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Alkali metals can explode if they are exposed to water. Lanthanide Metals + Cerium. The alkali metals exhibit many of the physical properties common to metals, although their densities are lower than those of other metals. Calcium. They react readily with nonmetals, particularly halogens. Alkali Metals. Alkali Metals DRAFT. They include lithium, sodium and potassium, which all react vigorously with water to produce an alkaline solution. The Alkali Metals are: Lithium; Sodium; Potassium; Rubidium; Cesium What are the trends as you go down the group? Alkali metals. CC BY-SA 3.0. http://en.wiktionary.org/wiki/alkali_metal As with dihydrogen the bonding is associated with the molecular orbital combination of the two valence s-orbitals (Figure $$\PageIndex{3}$$.9). This group lies in the s-block of the periodic table, as all alkali metals have their outermost electron in … http://en.wiktionary.org/wiki/alkali_metal, http://en.wikipedia.org/wiki/caesium%20atomic%20clock, http://en.wikipedia.org/wiki/Alkali_metals, http://en.wikibooks.org/wiki/General_Chemistry/Chemistries_of_Various_Elements/Group_1%23Alkali_Metals, http://en.wikipedia.org/wiki/Alkali_metal%23Periodic_trends, http://en.wikipedia.org/wiki/File:Lithium_paraffin.jpg, https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/. Learn about the properties of the alkali metals, one of the element groups. The Group 1 elements in the periodic table are known as the alkali metals. Transition Metals-Copper. The rate of reaction with oxygen, or with air, depends upon whether the metals are in the solid or liquid state , as well as upon the degree of mixing of the metals with the oxygen or air. The oxides react vigorously with water to form a hydroxide. Almost every salt of an alkali metal is highly soluble in water. The alkaline earth metals are found in group IIA of the periodic table, which is the second column of elements. The alkali metals provide the best example of group trends in properties in the periodic table, with elements exhibiting similar properties. Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. These metals are highly reactive and are found in ores of other metals and minerals. Their low ionization energies result in their metallic properties and high reactivities. Why and how Joe Biden won the White House. This group lies in the s-block of the periodic table as all alkali metals have their outermost electron in an s-orbital,bonding well with halogens a common example of that is table salt (sodium chloride). All alkali metals react with oxygen in the air to form... metal oxides. CC BY-SA 3.0. http://en.wikibooks.org/wiki/General_Chemistry/Chemistries_of_Various_Elements/Group_1%23Alkali_Metals Alkali metals can be easily distinguished from others as these metals have only 1 electron in their outer shell and can easily donate that electron to achieve a stable state. All of the alkali metals like to give up their single valence electron," says Dr. Chip Nataro, chemistry professor at Lafayette College in Easton, Pennsylvania. princesingh8662 princesingh8662 Answer: At the end of the chemical reaction, lithium gives lithium monoxide (LiO), sodium gives sodium peroxide (Na 2 O 2), and other alkali metals give superoxides (that is, each alkali metal atom forms bonds with two oxygen atoms). Home. All are solids at 0°C and melt below 200°C. The speed at which alkali metals react with oxygen in the air... increases as you go down group I. Lithium tarnishes... slowly. caesium atomic clockA primary frequency standard in which electronic transitions between the two hyperfine ground states of caesium-133 atoms are used to control the output frequency. Erbium. 94 times. Alkali metals are a group of metals that are rarely found in their pure form; they are highly reactive and are notorious for exploding when exposed to water. The alkali metals are a group of chemical elements from the s-block of the periodic table with similar properties: they appear silvery and can be cut with a plastic knife. All the halides readily dissolve in water except lithium fluoride (LiF). Periodic Table Study Guide - Introduction & History, Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College. There are alkali metals all around you right now. Wikipedia The alkaline earth metals are six chemical elements in group 2 of the periodic table.They are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). This particular resource used the following sources: http://www.boundless.com/ 3 months ago. All the alkali metals impart a characteristic colour to the flame. Exposing an alkali metal to a halogen will cause an extremely exothermic reaction that results in an ionic salt. The alkali metals are a group in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). All alkali metals are found naturally, but they do not occur freely because they are so reactive. Lead. CC BY-SA 3.0. http://en.wikipedia.org/wiki/caesium%20atomic%20clock Save. Experiments have been conducted to attempt the synthesis of ununennium (Uue), which is likely to be the next member of the group if the attempt is successful. Group IIA metals are hard and shiny and usually malleable and … Like the alkali metals, these elements are found in compounds rather than pure form. Most alkali metals have many different applications. 77% average accuracy. All the halides are ionic in nature except lithium iodide. Alkali metals are found in which group of the periodic table? Rather, they are found in ionic compounds, compounds consisting of ions held together by opposite charges. Why are they so reactive? She has taught science courses at the high school, college, and graduate levels. Lithium, sodium, potassium, rubidium, and cesium have a great many other properties in common. In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen. All alkali metals are placed under group number 1 and s-block of periodic table. Gold. This site uses cookies from Google and other third parties to deliver its services, to personalise adverts and to analyse traffic. The chemical reaction of the alkali metals with the halogens is: 2 M + X 2 2 MX (where M represents an alkali metal and X represents a halogen) Reaction with Oxygen. Photoelectric effect: Alkali metals (except Li) exhibits photoelectric effect. What are alkali metals? Alkali Metals … alkali metalAny of the soft, light, reactive metals of Group 1 of the periodic table; lithium, sodium, potassium, rubidium, cesium, and francium. By using ThoughtCo, you accept our, Location of the Alkali Metals on the Periodic Table, Properties of the Basic Metals Element Group, Transition Metals and the Properties of the Element Group. ThoughtCo uses cookies to provide you with a great user experience. Alkali metals react vigorously with all the halogens to form solid ionic halides with a definite crystal structure. The elements have very similar properties: they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure.. Wikipedia It is predicted that the next alkali metal after ununennium would be unhexpentium (Uhp), an element that has not yet received even attempts at synthesis due to its extremely high atomic number. They also can oxidize very easy when in contact with air. All alkali metals exhibit an oxidation state of 2 See answers akkant1978 akkant1978 Answer: Alkali metals show oxidation state of +1. Compounds of potassium as well as other alkali metals were obtained from wood ashes by early chemists. Wiktionary Reactivity decreases from fluorine to iodine . The pure alkali metal can also react directly with water. The alkali metals are softer than most other metals. They form conducting solutions, proving their ionic nature. lyeA strong caustic alkaline solution of potassium or sodium salts, obtained by leaching wood ashes. Wikipedia In this process, the alkali metal is said to be oxidized, and whatever takes the electron from the alkali metal is reduced. The alkali metals also have low densities. Recall the periodic trends observed in the alkali metals. Oxidation state: The alkali metals exhibit oxidation state of +1 in their compounds and are strongly electropositive in character. 0 votes . The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. The reactions release surprising amounts of heat and light. A common application of the compounds of sodium is the sodium-vapor lamp, which emits very efficient light. Lithium. For instance, when moving down the table, all known alkali metals show: In general, their densities increase when moving down the table, with the exception of potassium, which is less dense than sodium. A less-pinhole perovskite film is obtained by precisely controlling the doping dosage and element species, significantly reducing the defects. The alkali metals are silver-colored (cesium has a golden tinge), soft, low-density metals, which react readily with halogens to form ionic salts, and with water to form strongly alkaline hydroxides. Edit. In a chemical equation, alkali metals are represented with an M. Here are some sample reaction equations: $4{ M }_{ (s) }+{ O }_{ 2(g) }\rightarrow 2{ M }_{ 2 }O$, ${ M }_{ 2 }O_{ (s) }+H_{ 2 }{ O }\rightarrow 2MOH_{ (aq) }$, $2{ M }_{ (s) }+2{ H }_{ 2 }O\rightarrow 2MOH_{ (aq) }+{ H }_{ 2(g) }$, $2{ M }_{ (s) }+{ Cl }_{ 2(g) }\rightarrow 2MCl_{ (s) }$. (adsbygoogle = window.adsbygoogle || []).push({}); The alkali metals are a group of chemical elements in the periodic table with the following physical and chemical properties: They can all be cut easily with a plastic knife due to their softness, and their shiny surface tarnishes rapidly in air due to oxidation. highly reactive at standard temperature and pressure, readily lose their outermost electron to form cations with a charge of +1. Because of their high reactivity, alkali metals must be stored under oil to prevent reaction with air. The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Any alkali metal, on coming in contact with air or oxygen, starts burning and oxides are formed in the process. Under normal conditions, lithium is the least dense of all the solid elements and the lightest of all metals. How do they react with water, oxygen and chlorine? Preview this quiz on Quizizz. Gadolinium. The electropositive character increases from Li to Cs. All the alkali metals react with water, with the heavier alkali metals reacting more vigorously than the lighter ones. An alkali metal can easily lose its valence electron to form the univalent cation. 0. Wiktionary Post-Transition Metals + Bismuth. Herein, we demonstrate alkali metal ions including Li +, Na +, K +, and Rb + as mediators to regulate the crystal lattice and film quality of Cs 2 AgBiBr 6 perovskites. The alkali metals are the elements located in Group IA of the periodic table. The alkali metals are a group of elements in the periodic table.They are all in the first column of the periodic table. champions pay tribute to Trebek. This gives them the largest atomic radii of the elements in their respective periods. They are low enough for the first three (lithium, sodium and potassium) to float on water. Aluminium. Alkali Metals Lithium is stored in oil because of its high reactivity. Alkaline Earth Metals + Beryllium. Alkali metals make up … Alkali metals react with oxygen to form oxides, which have a duller appearance and lower reactivity. CC BY-SA 3.0. http://en.wiktionary.org/wiki/lye It is much used in making soap as well as in biodiesel. Thorium. We give a sorted list of metals. The alkali metals are so called because reaction with water forms alkalies (i.e., strong bases capable of neutralizing acids ). Most alkali metals have many different applications, such as rubidium and caesium atomic clocks, sodium-vapor lamps, and table salt. This produces a layer of dull oxide on the surface of the metal, called tarnish. Silver. Powered by Create your own unique website with customizable templates. The alkali metals consist of the chemical elements lithium (Li), sodium (Na), potassium (K), [note 1] rubidium (Rb), caesium (Cs), [note 2] and francium (Fr). Public domain. Two of the most well-known applications of the pure elements are rubidium and cesium atomic clocks, of which cesium atomic clocks are the most accurate representation of time known as of 2012. Chemistry. Periodic Trends of Alkali Metals. Alkali metals react violently with water, halogens, and acids. The oxides are much less reactive than the pure metals. The alkali metals are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs),and francium (Fr). The alkali metals are the elements located in Group IA of the periodic table. Actinide Metals + Californium. All the discovered alkali metals occur in nature. The alkali metals (lithium, sodium potassium., cesium, and francium) are the most reactive metals in the Periodic Table (Table 3.7) insofar as they react vigorously or even explosively with cold water, resulting in the displacmet of hydrogen, itself a flammable gas that can … Alkali metals are highly reactive at standard temperature and pressure and readily lose their outermost electron to form cations with charge +1. stanton_wertjes. CC BY-SA 3.0. http://en.wikipedia.org/wiki/Alkali_metal%23Periodic_trends Alkaline earths are reactive but less so than alkali metals. All the alkali metals form M 2 dimers in the vapor phase in an analogous manner to hydrogen. All the discovered alkali metals occur in nature. College releases autopsy results for student with COVID-19 Alkali metal, any of the six chemical elements that make up Group 1 (Ia) of the periodic table —namely, lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). CC BY-SA 3.0. http://en.wikipedia.org/wiki/Alkali_metals Plutonium. All of the alkaline earth metal atoms have a +2 oxidation state. Sodium is found in table salt, lithium in your phone battery and potassium in your bananas. All alkali metals articles in Chemistry World. When alkali metals react with halogens, halides are formed. The only element in the first column that is not usually considered an alkali metal is hydrogen.Hydrogen and the alkali metals make up the group 1 elements of the periodic table. Boundless Learning 8th - 10th grade. As a result, compounds of lithium are more soluble in organic solvents while those of other alkali metals are more soluble in water. Alkali metals have low electronegativities. In this case, the metal is a basic anhydride. Sodium hydroxide (NaOH), also called lye, is an industrial-strength base. All the alkali metals react directly with oxygen; lithium and sodium form monoxides, Li 2 O and Na 2 O, and the heavier alkali metals form superoxides, MO 2. All the alkali metals are soft and, except for Cs which is yellow, are silvery-gray in color. 1. Alkali metals have one electron in their outer shell, which is loosely bound. Boundless vets and curates high-quality, openly licensed content from around the Internet. Wikipedia Alkali metals are found in which group of the periodic table? Cesium and francium are the most reactive elements in this group. As with all metals, the alkali metals are malleable, ductile, and are good conductors of heat and electricity. Table salt, or sodium chloride, on the other hand, has been used since antiquity. All the alkali metals when heated with oxygen form different types of oxides for example, lithium forms lithium oxide, sodium forms sodium peroxide, while K, Rb and Cs form their respective superoxides (where M=K, Rb or Cs). With the loss of valence electron it attains the stable configuration of nearest inert gas. Edit. The alkali metals are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs),and francium (Fr). Wikibooks Alkaline Earth Metals . What Are the Properties of the Alkaline Earth Metals? This group lies in the s-block of the periodic table, as all alkali metals have their outermost electron in an s-orbital. The resulting hydroxides of these elements dissociate completely in water to form some of the strongest bases known. Want to know more about Alkali metals? Stars, 'Jeopardy!' CC BY-SA 3.0. http://en.wikipedia.org/wiki/File:Lithium_paraffin.jpg Gaseous hydrogen is released, which is flammable. Are all in the periodic table, with elements exhibiting similar properties very easy when in contact with air pure! To water analyse traffic low enough for the first three ( lithium, sodium potassium. 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# Proving Linear Independence of Sequences

Good morning,

my question is about proving the linear independence of sequences. In the theory of linear difference equations, one needs a fact, that for all distinct $\lambda_1,\ldots,\lambda_m \in \mathbb{C}$ and all $k_1,\ldots,k_m \in \mathbb{N}$, the functions (sequences) $$n^{0} \lambda_1^n, n^1 \lambda_1^n, \ldots, n^{k_1 - 1} \lambda_1^n,$$ $$n^{0} \lambda_2^n, n^1 \lambda_2^n, \ldots, n^{k_2 - 1} \lambda_2^n,$$ $$\ldots,$$ $$n^{0} \lambda_m^n, n^1 \lambda_m^n, \ldots, n^{k_m - 1} \lambda_m^n$$ are linearly independent. Although this fact is extensively used in literature, I haven't found any satisfactory proof and my attempts to prove this by myself weren't successful as well.

I know how to prove that $\lambda_1^n,\lambda_2^n,\ldots,\lambda_m^n$ are linearly independent for distinct $\lambda_1,\ldots,\lambda_m \in \mathbb{C}$ (using Vandermonde determinant). Moreover, I know how to prove that $n^0 \lambda_i^n, n^1 \lambda_i^n, \ldots, n^{k_i - 1} \lambda_i^n$ are linearly independent (if one supposes that they are dependent, one gets an equation that says that a polynomial with at least one non-zero coefficient is zero for all $n \in \mathbb{N}$, i.e., a contradiction). But I don't know how to combine these results. Simply the fact that both sets of sequences are linearly independent does not prove that their union is linearly independent. Any ideas would be helpful.

Thank you in advance.

You can get linear independence quite easily by looking at growth rates (analytic considerations rather than algebraic identities). Your list is very easily seen to fall into the category below.

Any set of (say, positive) functions $f_1(n), \ldots, f_k(n)$ satisfying $f_i = o(f_j)$ for all $1 \le i<j\le k$ must be linearly independent. Suppose there were a non-trivial linear combination $\sum_{i=1}^k c_i f_i = 0$, and WLOG $c_k=1$. Then choose $n$ large enough so that $f_k(n) > \sum_{i<k} |c_i|\, f_i(n)$. Contradiction.

EDIT: As pointed out by anon, the question calls for $\lambda_i \in \mathbb C$, in which case it's quite possible for members of this family to be comparable in size. However, this argument allows one to decompose the entire family into small sets with matching growth rates, and those can be handled by Vandermonde.

• Thanks for pointing this out! I didn't notice the $\lambda_i \in \mathbb C$. – Erick Wong Jun 21 '12 at 8:50
• You don't need $f_i$ to be positive; it's fine if they're complex-valued and $|f_i|=o\big(|f_j|\big)$ for $1\le i<j\le k$. Then we can get a contradiction on $$|f_k(n)|>\sum_{i<k}|c_if_i(n)|\ge\left|\sum_{i<k}c_if_i(n)\right|$$ (I say this because OP set $\lambda_i\in\Bbb C$.) – anon Jun 21 '12 at 8:50
• Well I see now. Thanks a lot! – 042 Jun 21 '12 at 9:04

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# A Classic

Cosmin is located at the origin of an $$xy$$-plane. Every minute, he will move up, down, left, or right by one unit. What is the probability that he will ever return to the origin?

×


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# The knapsack cryptosystem at work

This note is an example of the knapsack crypto-system at work.So before you got through this note I recommend reading the main one . Click here

Suppose that a typical user of this crypto-system selects as the secret key the super-increasing sequence $${3,5,11,20,41}$$ and the modulus $$m = 85$$ and the multiplier $$a = 44$$.Each member of the super-increasing sequence is multiplied by 44 and reduced modulo 85 to yield $${47,50,59,30,19}$$.This is the encryption key the user submit to the public directory.

Suppose someone wants to send a plaintext message to the user such as "HELP US". First we convert it into a string of 1's and 0's $M = 00111 00100 01011 10100 10010$ The string is then broken up into blocks of digits,in the current case,blocks of length 5.Using the listed public keys to encrypt,the sender transforms the successive blocks into $47*0+50*0+59*1+30*1+19*1 = 108$ $47*0+50*0+59*1+30*0+19*0 = 59$ $47*0+50*1+59*0+30*1+19*1 = 99$ $47*0+50*1+59*1+30*1+19*1 = 158$ $47*1+50*0+59*1+30*0+19*0 = 106$ $47*1+50*0+59*0+30*1+19*0 = 77$

The transmitted ciphertext consists of the following series of positive integers $${108,59,99,158,106,777}$$.

To read the message the legitimate receiver first solves the congruence $44x \equiv 1 \pmod{85}$,yielding $y \equiv 29 \pmod{85}$.Then each ciphertext number is multiplied by 29 and reduced modulo 85 to produce a super-increassing knapsack problem.

For example,$$108$$ is converted to $$72$$ as $108*29 \equiv 72 \pmod{85}$ and the corresponding knapsack problem is $3x_1 + 5x_2 + 11x_3 + 20x_4+41x_5 = 72$.

The procedure for solving super-increasing knapsack problems quickly yields the solution $$x_1 = x_2 =0$$,$$x_3 = x_4=x_5 = 1$$,in this way $$00111$$,the first block of the binary equivalent of the plaintext is obtained.

3 years, 6 months ago

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Home > Type 1 > Type 1 Error Type 2 Error Relationship

# Type 1 Error Type 2 Error Relationship

## Contents

Type III Errors Many statisticians are now adopting a third type of error, a type III, which is where the null hypothesis was rejected for the wrong reason.In an experiment, a Two types of error are distinguished: typeI error and typeII error. It does not mean the person really is innocent. When the sample size is increased above one the distributions become sampling distributions which represent the means of all possible samples drawn from the respective population. check over here

Recall also that we choose the probability of making a Type I error when we set Alpha and that if we decrease the probability of making a Type I error we False negatives may provide a falsely reassuring message to patients and physicians that disease is absent, when it is actually present. These error rates are traded off against each other: for any given sample set, the effort to reduce one type of error generally results in increasing the other type of error. If the consequences of making one type of error are more severe or costly than making the other type of error, then choose a level of significance and a power for https://en.wikipedia.org/wiki/Type_I_and_type_II_errors

## Type 1 Error Example

Usually a type I error leads one to conclude that a supposed effect or relationship exists when in fact it doesn't. Don't reject H0 I think he is innocent! The test requires an unambiguous statement of a null hypothesis, which usually corresponds to a default "state of nature", for example "this person is healthy", "this accused is not guilty" or Mosteller, F., "A k-Sample Slippage Test for an Extreme Population", The Annals of Mathematical Statistics, Vol.19, No.1, (March 1948), pp.58–65.

• Devore (2011).
• Thank you to...
• on follow-up testing and treatment.
• Justice System - Trial Defendant Innocent Defendant Guilty Reject Presumption of Innocence (Guilty Verdict) Type I Error Correct Fail to Reject Presumption of Innocence (Not Guilty Verdict) Correct Type II
• Optical character recognition (OCR) software may detect an "a" where there are only some dots that appear to be an "a" to the algorithm being used.
• ISBN1584884401. ^ Peck, Roxy and Jay L.
• Archived 28 March 2005 at the Wayback Machine.‹The template Wayback is being considered for merging.› References ^ "Type I Error and Type II Error - Experimental Errors".
• Like any analysis of this type it assumes that the distribution for the null hypothesis is the same shape as the distribution of the alternative hypothesis.
• Trying to avoid the issue by always choosing the same significance level is itself a value judgment.
• Example: In a t-test for a sample mean µ, with null hypothesis""µ = 0"and alternate hypothesis"µ > 0", we may talk about the Type II error relative to the general alternate

p.100. ^ a b Neyman, J.; Pearson, E.S. (1967) [1933]. "The testing of statistical hypotheses in relation to probabilities a priori". These questions can be understood by examining the similarity of the American justice system to hypothesis testing in statistics and the two types of errors it can produce.(This discussion assumes that For a given test, the only way to reduce both error rates is to increase the sample size, and this may not be feasible. Type 1 Error Calculator Retrieved 2016-05-30. ^ a b Sheskin, David (2004).

Such tests usually produce more false-positives, which can subsequently be sorted out by more sophisticated (and expensive) testing. Probability Of Type 1 Error This is why the hypothesis under test is often called the null hypothesis (most likely, coined by Fisher (1935, p.19)), because it is this hypothesis that is to be either nullified Negation of the null hypothesis causes typeI and typeII errors to switch roles. Home > Research > Methods > Type I Error - Type II Error . . .

A Type I error occurs when we believe a falsehood ("believing a lie").[7] In terms of folk tales, an investigator may be "crying wolf" without a wolf in sight (raising a Type 1 Error Psychology This means only that the standard for rejectinginnocence was not met. If the result of the test corresponds with reality, then a correct decision has been made. A typeII error (or error of the second kind) is the failure to reject a false null hypothesis.

## Probability Of Type 1 Error

This is why replicating experiments (i.e., repeating the experiment with another sample) is important. Of course, modern tools such as DNA testing are very important, but so are properly designed and executed police procedures and professionalism. Type 1 Error Example on follow-up testing and treatment. Probability Of Type 2 Error Marascuilo, L.A. & Levin, J.R., "Appropriate Post Hoc Comparisons for Interaction and nested Hypotheses in Analysis of Variance Designs: The Elimination of Type-IV Errors", American Educational Research Journal, Vol.7., No.3, (May

Moulton (1983), stresses the importance of: avoiding the typeI errors (or false positives) that classify authorized users as imposters. check my blog ISBN1-57607-653-9. The type II error is often called beta. A typeI error may be compared with a so-called false positive (a result that indicates that a given condition is present when it actually is not present) in tests where a Type 3 Error

This is represented by the yellow/green area under the curve on the left and is a type II error. Statistical significance The extent to which the test in question shows that the "speculated hypothesis" has (or has not) been nullified is called its significance level; and the higher the significance They also cause women unneeded anxiety. this content avoiding the typeII errors (or false negatives) that classify imposters as authorized users.

The null hypothesis has to be rejected beyond a reasonable doubt. Power Of A Test Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. p.54.

## Related terms See also: Coverage probability Null hypothesis Main article: Null hypothesis It is standard practice for statisticians to conduct tests in order to determine whether or not a "speculative hypothesis"

About the only other way to decrease both the type I and type II errors is to increase the reliability of the data measurements or witnesses. Remove Cancel × CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on Example 2 Hypothesis: "Adding fluoride to toothpaste protects against cavities." Null hypothesis: "Adding fluoride to toothpaste has no effect on cavities." This null hypothesis is tested against experimental data with a Misclassification Bias Distribution of possible witnesses in a trial showing the probable outcomes with a single witness if the accused is innocent or obviously guilty..

Related 18Comparing and contrasting, p-values, significance levels and type I error4Frequentist properties of p-values in relation to type I error1Error type I for $X_i \sim Exp(\theta)$1Hypothesis testing, find $n$ to limit The incorrect detection may be due to heuristics or to an incorrect virus signature in a database. Etymology In 1928, Jerzy Neyman (1894–1981) and Egon Pearson (1895–1980), both eminent statisticians, discussed the problems associated with "deciding whether or not a particular sample may be judged as likely to http://degital.net/type-1/type-1-and-type-2-error-statistics-examples.html It is asserting something that is absent, a false hit.

Medical testing False negatives and false positives are significant issues in medical testing. If I did not flip the coin n = 10 times, but n → ∞ times, the calculated true alpha would approach set alpha. Examples of type II errors would be a blood test failing to detect the disease it was designed to detect, in a patient who really has the disease; a fire breaking is never proved or established, but is possibly disproved, in the course of experimentation.

p.28. ^ Pearson, E.S.; Neyman, J. (1967) [1930]. "On the Problem of Two Samples". Similar considerations hold for setting confidence levels for confidence intervals. Obviously, there are practical limitations to sample size. Connection between Type I error and significance level: A significance level α corresponds to a certain value of the test statistic, say tα, represented by the orange line in the picture

Computers The notions of false positives and false negatives have a wide currency in the realm of computers and computer applications, as follows. Cambridge University Press. Retrieved Oct 29, 2016 from Explorable.com: https://explorable.com/type-i-error . In hypothesis testing the sample size is increased by collecting more data.

Various extensions have been suggested as "Type III errors", though none have wide use. The probability that an observed positive result is a false positive may be calculated using Bayes' theorem. This is one reason2 why it is important to report p-values when reporting results of hypothesis tests. Cambridge University Press.

Rogers AP Statistics | Physics | Insultingly Stupid Movie Physics | Forchess | Hex | Statistics t-Shirts | About Us | E-mail Intuitor ]Copyright © 1996-2001 Intuitor.com, all rights reservedon the Joint Statistical Papers. This is how science regulates, and minimizes, the potential for Type I and Type II errors.Of course, in non-replicatable experiments and medical diagnosis, replication is not always possible, so the possibility This is why the hypothesis under test is often called the null hypothesis (most likely, coined by Fisher (1935, p.19)), because it is this hypothesis that is to be either nullified

By using this site, you agree to the Terms of Use and Privacy Policy. The design of experiments. 8th edition. However, if the result of the test does not correspond with reality, then an error has occurred. Collingwood, Victoria, Australia: CSIRO Publishing.


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# Where did the extra dime come from? [duplicate]

This question already has an answer here:

In a snack bar there were two baskets of pears of different sorts, 60 pears in each. It was proposed to sell pears from the first basket at 4 for 30¢, and pears from the second basket at 6 for 50¢. Thus the two baskets of pears would be sold for \$9.50 in all. However, the cashier, in order to simplify her work, mixed together the pears of both kinds and proceeded to sell ten pears for 80¢. As a result of the sale of pears, there turned out 10¢ too much: not \$9.50 but \$9.60. Where did this extra dime come from? (Source: Lapses in Mathematical Reasoning by V. M. Bradis, V. L. Minkovskii, A. K. Kharcheva.) ## marked as duplicate by f'', Peregrine Rook, ffao, Deusovi♦, Mike EarnestApr 21 '16 at 4:25 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. • −1, unclear what the question is. "Where did this extra dime come from?" From selling 120 pears at ten for 80¢, of course. What's the question? ("As a result of the sale of pears, there turned out 10¢ too much: not \$9.50 but \$9.60." Well, yeah. If you change the price on things, then the total amount taken in will change.) – msh210 Apr 20 '16 at 22:44 • What's the word between extortion and bait-and-switch? Fraud; that's where. – Mazura Apr 20 '16 at 23:43 • @msh210. I was with you for a while wondering that the puzzling interest was here as the$9.60 answer seemed valid with no issues... I think people are thrown by the fact that if you added the 30¢ and the 50¢ you get 80¢, so the math should 'add up'. They're forgetting that you had more 30¢ units to sell. Suspect we just made the jump through that confusion directly (and with math intact). Maybe people with our er, 'logic units' will be stumped about something else in turn... :) – Lamar Latrell Apr 21 '16 at 2:12
• simply from selling cheap pear at a higher price... – njzk2 Apr 21 '16 at 2:59

Explanation:

The pears in the first basket cost 0.30/4 = 0.075 each. The pears in the second basket cost 0.50/6 = 0.0833... each. If we average these prices together, the mixed pears should have been sold at 0.079166... each (selling 120 of them at this price produces 9.50 in revenue). However, the cashier opted instead to sell them at 0.08 a piece, which is 1/12 of a cent more than their price should be. That's where the additional dime comes from.

It seems rather obvious to me. Am I missing something?

The extra 10 is from selling them at a higher average price.
The plan was to sell one basket at a price of 7 1/2 per pear (albeit in batches of 4), and the other at 8 1/3 per pear (in batches of 6).
This averages just under 8 per pear, which is what the cashier actually sells them for, making an extra 10 cents.

Edit to add this:
I see now where there could be an incorrect way to reason about it that some people (like the cashier maybe) might do. It would have been nice if that were more clearly stated in the question.
The cashier may have thought as follows: Putting together a batch of 4 at \$0.30 and 6 at \$0.50 gives 10 at \$0.80. Therefore I'll sell them as 10 for \$0.80 and get the same price.

This reasoning fails because:

There are more batches of 4 than of 6, and the cashier's method assumes that there are equally many. The batches of 4 are the cheaper ones, so those extra batches of 4 are essentially sold at the incorrect higher average price that the cashier arrives at.

• Both I and the original authors thought that the cashier’s (incorrect) line of reasoning did not need to be explained. – MJD Apr 20 '16 at 16:00

12 of the cheaper pears are sold as the more expensive pears.

She had 60 of each pear, which would be 15 sets of 4 cheap pears and 10 sets of 6 expensive pears. By selling them in groups of 10s, each purchase would yield (on average) 5 cheap pears and 5 expensive pears, but would be the price of 4 cheap pears and 6 expensive pears.

If we assume the first 10 sets sold are 4 cheap pears and 6 expensive pears, that would leave 2 sets of 10 cheap pears. Of those 20 pears, 6 out of 10 would be overpriced. So 12 of them are cheaper pears being sold for the price of the more expensive pears. $$\mathrm{12\ pears \times \left(\frac{50¢}{6\ pears} - \frac{30¢}{4\ pears}\right) = 10¢}$$


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### Primary 6 Problem Sums/Word Problems - Try FREE

Score :
(Single Attempt)

#### Question

Margaret observed that in a campus interview, only 75% of the candidates who were interviewed got selected.

120 girls were selected.

5/6 of those who did not get selected were girls.

The remaining number of people who did not get selected were 10 boys.

It was found that only boys and girls took part in the interview.

Of the total number of candidates interviewed, what was the percentage of boys who had been selected?

The correct answer is : 25
%


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17-29-053

downward light output ratio

DLOR

quotient of the downward flux of a luminaire, measured under specified practical conditions with its own lamps and equipment, and the sum of the individual luminous fluxes of the same lamps when operated outside the luminaire with the same equipment, under specified conditions

Note 1 to entry: The downward light output ratio has unit one.

Note 2 to entry: This entry was numbered 845-09-40 in IEC 60050-845:1987.

Note 3 to entry: This entry was numbered 17-352 and 17-344 in CIE S 017:2011.

Publication date: 2020-12


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• # ,

January 08, 2019, at 10:30 PM
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January 08, 2019, at 10:10 PM
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January 08, 2019, at 9:10 PM
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January 08, 2019, at 8:40 PM
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January 08, 2019, at 8:30 PM
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January 08, 2019, at 7:20 PM
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January 08, 2019, at 7:10 PM
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January 08, 2019, at 3:10 PM
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January 08, 2019, at 3:00 PM
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January 08, 2019, at 2:40 PM
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January 08, 2019, at 2:30 PM
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POPULAR ONLINE


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It is currently 16 Jul 2019, 06:43

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# By what percent did Pop downloads increase from 2008 to 2012

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Posts: 7322
Followers: 122

Kudos [?]: 1430 [0], given: 6569

Expert's post
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Question Stats:

80% (02:09) correct 20% (01:16) wrong based on 5 sessions
Attachment:

#GREpracticequestion By what percent did Pop downloads increase.jpg [ 117.41 KiB | Viewed 935 times ]

A. 43%

B. 57%

C. 75%

D. 86%

E. 133%
[Reveal] Spoiler: OA

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Re: By what percent did Pop downloads increase from 2008 to 2012 [#permalink]  15 Apr 2019, 11:52
How is it c?

Thanks for the response.
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Re: By what percent did Pop downloads increase from 2008 to 2012 [#permalink]  15 Apr 2019, 12:26
1
KUDOS
Expert's post
Hi LT2018

You need to look at both graphs.

The first one says that the downloads are about 10 % in total and that around 40 % of these are pop. So the downloads are 4 million

For 2012 the total downloads increased to 28 million and of these, 25 % (second graph) are pop. So in total 7 millions

$$\frac{7-4}{4} = \frac{3}{4} = 75$$ %

Regards
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text:
12 terms

Lecture #11
###### PLAY
term for processes in which no heat energy is added or removed, but in which the temperature changes
warm/less dense air (lighter)
rises
cold/denser air (heavier)
sinks
as a parcel of air rises
-it expands (volume changes)
-it comes under less and less pressure as it rises in the atmosphere and thus expands
-as the air expands, the molecular action (kinectic energy) decreases and thus the temperature decreases
expanding air
= temperature decreases
compressing air
= temperature increases
changing the temperature of the air without adding or subtracting heat (simply the result of compressing the air or allowing it to expand)
P (mb) = T ( ° K) x p (kg/m3) x C (2.87)
at the surface as temperature decreases, density and pressure increases; as the temperature increases, air density and pressure decreases
rate of adiabatic change
the rate at which the air temperature changes as it rises or falls through the atmosphere
dry adiabatic rate (DAR or DALR)
-when air is unsaturated (air temperature > dew point temperature; RH < 100% )
-the DAR is a constant rater of 1°C/100m or 10°C/1000m
-rising air cools at 1°C/100m (10°C/1000m)
-falling air warms at 1°C/100m (10°C/1000m)
wet or saturated adiabatic rate (SAR or SALR)
-when the air is saturated; air temperature = dew point temperature; RH = 100%
-the SAR is not a constant rate but a variable rate: 5°-9°C/100m
condensation level
height at which condensation occurs; where cloud formation begins (usually at the bottom of a cloud mass)


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## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeMay 18th 2010
• CommentRowNumber2.
• CommentAuthorDavid_Corfield
• CommentTimeMay 18th 2010

Is there any call for a factorization into a triple of systems, as in the stuff, structure, property story?

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMay 18th 2010

I don’t think I’ve ever seen that situation formalized, but it would be worth doing. I can think of at least one other example.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeMay 19th 2010

The other example being the combination of (hyperconnected, localic) and (surjection, inclusion) factorizations for topoi, which are very similar to the (eso+full, faithful) and (eso, fully faithful) factorizations for categories.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeMay 19th 2010
• (edited May 19th 2010)

In a recent email exchange with Andre Joyal he pointed out to me the factorisation systems

(epimorphisms of toposes, sub-toposes)

(connected morphisms of toposes, totally disconnected localic morphisms)

(1-connected morphisms of toposes, Galois morphisms)

Are these examples of factorisation systems in a 2-category (I mean interestingly so)? I believe the third system has only be looked at in the special case of morphisms $E\to \mathrm{Set}$. Marta Bunge I think has done the second, and the first one is ’old’.

• CommentRowNumber6.
• CommentAuthorDavid_Corfield
• CommentTimeMay 19th 2010

Mike @3: Does any notion of orthogonality exist for triple factorisations? Would that matter to their potential interest? Maybe what I’d really like to know is the whole point of factorisation systems.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMay 19th 2010
• (edited May 19th 2010)

DavidR #5, the first one is definitely a factorization system, aka (surjections, inclusions). I’m not sure about the others, but it seems quite possible. I know there’s an “almost” factorization system on toposes which consists of (connected+locally-connected morphisms, local homeomorphisms) but according to the Elephant the factorizations only exist for morphisms that are already known to be locally connected. Perhaps the second one you mention above is a version of this generalized to non-locally-connected morphisms.

@DavidC #6: Both examples of triple factorizations are “actually” just two pairs of ordinary “double” factorizations related in a certain way, so they have two resulting kinds of orthogonality.

The Whole Point of Factorization Systems would probably be a good topic for a blog post one day… (-: I’ll see if I can think of some good examples.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMay 19th 2010

The Whole Point of Factorization Systems would probably be a good topic for a blog post one day

Yes. Or also for an Idea-section in the relevant nLab entry. It is one of these concepts whose impact is in maybe surprising contrast to the simplicity of their definition. I am not sure if I understand The Whole Point of factorization systems. I think I understand some points of them.

• CommentRowNumber9.
• CommentAuthorzskoda
• CommentTimeMay 19th 2010
• (edited May 19th 2010)

@2: David, when I hear the word 2-categorical factorization system my first association is to split it into 3 as in the Postnikov tower for 2-types. As I was discussing before this could play a role in a possible discussion of model 2-categories which would model $\left(\infty ,2\right)$-categories. I am very interested in this idea which may be a blunder despite its intuitive plausibility.

@7: Urs, in joyalslab, there is a beautiful, original and extensive treatment of factorization systems and weak factorization systems.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeMay 19th 2010
• (edited May 19th 2010)

in joyalslab, there is a beautiful, original and extensive treatment of factorization systems and weak factorization systems.

That reminds me: we need to start putting links to Joyal’s CatLab into the relevant nLab enties.

• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeMay 20th 2010

@9: Zoran, so instead of the 3 classes for a model category forming weak factorisation systems $\left(C,F\cap W\right)$ and $\left(C\cap W,F\right)$ , there might be 5 classes with various combinations of intersections making weak factorisation triple systems? Or something like that.

• CommentRowNumber12.
• CommentAuthorDavidRoberts
• CommentTimeMay 20th 2010

@David #11

or 4 classes? Weak equivalences $W$ + three other classes $\left(A,B,C\right)$, the intersection of $W$ with each of the others in turn to form $\left(A\cap W,B,C\right)$ etc. to get the three classes needed to form a three-way factorisation?

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeOct 10th 2011

added to factorization system in a 2-category a reference (there are probably more canonical ones, though) and a remark that the system [essentially srujective + full] / [faithful] on Grpd is a special case of the n-connected/n-truncated factorization system. I’ll now add more details to that latter entry.


text:
Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 Bib Ind

### 12 Combinatorial representation theory

#### 12.1 Introduction

Here we introduce the implementation of the software package CREP initially designed for MAPLE.

#### 12.2 Different unit forms

##### 12.2-1 IsUnitForm
 ‣ IsUnitForm ( category )

The category for unit forms, which we identify with symmetric integral matrices with 2 along the diagonal.

##### 12.2-2 BilinearFormOfUnitForm
 ‣ BilinearFormOfUnitForm( B ) ( attribute )

Arguments: B -- a unit form.

Returns: the bilinear form associated to a unit form B.

The bilinear form associated to the unitform B given by a matrix B is defined for two vectors x and y as: x*B*y^T.

##### 12.2-3 IsWeaklyNonnegativeUnitForm
 ‣ IsWeaklyNonnegativeUnitForm( B ) ( property )

Arguments: B -- a unit form.

Returns: true is the unitform B is weakly non-negative, otherwise false.

The unit form B is weakly non-negative is B(x,y) ≥ 0 for all x≠ 0 in Z^n, where n is the dimension of the square matrix associated to B.

##### 12.2-4 IsWeaklyPositiveUnitForm
 ‣ IsWeaklyPositiveUnitForm( B ) ( property )

Arguments: B -- a unit form.

Returns: true is the unitform B is weakly positive, otherwise false.

The unit form B is weakly positive if B(x,y) > 0 for all x≠ 0 in Z^n, where n is the dimension of the square matrix associated to B.

##### 12.2-5 PositiveRootsOfUnitForm
 ‣ PositiveRootsOfUnitForm( B ) ( attribute )

Arguments: B -- a unit form.

Returns: the positive roots of a unit form, if the unit form is weakly positive. If they have not been computed, an error message will be returned saying "no method found!".

This attribute will be attached to B when IsWeaklyPositiveUnitForm is applied to B and it is weakly positive.

 ‣ QuadraticFormOfUnitForm( B ) ( attribute )

Arguments: B -- a unit form.

Returns: the quadratic form associated to a unit form B.

The quadratic form associated to the unitform B given by a matrix B is defined for a vector x as: frac12x*B*x^T.

##### 12.2-7 SymmetricMatrixOfUnitForm
 ‣ SymmetricMatrixOfUnitForm( B ) ( attribute )

Arguments: B -- a unit form.

Returns: the symmetric integral matrix which defines the unit form B.

##### 12.2-8 TitsUnitFormOfAlgebra
 ‣ TitsUnitFormOfAlgebra( A ) ( operation )

Arguments: A -- a finite dimensional (quotient of a) path algebra (by an admissible ideal).

Returns: the Tits unit form associated to the algebra A.

This function returns the Tits unitform associated to a finite dimensional quotient of a path algebra by an admissible ideal or path algebra, given that the underlying quiver has no loops or minimal relations that starts and ends in the same vertex. That is, then it returns a symmetric matrix B such that for x = (x_1,...,x_n) (1/2)*(x_1,...,x_n)B(x_1,...,x_n)^T = ∑_i=1^n x_i^2 - ∑_i,j dim_k Ext^1(S_i,S_j)x_ix_j + ∑_i,j dim_k Ext^2(S_i,S_j)x_ix_j, where n is the number of vertices in Q.

##### 12.2-9 EulerBilinearFormOfAlgebra
 ‣ EulerBilinearFormOfAlgebra( A ) ( operation )

Arguments: A -- a finite dimensional (quotient of a) path algebra (by an admissible ideal).

Returns: the Euler (non-symmetric) bilinear form associated to the algebra A.

This function returns the Euler (non-symmetric) bilinear form associated to a finite dimensional (basic) quotient of a path algebra A. That is, it returns a bilinear form (function) defined by
f(x,y) = x*CartanMatrix(A)^(-1)*y
It makes sense only in case A is of finite global dimension.

##### 12.2-10 UnitForm
 ‣ UnitForm( B ) ( operation )

Arguments: B -- an integral matrix.

Returns: the unit form in the category IsUnitForm (12.2-1) associated to the matrix B.

The function checks if B is a symmetric integral matrix with 2 along the diagonal, and returns an error message otherwise. In addition it sets the attributes, BilinearFormOfUnitForm (12.2-2), QuadraticFormOfUnitForm (12.2-6) and SymmetricMatrixOfUnitForm (12.2-7).

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text:
# Bras and Kets

1. Jul 26, 2005

### rayveldkamp

Hi,
I am trying to show that |g> = A|f> implies
<g| = <f|B

where A is an operator and B is its Hermitian conjugate.
I think my problem is with notation, but i have not been able to show this as yet.
Thanks

Ray

2. Jul 26, 2005

### OlderDan

In a matrix representation, you can write the original equation as a sum of products using matrix multiplication rules. Take the complex conjugate, and replace the conjugates of the elements of A with elements of B. Then from the relationship between elements of <g| and |g>, <f| and |f> you have all you need.

3. Jul 26, 2005

### HallsofIvy

Staff Emeritus
That's pretty much the definition of Hermitian conjugate, isn't it?

4. Jul 26, 2005

### Gokul43201

Staff Emeritus
As far as I'm aware, that is how the hermitian conjugate (or adjoint) is defined - though the dual correspondence for A|a> !

I guess you could take the matrix operation of finding the adjoint as definition, and "derive" this result as Older Dan suggests.

5. Jul 27, 2005

### rayveldkamp

Hi,
We have only just been introduced to Dirac notation, and have not had a lot of experience in linear analysis and dual vector spaces etc... I have figured out how to do it, i just contract |g> with an arbitrary bra <h|, then do the same with <g| and an arbitrary bra |h>, show the two are equal and hence the expression for <g| must be correct.
Thanks for the help guys, much appreciated


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> De Moivre’s and the nth Root Theorems | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Trigonometry - Second Edition Go to the latest version.

# 6.7: De Moivre’s and the nth Root Theorems

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Use De Moivre’s Theorem to find the powers of complex numbers in polar form.
• Find the $n^{th}$ roots of complex numbers in polar form.

## De Moivre’s Theorem

The basic operations of addition, subtraction, multiplication and division of complex numbers have all been explored in this chapter. The addition and subtraction of complex numbers lent themselves best to numbers expressed in standard form. However multiplication and division were easily performed when the complex numbers were in polar form. Another operation that is performed using the polar form of complex numbers is the process of raising a complex number to a power.

The polar form of a complex number is $r(\cos \theta + i \sin \theta)$. If we allow $z$ to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form. Recall #3 and #7 from the Review Questions in the previous section.

If $z = r(\cos \theta + i \sin \theta)$ and $z^2 = z \cdot z$ then:

$z^2 &= r(\cos \theta + i \sin \theta) \cdot r(\cos \theta + i \sin \theta)\\ z^2 &= r^2 [\cos (\theta + \theta) + i \sin (\theta + \theta)]\\ z^2 &= r^2 (\cos 2\theta + i \sin 2\theta)$

Likewise, if $z = r(\cos \theta + i \sin \theta)$ and $z^3 = z^2 \cdot z$ then:

$z^3 &= r^2(\cos 2\theta + i \sin 2\theta) \cdot r(\cos \theta + i \sin \theta)\\ z^3 &= r^3 [\cos (2\theta + \theta) + i \sin (2\theta + \theta)]\\ z^3 &= r^3 (\cos 3\theta + i \sin 3\theta)$

Again, if $z = r(\cos \theta + i \sin \theta)$ and $z^4 = z^3 \cdot z$ then

$z^4 = r^4(\cos 4\theta + i \sin 4\theta)$

These examples suggest a general rule valid for all powers of $z$, or $n$. We offer this rule and assume its validity for all $n$ without formal proof, leaving that for later studies. The general rule for raising a complex number in polar form to a power is called De Moivre’s Theorem, and has important applications in engineering, particularly circuit analysis. The rule is as follows:

$z^n = [r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$

Where $z = r(\cos \theta + i \sin \theta)$ and let $n$ be a positive integer.

Notice what this rule looks like geometrically. A complex number taken to the $n$th power has two motions: First, its distance from the origin is taken to the $n$th power; second, its angle is multiplied by $n$. Conversely, the roots of a number have angles that are evenly spaced about the origin.

Example 1: Find. $[2(\cos 120^\circ + i \sin 120^\circ)]^5$

Solution: $\theta=120^\circ=\frac{2\pi}{3} \ \text{rad}$, using De Moivre’s Theorem:

$z^n=[r(\cos \theta+i \sin \theta)]^n &= r^n(\cos n\theta + i \sin n\theta)\\[2(\cos 120^\circ + i \sin 120^\circ)]^5 &= 2^5 \left[\cos 5\frac{2\pi}{3}+i \sin 5\frac{2\pi}{3}\right]\\&= 32\left (\cos \frac{10\pi}{3}+i \sin \frac{10\pi}{3} \right )\\&= 32 \left(-\frac{1}{2}+-\frac{i\sqrt{3}}{2}\right)\\&=-16-16i \sqrt{3}$

Example 2: Find $\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10}$

Solution: Change into polar form.

$r &= \sqrt{x^2+y^2} && \theta = \tan^{-1} \left(\frac{\sqrt{3}}{2} \cdot -\frac{2}{1} \right)=-\frac{\pi}{3} + \pi=\frac{2\pi}{3}\\r &= \sqrt{\left(\frac{-1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\r &= \sqrt{\frac{1}{4}+\frac{3}{4}}\\r &= \sqrt{1}=1$

The polar form of $\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)$ is $1 \left(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3}\right)$

Now use De Moivre’s Theorem:

$z^n = [r(\cos \theta + i \sin \theta)]^n &= r^n(\cos n\theta+i \sin n\theta)\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1^{10}\left[\cos 10\left(\frac{2\pi}{3}\right)+i \sin 10 \left(\frac{2\pi}{3}\right)\right]\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= 1 \left(\cos \frac{20\pi}{3}+i \sin \frac{20\pi}{3}\right)\\\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)^{10} &= -\frac{1}{2}+i \frac{\sqrt{3}}{2} \rightarrow \ \text{Standard Form}$

## nth Roots

We have explored all of the basic operations of arithmetic as they apply to complex numbers in standard form and in polar form. The last discovery is that of taking roots of complex numbers in polar form. Using De Moivre’s Theorem we can develop another general rule –one for finding the $n^{th}$ root of a complex number written in polar form.

As before, let $z = r(\cos \theta + i \sin \theta)$ and let the $n^{th}$ root of $z$ be $v = s (\cos \alpha + i \sin \alpha)$. So, in general, $\sqrt[n]{z}=v$ and $v^n=z$.

$\sqrt[n]{z} &= v\\\sqrt[n]{r(\cos \theta+i \sin \theta)} &=s(\cos \alpha + i \sin \alpha)\\[r(\cos \theta+i \sin \theta)]^{\frac{1}{n}} &= s(\cos \alpha +i \sin \alpha)\\r^{\frac{1}{n}}\left(\cos \frac{1}{n} \theta+i \sin \frac{1}{n}\theta \right) &= s(\cos \alpha+i \sin \alpha)\\r^{\frac{1}{n}}\left(\cos \frac{\theta}{n}+i \sin \frac{\theta}{n} \right) &= s(\cos \alpha+i \sin \alpha)$

From this derivation, we can conclude that $r^{\frac{1}{n}}=s$ or $s^n=r$ and $\alpha=\frac{\theta}{n}$. Therefore, for any integer $k (0, 1, 2, \ldots n -1)$, $v$ is an $n^{th}$ root of $z$ if $s=\sqrt[n]{r}$ and $\alpha=\frac{\theta+2\pi k}{n}$. Therefore, the general rule for finding the $n^{th}$ roots of a complex number if $z = r(\cos \theta + i \sin \theta)$ is: $\sqrt[n]{r} \left(\cos \frac{\theta+2\pi k}{n}+i \sin \frac{\theta+2\pi k}{n}\right)$. Let’s begin with a simple example and we will leave $\theta$ in degrees.

Example 3: Find the two square roots of $2i$.

Solution: Express $2i$ in polar form.

$& r=\sqrt{x^2+y^2} && \cos \theta=0\\& r=\sqrt{(0)^2+(2)^2} && \qquad \theta=90^\circ\\& r=\sqrt{4}=2$

$(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{90^\circ}{2}+i \sin \frac{90^\circ}{2}\right)=\sqrt{2}(\cos 45^\circ +i \sin 45^\circ)=1+i$

To find the other root, add $360^\circ$ to $\theta$.

$(2i)^{\frac{1}{2}}=2^{\frac{1}{2}} \left(\cos \frac{450^\circ}{2}+i \sin \frac{450^\circ}{2}\right)=\sqrt{2}(\cos 225^\circ +i \sin 225^\circ)=-1-i$

Example 4: Find the three cube roots of $-2-2i \sqrt{3}$

Solution: Express $-2-2i \sqrt{3}$ in polar form:

$r &=\sqrt{x^2+y^2}\\r &= \sqrt{(-2)^2+(-2\sqrt{3})^2}\\r &= \sqrt{16}=4 && \theta = \tan^{-1} \left(\frac{-2\sqrt{3}}{-2}\right)=\frac{4\pi}{3}$

$& \sqrt[n]{r}\left( \cos \frac{\theta + 2\pi k}{n}+i \sin \frac{\theta + 2\pi k}{n}\right)\\\sqrt[3]{-2-2i \sqrt{3}} &= \sqrt[3]{4} \left(\cos \frac{\frac{4 \pi}{3} + 2\pi k}{3}+i \sin \frac{\frac{4\pi}{3} +2\pi k}{3}\right) k=0, 1, 2$

$z_1 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{0}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{0}{3}\right)\right] && k=0\\&= \sqrt[3]{4}\left[\cos \frac{4\pi}{9}+i \sin \frac{4\pi}{9}\right]\\z_2 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{2\pi}{3}\right)\right] && k=1\\&= \sqrt[3]{4}\left[\cos \frac{10\pi}{9}+i \sin \frac{10\pi}{9}\right]\\z_3 &= \sqrt[3]{4}\left[ \cos \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)+i \sin \left(\frac{4\pi}{9}+\frac{4\pi}{3}\right)\right] && k=2\\&= \sqrt[3]{4}\left[\cos \frac{16\pi}{9}+i \sin \frac{16\pi}{9}\right]$

In standard form: $z_1=0.276+1.563i, z_2=-1.492-0.543i, z_3=1.216-1.02i$.

## Solve Equations

The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the complex plane, the $n^{th}$ roots are equally spaced on the circumference of a circle.

Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to include complex numbers. The easiest way to explore the process is to actually solve an equation. The solution can be obtained by using De Moivre’s Theorem.

Example 5:

Consider the equation $x^5 - 32 = 0$. The solution is the same as the solution of $x^5 = 32$. In other words, we must determine the fifth roots of 32.

Solution:

$x^5-32 &= 0 \ \text{and} \ x^5=32.\\r &= \sqrt{x^2+y^2}\\r &= \sqrt{(32)^2+(0)^2}\\r &= 32\\\theta &= \tan^{-1} \left(\frac{0}{32}\right)=0$

Write an expression for determining the fifth roots of $32 = 32 + 0i$

$32^{\frac{1}{5}} &= [32( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{5}}\\&= 2\left(\cos \frac{2\pi k}{5}+i \sin \frac{2\pi k}{5}\right)k=0, 1, 2, 3, 4\\x_1 &= 2\left(\cos \frac{0}{5}+i \sin \frac{0}{5}\right) \rightarrow 2(\cos 0+i \sin 0)=2 && for \ k=0\\x_2 &= 2\left(\cos \frac{2\pi}{5}+i \sin \frac{2\pi}{5}\right) \approx 0.62 + 1.9i && for \ k=1\\x_3 &= 2\left(\cos \frac{4\pi}{5}+i \sin \frac{4\pi}{5}\right) \approx -1.62 + 1.18i && for \ k=2\\x_4 &= 2\left(\cos \frac{6\pi}{5}+i \sin \frac{6\pi}{5}\right) \approx -1.62-1.18i && for \ k=3\\x_5 &= 2\left(\cos \frac{8\pi}{5}+i \sin \frac{8\pi}{5}\right) \approx 0.62-1.9i && for \ k=4$

## The Geometry of Complex Roots

In the previous example, we have one real and four complex roots. Plot these in the complex plane.

The $n^{th}$ roots of a complex number, when graphed on the complex plane, are equally spaced around a circle. So, instead of having all the roots, all that is necessary to graph the roots is one of them and the radius of the circle. For this particular example, the roots are $\frac{2\pi}{5}$ or $72^\circ$ apart (look in the root equation in the example, $\theta$ increased by $\frac{2\pi}{5}$). This goes along with what we know about regular pentagons. The roots are $\frac{2\pi}{n}$ degrees apart.

Example 6: Calculate the three cube roots of 1 and represent them graphically.

Solution: In standard form, $1 = 1 + 0i \ r = 1$ and $\theta = 0$. The polar form is $1 + 0i = 1 [\cos (0 + 2\pi k) + i \sin (0 + 2\pi k)]$. The expression for determining the cube roots of $1 + 0i$ is:

$(1+0i)^{\frac{1}{3}}=1^{\frac{1}{3}} \left(\cos \frac{0+2 \pi k}{3}+i \sin \frac{0+2 \pi k}{3} \right)$

When $k = 0, k = 1$ and $k = 2$ the three cube roots of 1 are $1, -\frac{1}{2}+i \frac{\sqrt{3}}{2}, -\frac{1}{2} -i \frac{\sqrt{3}}{2}$. When these three roots are represented graphically, the three points, on the circle with a radius of 1 (the cubed root of 1 is 1), form a triangle.

## Points to Consider

• If the roots can be determined, will some form of De Moivre’s Theorem be used?
• If the root of a complex number in polar form can be determined, can the solution to an exponential equation be found in the same way?
• Does the number of roots have anything to do with the shape of the graph?

## Review Questions

1. Show that $z^3 = 1$, if $z=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$
2. Evaluate:
1. $\left[\frac{\sqrt{2}}{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)\right]^8$
2. $\left [3 \left (\sqrt{3}-i\sqrt{3} \right ) \right ]^4$
3. $(\sqrt{5}-i)^7$
4. $\left[3 \left(\cos \frac{\pi}{6}+i\sin \frac{\pi}{6}\right)\right]^{12}$
3. Rewrite the following in rectangular form: $[2(\cos 315^\circ + i \sin 315^\circ)]^3$
4. Find $\sqrt[3]{27i}$.
5. Find the principal root of $(1 + i)^{\frac{1}{5}}$. Remember the principal root is the positive root i.e. $\sqrt{9}=\pm 3$ so the principal root is +3.
6. Find the fourth roots of $81i$.
7. Solve the equation $x^4 + 1 = 0$. What shape do the roots make?
8. Solve the equation $x^3-64=0$. What shape do the roots make?

Feb 23, 2012

Dec 16, 2014


text:
# Universal rigidity on the line, point order

 Authors: Robert Connelly, Steven J. Gortler, and Louis Theran Preprint: 2104.01892, 2021 Full text: arXiv

We show that universal rigidity of a generic bar and joint framework $$(G,p)$$ in the line depends on more than the ordering of the vertices. In particular, we construct examples of one-dimensional generic frameworks with the same graph and ordering of the vertices, such that one is universally rigid and one is not. This answers, in the negative, a question of Jordan and Nguyen.


text:
In a discussion on Octodon, it was suggested to simply replace the HDD with a SSD and replace the 4GB RAM with 16GB.

Looking at my system information, trying to figure out how the HDD is connected to the laptop, I find this:

NVidia MCP89 AHCI:

Vendor:	NVidia
Product:	MCP89 AHCI
Physical Interconnect:	SATA
Description:	AHCI Version 1.30 Supported

Hitachi HTS545025B9SA02:

Capacity:	250.06 GB (250'059'350'016 bytes)
Model:	Hitachi HTS545025B9SA02
Revision:	PB2AC60W
Serial Number:	100806PBL200CSHY1MMN
Native Command Queuing:	Yes
Queue Depth:	32
Removable Media:	No
Detachable Drive:	No
BSD Name:	disk0
Rotational Rate:	5400
Medium Type:	Rotational
Partition Map Type:	GPT (GUID Partition Table)
S.M.A.R.T. status:	Verified
Volumes:
EFI:
Capacity:	209.7 MB (209'715'200 bytes)
File System:	MS-DOS FAT32
BSD Name:	disk0s1
Content:	EFI
Volume UUID:	85D67001-D93E-3687-A1C2-79D677F0C2E0
Macintosh HD:
Capacity:	249.2 GB (249'199'591'424 bytes)
Available:	37.12 GB (37'120'630'784 bytes)
Writable:	Yes
File System:	Journaled HFS+
BSD Name:	disk0s2
Mount Point:	/
Content:	Apple_HFS
Volume UUID:	EA695A3A-437E-39B5-A8B9-4B5883F265A1
Recovery HD:
Capacity:	650 MB (650'002'432 bytes)
File System:	Journaled HFS+
BSD Name:	disk0s3
Content:	Apple_Boot
Volume UUID:	7981E96A-0868-350E-B49D-6F474233EFFD

The website of a local shop has 36 candidates in the “SATA III” and 250GB range for me. Yikes! Which one to pick? I’m leaning towards the Crucial MX300 (525GB, 2.5”) for CHF 165.-

– Alex Schroeder 2017-11-23 19:36 UTC

What FSF’s giving guide doesn’t say yet (the “Sales Details” link is currently broken) is that if you use the voucher code “fsfftw” there’ll be a 5% discount and Vikings donates 15 EUR to the FSF for each sale. We also don’t charge extra for Swiss chargers...

Those laptops are also available from Technoethical (Romania) and Libiquity (U.S.A.)

May I add that Purism isn’t a good choice since their hardware comes with Intel ME. @RaptorEng: “Intel ME hardware is always present at high privilege level. Given the inability to completely disable the hardware, malware will always be able to be injected via physical access. Since signing keys in ME hardware can’t be changed, difficult-to-detect malware will remain possible in affected platforms forever.”

Quitter: https://quitter.se/vikings

Thomas 2017-11-24 13:30 UTC

Heh, oh well. I’ve been using X200 for quite some time, and only “upgraded” from it to X230 just a month or two ago. I did install libreboot on it once, but unfortunately it didn’t last for long because it seems that not all X200-s are alike, and in my particular case I wasn’t able to control the backlight (or maybe it was something else, never really figured it out). Sad, but at least with libreboot flashing the original bios is possible without taking your laptop apart again

In my opinion, we are fucked. I switched to x230 because shit has to be done, I can’t be using hardware that is 8-9 years old now. I mean, I can, but then I’m just wasting my time on waiting for things to finish.

Maybe there’s a light in a sense that if we manage to libre-ize some modern laptop, it will be possible to use it for many-many years to come. But x200 is not it, unfortunately.

Oh, and 300€ for a laptop… I don’t really understand that. Freedom my ass, but I just bought x230 for 120€ locally (used, great condition). So x200 currently is just a tad under 100€, with practically an infinite supply given than all local IT companies are getting rid of them all in bulk (although maybe nowadays it’s easier to buy x230 than x200). Yeah, you’ll have to flash libreboot yourself, but that’s a fun little weekend project.

– AlexDaniel 2017-12-04 04:04 UTC

Then again, there’s the fact that laptop assembly, flashing and installing is not my hobby so I’ll happily pay somebody else to do it... In this particular case I have already ordered the Purism Librem 13 and I will be getting an SSD for the old MacBook Pro.

– Alex 2017-12-04 07:51 UTC

Yesterday I went to a local shop to buy the SDD, RAM and screwdrivers required to replace the MacBook Pro disk. They told me that Apple did not allow them to sell the necessary screwdrivers (#00 Philips and Torx T6, apparently). I told them them that I could buy those at the mail order and that without them, there was no point in buying the other stuff. They said they couldn’t and they’re sorry.

I really don’t want to buy another Apple laptop. I hope the Purism laptop does all I need.

And now I’m ordering the stuff from the mail order shop.

– Alex Schroeder 2017-12-10 12:09 UTC

I ordered:

1 x OEM Tool Kit Schraubenzieher und Bit Set CHF 26.90 incl.
1 x Samsung 850 EVO Basic (500GB, 2.5”) CHF 177.00 incl.
1 x Corsair Memory (2×, 4GB, DDR3-1066, SODIMM 204) CHF 79.00 incl.

I’m going to pick this up in a few days.

– Alex Schroeder 2017-12-10 19:42 UTC

Got the stuff, made another backup to a USB disk using Time Machine, powered it off, exchanged disk and memory, powered it up, held down the Alt (“option”) key and restored it from the backup — no problem!

This is great.

– Alex 2017-12-15 23:28 UTC


text:
# Problem: Scheduling diverse teams while minimizing frequency

I have the following problem which I have been asked to solve. I like to think that a few years ago, I'd be able to do it myself, but these days I must admit that help is good :)

Given $n$ employees $E = \{e_1,\ldots,e_n\}$, an organisation wants to construct a plan for the next $w$ weeks: each week, $k<n$ employees must be chosen to be on a specific duty in a team.

For each employee $e_i$, we would like to maximize the time between any two consecutive weeks that $e_i$ is on duty. This goes for all $i$, so it should be fair in the sense that no employee is favored.

Further, we would like to have as diverse teams as possible, so for any $1 \leq i,j \leq n$, we would like to minimize the number of times $e_i$ and $e_j$ are on the same team.

As an additional hiccup, $m<n$ of the employees are not available from the beginning, so let's say that employees $\{e_i \mid 1 \leq i \leq m\}$ are only available from week $q_i$ onwards. These $m$ employees should not be "punished" in the sense that they have to make up for work they "missed" when they were not yet employed.

Any ideas and tips on how to model this, e.g. as a MIP would be great!

As a concrete example, $n=17$, $w=17$, $k=3$ and $m=4$ where $q_1=q_2=q_3=5$ and $q_4=9$.

Edit: I want to add that this is not a homework question, but a real life problem. There is no mathematically stringent way to pose my constraints; this is up to interpretation, as is how the constraints should combine and weigh against each other in an objective function. In this lies the heart of my question. As some approaches may be a lot simpler than others, I am willing to sacrifice some stringency in return of a simpler problem/model.

• I think you need to think more about your question formulation first. I don't see a single objective function you want to minimize. If you want to minimize the time-between-weeks for all employees, that is $n$ different objective functions, and minimizing the number for one employee might increase the number for a different one. Then you also add the goal of having teams as diverse as possible. Well, you'll need a way to identify a single objective function that you want to minimize, which captures how you want to trade off between those goals. – D.W. Feb 16 '18 at 20:02
• Anyway, once you've figured that out, cs.stackexchange.com/q/12102/755 has techniques that may be helpful here. – D.W. Feb 16 '18 at 20:07
• Hi @D.W. Thanks for your comments. The goal of incorporating my requirements into a meaningful single objective function is what I'm asking for help with. I apologize if that was unclear. I realize that my goals are themselves not clearly defined, in the sense that I did not specify e.g. how bad it would be to share team with 2 previous members compared to just 1 would be, let alone how the frequency constraint weighs against the diversity constraint. As this is not a homework question, but a real life problem, these are up for interpretation. – Martin Lauridsen Feb 19 '18 at 7:13
• Further, some approaches may be a lot simpler than others. I think we're willing to trade off quite some stringency on the constraints in offer for simplicity of the problem/model. – Martin Lauridsen Feb 19 '18 at 7:14
• OK. I don't think we can do that for you. That requires figuring out how you want to trade off these different goals, and we can't make that decision for you. Only you know what you'll consider meaningful. – D.W. Feb 19 '18 at 12:09


text:
533 views

What will be the output of the following program?

#include <stdio.h>
void f1(int p1, int *p2, int **p3)
{
p1 = 20;
*p2 = p1;
**p3 = *p2;
p1 = 10;
}
int main()
{
int a = 5, b = 5, *c = &b;
f1(a, &b, &c);
printf("%d %d %d", a, b, *c);
}
1. 5 5 5
2. 5 20 20
3. 10 20 20
4. 20 20 20

a is called by value so any modification in a will be lost.
b is call also call by value but being a pointer, the dereferenced effect is visible in caller function (call by pointer) so modification will be there.
same for c.

final values of a,b,c are 5,20,20.

Why the value of b and c is 20,20?

p1=20

*p2 (is the value of b) =p1 ===> b=20

**p3(is the value of b) = *p2 ===> b=b

note that 'c' is a pointer and *c=b

Option b .


text:
# 4.10 Frequency domain problems  (Page 4/6)

 Page 4 / 6

## Circuits filter!

A unit-amplitude pulse with duration of one second serves as the input to an RC-circuit having transfer function $H(f)=\frac{i\times 2\pi f}{4+i\times 2\pi f}$

1. How would you categorize this transfer function: lowpass, highpass, bandpass, other?
2. Find a circuit that corresponds to this transfer function.
3. Find an expression for the filter's output.

## Reverberation

Reverberation corresponds to adding to a signal its delayed version.

1. Assuming $\tau$ represents the delay, what is the input-output relation for a reverberation system? Is the systemlinear and time-invariant? If so, find the transfer function; if not, what linearity or time-invariancecriterion does reverberation violate.
2. A music group known as the ROwls is having trouble selling its recordings. The record company'sengineer gets the idea of applying different delay to the low and high frequencies and adding theresult to create a new musical effect. Thus, the ROwls' audio would be separated into two parts (oneless than the frequency ${f}_{0}$ , the other greater than ${f}_{0}$ ), these would be delayed by ${\tau }_{l}$ and ${\tau }_{h}$ respectively, and the resulting signals added. Drawa block diagram for this new audio processing system, showing its various components.
3. How does the magnitude of the system's transfer function depend on the two delays?

## Echoes in telephone systems

A frequently encountered problem in telephones is echo. Here, because of acoustic coupling between the ear pieceand microphone in the handset, what you hear is also sent to the person talking. That person thus not onlyhears you, but also hears her own speech delayed (because of propagation delay over the telephonenetwork) and attenuated (the acoustic coupling gain is less than one). Furthermore, the same problem appliesto you as well: The acoustic coupling occurs in her handset as well as yours.

1. Develop a block diagram that describes this situation.
2. Find the transfer function between your voice and what the listener hears.
3. Each telephone contains a system for reducing echoes using electrical means. What simple system couldnull the echoes?

## Effective drug delivery

In most patients, it takes time for the concentration of an administered drug to achieve a constant level in the blood stream. Typically, if the drug concentration in the patient's intravenous line is ${C}_{d}u(t)$ , the concentration in the patient's blood stream is ${C}_{p}(1-e^{-(at)})u(t)$ .

1. Assuming the relationship between drug concentration in the patient's drug and the delivered concentration can be described as a linear, time-invariant system, what is the transfer function?
2. Sometimes, the drug delivery system goes awry and delivers drugs with little control. What would the patient's drug concentration be if the delivered concentration were a ramp?More precisely, if it were ${C}_{d}tu(t)$ ?
3. A clever doctor wants to have the flexibility to slow down or speed up the patient's drug concentration. In other words, the concentration is to be ${C}_{p}(1-e^{-(bt)})u(t)$ , with $b$ bigger or smaller than $a$ . How should the delivered drug concentration signal be changed to achieve this concentration profile?

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles


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# Find the volume common to two spheres, each with radius $r$if the center of each sphere lies on the surface of the othersphere.

## $$\frac{5}{12} \pi r^{3}$$

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# What is the value of x^2 + y^2 ? (1) x^2+ y^2 = 2xy + 1 (2)

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What is the value of x^2 + y^2 ? (1) x^2+ y^2 = 2xy + 1 (2) [#permalink]

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05 Jun 2011, 09:16
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What is the value of$$x^2 + y^2$$?
(1) $$x^2+ y^2$$ = 2xy + 1
(2)$$x^2 + y^2$$= 4 - 2xy
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Re: Binomial DS [#permalink]

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05 Jun 2011, 09:39
x^2 + y^2 ?

Stmt1: x^2 + y^2 = 2xy+1 Insufficient
Stmt2: x^2+y^2 = 4-2xy Insufficient

Together,
2xy + 1 = 4 - 2xy
4xy=3
xy=3/4
So, x^2+y^2 = 2*3/4 + 1
= 5/2

OA C.
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Re: Binomial DS [#permalink]

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05 Jun 2011, 17:14
OA was E.

But i agree that it should be C. If anyone thinks OA is correct please post your explanation. Thanks!
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Re: Binomial DS [#permalink]

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05 Jun 2011, 19:44
I too think the answer is C. I did the following to verify the calcualtion:

(1) is insufficient

(x-y)^2 = 1

x-y = -1 or x-y = 1

(2) is insufficient

(x+y)^2 = 4

X+y = 2 or x+y = -2

So we have the following values of x and y

x-y = -1
X+y = 2

x = 1/2, y = 3/2

x-y = 1
X+y = 2

x = 3/2, y = 1/2

x-y = -1
x+y = -2

x = -3/2 y = -1/2

x-y = -1
x+y = -2

x = -3/2 y = -1/2

So value of x and y in each case should be same

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06 Jun 2011, 00:10
a x-y = +|- 1 not sufficient.

b x+y = +|- 2 not sufficient.

a+b solving we get x = +|- 1.5 and +|- 0.5 similarly y = +|- 1.5 and +|- 0.5.

hence x^2 + y^2 = (1.5)^2 + (0.5)^2 = 2.25 + 0.25 = 2.5 (for fixed set of x,y as solutions)

Thus C.
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Re: Binomial DS   [#permalink] 06 Jun 2011, 00:10
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# What is the value of x^2 + y^2 ? (1) x^2+ y^2 = 2xy + 1 (2)

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Force on a safety strap

Tags:
1. Jul 12, 2015

Postal Pete

1. The problem statement, all variables and given/known data
A 100 kg man is walking on a tight cable. His safety harness has a strap that is attached to an overhead cable that will allow him to fall 1 meter before stopping him from falling any further. What is the force on the strap when it stops his fall at 1 meter.

2. Relevant equations
v = at
Fd = KE = .5mv^2
a = 9.81 m/s^2
Weight = ma = m*9.81
3. The attempt at a solution
v = sqrt (2*9.81*1) = 4.43 m/s
KE = .5*100*(4.43)^2 = 981 J
Fd = F*1 = KE = 981 J
Force on the strap at the end of the 1 m fall is 981 N
The man's weight is ma = 100*9.81 = 981 N
Shouldn't the force on the strap after the one meter fall be greater than the man's weight?

2. Jul 12, 2015

andrewkirk

There's a missing piece of the question, which is how long the strap takes to arrest his fall. That will be a function of how much elasticity there is in the strap and in the cable from which it hangs.

Note that it is impossible for them to be infinitely stiff, and hence the fall cannot be arrested instantaneously. If it were, the deceleration, and hence the force, would be infinite, and the man would be pulverised.

In practice the force on the strap will vary over time, going from near zero when it first begins to take his weight, increasing to a maximum, and then reducing until it just equals his weight.

What they probably want you to give is the average force, given a period of time t for the deceleration. THat force will be the man's weight, plus an extra force that decelerates him.

If you work out the momentum he has when he has fallen one metre then you can work out the average force using the equation p (momentum) = F t.

3. Jul 13, 2015

haruspex

I concur with andrewkirk (except that there may be some bounce, so the force might oscillate before settling out at equal to the man's weight).
But you may be puzzled as to what was wrong with your own solution
It will be clearer if we stick to symbols. You computed v from $\frac 12mv^2=mgh$, then set $\frac 12mv^2=Fh$. Clearly that will produce F=mg. You effectively treated the retarding force as constant over the whole metre of descent.

4. Jul 13, 2015

Postal Pete

Thanks for your response. I had not thought about the elasticity factor. The reason that I was interested in this problem comes from being on what was called an "Adventure Ropes Course" (although you were actually walking on steel cables) and being told by the guy who got you into your harness that the breaking point of the strap was 5000 pounds and you don't weigh 5000 pounds so don't worry about the strap breaking. But I knew that if I did fall the length of the strap the force on the strap would be greater than my weight. The time of fall would be from d = .5at^2. Solving for t, t = sqrt (2d/a) = sqrt (2*1/9.81) = .45 s. From p = mv = Ft, solving for F, F = mv/t = 100*4.43/.45 = 984 N, which is only 3 N more than the man's static weight but which I think is correct.

5. Jul 13, 2015

Postal Pete

Thanks for your response. Expressing the problem using just symbols clarifies my error.

6. Jul 13, 2015

andrewkirk

No it's much more than that. There are two errors in the calc.

Firstly, the time you must divide the momentum by to get the deceleration force is the time of deceleration, not the time of the fall. The deceleration time will be much less, unless there's a lot of give in the overhead cable. I would guess at a typical deceleration time of about one-third of the fall time - ie about 0.15s rather than 0.45s - so you need to triple your 984N.

Secondly, you have only included the force of deceleration. You need to add to that the man's weight to get the total force.

So the final force may be about four times the man's weight. Fortunately, that's still a very long way short of the breaking strain of the strap!

7. Jul 13, 2015

Postal Pete

8. Jul 13, 2015

Postal Pete

Thanks for your interest and help with this problem. I can see why I could not find the solution in my first year physics text.