The mathematical problems underlying Information Security
Modern information protection methods are based on a couple of mathematical problems that are believed to be hard to solve. In this note these problems and their cryptographic applications are discussed.
Many people prefer to keep their personal records confidential. To maintain confidentiality while sending personal data across the Internet -while e-banking, for instance-the data needs to be protected. Many common web-based applications do indeed provide protection, often without the user even being aware of it.
Information protection methods use encryption : before it is sent, the data is garbled into an incomprehensible looking string of symbols, in such a way that only the intended recipient can make sense of it and recover the original information. Anyone who eavesdrops on the communication does not learn anything, except that communication is taking place. Similar methods are used to establish a user’s identity.
This does not imply that encryption suffices. Miscreants may try to gain improper access to information using any of an ever greater variety of other methods. Weaknesses in operating systems can be exploited to get remote access to a user’s computer, a user’s password may be stolen or guessed, or a user may be tricked into revealing passwords using social engineering, etc. Thus, encryption alone does not guarantee confidentiality of a user’s data. But it is a necessary part of the protection chain.
This note concentrates on the basic mathematical tools that underly encryption and the corresponding decryption. How these tools are embedded in actual practical systems, and the many precautions that must be taken when doing so, is not discussed.
Number theory plays an important role in all widely accepted basic encryption tools .It is the branch of mathematics that studies, roughly speaking, integer numbers such as 0,1,2,3, ....An entertaining activity to some, but long thought to be mostly useless by most others. That changed when cryptographic applications of prime numbers were discovered. A positive integer is prime if it is evenly divisible only by 1 and itself. For instance, 17 is a prime number, since it is evenly divisible only by 1 and by 17 itself. But 91 is not prime because 91 = 7 *13, so 91 is evenly divisible by a number other than 1 and 91, namely 7 (or 13). The number 1 is not a prime. Number theorists have studied prime numbers for ages, but their cryptographic potential was discovered only relatively recently (approximately in the 1970s, depending on whose claims one is willing to believe). These cryptographic applications cleverly exploit two problems involving primes :
- Factoring integers, and
- Computing discrete logarithms.
These problems have been studied extensively, in particular since the discovery of their cryptographic relevance. Only little progress was made, and it is widely believed that the two problems are ’hard’. This ’hardness’ is precisely what is required for the cryptographic applications, so from that point of view it is to be hoped that the problems remain unsolved. It should be kept in mind, however, that progress is conceivable and that it could in principle go either way : a proof of hardness or an efficient solution method. The former would provide the currently missing theoretical justification for the security of the cryptographic applications. The latter would mean that all widely accepted basic encryption tools can efficiently be broken : hardly a desirable outcome but unfortunately a possibility that cannot be discounted yet. One smart idea can still change the world-a dangerous or enviable proposition, depending on one’s point of view.
In the next sections, prime numbers and the two problems mentioned above are briefly discussed along with their cryptographic applications,
Examples of prime numbers abound. There arealready 25 primes less than 100 :
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
It can easily be argued that there are infinitely many prime numbers. If there would be a finite list of primes, then the product over all numbers on the list (say P ) is obviously evenly divisible by each of the prime numbers on the list. On the other hand P +1 is not evenly divisible by any of the numbers on the list. So, P +1 must be a prime number itself (and not on the list) or be evenly divisible by a prime smaller than P +1 that is not on the list. Because any finite list of primes can be extended with at least one other prime there are infinitely many primes.
This simple observation that there are infinitely many primes is made more explicit by a less straightforward result, the Prime Number Theorem. A precise statement of this theorem is not required here. Roughly speaking, it says that the chance that a randomly selected d-digit number is prime is proportional to 1/d. This allows to make estimates such as : ’there are more than 1096 primes that have one hundred decimal digits’. The fact that there are so ’many’ primes is important for cryptographic applications. For instance, it implies that if two different persons each select at random a prime of, say, one hundred decimal digits, the chance is negligible that they pick the same prime.
This example raises three questions. Why would anyone want to generate such huge primes ? And why would it be important that different people most likely select different primes ? These two questions are addressed in the next section. The final question, how one would go about selecting a random prime of any size, is addressed in the remainder of this section.
Prime numbers can easily be distinguished from numbers that are not prime (a number that is not a prime number is called composite). To test whether or not a positive integer n is a prime, the ’obvious’ approach would be to check if n is evenly divisible by a positive integer less than n (note that it suffices to restrict the search to the primes that are at most √n). Although for some composite n’s this method may work quickly, in general it cannot be expected to work quickly-even worse, for large n it cannot be expected to work at all because it would take too much time. A less obvious but efficient method to recognize if n is composite is based on Fermat’s little theorem. This theorem says that if p is a prime number then for any integer number a the number ap a is evenly divisible by p. Thus, if an integer a can be found such that n - does not divide an a, then n cannot be prime (because, if it were, n would n - divide an - a due to Fermat’s little theorem).
Fermat’s little theorem leads to a fast and practical method to recognize composite numbers if two additional facts are taken into account. In the first place, and using a slightly sharpened version of the theorem, if n is composite then most a-values in 2, 3,...,n −1would attest to that fact (namely, because n would not divide an a). Furthermore, even if n counts thousands of digits, it can be verified quickly on any current computer whether or not n divides a−a. This is based on a computational trick called modular exponentiation. This trick has many applications in cryptography. Developing ever faster methods for modular exponentiation is a favorite research topic. More precisely, what modular exponentiation does for any integers m, b, and e ≥0 is calculating be mod m, which is a shorthand for ’the remainder upon division by m of the number be ’. The integers m, b, and e are often referred to as the modulus, base, and exponent, respectively. To check if n divides an a, modular exponentiation is applied to modulus n, base a, and exponent n, and for the resulting value a mod n it is checked if (an mod n)−a is evenly divisible by n.
Thus, to prove that n is composite, randomly select values a from2, 3,...,n−1and use modular exponentiation to check if n divides an −a, until an a is found for which it does not. If, after repeated attempts with randomly selected a-values, n turns out to divide all corresponding (an a) -values, the compositeness proof for n fails. In that case, because the proof would most likely not have failed if n is composite, it must be assumed that n is prime. The latter is just an assumption. It is not a mathematical proof that n is prime. The ’alleged primes’ resulting from a failed compositeness proof are sometimes referred to as industrial primes. For all practical purposes, an industrial prime is prime, except that it has not been proved that the number is prime.
Combining the possibility of fast recognition of composites (and thus of industrial primes) with the relatively high density of primes (the Prime Number Theorem), leads to a relatively fast method to generate random (industrial) primes of any number of digits d : repeatedly pick a random d-digit number until one is found that is an industrial prime. From a practical point of view this answers the third question raised above.
The above method to generate primes trades mathematical rigor for practicality. A mathematically rigorous proof that the result is indeed prime can be obtained using entirely different methods. For the vast majority of industrial primes used in practical information protection scenarios, however, such proofs have not been obtained, and would not add much value either.
Integer Factorization and RSA
Multiplication of integer numbers is easy. School kids memorize facts such as ’3 times 5 equals 15’, and learn how to use the elementary tables of multiplication to solve more challenging multiplication problems such as 123 * 456 = 56088 using the schoolbook method of multiplication. In this method each digit of the first number is multiplied by each digit of the second number, and the product is put together using simple additions. Numbers of any length can quickly be multiplied in the same way : on a computer numbers of thousands of decimal digits can be multiplied in a fraction of a second. Much faster methods exist, but they are beyond the scope of this note.
Doing a multiplication is easy, but undoing it seems to be hard : given a positive composite integer, say n, find two integers greater than 1 that when multiplied together produce n. These integers are called factors of n. For ex-ample, given n = 15, find 3 and 5 (since 3 *5 = 15). This problem is known as integer factorization. In general it is believed to be a hard problem. As mentioned in Section 1, however, the hardness of integer factorization has not been proved. The problem is easy on a quantum computer, but so far no one knows how to build one. In the remainder of this section it is shown how the alleged difficulty of integer factorization can be exploited for cryptographic purposes, and the state of the art of integer factorization algorithms is briefly discussed. Cryptographic application. Suppose that someone, say Alice, wants to en-able anyone else to send encrypted messages to her. An elegant way to do this is the RSA public key crypto system, named after its inventors Ronald Rivest, Adi Shamir, and Leonard Adleman. It is a called a public key cryptosystem because Alice has to publish certain key material, allowing anyone to encrypt messages intended for her, while keeping secret a corresponding private key that allows her to decrypt the encrypted messages sent to her. Obviously, for such a scheme to work it should be impossible to derive Alice’s private key from the published public key, since otherwise anyone could perform the decryption. For RSA this impossibility requires the hardness of integer factorization. It works as follows.
As mentioned in Section 2, Alice can reasonably quickly generate random primes of virtually any size. Let pA and qA be two distinct random primes that she generated, such that factoring their product nA = pA *qA is believed to be infeasible. This means that Alice can safely publish nA while keeping secret ’her’ primes pA and qA : even though pA and qA are uniquely determined by nA there is simply no efficient (known) way to find pA and qA based on the knowledge of just nA (if pA and qA are chosen appropriately). Except for pA and qA Alice also determines an integer eA and calculates an integer dA such that eA * dA −1 is evenly divisible by both pA −1 and qA −1. Given eA, pA, and qA the computation of dA can be done very quickly using another important computational trick, the extended Euclidean algorithm. The integer eA is published by Alice (along with nA), but she keeps secret dA (alongwith pA and qA). The pair (nA,eA) is Alice’s public key, consisting of Alice’s public modulus nA and her public exponent eA. Alice’s corresponding secret or private key consists of dA, the secret or private exponent.
The primes pA and qA can in principle be discarded after the set-up phase, but because they can be used to derive the private exponent dA (since eA is publicly known), they should in any case not be made public. It follows that, for the security of the RSA cryptosystem, factoring the public modulus nA should be difficult. It has however not been proved that the factorization of nA is needed to derive dA from eA and nA.
The set-up may look cumbersome. But, Alice has to do it just once. And, once it has been done, anyone who knows Alice’s public key can easily encrypt messages resulting in ciphertexts that only Alice can decrypt : for a message M, which is a positive integer less than nA, the encryption E(M) equals MeA mod nA.The corresponding decryption looks just as simple : to decrypt E(M) Alice calculates (E(M))dA mod nA which is equal to the original M. Both processes make us of modular exponentiation (cf.previous section).
If RSA is used to exchange plain text messages, they need to be converted to integer values in some agreed upon fashion. There are lots of ways in which this can be done. For instance, one could encode ’a’ as ’01’, ’b’ as ’02’, etc., until ’z’ as ’26’(andthus ’secret’ as ’190503180520’), or use any other more efficient scheme.
For readers interested in more details it is shown that the decryption
indeed produces M. Substituting the identity
E(M)= MeA mod nA
into the decryption
It follows that the decryption equals
MeA*dA mod nA.
For the decryption to work, this should be proved equal to M. The exponentseA and dA were constructed such that eA *dA −1is evenly divisible by both pA −1 and qA −1. In particular, there is an integer value k such that eA *dA −1= k *(pA −1), which can be written as eA *dA =1+k *(pA −1). This implies that the decryption equals
M1+k*(pA−1) mod nA,
which is the same as
M1 *Mk*(pA−1) mod nA.
It follows from Fermat’s little theorem (cf. previous section) and the fact that pA is prime that pA divides MpA −M, and thus M *(MpA& minus ;1 −1). Because pA is prime and because it may safely be assumed that pA does not divide M, it follows that pA divides MpA& minus ;1 −1. This is the same as saying that MpA& minus ;1 mod pA equals 1. But if MpA& minus ;1 mod pA equals 1, then Mk*(pA−1) mod pA equals 1k and thus 1. Therefore, the remainder upon division by pA of the decryption M1 *Mk*(pA−1) mod nA (while using the fact that pA divides nA)equals
M1 *1 mod pA
It is proved similarly (but with a possibly different k-value) that the remainder upon division by qA of the decryption equals
M mod qA.
Because the remainder of the decryption upon division by pA and qA equals M mod pA and M mod qA, respectively, the Chinese remainder theorem now learns that the remainder upon division by the product pA *qA = nA of the decryption equals M mod(pA *qA)= M mod nA, but because M is positive and less than nA,it follows that the decryption equals M itself.
Despite the deceptively simple formulation, and the fact that a lot of effort has been invested to make modular exponentiation as fast as possible, encryption and decryption are too slow for real-time encryption and decryption of large volumes of data. In practice the method is typically used to exchange keys which are then used along with much faster symmetric key encryption. An other important application of the RSA cryptosystem is digital signatures. Without further precautions ,RSA as described here is vulnerable to all kinds of attacks. All these issues are beyond the scope of this note.
The efficiency of the RSA cryptosystem decreases rapidly with larger modulus sizes. Therefore, in practice one tries to get away with the smallest modulus size that still offers an acceptable level of security : obviously a modulus such as n =15 does not offer any security (since its factors 3 and 5 are easy to find), but a very secure modulus of millions of digits is practically infeasible. Finding the right middle ground between the security and practicality requirement depends on the state of the art of integer factoring algorithms, and is further commented on below.
Suppose Alice has received a confidential note from Bob in the manner described above. How does she reply to Bob in a confidential manner ? Simply by using the above scheme along with Bob’s public key (nB,eB). Thus, Bob too has to go through the set-up phase, generating his random primes pB and qB and his public and private exponents. For more general communications involving more parties, each party has to generate its own public and private key, and everyone’s public key has to be made publicly available. This leads to the need for so-called Public Key Infrastructures (PKIs) and Certification Authorities (CAs) to certify the link between parties and their public keys. All these issues are beyond the scope of this note.
One remaining issue is how one assures that different parties select different primes.If any two parties, say Alice and Bob, happen to select even one identical prime, then the greatest common divisor (which can very quickly be computed using the classical Euclidean algorithm) of their public moduli nA and nB would reveal this common prime -and thus the secret keys of both Alice and Bob- to anyone who cares to perform this simple calculation. To address this concern it suffices to notice that there are so many primes (assuming appropriate size) that if Alice and Bob use proper random number generators the chance is negligible that this occurs : the chance to win the lottery ten times in a row is higher, even if many more than two parties are involved.
Integer factorization algorithms. There are two types of integer factorization algorithms : special purpose and general purpose methods. For the former ones the time to success depends mostly on special properties of the factor one finds, for the latter it depends just on the size of the number being factored. Because moduli as used in the RSA cryptosystem cannot be expected to have factors with special properties, special purpose factoring methods are not relevant for this note.
The most common special purpose factoring methods are the following. If an integer of d decimal digits has a factor of k decimal digits, the factor can be found using trial division in time proportional to 10k *d, which is acceptable if k is small. Pollard’s rho method finds the same factor in expected time proportional to 10k/2 *d2, which is much faster than trial division if k gets large. Pollard’s p −1 method finds a prime factor p in expected time proportional to B *d2, where B is the largest prime factor in p −1. Currently the fastest special purpose factoring method is the Elliptic Curve Method. It finds prime factors larger than 50 decimal 8 digits on a regular basis. At this point the largest factor it has found is a prime factor of 74 decimal digits of a 162 decimal digit composite.
An accessible description of general purpose factoring algorithms is Carl Pomerance’s A tale of two sieves[see www.ams.org/notices/199612/pomerance.pdf ]]. All currently known methods arebased on the same simple idea, but no substantial progress has been made since the invention in the late 1980s of the Number Field Sieve, the current fastest general purpose integer factoring algorithm. This lack of progress may be an indication that this ’same simple idea’ that factoring researchers have been pounding on since the mid 1970s is a dead end and that an entirely new idea is needed to make further progress.
All that is known-or can reasonably be predicted, even though making predictions of this sort is risky-is the following. In the late 1970s an RSA modulus of 129 decimal digits was published, along with the claim that factoring it would take 40 quadrillion years. Back then, this sounded reasonable. In 1994 this 129-digit modulus was factored. In the mid 1990s 155-digit moduli were widely used in actual implementations of the RSA cryptosystem. In 1999 the first 155-digitRSA modulus was factored, thereby showing that this widely used size no longer offers adequate security. RSA moduli having about 200 decimal digits have recently been factored, but this requires a substantial effort involving hundreds of computers and months of computing time. Assuming somewhat refined versions of current algorithms (and more computing power) one may expect that moduli of about 300 decimal digits can be factored within the next 10 or 20 years. The current practical use of RSA moduli of about 300 decimal digits thus seems fine, but they cannot be relied upon for long term use. Moduli of 500 or more decimal digits look out of reach for current methods. But how long before a single new bright idea affects those moduli as well ? Only the future will tell.
Discrete Logarithms and the Diffe-Hellman Protocol
As mentioned in Section 2, modular exponentiation can be use to efficiently calculate be mod m, the remainder upon division by m of be, for any modulus m, base b, and exponent e. Undoing a modular exponentiation is the problem where a modulus m, base b, and a positive integer y that is less than m are given and where an exponent e must be calculated such that y = be mod m. This problem is commonly referred to as the discrete logarithm problem. There are many remarkable parallels between the discrete logarithm problem and integer factorization : for properly chosen m and b it is believed to be hard, the hardness has not been proved, it is easy on a quantum computer, the alleged hardness can be used for cryptographic purposes, and all known algorithms to compute discrete logarithms are very similar to integer factorization algorithms. In this section a famous cryptographic application and some discrete logarithm algorithms are briefly discussed.
The Diffe-Hellman protocol. The Diffe-Hellman protocol is a key agreement protocol. It allows two parties, say Alice an Bob, to agree on a secret piece of information, commonly referred to as a key, by communicating over a public channel. Afterwards, the resulting key can be used by Alice and Bob to communicate confidentially (and efficiently) over the same public channel, using symmetric key encryption.
In the Diffe-Hellman protocol it is assumed that a suitable modulus m and base b are known system-wide. Alice simply picks an exponent eA ≥0at random, uses modular exponentiation to calculate beA mod m, and transmits the resulting value beA mod m to Bob. Bob, similarly, picks an exponent eB 0 at random, ≥ uses modular exponentiation to calculate beB mod m, and transmits the resulting value beB mod m to Alice. Alice then uses her exponent eA and the value beB mod m received from Bob to calculate (beB mod m)eA mod m, while Bob uses his exponent eB and the value beA mod m received from Alice to calculate (beA mod m)eB mod m. But the value (beB mod m)eA mod m calculated by Alice equals beB*eA mod m, and so does the value (beA mod m)eB mod m calculated by Bob (since eB *eA equals eA *eB). So, at this point Alice and Bob share the ’key’ beA*eB mod m.
If the value eA can be derived from the system parameters m and b and the value beA mod m that is transmitted by Alice over the public channel, then the shared key beA*eB mod m can be computed using modular exponentiation from m, eA, and the value beB mod m transmitted by Bob. So, the security of the resulting key relies on the difficulty of computing the discrete logarithm eA of the transmitted value beA mod m. Thus, m and b must be chosen in such a way that computing this discrete logarithm is too hard. Vice versa, the security also depends on the difficulty of computing the discrete logarithm eB of the other transmitted value(beB mod m). Though computing a discrete logarithm suffices to derive the key beA*eB mod m from the two transmitted values beA mod m and beB mod m (and the public parameters m and b) it has not been proved that computing a discrete logarithm is necessary : in principle there may be a shortcut that leads to the key beA*eB mod m by combining the available values m, b, beA mod m, and beB mod m in some other way. The problem of finding beA*eB mod m given m, b, beA mod m, and beB mod m is known as the Diffe-Hellman problem. It is at most as hard as the discrete logarithm problem, but may be easier. Nevertheless, it is not uncommon to say that the security of the Diffe-Hellman protocol relies on the difficulty of the discrete logarithm problem (though it mayactually be easier), since if the discrete logarithm problem can be solved, then the Diffe-Hellman prototol can be ’broken’ (i.e., the key can be found) by anyone eavesdropping on the communication.
As usual, the elementary description as given above is vulnerable to all kinds of attacks unless precautions are taken. The most infamous (and common) one is the man in the middle attack : if Alice and Bob do not make sure that they actually communicate with each other they could each be setting up separate secret keys with a third party, which could then have easy access to all confidential data that Alice and Bob want to exchange.
Discrete logarithm algorithms. The difficulty of the discrete logarithm problem depends to a large extent on properties of the modulus m and the base b other than just sheer size. It is assumed that m and b are carefully chosen to avoid ’easy’ cases.
discrete logarithm problem.
If the modulus m is composite then, as a consequence of the Chinese remainder theorem, the discrete logarithm problem can be reduced to discrete logarithm problems involving the prime factors of m. In some applications m is constructed in such a way that its factorization is hard to find, but it is more common to select a prime modulus m. Here it is assumed that m is prime. An important number is the smallest positive integer ℓ for which bℓ mod m equals 1 : the order of b modulo m. Solving the discrete logarithm problem for some value y consists of finding a value e such that be mod m equals y. Writing e = k *ℓ + r for some integer r with 0 ≤r<ℓ, it follows that br mod m equals y because bℓ mod m equals 1. Therefore, the discrete logarithm problem can be solved by comparing y to the ℓ values b0 mod m, b1 mod n, ..., bℓ−1 mod m, one of which must be a match if the problem allows a solution. To make this exhaustive search infeasible the order ℓ of b modulo m must be sufficiently large. Similar to the trial division method of integer factorization, the time required for this method is proportional to 10k if e turns out to be a k-digit integer. A discrete logarithm variant of Pollard’s rho method applies too, reducing the 10k to expected 10k/2. But, this is not the end of the story, as far as the order ℓ of b modulo m is concerned. The Pohlig-Hellman method implies that the difficulty of the discrete logarithm problem is not determined by the size of ℓ itself but by the size of the largest prime factor of ℓ. Therefore, the base b is usually chosen with the additional requirement that the order ℓ of b modulo m is prime. Complicating the picture even further is the fact that the order ℓ of b modulo m is a factor of m −1. This is a consequence of Fermat’s little theorem and the assumption that m is prime, To summarize, the modulus m is chosen as a prime such that m −1 has a k-digit prime factor ℓ where k is large enough to make a computational effort proportional to 10k/2 infeasible, and b is chosen such that its order modulo n equals ℓ. Now, if the exponent e is chosen as some random integer in the range from 0 to ℓ −1, then the discrete logarithm computation of the number y =(be mod m) will-most likely-not be unexpectedly easy.
Under the assumption that the discrete logarithm problem cannot be solved using special properties of m or b, the difficulty of the problem is determined by the size of m. The best methods known in this case are all very similar to general purpose integer factoring methods. For n and m of approximately the same size, the effort required to factor an RSA modulus n is comparable to solving a (general) discrete logarithm problem with prime modulus m. Thus, parameters of about the same size should be used for RSA and the Diffe-Hellman protocol as discussed here. As is the case for integer factorization, there is no mathematical proof that any practically applicable discrete logarithm problem is indeed hard.
Other discrete logarithm problems. The first proposed cryptographic applications of the discrete logarithm problem all used the simplest case described above based on plain integer numbers. It was later realized that there are instances of the discrete logarithm problem involving other mathematical structures that are suitable for cryptographic applications as well and that may even offer certain advantages compared to the more traditional integer approach. A popular example is Elliptic Curve Cryptosystems which involve the use of advanced mathematical structures called elliptic curves. Unfortunately, as in the integer case, there is no proof of the hardness of the elliptic curve discrete logarithm problem.
To an outsider it may seem strange that, deep down, the security of virtually all electronic communications is based on a couple of mathematical problems that are not well understood. The security argument relies solely on years of failed solution attempts, not on rigorous mathematical reasoning. This is worrying to some, but does not seem to be a concern to practitioners. Other method, relying on different mathematical problems, have been proposed. But they have never reached the level of popularity of factoring or discrete logarithm based systems, and most problems are regarded with even more skepticism than the ones discussed here. All we can do at this point is keep looking for better alternatives and hope that quantum computers cannot be built.
Cherchez ...- dans tous les Flash informatique
(entre 1986 et 2001: seulement sur les titres et auteurs)
- par mot-clé
Cette page est un article d'une publication de l'EPFL.
Le contenu et certains liens ne sont peut-être plus d'actualité.
ResponsabilitéLes articles n'engagent que leurs auteurs, sauf ceux qui concernent de façon évidente des prestations officielles (sous la responsabilité du DIT ou d'autres entités). Toute reproduction, même partielle, n'est autorisée qu'avec l'accord de la rédaction et des auteurs.
Archives sur clé USBLe Flash informatique ne paraîtra plus. Le dernier numéro est daté de décembre 2013.
Taguage des articlesDepuis 2010, pour aider le lecteur, les articles sont taggués:
- tout public
que vous soyiez utilisateur occasionnel du PC familial, ou bien simplement propriétaire d'un iPhone, lisez l'article marqué tout public, vous y apprendrez plein de choses qui vous permettront de mieux appréhender ces technologies qui envahissent votre quotidien
- public averti
l'article parle de concepts techniques, mais à la portée de toute personne intéressée par les dessous des nouvelles technologies
le sujet abordé n'intéresse que peu de lecteurs, mais ceux-là seront ravis d'approfondir un thème, d'en savoir plus sur un nouveau langage.


Welcome! I am a fifth year Ph.D. student in mathematics at the University of Georgia. I visited the Max Planck Institute for Mathematics in Bonn during the 2019-2020 academic year. I earned my B.A. in mathematics at Saint Louis University in 2016.
I can be reached at seblackwell (at) uga (dot) edu.
I co-organized a virtual mini-conference series for graduate students in topology and geometry, held on April 25, June 6, and July 18. See details here.


Clojure.java contains support for Ratio types. It would be nice to have them in Clojurescript as well, but as far as I can tell this would be new development (please comment if I'm wrong). That is, there is no implementation of Ratio types in GClosure so this feature would need to be implemented from the ground up. In additional to the Ratio type, the following corresponding functions would also need implementations:
Plus the ratio type would need to be rolled into the existing mathematical frameworkings.
Imported from github issue #66


Phone: 856-378-5020 x 816
Degrees and Certifications:
Masters in Elementary Education (K-8) Masters in Mathematics Curriculum & Instruction Special Education Certified
Mrs. Selena Mohan
Greetings! My name is Mrs. Mohan and I am the middle school mathematics teacher and also teach social studies to my homeroom students. I have my masters in education and mathematics curriculum and instruction and am special education certified teacher. I also teach college mathematics in the evening and private tutor to students in K - 8. This is my second year at Pineland Learning Center. Prior to coming to Pineland I ran the developmental mathematics program at Cumberland County College for 5 years and also taught there for a total of 10 years. Entering the teaching field I began my classroom experience at a Catholic School in Bridgeton, which was unfortunately closed due to changes in the Diocese, but I also taught middle school mathematics for 2 years there as well science for 1 year.
My educational goal is to help my students realize that math is important, although they might not use every concept that is taught to them every day, they are using math each day. I also want my students to challenge themselves and not give up if they stumble.


Textron Systems Team Wins Big Stem Challenge
May 30, 2018
In May, Greg Dubicki, Shannon Dubicki, Jenny Harper, Ken Powers and Melinda Saraullo of Textron Systems competed in a local Corporate Science, Technology, Engineering and Mathematics (STEM) Challenge event hosted by Junior Achievement of South Central Pennsylvania. The challenge was divided up into five areas of competition: Electrical, Relay, Math, Civil Engineering, and Robotics.
Points are awarded based on assignments completed in the time allotted. After a day filled with difficult challenges, the Textron Systems team won first place in both the Electrical Competition and Civil Engineering Competition, second place in the Relay Competition and were selected as the overall winners of the STEM challenge. Congratulations to our winners!


Table of Contents
An algorithm is a well-ordered collection of unambiguous and effectively computable operations that when executed produces a result and halts in a finite amount of time.
Characteristics of Algorithms
- Algorithms are well-ordered.
- Algorithms have unambiguous operations.
- Algorithms have effectively computable operations.
- Algorithms produce a result.
- Algorithms halt in a finite amount of time.
These characteristics need a little more explanation, so we will look at each one in detail.
Algorithms are well-ordered
Since an algorithm is a collection of operations or instructions, we must know the correct order in which to execute the instructions. If the order is unclear, we may perform the wrong instruction or we may be uncertain which instruction should be performed next. This characteristic is especially important for computers. A computer can only execute an algorithm if it knows the exact order of steps to perform.
Algorithms have unambiguous operations
Each operation in an algorithm must be sufficiently clear so that it does not need to be simplified. Given a list of numbers, you can easily order them from largest to smallest with the simple instruction “Sort these numbers.” A computer, however, needs more detail to sort numbers. It must be told to search for the smallest number, how to find the smallest number, how to compare numbers together, etc. The operation “Sort these numbers” is ambiguous to a computer because the computer has no basic operations for sorting. Basic operations used for writing algorithms are known as primitive operations or primitives. When an algorithm is written in computer primitives, then the algorithm is unambiguous and the computer can execute it.
Algorithms have effectively computable operation
Each operation in an algorithm must be doable, that is, the operation must be possible to do. Suppose you were given an algorithm for planting a garden where the first step instructed you to remove all large stones from the soil. This instruction may not be done if there is a four-ton rock buried just below ground level. For computers, many mathematical operations such as division by zero or finding the square root of a negative number are also impossible. These operations are not effectively computable so they cannot be used in writing algorithms.
Algorithms produce a result
In our simple definition of an algorithm, we stated that an algorithm is a set of instructions for solving a problem. Unless an algorithm produces some result, we can never be certain whether our solution is correct. Have you ever given a command to a computer and discovered that nothing changed? What was your response? You probably thought that the computer was malfunctioning because your command did not produce any type of result. Without some visible change, you have no way of determining the effect of your command. The same is true with algorithms. Only algorithms which produce results can be verified as either right or wrong.
Algorithms halt in a finite amount of time
Algorithms should be composed of a finite number of operations and they should complete their execution in a finite amount of time. Suppose we wanted to write an algorithm to print all the integers greater than 1. Our steps might look something like this.
- Print the number 2.
- Print the number 3.
- Print the number 4.
While our algorithm seems to be pretty clear, we have two problems. First, the algorithm must have an infinite number of steps because there are an infinite number of integers greater than one. Second, the algorithm will run forever trying to count to infinity. These problems violate our definition that an algorithm must halt in a finite amount of time. Every algorithm must reach some operation that tells it to stop.
Advantages of Algorithm
- It is a step-by-step rep. of a solution to a given problem ,which is very easy to understand.
- It has got a definite procedure.
- It easy to first develop an algorithm, then convert it into a flowchart & then into a computer program.
- It is independent of programming language.
- It is easy to debug as every step is got its own logical sequence.
Disadvantages of Algorithm
- It is time consuming & cumbersome as an algorithm is developed first which is converted into flowchart & then into a computer program.
Algorithm to find out number is odd or even
step 1 : start
step 2 : input number
step 3 : rem=number mod 2
step 4 : if rem=0 then
print “number even”
print “number odd”
step 5 : stop


My name is Christie Williamson. I have worked in the Carroll County School system for twenty years. The last ten years, I have had the pleasure of teaching at Roopville Elementary.
I graduated from University of West Georgia with a Bachelor degree in Education in 2000 and again in 2005 with an Educational Leadership and Supervision degree. Math is my passion and I get to work with students daily in fifth grade math. I am the sponsor for Jr. Beta Club and co-sponsor for math and technology teams. I also recently worked diligently with our STEM educators to earn Carroll County and National STEM certification.


By Michael HartleyFrom this web page, you can get common denominator examples - examples showing how to put two fractions over a common denominator. To get the examples, follow these steps -
- Firstly, click Generate Example below. You'll be given two fractions, and asked to put them over a common denominator.
- When you've tried this, click Show Solution to check your answer.
- If you want to see how to get the answer shown, click Show Method, and you'll see the steps.
3953 Examples Generated So Far,
If you like what you've just read, sign up for this site's free newsletters:


You burn about 0.17 calories for every step you climb, so you burn roughly a calorie and a half for every 10 upward steps. 4. You also burn calories going down. Every stair descended burns about 0.05 calories, so you burn 1 calorie for every 20 steps down.
Example. 10 steps going up = 1.7 calories meaning 20 steps (1st floor) = 3.4 calories and if you office is on the 5th floor it equates to 17 calories. And coming down 100 steps (5th floor) = 5 calories. So in total you burn about 22 calories going up to the 5th floor and back down. If you do this 5 times a day you will burn around 110 calories – more than a bag of salt and vinegar walkers crisps!!


It's that, not matter what technique, ruler, or method I use -- the "wings" never come out even across the top.
Today, I want to share with you a method of making Flying Geese that requires no math, no measuring, and makes four geese at a time.
The Lazy Girl's "Flying Geese X 4 No Math Ruler" doesn't make sewing Flying Geese blocks easier. It just makes it faster.
The Flying Geese X 4 ruler is marked for twelve sizes of geese. Simply select the size geese you need, and cut one square on the corresponding solid line:
This will make the "body" part of the geese.
These are the "wings."
For the "Easy Street" mystery, I wanted "scrappy" geese, so I cut the two larger squares from different fabrics, and used six different blue fabrics for the wings.
Place the smaller squares on opposite corners of the large square and draw a line diagonally from corner to corner. Then sew a scant 1/4" on either side of the line:
Cut on the line and press toward the small triangles (the "wings" of your geese):
Again, because I wanted "scrappy" geese, I made two sets at once. That way, I will get eight different geese.
Next, place a third smaller square at the "empty" corner of the geese "body," matching up the raw edges.
Draw a diagonal line corner to corner on the smaller square and sew a scant 1/4" seam on either side of the line. It is important to begin your stitching line exactly where the small squares meet -- this will improve the chances of your geese's "wings" being even:
Cut on the line and press toward the smaller triangle:
Remember, each large square makes four geese units, so count accordingly!
It will be easier to sew your geese together if they are trimmed to the exact size. Eleanor Burn's small Flying Geese template is the right size to trim the geese for Easy Street.
Even taking all these photos, I had my 64 geese in less than one hour!
Did you know a group of geese is called a gaggle?
Tomorrow is WIP Wednesday, and I have another Designer Mystery BOM block to share, along with some of the setting blocks.
From the desk of your


3 edition of A treatise on the theory of algebraical equations. By J. Hymers. found in the catalog.
December 20, 2005
by Scholarly Publishing Office, University of Michigan Library
Written in English
|The Physical Object|
|Number of Pages||301|
Find link is a tool written by Edward Betts.. searching for Spherical trigonometry 73 found ( total) alternate case: spherical trigonometry José Anastácio da Cunha (89 words) exact match in snippet view article find links to article work on the theory of equations, algebraic analysis, plain and spherical trigonometry, analytical geometry, and differential calculus. Theory and solution of algebraical equations of the higher orders SCM A key to Bland's algebraical problems SCM An elementary treatise on the geometrical and algebraical investigation of maxima and minima SCM Hemming, George Wirgman SCM Price, Bartholomew A treatise on the differential calculus, and its application to.
Airy, at p. , Note ofthe Undulatory Theory, gives as the equations for determining the small disturbances of an elastic medium in three dimensions d u 2 d (du dv dw\ = la d + +), dt2"' dx\dx dy dz) d2v 2 d (du dv dw\ dt2a dy \dx dy dz) ' d2w 2 d du dv dw t — a- - + - + i dt dz \dSc dy dz J We might eliminate v and w, by cross. Hymers J. A treatise on the theory of algebraical equations [Cambridge, ] Manin Yu.I. Algebraic aspects of nonlinear differential equations ( 11, , ) The variational bicomplex [book draft, free web version, Utah, ]. Rolle's theorem is found in English in in A treatise on the theory of algebraical equations by John Hymers [Univesity of Michigan Historical Math Collection]. ROMAN NUMERAL is found in in Phil. Trans. xxxix, (OED2).
Bibliography of the History of Science and Technology. This is the really fun section. It covers gizmos, inventions, mechanisms, whirligigs and all sorts of scientific, medical and engineering apparatus. If you want to find and/or build some neat gadget from the s, chances are you'll find it . Treatise on algebra: in practice and theory. Vol 2 Treatise on algebra: in practice and theory. Vol 2 Introduction to the study of the mathematical principles of natural philosophy. Vol 1 [Greek] Eukleidou Stoichei*on bibl. 15 ek t*on The*onos synousi*on New ed 1st ed 9th ed 17th ed 34th ed 52nd ed Revised ed 36th ed 25th ed Edinburgh ed London. Full text of "Catalogue of the Scientific Books in the Library of the Royal Society" See other formats.
Tony Sargs alphabet
Conspiracy on the high seas
Choking on Marlon Brando
The presidents words
Contemporary East European Poetry
Recent progress in biology and medicine, its social and ethical implications
The babysitter takes dogs
The unorthodox imagination in late medieval Britain
Conical lens for 5/54 gun launched missile
[Theatre Royal Bristol
A Survival manual for medical students
A discourse delivered in the chapel of Harvard College, June 19. 1798.
Lotus 1-2-3 release 4 for Windows
brief submitted to the Royal Commission on Co-operatives ...
SyntaxTextGen not activatedA Treatise on Projective Geometry, by George Pdf Jones (page images at Cornell) A Pdf on the Differential Geometry of Curves and Surfaces (Boston et al.: Ginn and Co., c), by Luther Pfahler Eisenhart (page images at Cornell) A Treatise on the Geometry of Surfaces, by Alfred Barnard Bassett (page images at Cornell).A treatise on the theory of algebraical equations John Hymers A Treatise on the Theory of Bessel Functions G.
N. Watson A treatise on universal algebra with applications Alfred North Whitehead.A Treatise on the Theory of Algebraical Equations by John Hymers The Laws ebook Algebra An Elementary Course in Algebraic Ebook by Alfred George Cracknell A Treatise on the Stability of a Given State of Motion, Particularly Steady Motion Being the Essay to Which the Adams Prize Was Adjudged inin the University of Cambridge by Edward John Routh.


Inscribed Sphere Radius
A cube with a side length (S) of 4 has an inscribed sphere with a radius (R) of 2.
Inscribed Sphere Radius Formula
for cubes with a side length S find the radius R of the inscribed sphere
- R = S/2
substitute the side length S with the measured value, in this example lets use a side length of 4
- R = 4/2
divide the side length in half to find the radius of an inscribed sphere
- R = 2
Given a cube with a side length S the radius (R) of an inscribed sphere can be found by dividing the cube side length in half.


Computer Skills Tutors in Festus, MO
Find Private & Affordable Computer Skills Tutoring in the Festus Area!
Rhodes College - Bachelors, Mathematics & Economics , Georgia Institute of Technology-Main Campus - Masters, Computer Science
Virginia Tech - BA, Communication
BA Information Technology MA Education in Mathematics at Webster University Studied how to teach math not just how to do math. I know...
Psychology/Journalism Major at Lindenwood University previously Junior Psychology/Journalism Major at University of Phoenix online
Bachelor's in Music Education (K-12) and certified Elementary Education, Missouri State University, 2010
Graduate of University of Maryland, College Park, 4.0, with a Bachelor's of Science in Elementary Education. Currently working on...
B.S. in Mathematics


Velocity Value Growth is an extension of Velocity Profit Enhancement and a precursor to leaving a greater legacy. When valuing a business, consider an arithmetic approach:
FACTOR A x FACTOR B = Value of a Business
Factor A: A function or derivative of profitability
Factor B: A multiplier of the value determined by industry and quality of company
The sales multiple (multiplier) in Factor B stems from business value drivers. Following a comprehensive assessment of a business’ current status in relation to these drivers, we establish the areas for improvement. Closing the value gap will have the greatest impact on the attractiveness and readiness of the business for growth, acquisition, or sale, thus maximizing legacy opportunities.


False positive risk calculationsThis web app was written by Colin Longstaff and David Colquhoun with help from Brendan Halpin. And in ver 1.7, help from Dr Will Parry to fix default for nsamp2.
Statistical considerationsThe question which we ask in refs[1 - 4, 6, 7] is as follows. If you observe a "significant" P value after doing a single unbiased experiment, what is the probability that your result is a false positive? An account of the precise assumptions that underlie the calculations is given in ref . "False positive risk" (FPR) is defined here as the probability that a result which is "significant" at a specified P value, is a false positive result. It is defined and explained in refs [3, 7 and at 26' in ref 6]. The same thing was called "false discovery rate"in refs  and , and it was called "false positive rate" in earlier drafts of refs [3, 7] The notation in this field is a mess and it's important to check the definitions in each paper There are two different ways to calculate FPR. These are explained in detail in section 10 of ref , and, more carefully, in section 3 of ref . They can be called the p-equals method, and the p-less-than method. The latter definition is used most frequently (eg by Ioannidis and by Wacholder), but the former method is more appropriate for answering our question. All three options give results that are calculated with both methods. The results with the p-equals method, give a higher false positive risk, for any given P value, than the other method (see Fig 2 in ref ), but they are the appropriate way to answer the question.
How to run calculationsClick on the calculations tab, and choose which calculation to do by selecting one of the three radio buttons (top left) The input boxes that are appropriate for the calculation will appear. There are three variables, observed P value, FPR and the prior probability that the null hypothesis is false. The calculator will work out any one of these, given numbers for the other two. All three calculations require also the number of observations in each sample, and the effect size, expressed as a multiple of the standard deviation of the observations (default value 1.0) The default number per sample is 16 which gives a power of 0.78 for P = 0.05 and effect size = 1 -see refs  and  for more details. Note that all that matters is the effect size expressed as a multiple of the standard deviation of the original observations (sometimes known as Cohen's d ). The true mean of sample 1 is always 0 (null hypothesis), The true mean of sample 2 is set to the normalised effect size so the true standard deviation can always be set to 1, with no loss of generality.
A real life exampleFully worked examples are given in section 8 of ref .
A study of transcranial electromagnetic stimulation, published In Science concluded that it "improved associative memory performance", P = 0.043. If we assume that the experiment had adequate power (the sample size of 8 suggests that might be optimistic) then, in order to achieve a false positive risk of 5% when we observe P = 0.043, we would have to assume a prior probability of 0.85 that the effect on memory was genuine (found from radio button 1). Most people would think it was less than convincing to present an analysis based on the assumption that you were almost certain (probability 0.85) to be right before you did the experiment.
Another way to express the strength of the evidence provided by P = 0.043 is to note that it makes the existence of a real effect only 3.3 times as likely as the existence of no effect (likelihood ratio). This would correspond to a minimum false positive risk of 23% if we were willing to assume that non-specific electrical zapping of the brain was as likely as not to improve memory (prior probability of a real effect was 0.5) (found via radio button 3).
The radio button 2 option shows that in the most optimistic case (prior = 0.5), you need to have P = 0.008 to achieve an FPR of 5 percent. (Example from refs  and .)
Matching powerMuch the same results are found for FPR if the power is kept constant. This is explained and exemplified in section 5 and Figure 1 of ref .
For example effect size = 1 and n = 16 gives power = 0.78. For an effect size of 0.5 SD, n = 61 gives similar power and also a similar FPR etc. And for an effect size of 0.2 SD, a power of 0.78 requires n = 375, and again this gives similar FPR etc. See ref  for more details. So choose n so that the calculated power matches that of your experiment. There is a popular account of the logic involved in ref . And ref  has, in section 9, a response to the recent 72 author paper, Benjamin et al , on related topics. There is a more technical account of the assumptions in ref . And a defence of those assumptions in ref .
VersionsFrom ver 1.1 onwards, the effect size (expressed as a multiple of the standard deviation of the observations) can be entered. From ver 1.3 onwards, the values of power that are printed out are calculated for P = 0.05 and the specified effect size (expressed as a multiple of the standard deviation of the observations). In earlier versions they were calculated using the observed P value). Ver 1.4 has updated help notes. Ver 1.5 has updated help notes and references. Ver 1.6 is unchanged apart from the default radio button selected at start-up is now button 3, rather than button 1. Ver 1.7 allows unequal sample sizes (but still assumes same variance for both samples).
References1. Colquhoun D.(2014) An investigation of the false discovery rate and the misinterpretation of p-values. Royal Society Open Science 1(3):140216. doi: 10.1098/rsos.140216. Click for full text
2. Colquhoun D. False discovery rates: the movie (now superseded by ref ) Click for YouTube
3. Colquhoun D. (2017). The reproducibility of research and misinterpretation of P values. Royal Society Open Science 4 (12), doi: ttp://dx.doi.org/10.1098/rsos.171085 Click for full text
4. Colquhoun D. (2016). The problem with p-values. Aeon Magazine Click for full text
5. Benjamin, D. et al. (2017) Redefine Statistical Significance. PsyArXiv Preprints, July 22, 2017. Click for full text
6. Colquhoun, D. (2018) Colquhoun D. The false positive risk: a proposal concerning what to do about p-values (version 2) [talk based on that given at EvidenceLive, 2018] Click for YouTube
7. Colquhoun, D. (2019) The false positive risk: a proposal concerning what to do about p values. American Statistician . Click for full text
8. Colquhoun, D. (2019b) A response to critiques of "The reproducibility of research and the misinterpretation of p-values". Royal Society Open Science . Click for full text
A list of all of DC's publications on p values can be found at Some papers about p values.


Playing With Numbers
CBSE Class 6 Maths
Q1: Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
(a) ✗ (e.g. 15 is divisible by 3 but not by 9)
(e) ✗ (e.g. 8 and 9 are co-primes and none of these are prime. )
Q2: Here are two different factor trees for 60. Write the missing numbers.


Tensor Analysis on Manifolds (Dover Books on Mathematics): A Mathematical Tool for Physics and Geometry
Tensor Analysis on Manifolds (Dover Books on Mathematics) Book PDF
If you are looking for a comprehensive and accessible introduction to tensor analysis on manifolds, you might want to check out the book Tensor Analysis on Manifolds by Richard L. Bishop and Samuel I. Goldberg. This book is a classic in the field of differential geometry and provides a solid foundation for further studies in mathematics and physics. In this article, we will give you an overview of the main topics covered in the book, as well as some applications and examples of tensor analysis on manifolds. We will also provide you with a link to download the book in PDF format for free.
Tensor Analysis on Manifolds (Dover Books on Mathematics) book pdf
Before we dive into the details of tensor analysis on manifolds, let us first review some basic concepts and definitions that will help us understand the subject better.
What is tensor analysis?
A tensor is a mathematical object that can be used to describe various physical phenomena, such as forces, stresses, strains, electric fields, magnetic fields, etc. A tensor can be thought of as a generalization of scalars, vectors, and matrices, which are special cases of tensors. A scalar is a tensor of rank zero, a vector is a tensor of rank one, and a matrix is a tensor of rank two. A tensor of rank n can be represented by an array of n indices, each ranging from 1 to some dimension d.
For example, a scalar can be written as a = a1, where the index 1 indicates that it has only one component. A vector can be written as b = bi, where i ranges from 1 to d. A matrix can be written as c = cij, where i and j range from 1 to d. A general tensor of rank n can be written as t = ti1i2...in, where each index ranges from 1 to d.
Tensors can be manipulated by various operations, such as addition, subtraction, multiplication, contraction, etc. For example, two tensors of the same rank can be added or subtracted component-wise: t + s = (ti1i2...in + si1i2...in). A tensor can be multiplied by a scalar: k * t = (k * ti1i2...in). A tensor can be contracted by summing over repeated indices: ti = t11 + t22 + ... + tdd. A tensor can also be multiplied by another tensor, resulting in a new tensor of higher or lower rank, depending on the number of indices involved.
What is a manifold?
A manifold is a mathematical space that locally resembles a Euclidean space of some dimension, but globally may have a more complicated shape or structure. For example, the surface of a sphere is a two-dimensional manifold, because every small patch of it looks like a flat plane, but the whole surface is curved and finite. A torus (donut-shaped surface) is another example of a two-dimensional manifold, which has a different topology than a sphere. A manifold can also have more than two dimensions, such as a hypersphere or a hyperboloid.
A manifold can be equipped with various structures that allow us to measure distances, angles, areas, volumes, etc. on it. For example, a metric tensor is a tensor field that assigns a positive-definite symmetric matrix to each point of the manifold, which defines the inner product and the norm of vectors at that point. A metric tensor can be used to compute the length of curves, the angle between vectors, the area of surfaces, the volume of regions, etc. on the manifold. A metric tensor also induces a notion of curvature on the manifold, which measures how much the manifold deviates from being flat.
Why study tensor analysis on manifolds?
Tensor analysis on manifolds is a powerful tool that can be used to study various aspects of geometry, topology, physics, and engineering. By using tensors, we can express quantities and equations that are invariant under coordinate transformations, which means that they do not depend on the choice of coordinates used to describe the manifold. This makes tensors suitable for describing natural phenomena that obey universal laws, such as gravity, electromagnetism, fluid dynamics, etc.
Tensor analysis on manifolds also allows us to generalize concepts and results from Euclidean spaces to more general spaces that may have curvature or other features. For example, we can define derivatives of functions and vector fields on manifolds using covariant derivatives, which take into account the connection and curvature of the manifold. We can also define integrals of functions and differential forms on manifolds using integration on chains, which generalize the notions of line integrals, surface integrals, and volume integrals.
Main Concepts and Theorems
In this section, we will briefly introduce some of the main concepts and theorems that are covered in the book Tensor Analysis on Manifolds. We will not go into too much detail or provide proofs here, but we will give some examples and intuition for each topic. For more details and rigor, we refer you to the book itself.
Tensors and tensor fields
Definition and examples
A tensor on a manifold M is an n-linear map from n copies of the tangent space TpM at some point p to the real numbers R. That is, a tensor T takes n vectors v, v, ..., v at p and produces a real number T(v, v, ..., v). The number n is called the rank or order of the tensor. A tensor field is a function that assigns a tensor to each point of the manifold.
For example, a scalar function f: M -> R is a tensor field of rank zero, because it takes zero vectors at each point and produces a real number f(p). A vector field X: M -> TM is a tensor field of rank one, because it takes one vector at each point and produces a real number X(p)(v) = , where is the inner product induced by the metric tensor. A bilinear form B: M -> T*M x T*M -> R is a tensor field of rank two, because it takes two vectors at each point and produces a real number B(p)(v,w) = B(p)(v)(w), where B(p)(v) is a linear map from TpM to R.


Mar 01 2013
Social Network Analysis and Mining (SNAM) is a multidisciplinary journal serving researchers and practitioners in academia and industry. It is the main venue for a wide range of researchers and readers from computer science, network science, social sciences, mathematical sciences, medical and biological sciences, financial, management and political sciences. We solicit experimental and theoretical work on social network analysis and mining using a wide range of techniques from social sciences, mathematics, statistics, physics, network science and computer science.
Read, download and save these articles online. They are available online for free until April 15, 2013:
- Social networks and online environments: when science and practice co-evolve – Devan Rosen, George A. Barnett, Jang Hyun Kim
- Social network analysis: developments, advances, and prospects – John Scott
- Supervised methods for multi-relational link prediction – Darcy Davis, Ryan Lichtenwalter, Nitesh V. Chawla
- Trends in science networks: understanding structures and statistics of scientific networks – Miray Kas, Kathleen M. Carley, L. Richard Carley
- Modeling blogger influence in a community – Nitin Agarwal, Huan Liu, Lei Tang, Philip S. Yu


Most kids today think of math as hard work. No matter how much they dislike it they will have to use it in school and everyday life. The studies have shown that kids who start learning math at an early age will later become better at problem solving. They are also more likely to choose a career in the fields of technology or finances. But making them practice math can be quite a trouble. That is why you should do your best to make math learning fun and interesting for them. Here are some ways to do so.
Use math for fun activities
One of the best ways to get your children to practice math is to have them use it while doing something they like. Think of some fun activities they enjoy doing and try to make them implement math in them. Cooking is one of the things children enjoy helping you with and can also practice math while doing so. Let them do all the counting and measuring. This way they will feel proud because they are helping and won’t even think of counting and measuring as a form of math. Another place you can have them practice math is your car. Every time they ask you “are we there yet?” you can tell them how far your destination is and how fast you are going and have them do the counting by themselves.
Understanding the concept of math
In order for your child to become aware of what he or she is doing subconsciously, they will need to understand the concept of math. Although at this point it might sound too raw, there are ways to do it through play as well. The traditional Montessori approach to learning has developed and applied great techniques for learning math. It has been divided into levels, according to child abilities at each age. The main thing to know is that it doesn’t push or forces your child towards improvement, but rather the child himself progresses at its own pace. What is also important is that your child is playing and learning at the same time but without feeling oppressed.
Have them use it in everyday activities
Your children are surrounded by the things that require some math skills. Counting money or time and measuring things are all interesting options for your children to practice math. One of the ways to make them use math is to take your children shopping with you. Ask them to tell you the prices of the products on the bottom shelves and have your children compare them. You can also ask them to calculate how much you will pay at the counter and get them a prize if they get it right. Make this a game and do it every time you go shopping. You can also find some other interesting shopping games to play with your children. Boys also like playing with tools. Give them a tape measure and ask them how long something is.
Have them use math while playing
The best way for children to solve math problems is while playing games. There are many ways you can help them do this. There are many trading card games today that children simply adore. Look for these online and buy them cards. Explain them the rules and play the game with them. Most of these include a lot of counting and comparing numbers. They also have amazing artwork which makes it very interesting for children. You can also use your PC to have children practice math. There are many cool math games online that your children can play. Help them in the beginning and let them play on their own later. All of the games like this will improve their math and problem solving skills.
The most important thing to do is to make math look interesting for children. Once you manage to make math a part of the activities and games your children like you will have no trouble making them practice it. Just get involved and try to help them implement math in their daily lives, one day at a time.
Emma is a teacher, constantly improving her skills both as a teacher and as a parent. She is passionate about writing and learning new things that can help you to lead a quality life. She is a regular contributor to High Style Life. You can follow her on Twitter @EmmahLawson


Definitions from a mathematical mouse
addition: the process of putting together one number/amount with at least one other number/amount, which finds the total or sum of them.
My Youtube videos below demonstrate how to understand formal methods for addition. The great thing is you can watch them as many times as you need and if you ever forget part of the method, you can simply watch it again anytime you want...but not when you're supposed to be sleeping in bed!
Of course, all great mousematicians know that sometimes using mental methods can be the best strategy and sometimes using formal methods is the best method. Don't always use the same approach - use the best approach!
TdM Brie-Anna de Mouse, creator and star of Maths with a Mouse
Before we get started with formal methods, perhaps you need to brush up on your mental addition and subtraction skills or simply want to play some great addition and subtraction games. If you do, push the big yellow button now!
Addition without carrying
Addition with carrying
Ready to use what you have learned? Have a go at these great Brain Workouts from mathplayground.com. Try to use your understanding of formal addition to help find solutions.
Warning: it might not be that easy!
Addition with decimals


How Much Water Volume is in My Pond?
Water Volume Calculation
Using Salt and a Salinity Meter
Measuring water volume in ponds is difficult to do using simple volume calculations. Simple "water volume calculators" only "estimate" water volume. Most ponds have varying shapes and depths. They include water within the plumbing, the filtration, waterfalls, and streams. So the best way to calculate water volume in all that is to use dissolved salt measurements. We measure the salinity, then add salt and measure the salinity again. The difference in these measurements coupled with how much salt we added will tell us how much total water volume we have in our pond.
The formula for using salinity to measure water volume is based upon a common multiplier:
One PPT (parts per thousand) = 1 pound of salt added to 1,000 gallons of pond water. Pond water weighs 8.34 pounds per gallon.
1,000 gallons ÷ 8.34 pounds = 119.90 or 120 for easier math. 120 is the multiplier.
The procedure entails measuring the current salinity of the water in the pond, adding a known amount of 99+% pure crystal salt (by weight), measuring the difference this addition made to the salinity, and, with these figures, we can calculate the total water volume in the pond, waterfalls, plumbing, and filtration system.
- 1. Estimate the pond's water volume by measuring the length, width, and average depth. L x W x AD x 7.48 = estimated water volume
- 2. For the safety of your fish and water plants, divide your estimated water volume by 3 (if your estimate is 1,500 gallon, divide that by 3 for an adjusted estimate of 500 gallons for adding salt - most people over estimate how much water their pond holds)
- 3. Measure the pond water salinity in PPT (parts per thousand) with a salinity meter to create a "base-point"
- 4. Using your adjusted estimated water volume, add salt at a rate of 1 pound per 100 gallons of pond water - or 10 pounds per 1,000 gallons of pond water
- 5. After the salt has completely dissolved and thoroughly circulated through the pond for 24 hours measure the salinity of the pond once more
Measure your pond with the salinity meter before adding salt to get a base point. If you've never added salt to your pond, more than likely your base reading will be 0.0 PPT, if it reads another number we simply use subtraction.
Example: Pond salinity base point of 0.2 PPT. If the adjusted water volume estimate is 500 gallons, add 5 pounds of salt. After the salt has thoroughly dissolved and been circulated throughout the pond for 24 hours, measure the salinity again. Let's say that the salinity meter now reads at 0.7 PPT.
Here is the formula:
- Adjusted Estimated Pond Water Volume is 500 Gallons
- Baseline salinity test is 0.2 PPT
- We added 5 pounds of 99+% pure crystal salt to the pond and let dissolve and circulate for 24 hours
- New salinity test reads 0.7 PPT
- 0.7 PPT - 0.2 PPT = 0.5 PPT
- 120 (multiplier number) x 5 pounds of salt added = 600
- 600 ÷ 0.5 PPT =
Top of Page


In all the versions of online baccarat that exist, some rules are always the same. In each game, the player’s hand competes against the dealer’s hand and there are three possible outcomes: point, bank, or tie. Each card is assigned a point value.
|50 Freespins on 1st. deposit
|50£ Bonus + 50 Extra spins
|100% Bonus up to 100£
|100% Bonus up to 100£
|100% Bonus up to 300£
Online baccarat- cards value
2 to 9 – the cards are worth the number of points each card says
10, J, Q, K – worth 0 points
Aces – worth 1 point
To calculate the value of a hand, all the cards are added and the value of the digit to the right is observed. For example, a hand with a 5 and a 9 would be worth 4 points (since they both add up to 14 and the digit to the right of 14 is 4). If the cards were a 2 and a 6, then the value of the hand would be 8. As a result of this way of scoring points, the highest value a baccarat hand can reach is 9.
Point And Banking
In this variant, the most popular throughout the best casino online across the United States, the United Kingdom, Canada and other countries, the dealer plays both hands according to certain card-drawing rules. One hand is called a “point” and the other is called a “banker.” As mentioned above, participants can choose to bet that one of the parties will win or that a tie will occur.
During a round, each hand receives two face-up cards from a 4, 6, or 8 deck shoe. If any of the parties add a total of 8 or 9, it is declared the winner. If both parties add up to 8 or 9, a tie is declared. If the round does not end here, then the card drawing rules are consulted.
If the point has a total of 6 or 7, do not take additional cards. If he is between 0 and 5, he receives another letter.
If the point did not receive a third card, the same card-taking rules apply to the dealer. But if, on the contrary, the point did receive a letter, the bank must do the following:
The dealer draws a card if his total is 2 or less.
Once the bets are placed, the cards will be dealt and then discovered. If the total sum of the player’s cards is 6, 7, 8 or 9, no more cards are dealt to him. On the other hand, a third card will be awarded to him if the sum of his two cards is 5 or less.
The dealer may also receive an additional card depending on his hand, but also that of the player, as you will see below in the table perfectly summarizing the third card baccarat rule.
At the end of the draws, the hands are counted to find out which one will be the winner. Player and dealer bets have a ratio of 1 to 1. If you bet on a tie, however, and it turns out to be the result, the ratio becomes 8 to 1.
The winning hand is then determined and the winners (if any) are paid out. Bets on the point hand generally pay twice the stake, while bets on the dealer’s hand deduct a 5% commission. A tie is paid 8 to 1 and the money from the bets to the point and the bank is accumulated for the next round.
Advantages of online baccarat
Enjoying free games in sites recommended by www.newcasinogames.net, such as online sic bo or blackjack has always offered a greater advantage to the player in the long run; which goes beyond being able to have fun without having to spend any money in online baccarat.


For example in the transaction entry screen where you enter the total, if I can put in 45*15.3 and it calculates the value for me automatically and puts that value in the total upon leaving the control. That will help reduce the need to switch to a calculator to figure out general math calculations and allow for speedy input. Going back into that control shows the calculation rather than the total so the user can see what they inputted. This could also work well for anyone who wants to remember the cost / unit value and not just the total.
This came up during a usability test who said the accounting tool she uses has that feature and it's very efficient and handy.


Published at Saturday, April 25th, 2020 - 15:46:31 PM. Coloring Page. By Alaric Winkler.
As already mentioned free coloring pages are a great way to teach your child all sorts of things, numbers for example. There are different ways this can be done, one way would be to simply have the number somewhere on the page, such as the number 10, and then somewhere else on the page would be an object or animal cleverly twisted into that number. There could also simply be 10 of something on the page, you can get your child to color the page in and then count out how many apples for example there are, which will obviously correspond to the number on the page.
Any content, trademark’s, or other material that might be found on the Niobrarachalk website that is not Niobrarachalk’s property remains the copyright of its respective owner/s. In no way does Niobrarachalk claim ownership or responsibility for such items, and you should seek legal consent for any use of such materials from its owner.


It'll always give you kind of the positive distance. We can see that, okay, this x minus one squared. Directrix is y is equal to six and a half. I don't know, my brain just processes things better that way. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. Find the equation of the parabola in the example above. I might be careful with my language.
Equate these two distance equations and the simplified equation in x 0 and y 0 is equation of the parabola. A parabola is the geometric place of points in the coordinate axes that have the property that they are equidistant from a fixed point called the focus and a line called the directrix. Well, we've already seen the technique where, look, we can see the different parts. Y is equal to six and a half and the focus, well, we know the x coordinate of the focus, a is going to be equal to one and b is going to be three-fourths less than the y coordinate of the directrix. If you have any doubt please through mail we will help you to clear your doubt. If it is downward opening, it's going to be this maximum point. Recall that the focus and the vertex of a parabola are on the same line of symmetry.
So our actual parabola is going to look is going to look something it's gonna look something like this. When x equals one, you get one minus one squared. Remember, the vertex, if the parabola is upward opening like this, the vertex is this minimum point. Well, when does this equal zero? So the directrix might be something like this. Thus the axis of the parabola is vertical.
Actually, let me write that as a. The focus is a,b and the directrix is y equals k and this is gonna be the equation of the parabola. Applications The parabola has countless applications in Physics, because of the way the gravity force and Newton's laws operate, the trajectory of most bodies that are thrown out will follow a parabolic trajectory. So 23 over four minus three-fourths. So you got b minus k equals something. Times x minus one squared plus b plus k.
Instructors are independent contractors who tailor their services to each client, using their own style, methods and materials. A parabola can open up or down if x is squared or open left or right if y is squared. So the first thing I like to do is solve explicitly for y. This enables us to identify the direction which the required parabola opens. So this quantity over here is either going to be zero or negative. One over two times b minus k needs to be equal to negative one-third. Apollonius and other mathematicians discovered that when you cut a cone with a plane, depending on the relative angle of the cone and the plane, the cone is cut in a way that the section has different shapes.
This distance has to be the same as this distance right over here and what's another way of thinking about this entire distance? And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four five, six, seven. Figure 1 shows a picture of a parabola Notice that the distance from the focus to point x 1, y 1 is the same as the line perpendicular to the directrix, d 1. This would actually always work. In this page parabola-focus , we have discussed how to find the focus, equation of directrix, vertices and length of the latus rectum. We will discus how to find the above in little different form of the equation. A Greek mathematicians named Apollonius is credit with having contributed with the modern version, using coordinate systems, of the conic sections. To find the distance from the vertex to the directrix find the absolute value of the change between their x-coordinates.
The point is called the focus of the parabola and the line is called the. We're gonna see, we're gonna go to one. Actually, let me do this in a different color. The General Equation of the Parabola There are simple derivations to get the equation of a parabola based on the location of a directrix and the focus, but we will skip the derivation in this introduction. It's not going to add to 23 over four, it's either gonna add nothing or take away from it. It turns out that many maximization and minimization problems have a quadratic function to maximize, and geometrically, the maximum or minimum depending on if the parabola opens up or down is achieved at the vertex.
We could take the reciprocal of both sides and we get two times b minus k is equal to, is equal to three, is equal to three. That's the focus, one comma five. Remember, this coordinate right over here is a, b and this is the line y is equal to k. So this thing's going to hit a maximum point, when this thing is zero, when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So let's think about the vertex of this parabola right over here. This curve can be a parabola. So we don't know just yet where the directrix and focus is, but we do know a few things.
So I could say the directrix, so let me see, I'm running out of space, the directrix is gonna be y is equal to the y coordinate of the focus. Use our free online Parabola calculator to solve your academic mathematical and engineering problems. Here is a simple online Directrix calculator to find the parabola focus, vertex form and parabola directrix. Varsity Tutors does not have affiliation with universities mentioned on its website. . Equate the two distance expressions and square on both sides.


Who said learning should be boring? It’s time to get out of that classroom and experience science and math the right way—through real world experiments. SMILE Camp is a Science and Math Interactive Learning Experience. Instead of memorizing charts and graphs, your child will participate in hands on experimentation and adventure.
SMILE Camp students are surrounded by intellectual peers and are challenged during every session to reach to their full potential. SMILE Camp endeavors to create and nurture a life-long passion for STEAM fields (Science, Technology, Engineering, Arts and Math) in the next generation.
Some of the practical math and science applications taught in our camps include building and testing a working electric vehicle, learning 3D printing and prototyping, metallurgy, working with liquid crystals (as used in LCD screens), molecular gastronomy, and much more.
Our instructors are Engineering students from NC State with a passion for STEAM. SMILE Camp has over 30 options for prospective students to choose from and each option includes real hands-on experiments. For more information head on over to smilecamp.org.


Professorial Lecturer or Instructor in Data Science
Professorial Lecturer or Instructor in Data ScienceAmerican University, Department of Mathematics & Statistics, Washington DC The Department of Mathematics and Statistics in the College of Arts and Sciences at American University invites applications for a full-time teaching faculty position beginning Fall 2023. The appointment is a 9-month term (non-tenure track) position and will commence on August 28, 2023. Rank will depend on terminal degree, and if appropriate, stature in the field. Review of submitted materials is guaranteed if the application is received by March 31, 2023. The search will continue until the position is filled, or until August 13, 2023, whichever comes first. The position will require
This job listing is no longer active.
Check the left side of the screen for similar opportunities.


* The preview only shows a few pages of manuals at random. You can get the complete content by filling out the form below.
13. Power A. Introduction B. Example Calculations C. Factors Affecting Power D. Exercises
Introduction to Power by David M. Lane Prerequisites • Chapter 11: Significance Testing • Chapter 11: Type I and Type II Errors • Chapter 11: Misconceptions Learning Objectives 1. Define power 2. Identify situations in which it is important to estimate power Suppose you work for a foundation whose mission is to support researchers in mathematics education and your role is to evaluate grant proposals and decide which ones to fund. You receive a proposal to evaluate a new method of teaching high-school algebra. The research plan is to compare the achievement of students taught by the new method with the achievement of students taught by the traditional method. The proposal contains good theoretical arguments why the new method should be superior and the proposed methodology is sound. In addition to these positive elements, there is one important question still to be answered: Does the experiment have a high probability of providing strong evidence that the new method is better than the standard method if, in fact, the new method is actually better? It is possible, for example, that the proposed sample size is so small that even a fairly large population difference would be difficult to detect. That is, if the sample size is small, then even a fairly large difference in sample means might not be significant. If the difference is not significant, then no strong conclusions can be drawn about the population means. It is not justified to conclude that the null hypothesis that the population means are equal is true just because the difference is not significant. Of course, it is not justified to conclude that this null hypothesis is false. Therefore, when an effect is not significant, the result is inconclusive. You may prefer that your foundation's money be used to fund a project that has a higher probability of being able to make a strong conclusion. Power is defined as the probability of correctly rejecting a false null hypothesis. In terms of our example, it is the probability that given there is a difference between the population means of the new method and the standard method, the sample means will be significantly different. The probability of failing 449
to reject a false null hypothesis is often referred to as β (the Greek letter beta). Therefore power can be defined as: power = 1 - β. It is very important to consider power while designing an experiment. You should avoid spending a lot of time and/or money on an experiment that has little chance of finding a significant effect.
Example Calculations by David M. Lane Prerequisites • Chapter 5: Binomial Distribution • Chapter 12: Testing a Single Mean • Chapter 13: Introduction to Power Learning Objectives 1. Compute power using the binomial distribution 2. Compute power using the normal distribution 3. Use a power calculator to compute power for the t distribution In the “Shaking and Stirring Martinis” case study, the question was whether Mr. Bond could tell the difference between martinis that were stirred and martinis that were shaken. For the sake of this example, assume he can tell the difference and is able to correctly state whether a martini had been shaken or stirred 0.75 of the time. Now, suppose an experiment is being conducted to investigate whether Mr. Bond can tell the difference. Specifically, is Mr. Bond correct more than 0.50 of the time? We know that he is (that's an assumption of the example). However, the experimenter does not know and asks Mr. Bond to judge 16 martinis. The experimenter will do a significance test based on the binomial distribution. Specifically, if a one tailed test is significant at the 0.05 level, then he or she will conclude that Mr. Bond can tell the difference. The probability value is computed assuming the null hypothesis is true (π = 0.50). Therefore, the experimenter will determine how many times Mr. Bond is correct, and compute the probability of being correct that many or more times given that the null hypothesis is true. The question is: what is the probability the experimenter will correctly reject the null hypothesis that π = 0.50? In other words, what is the power of this experiment? The binomial distribution for N = 16 and π = 0.50 is shown in Figure 1. The probability of being correct on 11 or more trials is 0.105 and the probability of being correct on 12 or more trials is 0.038. Therefore, the probability of being correct on 12 or more trials is less than 0.05. This means that the null hypothesis will be rejected if Mr. Bond is correct on 12 or more trials and will not be rejected otherwise. 451
0.2 0.15 0.1 0.05 0 0
Figure 1. The binomial distribution for N = 16 and π = 0.50. We know that Mr. Bond is correct 0.75 of the time. (Obviously the experimenter does not know this or there would be no need for an experiment.) The binomial distribution with N = 16 and π = 0.75 is shown in Figure 2. 0.25
0.2 0.15 0.1 0.05 0 0
Figure 2. The binomial distribution for N = 16 and π = 0.75.
The probability of being correct on 12 or more trials is 0.63. Therefore, the power of the experiment is 0.63. To sum up, the probability of being correct on 12 or more trials given that the null hypothesis is true is less than 0.05. Therefore, if Mr. Bond is correct on 12 or more trials, the null hypothesis will be rejected. Given Mr. Bond's true ability to be correct on 0.75 of the trials, the probability he will be correct on 12 or more trials is 0.63. Therefore power is 0.63. In the section on testing a single mean for significance in Chapter 12, the first example was based on the assumption that the experimenter knew the population variance. Although this is rarely true in practice, the example is very useful for pedagogical purposes. For the same reason, the following example assumes the experimenter knows the population variance. Power calculators are available for situations in which the experimenter does not know the population variance. Suppose a math achievement test were known to have a mean of 75 and a standard deviation of 10. A researcher is interested in whether a new method of teaching results in a higher mean. Assume that although the experimenter does not know it, the population mean for the new method is 80. The researcher plans to sample 25 subjects and do a one-tailed test of whether the sample mean is significantly higher than 75. What is the probability that the researcher will correctly reject the false null hypothesis that the population mean for the new method is 75 or lower? The following shows how this probability is computed. The researcher assumes that the population standard deviation with the new method is the same as with the old method (10) and that the distribution is normal. Since the population standard deviation is assumed to be known, the researcher can use the normal distribution rather than the t distribution to compute the p value. Recall that the standard error of the mean (σM) is
= which is equal to 10/5 = 2 in this example. As can be seen in Figure 3, if the null hypothesis that the population mean equals 75 is true, then the probability of a sample mean being greater than or equal to 78.29 is 0.05. Therefore, the
experimenter will reject the null hypothesis if the sample mean, M, is 78.29 or larger.
Figure 3. The sampling distribution of the mean if the null hypothesis is true. The question, then, is what is the probability the experimenter gets a sample mean greater than 78.29 given that the population mean is 80? Figure 4 shows that this probability is 0.80.
Figure 4. The sampling distribution of the mean if the population mean is 80. The test is significant if the sample mean is 78.29 or higher. Therefore, the probability that the experimenter will reject the null hypothesis that the population mean for the new method is 75 or lower is 0.80. In other words, power = 0.80.
Calculation of power is more complex for t tests and for Analysis of Variance. There are many programs that compute power.
Factors Affecting Power by David M. Lane Prerequisites • Chapter 11: Significance Testing • Chapter 11: Type I and Type II Errors • Chapter 11: One- and Two-Tailed Tests • Chapter 13: Introduction to Power • Chapter 13: Example Calculations Learning Objectives 1. State five factors affecting power 2. State what the effect of each of the factors is Several factors affect the power of a statistical test. Some of the factors are under the control of the experimenter, whereas others are not. The following example will be used to illustrate the various factors. Suppose a math achievement test were known to be normally distributed with a mean of 75 and a standard deviation of σ. A researcher is interested in whether a new method of teaching results in a higher mean. Assume that although the experimenter does not know it, the population mean μ for the new method is larger than 75. The researcher plans to sample N subjects and do a one-tailed test of whether the sample mean is significantly higher than 75. In this section, we consider factors that affect the probability that the researcher will correctly reject the false null hypothesis that the population mean is 75. In other words, factors that affect power. Sample Size Figure 1 shows that the larger the sample size, the higher the power. Since sample size is typically under an experimenter's control, increasing sample size is one way to increase power. However, it is sometimes difficult and/or expensive to use a large sample size.
Figure 1. The relationship between sample size and power for H0: μ = 75, real μ = 80, one-tailed α = 0.05, for σ's of 10 and 15. Standard Deviation Figure 1 also shows that power is higher when the standard deviation is small than when it is large. For all values of N, power is higher for the standard deviation of 10 than for the standard deviation of 15 (except, of course, when N = 0). Experimenters can sometimes control the standard deviation by sampling from a homogeneous population of subjects, by reducing random measurement error, and/ or by making sure the experimental procedures are applied very consistently. Difference between Hypothesized and True Mean Naturally, the larger the effect size, the more likely it is that an experiment would find a significant effect. Figure 2 shows the effect of increasing the difference between the mean specified by the null hypothesis (75) and the population mean μ for standard deviations of 10 and 15.
Figure 2. The relationship between μ and power for H0: μ = 75, one-tailed α = 0.05, for σ's of 10 and 15. Significance Level There is a trade-off between the significance level and power: the more stringent (lower) the significance level, the lower the power. Figure 3 shows that power is lower for the 0.01 level than it is for the 0.05 level. Naturally, the stronger the evidence needed to reject the null hypothesis, the lower the chance that the null hypothesis will be rejected.
Figure 3. The relationship between significance level and power with one-tailed tests: μ = 75, real μ = 80, and σ = 10. One- versus Two-Tailed Tests Power is higher with a one-tailed test than with a two-tailed test as long as the hypothesized direction is correct. A one-tailed test at the 0.05 level has the same power as a two-tailed test at the 0.10 level. A one-tailed test, in effect, raises the significance level.
Statistical Literacy by David M. Lane Prerequisites • Chapter 13: A research design to compare three drugs for the treatment of Alzheimer's disease is described here. For the first two years of the study, researchers will follow the subjects with scans and memory tests. What do you think? 458
The data could be analyzed as a between-subjects design or as a within-subjects design. What type of analysis would be done for each type of design and how would the choice of designs affect power? For a between-subjects design, the subjects in the different conditions would be compared after two years. For a withinsubjects design, the change in subjects' scores in the different conditions would be compared. The latter would be more powerful.
Exercises Prerequisites 1. Define power in your own words. 2. List 3 measures one can take to increase the power of an experiment. Explain why your measures result in greater power. 3. Population 1 mean = 36 Population 2 mean = 45 Both population variances are 10. What is the probability that a t test will find a significant difference between means at the 0.05 level? Give results for both one- and two-tailed tests. Hint: the power of a one-tailed test at 0.05 level is the power of a two-tailed test at 0.10. 4. Rank order the following in terms of power. Population 1 Mean
Population 2 Mean
5. Alan, while snooping around his grandmother's basement stumbled upon a shiny object protruding from under a stack of boxes . When he reached for the object a 459
genie miraculously materialized and stated: “You have found my magic coin. If you flip this coin an infinite number of times you will notice that heads will show 60% of the time.” Soon after the genie's declaration he vanished, never to be seen again. Alan, excited about his new magical discovery, approached his friend Ken and told him about what he had found. Ken was skeptical of his friend's story, however, he told Alan to flip the coin 100 times and to record how many flips resulted with heads. (a) What is the probability that Alan will be able convince Ken that his coin has special powers by finding a p value below 0.05 (one tailed). Use the Binomial Calculator (and some trial and error) (b) If Ken told Alan to flip the coin only 20 times, what is the probability that Alan will not be able to convince Ken (by failing to reject the null hypothesis at the 0.05 level)?