-up shown in Fig.3, observe the distribution of light at the observing screen
when there is no diffracting object (i.e. I(x)=|A(x)|2) with the 1m focal length lens in place and record
this in your labbook. Now tape a hair over one of the blank slides provided and observe its
diffraction pattern. Use Eqs.(1-3) to explain the form of the pattern you see and also to find the
diameter of the hair. Quote an error value on the diameter. Try using a micrometer to measure the
any assumptions to produce your estimate of the diameter of the hair using the diffraction method? If
so, make sure these are stated clearly in your lab book.
6
Experiment 2: YOUNG’S DOUBLE SLIT INTERFERENCE &
DIFFRACTION GRATINGS
In this 3-hour laboratory session, you will explore more complex features of interference and
diffraction. In the first part you will investigate a two-slit diffraction pattern. In the later part you
will explore how gratings, made from multiple slits, diffract light.
Use the double slit slide, labelled "Slit 2" as the object and place it immediately after the laser output
as in Figure 3. The two slits are each supposedly of the same width (nominally that of the single slit
in Expt.l), separated by roughly 4 times the width. Observe the pattern on the screen using the 1.0m
focal length lens and observing on the screen placed 1m from the lens. Draw a labelled sketch (with
a scale marked) of this diffraction pattern. Using the scale on a ruler, or otherwise, make a crude
estimate of the spacing between successive minima of intensity.
P
r1
x
S1
r2
d
I
S2 dsin
L
Figure 4: Geometry for Young’s double slit interference phenomenon: for large L, the path
difference to point P is approximately d sin , where d is the slit separation.
(ii) Figure 4 shows the geometry used to derive the form of the interference pattern produced by two
infinitely narrow (i.e. idealised) slits. If the slits are infinitely narrow, then the difference between
the paths S2P and S1P is (for a distant screen) given by
= r2 - r1 d sin
Since the total light pattern at point P is found by summing the contributions from the two slits (with
due account of the phases) the positions of the bright and dark-fringes are given by constructive or
destructive interference conditions:
MAXIMA: d sin = m
1
MINIMA: d sin = m
2
where m= 0, 1, 2, 3,....
In the experiment, we observe in the focal plane of the focussing lens f, and for small angles, sin
= x/f , so we can write the position of the bright and dark fringes as:
f
MAXIMA: x m m (4a)
d
1 f
MINIMA: x m m (4b)
2 d
where m= 0, 1, 2, 3,....
7
The spacing between fringes (between successive maxima or successive minima) is therefore f/d.
For slits of negligible width, the form of the intensity distribution of the interference pattern is, for
small angles:
dx
I( x ) 4I 0 cos 2 (5)
f
where I0 is the intensity due to a single source and is assumed to be uniform across the screen.
Does equation (5) give a good description of the pattern you observed in part (i)? Describe any
In fact, when the slits have a finite width a, the intensity is equal to the product of the distribution of
the single-slit equation (1) with equation (5) and may be written as:
sin ax
2
dx
I( x ) 4I0 axf cos2
f
(6)
f
where a is the slit width and d is the separation of the slits. The pattern is sketched in Figure 5 for the
case where d = 4a.
1.0
0.8
Intensity
0.6
0.4
0.2
Figure 5: Young’s double slit fringes for the case where the separation of the slits equals 4 times
0.0 to
their widths (assumed -10 be equal). -5 0 5 10
Distance
(iii) In this section you are asked to verify equation (4b) for the positions of the minima. You should
use the slit/photodiode assembly to locate the minima of intensity. Use the arrangement of Figure 3
with the double slit as the diffracting object and the f = 0.5 m focussing lens. Make sure that the
distance from the lens to the measuring slit on the photodiode stage is as close as possible to 500
mm, to within a tolerance of 2mm. Start near the centre of the pattern and work your way out,
searching for each minimum: note the micrometer reading x for as many orders as you can (at least
5). You will therefore end up with a table with the following headings:
Order m Value of xm
Use the computer program CurveExpert to draw a graph of the values of xm against order m. The
graph should be a straight line whose slope is the separation between orders and according to the
theory it should equal f/d. Compute the slope and standard error of the slope of the least squares
best-fit straight line through the experimental data. Instructions for the use of this program are given
on colour cards near the computers.
Using = 670 1nm, d =230 20m and f = 500 1 mm, calculate the theoretical value of the
gradient and its standard error. Compare the experimentally determined and calculated values and
their standard errors. Have you verified the theory to within the limits of random experimental error?
Experiment 2(b): N-slit diffraction and diffraction gratings
You are provided with slides marked “multiple slits 3”, “multiple slits 4”, etc up to 6 slits. Take the
3-slit slide and observe the diffraction pattern formed by it carefully. You will probably find it a lot
8
easier to see the pattern if you use the eyepiece provided to magnify the patterns as shown in Fig. 6.
Make a clear record in your lab book. Do the same for the slide marked 4-slits.
bi-convex lens observing
f=1000mm eyepiece screen
LASER
f
diffracting object
Figure 6: Arrangement for observation of magnified interference and diffraction patterns using
an eyepiece: observations are made on the screen.
Figure 7 shows the interference patterns for 2, 3 and 4 infinitely narrow slits, all on the same scale. Is
this what you see (be honest!)?
18
16
14
N=4
12
Intensity (I /I0 )
10 N=3
8
6
4
N=2
2
0
-4 -2 0 2 4
Distance
Figure 7: The intensity interference patterns for 2, 3 and 4 infinitely narrow slits of equal separation.
Now quickly look at the interference patterns for 5-slits and then 6-slits. Can you guess what the
pattern will look like as the number of slits N ?
A diffraction grating consists of a very large number of slits. You are provided with three coarse
gratings, labelled “1”, “2” and “3”. Observe the diffraction pattern from each one carefully. You
might find it useful to compare the patterns with that from the two slit object. For each grating find
1. The width a of each slit (they are nominally all the same).
2. The centre-to-centre separation d of adjacent slits.
9
Experiment 3. IMAGING WITH LENSES
The vast majority of instruments that incorporate some optics have simple items such as lenses and
mirrors. The function of these elements is usually to provide even, or controlled, illumination of an
area (the reflector behind a torch bulb must be one of the "simplest" examples of this) or to relay an
image of an object to a specified plane or surface (e.g. a camera lens). In this series of experiments,
you will study the imaging properties of lenses using the concepts of geometrical optics. First we
start by imaging with a single lens and test the lens formulae that predict perfect images, free of any
kind of defect or aberration. Finally we will examine the quality of images using a resolution test
chart.
Simple Theory of Thin Lens Imaging. Figure 8 is a ray diagram illustrating the imaging action of a
lens. The object distance s, the image distance s‟ are all measured from the centre of a thin
converging lens. (A similar diagram can be made for a diverging lens but we will not be using one in
this experiment so will not consider it here). The diagram is an oversimplification and we will later
consider some of the features of real lenses, including the quality of the image formation.
F‟ I
O F
s s‟
f
Figure 8: A lens forms an image (I) of an object (O) where s is the object distance and s’ is the
image distance.
For small angles, the "thin lens formula" can be derived from this diagram:
1 1 1
(7)
s s f
with the following sign convention:
Light travels from left to right: s is positive (real) on the left, negative (virtual) on the right: s‟ is
positive on the right, negative on the left: f is positive for a converging lens, negative for a
diverging lens. (You should be aware that there are other sign conventions).
The linear magnification m is defined, in this sign convention, as
s
m . (8)
s
List of Equipment
Incandescent light source, post-mounted, with power supply. Maximum voltage is 5 V!
Plano-convex lens, nominal focal length 100 mm
Plano-convex lens, nominal focal length 160 mm
Achromatic doublet lens, nominal focal length 160 mm
Ground glass viewing screen, 40 mm diameter
Set of object slides including a resolution test chart slide and green filter slide
CCD camera
10
Plane mirror, 40 mm diameter
Experiment 3a: Testing the lens Formula
The lens formula, Eq.(7), can be tested and used to find the focal length, by plotting the reciprocal of
the image distance, 1/s‟, against the reciprocal of the object distance, 1/s. The intercept of this graph
on the 1/s‟ axis is the reciprocal of the focal length (1/f).
f image, observed on
object ground-glass screen
Light
source
s s‟
Figure 8: Experimental system for imaging with a positive lens.
Figure 8 shows a simple set-up for imaging with a positive focal length lens. Switch on the light
source – the voltage on the supply should be at voltage 5V (max.) - and use the slide with a pin-hole
as an object placed in the slot-holder in front of the light source. Use the plano-convex lens of
nominal focal length 100mm. Observe the image of the object slide on the ground-glass screen by
looking at the back of the screen and adjusting its position for sharpest image. It may be difficult to
see the image until you are close to the image plane and you may need to centre the image on the
screen by adjusting the height of the lens.
Use your ruler to measure the s and s‟ values over as wide a range as the length of the bench allows.
Start with the smallest possible separation (s+s‟) between the object and image planes (how is this
distance related to f?) and then use 5 or 6 more object-to-image distances, using the whole length of
the optical bench. For each object-to-image distance, there will be two positions of the lens that yield
a sharp image, so your graph will consist of 10-12 points.
Now use computer program CurveExpert to plot a graph of 1/s‟ versus 1/s and also to determine the
focal length and its standard error. Instructions on how to use this program are given in Appendix l.
Print out a plot of the graph and paste it in your lab-book.
Experiment 3b: Image Quality and Resolution of Lenses
Previously, we considered imaging with “thin” lenses and within a small-angle geometrical optics
(ray) approximation. For real lenses a fuller treatment needs to consider also the physical optics
(wave) picture. There are two key reasons for loss of quality by imaging with a real lens: i) the
finite diameter D of the lens and ii) lens aberrations due to details of a real (“thick”) lens. Consider
light from a point object (e.g. point object A in Figure 9) that expands as a spherical wave (only the
centre ray is illustrated in Figure 9). Only the part of this wave, that is within the lens diameter, is
imaged. The lens, therefore, acts as a limiting (circular) aperture that will diffract light as well as
imaging it. The result is that the point object A is not imaged to a point but to an image spread
function A‟. For a lens with no aberrations the characteristic angular size of this image is:
1.22 D (9)
for light of wavelength . (The 1.22 is a geometrical factor due to the circular shape of the
diffracting object). As a consequence, if two point objects (e.g. A and B) subtend an angle of <
the two images will be merged and will not be resolved as two separate objects. They are said to be
unresolved. (Resolution is commonly defined as the Rayleigh criterion and you may find it helpful
minimum resolvable object size (x) can be found by using the (small-angle) relationship = x/L.
11
Equation 9 is the “diffraction-limited” resolution of a lens, but lens aberrations further degrade and
spread out the image, and hence lead to poorer resolution. A full discussion of lens aberration is
outside the scope of this experiment but some details of types of aberrations are made at the end of
the experimental section.
A B‟
x
A‟
B L L
Fig.9 Resolution limit of a lens due to its finite diameter D (and aberrations)
The purpose of this experiment is to give you some idea of the imaging quality and resolving power
of lenses. In this experiment, we measure imaging quality using a resolution test based on the three
bar chart. This expresses resolution in line pairs per millimetre. For a line in a three-bar element to
be just resolved gives the relationship: smallest resolvable detail (in mm) = (0.5).(line pairs per mm)-
1
.
One of the slides provided is the three-bar chart. Examine it visually. It consists of 8
groups numbered 0 - 7, with 6 elements per group, an element consisting of two
target patterns of three lines at 90 degrees to each other, the line-to-space ratio being
equal. The number of line pairs per mm equals 1.00 for the largest element on group
0 and increases by 6 2 1.12 per element. The table gives the line pairs per mm for
all elements in groups 1 to 7 in the chart. Look at the chart and try to identify the
positions of the largest groups.
Element Number Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7
1 2.00 4.00 8.00 16.0 32.0 64.0 128.
2 2.24 4.49 8.98 17.95 36.0 71.8 144.
3 2.52 5.04 10.1 20.16 40.3 80.6 161.
4 2.83 5.68 11.3 22.62 45.3 90.5 181.
5 3.17 6.35 12.7 25.39 50.8 102. 203.
6 3.56 7.13 14.3 28.51 57.0 114. 228.
The following two experiments illustrate the use of this resolution chart for assessing the quality of
imaging systems.
(i) Resolution of the eye.
Put the resolution chart slide into the slide holder on the light source and sit in front of the source
with the slide at approximately 250mm, the near distance of the eye. Decide which is the finest
group you can resolve (i.e. just see the bars on the target). You might wish to do two measurements,
one for the vertical bars and a second for the horizontal bars. Your partner should also do this
experiment. Take 3 or 4 independent readings each and find the average resolution, expressed in line
pairs per millimetre.
(If you wear glasses or contact lenses, you might want to try doing the experiment with and without
these).
12
Is your result consistent with equation 9? (Take eye‟s pupil diameter D=2mm and wavelength
=550nm).
(ii) Resolution of simple lenses.
Using the three-bar resolution chart as an object, form an image of the chart directly onto the CCD1
camera with the doublet lens provided (f = 160mm), making sure its flatter surface is towards the
object. The CCD camera should be mounted on the adjustable saddle so that it can be moved easily
in both the vertical and horizontal directions. The magnification required is about 4 (or more), giving
an image of the central groups (2 and higher) that fills the TV screen. [You will need to use eqns 7
and 8]. It can be quite tricky to align and focus the system: in particular, make sure that the doublet is
not twisted and that it is exactly perpendicular to the optica1 axis. It is also important to make sure
there is no stray light from desk lamps, etc. When everything is a1igned properly, you should be able
to see targets in Group 6. Make careful observations of the appearance of the image, both its
resolution and contrast, for the following lenses, keeping the overall object-to-image distance
approximately constant and making sure the lens is perpendicular to the optical axis, i.e. it is imaging
the object on axis.
(a) Doublet, f 160mm, flatter side towards object.
(b) Doublet, f 160mm, more curved side towards object.
(c) Plano-convex singlet, f 160mm, flat side towards object.
(d) Plano-convex singlet, f 160mm, curved side towards object.
Rank the four cases in order of image quality.
An aside about lens aberrations - You will have noted that all lenses have the same diameter so
that the difference in each lens‟ imaging quality must be due to additional aberrations, over and
above the finite diameter of the lens. Full discuss of lens aberration is beyond the scope of this
script, however it is noted that two key types of aberration, spherical and chromatic, will be
important in these cases. The lens formula of eqn 7 assumes thin lenses and small angles.
Spherical aberration arises from „errors‟ due to rays far from axis (and hence larger angles) not
imaging to the same point as rays close to the axis (small angle, or paraxial, rays). Since refraction
occurs at both lens surfaces, the strength of this aberration is more severe when a plano-convex lens
is placed one way around compared to the other.
Chromatic aberration arises from the wavelength dependence of the refractive index (e.g. this
produces the rainbow-like dispersion by a prism). Since the focal length of the lens depends on the
1
CCD stands for Charge Coupled Device and refers to the readout mechanism in the detector chip. The CCD camera has a total image
area of 6.4 by 4.4mm (H x V), within which there are 542 by 582 pixels (i.e. the pixels on the chip are approximately 12.2 by 7.6 m).
13
refractive index it is clear that different wavelengths will image at slightly different positions – and
this can be quite severe for a white light source. For example, if n(blue)>n(red) then f(blue)<f(red)
Blue
The doublet lens is made of two glass materials with refractive index properties that partially
compensate for this aberration.
Red
Blue
14
Experiment 4. POLARISATION OF LIGHT
The purpose of this session is to give you an introduction to the concept of the polarisation of light.
Light from a normal incandescent or fluorescent lamp is usually "unpolarised" whereas laser light is
usually linearly polarised - to understand what these terms mean, we have to consider in some detail
the nature of light waves. Light is a transverse electro-magnetic vector wave. For light travelling in
the +z direction (say) there is a time- and space-varying electric field E(x,t) E 0 cos(t kz) and
magnetic field B(x,t) B 0 cos(t kz) , where E and B are vector quantities, transverse to the
propagation direction (i.e. they lie in the x-y plane). In free space, there is a simple relation between
E and B, so that specifying one of them - usually we choose E - is still a complete description of the
field.
We can think of E as a two-dimensional vector, with components- Ex(z, t) and Ey(z, t) along mutually
perpendicular x and y directions at each point in space (z) and time t. These components oscillate
sinusoidally in space and time, with amplitudes E0,x and E0,y and, in the general case with a phase
difference . Different states of polarisation are the result of different values of E 0,x, E0,y and
particularly , the phase difference. A simple mathematical representation of the field (at z=0,
say) is:
E0, x
E E cos(t )
(10)
0, y
Since the field components oscillate sinusoidally, the resultant field is simply the sum of two simple
harmonic motions of the same frequency applied in perpendicular directions. Consider the case when
Ex and Ey oscillate in phase (=0) and we get linearly polarised light. Figure 10 shows the case of
light linearly polarised (a) along the x-axis, and (b) when it is polarised at angle to the x-axis.
Figure 10 also shows the case in which E0,x = E0,y and (c) = 45 deg (elliptically-polarised light), and
(d) = 90 deg (circularly-polarised light). [In the last case, the circle can be seen to be
circumscribed by summing the perpendicular components Ex=E0cos(t) and Ey=E0cos(t-
900)=E0sin(t)].
a b c d
1 cos( ) E 0 cos(t ) E0 cos(t )
E E0 cos(t ) E
E
0 sin( ) cos(t )
E E0
2 cos(t / 4)
2 sin(t )
Figure 10. Representation of polarised light states. (a) linearly-polarised along x-axis, (b) linearly-
polarised angle to x-axis (c) elliptical polarisation, = 450, (d) circular polarisation, = 900.
In natural light (e.g. light from a blackbody), the amplitudes E0,x and E0,y and the phase difference
all vary randomly in time and therefore no fixed state of polarisation is observed: we say it is
"unpolarised". Laser light is produced in a cavity whose geometry usually forces a single polarisation
state, typically linear polarisation. Light that is initially unpolarised can be made linearly polarised by
a variety of means. In the laboratory experiment, we use sheet "Polaroid" in which long-chain
molecules are aligned in such a way as to transmit only the component of the electric field along one
direction. We refer to this device as a polariser.
Law of Malus. If linearly polarised light of unit amplitude (and hence unit intensity) is incident
on a polariser whose transmission axis is the same as that of the incident light, then the amplitude
15
and intensity of the transmitted wave is also unity. In practice, there are bound to be losses from true
absorption or from surface reflections and the transmitted intensity will be less than unity. A
polariser used to study already polarised light is usually called an analyser (obviously, the same
device can be viewed as a polariser or analyser depending on its use). On the other hand, if the
linearly polarised light has its electric field component perpendicular to the transmission axis of the
analyser, then in principle the amplitude and intensity of the transmitted light is zero. Again, in
practice, it will have some small, non-zero value because of imperfections in the analyser.
If the angle between the direction of the electric field of the linearly polarised light and the
transmission axis of the analyser is , then the transmitted amplitude (for a unit amplitude incident
wave) is cos (consider for example the input field has polarisation as shown in Figure 10b and that
the analyser polariser is aligned to x-axis). The transmitted intensity is cos2 (since intensity is
proportional to square of electric field). This is usually written in the form,
I=I0 cos2
where I0 is the incident intensity of the linearly polarised incident beam. Eq. (20) is known as the
Law of Malus.
List of Equipment
Lathe-type optical bench and fittings used for lenses section
Red LED light source, with power supply
1 linear polariser ("Polaroid') 40 x 40 mm
1 analyser (i.e. another linear polariser) in rotation post mount
Photodiode (as used in diffraction experiments with slit removed) on translation stage and power
supply.
Experiment 7. Verification of Malus' Law
bi-convex lens Photodiode on
f=160mm translation stage
Red LED
light source
linear analyser in
polariser rotation mount
Figure 11: Set up for verification of Malus' Law.
Check that the power supply to the photodiode is switched on before taking any measurements. [This
is essential as the photodiode will give a reading even without power supply on but it will not be
linear with intensity]. Before placing the polariser and analyser in position, use the f = +160mm
focal length singlet lens to focus the central bright spot of the red LED light source onto the
photodiode at a magnification of approximately 0.25 (i.e. use eqn 7 and 8). Place the 40 x 40mm
linear polariser and the analyser in its rotation mount immediately in front of the detector, as shown
in Figure 11. Be careful with the alignment: make sure that the light from the LED is passing
centrally through both the polariser and analyser.
If you rotate the analyser, you will find that the signal on the photodiode varies, and in principle
should be varying as cos2. In order to reduce any background illumination you can make sensible
use of black card and switch off your "reading" light if this is practical (but do not keep switching it
on and off as this is going to affect your neighbours' results). Once you have minimised the
background illumination, record the intensity I as a function of the angle marked on the rotation
mount, for increments of of say 10deg between 0 and 360deg. The angle = + , where is an
16
unknown offset which depends on angle of the transmission axis of the linear polariser (this depends
on how you mounted the polariser).
Plot a graph of I versus . Your graph should be cos2 in form (but not necessarily equal to a
maximum at = 0 deg) on top of a background level. Determine the offset angle and the mean
background level Iback.
Plot a second graph of I – Iback versus cos2 () using the computer and determine the least squares
straight line fit. Have you verified Malus law?
17
```
To top | 10,624 | 41,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-18 | longest | en | 0.926448 |
http://www.drumtom.com/q/2yy-1-y-2-solve-the-de | 1,477,159,984,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719033.33/warc/CC-MAIN-20161020183839-00271-ip-10-171-6-4.ec2.internal.warc.gz | 412,758,374 | 8,913 | # 2yy''=1+(y')^2, solve the DE?
• 2yy''=1+(y')^2, solve the DE?
Solution for Y+2=y equation: Simplifying Y + 2 = y Reorder the terms: 2 + Y = y Solving 2 + Y = y Solving for variable 'Y'. Move all terms containing Y to ...
Positive: 31 %
Solution for Y+2=y equation: Simplifying Y + 2 = y Reorder the terms: 2 + Y = y Solving 2 + Y = y Solving for variable 'Y'. Move all terms containing Y to ...
Positive: 28 %
### More resources
Wolfram Alpha ® Computational Knowledge Engine ... Gaussian blur with radius 2. ... y'' + y = 0. Email ...
Positive: 31 %
Solve the initial value problem y' = (x^2 - y^2) / (xy) with y(2) = 2. Note that this involves a homogeneous differential equation. i know you need to ...
Positive: 26 %
... substitute 2y2 º2for x in Equation 1 and solve for y. x2+ 4y2º 4 ... +4+y2º10=0 49.4x2ºy2 10 + ++ + + + + + distance + Solving Quadratic Systems 2 | 296 | 877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2016-44 | longest | en | 0.708744 |
http://math.stackexchange.com/questions/1862/what-is-the-rigorous-definition-of-polyhedral-fan-what-are-some-good-resources | 1,469,549,599,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824995.51/warc/CC-MAIN-20160723071024-00193-ip-10-185-27-174.ec2.internal.warc.gz | 149,946,245 | 17,920 | # What is the rigorous definition of polyhedral fan? What are some good resources to learn about them? What context do they arise naturally in?
I've been reading about tropical geometry and many papers reference polyhedral fans. I feel like I have a decent intuitive picture of what they are from reading articles but I still haven't been able to guess the general definition. All the ones I've encountered have been systems of linear inequalities, so that is my best guess at a general definition.
Any comments on where they appeared first historically or links/books to general resources on learning about them would be appreciated. Also, I'm curious to know what other areas of math these show up in?
-
A polyhedral cone is a subset of a real vector space which is the intersection of finitely many closed half spaces. (The defining planes of these half spaces must pass through $0$.) A fan is a finite set $F$ of polyhedral cones, all living in the same vector space, such that
(1) if $\sigma$ is a cone in $F$, and $\tau$ is a face of $\sigma$, then $\tau$ is in $F$.
(2) if $\sigma$ and $\sigma'$ are in $F$, then $\sigma \cap \sigma'$ is a face of both $\sigma$ and of $\sigma'$.
For a pure combinatorics reference, have you tried Chapter 2 of De Loera, Rambau and Santos? They focus on polyhedral complexes, which is the more general setup where you don't require that the half spaces pass through $0$, but they talk about fans as well. I haven't had a chance to look at it yet but, based on my knowledge of the authors, I expect it is very good. | 369 | 1,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | latest | en | 0.963443 |
http://www.convertunits.com/from/20417+days/to/minutes | 1,386,331,366,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163051516/warc/CC-MAIN-20131204131731-00059-ip-10-33-133-15.ec2.internal.warc.gz | 291,320,974 | 4,365 | # ››Convert day to minute
days
minutes
How many days in 1 minutes? The answer is 0.000694444444444.
We assume you are converting between day and minute.
You can view more details on each measurement unit:
days or minutes
The SI base unit for time is the second.
1 second is equal to 1.15740740741E-5 days, or 0.0166666666667 minutes.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between days and minutes.
Type in your own numbers in the form to convert the units!
# ››Date difference between calendar days
You may also want to find out how many days are between two dates on the calendar. Use the date calculator to get your age in days or measure the duration of an event.
# ››Want other units?
You can do the reverse unit conversion from minutes to days, or enter any two units below:
Convert to
I'm feeling lucky, show me some random units.
# ››Definition: Day
A continous period of 24 hours which, unless the context otherwise requires, runs from midnight to midnight.
# ››Definition: Minute
A minute is: * a unit of time equal to 1/60th of an hour and to 60 seconds. (Some rare minutes have 59 or 61 seconds; see leap second.)
# ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 411 | 1,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-48 | latest | en | 0.877353 |
https://www.jiskha.com/display.cgi?id=1330704661 | 1,503,504,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.75/warc/CC-MAIN-20170823152006-20170823172006-00204.warc.gz | 914,540,577 | 3,920 | # bio chemistry
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calculate the PH of each of the following buffers prepared by placing in 1.0 L of solution 0.10 mol NH3 and 0.10 mol of NH4CL
• bio chemistry -
Use the Henderson-Hasselbalch equation.
## Similar Questions
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A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50 mL of 0.300 NH4Cl. The pKb of NH3 is 4.74. NH3 + H2O-> NH4+ +OH- 7.50 mL of 0.125 M HCl is added to the 100 mL of the buffer solution. Calculate the concentration …
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Question
If the cardinal number of a set A is 16 then find the number of elements in A.
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Solution
Cardinal no. =16Cardinal no = No. of elements in set.∴ No. of elements in set =16
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